universidad de chile facultad de ciencias físicas y matem´aticas ...

A mis padres Benjamín y Angélica, por su amor y apoyoincondicional

AgradecimientosEn primer lugar, agradezco sinceramente a todas aquellas personas que de una u otra formahan sido parte de este importante proceso de aprendizaje y crecimiento.Por su dedicada orientación, su apoyo generoso, sus valiosos consejos y sus críticas tremendamenteconstructivas, agradezco de corazón a Marcos Kiwi, mi director de tesis. Muchasgracias por que cada reunión que tuvimos fue siempre una ganancia profesional y una motivaciónpersonal para continuar con esta investigación. Estoy segura que sin su ayuda y tiempoeste sueño no sería realidad.Quiero agradecer a Martin Loebl, quien informalmente guió este trabajo de tesis y a quienconsidero un pilar fundamental en mi crecimiento matemático. Gracias por todo el apoyo queme ha brindado, por creer en mi, por darme la oportunidad de trabajar juntos e integrarme ensu comunidad científica. Gracias por enseñarme que un matemático brillante es también unagran persona. Gracias infinitas por su amistad y paciencia.A Mihyun Kang agradezco por motivarme a seguir aprendiendo y a continuar por la sendade la investigación en los momentos de debilidad e inseguridad. Gracias por confiar en miscapacidades, por el gran apoyo, por cada una de las reuniones de trabajo y las conversacionesque tuvimos.A todos los docentes del DIM y miembros del CMM con quienes he tenido el privilegio detener clases y compartir seminarios, agradezco por inculcarme el amor por las matemáticas.También quisiera agradecer a los profesores del KAM y miembros del ITI por acogerme generosay cordialmente en su comunidad. Muchas gracias a todos por darme la oportunidad deaprender de sus enseñanzas y experiencias.A Gelasio Salazar y a Martín Matamala agradezco por aceptar ser parte de la comisión deeste trabajo de tesis, por su tiempo, disposición, comentarios y correcciones.Agradezco especialmente a Michel Curé, quien hace poco más de cuatro años fue la personaque me incentivó a continuar estudios doctorales.

Así también deseo agradecer el apoyo económico que me han brindado distintas institucionesy proyectos. Entre otros, agradezco al Programa MECESUP que ha financiado mis cuatro añosde estudio doctoral, a CONICYT vía Programa Basal en Modelamiento Matemático, a FON-DECYT 1090227 y al Institute for Theoretical Computer Science Charles University, Prague. ACONICYT por mi beca de pasantía doctoral en el extranjero y al Núcleo Milenio Informacióny Coordinación en Redes ICM/FIC P10-024F.De corazón agradezco a todos los funcionarios del DIM, en especial a Eterin, Silvia, Regina,Don Oscar y Don Luis, quienes con su gran vocación y buena voluntad, han estado siempredispuestos a ayudar de la mejor manera cuando uno lo necesita.Agradezco sinceramente a todas las maravillosas personas que conocí mientras estuve enPraga. A Eva, Tomas, Honza H., Honza B., Andrew y Mayumi mil gracias por todos los momentoscompartidos, por la hermosa Navidad en Tábor, las cenas, los paseos, los bailes, lasfiestas, las risas y sobretodo por su amistad que no sabe de tiempo ni de lugar. A Nana, quetal vez nunca se entere de estos agradecimientos, le agradezco su bondad, cariño y gran voluntad.A mis amigos y compañeros del doctorado, agradezco por los buenos momentos que hemosvivido juntos, por las conversaciones y almuerzos. A Naty, Maxy, Lucho, Flavio, José Aliste,José Zamora, Alvaro Daniel, Alvarito y Pablo agradezco por su ayuda y amistad. A Claritaagradezco por ser mi gran amiga, por su lealtad, sinceridad y su apoyo incondicional. A Jairoagradezco por nuestra amistad y por todas aquellas veces que me subió el ánimo con su alegría.Quiero agradecer a Oscar, quien a sido un apoyo fundamental en todo sentido. Gracias portu amor y comprensión, por levantarme en los malos momentos, por escucharme, acompañarmey por la paciencia, por vivir y disfrutar junto a mi los momentos felices.Finalmente, deseo agradecer a aquellas personas que han estado siempre en mi vida, mifamilia. Les doy gracias por apoyarme, por comprender y respetar mi poco tiempo, por darmetantos momentos de alegría y por ser mi fuente de energía cuando flaqueo. A mis abuelos Nenay Tata, agradezco su enseñanza de amor y respeto para toda la vida. A mi hermano Jaime,agradezco su generosidad, nobleza y gran corazón. A mi herma Paula, agradezco su amistad, sugran valentía y por darme la felicidad de tener dos sobrinos maravillosos. A mi papá, agradezcopor tanto amor que me ha entregado, por sus sacrificios para darnos siempre lo mejor, por suapoyo constante y por estar siempre presente. A mi mamá, agradezco por ser mi mejor amiga,por su amor infinito, por su vocación de madre, por su optimismo, su apoyo imprescindible, portodas las conversaciones y por tener siempre las palabras justas que reconfortan mi espíritu.

Índice general1. Introducción 12. Marco teórico y resultados 62.1. Preliminares . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62.1.1. El modelo de Ising . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62.1.2. Estados satisfactorios . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82.1.3. Emparejamientos perfectos: dual geométrico de los estados satisfactorios 92.2. Método de la matriz de transferencia . . . . . . . . . . . . . . . . . . . . . . . . 112.2.1. Triangulaciones de un n-ágono . . . . . . . . . . . . . . . . . . . . . . . . 122.2.2. Triangulaciones apiladas . . . . . . . . . . . . . . . . . . . . . . . . . . . 122.3. Complejidad computacional . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132.4. Estado fundamental no degenerado . . . . . . . . . . . . . . . . . . . . . . . . . 143. Conclusiones 15Bibliografía 18A. Estados satisfactorios en triangulaciones de un n-ágono 20B. Modelo de Ising antiferromagnético en triangulaciones aplicado a contar emparejamientosperfectos 38C. Dificultad computacional del problema de enumerar estados satisfactorios entriangulaciones 62D. Estado fundamental no degenerado en el modelo de Ising antiferromagnéticoen triangulaciones 83

Capítulo 1IntroducciónEste trabajo de tesis se centra en el estudio del comportamiento de los así llamados estadossatisfactorios del modelo de Ising antiferromagnético en triangulaciones incrustadas en superficiescerradas y orientables. Estos estados, como se verá más adelante, corresponden a unaclase particular de estados fundamentales o de mínima energía del modelo de Ising antiferromagnético,uno de los tópicos más extensamente estudiado en el área de la física estadística. Lasrazones para estudiar estos elementos son diversas, una de ellas es que explícitamente conectandos notables ramas de la ciencia: la matemática discreta y como ya ha sido mencionado, la físicaestadística. En este sentido y de una manera precisa que se establecerá rigurosamente en lospreliminares de este trabajo, los estados satisfactorios del modelo de Ising antiferromagnético entriangulaciones, se identifican con los emparejamientos perfectos en grafos cúbicos sin puentes.Un estado satisfactorio no necesariamente existe, pero si existe entonces todo estado demínima energía corresponde a un estado satisfactorio. En física estadística, al número de estadosde mínima energía se le conoce como la degenerancia del estado fundamental y es unparámetro vastamente estudiado y de gran interés físico, debido a que, entre otros, determinala entropía del sistema. En particular, el modelo de Ising antiferromagnético en triangulacionesestá fuertemente caracterizado por una alta degenerancia del estado fundamental, lo cual setraduce en entropía no nula.Una incrustación de un grafo en una superficie es un dibujo del grafo en la superficie (i.e. losvértices están representados por puntos y las aristas por curvas continuas) tal que cada aristaconecta los vértices que la componen y el interior de cada arista es disjunto del resto del dibujo.Además, una triangulación incrustada en una superficie cerrada y orientable Ω, o simplementeuna triangulación, es un grafo incrustado en Ω con cada cara acotada por un 3-ciclo (triángulo)del grafo.En el contexto de dualidad geométrica en incrustaciones de grafos, la degenerancia del estadofundamental en triangulaciones que admiten un estado satisfactorio, se conecta con una1

famosa conjetura de Lovász y Plummer, recientemente demostrada, la cual afirma que el númerode emparejamientos perfectos en grafos cúbicos sin puentes es exponencial en función de lacantidad de vértices del grafo. En este aspecto y de acuerdo a un resultado que estableceremosen este trabajo, se tiene que la cantidad de estados satisfactorios que admite una triangulaciónes a lo más dos veces el número de emparejamientos perfectos que tiene su dual geométrico, elcual es siempre un grafo cúbico sin puentes. Además, esta relación es una igualdad cuando seconsidera la clase de triangulaciones planas.De manera resumida y en palabras, las principales contribuciones de esta tesis, se puedendividir en tres tópicos distintos, los cuales están relacionados. En una primera instancia se desarrollala adaptación de la técnica de la matriz de transferencia para la obtención de cotas inferiorespara el valor del número de estados satisfactorios de dos familias distintas de grafos planosirregulares. El método de la matriz de transferencia, es la base de diversas técnicas de cálculoque estudian fenómenos críticos en la física estadística. De forma intuitiva y a grandes rasgos, sisuponemos que un sistema de partículas puede ser descompuesto en bloques conexos que interactúansolamente con aquellos bloques que intersectan su frontera, es posible definir una matrizque codifica las propiedades locales de cada bloque, de manera que determinadas operacionessobre esta matriz entregan la información global del sistema. Esta herramienta ha sido utilizadaprincipalmente para solucionar problemas en donde la distribución de las partículas (del sistemaanalizado) describe estructuras regulares, típicamente reticulados. Aquí justamente radica laimportancia de la adaptación desarrollada en este trabajo, puesto que se presenta una forma deemplear el método de la matriz de transferencia en estructuras geométricamente más complejas.La dificultad para aplicar el método recién descrito en los casos que se detallan en estetrabajo, radica en que no es posible fijar una única matriz para codificar localmente la información,y más aún, dependiendo de la configuración local de la estructura, ni siquiera es factiblela definición de una matriz. A cambio, para alcanzar el objetivo, la adaptación del método seapoya fuertemente en la construcción recursiva de las estructuras en cuestión. Básicamente, aeste procedimiento constructivo se le asocian operaciones y vectores que transfieren información,los cuales codifican la cantidad de estados satisfactorios de la estructura obtenida en cadapaso de la construcción.Como ya se ha dicho, el análisis de sistemas físicos en donde las partículas describen estructurasgeométricamente complejas, es un tema poco dominado tanto por físicos como pormatemáticos. Ciertamente, solamente se conocen soluciones del modelo de Ising, lo cual significacalcular su función de partición, para algunos reticulados planos. Por esto, se hace sumamenteinteresante, explorar el modelo de Ising en triangulaciones incrustadas en superficies orientables,cerradas y de género positivo.2

Por lo ya comentado, no es sorprendente, que el segundo tópico de este trabajo se enfoqueen clasificar, según su dificultad computacional, los problemas derivados de la existencia ycantidad de estados satisfactorios en triangulaciones incrustadas en superficies orientables, cerradasy de género positivo. En conjunto, se estudia el problema de decidir si una triangulaciónadmite o no un estado satisfactorio y el problema de enumerar estados satisfactorios en unatriangulación, donde las triangulaciones están incrustadas en superficies orientables, cerradas,de género positivo y arbitrario. En este punto, la arbitrariedad del género de la superficie, juegaun rol esencial, pues, se sabe que ambos problemas se pueden resolver en tiempo polinomial sise fija el género de la superficie. Acá, se muestra que tanto el problema de decisión, como elproblema de enumeración son computacionalmente difíciles cuando la triangulación está incrustadaen una superficie de género arbitrario. En líneas generales, se prueba que decidir si unatriangulación admite o no un estado satisfactorio, es un problema NP-completo y que enumerardichos estados es un problema #P-completo. Ambos resultados, se establecen mediante unareducción débilmente parsimoniosa desde la variante positiva del 3-SAT de “no todos iguales”,comúnmente abreviado como Positive-NAE-3SAT.Para la confección de las reducciones mencionadas, fue necesario el diseño de gadgets, esdecir de triangulaciones incrustadas en superficies orientables y con bordes que satisfacierancondiciones no triviales de existencia y de unicidad de estados satisfactorios. Las propiedadesde unicidad en un principio de la investigación no se pensaron factibles, debido a la alta degeneranciadel estado fundamental que caracteriza a estos sistemas. En efecto, contrariamente a lointuido por la física estadística, la elaboración de estos gadgets, guía sutilmente la construcciónde triangulaciones con una cantidad arbitraria de vértices y pocos estados fundamentales. Eneste contexto, este trabajo aporta en distintas direcciones, entre ellas, con el diseño particulary global de las gadgets dotadas de delicadas y “rebuscadas”propiedades físicas que puedeneventualmente ser útiles para estudiar la dificultad computacional de otros problema de tipoIsing relacionados y además, como veremos a continuación, da un primer paso para una mejorcomprensión de la intrincada relación entre la frustración geométrica 1 y la degenerancia delestado fundamental en física estadística.El último tema que se trabaja en esta tesis, nace como consecuencia de la confección delas recién mencionadas gadgets. Dado que en dicho estudio se encontraron triangulaciones detamaño arbitrario y una cantidad pequeña de estados satisfactorios, surgió la inquietud de sabersi el mismo fenómeno se tiene cuando se fija el género de la superficie en la que las triangulacionesestán incrustadas. Nuevamente, según los postulados de la física estadística, en estos casos, seespera que la degenerancia del estado fundamental sea exponencial en función del número de1 propiedad física que exhiben algunos sistemas de partículas, comúnmente caracterizada por una altadegenerancia del estado fundamental; en particular, las triangulaciones analizadas en este trabajo presentanfrustración geométrica.3

vértices de la triangulación, lo cual implica entropía positiva en el límite termodinámico ycoincide con la caracterización de los sistemas geométricamente frustrados. Cuando el génerode la superficie es cero, se tiene dicha exponencialidad como consecuencia de la conocida,previamente mencionada, exponencialidad de los emparejamientos perfectos en grafos cúbicos.En este trabajo se muestra que un fenómeno más complejo e inesperado, surge cuando el génerode la superficie se incrementa. Concretamente, se prueba que para toda superficie fija, cerrada,orientable Ω y de género arbitrario, existe una secuencia vértice creciente de triangulacionesincrustadas en Ω sin degeneración del estado fundamental, es decir que admiten exactamenteun par de estados de mínima energía. Además, se da a conocer una estrategia para construirtales secuencias justamente basadas en la unicidad de pares de estados satisfactorios. Estesorprendente resultado deja muchas preguntas por responder y problemas abiertos, no soloen física sino también en matemática. Asimismo, se espera que tanto la construcción de estassecuencias de triangulaciones como la estrategia para obtenerlas aporte notablemente al estudiode los sistemas geométricamente frustrados y encuentren aplicación en otras áreas.Estructura de la TesisEsta tesis consta de tres capítulos, el primero es introductorio, el segundo contiene el marcoteórico y los resultados obtenidos a partir del trabajo de tesis y el último, las conclusionesglobales y particulares de este estudio, así como también las posibles líneas de investigación aseguir.En la Sección 2.1 enunciamos formalmente el modelo de Ising en triangulaciones, introduciendodefiniciones y propiedades fundamentales que ayuden al lector a familiarizarse con elmodelo y a comprender la relevancia del trabajo que se realiza en esta tesis. En esta mismasección, se presenta una relación entre los estados satisfactorios en triangulaciones y los emparejamientosperfectos en grafos cúbicos, junto con una breve discusión acerca de la conjetura deLovász y Plummer. Esto último si bien no es explícitamente parte de las secciones posteriores,es otra de las motivaciones para este estudio, que vale la pena incluir y tener en cuenta.La Sección 2.2 de esta tesis se enfoca en la adaptación del método de la matriz de transferencia.Básicamente, esta sección se divide en dos partes: la Subsección 2.2.1 en donde seexplora la cantidad de estados satisfactorios de la familia de triangulaciones de un n-ágono y laSubsección 2.2.2 que da a conocer la adaptación del método de la matriz de transferencia paracalcular una cota inferior para la degenerancia del estado fundamental de la clase de triangulacionesapiladas. Cada apartado está compuesto por un breve comentario y por un apéndice(Apéndices A y B respectivamente) en donde se adjunta un artículo científico que contiene eldesarrollo de la teoría.4

La Sección 2.3 tiene por objetivo el estudio de la complejidad computacional de los problemasde decidir existencia de estados satisfactorios en triangulaciones y enumerar dichos estados.Al igual que en la sección anterior, se añade un artículo científico, el cual se encuentra en elApéndice C, con el trabajo realizado y los resultados pertinentes, precedido por una debidadiscusión acerca del tema.En la Sección 2.4 se presenta la construcción de las secuencias de triangulaciones dotadas deun estado fundamental no degenerado. De la misma forma que en los capítulos anteriores, estetrabajo se exhibe mediante la incorporación de un artículo científico en un apéndice, a saber elApéndice D.Finalmente, en el Capítulo 3 se presenta un breve resumen del trabajo logrado en esta tesisy se concluye con el planteamiento de problemas abiertos y las posibles líneas de investigaciónque los resultados obtenidos y las técnicas desarrolladas en este trabajo arrojan.5

Capítulo 2Marco teórico y resultados2.1. Preliminares2.1.1. El modelo de IsingEl modelo de Ising es uno de los modelos de partículas que interactúan entre sí más estudiadosen física estadística. Históricamente, el modelo de Ising, ha jugado un papel fundamentalen el desarrollado de la comprensión del ferromagnetismo, del antiferromagnetismo y de lastransiciones de fase. Su aparición data del año 1925, fecha en que se publicó la tesis doctoralde Ising [11]. A pesar de la simpleza y naturalidad del planteamiento del modelo de Ising, susolución, es decir la obtención de su así llamada función de partición, está lejos de ser conocida,salvo para casos especiales de reticulados planos [3, 11, 22]. Hoy en día, el modelo de Isingy sus generalizaciones se emplean para explicar fenómenos no solamente físicos, sino tambiénbiológicos y sociales [10, 17].Por otro lado, el estudio del modelo de Ising (y en general de la física estadística) ha estadofuertemente asociado a las matemáticas discretas. Técnicas, métodos y avances desarrolladosen el área de las matemáticas discretas han sido de gran utilidad para atacar problemas dela física estadística y vice versa (ver por ejemplo [9, 19]). Dicha inter-relación es una de lasrazones que motivan este trabajo. En efecto, en la siguiente sección se establece una particularconexión entre el modelo de Ising y los emparejamientos perfectos en grafos cúbicos.Típicamente, para estudiar el modelo de Ising, las partículas se sitúan en los vértices de ungrafo y el tipo de interacción entre ellas queda determinado por la existencia y peso de las aristasde tal grafo. En esta tesis, se analiza el modelo de Ising en donde las partículas y su interaccióndescriben triangulaciones incrustadas en superficies cerradas orientables con aristas con peso -1.Vamos ahora a describir el modelo de Ising en triangulaciones. Dada una triangulación T ,un estado de T en el modelo de Ising es una función que asigna a cada vértice de la triangula-6

ción T un valor en el conjunto {1, -1}. Los valores 1 y -1 son comúnmente llamados espines.Si V (T ) denota el conjunto de vértices de T , un estado de T puede ser representado por unelemento del conjunto {1, -1} |V (T )| .Para cada estado σ ∈ {1, -1} |V (T )| , la energía o Hamiltoniano del modelo de Ising en unatriangulación T , está dada por la ecuación∑H(σ) = - J uv σ u σ v (2.1)uv∈E(T )donde E(T ) denota el conjunto de aristas de la triangulación T y para cada arista uv en E(T )el parámetro J uv es llamado constante de acoplamiento (J uv básicamente define el tipo de interacciónentre los nodos u y v). En general, la constante de acoplamiento puede variar depositiva a negativa dependiendo del sistema que se quiera estudiar. En particular, como se hamencionado previamente, en este trabajo se estudia el tipo de interacción antiferromagnética,para lo cual se considera una constante de acoplamiento igual a -1 para cada arista. A estavariante del modelo de Ising se le conoce como modelo de Ising antiferromagnético.Dos problemas matemáticos surgen naturalmente a partir del modelo de Ising. El primeroes el estudio de los estados que proveen la mínima energía posible al sistema, conocidos comoestados fundamentales, y el segundo es el cálculo de la función de partición, la cual encapsulatoda la información física del sistema en cuestión. La función de partición del modelo de Isingen un grafo G, se define como∑f(t, G) =exp [−H(σ)/Kt], (2.2)σ∈{1,-1} |V (G)|donde K es la constante de Boltzman y t la temperatura. Este trabajo se enfoca en el primertópico, es decir en el estudio de estados fundamentales, por lo cual no vamos a entrar en detallesacerca de la función de partición. Sin embargo, vale la pena al menos conocer su definición paratener una visión de la dificultad que implica su cálculo exacto. Relacionado al análisis de losestados de mínima energía, se encuentra la degenerancia del estado fundamental, que por definicióncorresponde al número de estados de mínima energía que admite un sistema. Vale notarque el cálculo de la función de partición determina inmediatamente la degenerancia del estadofundamental. Por diversos motivos, la degenerancia del estado fundamental es una cantidadde gran interés en la física estadística (ver [25] para mayor información y más referencias), loque ha implicado que por muchos años se desarrollen e implementen técnicas que permitan sucálculo [16, 20]. En este trabajo, veremos la adaptación de uno de los métodos más renombradosen dicha área, llamado el método de la matriz de transferencia, para determinar cotas inferioresde la degenerancia del estado fundamental de dos clases de triangulaciones planas.7

En el modelo de Ising antiferromagnético, dado un estado σ ∈ {1, -1} |V (T )| , se dice quela arista uv ∈ E(T ) es frustrada por σ o que σ frustra uv si σ u = σ v . Cualquier estadoσ ∈ {1, -1} |V (T )| frustra al menos una arista de cada cara de la triangulación, ya que cada caraes acotada por un 3-ciclo. Los sistemas con la propiedad de que todo estado frustra al menosuna arista de cada cara, se conocen como sistemas geométricamente frustrados, y están caracterizadospor una alta degenerancia del estado fundamental (ver [21] para más detalles acercadel término físico frustración geométrica y sus propiedades). En otras palabras, en sistemasgeométricamente frustrados la degenerancia del estado fundamental es típicamente exponencialen función de la cantidad de vértices del grafo subyacente (ver [21, 23, 28]; en [23] seencuentran más referencias). Dentro de este contexto, en este trabajo de tesis se exhibe unfenómeno inusual e inesperado, específicamente, se presentarán sistemas geométricamente frustradosque tienen un estado fundamental no degenerado (i.e. el sistema admite solamente un parde estados fundamentales: si σ es un estado fundamental de un sistema que presenta un estadofundamental no degenerado, entonces los únicos estados fundamentales del sistema son σ y -σ).En el siguiente apartado vamos a definir un tipo especial de estado fundamental, el cualserá crucial no solamente para comprender el trabajo realizado y las técnicas desarrolladas paralograr cada uno de los resultados que presentamos en esta tesis sino también para introduciruna interesante relación que conecta la degenerancia del estado fundamental con la cantidadde emparejamientos perfectos en grafos cúbicos.2.1.2. Estados satisfactoriosEn el modelo de Ising antiferromagnético, se tiene que la energía de un estado σ de latriangulación T está dado por la expresión ∑ uv∈E(T ) σ uσ v . La máxima energía es alcanzadapor el estado que frustra todas las aristas de la triangulación, es decir el estado que asigna acada vértice espín 1. En el otro extremo, la mínima energía la entrega un estado que frustre lamínima cantidad posible de aristas. Como ya sabemos, cada estado antiferromagnético de unatriangulación T frustra al menos una arista de cada una de sus caras. Dicho esto, si un estadoσ frustra exactamente una arista de cada cara de T , es un estado fundamental (de mínimaenergía), a este tipo de estados fundamentales los llamamos estados satisfactorios. A saber, σes un estado satisfactorio de T si y solo si frustra exactamente |F (T )|/2 aristas, donde F (T )denota el conjunto de caras de la triangulación T .Si una triangulación admite un estado satisfactorio (veremos que no toda triangulaciónadmite uno), entonces todo estado fundamental corresponde a un estado satisfactorio. Por lotanto, si una triangulación admite un estado satisfactorio, la degenerancia del estado fundamentales justamente la cantidad de estados satisfactorios.8

En esta tesis utilizaremos también la definición de estado satisfactorio en pseudo-triangulaciones,es decir en grafos planos tales que cada cara excepto la cara exterior está acotada por un3-ciclo o triángulo (que será el caso de las triangulaciones de un n-ágono en la Subsección 2.2.1de la Sección 2.2) o bien en triangulaciones incrustadas en superficies orientables con bordes(lo cual se verá en el diseño de las gadgets de la Sección 2.3 y en la construcción de lastriangulaciones de la Sección 2.4).2.1.3. Emparejamientos perfectos: dual geométrico de los estadossatisfactoriosEn esta sección, se expone una relación vía dualidad geométrica entre la degenerancia delestado fundamental del modelo de Ising antiferromagnético en triangulaciones y la cantidad deemparejamientos perfectos en grafos cúbicos sin puentes.Emparejamientos perfectos en grafos cúbicosUn grafo se dice cúbico si cada uno de sus vértices tiene tres aristas incidentes. Un puenteen un grafo es una arista tal que su eliminación incrementa el número de componentes delgrafo. Además, un emparejamiento perfecto de un grafo es una colección de aristas del grafoque incide en cada vértice del grafo exactamente una vez.Un clásico teorema de Petersen del año 1891 establece que todo grafo cúbico sin puentes admiteun emparejamiento perfecto. A mediados de los años setenta, Lovász y Plummer afirmanque el número de emparejamientos perfectos en grafos cúbicos sin puentes es exponencial enfunción de la cantidad de vértices del grafo. Esta conjetura se mantuvo abierta durante cuatrodécadas y hubieron pocos pero significativos avances parciales durante ese período.La veracidad de la conjetura para grafos bipartitos fue demostrada por Voorhoeve en1979 [27], quien probó que todo grafo cúbico bipartito con n vértices tiene al menos 6(4/3) n 2 −3emparejamientos perfectos. Luego Schrijver en 1998 extendió este resultado a la clase de grafosbipartitos k-regulares [24].Uno de los resultados más importantes en el avance de esta conjetura fue obtenido porChudnovsky y Seymour en el 2008 [4]. Ellos probaron la conjetura para la clase de grafos planares.Para precisar, mostraron que todo grafo planar, cúbico sin puentes con n vértices admiteal menos 2 cn emparejamientos perfectos, donde c = 1/655978752. A pesar de la relevancia desu resultado, la prueba es considerada extremadamente compleja por las técnicas utilizadas ypor el empleo del teorema de la cuatro coloración en grafos planares.9

Además, como parte de los avances en la conjetura algunas cotas inferiores fueron obtenidas,pero ninguna exponencial, sino hasta comienzos del 2011. Esperet, Kardoš, King, Král y Norine[7], presentaron una prueba para la conjetura de Lovász y Plummer. Ellos demostraron quetodo grafo cúbico sin puentes con n vértices tiene al menos 2 n/3656 emparejamientos perfectos.Su demostración utiliza técnicas probabilistas para probar existencia y se basa fuertemente enla caracterización del polítopo de emparejamientos perfectos de Edmonds.Dualidad geométricaDada la incrustación de un grafo G en una superficie Ω, se define su dual geométrico, el cualcorresponde a un grafo G ∗ incrustado en la misma superficie Ω obtenido de la siguiente manera:dentro de cada cara f de la incrustación de G en Ω, se dibuja un vértice f ∗ de G ∗ , luego por cadaarista e de G perteneciente a las caras f 1 y f 2 de la incrustación de G en Ω (donde f 1 y f 2 no sonnecesariamente caras distintas) se dibuja una arista de G ∗ conectando los vértices f1 ∗ y f2 ∗ . Unaincrustación de un grafo en una superficie es llamada fuerte si es 2-celular, es decir si cada carade la incrustación es homeomorfa a un disco abierto y si además, cada cara es acotada por unciclo del grafo. En particular, en una incrustación fuerte cada arista pertenece a exactamentedos caras distintas de la incrustación y el dual geométrico de tal grafo incrustado no tiene loops.En 1985, Jaeger conjeturó que todo grafo 2-conexo admite una incrustación fuerte en unasuperficie cerrada orientable [12] — esta conjetura es equivalente a la versión dirigida de lafamosa conjetura del ciclo de doble cobertura para la clase de grafos cúbicos sin puentes.Asumiendo que la conjetura de Jaeger es cierta, todo grafo cúbico sin puentes admite talincrustación y por lo tanto su dual geométrico es una triangulación. Por otro lado, el dualgeométrico de una triangulación es siempre un grafo cúbico sin puentes.Emparejamientos perfectos y estados satisfactoriosUn conjunto M de aristas de una triangulación se dice que es un conjunto de intersecciónde aristas si contiene exactamente una arista de cada cara (triángulo) de la triangulación.En general, asumiendo la conjetura de Jaeger tenemos que, dado G un grafo cúbico sinpuentes, M es un conjunto de intersección de aristas de G ∗ si y solo si M es un emparejamientoperfecto de G. En particular, la relación se tiene para grafos cúbicos planares, sin necesidad deutilizar la conjetura de Jaeger.Por otro lado, en una triangulación, el conjunto de aristas frustradas por un estado satisfactorioes un conjunto de intersección de aristas. Observemos que, dos estados satisfactoriosdistintos, definen el mismo conjunto de aristas frustradas si y solamente si uno es igual a menosel otro. Dado esto, el número de estados satisfactorios que admite una triangulación es a lo más10

dos veces la cantidad de conjuntos de intersección de aristas. En general, esta relación no es unaidentidad. Para triangulaciones planas la equivalencia es cierta: el grafo obtenido a partir de laeliminación de un conjunto de intersección de aristas de una triangulación plana, es un grafobipartito pues cada cara del grafo queda acotada por un ciclo de largo cuatro. La biparticióndetermina precisamente un par de estados satisfactorios, simplemente asignando espines igualesy contrarios a cada bipartición.En resumen, se tiene que si un estado satisfactorio existe, entonces la degenerancia del estadofundamental de una triangulación T es a lo más dos veces el número de emparejamientosperfectos que admite su dual geométrico T ∗ . Siendo esta relación una igualdad cuando se añadela condición de planaridad.Esta conexión entre estados satisfactorios y emparejamientos perfectos es en un principiouna de las motivaciones para el estudio que realizamos y nos abre paso hacia el desarrollo de unainteresante teoría entorno al comportamiento de los estados satisfactorios en triangulaciones.2.2. Método de la matriz de transferenciaEl método de la matriz de transferencia es una técnica clásica que se ha utilizado extensamenteen la física estadística. Este método ha sido de particular relevancia en el desarrolloteórico de los modelos de partículas que interactúan entre sí, entre otros, hace manejable lasecuaciones fundamentales de las teorías cuántica y electromagnética. Sin ir más lejos, el modelode Ising en una dimensión (cadena en línea recta), fue resuelto por Ising [11] utilizando comotécnica principal el método de la matriz de transferencia y luego, en diversas áreas de la ciencia,una gran cantidad de problemas en donde las partículas que interactúan entre sí describenreticulados, se han resuelto empleando críticamente este método (ver por ejemplo [6, 18]). Asimismo,se ha aplicado esta herramienta para atacar el problema de determinar la degeneranciadel estado fundamental [20]. En general, las técnicas, métodos y adaptaciones del método dela matriz de transferencia desarrollados para determinar cotas de la degenerancia del estadofundamental han sido centradas y exitosas en estudios de reticulados planos, no así en sistemasmás complejos, probablemente debido a la mayor dificultad que implica trabajar con estructuras(grafo subyacente) que presentan una geometría más irregular.En esta sección se discute la adaptación del método de la matriz de transferencia para estudiarla cantidad de estados satisfactorios de una familia de pseudo-triangulaciones planas y ladegenerancia del estado fundamental de una familia de triangulaciones planas. Cabe mencionar,que para cada clase de grafos estudiada la implementación del método es diferente, básicamenteen una primera instancia se adecuan los fundamentos de la técnica de la matriz de transferenciaa cada tipo de estructura a analizar y luego, se aplica una estrategia particular para cada caso.11

2.2.1. Triangulaciones de un n-ágonoLa primera familia de grafos planos que se explora, corresponde a la clase de triangulacionesde un n-ágono. A partir de este primer estudio, se escribió un artículo que lleva por título SatisfyingStates of triangulations of a Convex n-gon, publicado en marzo del 2010 por la revista TheElectronic Journal of Combinatorics y que es incluido como parte de esta tesis en el Apéndice A.Mediante una adaptación no trivial del método de la matriz de transferencia, se obtiene primerouna fórmula exacta para el número de estados satisfactorios de una subclase de las triangulacionesde un n-ágono, la cual está formada por un tipo particular de pseudo-triangulacionesque llamamos cadenas de triángulos y luego, una cota inferior exponencial para la cantidad deestados satisfactorios de la familia de triangulaciones de un n-ágono. Específicamente, denotandopor F k el k-ésimo número en la serie de Fibonacci y ϕ = (1 + √ 5)/2 ≈ 1,61803 la razónde oro, se establecen los siguientes resultados:Teorema 2.1 Si T es una cadena de triángulos con |V (T )| = n, entonces la cantidad deestados satisfactorios de T en el modelo de Ising antiferromagnético es igual a 2F n+1 .Teorema 2.2 Si T es una triangulación de un n-ágono, entonces la cantidad de estadossatisfactorios de T en el modelo de Ising antiferromagnético es al menos ϕ 2 ( √ ϕ) n . Donde,√ ϕ ≈ 1,27202.2.2.2. Triangulaciones apiladasEn esta parte se desarrolla una técnica para obtener una cota inferior para la degeneranciadel estado fundamental de la clase de triangulaciones apiladas, la cual corresponde a una familiade triangulaciones planas. Esta técnica en un principio tiene sus bases en el método de lamatriz de transferencia, para luego pasar a ser una técnica autónoma que puede eventualmenteser aplicada para calcular cotas en otros tipos de estructuras irregulares.Este estudio también generó un trabajo escrito, titulado Antiferromagnetic Ising model intriangulations with applications to counting perfect matchings. Dicho trabajo fue sometido parapublicación en noviembre del 2011 a la revista Discrete Applied Mathematics y también está incluidocomo parte de este trabajo en el Apéndice B.Denotando por ϕ = (1 + √ 5)/2 ≈ 1,61803 la razón de oro, se obtuvieron los siguientesresultados:Teorema 2.3 Sea T una triangulación apilada con |V (T )| = n, la degenerancia del estadofundamental de T en el modelo de Ising antiferromagnético es al menos 6ϕ 136 (n+3) .Además, como una consecuencia directa del teorema precedente, se tiene lo siguiente.12

Corollary 2.2.2.1 El número de emparejamientos perfectos de un grafo cúbico G, cuyo grafodual es una triangulación apilada es al menos 3ϕ 1 |V (G)|722.3. Complejidad computacionalMuchos problemas en física estadística son computacionalmente difíciles. En particular, losproblemas de tipo Ising son comúnmente problemas pertenecientes a la famosa clase NP o biena su análoga de enumeración, la clase #P [2, 5]. Informalmente, un problema de decisión pertenecea la clase NP si existe una máquina de Turing M no determinista y a tiempo polinomialde manera que el problema de decisión admite una solución x si y solo si M acepta x; deforma análoga, un problema de enumeración pertenece a la clase #P si existe una máquina deTuring M no determinista y a tiempo polinomial de manera que el problema de enumeraciónse resuelve contando el número de aceptaciones de M (definiciones formales de las clases NP y#P se encuentran en [1, Capítulos 2 y 17]). Por ejemplo, en este trabajo veremos que contar estadossatisfactorios en una triangulación está en #P puesto que es posible reconocer en tiempopolinomial si una asignación o configuración de espines dada es o no satisfactoria. Al igual queen NP, los problemas más difíciles de la clase #P son los llamados problemas #P-completos.En este contexto y a modo de ejemplo, en general, el problema de contar emparejamientosperfectos en un grafo es un clásico problema #P-completo, dado que se reduce a determinarla permanente de una matriz [26]. Paradójicamente, el problema de contar emparejamientosperfectos en grafos planares puede ser ejecutado en tiempo polinomial pues se reduce a evaluarel Pfaffian de una matriz, lo cual equivale a calcular un determinante [14, 15].Por otro lado, es sabido que la función de partición (ver ecuación 2.2) del modelo de Isingen un grafo se puede reescribir en términos del polinomio de Tutte de dicho grafo evaluadoen una hipérbola [8]. En general, determinar el polinomio de Tutte de un grafo e incluso laevaluación del polinomio en un punto en particular es un problema #P-difícil excepto en casosmuy especiales donde la evaluación puede ser ejecutada en tiempo polinomial [13]. En relación aesto, dada una superficie fija, cerrada y orientable Ω, la función de partición del modelo de Isingantiferromagnético en grafos incrustados en Ω puede ser calculada en tiempo polinomial [9].Esto implica que decidir existencia de estados satisfactorios y calcular la degenerancia del estadofundamental (implícitamente contar estados satisfactorios) en triangulaciones incrustadasen Ω son problemas que se pueden resolver en tiempo polinomial.Como ya anticipamos, en este capítulo estudiamos la complejidad computacional del problemade decidir si una triangulación (incrustada en una superficie cerrada orientable arbitraria)admite o no un estado satisfactorio y además, el problema de enumeración asociado.Concretamente, probamos que es NP-completo decidir si una triangulación admite o no unestado satisfactorio y que es #P-completo determinar el número de estados satisfactorios de13

una triangulación. Junto con esto, estudiamos también la dificultad computacional de calcularla degenerancia del estado fundamental en triangulaciones, demostrando que es un problema#P-difícil. Además, a consecuencia de los resultados obtenidos en este trabajo y debido a labiyección entre pares de estados del modelo de Ising antiferromagnético y cortes de aristas enuna triangulación (en particular, los estados fundamentales corresponden a cortes de aristasmaximales), se tiene que el problema de determinar el número de cortes de aristas maximales,conocido como MAX-CUT en triangulaciones es #P-difícil.Producto de la investigación relacionada con la complejidad computacional de los problemasasociados a los estados satisfactorios en triangulaciones, se escribió el trabajo que tiene por títuloComputational Hardness of Enumerating Satisfying Spin-Assignments in Triangulations el cualsometimos para publicación en Julio del 2011 a la revista Theoretical Computer Science. Dichoartículo es incorporado en el Apéndice C.2.4. Estado fundamental no degeneradoDiscutiremos ahora sobre una estrategia para construir triangulaciones incrustadas en superficiescerradas orientables con un estado fundamental no degenerado. Como ya fue mencionadoen el Capítulo 2.1, debido a la biyección que existe entre pares de estados fundamentales entriangulaciones planas y emparejamientos perfectos en grafos planares cúbicos sin puentes, ladegenerancia del estado fundamental en triangulaciones planas es exponencial en función de lacantidad de vértices de la triangulación. Al principio de esta investigación, motivados por laConjetura de Lovász y Plummer y en gran parte por la hipótesis de la física estadística acercade la alta degenerancia del estado fundamental en sistemas geométricamente frustrados [21], setenía como finalidad desarrollar técnicas para mostrar la exponencialidad de la degeneranciadel estado fundamental en triangulaciones incrustadas en superficies cerradas, orientables y degénero positivo que admiten estados satisfactorios.Sorprendentemente, este estudio dio un giro cuando al concluir el trabajo de complejidadcomputacional notamos que a partir de la construcción de los gadgets involucrados era posibleobtener triangulaciones con pocos estados satisfactorios y una gran cantidad de vértices. Luegode esto, rápidamente surgió la interrogante de si el mismo fenómeno se tiene al fijar la superficieen la cual las triangulaciones están incrustadas. El trabajo que se adjunta en el Apéndice Dresponde a esta pregunta afirmativamente y actualmente se encuentra disponible “on-line” enel repositorio científico ArXiv.14

Capítulo 3ConclusionesLas conclusiones de esta tesis se dividen en tres secciones: resultados, contribuciones yfuturas líneas de investigación.ResultadosEn la Figura 3.1, se muestra una tabla comparativa que exhibe el ya conocido comportamientode los emparejamientos perfectos en grafos cúbicos sin puentes junto con los resultadosque se han obtenido en este trabajo de tesis concernientes al comportamiento de los estadossatisfactorios en triangulaciones incrustadas en superficies cerradas y orientables.Relación Grafo Primal Dual geométricotipos de grafos triangulación grafo cúbico sin puentesobjetos estado satisfactorio emparejamiento perfectoexistencia de objetoscantidad de objetosdificultad computacionaldel problema deexistencia asociadodificultad computacionaldel problema deconteo asociadono siempre existe un estado satisfactorio;salvo en triangulaciones planasexponencial en el número de vérticespara triangulaciones planas; existenfamilias de triangulaciones que admitenuna cantidad constante de estadossatisfactorios.NP-completo (género arbitrario)#P-completo (género arbitrario)todo grafo cúbico sin puentes admiteun emparejamiento perfectoexponencial en el número de vérticesP#P-completoFigura 3.1: Estados satisfactorios contra emparejamientos perfectos15

ContribucionesLas contribuciones más relevantes de este trabajo de tesis se pueden resumir concretamenteen los siguientes cuatro ítems:Planteamiento de una relación entre emparejamientos perfectos en grafos cúbicos sinpuentes y estados satisfactorios en triangulaciones, junto con una completa descripciónacerca de las diferencias y similitudes que caracterizan dicha relación.Desarrollo de adaptaciones no triviales del método de la matriz de transferencia paracontar estados satisfactorios en familias de triangulaciones irregulares.Elaboración de exclusivas y originales gadgets que permiten la clasificación, según su dificultadcomputacional, de los problemas asociados a la existencia y cantidad de estadossatisfactorios en triangulaciones, así como también del problema de calcular la degeneranciadel estado fundamental en triangulaciones.Descubrimiento y diseño de una estrategia para la construcción de triangulaciones con unestado fundamental no degenerado.Trabajo a futuroEn los siguientes puntos se discuten las posibles líneas de investigación y los problemasabiertos que surgen a partir de este trabajo de tesis.En cuanto a las técnicas desarrolladas para contar estados satisfactorios en triangulaciones,sería interesante buscar familias de triangulaciones incrustadas en superficies cerradas,orientables y de género positivo, tales que, el método que se utilizó en este trabajopueda ser extendido y aplicado con el fin de demostrar exponencialidad de la degeneranciadel estado fundamental para dichas familias de triangulaciones. A grandes rasgos, unprimer requisito que debieran cumplir estas triangulaciones es que admitan un tipo deconstrucción recursiva.Relativo a la clase de grafos cúbicos sin puentes con un dual geométrico que no admite unestado satisfactorio, una pregunta que surge naturalmente es si esta clase de grafos cúbicoscorresponde o no a alguna familia especial de grafos y/o se pueden describir mediantealguna propiedad combinatorial no trivial.Un problema abierto que nace a partir de los resultados de este trabajo de tesis, es encontraruna completa caracterización de las triangulaciones que tienen un estado fundamentalno degenerado. Junto con esto, es también interesante estudiar el otro extremo, es decirexplorar la posibilidad de dar una caracterización de las triangulaciones que admiten unacantidad exponencial de estados satisfactorios.16

Dado que en este trabajo se ha demostrado que en general la degenerancia del estadofundamental en triangulaciones no es exponencial, ni siquiera polinomial en el númerode vértices de la triangulación, queda abierto saber que sucede con la degenerancia delestado fundamental en triangulaciones aleatorias.Un problema algo más particular, es estudiar, en aquellas triangulaciones que tienenun estado fundamental no degenerado, el comportamiento de la degenerancia del estadofundamental si se cambia la constante de acoplamiento (de -1 a 1) en exactamente unaarista.La última discusión está motivada justamente por el punto anterior y por los sistemasllamados vidrios de espín. En los vidrios de espín tipo Ising, las constantes de acoplamientoson aleatoriamente distribuidas; típicamente, cada constante de acoplamiento es 1 o-1 con igual probabilidad. En este sentido, el caso del modelo de Ising antiferromagnéticoes el caso para el cual cada constante de acoplamiento es -1 con probabilidad 1,i.e. P[J e =-1] = 1 para toda arista e. Luego, un tema interesante para analizar es elmodelo de Ising en triangulaciones dentro de este contexto aleatorio. Más precisamente,es de particular relevancia investigar el comportamiento de la degenerancia del estadofundamental en las triangulaciones que tienen un estado fundamental antiferromagnéticono degenerado, a medida que la probabilidad P[J e =-1] decrece. Por ejemplo, que sucedecon la degenerancia del estado fundamental si P[J e =1]=1/n, donde n denota el númerode vértices de la triangulación en cuestión.17

Bibliografía[1] S. Arora and B. Barak. Computational Complexity: A Modern Approach. CambridgeUniversity Press, New York, NY, USA, 1st edition, 2009.[2] F. Barahona. On the computational complexity of Ising spin glass models. Journal ofPhysics A: Mathematical and General, 15(10):3241–3253, 1982.[3] R. Baxter. Exactly Solved Models in Statistical Mechanics. Dover Publications, 2008.[4] M. Chudnovsky and P. Seymour. Perfect matchings in planar cubic graphs. Manuscript,2008.[5] B. Cipra. The Ising model is NP-complete. SIAM News, 33(6):1–3, 2000.[6] M. Creutz. Gauge fixing, the transfer matrix, and confinement on a lattice. PhysicalReview D, 15(4):1128–1136, 1977.[7] L. Esperet, F. Kardos, A. King, D. Král, and S. Norine. Exponentially many perfectmatchings in cubic graphs. Advances in Mathematics, 227(4):1646–1664, 2011.[8] C. Fortuin and P. Kasteleyn. On the random-cluster model: I. Introduction and relationto other models. Physica, 57(4):536 – 564, 1972.[9] A. Galluccio and M. Loebl. On the theory of Pfaffian orientations. I. Perfect matchingsand permanents. The Electronic Journal of Combinatorics, 6, 1998.[10] A Grabowski and R Kosinski. Ising-based model of opinion formation in a complex networkof interpersonal interactions. Physica A: Statistical Mechanics and its Applications,361(2):651–664, 2006.[11] E. Ising. Beitrag zur Theorie des Ferromagnetismus. Zeitschrift fur Physik, 31(1):253–258,1925.[12] F. Jaeger. A survey of the cycle double cover conjecture. In Cycles in Graphs, Annalsof Discrete Mathematics, volume 27 of North-Holland Mathematics Studies, pages 1–12.1985.18

[13] F. Jaeger, D. Vertigan, and D. Welsh. On the computational complexity of the Jonesand Tutte polynomials. Mathematical Proceedings of the Cambridge Philosophical Society,108(1):35–53, 1990.[14] P. Kasteleyn. Dimer statistics and phase transition. Journal of Mathematical Physics,4(2):287–293, 1963.[15] P. Kasteleyn. Graph Theory and Crystal Physics. In Graph Theory and TheoreticalPhysics, pages 43–110. Academic Press, 1967.[16] S. Kirkpatrick. Frustration and ground-state degeneracy in spin glasses. Physical ReviewB, 16(10):4630–4641, 1977.[17] M. Kitzbichler, M. Smith, S. Christensen, and E. Bullmore. Broadband criticality of humanbrain network synchronization. PLoS Computational Biology, 5(3):e1000314, 2009.[18] A. Kloczkowski, T. Sen, and R. Jernigan. The transfer matrix method for lattice proteins—anapplication with cooperative interactions. Polymer, 45(2):707–716, 2004.[19] M. Loebl. Discrete Mathematics in Statistical Physics. Advanced Lectures in Mathematics.Vieweg and Teubner, 2009.[20] M. Loebl and J. Vondrák. Towards a theory of frustrated degeneracy. Discrete Mathematics,271(1-3):179–193, 2003.[21] R. Moessner and A. Ramirez. Geometrical Frustration. Physics Today, 59(2), 2006.[22] L. Onsager. Crystal statistics. I. A two-dimensional model with an order-disorder transition.Physical Review, 65(3-4):117–149, 1944.[23] K. Raman. Geometry, frustration, and exotic order in magnetic systems. PhD thesis,Princeton University, 2005.[24] A. Schrijver. Counting 1-factors in regular bipartite graphs. Journal of CombinatorialTheory, Series B, 72(1):122–135, 1998.[25] J. Sethna. Statistical Mechanics: Entropy, Order Parameters and Complexity (OxfordMaster Series in Physics). Oxford University Press, USA, 2006.[26] L. Valiant. The complexity of computing the permanent. Theoretical Computer Science,8(2):189–201, 1979.[27] M. Voorhoeve. A lower bound for the permanents of certain (0, 1)-matrices. IndagationesMathematicae, 82(1):83–86, 1979.[28] G. Wannier. Antiferromagnetism. The triangular Ising net. Physical Review, 79(2):357–364, 1950.19

Apéndice AEstados satisfactorios entriangulaciones de un n-ágono20

Satisfying states of triangulations of a convex n-gonA. Jiménez ∗Departamento de Ingeniería MatemáticaUniversidad de Chileajimenez@dim.uchile.clM. Kiwi †Departamento de Ingeniería Matemática &Centro de Modelamiento Matemático UMI 2807, CNRS-UChileUniversidad de Chilewww.dim.uchile.cl/∼mkiwi/M. Loebl ‡Department of Applied Mathematics &Institute for Theorical Computer ScienceCharles Universitykam.mff.cuni.cz/∼loebl/Submitted: Dec 23, 2009; Accepted: Feb 26, 2010; Published: Mar 8, 2010Mathematics Subject Classification: 05C30AbstractIn this work we count the number of satisfying states of triangulations of aconvex n-gon using the transfer matrix method. We show an exponential (in n)lower bound. We also give the exact formula for the number of satisfying states ofa strip of triangles.1 IntroductionA classic theorem of Petersen claims that every cubic (each degree 3) graph with nocutedge has a perfect matching. A well-known conjecture of Lovasz and Plummer from the∗ Gratefully acknowledges the support of Mecesup via UCH0607 Project, CONICYT via Basal-FONDAP in Applied Mathematics, FONDECYT 1090227 and the partial support of the Czech ResearchGrant MSM 0021620838 while visiting KAM MFF UK.† Gratefully acknowledges the support of CONICYT via Basal-FONDAP in Applied Mathematics andFONDECYT 1090227.‡ Partially supported by Basal project Centro de Modelamiento Matemático, Universidad de Chile.the electronic journal of combinatorics 17 (2010), #R39 1

mid-1970’s, still open, asserts that for every cubic graph G with no cutedge, the numberof perfect matchings of G is exponential in |V(G)|. The assertion of the conjecture wasproved for the k−regular bipartite graphs by Schrijver [Sch98] and for the planar graphsby Chudnovsky and Seymour [CS08]. Both of these results are difficult. In general, theconjecture is widely open; see [KSS08] for a linear lower bound obtained so far.We suggest to study the conjecture of Lovasz and Plummer in the dual setting. Thisrelates the conjecture to a phenomenon well-known in statistical physics, namely to thedegeneracy of the Ising model on totally frustrated triangulations of 2−dimensional surfaces.In order to explain this we need to start with another well-known conjecture, namelythe directed cycle double cover conjecture of Jaeger (see [Jae00]): Every cubic graph withno cutedge can be embedded in an orientable surface so that each face is homeomorphicto an open disc (i.e., the embedding defines a map) and the geometric dual has no loop.By a slight abuse of notation we say that a map in a 2−dimensional surface is atriangulation if each face is bounded by a cycle of length 3 (in particular there is noloop); hence we allow multiple edges. We say that a set S of edges of a triangulation T isintersecting if S contains exactly one edge of each face of T.Assuming the directed cycle double cover conjecture, we can reformulate the conjectureof Lovasz and Plummer as follows: Each triangulation has an exponential number ofintersecting sets of edges.We next consider the Ising model. Given a triangulation T = (V, E), we associate thecoupling constant c(e) = −1 with each edge e ∈ E. A spin-assignment of U ⊆ V is afunction σ : U → {+, -} where + denotes 1 and - denotes −1. Each spin-assignment of Uis naturally identified with an element from {+, -} |U| . A state of the Ising model is anyspin-assignment of V. The energy of a state s is defined as − ∑ {u,v}∈Ec(uv) · σ(u) · σ(v).The states of minimum energy are called groundstates. The number of groundstates isusually called the degeneracy of T, denoted g(T), and it is an extensively studied quantity(for regular lattices T) in statistical physics (see for example [LV03]). Moreover, a basictool in the degeneracy study is the transfer matrix method.We further say that a state σ frustrates edge {u, v} if σ(u) = σ(v). Clearly, eachstate frustrates at least one edge of each face of T, and a state is a groundstate if itfrustrates the smallest possible number of edges. We say that a state σ is satisfying if σfrustrates exactly one edge of each face of T. Hence, the set of the frustrated edges of anysatisfying state is an intersecting set defined above, and we observe: The number of thesatisfying states is at most twice the number of the intersecting sets of edges. Moreover,the converse also holds for planar triangulations: if we delete an intersecting set of edgesfrom a planar triangulation, we get a bipartite graph and its bipartition defines a pair ofsatisfying states.We finally note that a satisfying state does not need to exist, but if it exists, then theset of the satisfying states is the same as the set of the groundstates.Summarizing, half the number of satisfying states is a lower bound to the number ofintersecting sets. We can also formulate the result of Chudnovsky and Seymour by: Eachplanar triangulation has an exponential degeneracy. This motivates the problem we studythe electronic journal of combinatorics 17 (2010), #R39 2

as well as the (transfer matrix) method we use.Given C n a convex n-gon, a triangulation of C n is a plane graph obtained from C n byadding n−3 new edges so that C n is its boundary (boundary of its outer face). We denoteby ∆(C n ) the set of all triangulations of C n . An almost-triangulation is a plane graph sothat all its inner faces are triangles. Note that if n 3, then ∆(C n ) is a subset of theset of almost-triangulations with n − 2 inner faces. For T an almost-triangulation, we saythat a state σ is satisfying if σ frustrates exactly one edge of each triangular face of T. Wedenote by s(T) the number of satisfying states of an almost-triangulation T. The maingoal of this work is to show that the number of satisfying states of any triangulation of aconvex n-gon is exponential in n.Organization: We first recall, in Section 2, a known and simple bijection between triangulationsof a convex n-gon and plane ternary trees with n − 2 internal vertices. We thenformally state the main results of this work. In Section 3 we give a constructive step bystep procedure that given a plane ternary tree Γ with n − 2 internal vertices, sequentiallybuilds a triangulation T of a convex n-gon by repeatedly applying one of three differentelementary operations. Finally, in Section 4 we interpret each elementary operation interms of operations on matrices. Then, we apply the transfer matrix method to obtain,for each triangulation of a convex n-gon T, an expression for a matrix whose coordinatesadd up to the number of satisfying states of T. We then derive a closed formula for thenumber of satisfying states of a natural subclass of ∆(C n ); the class of “triangle strips”.Finally, we establish an exponential lower bound for the number of satisfying states oftriangulations of a convex n-gon. Future research directions are discussed in Section 5.2 Structure of the class of triangulations of a convexn-gonLet T be a triangulation of a convex n-gon. Denote by F(T) the set of inner faces of Tand let {I(T), O(T)} be the partition of F(T) such that ∆ ∈ I(T) if and only if no edge of∆ belongs to the boundary of T (i.e. to C n ). We henceforth refer to the elements of I(T)by interior triangles of T. Consider now the bijection Γ between ∆(C n ) and the set of allplane ternary trees with n − 2 internal vertices and n leaves that maps T to Γ T so that:(i) {γ ∆ , γ ∆ ′} is an edge of Γ T if and only if ∆ and ∆ ′ are inner faces of T that sharean edge, and(ii) e is a leaf of Γ T adjacent to γ ∆ if and only if e is an edge of C n that belongs to ∆.(See Figure 1 for an illustration of how Γ acts on an element of ∆(C n ).) The bijectionΓ induces another bijection, say γ, from the inner faces of T (i.e. F(T)), to the internalvertices of Γ T . In particular, inner faces ∆ and ∆ ′ of T share an edge if and only if{γ ∆ , γ ∆ ′} is an edge of Γ T which is not incident to a leaf. Hence, γ identifies interiortriangles of T with internal vertices of Γ T that are not adjacent to leaves.the electronic journal of combinatorics 17 (2010), #R39 3

010000 1111 010000 11110000 11110000 11110000 1111010000 11110000 1111 0100 11 0000 1111010000 1111 0000 11110000 1111 0000 111100 1101000 1110000000000000011111110000 1111 0000 1111 1111111010000 1111000 1110000 1111000 1110000 1111000 1110000 111101000 111 00 110000 1111000 111 00 110000 1111000 1110000 1111000 11111000011110000 1111000 1110000 11110000 111101 000 1110000 11110000 111100 11 0000000111111101000 111 000 111 00 110000 1111 00 11 00000001111111000 111 000 111 000 1110000 1111 00 11000 111 000 1110000 1111 00 11000 111 000 1110000 1111 00 11000 1110100 11011100000000000 1110000 1111000 111 1110000 1111000 111000 111 11100 111110001110000111100111000111000011110011100011100001111001110001110000111100111000011110000111100111000000 111 011110000111100000 111 000 111 000 1110000 111100000 111 000 111010000111100110001110100 1101110001110100 11000 1110100 11000 11101000 11101000 11101000 111 01 01000 11100 110101 00 1101Figure 1: A triangulation of a convex 9-gon T and the associated tree Γ T .2.1 Main resultsSay a triangulation of a convex n-gon T is a strip of triangles provided |I(T)| = 0.Our first result is an exact formula for the number of satisfying states of any strip oftriangles. Our second main contribution gives an exponential lower bound for the numberof satisfying states of any triangulation of a convex n-gon. Specifically, denoting by F kthe k-th Fibonacci number and ϕ = (1 + √ 5)/2 ≈ 1.61803 the golden ratio, we establishthe following results:Theorem 1 If T is a triangulation of a convex n-gon with |I(T)| = 0, then s(T) = 2F n+1 .Theorem 2 If T is a triangulation of a convex n-gon, then s(T) ϕ 2 ( √ ϕ) n . Moreover,√ ϕ ≈ 1.27202.3 Construction of triangulations of a convex n-gonIn this section we discuss how to iteratively construct any triangulation of a convex n-gon.First, we introduce two basic operations whose repeated application allows one to buildstrips of triangles. Then, we describe a third operation which is crucial for recursivelybuilding triangulations with a non-empty set of interior triangles from triangulations withfewer interior triangles.3.1 Basic operationsLet T = (V, E) be a triangulation of a convex n-gon. We will often distinguish a boundaryedge of T to which we shall refer as bottom edge of T and denote by ⌊T⌋.We now define two elementary operations (see Figure 2 for an illustration):the electronic journal of combinatorics 17 (2010), #R39 4

Operation WInput: (T, ⌊T⌋) where T ∈ ∆(C n ) and ⌊T⌋ = (β 1 , β 2 ).Output: (̂T, ⌊̂T⌋), where ̂T ∈ ∆(C n+1 ) is a triangulation obtained fromT by adding a new vertex ̂β 1 to T and two new edges {̂β 1 , β 1 }and {̂β 1 , β 2 }. Moreover, ⌊̂T⌋ = (̂β 1 , β 2 ).Operation ZInput: (T, ⌊T⌋) where T ∈ ∆(C n ) and ⌊T⌋ = (β 1 , β 2 ).Output: (̂T, ⌊̂T⌋), where ̂T ∈ ∆(C n+1 ) is a triangulation obtained fromT by adding a new vertex ̂β 2 to T and two new edges {β 1 , ̂β 2 }and {̂β 2 , β 2 }. Moreover, ⌊̂T⌋ = (β 1 , ̂β 2 ).Henceforth, we also view operations W and Z as maps from inputs to outputs. Abusingterminology, we consider two nodes joined by an edge to be a degenerate triangulationwhose bottom edge is its unique edge. Let T 0 be a degenerate triangulation. Say that ⌊T 0 ⌋is the top edge of T, denoted ⌈T⌉ (see Figure 2), if there is a sequence R 1 , . . . , R l ∈ {W, Z}such that (T, ⌊T⌋) is obtained by evaluating R l ◦ · · · ◦ R 2 ◦ R 1 at (T 0 , ⌊T 0 ⌋). When bottomedges are clear from context, we shall simply writeT = R l ◦ · · · ◦ R 2 ◦ R 1 (T 0 ) .WZ⌈T⌉α 1 α 2α 1α 2⌈T⌉⌊T⌋β 1 β 2β 1 β 2⌊T⌋bβ 1b β2Figure 2: An arbitrary strip of triangles T with ⌈T⌉ = (α 1 , α 2 ) and ⌊T⌋ = (β 1 , β 2 ).Operations W and Z evaluated at (T, ⌊T⌋).3.2 The |I(T)| = 0 caseOur goal in this section is to show that any triangulation of a convex n-gon with nointerior triangles can be obtained by sequentially applying basic operations of type Wand Z starting from a degenerate triangulation.Let T be a triangulation such that |I(T)| = 0. Note that each internal vertex of Γ Tis adjacent to at least one leaf. Hence, Γ T has two internal vertices each one adjacentthe electronic journal of combinatorics 17 (2010), #R39 5

to exactly two leaves, and n − 4 internal vertices adjacent to exactly one leaf. Thisimplies that Γ T is made up of a path P = γ ∆ 1 . . . γ ∆ n−2 with two leaves connected toeach γ ∆ 1 and γ ∆ n−2, and one leaf connected to each internal vertex of the path P (seeFigure 3). To obtain T from Γ T we choose one of the two endnodes of the path (say γ ∆ 1)and sequentially add the triangles ∆ 1 , . . . , ∆ n−2 one by one, according to the bijectionγ, starting from γ ∆ 1 and following the trajectory of the path P . Consequently, we canconstruct T from a pair of vertices (α 1 , α 2 ) of ∆ 1 by applying a sequence of n−2 operationsR 1 , R 2 , . . . , R n−2 ∈ {W, Z}, where the choice of each operation depends on the structureof Γ T . For example, for the triangulation in Figure 3, provided ⌈T⌉ = (α 1 , α 2 ) and⌊T⌋ = (β 1 , β 2 ), we have that R 1 = W, R 2 = Z, R 3 = Z, and so on and so forth.˜Γ˜T00 11 010101γ ∆ 20101γ ∆ 10101γ ∆ 3 0100 11 0100 11 01γ01∆ 40101010100 11 010100 11 0101γ ∆ n−5γ ∆ n−4010101010100 1101 γ ∆ n−30101γ ∆ n−2β 1∆ 101 00 11∆ 200 1100 11∆ 3∆ 400 1100 11010101 00 11∆ n−50101∆ n−401∆ n−3∆ n−2α 1α 200 1100 1100 11010100 1100 11β 2Figure 3: A tree ˜Γ in the range of bijection Γ and construction of triangulation ˜T suchthat Γ eT = ˜Γ.The next result summarizes the conclusion of the previous discussion.Lemma 3 For any T ∈ ∆(C n ) it holds that |I(T)| = 0 if and only if there is a degeneratetriangulation T 0 and basic operations R 1 , R 2 , . . . , R n−2 ∈ {W, Z} such thatT = R n−2 ◦ · · · ◦ R 2 ◦ R 1 (T 0 ) .In fact, there are non–negative integers w 1 , . . . , w m , z 1 , . . . , z m adding up to n − 2 suchthat w j 1 for j ≠ 1, z j 1 for j ≠ m, andT = Z zm ◦ W wm ◦ · · · ◦ Z z 2◦ W w 2◦ Z z 1◦ W w 1(T 0 ) .the electronic journal of combinatorics 17 (2010), #R39 6

3.3 The |I(T)| 1 caseWe now consider the following additional basic operation (see Figure 4 for an illustration):Operation •Input: (T i , ⌊T i ⌋) where T i ∈ ∆(C ni ), i ∈ {1, 2} and ⌊T i ⌋ = (β1, i β2).iOutput: (T, ⌊T⌋), where T ∈ ∆(C n1 +n 2 −1) is a triangulation obtainedfrom T 1 and T 2 by identifying β1 2 with β2 1 and adding the edge{β1, 1 β2}. 2 Moreover, ⌊T⌋ = (β1, 1 β2).200 11 00000111110100 1101 01 0000 1111 00 110101010101010101010101010101010000011110100100101111 000000011111001111T 1 • T 201010101β = β010101012 1 = β2 101010101010101010101010101010100 11010000011111 00 1101010000 1111 00 110100 11 00 110100 11β11 β12 β21 β22T = T 1 • T 2000000111111 00000011111100 11000000111111 000000111111000 11100 11 000000111111 000000111111 000 11100 11 000000111111 000000111111 000 11100 11 000000111111 000000111111 000 1110000001111110100 11 000000111111 000000111111000000111111000 1110100000011111100 11 000000111111 000000111111000000111111000 111000000111111 000000111111 000000111111000000111111000000111111 000000111111 000000111111000000111111000000111111 000000111111 T 1 00000011111100 11 000000111111T000000111111000000111111 2000000111111000000111111 000000111111000000111111 000000111111000000111111 00000011111100000011111101 0000001111110100000011111101 00000011111101β11 β22Figure 4: Building an interior triangle by means of operation •.Assume T is such that |I(˜T)| = 1. In particular, let I(T) = {∆}. Clearly, the tree Γ Tcontains exactly one internal vertex that is not adjacent to a leaf. Hence, in Γ T there mustbe three internal vertices each of them adjacent to two leaves, and n − 6 internal verticesadjacent to exactly one leaf. Thus, we can identify in Γ T three paths P 1 = γ ∆ 11. . . γ ∆ 1 n1,P 2 = γ ∆ 21. . . γ ∆ 2 n2, and P 3 = γ ∆ 3 n3. . . γ ∆ 31with end-vertices γ ∆ 1 n1= γ ∆ 2 n2= γ ∆ 3 n3= γ ∆ , andsuch that: (1) n 1 + n 2 + n 3 = n and n 1 , n 2 , n 3 2, (2) each γ ∆j with j ∈ {1, 2, 3} is1adjacent to two leaves of Γ T , and (3) each γ ∆j with j ∈ {1, 2, 3} and i j ∈ {2, . . . , n j − 1}i jis adjacent to a single leaf of Γ T .Given Γ T , we can construct T by means of the following iterative step by step procedure:1. For i ∈ {1, 2}, add triangles ∆ i 1, . . . , ∆ i n i −1 according to the bijection following thetrajectory from γ ∆ i1to γ ∆ ini −1 given by P i, thus obtaining a triangulation T i suchthat Γ Ti is the minimal subtree of Γ T containing P i \ γ ∆ . Moreover, note thatT i ∈ ∆(C ni +1) is such that |I(T i )| = 0, and that there is a degenerate triangulationT i,0 which is an edge of triangle ∆ i 1, and basic operations R i 1, . . . , R i n i −1 ∈ {W, Z}such thatT i = R i n i −1 ◦ . . . ◦ R i 2 ◦ R i 1(T i,0 ) .Also, note that ⌊T i ⌋ is an edge of ∆ i n i −1.2. Apply operation • in order to construct ̂T = T 1 • T 2 ∈ ∆(C n1 +n 2 +1). Note that∆ ∈ F(̂T) and ⌊̂T⌋ is the unique edge of ∆ which is in the boundary of ̂T.the electronic journal of combinatorics 17 (2010), #R39 7

00 110100 1100 1100 1101010101 00 1100 1100 1100 1100 110100 1100 1100 1100 11 01010100 1101010101010100 110100 1100 1100 110101010100 1100 1100 11Steps 1 and 20100 1100 110101010100 1100 1101010101010100 1100 1100 1100 1100 1100 11010100 11010100 11010101010101010101010101010101010101010101010100 11 00 11Step 3010100 11 00 1100 1100 1100 1100 1100 1100 110101010100 1100 1100 1100 11010101010100 110100 1100 1100 1100 1100 1100 1100 1100 1100 1100 1100 1100 1100 1100 11Figure 5: Sketch of construction of an arbitrary T with |I(T)| = 1.3. Finally, starting from ̂T add triangles associated to vertices of the path P 3 . This isdone by performing a sequence of n 3 − 1 operations W and Z along P 3 \ γ ∆ startingfrom (̂T, ⌊̂T⌋). Given that ̂T ∈ ∆(C n1 +n 2 +1), we obtain T ∈ ∆(C n1 +n 2 +n 3) (recallthat n 1 + n 2 + n 3 = n).We summarize the previous discussion as follows:Lemma 4 Let T be a triangulation of a convex n-gon such that |I(T)| = 1. For somen 1 , n 2 , n 3 2 such that n 1 +n 2 +n 3 = n, there are triangulations T 1 and T 2 of convex (n 1 +1) and (n 2 + 1)-gons such that |I(T 1 )| = |I(T 2 )| = 0, and basic operations R 1 , . . . , R n3 −1 ∈{W, Z} such thatT = R n3 −1 ◦ · · · ◦ R 2 ◦ R 1 (T 1 • T 2 ) .Now, we state the main result concerning the recursive construction of an arbitrarytriangulation of a convex n-gon that we will need.Lemma 5 Let T be a triangulation of a convex n-gon such that |I(T)| = m 2. Then,there are ̂n 5, ñ 3 and l 1 such that ñ+̂n+l−1 = n, and triangulations ˜T ∈ ∆(C en )and ̂T ∈ ∆(C bn ) satisfying:the electronic journal of combinatorics 17 (2010), #R39 8

1. |I(˜T)| = 0,2. (̂T, ⌊̂T⌋) is either:(a) The output of operation W or Z and |I(̂T)| = m − 1, or(b) The output of operation • and |I(̂T)| = m − 2.3. There are basic operations R 1 , . . . , R l ∈ {W, Z} for which T = R l ◦· · ·◦R 2 ◦R 1 (˜T•̂T).Proof: Observe that there must be an internal vertex of Γ T , say γ ∆ , such that if Γ bT , Γ eTand Γ Tl+2 are the three sub-trees of Γ T rooted in γ ∆ , then all internal vertices of Γ eT \ γ ∆and Γ Tl+2 \ γ ∆ are adjacent to at least one leaf. In particular, |I(˜T)| = |I(T l+2 )| = 0, andcondition 1 of the statement of the lemma is satisfied.Let γ b∆ be the neighbor of γ ∆ in Γ bT . Note that one of the following two situations mustoccur:Case 1: In Γ bT \ γ ∆ , the vertex γ b∆ is adjacent to a leaf (see Figure 6.(a)). Inparticular, Γ bT has exactly m−1 internal vertices which are not adjacent to any leaf,orCase 2: None of the neighbors of γ b∆ in Γ bT \ γ ∆ are adjacent to leaves (see Figure6.(b)). In particular, Γ bT has exactly m−2 internal vertices which are not adjacentto any leaf.00 1101Γ00 11 eT Γ00 110100 11 eT00 1100 11110000 1100 1100 11Γ00 11bT00 11010 0 1 1 00 11 00 11 γ 00 1100 11∆ b 00 11γ 101∆ b0100 1100 11γ ∆00 110 1 00 11γ00 11 0101 γ∆ 00 11∆ b0100 11 010100110100 110100 11 γ∆ b 201 00 1101 01 00 1100 11 00 110100 1100 110100 110100 1100 11 010100 1100 11 0100 11Γ 00 11Tl+2 Γ 01Tl+200 11 00 11 00 11 00 1100 11Γ bT(a)(b)Figure 6: Structure of Γ T depending on the one of subtree Γ bT .Assume that the first case holds. Recall that |I(̂T)| = m − 1. Let ̂T 0 be the triangulationsuch that Γ bT0 is the ternary tree obtained from Γ bT \ γ ∆ by deleting the neighbor of γ b∆which is a leaf. Let ⌊̂T 0 ⌋ be the edge of ̂T 0 corresponding to the unique edge incident tothe electronic journal of combinatorics 17 (2010), #R39 9

γ b∆ in Γ bT0 . Note that applying one basic operation of type W or Z we can obtain (̂T, ⌊̂T⌋)from (̂T 0 , ⌊̂T 0 ⌋). Therefore, (̂T, ⌊̂T⌋) satisfies condition 2a of the statement of the lemma.Suppose now that the second case holds. Recall that |I(̂T)| = m − 2. Let γ b∆ 1 and γ b∆ 2 bethe vertices in Γ bT \ γ ∆ that are neighbors of γ b∆ . Let Γ bT,1 and Γ bT,2 be the trees obtainedfrom Γ bT\γ∆ by removing the trees rooted at γ b∆ 2 and γ b∆ 1, respectively. Consider i ∈ {1, 2}and note that Γ bT,i is a ternary tree since by hypothesis neither γ b∆ 1 nor γ b∆ 2 are adjacentto leaves of Γ bT \ γ ∆ . Let ̂T i be the triangulation that is in bijective correspondence withΓ bT,i . Define ⌊̂T i ⌋ to be the edge of triangulation ̂T i which is in bijection with the edge(γ b∆ , γ b∆ i) of Γ bT,i . Note that (̂T, ⌊̂T⌋) may be obtained as ̂T 1 • ̂T 2 . Therefore, (̂T, ⌊̂T⌋)satisfies condition 2b of the statement of the lemma.To finish the construction of T it suffices to apply an appropriate sequence of l operationsfrom the set {W, Z} starting from (˜T • ̂T, ⌊˜T • ̂T⌋). The result follows.4 Satisfying StatesIn this section we first present a technique, the so called Transfer Matrix Method. Thetechnique is usually applied in situations where there is an underlying regular lattice, andgives formulas for its degeneracy. We adapt the technique to the context where instead ofa lattice there is a triangulation of a convex n-gon T and use it to determine s(T). Then,we apply the method to derive an exact formula for the number of satisfying states ofstrips of triangles. Finally, we extend our arguments in order to establish an exponentiallower bound for s(T) of any T triangulation of a convex n-gon.4.1 Transfer matrices and satisfying matrixHenceforth, the index of rows and columns of all 4 × 4 matrices we consider will beassumed to belong to {+, -} 2 . Let T be a triangulation of a convex n-gon such that|I(T)| = 0. From now on, let 1 denote the 4 × 1 vector all of whose coordinates are 1,i.e. 1 = (1, 1, 1, 1) t . Our immediate goal is to obtain a matrix M = M(T) of type 4 × 4that satisfies the following two conditions:Condition 1: Columns and rows of M are indexed by spin-assignments of the topand bottom node pairs of T, respectively.Condition 2: For φ, ψ ∈ {+, -} 2 , the value M[φ, ψ] is equal to the number ofsatisfying states of T if the spin-assignments of the top and bottom node pairs of Tare ψ and φ, respectively.Matrix M is called the satisfying matrix of T. It immediately follows thats(T) = 1 t · M · 1 .the electronic journal of combinatorics 17 (2010), #R39 10

By Lemma 3, each triangulation T ∈ ∆(C n ) such that |I(T)| = 0 may be constructedby applying a sequence of n − 2 operations of type W or Z starting from T’s top edge.To each operation R ∈ {W, Z} we associate a so called transfer matrix of type 4 × 4, sayR ∈ {W, Z} such that:• Columns of R are indexed by spin-assignments of the bottom node pair of T.• Rows are indexed by spin-assignments of the bottom node pair of R(T).• For φ, ψ ∈ {+, -} 2 , matrix R satisfies⎧1 , if by setting the spin-assignments of the bottom node⎪⎨pairs of T and R(T) to ψ and φ respectively, the state ofR[φ, ψ] =the triangle created by the application of R is satisfying,⎪⎩0 , otherwise.Proposition 6 Let n 3 and T 0 be a degenerate triangulation. Let T ∈ ∆(C n ) besuch that T = R n−2 ◦ · · · R 2 ◦ R 1 (T 0 ). If R i denotes the transfer matrix associated toR i ∈ {W, Z}, then M(T) = R n−2 · · · R 2 · R 1 .Proof: We proceed by induction on n. If n = 3 we have that T = R 1 (T 0 ) and thestatement follows by definition of M(T) and R. Assume n > 3. By inductive hypothesisthe satisfying matrix of the triangulation ̂T = R n−3 ◦ · · · ◦ R 2 ◦ R 1 (T 0 ) ∈ ∆(C n−1 ) isM(̂T) = R n−3 · R n−4 · · · R 2 · R 1 .The matrix R n−2 · M(̂T) satisfies Condition 1 since columns of the matrix M(̂T) areindexed by the spin-assignment of ⌈̂T⌉ = ⌈T⌉ and the rows of matrix R n−2 by the spinassignmentof ⌊T⌋.We still need to show that R n−2 · M(̂T) satisfies Condition 2. By inductive hypothesis,we have that M(̂T)[χ, ψ] is the number of satisfying states of ̂T if the spin-assignmentsfor ⌊̂T⌋ and ⌈̂T⌉ are χ and ψ, respectively. By definition, R n−2 [φ, χ] may be 1 or 0depending on whether or not the application of R n−2 to (̂T, ⌊̂T⌋) creates a triangle forwhich a satisfying state is obtained by setting the spin-assignments of ⌊T⌋ and of ⌊̂T⌋equal to φ and χ, respectively. Therefore, R n−2 [φ, χ] = 1 if and only if each satisfyingstate in ̂T with spin-assignment χ and ψ for ⌊̂T⌋ and ⌈̂T⌉ respectively, is a satisfying statein T with spin-assignment φ and ψ for ⌊T⌋ and ⌈T⌉ respectively. By definition of M(T),it immediately follows thatM(T)[φ, ψ] =∑()R n−2 [φ, χ] · M(̂T)[χ, ψ] = R n−2 · M(̂T) [φ, ψ] ,χ∈{+,-} 2and that M(T) = R n−2 · M(̂T), thus concluding the inductive proof.the electronic journal of combinatorics 17 (2010), #R39 11

4.2 Satisfying states of strips of trianglesWe now apply the transfer matrix method to count the number of satisfying states in anytriangulation of a convex n-gon T satisfying the condition |I(T)| = 0. First, we observethat the matrices W and Z associated to operations W and Z, respectively, are given by:W =⎛⎜⎝0 0 1 00 1 0 11 0 1 00 1 0 0⎞⎟⎠ ,⎛Z = ⎜⎝0 1 0 01 1 0 00 0 1 10 0 1 0Note that W = Π · Z · Π where Π is the following permutation matrix:⎛⎞1 0 0 0Π = ⎜ 0 0 1 0⎟⎝ 0 1 0 0 ⎠ .0 0 0 1Since Π −1 = Π, for any k 0 we get that⎞⎟⎠ .W k = (Π · Z · Π) k = Π · Z k · Π . (1)Theorem 7 Let T 0 be a degenerate triangulation, w 1 , . . . , w m , z 1 , . . . , z m be a sequenceof non–negative integers adding up to n − 2 such that w j 1 for j ≠ 1 and z j 1 forj ≠ m. If T = Z zm ◦ W wm ◦ . . . ◦ Z z 1◦ W w 1(T 0 ) and M = M(T), thenM = Z zm · Π · Z wm · Π · · · Π · Z z1 · Π · Z w1 · Π .Moreover, if F k denotes the k-th Fibonacci number, then⎛ ⎞F n−1M · 1 = ⎜ F n⎟⎝ F n⎠ .F n−1Proof:From Proposition 6 we haveM = Z zm · W wm · Z z m−1· W w m−1· · · Z z2 · W w2 · Z z1 · W w 1.By (1), the first stated identity immediately follows.Now, for the second part, let k 1. Observe that( 0 11 1) k=( )Fk−1 F k.F k F k+1the electronic journal of combinatorics 17 (2010), #R39 12

It follows that,Z k · 1 =⎛⎜⎝⎞ ⎛F k−1 F k 0 0F k F k+1 0 0⎟0 0 F k+1 F k⎠ · 1 = ⎜⎝0 0 F k F k−1⎞F k+1F k+2F k+2F k+1⎟⎠ . (2)The first stated identity, the fact that Π · Z k · 1 = Z k · 1, and observing that Π · 1 = 1,we get thatM · 1 = Z zm · Π · Z wm · Π · · · Z z1 · Π · Z w1 · Π · 1= Z zm · Z wm · · · Z z1 · Z w1 · 1 .Since ∑ mi=1 (z i + w i ) = n − 2, the desired conclusion follows from (2).Proof of Theorem 1: By hypothesis and Lemma 3 we have that for some degeneratetriangulation T 0 there are non–negative integers w 1 , . . . , w m , z 1 , . . . , z m adding up to n−2such that w j 1 if j ≠ 1, z j 1 if j ≠ m, andT = Z zm ◦ W wm ◦ · · · ◦ Z z 2◦ W w 2◦ Z z 1◦ W w 1(T 0 ) .By Theorem 7, we get that s(T) = 1 t · M(T) · 1 = 2(F n + F n−1 ) = 2F n+1 .We now obtain some intermediate results that we will need to prove Theorem 2: LetT ∈ ∆(C n ) and {β 1 , β 2 } be an edge belonging to the boundary of T. The satisfyingvector of T associated to node pair (β 1 , β 2 ) denoted by v T ((β 1 , β 2 )) is a vector indexedby the spin-assignments {+, -} 2 of (β 1 , β 2 ), so that v T ((β 1 , β 2 ))[ψ] is equal to the numberof satisfying states of T if the spin-assignment of (β 1 , β 2 ) is equal to ψ. For instance, byTheorem 7, for every triangulation T of a convex n-gon with no interior triangles,⎛ ⎞F n−1v T (⌊T⌋) = ⎜ F n⎟⎝ F n⎠ .F n−1Clearly, for every T ∈ ∆(C n ) we have thatv T [++] = v T [--] , v T [+-] = v T [-+] . (3)Note that for edges (β 1 , β 2 ) ≠ (̂β 1 , ̂β 2 ) belonging to the boundary of T, if⎛ ⎞⎛ ⎞x̂xv T ((β 1 , β 2 )) = ⎜ y⎟⎝ y ⎠ , v T(( ̂β 1 , ̂β 2 )) = ⎜ ŷ⎟⎝ ŷ ⎠ ,x̂xthen 2(x + y) = 2(̂x + ŷ), or equivalently x + y = ̂x + ŷ.the electronic journal of combinatorics 17 (2010), #R39 13

Proposition 8 If R ∈ {W, Z}, ̂T ∈ ∆(C bn ), and T = R(̂T), thenv T (⌊T⌋) = R · v bT (⌊̂T⌋) .Proof: Implicit in the proof of Proposition 6.We now define a useful operation on satisfying vectors. Let • be the binary operatorover N 4 defined by⎛ ⎞ ⎛ ⎞ ⎛⎞x 1 y 1x 2 y 3⎜ x 2⎟⎝ x 3⎠ • ⎜ y 2⎟⎝ y 3⎠ = ⎜ x 1 y 2 + x 3 y 4⎟⎝ x 4 y 3 + x 3 y 1⎠ .x 4 y 4 x 3 y 2Proposition 9 Let T 1 ∈ ∆(C n1 ) and T 2 ∈ ∆(C n2 ) be such that ⌊T 1 ⌋ = (β 1 1, β 1 2) and⌊T 2 ⌋ = (β 2 1, β 2 2). Then,v T1 •T 2((β 1 1, β 2 2)) = v T1 ((β 1 1, β 1 2)) • v T2 ((β 2 1, β 2 2)) .Proof: To simplify the notation we henceforth denote v T1 •T 2((β1, 1 β2)), 2 v T1 ((β1, 1 β2)) 1 andv T2 ((β1, 2 β2)) 2 by v β 11 β2 2, v β1 1 , and v β1 2 β1 2 , respectively. For i ∈ {1, 2}, we know that v β2 2 β1 i [ψ] βi 2is equal to the number of satisfying states of T i if ψ ∈ {+, -} 2 is the spin-assignment for(β1, i β2). i We consider the following cases depending on the spin-assignment of (β1, 1 β2).2• Spin-assignment of (β 1 1, β 2 2) is ++: Since +++ is not a satisfying assignment for thetriangle (β 1 1, β, β 2 2) of T, if the spin-assignment of β = β 2 1 = β 1 2 is +, then the state ofT is not satisfying. If the spin assignment of (β 1 1, β, β 2 2) is +-+, each satisfying stateof T 1 and T 2 (with spin-assignment for (β 1 1, β 1 2) equal to +- and spin-assignment for(β 2 1, β 2 2) equal to -+) is a satisfying state for T, andv β 11 β 2 2 [++] = v β 1 1 β1 2 [+-]v β 2 1 β2 2 [-+] .• Spin-assignment of (β1, 1 β2) 2 is +-: Note that the triangle (β1, 1 β, β2) 2 with spinassignment++- fulfills the condition of satisfying state. Hence, each satisfying stateof T 1 and T 2 (with spin-assignment for (β1, 1 β2) 1 equal to ++ and spin-assignment for(β1, 2 β2) 2 equal to +-) is a satisfying state for T. Analogously, if the spin-assignmentof β is equal to -, each satisfying state of T 1 and T 2 (with spin-assignment for(β1, 1 β2) 1 equal to +- and spin-assignment for (β1, 2 β2) 2 equal to --) is a satisfyingstate for T. It follows thatv β 11 β 2 2 [+-] = v β 1 1 β1 2 [++]v β 2 1 β2 2 [+-] + v β 1 1 β1 2 [+-]v β 2 1 β2 2 [--] .Finally, by a symmetry argument, we also have that v β 11 β2 2[--] = v β1 1 [-+]v β1 2 β1 2 [+-] andβ2 2v β 11 β2 2[-+] = v β1 1 [-+]v β1 2 β1 2 [++] + v β2 2 β1 1 [--]v β1 2 β1 2 [-+]. β2 2the electronic journal of combinatorics 17 (2010), #R39 14

We now recall some basic well known facts about Fibonacci numbers. Let ϕ denotethe golden ratio. If F n denotes the n-th Fibonacci number, it is well known that F n+1 =F n + F n−2 for all n 1, and thatF n = ϕn − (− 1 ϕ )n√5.It immediately follows that for all n 1,ϕ n−2 F n ) 2(11 +ϕ√5ϕ n ϕ n . (4)Lemma 10 If T is a triangulation of a convex n-gon, then ϕ n−|I(T)| ϕ 2 ( √ ϕ) n .Proof: Since |O(T)| |I(T)| + 2 and |O(T)| + |I(T)| = n − 2, we get that |I(T)| (n/2) − 2. The claimed result immediately follows.Proof of of Theorem 2: We claim that for any triangulation of a convex n-gon T suchthat |I(T)| = m it holds that s(T) ϕ n−m . To prove this claim we proceed by inductionon m. If m = 0, by Theorem 1 we have that s(T) = 2F n+1 . Using the lower bound in (4)we obtain s(T) 2ϕ n−1 ϕ n . If m = 1, by Lemma 4 we know that for some n 1 , n 2 , n 3 2such that n 1 + n 2 + n 3 = n there are triangulations T 1 ∈ ∆(C n1 +1) and T 2 ∈ ∆(C n2 +1)such that |I(T 1 )| = |I(T 2 )| = 0, and basic operations R 1 , . . . , R n3 −1 ∈ {W, Z} such thatBy Theorem 7, for i ∈ {1, 2} we know thatT = R n3 −1 ◦ · · · ◦ R 2 ◦ R 1 (T 1 • T 2 ) .v Ti (⌊T i ⌋) =⎛⎜⎝F niF ni +1F ni +1F niNow, denote v T1 •T 2(⌊T 1 • T 2 ⌋) by v. Observe that Proposition 9 and the definition of •imply that⎛ ⎞ ⎛ ⎞ ⎛⎞F n1 F n2F n1 +1F n2 +1v = ⎜ F n1 +1 ⎟⎝ F n1⎠ • ⎜ F n2 +1 ⎟⎝+1 F n2⎠ = ⎜ F n1 F n2 +1 + F n1 +1F n2⎟⎝+1 F n1 F n2 +1 + F n1 +1F n2⎠ .F n1 F n2F n1 +1F n2 +1Repeated application of Proposition 8 yields that⎞⎟⎠ .s(T) = 1 t · R n3 −1 · · · R 2 · R 1 · v .the electronic journal of combinatorics 17 (2010), #R39 15

By (3) and due to the block structure of Z, we have that Π · v = v and Π · Z q · v = Z q · v,for every q 0. Therefore, since W = Π · Z · Π, the last displayed identity may berewritten as s(T) = 1 t · Z n 3−1 · v. Hence,⎛⎞F n3 −2 F n3 −1 0 0s(T) = 1 t · ⎜ F n3 −1 F n3 0 0⎟⎝ 0 0 F n3 F n3⎠ · v−10 0 F n3 −1 F n3 −2= 2 (F n3 F n1 +1F n2 +1 + F n3 +1(F n1 F n2 +1 + F n1 +1F n2 )) .Since Fibonacci numbers satisfy the identity F p+q = F p F q−1 + F p+1 F q , we get thats(T) = 2 (F n3 (F n1 +1F n2 +1 + F n1 F n2 +1 + F n1 +1F n2 ) + F n3 −1(F n1 F n2 +1 + F n1 +1F n2 ))= 2 (F n3 (F n1 +2F n2 +1 + F n1 +1F n2 ) + F n3 −1(F n1 +2F n2 + F n1 +1F n2 −1 − F n1 −1F n2 −1))= 2(F n3 F n1 +n 2 +2 + F n3 −1(F n1 +n 2 +1 − F n1 −1F n2 −1))= 2(F n1 +n 2 +n 3 +1 − F n3 −1F n1 −1F n2 −1) .Since n = n 1 + n 2 + n 3 , 2 > ϕ and ϕ 2 − 1 = ϕ, by (4) it follows thats(T) 2ϕ n 1+n 2 +n 3 −1 ( 1 − ϕ −2) ϕ n−1 .Now, suppose the claim holds for every triangulation T ∈ ∆(C n ) such that |I(T)| < m.Let T ∈ ∆(C n ) be such that |I(T)| = m.We know from Lemma 5 that there is a ˜T ∈ ∆(C en ) such that |I(˜T)| = 0, a ̂T ∈ ∆(C bn )satisfying condition 2 of Lemma 5, basic operations R 1 , . . . , R l ∈ {W, Z} where l 1, andn = ̂n + ñ + l − 1 such thatT = R l ◦ · · · ◦ R 1 (˜T • ̂T) .By an argument similar to the one used to handle the m = 1 case, we have thats(T) = 1 t · R l · · · R 2 · R 1 · v eT• b T(⌊˜T • ̂T⌋) .Since ˜T ∈ ∆(C en ) is such that |I(˜T)| = 0, by Theorem 7 we have that⎛ ⎞F en−1v eT (⌊˜T⌋) = ⎜ F en⎟⎝ F en⎠ .F en−1Let ̂x and ŷ denote v bT (⌊̂T⌋)[++] and v bT (⌊̂T⌋)[+-] respectively. Observe that (3) impliesthat v bT (⌊̂T⌋)[-+] = ŷ and v bT (⌊̂T⌋)[--] = ̂x. Hence, by Proposition 9,⎛ ⎞ ⎛ ⎞ ⎛⎞F en−1 ̂xŷF env eT• T b(⌊˜T • ̂T⌋) = ⎜ F en⎟⎝ F en⎠ • ⎜ ŷ⎟⎝ ŷ ⎠ = ⎜ ̂xF en + ŷF en−1⎟⎝ ̂xF en + ŷF en−1⎠ .F en−1 ̂xŷF enthe electronic journal of combinatorics 17 (2010), #R39 16

Denoting v = v eT• T b(⌊˜T • ̂T ⌋) we again observe that (3) implies that Π · v = v andΠ · Z q · v = Z q · v for all q 0. Putting everything together we conclude that⎛⎞ŷF ens(T) = 1 t · Z l · ⎜ ̂xF en + ŷF en−1⎟⎝ ̂xF en + ŷF en−1⎠ŷF en= 2(̂xF l+2 F en + ŷ(F l+1 F en + F l+2 F en−1 )) .The lower bound for Fibonacci numbers given in (4) and the fact that 2 > ϕ imply thats(T) 2 (̂xϕ l+en−2 + 2ŷϕ l+en−3) 2(̂x + ŷ)ϕ l+en−2 .Recalling that s(̂T) = 2(̂x+ŷ) and observing that conditions 1 and 2 of Lemma 5 guaranteethat |I(̂T)| is equal to m − 1 or m − 2, from the inductive hypothesis we obtain thats(̂T) ϕ bn−(m−1) . It follows that s(T) ϕ bn+en+l−2−(m−1) = ϕ n−m . This concludes theinductive prove of the claim. Lemma 10 immediately implies the desired result.5 ConclusionWe have established that the number of satisfying states of any triangulation of a convexn-gon is exponential in n. It would be of interest to generalize this result to more generaltriangulations. Two natural cases to address next are triangulations that are embedableover low genus surfaces and k-trees.References[CS08] M. Chudnovsky and P. Seymour. Perfect matchings in planar cubic graphs.manuscript, 2008.[Jae00] F. Jaeger. A survey of the cycle double cover conjecture. Discrete AppliedMathematics, 99(1):71–90, 2000.[KSS08] D. Král, J.-S. Sereni, and M. Stiebitz. A new lower bound on the number ofperfect matchings in cubic graphs. manuscript, 2008.[LV03][Sch98]M. Loebl and J. Vondrák. Towards a theory of frustrated degeneracy. DiscreteMathematics, 271(1-3):179–193, 2003.A. Schrijver. Counting 1-factors in regular bipartite graphs. Journal of CombinatorialTheory, Series B, 72(1):122–135, 1998.the electronic journal of combinatorics 17 (2010), #R39 17

Apéndice BModelo de Ising antiferromagnético entriangulaciones aplicado a contaremparejamientos perfectos38

Antiferromagnetic Ising model in triangulations withapplications to counting perfect matchingsAndrea Jiménez ∗ Marcos Kiwi †Submitted to Discrete Applied Mathematics, November 2011AbstractIn this work we give a lower bound for the groundstate degeneracy of the antiferromagneticIsing model in a class of triangulations. As consequence, we show that every cubicplanar graph G whose geometric dual graph is a stack triangulation has at least 3ϕ|V (G)|/72distinct perfect matchings, where ϕ is the golden ratio. Our work builds on a novel approachrelating Lovász and Plummer’s theorem and the number of so called groundstatesof the widely studied Ising model from statistical physics.Keywords: Ising model in triangulations; Groundstates; Transfer matrix method; Cubic graphs; Numberof perfect matchings.1 IntroductionThe Ising model is one of the most studied models in statistical physics. The study of the Isingmodel in a system (graph) helps to understand physical phenomena associated to its thermodynamicproperties [11]. In particular, antiferromagnetic systems are interesting, in part becauseof their lack of order at zero temperature [8]. So far, the Ising model has been widely exploredfor large scale and mostly regular (lattice like) systems using various different methods [1]. Inthis context, our work contributes in two ways: (1) to the understanding of the Ising model’s behaviorfor irregular systems and not necessarily of small size, and (2) to the development of an∗ Depto. Ing. Matemática, U. Chile & Depart. of Applied Mathematics, Charles U. Web:www.dim.uchile.cl/∼ajimenez. Gratefully acknowledges the support of MECESUP UCH0607, CONI-CYT and the partial support of the Czech Research Grant MSM 0021620838.† Depto. Ing. Matemática & Ctr. Modelamiento Matemático UMI 2807, U. Chile. Web:www.dim.uchile.cl/∼mkiwi. Gratefully acknowledges the support of CONICYT via Basal in AppliedMathematics and FONDECYT 1090227.1

adaptation of the extensively employed transfer matrix method [9] used by statistical physicists,and thus widen the scope of its applicability.Let us now precisely describe the antiferromagnetic Ising model. By a slight abuse of notationwe say that a map in a 2−dimensional surface is a triangulation if each face is bounded bya cycle of length 3 (in particular there is no loop); hence we allow multiple edges. Given atriangulation ∆ = (V, E) we associate the coupling constant c(e) = −1 with each edge e ∈ E.For any W ⊆ V , a spin assignment of W is any function s : W → {1, −1} and 1, −1 arecalled spins. A state of ∆ is any spin assignment of V . The energy of a state s is definedas − ∑ e={u,v}∈Ec(e)s(u)s(v). The states of minimum energy are called groundstates. Thenumber of groundstates is usually called the groundstate degeneracy of ∆, denoted g(∆).A graph is said to be cubic if each vertex has degree 3 and bridgeless if it contains no cutedges.In the mid-1970’s, Lovász and Plummer asserted that for every cubic bridgeless graph with nvertices, the number of perfect matchings is exponential in n. For bipartite graphs, the assertionwas positively solved by Voorhoeve [12]. Chudnovsky and Seymour showed that it holds forplanar graphs [3]. Independently and after the announcement of our work, Esperet, Kardos,King, Král and Norine [4] announced the positive resolution of the full conjecture; the mainnon-elementary ingredient of the proof is Edmonds’ perfect matching polytope theorem.We suggest to study the theorem of Lovász and Plummer in the dual setting. This relates theproblem of counting perfect matchings to the groundstate degeneracy of the antiferromagneticIsing model on triangulations. In order to explain this, lets recall another well-known conjecture,namely the directed cycle double cover conjecture of Jaeger (see [6]): Every cubicbridgeless graph can be embedded in an orientable surface so that each face is homeomorphicto an open disc (i.e., the embedding defines a map) and the geometric dual has no loop.A set M of edges of a triangulation ∆ is intersecting if M contains exactly one edge of eachface of ∆. Assuming the directed cycle double cover conjecture, M is an intersecting set of G ∗if and only if M is a perfect matching of G. We can now reformulate the Lovász and PlummerTheorem as follows: Each triangulation has an exponential number of intersecting sets ofedges.Given a state s of ∆ we say that edge {u, v} is frustrated by s or that s frustrates edge {u, v}if s(u) = s(v). Note that each state frustrates at least one edge of each face of ∆. A state isa groundstate if it frustrates the smallest possible number of edges. Clearly, if there is a statewhich frustrates exactly one edge of each face of ∆, it is a groundstate and the set of edgesfrustrated is an intersecting set and in this case, the number of groundstates is at most twice thenumber of intersecting sets of edges. Note that this always happens for planar triangulationsand not necessarily in general.The converse also holds for planar triangulations: if we delete an intersecting set of edges froma planar triangulation, then we get a bipartite graph and its bipartition determines a ground-2

state. Hence, Chudnovsky and Seymour’s result can be reformulated as follows: Each planartriangulation has an exponential (in the number of vertices) groundstate degeneracy.The result we derive in this work, from the lower bound groundstate degeneracy of the antiferromagneticIsing model on stack triangulations, applies to a subclass of graphs for whichboth [3, 4] already establish the validity of Lovász and Plummer’s claim, albeit for a muchsmaller rate of exponential growth and arguably by more complicated and involved arguments.Nevertheless, we believe that the relevance of this result (Corollary 2), is that it validates thefeasibility of the alternative approach, proposed in [7], for addressing Lovász and Plummer’stheorem.Specifically, letting ϕ = (1 + √ 5)/2 ≈ 1.61803 denote the golden ratio, we establish the following:Theorem 1 The groundstate degeneracy of the antiferromagnetic Ising model in a stack triangulation∆ with |∆| vertices is at least 6ϕ 1 36 (|∆|+3) .As a rather direct consequence of the preceding theorem we obtain the following result.Corollary 2 The number of perfect matchings of a cubic graph G whose dual graph is a stacktriangulation is at least 3ϕ 172 |V (G)| .1.1 OrganizationThe paper is organized as follows. We provide some mathematical background in Section 2.Then, in Section 3, we describe a bijection between rooted stack triangulations and coloredrooted ternary trees — this bijection allows us to work with ternary trees instead of triangulations.In Section 4, we first introduce the concept of degeneracy vector in stack triangulations.This vector satisfies that the sum of its coordinates is the number of pseudo-groundstates of thestack triangulation. We also introduce the concept of root vector of a ternary trees and show thatvia the aforementioned bijection, the degeneracy vector of a stack triangulation ∆ is the sameas the root vector of the associated colored rooted ternary tree. In Section 5, we adapt to oursetting the transfer matrix method as used in statistical physics in the study of the Ising model.Some essential results are also established. In Section 6, we prove the main results of this work.In Section 7, we conclude with a brief discussion and comments about possible future researchdirections.2 PreliminariesWe now introduce the main concepts and notation used throughout this work.3

2.1 Stack triangulationsLet ∆ 0 be a triangle. For i ≥ 1, let ∆ i be the plane triangulation obtained by applying thefollowing growing rule to ∆ i−1 .growing rule: Given a plane triangulation ∆,1. Choose an inner face f from ∆,2. Insert a new vertex u at the interior of f.3. Connect the new vertex u to each vertex of the boundary of f.Clearly, the number of vertices of ∆ n is n + 3. The collection of ∆ n ’s thus obtained are calledstack triangulation. Among others, the set of stack triangulations coincides with the set ofplane triangulations having a unique Schnyder Wood (see [5]) and is the same as the collectionof planar 3-trees (see [2, page 167]).Consider now the stack triangulation ∆ 1 and for i ≥ 2, let ∆ i be the plane triangulation obtainedby applying the growing rule to ∆ i−1 restricting Step 1 so the face chosen is one of the threenew faces obtained by the application of the growing rule to ∆ i−2 . For n ≥ 1, we say that ∆ nis a stack-strip triangulation (for an example see Figure 1). Clearly, stack-strip triangulationsare a subclass of stack triangulations.2465 13Figure 1: Example of stack-strip triangulation (numbers correspond to the order in which nodesare added by the growing rule).Let ∆ n be a stack triangulation with n ≥ 0 and ∆ 0 be the starting plane triangle in its construction.If we prescribe the counterclockwise orientation to any edge of ∆ 0 , we say that ∆ n is arooted stack triangulation (see Figure 2).2.2 Ternary treesA rooted tree is a tree T with a special vertex v ∈ V (T ) designated to be the root. If v is theroot of T , we denote T by T v . A rooted ternary tree is a rooted tree T v such that all its vertices4

vuvuFigure 2: A stack triangulation (left) and the rooted stack triangulation obtained by prescribingthe counterclockwise orientation to the edge {v, u} (right).have at most three children. From now on, let X be an arbitrary set with three elements. Wesay that a rooted ternary tree T v is colored by X (or simply colored) if; (1) each non-root vertexis labeled by an element of X, and (2) for every vertex of V (T ) all its children have differentlabels.3 From stack triangulations to ternary treesIt is well known that stack triangulations are in bijection with ternary trees (see [10]). Forour purposes, the usual bijection is not enough (we need a more precise handle on the way inwhich triangular faces touch each other). The main goal of this section is to precisely describea one-to-one correspondence better suited for our purposes.3.1 BijectionLet ∆ n be a rooted stack triangulation with n ≥ 1 and ∆ 0 be the starting plane triangle in itsconstruction. We will show how to construct a colored rooted ternary tree T (∆ n ) which will bein bijective correspondence with ∆ n .Throughout this section, the following concept will be useful.Definition 1 Let ∆ be a rooted stack triangulation. Let ˜∆ be the rooted stack triangulationobtained by prescribing the counterclockwise orientation to exactly one edge of each inner faceof ∆. We refer to ˜∆ as an auxiliary stack triangulation of ∆.Note that in an auxiliary stack triangulation of ∆, we allow inner faces of ∆ to have edgesoriented clockwise as long as exactly one of its edges is oriented counterclockwise. It is alsoallowed to have edges with both orientations.We now, describe the key procedure in the construction of T (∆ n ). For i ∈ {1, . . . , n}, letf i , u i and ∆ i , denote the chosen face, the new vertex and the output corresponding to the i-thapplication of the growing rule in the construction of ∆ n . The procedure recursively constructs5

an auxiliary stack triangulations ˜∆ i of ∆ i . Initially, i = 1 and ˜∆ 0 is ∆ 0 with one of its edgesoriented counterclockwise.Labeling procedure:Step 1: Let ⃗e fi be the counterclockwise oriented edge of f i . The orientation of ⃗e fiinduces a counterclockwise ordering of the three new faces around u i starting bythe face that contains ⃗e fi , say f i (1). Let f i (2) and f i (3) denote the second and thirdnew faces according to the induced order. For each j ∈ {1, 2, 3}, we say that f i (j)is in position j or that j is the position of f i (j). (See Figure 3.)Step 2: For each j ∈ {2, 3}, take the unique edge e fi (j) in E(f i ) ∩ E(f i (j))and prescribe the counterclockwise orientation to this edge (see Figure 3). For allother faces of ∆ i not contained in f i , keep the same counterclockwise orientededge. (Observe that for each j ∈ {1, 2, 3}, the triangle f i (j) has a prescribedcounterclockwise orientation in one of its three edges. Moreover, note that ⃗e fi =⃗e fi (1).)⃗e f (3) ⃗e f (2)f⃗e fe f (3)f(3) f(2)vf(1)⃗e f = ⃗e(1)e f (2)f(3)f(1)⃗e f (1)f(2)Figure 3: Labeling procedure. Left to center, step 1. Center to right, step 2.The set Θ ∆n = {(f i , u i , f i (1), f i (2), f i (3))} i∈{1,...,n} will be henceforth referred to as the growthhistory of ∆ n . Note that, for j ∈ {1, 2, 3}, each face f 1 (j) together with its oriented edgeinduce a rooted stack triangulation, henceforth denoted ∆ j n , on the vertices of ∆ n that lie on theboundary and interior of f 1 (j).We are ready to describe T (∆ n ) in terms of the growth history of ∆ n :Combinatorial description of T (∆ n ): Let X = {1, 2, 3}. Let V (T (∆ n )) ={u 1 , . . . , u n }. Let u 1 be the root of T (∆). For i ∈ {2, . . . , n}, u i is a child ofvertex u j if there is a k ∈ {1, 2, 3} such that f i = f j (k). The label of u i is k. Foran example see Figure 4.In particular, we have proved the following result.6

1 2 32 13Figure 4: Example of the bijection between rooted stack triangulations and colored rootedternary trees.Proposition 3 Let ∆ n be a rooted stack triangulation. The colored ternary tree T (∆ n ) rootedon v, satisfies the following statements:1. If ∆ i n has 3 vertices for all i ∈ {1, 2, 3}, then T (∆ n) has exactly one vertex v (its root).2. If there are i, j ∈ {1, 2, 3} with i ≠ j such that ∆ i n and ∆j n have 3 vertices and ∆k n withk ∈ {1, 2, 3}\{i, j} has at least 4 vertices, then the root v has exactly one child w labeledby k. Moreover, the root of T (∆ k n) is w, where T (∆ k n) is the colored sub-ternary tree ofT (∆ n ) induced by w and its descendants.3. If there is an i ∈ {1, 2, 3} such that ∆ i n has 3 vertices and j, k ∈ {1, 2, 3} \ {i} withj ≠ k such that ∆ j n and ∆k n have at least 4 vertices, then the root v has exactly twochildren w j and w k labeled by j and k, respectively. Moreover, for every t ∈ {j, k}, theroot of T (∆ t n ) is w t, where T (∆ t n ) is the colored sub-ternary tree of T (∆ n) induced byw t and its descendants.4. If ∆ i n , i ∈ {1, 2, 3}, has at least 4 vertices, then the root v has three children w 1, w 2and w 3 labeled by 1, 2 and 3, respectively. Moreover, for every i ∈ {1, 2, 3}, the root ofT (∆ i n ) is w i, where T (∆ i n ) is the colored sub-ternary tree of T (∆ n) induced by w i andits descendants.4 Transfer MethodThe main tool we use to carry out our work, is an adaptation of a method (well known amongphysicist) called the transfer matrix method. In order to adapt the method, we need some extradefinitions. Say that a state s is satisfying for a face f of a planar triangulation ∆, if there isexactly one edge in the boundary of f that is frustrated by s. Moreover, say that s is a satisfying7

state of ∆ if s is satisfying for every inner face. Clearly, a satisfying state of ∆ frustrates itsouter face if and only if is a groundstate.In [7], we directly apply the transfer matrix method to obtain the number of satisfying states oftriangulations of a convex n-gon. In this work we develop the technique further by consideringtransfer vectors instead of transfer matrices.4.1 MethodologyIn general terms, our aim is to obtain for each stack triangulation ∆ a vector v ∆ in R 4 such thatthe sum of its coordinates equals twice the number of satisfying states of ∆. We now elaborateon this. Let n ≥ 1 and ∆ n be a rooted stack triangulation. Let ∆ 0 = (v 1 , v 2 , v 3 ) denote thestarting triangle in the construction of ∆ n such that {v 1 , v 2 } is the oriented edge with v 1 thetail and v 2 the head. We wish to construct a vector v ∆n ∈ R 4 such that its coordinates areindexed by the ordered set I = {+ + +, + + −, + − +, − + +}. For every φ ∈ I, the φ-thcoordinate of v ∆n , denoted ∆ n [φ], is defined as the number of satisfying states of ∆ n whenthe spin assignment of (v 1 , v 2 , v 3 ) is equal to φ. The vector v ∆n will be called the degeneracyvector of ∆ n . In particular, v ∆0 = (0, 1, 1, 1) t is the degeneracy vector of a triangle. Clearly,for every φ ∈ I we have the relation∆ n [φ] = ∆ n [−φ] (1)Let Θ ∆n = {(f i , u i , f i (1), f i (2), f i (3))} i∈{1,...,n} be the growth history of ∆ n . Let v denote u 1 .Recall that f 1 (j) induces a rooted stack triangulation ∆ j n according to the growth history of ∆ n,(see Subsection 3.1): the oriented edge of ∆ 1 n is {v 1, v 2 } with v 1 its tail and v 2 its head; theoriented edge of ∆ 2 n is {v 2 , v 3 } with v 2 its tail and v 3 its head; and the oriented edge of ∆ 3 n is{v 3 , v 1 } with v 3 its tail and v 1 its head.The following result shows how to express the degeneracy vector of ∆ n in terms of the degeneracyvectors v ∆ 1 n, v ∆ 2 n, and v ∆ 3 n.Proposition 4 For each j ∈ {1, 2, 3}, let v ∆jn= (vj k ) k∈{0,1,2,3} . Then,⎛v1 0 v0 2 v0 3 + ⎞v1 1 v1 2 v1 3v1 0 v ∆n = ⎜v2 2 v3 3 + v1 1 v3 2 v2 3⎝v1 2v3 2 v0 3 + ⎟v3 1 v2 2 v1 3⎠ .v1v 2 2v 1 3 3 + v1v 3 2v 0 32Proof: Let φ ∈ I. Note that v ∆n [φ] equals the sum of the number of satisfying statesof ∆ n when (v 1 , v 2 , v 3 , v) are assigned spins (φ, +) and (φ, −). For a given spin assignmentto (v 1 , v 2 , v 3 , v), the number of satisfying states of ∆ n , is obtained by multiplying the number8

of satisfying states of each ∆ i n when the spin assignment of its outer faces agree with the fixedspins assigned to (v 1 , v 2 , v 3 , v).First, consider the case where φ = + + +. If v’s spin is +, then∆ 1 n [+ + +] · ∆2 n [+ + +] · ∆3 n [+ + +] = v0 1 v0 2 v0 3 .If v’s spin is −, then∆ 1 n [+ + −] · ∆2 n [+ + −] · ∆3 n [+ + −] = v1 1 v1 2 v1 3 .Hence, ∆ n [φ] = v1 0v0 2 v0 3 + v1 1 v1 2 v1 3 .Now, consider the case where φ = + + −. If v’s spin is +, then∆ 1 n [+ + +] · ∆2 n [+ − +] · ∆3 n [− + +] = v0 1 v2 2 v3 3 .Recalling that by identity (1) we have that ∆ 2 n [+−−] = ∆2 n [−++] and ∆3 n [−+−] = ∆3 n [+−+],if v’s spin is −, then∆ 1 n [+ + −] · ∆2 n [+ − −] · ∆3 n [− + −] = v1 1 v3 2 v2 3 .Hence, ∆ n [φ] = v1 0v2 2 v3 3 + v1 1 v3 2 v2 3 .The other two remaining cases, where φ equals + − + and − + +, can be similarly dealt withand left to the interested reader.4.2 Root vectors of ternary treesWe will now introduce the concept of root vector of a colored rooted ternary tree. Then, we willsee that v ∆ is the degeneracy vector of the rooted stack triangulation ∆ if and only if v ∆ is theroot vector of the colored rooted ternary tree T (∆).Let T be a colored rooted ternary tree. For any node u of T \ {v}, we denote by l u ∈ {1, 2, 3}its label.Definition 2 Let T be a colored ternary tree rooted at v. We recursively define the root vectorv ∈ R 4 of T associated to v according to the following rules:Rule 0: v = (1, 1, 1, 1) t when v does not have any children.Rule 1: If v has exactly one child u with u = (u s ) s=0,...,3 , then v ∈ [u] where[u] = { (u 1 , u 0 + u 1 , u 3 , u 2 ) t , (u 1 , u 3 , u 2 , u 0 + u 1 ) t , (u 1 , u 2 , u 0 + u 1 , u 3 ) t} .The choice of v depends on the label of u; if l u = i, v is the i-th vector in [u].9

Rule 2: If v has two children u and w with u = (u s ) s=0,...,3 , w = (w s ) s=0,...,3 , and(l u , l w ) ∈ {(1, 2), (2, 3), (3, 1)}, then⎧⎛⎞ ⎛ ⎞ ⎛ ⎞ ⎫u 1 w 1u 1 w 1u 1 w 1⎪⎨u 0 w 2 +u 1 w 3v ∈ ⎜ ⎟⎝ u 3 w 2 ⎠ , u 3 w 2⎜ ⎟⎝u 3 w 0 +u 2 w 1 ⎠ , u 3 w 0 +u 2 w ⎪⎬1⎜ ⎟ .⎝u 0 w 2 +u 1 w 3 ⎠⎪⎩⎪⎭u 2 w 1 +u 3 w 0 u 1 w 3 +u 0 w 2 u 3 w 2The choice of v depends on (l u , l w ); if l u = i, v is the i-th vector in the last set.Rule 3: If v has three children u, w and z with u = (u s ) s=0,...,3 , w = (w s ) s=0,...,3 ,z = (z s ) s=0,...,3 , and (l u , l w , l z ) = (1, 2, 3), then⎧⎛⎞⎫u 0 w 0 z 0 +u 1 w 1 z 1⎪⎨u 0 w 2 z 3 +u 1 w 3 z ⎪⎬2v ∈ ⎜⎟ .⎝u 2 w 3 z 0 +u 3 w 2 z 1 ⎠⎪⎩⎪⎭u 2 w 1 z 3 +u 3 w 0 z 2The following result establishes that determining the degeneracy vector of rooted stack triangulationsis equivalent to determining the root vector of colored rooted ternary trees.Lemma 5 Let n ≥ 1 and ∆ n be the rooted stack triangulation ∆ n . Then, the root vector of thecolored ternary tree T (∆ n ) in bijection with ∆ n equals the degeneracy vector of ∆ n .Proof: By induction on n. For the base case n = 1; the stack triangulation ∆ 1 is isomorphic toK 4 and T (∆ 1 ) is a vertex. It is clear that ∆ 1 [φ] = 1 for all φ ∈ I, and the root vector of T (∆ 1 )is obtained by Rule 0 in Definition 2.Now, let ∆ n be a rooted stack triangulation with n > 1. We denote by v the root of T (∆ n ). Weseparate the proof in cases according to the number of vertices of the rooted stack triangulations∆ ni= ∆ i n with i ∈ {1, 2, 3}. We note that if n i = 0 for every i ∈ {1, 2, 3}, then n = 1. Thus,we can assume that n i ≥ 1 for at least one index i ∈ {1, 2, 3}. We now consider three possiblesituations.First, assume there are i, j ∈ {1, 2, 3} with i ≠ j and k ∈ {1, 2, 3}\{i, j} such that n i = n j = 0and n k ≥ 1. By definition of the degeneracy vector, we have that v ∆ni = v ∆nj = (0, 1, 1, 1) t .Let v ∆nk= (v t k ) t∈{0,1,2,3}. According to Proposition 4, we have that⎧⎛⎞ ⎛ ⎞ ⎛ ⎞ ⎫v11 v21 v31 ⎪⎨v1 0 v ∆n ∈ ⎜+ v1 1⎟⎝ v1⎪⎩3 ⎠ , v23 ⎜ ⎟⎝ v22 ⎠ , v 2 ⎪⎬3⎜ ⎟ ,⎝v3 0 + v31 ⎠v12 v2 1 + ⎪⎭v0 2 v3310

where v ∆n is the k-th vector in the set above. Statement 2 of Proposition 3 says that T (∆ nk )is labeled by k and rooted on w, where w is the unique child of v. Given that 1 ≤ n k < n,by induction we get that w = v ∆nk . By Definition 2, we know that v is obtained from w byapplication of Rule 1. Hence, v = v ∆n .Assume now that there is an i ∈ {1, 2, 3} such that n i = 0 and j, k ∈ {1, 2, 3} \ {i} with j ≠ ksuch that n j , n k ≥ 1. We have that v ∆ni = (0, 1, 1, 1) t . Consider v ∆nj = (v t j ) t∈{0,1,2,3} andv ∆nk= (v t k ) t∈{0,1,2,3}. Proposition 4 implies that⎧⎛⎞ ⎛ ⎞ ⎛ ⎞⎫v2 1 ⎪⎨v1 3v1 1 v 3v ∆n ∈ ⎜2v32 v1 3v1 1 ⎝v2 ⎪⎩3v0 3 + ⎟v2 2 v1 3⎠ , v⎜1v 0 3 3 + v1v 1 2 v1 23⎝v1 2v0 3 + ⎟v3 1 v1 3⎠ , v⎜1v 0 2 2 + v1v 1 3 ⎪⎬2⎟ ,⎝ v1 3v2 2⎠⎪⎭v2v 1 3 3 + v2v 0 32 v1v 2 33 v1v 2 2 1 + v1v 3 20where v ∆n is the i-th vector in the set above. Statement 3 of Proposition 3 guarantees that theroot v of T (∆) has exactly two children w and u labeled j and k, respectively. Moreover, T (∆ j n )and T (∆ k n) are rooted on w and u, respectively. We know that 1 ≤ n j < n and 1 ≤ n k < n,then by induction, w = v ∆nj and u = v ∆nk . If we now apply Rule 2 of Definition 2, we getv = v ∆n .Finally, assume that n > n j ≥ 1 for every j ∈ {1, 2, 3}. Suppose that v ∆nj = (vj t) t∈{0,1,2,3}for each j ∈ {1, 2, 3}. Statement 4 of Proposition 3 and the induction hypothesis imply that theroot v of T (∆) has three children w 1 , w 2 and w 3 such that w j = v ∆nj for each j ∈ {1, 2, 3}. ByProposition 4 and since v is derived by applying Rule 3 of Definition 2, the desired conclusionfollows.5 Colored Rooted Ternary TreesThe goal of this section is to prove a result that we should refer to as the Main Lemma whichshows that the groundstate degeneracy of stack triangulations is exponential in the number ofits nodes.We now introduce notation that will be useful when dealing with rooted ternary trees. We denoteby |T | the number of vertices of the ternary tree T . For any node u of T , we denote by T u thecolored rooted sub-ternary tree of T rooted at u and induced by u and its descendants. If v isthe father of u in T , we say that T u is a component of v.Also, we denote by P ˜w,w any path with end nodes ˜w and w. Moreover, ||P ˜w,w || = |P ˜w,w | − 1denotes the length of P ˜w,w .11

5.1 RemaindersIn this subsection we introduce the concept of remainder of a rooted ternary tree and provesome useful and fundamental claims related to this concept. We will show that after removingremainders from a rooted ternary tree we are still left with a tree of size at least a third ofthe original one. The root vertex of the derived remainder-free tree will provide a componentwise lower bound on the components of the root vertex of the original rooted ternary tree. Theunderlying motivation for this section is that lower bounding the components of a root vertex issignificantly easier for remainder-free rooted ternary trees.Definition 3 Let v be a leaf of T and w be its father. Consider the following cases:I.- If u ≠ v is a child of w, then |T u | ≥ 3.II.- If T w is just the edge wv, then the father of w, say y, has two children w and u, where|T u | ≥ 3.If Case I holds, we say that {v} is a remainder of T and that w is the generator of {v}. IfCase II holds we say that {v, w} is a remainder of T and that y is its generator. We say that Tis remainder-free if it does not contain any remainder. We denote the set of remainders of T byR(T ) and by G(R(T )) the set of its generators.See Figures 5 and 6 for an illustration of the distinct situations encompassed by each of thepreceding definition’s cases.wwvuzvu|T u | ≥ 3 |T z | ≥ 3|T u | ≥ 3Figure 5: Structure of T w ⊆ T having a remainder v of T with generator w. Case where w hasthree children (left) and two children (right).Proposition 6 Let T be a rooted ternary tree. Then, |R(T )| = |G(R(T ))|.Proof: It is enough to show that any vertex w ∈ G(R(T )) is the generator of exactly oneremainder of T . For the sake of contradiction, suppose that w is the generator of at least tworemainders of T , say S 1 and S 2 . We consider three possible cases which cover all possible12

ywvu|T u | ≥ 3Figure 6: Structure of T y ⊆ T having a remainder {v, w} with generator y.scenarios: (i) S 1 = {v} and S 2 = {u}, (ii) S 1 = {v, ṽ} and S 2 = {u, ũ}, and (iii) S 1 ={v}, S 2 = {u, ũ}.If S 1 = {v} and S 2 = {u}, then by Case I of Definition 3, we get that |T v | ≥ 3. If S 1 = {v, ṽ}and S 2 = {u, ũ}, then by Case II of Definition 3, we get that |Tṽ| ≥ 3. If S 1 = {v} andS 2 = {u, ũ}, then by Case II of Definition 3, we have that |T v | ≥ 3. Hence, all feasible caseslead to contradictions.Let V R(T ) denote the subset of vertices of T which belong to the elements of R(T ),i.e. V R(T ) = ∪ S∈R(T ) {v : v ∈ S}.Lemma 7 Let T be a rooted ternary tree. Then, ˜T = T \ V R(T ) is remainder-free.Proof: For the sake of contradiction, assume S is a remainder of ˜T . We consider three scenariosdepending on which case of Definition 3 holds for S.First, assume S = {v} satisfies Case I of Definition 3 and the father of v in ˜T has tree childrenv, u, z with | ˜T u |, | ˜T z | ≥ 3. Clearly, |T u | ≥ | ˜T u | and |T z | ≥ | ˜T z |. Since v has no children in ˜T , itfollows that v /∈ G(R(T )). Thus, v is a leaf of T and {v} ∈ R(T ).Assume now that S = {v} satisfies Case I of Definition 3 and v’s father in ˜T , say w, has twochildren v, u with | ˜T u | ≥ 3. We have |T u | ≥ | ˜T u | ≥ 3. Moreover, since v is a leaf of ˜T , it mustalso hold that v is a leaf of T (otherwise, all of v’s children in T must belong to some remainder,a situation that is not possible). If w has three children in T , say v, u, z, then {z} ∈ R(T ). Thisimplies that |T v | ≥ 3, contradicting the fact that v is a leaf of T . Hence, w has two children inT . It follows that {v} ∈ R(T ).Finally, assume S = {v, ṽ} satisfies Case II of Definition 3. Let w be the generator of S andthe father of ṽ in ˜T . Then, w has two children ṽ, u in ˜T with | ˜T u | ≥ 3. We again have that|T u | ≥ | ˜T u | ≥ 3 and that v is a leaf of T . Assume w has three children in T , say ṽ, u, z. Then,{z} ∈ R(T ), implying that |Tṽ| ≥ 3, and hence ṽ ∈ G(R(T )). Therefore, |T v | ≥ 3, but thiscannot happen because v is a leaf of T . Thus, w must have only two children in T . If ṽ hasexactly two children in T , then ṽ ∈ G(R(T )) and |T v | ≥ 3, contradicting again the fact that v13

is a leaf. If ṽ has only one child, then {v, ṽ} ∈ R(T ), which contradicts the fact that ṽ is a nodeof ˜T .Since all possible scenarios lead to a contradiction, the desired conclusion follows.Lemma 8 Let ˜T = T \ V R(T ) . Then, | ˜T | ≥ |T |/3.Proof: Follows from the fact that G(R(T )) and R(T ) are disjoint, that each element S ∈ R(T )is of cardinality at most 2, and Proposition 6.5.2 Counting satisfying statesIn this section, we establish properties of the root vectors of colored rooted ternary trees andrelate them to characteristics of colored tree. Informally, for some special classes of coloredrooted ternary trees, we obtain lower bounds for the sum of the coordinates of its associatedrooted vectors.Recall that ϕ = (1+ √ 5)/2 ≈ 1.61803 denotes the golden ratio. For s ∈ { 0, . . . , 3 }, let e s ∈ Nand e = (ϕ es ) s=0,...,3 . Define3∑Ψ(e) = 2 e j , and Φ(e) = Ψ(e) − |{ s | e s > e 0 }| .j=1Henceforth, for a vector v we let v denote the collection of all vectors obtained by fixing thefirst coordinate of v and permuting the remaining coordinates in an arbitrary way. Note that ife = (ϕ es ) s=0,...,3 with e 0 , e 1 , e 2 , e 3 ∈ N, then for all ẽ ∈ e we have that Ψ(ẽ) = Ψ(e) andΦ(ẽ) = Φ(e). For a set S of vectors, we let S denote the union of the sets v where v variesover S.Given vectors x = (x s ) s=0,...,3 and y = (y s ) s=0,...,3 , we write x ≥ y if x s ≥ y s for all s ∈{0, . . . , 3}.Proposition 9 Let T v be a colored rooted ternary tree with |T v | = 2. Then, there aree 0 , e 1 , e 2 , e 3 ∈ N such that v ≥ e = (ϕ es ) s=0,...,3 and Ψ(e) = 2.Proof: Clearly T v is a rooted tree on v with exactly one child w which is a leaf of T v . In otherwords, T v = P w,v with ||P w,v || = 1. We observe that by applying Rules 0 and 1, we get thatw = (1, 1, 1, 1) t and v ∈ (1, 2, 1, 1) t . Given that 1 = ϕ 0 and 2 ≥ ϕ 1 , it is easy to see thatthe desired vector e belongs to (ϕ 0 , ϕ 1 , ϕ 0 , ϕ 0 )Proposition 10 Let T v be a colored rooted ternary tree with |T v | = 3. Then, there aree 0 , e 1 , e 2 , e 3 ∈ N such that v ≥ e = (ϕ es ) s=0,...,3 and Ψ(e) = 4.14

Proof: Since |T v | = 3, either T v = P w,v with ||P w,v || = 2, or v has exactly two children w andu, which are leaves of T v .In the first scenario, applying Rule 0 once and Rule 1 twice, we get that v ∈ (2, 3, 1, 1) t , (1, 2, 2, 1) t .Given that 1 = ϕ 0 , 2 ≥ ϕ 1 and 3 ≥ ϕ 2 , we can take e ∈ (ϕ 1 , ϕ 2 , ϕ 0 , ϕ 0 ) t , (ϕ 0 , ϕ 1 , ϕ 1 , ϕ 0 ) t satisfying the statement.In the second scenario, applying Rule 0, we get that w and u are vectors all of whose coordinatesare 1. Applying Rule 2, we see that v ∈ (1, 2, 1, 2) t . Given that 1 = ϕ 0 and 2 ≥ ϕ 1 , thedesired vector e may be chosen from the set (ϕ 0 , ϕ 1 , ϕ 0 , ϕ 1 ) t .Proposition 11 Let T v be a colored rooted ternary tree with |T v | = 4. Then, there aree 0 , e 1 , e 2 , e 3 ∈ N such that v ≥ e = (ϕ es ) s=0,...,3 and Ψ(e) ≥ 6.Proof: The tree T v may be one of the four trees depicted in Figure 7. Each case is analyzedseparately below (in the order in which they appear in Figure 7).vvvvwuwwuz˜wuzwFigure 7: All rooted ternary trees with 4 vertices.For the first case, note that by Rule 0 we have that w, u and z are vectors all of whosecoordinates are 1. Thus, by Rule 3, we get that v = (2, 2, 2, 2) t . Hence, v ≥ e wheree = (ϕ 1 , ϕ 1 , ϕ 1 , ϕ 1 ) t .For the second case, by Rule 0 we have that all coordinates of u and ˜w are 1. Thus, by Rule 1,w ∈ (1, 2, 1, 1) t . Then, by Rule 2, we get that v ∈ (2, 3, 1, 2) t , (1, 2, 1, 3) t , (1, 3, 2, 2) t .Given that 1 = ϕ 0 , 2 ≥ ϕ 1 and 3 ≥ ϕ 2 the result follows.For the third case, note that |T w | = 3 and that the structure of T w is the same as the second oneconsidered in the proof of Proposition 10. Hence, we know that w ∈ (1, 2, 1, 2) t . By Rule 1,we get that v ∈ (2, 3, 2, 1) t , (1, 2, 2, 2) t . Given that 1 = ϕ 0 , 2 ≥ ϕ 1 and 3 ≥ ϕ 2 the claimedresult follows.We leave the last case to the interested reader.15

vwu˜wũFigure 8: The tree T v of Proposition 12.Proposition 12 Let T v be a colored rooted ternary tree with |T v | = 5 and where v has twochildren which are not leaves. Then, there are e 0 , e 1 , e 2 , e 3 ∈ N such that v ≥ e = (ϕ es ) s=0,...,3and Ψ(e) ≥ 8.Proof: Assume T v is as depicted in Figure 8. Clearly, w, u ∈ (1, 2, 1, 1) t . By Rule 2 we getthat v ∈ (4, 3, 1, 3) t , (1, 3, 3, 2) t , (2, 4, 2, 2) t , (2, 2, 1, 5) t , (1, 3, 1, 3) t , (1, 3, 4, 3) t . The desiredconclusion follows since 1 = ϕ 0 , 2 ≥ ϕ 1 , 3 ≥ ϕ 2 , 4 ≥ ϕ 2 and 5 ≥ ϕ 3 .Proposition 13 Let T v be a colored rooted ternary tree such that v has three children u, w andz. Suppose that 1 ≤ |T u | ≤ 3 and 1 ≤ |T w | ≤ 3. Then,• If |T z | = 2, there are e 0 , e 1 , e 2 , e 3 ∈ N such that v ≥ e = (ϕ es ) s=0,...,3 and Ψ(e) ≥ 8.• If |T z | = 3, there are e 0 , e 1 , e 2 , e 3 ∈ N such that v ≥ e = (ϕ es ) s=0,...,3 and Ψ(e) ≥ 10.Proof: We first note that if |T x | ≥ 2, then x ≥ (1, 1, 1, 1) t . This implies that it is enough toprove both statements for the case |T u | = 1 and |T w | = 1. Observe that by Rule 0, we have thatu = w = (1, 1, 1, 1) t .For the first statement, assume |T z | = 2. By Rule 1, we have that z ∈ (1, 2, 1, 1) t . Then, byRule 3 we have that v ∈ (3, 3, 2, 2) t , (2, 3, 3, 2) t . The result follows, since 2 ≥ ϕ 1 and 3 ≥ ϕ 2 .Assume now that |T z | = 3. From the proof of Proposition 10 we know thatz ∈ (2, 3, 1, 1) t , (1, 2, 2, 1) t . Then, by Rule 3 we have thatv ∈ (5, 2, 5, 2) t , (3, 4, 3, 4) t , (3, 3, 3, 3) t , (2, 4, 2, 4) t . The desired conclusion follows since2 ≥ ϕ 1 , 3 ≥ ϕ 2 , 4 ≥ ϕ 2 and 5 ≥ ϕ 3 .Lemma 14 Let T = T v be a colored rooted ternary tree, such that T v = Tṽ ∪ Pṽ,v where Pṽ,vis non-trivial. If ṽ ≥ ẽ = (ϕẽs ) s=0,...,3 with ẽ 0 , ẽ 1 , ẽ 2 , ẽ 3 ∈ N, then there are e 0 , e 1 , e 2 , e 3 ∈ Nsuch that v ≥ e = (ϕ es ) s=0,...,3 and Φ(e) ≥ Φ(ẽ) + ||Pṽ,v ||.Proof:It is enough to prove the result for Pṽ,v of length 1. By Rule 1, we get that v ∈ [ṽ]where ṽ ≥ ẽ for some ẽ ∈ (ϕẽ1 , ϕẽ1 + ϕẽ0 , ϕẽ3 , ϕẽ2 ) t . Assume ẽ is the vector within thedouble brackets (the other cases are similar). We now consider several scenarios:16

• Case ẽ 1 > ẽ 0 + 1: Clearly, v ≥ e = , , , ) t . Moreover, Ψ(e) = Ψ(ẽ) and(ϕẽ1 ϕẽ1 ϕẽ3 ϕẽ2|{ s | e s > e 0 }| ≤ |{ s | ẽ s > ẽ 0 }| − 1. Hence, Φ(e) ≥ Φ(ẽ) + 1.• Case ẽ 1 ∈ {ẽ 0 , ẽ 0 + 1}: If ẽ 1 = ẽ 0 , then + = ≥ ϕẽ1 ϕẽ0 2ϕẽ1 ϕẽ1+1 . Since 1 + ϕ = ϕ 2 , ifẽ 1 = ẽ 0 + 1, then + = ϕẽ1 ϕẽ0 ϕẽ1+1 . Hence, v ≥ e = , (ϕẽ1 ϕẽ1+1 , , ) t . Moreover,ϕẽ3 ϕẽ2Ψ(e) = Ψ(ẽ) + 2 and |{ s | e s > e 0 }| ≤ |{ s | ẽ s > ẽ 0 }| + 1. Hence, Φ(e) ≥ Φ(ẽ) + 1.• Case ẽ 1 ≤ ẽ 0 − 1: Since 1 + ϕ = ϕ 2 , if ẽ 1 = ẽ 0 − 1, then + = ϕẽ1 ϕẽ0 ϕẽ1+2 . Ifẽ 1 ≤ ẽ 0 − 2, then + ≥ ϕẽ1 ϕẽ0 ϕẽ1+2 . Hence, v ≥ e = , (ϕẽ1 ϕẽ1+2 , , ) t . Moreover,ϕẽ3 ϕẽ2Ψ(e) = Ψ(ẽ) + 4 and |{ s | e s > e 0 }| ≤ |{ s | ẽ s > ẽ 0 }| + 3. Hence, Φ(e) ≥ Φ(ẽ) + 1.The following result is an immediate consequence of Lemma 14.Corollary 15 Let T = T v be a colored rooted ternary tree, such that T v = Tṽ ∪ Pṽ,v where Pṽ,vis non-trivial. If ṽ ≥ ẽ, then an e exists such that v ≥ e andΨ(e) ≥ Ψ(ẽ) + max{||Pṽ,v || − 3, 0} .Lemma 16 Let T v be a colored rooted ternary tree, such that v has two children w and u.If w ≥ e w = (ϕ ew s )s=0,...,3 , with e w 0 , ew 1 , ew 2 , ew 3 ∈ N and u ≥ e u = (ϕ eu s )s=0,...,3 , withe u 0, e u 1, e u 2, e u 3 ∈ N, then there are e 0 , e 1 , e 2 , e 3 ∈ N such that v ≥ e = (ϕ es ) s=0,...,3 andΨ(e) = Ψ(e w ) + Ψ(e u ).Proof: Since w ≥ e w and u ≥ e u , by Rule 2 we have that v ≥ ṽ where⎧⎛⎞ ⎛⎞ ⎛ϕ ew 1 +eu 1ϕ ew 1 +eu 1ϕ ew 1 +eu 1⎪⎨ϕ ew 0 +eu 2 + ϕe w 1 +eu 3ṽ ∈ ⎜⎟⎝ ϕ⎪⎩ew 3 +eu 2 ⎠ , ϕ ew 3 +eu 2⎜⎟⎝ ϕ ew 2 +eu 1 + ϕe w 3 +eu 0 ⎠ , ϕ ew 3 +eu 0 + ϕe w 2 +eu 1⎜⎝ ϕ ew 0 +eu 2 + ϕe w 1 +eu 3ϕ ew 2 +eu 1 + ϕe w 3 +eu 0 ϕ ew 1 +eu 3 + ϕe w 0 +eu 2ϕ ew 3 +eu 2⎞⎫⎪⎬⎟ .⎠⎪⎭Moreover, ϕ ew 0 +eu 2 + ϕe w 1 +eu 3 ≥ ϕe w 1 +eu 3 and ϕe w 2 +eu 1 + ϕe w 3 +eu 0 ≥ ϕe w 2 +eu 1 , so depending on thevalue of ṽ we can take⎧⎛⎞ ⎛ ⎞ ⎛ ⎞⎫ϕ ew 1 +eu 1 ϕ ew 1 +eu 1 ϕ ew 1 +eu 1⎪⎨ϕ ew 1 +eu 3e ∈ ⎜ ⎟⎝ϕ ⎪⎩ew 3 +eu 2 ⎠ , ϕ ew 3 +eu 2⎜ ⎟⎝ϕ ew 2 +eu 1 ⎠ , ϕ ⎪⎬ew 2 +eu 1⎜ ⎟ ,⎝ϕ ew 1 +eu 3 ⎠⎪⎭ϕ ew 2 +eu 1ϕ ew 1 +eu 3and obtain that v ≥ e and Ψ(e) = Ψ(e w ) + Ψ(e u ).ϕ ew 3 +eu 217

Lemma 17 Let T v be a colored rooted ternary tree, such that v has three children w, u and z. Ifw ≥ e w = (ϕ ew s )s=0,...,3 with e w 0 , ew 1 , ew 2 , ew 3 ∈ N, u ≥ eu = (ϕ eu s )s=0,...,3 with e u 0 , eu 1 , eu 2 , eu 3 ∈ Nand z ≥ e z = (ϕ ez s )s=0,...,3 with e z 0, e z 1, e z 2, e z 3 ∈ N, then there are e 0 , e 1 , e 2 , e 3 ∈ N such thatv ≥ e = (ϕ es ) s=0,...,3 and Ψ(e) = Ψ(e w ) + Ψ(e u ) + Ψ(e z ).Proof: By Rule 3 we have that⎛⎞ ⎛ϕ ew 0 +eu 0 +ez 0 + ϕe w 1 +eu 1 +ez 1ϕ ew 0 +eu 3 +ez 2 + ϕe w 1 +eu 3 +ez 2v = ⎜⎟⎝ϕ ew 2 +eu 3 +ez 0 + ϕe w 3 +eu 2 +ez 1⎠ ≥ ⎜⎝ϕ ew 2 +eu 1 +ez 3 + ϕe w 3 +eu 0 +ez 2ϕ ew 1 +eu 1 +ez 1ϕ ew 1 +eu 3 +ez 2ϕ ew 3 +eu 2 +ez 1ϕ ew 2 +eu 1 +ez 3Let e be the last vector in the preceding expression and note that Ψ(e) = Ψ(e w )+Ψ(e u )+Ψ(e z ).⎞⎟⎠ .5.3 Main LemmaThe main result of this work, i.e. Theorem 1, will follow almost directly from the next keyclaim which roughly says that the root vector v of a colored rooted ternary remainder-free treeT = T v either has large components relative to the size of T v , or T v corresponds to a short pathPṽ,v and a tree Tṽ whose root vertex ṽ has large components relative to the size of Tṽ.Lemma 18 Let T = T v be a colored rooted ternary remainder-free tree such that |T | ≥ 4.Then, there is a path Pṽ,v such that T v = Tṽ ∪ Pṽ,v with 0 ≤ ||Pṽ,v || ≤ 5 (if ||Pṽ,v || = 0, thenṽ = v and Tṽ = T v ) and there are eṽ0 , eṽ1 , eṽ2 , eṽ3 ∈ N such thatṽ ≥ eṽ = (ϕ eṽs ) s=0,...,3 , and Ψ(eṽ) ≥ |T ṽ| + 72. (2)Proof: We proceed by induction on |T |. For the base case |T v | = 4, by Proposition 11, thereexists an e ≤ v such that Ψ(e) ≥ 6 > (|T v | + 7)/2. Let T v be a colored rooted ternary treeremainder-free with |T v | ≥ 5. We separate the proof in three cases depending on the number ofchildren of the root v. It is clear that for any node u of T v , the tree T u is a colored rooted ternaryremainder-free tree.Case 1 (v has one child w): We have |T w | = |T v | − 1 ≥ 4. By induction T w = T ˜w ∪ P ˜w,wwith 0 ≤ ||P ˜w,w || ≤ 5 and ˜w satisfying (2). If ||P w, ˜w || < 5, then T v = T ˜w ∪ P ˜w,v and thusit satisfies the desired property. By Corollary 15, we know that there is and e ≤ v suchthat Ψ(e) ≥ Ψ(e ˜w ) + ||P ˜w,v || − 3. Given that |T v | = |T ˜w | + ||P w, ˜w || + 1, if ||P w, ˜w || = 5,then there is an e ≤ v such thatΨ(e) ≥ Ψ(e ˜w ) + ||P ˜w,v || − 3 ≥ |T ˜w| + 72Therefore, T v satisfies the desired property.18+ ||P ˜w,v || − 3 = |T v| + 72.

Case 2 (v has two children w and u): First, note that T w and T u have size at least 2 (otherwisewe would have, say |T w | = 1 and |T u | = |T v | − |T w | − 1 ≥ 3, implying that w is aremainder of T , a contradiction). If |T w | = 2, then |T u | = 2 (otherwise, |T u | ≥ 3,implying that there is a remainder S of T such that w ∈ S, a contradiction). Since|T w | = |T u | = 2, by Proposition 12 we have that there is an e ≤ v such that Ψ(e) = 8 >(|T v | + 7)/2.Hence, we assume that |T w |, |T u | ≥ 3. If |T w | = |T u | = 3, by Proposition 10 andLemma 16, we get that there is an e ≤ v such that Ψ(e) = 8 > (|T v | + 7)/2.We now assume that |T w | = 3 and |T u | ≥ 4. By induction, T u = Tũ ∪ Pũ,u with0 ≤ ||Pũ,u || ≤ 5 and ũ satisfying (2). By Lemma 16, there is an e ≤ v such thatΨ(e) = Ψ(e w ) + Ψ(e u ). By Proposition 10, Corollary 15, and the fact that |T v | =|Tũ| + ||Pũ,u || + 4,Ψ(e) ≥ Ψ(e w ) + Ψ(eũ) + max{||Pũ,u || − 3, 0}≥ 4 + |T ũ| + 7+ max{||Pũ,u || − 3, 0}2= |T v| + 72≥ |T v| + 72≥ |T v| + 72Hence, T v satisfies the desired property.+ 4 − ||P ũ,u||2+ max{||Pũ,u || − 3, 0}+ 1 2 max{3 − ||P ũ,u||, ||Pũ,u || − 3, 0}.Finally, we assume that |T w |, |T u | ≥ 4. By induction, T w = T ˜w ∪P ˜w,w and T u = Tũ ∪Pũ,uwhere 0 ≤ ||P ˜w,w ||, ||Pũ,u || ≤ 5 and ũ, ˜w satisfying (2). By Lemma 16, there is an e ≤ vsuch that Ψ(e) = Ψ(e w ) + Ψ(e u ). By Corollary 15 and given that |T v | = |T ˜w | + |Tũ| +||P ˜w,w || + ||Pũ,u || + 1,Ψ(e) ≥ Ψ(e ˜w ) + Ψ(eũ) + + max{||P ˜w,w ||−3, 0} + max{||Pũ,u ||−3, 0}≥ |T ˜w| + 7+ |T ũ| + 7+ max{||P ˜w,w ||−3, 0} + max{||Pũ,u ||−3, 0}2 2= |T v| + 72≥ T v + 72.Hence, T v satisfies the desired property.+ 1 2 max{||P ˜w,w||−3, 3−||P ˜w,w ||} + 1 2 max{||P ũ,u||−3, 3−||Pũ,u ||}Case 3 (v has three children w, u and z): Since |T v | ≥ 5, it can not happen that |T u | = |T u | =|T z | = 1. If 1 ≤ |T w | ≤ 3, 1 ≤ |T u | ≤ 3 and 2 ≤ |T z | ≤ 3, we have that: if |T z | = 2, then|T v | ≤ 9 and by the first statement of Proposition 13 there is a vector e ≤ v such that19

Ψ(e) ≥ 8 = 16/2 ≥ (|T v |+7)/2; if |T z | = 3, then |T v | ≤ 10 and by the second statementof Proposition 13 there is a vector e ≤ v such that Ψ(e) ≥ 10 > 17/2 ≥ (|T v | + 7)/2.Therefore, T v satisfies the desired property.We now assume that at least one of the children of v induces a subtree with at least 4vertices.• If 1 ≤ |T w |, |T u | ≤ 2 and |T z | ≥ 4, then by Rules 0, 1 and 3, we have⎛ ⎞⎛ ⎞⎛ ⎞ϕ ez 0 + ϕe z 1ϕ ez 0 + ϕe z 1ϕ ez 0 + ϕe z 1ϕ ez 3 + ϕe z 2v ≥ ⎜ ⎟⎝ϕ ez 0 + ϕe z 1⎠ , or v ≥ ϕ ez 2 + ϕe z 3⎜ ⎟⎝ϕ ez 3 + ϕe z 2⎠ , or v ≥ ϕ ez 0 + ϕe z 1⎜ ⎟⎝ϕ ez 2 + ϕe z 3⎠ .ϕ ez 3 + ϕe z 2ϕ ez 1 + ϕe z 0If e z 3 = e z 2, given that 2 > ϕ, we may choose the vector e from the set⎧⎛⎞ ⎛ϕ ez 1⎪⎨ϕ ez 3 +1⎜ ⎟⎝ ϕ⎪⎩ez 1 ⎠ , ⎜⎝ϕ ez 2 +1ϕ ez 1ϕ ez 3 +1ϕ ez 2 +1ϕ ez 1⎞ ⎛ ⎞⎫ϕ ez 1⎟⎠ , ϕ ⎪⎬ ez 1⎜ ⎟ .⎝ϕ ez 2 +1 ⎠⎪⎭ϕ ez 3 +1ϕ ez 2 + ϕe z 3If not, we have e z 3 ≥ e z 2 + 1 (analogously e z 2 ≥ e z 3 + 1) and given that ϕ + 1 = ϕ 2 ,we may choose the vector e from the set⎧⎛⎪⎨⎜⎝⎪⎩ϕ ez 1ϕ ez 2 +2ϕ ez 1ϕ ez 3⎞ ⎛⎟⎠ , ⎜⎝ϕ ez 1ϕ ez 3ϕ ez 2 +2ϕ ez 1⎞ ⎛ ⎞⎫ϕ ez 1⎟⎠ , ϕ ⎪⎬ ez 1⎜ ⎟ .⎝ ϕ ez 3 ⎠⎪⎭ϕ ez 2 +2Therefore, for any choice of e we get that Ψ(e) = Ψ(e z ) + 4. By induction, T z =T˜z ∪ P˜z,z with 0 ≤ ||P˜z,z || ≤ 5 and ˜z satisfying (2). Since |T v | ≤ |T˜z | + ||P˜z,z || + 5,by Corollary 15,Ψ(e) ≥ 4 + Ψ(e˜z ) + max{||P˜z,z || − 3, 0}≥ |T˜z| + 7+ 4 + max{||P˜z,z || − 3, 0}2≥ |T v| + 72= |T v| + 72≥ |T v| + 72Hence, T v satisfies the desired property.+ 3 − ||P˜z,z||2+ max{||P˜z,z || − 3, 0}+ 1 2 max{3 − ||P˜z,z||, ||P˜z,z || − 3}.20

• If 2 ≤ |T w |, |T u | ≤ 3 and |T z | ≥ 4. By Lemma 17, there is an e ≤ v such thatΨ(e) = Ψ(e w ) + Ψ(e u ) + Ψ(e z ). By Proposition 9 and Proposition 10, we haveΨ(e w ) = 2(|T w | − 1) and Ψ(e u ) = 2(|T u | − 1). By induction, T z = T˜z ∪ P˜z,z with0 ≤ ||P˜z,z || ≤ 5 and ˜z satisfying (2). Since |T v | = |T w | + |T u | + |T˜z | + ||P˜z,z || + 1,by Corollary 15,Ψ(e) ≥ 2(|T w | − 1) + 2(|T u | − 1) + Ψ(e˜z ) + max{||P˜z,z || − 3, 0}≥ |T˜z| + 7+ 2(|T w | + |T u |) − 4 + max{||P˜z,z || − 3, 0}2= |T v| + 72≥ |T v| + 72= |T v| + 72≥ |T v| + 72+ 3 2 (|T w| + |T u |) − ||P˜z,z||2+ 3 − ||P˜z,z||2Hence, T v satisfies the desired property.+ max{||P˜z,z || − 3, 0}+ 1 2 max{3 − ||P˜z,z||, ||P˜z,z || − 3}.− 9 2 + max{||P˜z,z|| − 3, 0}• The case |T w | = 1, |T u | = 3, and |T z | ≥ 4 can not happen, since it would imply that{w} is a remainder of T .• If 2 ≤ |T w | ≤ 3 and |T u |, |T z | ≥ 4. By Lemma 17, there is an e ≤ v such thatΨ(e) = Ψ(e w ) + Ψ(e u ) + Ψ(e z ). By Proposition 9 and Proposition 10, we haveΨ(e w ) = 2(|T w | − 1). By induction, T z = T˜z ∪ P˜z,z and T z = T˜z ∪ P˜z,z with0 ≤ ||Pũ,u ||, ||P˜z,z || ≤ 5 and ũ, ˜z satisfying (2). Since |T v | = |T w | + |Tũ| + |T˜z | +||Pũ,u || + ||P˜z,z || + 1, by Corollary 15,Ψ(e) ≥ 2(|T w | − 1) + Ψ(eũ) + Ψ(e˜z ) + max{||Pũ,u || − 3, 0} + max{||P˜z,z || − 3, 0}≥ |T ũ| + 7+ |T˜z| + 7+ 2(|T w | − 1)2 2+ max{||Pũ,u || − 3, 0} + max{||P˜z,z || − 3, 0}= |T v| + 7+ 3 2 2 |T w| − 2 + 6 − ||P ũ,u|| − ||P˜z,z ||2+ max{||Pũ,u || − 3, 0} + max{||P˜z,z || − 3, 0}≥ |T v| + 7+ 6 − ||P ũ,u|| − ||P˜z,z ||22+ max{||Pũ,u || − 3, 0} + max{||P˜z,z || − 3, 0}= |T v| + 7+ 1 2 2 max{3 − ||P˜z,z||, ||P˜z,z || − 3}≥ |T v| + 7.2Hence, T v satisfies the desired property.21

• If |T w |, |T u |, |T z | ≥ 4. Similar to the preceding case.6 Proof of Main ResultsProof of Theorem 1: Recall that T (∆ n ) is a colored rooted ternary tree on |∆ n |−3 nodes suchthat its root vector v is equal to the degeneracy vector of ∆ n . By Lemmas 7 and 8, the rootedcolored ternary tree ˜T (∆ n ) = T (∆ n ) \ V R(T (∆n)) is remainder-free and | ˜T (∆ n )| ≥ |T (∆ n )|/3.Clearly, the root vector ṽ of ˜T (∆ n ) is such that v ≥ ṽ. The Main Lemma guarantees that thereare e 0 , e 1 , e 2 , e 3 ∈ N such that v ≥ e = (ϕ es ) s=0,..,3 andΨ(e) ≥ (| ˜T (∆ n )| − 5) + 72= | ˜T (∆ n )| + 22≥ |T (∆ n)| + 612= |∆ n| + 312.Moreover, we know that ∆ n [φ] = ∆ n [−φ] for all φ ∈ {+, −} 3 . Hence, the groundstate degeneracyof ∆ n is at least 2 ∑ 3s=1 ϕes ≥ 6ϕ 1 3 Ψ(e) ≥ 6ϕ (|∆n|+3)/36 .Proof of Corollary 2: Let G be a cubic planar graph such that its geometric dual graph is thestack triangulation ∆. We know that the number of perfect matchings of G is equal to half ofthe groundstate degeneracy of ∆. From Euler’s formula we get that 2|∆| = |G| − 4. Therefore,by Theorem 1 we have that the number of perfect matchings of G is at least 3ϕ |G|/72 .7 Final CommentsThe approach followed throughout this work seems to be specially well suited for calculatingthe groundstate degeneracy of triangulations that have some sort of recursive tree like construction,e.g. 3-trees. It would be interesting to identify other such families of triangulations wheresimilar methods allowed to lower bound their groundstate degeneracy. Of particular relevancewould be to show that the approach we follow in this work can actually be successfully appliedto obtain exponential lower bounds for non-trivial families of non-planar bridgeless cubicgraphs.As already mentioned, our arguments are motivated by the transfer matrix method as used bystatistical physicists. We believe that most of the arguments we developed throughout thiswork can be stated in more combinatorial terms, except maybe for our Main Lemma. It mighteventually be worthwhile to clarify the implicit combinatorial structure of our proof arguments.22

AcknowledgementThe authors thank Martin Loebl for his encouragement, comments, and many insightful discussions.References[1] R. J. Baxter. Exactly Solved Models in Statistical Mechanics. Dover Publications, 2008.[2] A. Brandstädt, V. B. Le, and J. P. Spinrad. Graph classes: A survey. SIAM Monographsin Discrete Mathematics and Applications. SIAM, 1999.[3] M. Chudnovsky and P. Seymour. Perfect matchings in planar cubic graphs. Manuscript,2008.[4] L. Esperet, F. Kardos, A. King, D. Král, and S. Norine. Exponentially many perfectmatchings in cubic graphs. arXiv:1012.2878v1, 13 Dec 2010.[5] S. Felsner and F. Zickfeld. On the number of planar orientations with prescribed degrees.The Electronic Journal of Combinatorics, 15(1), 2008.[6] F. Jaeger. A survey of the cycle double cover conjecture. Discrete Applied Mathematics,99(1):71–90, 2000.[7] A. Jimeńez, M. Kiwi, and M. Loebl. Satisfying states of triangulations of a convex n-gon.The Electronic Journal of Combinatorics, 17(1), 2010.[8] R. Liebmann. Statistical Mechanics of Periodic Frustrated Ising Systems, volume 251 ofLecture Notes in Physics.[9] M. Loebl and J. Vondrák. Towards a theory of frustrated degeneracy. Discrete Mathematics,271(1-3):179–193, 2003.[10] J.-F. Marckert and M. Albenque. Some families of increasing planar maps. ElectronicJournal of Probability, 13:1624–1671, 2008.[11] J. P. Sethna. Statistical Mechanics: Entropy, Order Parameters and Complexity (OxfordMaster Series in Physics). Oxford University Press, USA, 2006.[12] M. Voorhoeve. A lower bound for the permanents of certain (0, 1)-matrices. IndagationesMathematicae, 82(1):83–86, 1979.23

Apéndice CDificultad computacional del problemade enumerar estados satisfactorios entriangulaciones62

Computational Hardness of Enumerating SatisfyingSpin-Assignments in TriangulationsAndrea Jiménez ∗ Marcos Kiwi †Submitted to Theoretical Computer Science, July 2011AbstractSatisfying spin-assignments in triangulations of a surface are states of minimum energy ofthe antiferromagnetic Ising model on triangulations which correspond (via geometric duality)to perfect matchings in cubic bridgeless graphs. In this work we show that it is NP-completeto decide whether or not a surface triangulation admits a satisfying spin-assignment, and thatit is #P-complete to determine the number of such assignments. Both results are derived viaan elaborate (and atypical) reduction that maps a Boolean formula in 3-conjunctive normalform into a triangulation of an orientable closed surface.Keywords: Ising model; Triangulations; Groundstates; Parsimonious reduction; #P-complete.1 IntroductionThe Ising model is one of the most studied models in statistical physics.Characterizing itsbehavior on a system (graph) helps to understand physical phenomena associated to its thermodynamicproperties [11]. The Ising model has been widely studied in lattices and regularstructures (see for example [1, 8] and references therein). In contrast, irregular systems havereceived much less attention, probably due to the difficulty of deriving meaningful analyticalresults.The number of distinct groundstates of the antiferromagnetic (negative coupling constant)Ising model of a system is called groundstate degeneracy and is typically exponentially large as afunction of a parameter that measures the system’s size (the number of nodes of the underlyinggraph). The latter translates to nonzero entropy at zero temperature when the system size goesto infinity, which in physical terms means that in the thermodynamical limit the spin arrangementsof particles in the system is disordered. This partly explains the considerable attention∗ Depto. Ing. Matemática, U. Chile. Web: www.dim.uchile.cl/∼ajimenez. Gratefully acknowledges thesupport of MECESUP UCH0607, and CONICYT via Basal in Applied Mathematics and FONDECYT 1090227.† Depto. Ing. Matemática & Ctr. Modelamiento Matemático UMI 2807, U. Chile. Web:www.dim.uchile.cl/∼mkiwi.Mathematics and FONDECYT 1090227.Gratefully acknowledges the support of CONICYT via Basal in Applied1

physicist have given to developing techniques for approximating the groundstate degeneracy ofa system.Typically, researchers have focussed on developing techniques for bounding the groundstatedegeneracy of a system, for example the Transfer Matrix Method [7, §6.6]. Instead, informallyspeaking, in this work we focus on the following two associated computational complexity problems;(1) hardness of deciding whether or not a given system admits a satisfying state, and (2)hardness of enumerating satisfying states (equivalently, computing the groundstate degeneracyof systems that admits satisfying states) of a given system.We show that the former problem is NP-complete and the latter is #P-complete.We now make precise the notions discussed above and formally state our main results.First, we describe the antiferromagnetic Ising model. We say that an embedding of a graphin an orientable closed surface is a surface triangulation if each face is bounded by a cycle oflength 3 (in particular there is no loop) — multiple edges allowed. Given a triangulation T ,let V (T ) and E(T ) denote the node and edge set of T . A mapping s : V (T ) → { −1, +1 } willbe called a spin-assignment (state) to T . We refer to −1 and +1 as spins. The energy of aspin-assignment s of the antiferromagnetic Ising model is defined as ∑ uv∈E(T )σ(u) · σ(v). Agroundstate is a spin-assignment of minimum energy. The number of distinct groundstates thata triangulation T admits is often referred to as the groundstate degeneracy of T . Clearly, underany spin-assignment to T the ends of at least one edge of each face of a surface triangulation Tare both assigned either -1 or +1. Moreover, a spin-assignment is a groundstate if it has thesmallest possible number of edges with both its ends being assigned the same spin. A face △ of asurface triangulation T is said to be frustrated under assignment s, if s restricted to V (△) is notidentically -1 or +1. A spin-assignment s to T is said to be satisfying (or feasible) if every face ofT is frustrated under s. Obviously a satisfying spin-assignment is a groundstate. The converse istrue for triangulations that can be embedded in the plane [6]. Nevertheless, the equivalence doesnot hold in general (the reader can verify that the toroidal triangulation depicted in Figure 1does not have satisfying spin-assignments). However, note that when satisfying spin-assignmentsexist, then a groundstate is necessarily a satisfying spin-assignment.uww ′uvvv ′ v ′uww ′uFigure 1: A triangulation of the torus with no satisfying spin-assignment.2

In [6] a relation was established between the groundstate degeneracy of the antiferromagneticand the number of perfect matchings in cubic bridgeless graphs. Specifically, let T betriangulation of a orientable closed surface and T ∗ its geometric dual. In [6], it is shown thatthe set of edges whose ends are assigned the same spin under a given satisfying spin-assignmentto T correspond to a perfect matching of the cubic bridgeless graph T ∗ . Moreover, it is shownthat if T admits a satisfying spin-assignment, then the groundstate degeneracy of T is at mosttwice the number of perfect matchings of T ∗ . Thus, lower bounds on the groundstate degeneracyof T provide lower bounds on the number distinct of perfect matchings of the cubic bridgelessgraph T ∗ . An old and famous conjecture of Lovász and Plummer, recently positively settled [4],claimed that the number of distinct perfect matchings of T ∗ is exponential in the size (numberof nodes) of T ∗ . The relation between number of satisfying spin-assignments of a surfacetriangulation and the number of perfect matchings of bridgeless cubic graphs is another one ofour motivations for considering the problem of computing the groundstate degeneracy of surfacetriangulations.In this work we show that the problem of deciding if a triangulation admits a satisfyingspin-assignment is NP-complete. We also establish that computing the groundstate degeneracyof surface triangulations that admit satisfiable spin-assignments is #P-complete.We observe that arbitrariness of the surface where triangulations are embedded plays anessential role, because it is known that both problems can be performed in polynomial timewhen triangulations are embedded in a fixed orientable closed surface [5].1.1 ContributionsLet T be a surface triangulation. For each v in V (T ) let δ(v) denote the set of edges incidentto v. The map π v : δ(v) → δ(v) is called cyclic permutation of the edges incident to v if for everye incident to v the edge π v (e) is the successor of e in the clockwise ordering around v definedby the surface embedding of T . The tuple π = (π v : v ∈ V (T )) is called the rotation system ofT . A direct consequence of the Heffter-Edmonds-Ringel rotation principle, is that every surfacetriangulation is uniquely determined, up to homeomorphism, by its rotation system [9, §3.2].Let SatAssign be the collection of (encodings of) rotation systems of surface triangulationsthat admit a satisfying spin-assignment. Also, let #SatAssign be the function mapping (encodingsof) rotation systems of surface triangulations to its number of satisfying spin-assignments.To see that SatAssign is in NP, first recall that in order to check that π = (π v : v ∈ V (T )) isan instance of SatAssign we need not start with a surface. Indeed, it suffices to check for everyv ∈ V (T ) that π v is a cyclic permutation of δ(v) = {uv : uv is in π u ’s domain}, a task that canbe performed in time quadratic in |V (T )| time in the Random Access Model. Then, observethat a certificate of membership in SatAssign of a rotation system of a surface triangulation Tis simply a spin-assignment s : V (T ) → { -1, +1 } and that verifying that such an assignmentis satisfying amounts to checking that each face ∆ of the surface triangulation T is frustratedunder s (which can be checked in O(1) time per face in the Random Access Model).In this work we establish the following results.3

Theorem 1 SatAssign is NP-complete.Theorem 2 #SatAssign is #P-complete.Both of the stated results follow from an elaborate weakly parsimonious reduction [3, Definition2.27] that maps a Boolean function in 3-conjunctive normal form to a rotation system of a triangulation(equivalently, to a triangulation embedded on a surface). As far as we are aware, thisseems to be an atypical reduction, whose underlying ideas (e.g. gadgets) might be of independentinterest due to their potential usefulness in the study of the computational hardness of otherrelated spin glass problems.2 Reduction idea and gadgetsThe two main results of this work follow from reductions from a well known variant of the standardnot-all-equal 3-satisfiability (abbreviated NAE-3SAT) problem which is known to be NPcompleteeven in the absence of negated variables [10], a variant we denote Positive-NAE-3SAT.For completeness sake, we recall in Figure 2 the precise definition of Positive-NAE-3SAT.Moreover, the counting version of Positive-NAE-3SAT, namely #Positive-NAE-3SAT, is #Pcomplete[2]. See Figure 3 for the precise definition of #Positive-NAE-3SAT.ProblemInputOutputPositive-NAE-3SATA Boolean formula ϕ in 3-conjunctive normal form such that each of itsclauses C 1 , . . . , C m has exactly three (all non-negated) literals.True if there is a truth assignment to ϕ such that for each clause C i not allof its variables are assigned the same truth value.Figure 2: Positive-NAE-3SAT.ProblemInputOutput#Positive-NAE-3SATA Boolean formula ϕ in 3-conjunctive normal form such that each of itsclauses C 1 , . . . , C m has exactly three (all non-negated) literals.A (binary encoding) of the number of distinct truth value assignments to ϕsuch for each clause C i not all of its variables are assigned the same truthvalue.Figure 3: #Positive-NAE-3SAT.The overall strategy we will follow in proving Theorems 1 and 2 is fairly standard, i.e. wedesign gadgets where truth values of variables are set (choice gadgets) and gadgets where thetruth value of clauses are evaluated (clause gadgets). We need to “carry” truth values fromchoice gadgets to clause gadgets, and make as many copies of the truth values taken by a literalas times they appear in all clauses. To achieve this task we build so called replicator gadgets.4

However, the construction of the aforementioned gadgets is quite delicate and non-obvious. Ingeneral, the main aspects we take care of in the construction of each gadget are existence anduniqueness of satisfying spin-assignments. However, there are subtle issues that need to beproperly handled when building and piecing together the different gadgets. Below, we describein separate sections each of the gadgets we will require for the reduction and establish that theysatisfy certain properties. First, we introduce some additional terminology and conventions wewill use throughout the remaining part of this work.2.1 PreliminariesNote that given a triangulation T , a spin-assignment s to T is satisfying if and only if -s is alsoa satisfying spin-assignment to T . We shall refer to this fact as duality. We will repeatedly useit in order to reduce the number of cases that need to be analyzed in order to establish some ofthe claims we will make. If T admits exactly two satisfying spin-assignments s and -s, we saythat s (-s respectively) is unique up to duality.A 3-cycle in a triangulation will be called positive for a spin-assignment if at least two of itsvertices are assigned spin +1. Otherwise, it will be called negative. This concept will be referredto as the sign of a 3-cycle.Henceforth, if s assigns the same spin to all nodes of a subgraph H of T (respectively allelements of S ⊆ V (T )), we say that H (respectively a subset S) is monochromatic under sSimilarly, we say that an edge is monochromatic (respectively non-monochromatic) under s if sassigns the same (respectively distinct) spins to both ends of the edge. Monochromatic and nonmonochromaticfaces are defined analogously depending on whether or not its circumscribingcycle is either monochromatic or non-monochromatic. An edge e in E(T ) will be called seriousif and only if e is monochromatic under every satisfying spin-assignment to T .The gadgets we build in this work are embedded graphs in orientable closed surfaces withsome removed disks (with holes) so that each face is bounded by a 3-cycle and each hole iscircumscribed by a 3-cycle. In other words, every gadget may be obtained from a triangulationby cutting along the boundary of some of its faces (triangles). Thence, every term defined forsurface triangulations is naturally adapted to gadgets so they will be reformulated only in caseit is needed.Throughout this work, serious edges are depicted as thicker lines and surface holes aredepicted as gray areas.2.2 Choice gadgetIn this section we describe a gadget (a triangulation of a surface) that we will associate to Booleanvariables in such a way that satisfying spin-assignments can be unambiguously interpreted astruth assignments to the Boolean variables. A choice gadget is a triangulation as depictedin Figure 4 embedded in a toroidal surface with one hole. The cycle with node set {u, v, w}circumscribing the removed triangle of the choice gadget L, henceforth denoted by C L , will be5

eferred to as the variable cycle of L and the edge uw will be called the fundamental edge of L(see Figure 4).Our reduction will associate to each variable x i a choice gadget L i . A satisfying spinassignmentwill be interpreted as setting x i to True if C Li ends up being monochromatic, andFalse otherwise. The key functionality that we will show a choice gadget provides is that ithas a unique up to duality satisfying spin-assignment where the variable cycle is monochromatic(respectively, non-monochromatic). Furthermore, choice gadgets will also be used as auxiliarybuilding blocks in the construction of another type of gadget we will soon encounter.vw w ′vvw w ′ vC Luuu ′ u ′Figure 4: Choice gadget (region in gray depicts a surface hole).The following result encapsulates the most relevant properties of choice gadgets.Proposition 3 Let L be a choice gadget. The fundamental edge of the choice gadget is serious.Moreover, there exists a unique up to duality feasible spin-assignment to L where the variablecycle C L of L is monochromatic (respectively, non-monochromatic).Proof: To prove the first statement, by duality, it suffices to show that there is no feasiblespin-assignment extension to L when node v (node labels as in Figure 4) is assigned spin +1and the fundamental edge uw is assigned spins +− or −+. In Figure 5(a) we work out thecase where uw is assigned spins +−; a subindex i accompanying a + or − sign indicates thatthe spin is forced by the spin-assignments with smaller indices in order for the assignment tobe satisfiable — if spins assigned to the vertices of a triangle are forced to be all of the samesign, then no satisfying assignment can exist under the given initial conditions. The case whenuw is assigned spins −+ is dealt with in the same way and worked out in Figure 5(b). Thisestablishes that the fundamental edge of L is serious.We now establish the claimed existence and uniqueness. Since the fundamental edge of L isserious, if s is a satisfying spin-assignment to L, then L’s fundamental edge is monochromaticunder s. Therefore, again by duality, it is enough to prove that in the following two cases thereexist exactly one feasible spin-assignment extension: (a) when the variable cycle C L (i.e. uwv)of L is assigned spin + + + (the monochromatic case), and (b) when it is assigned + + − (the6

+ 0− 0 − 1+ 0+2+ 0 − 0 − 1+ 0+ 0+ 2+ 0+ 0− 3 − 3 + 2→←+ 0 + 0− 0+ 2 →← − 0− 1+ 2 + 2+ 0+ 0 − 3+ 0− 3− 3C LC L(a) Assignment forced by fixing uw to+−. Forced monochromatic triangularfaces are labeled by →←.(b) Assignment forced by fixing uwto −+. Forced monochromatic triangularfaces are labeled by →←.Figure 5:non-monochromatic case).nodes of its variable cycle are assigned +1 is exhibited.In Figure 6(a), the unique satisfying spin-assignment to L when+ 0+ 0+ 0+ 0 − 1C L+ 0− 1+ 0 + 0 − 1 + 0− 3− 1 + 2− 3(a) Unique satisfying spinassignmentforced by thespin-assignment + + + to thevariable cycle of L.− 0+ 0 + 1− 0C L+ 0− 1 + 0− 0 + 1 − 2 − 0− 0 + 0 + 1 − 0(b) Unique satisfying spinassignmentto L, up to duality,when L’s variable cycle isnon-monochromatic.Figure 6:− 0+ 0− 0+ 0 − 1C L→←+ 0− 1− 0 + 0 − 1− 0→←+ 0− 1 − 1 + 0(c) Spin assignments forced bythe assignment of spin −1 tou ′ (forced monochromatic facesare labeled by →←).For the non-monochromatic case, by duality and since L’s fundamental edge is serious, it sufficesto consider the situation where L’s fundamental edge is assigned spins ++. Two subcases arise,depending on whether or not the spin +1 is assigned to node u ′ (node labels as in Figure 4) —each subcase is worked out separately in Figures 6(b) and 6(c).2.3 Replicator gadgetA variable may appear several times in different clauses (or even multiple times in the sameclause) of a Boolean formula in three conjunctive normal form. Given that the truth value avariable, say x i , will be unambiguously set by the values taken by a satisfying spin-assignment at7

the associated choice gadget, say L i , we require a way of “replicating” the encoding of the truthvalue of x i as many times as x i appears in the collection of formula clauses. To achieve this goal,to every choice gadget we will connect a special gadget, namely a k-replicator gadget. When thevalue k is clear from context or is not relevant, we will just write replicator gadget instead ofk-replicator gadget. For each k > 0, a k-replicator gadget will be a triangulation embedded in asurface of large genus (depending on k) with 2 k + 1 holes. The purpose of a k-replicator gadgetis to generate 2 k copies of the truth value encoded by a satisfying spin-assignment to the choicegadget to which the replicator gadget is connected.To construct a k-replicator gadget we will glue together (in a particular way soon to bediscussed) 2 k −1 so called block-replicator gadgets. A block-replicator gadget is a triangulation Rof an orientable closed surface of genus 4 with three holes. A block-replicator gadget is built bygluing together three choice gadgets and the surface triangulation of the torus with six removedtriangles depicted in Figure 7. 1 Specifically, the construction takes three choice gadgets, say ˆL,¯L and ˜L as depicted in Figure 4, and identifies the variable cycle CˆL(respectively, C¯L and C˜L)of the choice gadget ˆL (respectively, ¯L and ˜L) with the cycle Ĉ = u′ xu (respectively, ¯C = xywand ˜C = zw ′ v) of the block-replicator gadget R as depicted in Figure 7. The identification isdone in such a way that edge u ′ x (respectively, xy and zw ′ ) of the cycle Ĉ (respectively, ¯C and˜C) coincides with the fundamental edge of the choice gadget ˆL (respectively, ¯L and ˜L). Clearly,vuv ′vwww ′¯Cx yu ′ Ĉz ˜Cw ′vuv ′vFigure 7: Block-replicator gadget.under this construction, each block-replicator gadget has exactly three holes, because after gluingthe choice gadgets to the surface triangulation depicted in Figure 7, the holes circumscribed bythe cycles Ĉ, ¯C and ˜C disappear. The length 3 cycle uvw circumscribing one of R’s hole is referredto as the incoming cycle (node labels as in Figure 7). The length 3 cycles circumscribing theother two holes of R will be called the outgoing cycles. Moreover, edges vw, v ′ u ′ and xy will bereferred to as fundamental edges of R (see again Figure 7 for node labeling scheme).1 Here, choice gadgets are used as auxiliary gadgets. This auxiliary gadgets will not be associated to Booleanvariables. The reason why we rely on this auxiliary choice gadgets is solely because of one of the properties wehave shown they exhibit. Specifically, the fact that fundamental edges of choice gadgets are serious.8

The attentive reader might wonder whether the described block-replicator gadget is indeeda surface triangulation. Specifically, whether indeed every “surface” point has an open neighborhoodhomeomorphic to some open subset of the Euclidean plane. This is indeed the case.Moreover, a block-replicator gadget has the following key property, henceforth referred to as intersectionproperty: the incoming and outgoing cycles of a block-replicator gadget do not sharevertices. The intersection property implies that the surface on which the block-replicator gadgetis embedded can be smoothly deformed into the one depicted in Figure 8.Outgoing cyclesIncoming cycleFigure 8: Block-replicator gadget sketch.The key purpose of each block-replicator gadget is to enforce that if the incoming cycle ismonochromatic (respectively, non-monochromatic), then both outgoing cycles will be monochromatic(respectively, non-monochromatic). Moreover, we will see that the block-replicator gadgetinverts the sign of the incoming cycle, namely if the incoming cycle is positive (respectively, negative)the outgoing cycles are negative (respectively, positive). Actually, much more is true.Formally, we have the following results concerning block-replicator gadgets.Proposition 4 Let R be a block replicator gadget. Fundamental edges of R are serious. Inparticular, in every satisfying spin-assignment to R both outgoing cycles have the same sign andopposite to the sign of the incoming cycle.Proof: Given that xy is the fundamental edge of the variable cycle C¯L = xyw of the choicegadget ¯L, by Proposition 3, we have that xy is serious. To prove that vw is serious, by duality,it suffices to show that when vw is assigned +− and u is assigned + and −, there is no feasiblespin-assignment extension to R. These two situations are worked out in Figure 9.We proceed as above to prove now that u ′ v ′ is serious. In Figure 10, we show that in the caseswhere u ′ v ′ is assigned −+ and w ′ is assigned + and − there is no feasible spin-assignmentextension to R.Proposition 5 Let R be a block-replicator gadget. Then, there exists a unique up to dualitysatisfying spin-assignments to R for which the incoming cycle is monochromatic (respectively,non-monochromatic). Moreover, if s a satisfying spin-assignment to R, one of the two followingstatements hold:(i). the incoming and outgoing cycles are all monochromatic with the incoming cycle positive(respectively, negative) and both outgoing cycles negative (respectively, positive), or9

+ 0− 0 − 0+ 0+ 0˜C+ 0¯C− 1−1− 1Ĉ+ 0 + 0→←− 0 + 0− 2− 0+ 0− 0 − 0+ 0¯C+ 1+ 1− 2− 2Ĉ˜C+ 1→←+ 0(a)(b)Figure 9: In (a), satisfying spin-assignment forced by fixing the outgoing cycle uvw ′ to − + +and in (b) to − + −. Serious edges are shown as thick lines. Note that fixing the spin of an endof a serious edge immediately forces the spin of its other end.(ii). the incoming and outgoing cycles are all non-monochromatic with the incoming cycle positive(respectively, negative) and both outgoing cycles negative (respectively, positive).Proof: To prove existence and uniqueness of the satisfying spin-assignment when the incomingcycle is monochromatic, by duality, it is enough to prove that if all nodes in the incoming cycleuvw are assigned +1, then there exists a unique feasible spin-assignment extension to R (nodelabels as in Figure 7). This situation is worked out in Figure 11(a). On the other hand, byProposition 4, if s is a satisfying spin-assignment to R, the edge vw belonging to the incomingcycle is monochromatic (because it is serious). Thus, by duality, we can assume that s assigns tothe incoming cycle uvw spins + + + or − + +. Therefore, to establish existence and uniquenessof the satisfying spin-assignment when the incoming cycle is non-monochromatic, it will sufficeto show that there exists unique satisfying spin-assignment extension to R when the incomingcycle uvw is assigned spin − + +. This case is studied in Figure 11(b). We need to check thateach of the spin-assignments depicted in Figures 11(a) and 11(b) have a unique extension to theblock-replicator gadget, even when the auxiliary choice gadgets are glued to the block-replicatorgadget via proper identification of Ĉ, ¯C, ˜C (see labels in Figure 8) and the variable cycles ofthe auxiliary choice gadgets. Proposition 3 and the fact that the spin-assignments depicted inFigures 11(a) and 11(b) completely determine the spins of the nodes of Ĉ, ¯C, ˜C imply that thespin-assignment extensions to the whole block-replicator gadget are indeed feasible and unique.The remaining part of the claimed result can be ascertained by inspecting in Figures 11(a)and 11(b) the satisfying spin-assignments forced by the (non) monochromaticity of the incomingcycles.We are now ready to describe the construction of a k-replicator gadget. Take 2 k − 1 blockreplicatorgadgets R 1 , R 2 , R 3 , . . . , R 2 k −1. Identify the outgoing cycles of R 1 with the incoming10

+ 0− 1− 1 + 0− 1− 1 − 1 + 0− 1→←˜C¯C− 0 − 0+ 0 − 0+ 0+ 1 + 1Ĉ+ 1 + 1 + 0 + 1→←+ 1 + 1¯C− 0 − 0− 0 − 0 − 0Ĉ˜C− 0+ 1+ 0 + 1+ 1(b)(a)Figure 10: In (a), satisfying spin-assignment forced by fixing the outgoing cycle u ′ v ′ w ′ to − + +and in (b) to − + −. Serious edges are shown as thick lines. Note that fixing the spin of an endof a serious edge immediately forces the spin of its other end.cycles of the block replicator gadgets R 2 and R 3 so that the fundamental edges of R 1 thatbelong to the outgoing cycles and the fundamental edges of R 2 and R 3 that belong to theincoming cycles coincide. Continue in this way piecing together new block replicator gadgets andidentifying fundamental edges, and construct a “rooted binary tree of depth k” type structureof block-replicator gadgets. Let R 1 denote the block-replicator gadget at the “root” of the tree,and let R 2 k−1, R 2 k−1 +1, . . . , R 2 k −1 denote the block-replicator gadgets at the “leaves” of the tree.The incoming cycle of R 1 will be referred to as the starting cycle of the k-replicator gadget R kand the outgoing cycles of the block replicator gadgets R 2 k−1, R 2 k−1 +1, . . . , R 2 k −1 will be calledend cycles of R k . Moreover, fundamental edges of the block-replicator gadget belonging to theincoming cycle of R 1 and to the outgoing cycles of R 2 k−1, R 2 k−1 +1, . . . , R 2 k −1 will be referred toas fundamental edges of the k-replicator gadget.Note that each k-replicator gadget is a triangulation of an orientable closed surface of genus4 · (2 k − 1) with 2 k + 1 holes. Furthermore, the intersection property of the block-replicatorgadgets is trivially transferred to k-replicator gadgets; namely, the starting and end cycles of ak-replicator gadget do not share vertices.In our reduction, the starting cycle of each k-replicator gadget R k will be identified with thevariable cycle of a choice gadget, say L. By Proposition 5, this guarantees that the end cycles ofR k will be monochromatic if and only if the variable cycle of L is monochromatic. It is somewhatunfortunate that the block-replicator gadgets generate, at its outgoing cycles, encodings ofopposite signs as the one of its incoming cycle. By taking k even, we can guarantee that eachof the end cycles of R k will have the same chromaticity (monochromatic or nonmonochromatic)and sign as the variable cycle of the choice gadget L. The following result captures all relevantproperties we will need that are satisfied by replicator gadgets. The reader can easily checkthat the claimed properties are immediately inherited from those satisfied by block-replicator11

+ 0− 1+ 0+ 0 − 1+ 0− 1+ 0Ĉ+ 0+ 0− 1+ 0¯C− 1− 1− 1− 1 ˜C(a) Case where the incoming cycle is positiveand monochromatic. Note that theoutgoing cycles are forced to be negativeand monochromatic.− 1+ 0+ 0− 0¯C− 1 − 1+ 2 − 1 + 2+ 0− 0 − 1+ 0+ 0 + 0Ĉ+ 2 ˜C(b) Case where the incoming cycle is positiveand non-monochromatic. Note thatoutgoing cycles are forced to be negativeand non-monochromatic.Figure 11:gadgets.Corollary 6 Let k be a positive integer and let R k be a k-replicator gadget.statements hold:The following(i). Fundamental edges of R k are serious.(ii). For any satisfying spin-assignment to R k , the starting cycle and the end cycles have thesame sign.(iii). For any satisfying spin-assignment to R k , the starting cycle and the end cycles are alleither monochromatic or non-monochromatic.(iv). There is a unique up duality satisfying spin-assignment to R k so that the starting cycleand the end cycles are all monochromatic (respectively, non-monochromatic).As we have already mentioned, the starting cycle of a replicator gadget, say R, will beidentified with a variable cycle of a choice gadget, say L. Assuming that L is in turn associatedto a formula variable, say x, it follows that in any satisfying spin-assignment all end cycles of Rencode the same truth value of x encoded by the variable cycle of L. Eventually, some endcycles of R will be identified with cycles of the (next to be described) clause gadgets associatedto formula clauses where x appears. If the total number of appearances of x in formula clausesis t, then R will be a k-replicator gadget where k is the smallest positive even integer greateror equal than log 2 t. Thus, after identifying end cycles of R with cycles in clause gadgets, wemight end up with non-identified end cycles (a situation that occurs whenever log 2 t is not apositive even integer). The holes circumscribed by such end cycles will need to be “capped” inorder so at the end of our reduction we do indeed generate a surface triangulation. Moreover,12

holes will need to be “capped” in such a way that the properties satisfied by replicator gadgetsare preserved. To achieve this goal, when necessary, we will identify an end cycle with the outercycle of a cap gadget as depicted in Figure 12.Figure 12: A cap.The following statement is trivial.Proposition 7 For any spin-assignment to the outer cycle of a cap gadget, there exists a uniquesatisfying spin-assignment extension to the whole cap.2.4 Clause gadgetA clause gadget is a toroidal triangulation with three holes as depicted in Figure 13. The cyclescircumscribing the holes of the clause gadget will be called literal cycles. Moreover, edges uw,v ′ w and v ′ u will be referred to as fundamental edges of the clause gadget (depicted as thickerlines in Figure 13).uv v ′ uwwzu ′w ′zu v v ′ uFigure 13: Clause gadget.As already mentioned in the preceding section, we will eventually identify end cycles ofreplicator gadgets with literal cycles in such a way that fundamental edges of the replicator andclause gadgets coincide. In our reduction, replicator gadgets will “carry” from choice gadgetstowards clause gadgets the encodings of the truth values of formula variables. The clause gadgetis built in such a way as to allow a unique up to duality satisfying spin-assignment extension ifand only if not all the truth value encodings “arriving” to the clause gadget represent the sametruth value.13

Unfortunately, fundamental edges of clause gadgets are not serious. However, once everyliteral cycle of a clause gadget is identified with an end cycle of a replicator gadget, fundamentaledges of the clause gadget will become serious in the triangulation thus formed — since fundamentaledges of the replicator gadgets are serious, and because fundamental edges of the clauseand replicator gadgets will be identified. This explains why when stating the following claimswe assume seriousness of fundamental edges of the clause gadgets. The functionality providedby a clause gadget is summarized by the next results, the first of which is obvious.Proposition 8 Let C be a clause gadget. Assume that fundamental edges of C are serious.Then, for any satisfying spin-assignment, all literal cycles of C have the same sign.Proposition 9 Let C be a clause gadget. Assume that fundamental edges of C are serious. Inthe following cases there is no satisfying spin-assignment extension to C:(i). when all literal cycles of C are monochromatic, and(ii). when all literal cycles of C are non-monochromatic.Proof: To prove the first claim, by duality and Proposition 8, it suffices to show that if allnodes in each literal cycle are assigned +1, then there is no feasible spin-assignment extensionto C. This case is worked out in Figure 14(a).+ 0+ 0 + 0+ 0+ 0+ 0+ 0 + 0→←+ 0 + 0 + 0+ 0(a) Case where all literal cycles aremonochromatic.+ 0− 0 + 0+ 0 − 0 + 0 + 0+ 0+ 0+ 0− 0 − 0→←(b) Case where all literal cycles are nonmonochromatic.Figure 14:To establish the second claim, it suffices to show that the same conclusion holds when the literalcycles uwv, v ′ wu ′ and v ′ uw ′ are all assigned + + − (node labels as in Figure 13). This case isworked out in Figure 14(b).Proposition 10 Let C be a clause gadget. Assume that the fundamental edges of C are serious.If exactly one literal cycle of C is monochromatic (respectively, non-monochromatic) there is aunique up to duality satisfying spin-assignment extension to C.14

Proof: By duality and Proposition 8 the monochromatic case holds if we show that in each ofthe following situations there is exactly one feasible spin-assignment extension to C (node labelsas in Figure 13): (a) when the literal cycle uwv is assigned + + + and spins + + − are assignedto cycles v ′ wu ′ and v ′ uw ′ , (b) when the literal cycle v ′ wu ′ is assigned + + + and spins + + −are assigned to cycles uwv and v ′ uw ′ , and (c) when the literal cycle v ′ uw ′ is assigned + + +and spins + + − are assigned to cycles uwv and v ′ wu ′ . Each case is worked out separately inFigures 15(a), 15(b), and 15(c).+ 0 + 0 + 0+ 0+ 0+ 0+ 0 − 0 + 0+ 0+ 0+ 0+ 0 − 0 + 0+ 0+ 0+ 0+ 0 + 0 + 0 + 0(a)+ 0 − 0 + 0 + 0(b)+ 0 − 0 + 0 + 0(c)Figure 15: Unique forced satisfying spin-assignments to a clause gadget when exactly one literalcycle is monochromatic. Each shown spin-assignment encodes a truth value assignment to thevariables of a clause where not all truth values are equal.+ 0+ 0 − 0 + 0+ 0− 1 + 0 − 1+ 0 + 0+ 0+ 0 + 0 + 0+ 0− 1 − 0 − 1+ 0 + 0− 0+ 0 + 0 + 0+ 0− 1 + 0 − 1+ 0 + 0+ 0 − 0 + 0 + 0(a)+ 0 + 0 + 0 + 0(b)+ 0 + 0 + 0 + 0(c)Figure 16: Unique forced satisfying spin-assignments to a clause gadget when exactly one literalcycle is non-monochromatic. Each shown spin-assignment encodes a truth value assignment tothe variables of a clause where not all truth values are equal.For the non-monochromatic case, we proceed in the same way. Again, By duality and Proposition8, it suffices to examine the following situations (node labels as in Figure 13): (a) whenthe literal cycle uwv is assigned + + − and spins + + + are assigned to cycles v ′ wu ′ and v ′ uw ′ ,(b) when the literal cycle v ′ wu ′ is assigned + + − and spins + + + are assigned to cycles uwvand v ′ uw ′ , and (c) when the literal cycle v ′ uw ′ is assigned + + − and spins + + + are assignedto cycles uwv and v ′ wu ′ . Each case is worked out in Figures 16(a), 16(b), and 16(c).15

3 The reductionWe now describe the reduction from Positive-NAE-3SAT to SatAssign.Let ϕ be a Booleanformula in conjunctive normal form, where each clause has exactly three (non-negated) literals.Let x 1 , . . . , x n be the variables and Cl 1 , . . . , Cl m be the clauses of ϕ. Let t i denote the number oftimes variable x i appears in the collection of clauses (multiple occurrences are counted multipletimes). Definek i ={2, if t i = 1,2⌈(1/2) log 2 (t i )⌉, if t i > 1.To each variable x i we associate a choice gadget L i . To each clause Cl j we associate a clausegadget C j . For i = 1, . . . , n, we identify the starting cycle of a k i -replicator gadget R k iwith thevariable cycle of the choice gadget L i so that the fundamental edge of L i that belongs to L i ’svariable cycle and the fundamental edge of R k ithat belongs to its starting cycle coincide. Notethat the number of end cycles of R k iis at least t i . For i = 1, . . . , n, identify t i end cycles of R k iwith literal cycles of the clause gadgets C 1 , . . . , C m where variable x i appears in such a way thatthe fundamental edges of the end cycles of R k iand the fundamental edges of the literal cyclesof the clause gadgets coincide. Identify the remaining 2 k i− t i end cycles of R k i(if any) with theouter cycle of a cap gadget. Denote by T ϕ the surface triangulation thus obtained.We first make a simple observation.Lemma 11 The fundamental edges of each clause gadget of T ϕ are serious.Proof:Just observe that by construction of T ϕ , fundamental edges of clause gadgets are identifiedwith fundamental edges of replicator gadgets which are known to be serious, as establishedby Corollary 6 (i).Note that, by construction, the surface on which T ϕ is embedded is an orientable closed surfaceof genus at least m + n + 4 ∑ ni=1 (2k i− 1) (1 due to each choice gadget L 1 , . . . , L n , another1 due to each clause gadget C 1 , . . . , C m , and 4(2 k i− 1) due to each replicator gadget R k i,i = 1, . . . , n).Clearly, since each 3-cycle circumscribing gadget holes were identified with a3-cycle circumscribing another gadget hole, the surface on which T ϕ is embedded does not haveholes, i.e. its a closed surface. Moreover, since the construction process dos not create additionalfaces, each face of T ϕ is a face of some gadget. Thence, each face is bounded by a 3-cycle, so T ϕis a triangulation. Finally, note that since each of the gadgets used in the construction of T ϕ isembeddable in an orientable surface with holes, the resulting surface on which T ϕ is embeddedis also an orientable surface. Summarizing, T ϕ is a triangulation of an orientable closed surface.We say that ϕ is connected if for every non-trivial partition {S, ¯S} of the clauses of ϕ(i.e. S, ¯S ≠ ∅, S ∩ ¯S = ∅, and S ∪ ¯S equals the set of clauses of ϕ) there is at least one variablethat appears in one of the clauses in S and in one of the clauses of ¯S.We now make a couple of useful observations.Lemma 12 Let ϕ be an instance of Positive-NAE-3SAT. If ϕ is connected, then the surface inwhich T ϕ is embedded is also connected.16

Proof: Assume T ϕ is embedded in a non-connected surface. Consider the set S of clauseswhose associated clause gadgets are embedded in one of the connected surface components, sayS. Let ¯S be the collection of clauses not in S. Note that {S, ¯S} is non-trivial. Moreover, theset of variables that appear in clauses in S (respectively, in ¯S) correspond to those variablesassociated to choice gadgets embedded in S (respectively, not in S). Both of theses collectionof variables must be disjoint, contradicting the fact that ϕ is connected.Lemma 13 Consider an instance ϕ of Positive-NAE-3SAT. Let Cl be a clause of ϕ and x 1 ,x 2 , and x 3 the (not necessarily distinct) variables appearing in Cl. Let L 1 , L 2 , and L 3 be the(not necessarily distinct) choice gadgets of T ϕ associated to x 1 , x 2 and x 3 , respectively. Let Cbe the clause gadget of T ϕ associated to Cl. Then, for every satisfying spin-assignment s to T ϕ ,the literal cycles of C and the variable cycles of L 1 , L 2 , and L 3 have the same sign. Moreover,if ϕ is connected, then all literal cycles and variable cycles of T ϕ have the same sign.Proof: By Proposition 8 and Lemma 11, under any satisfying spin-assignment to T ϕ all literalcycles of C have the same sign. Without loss of generality, we can assume that all literal cyclesof C are positive. Hence, the end cycles of the replicator gadgets, say R k 1, R k 2, and R k 3, whichare identified with the literal cycles of C, must all be positive. Since k 1 , k 2 and k 3 are even, byCorollary 6 (ii), the starting cycles of R k 1, R k 2and R k 3are positive. Given that the variablecycles of L 1 , L 2 , and L 3 are identified in T ϕ with the starting cycles of R k 1, R k 2and R k 3, thefirst stated claim follows.The last statement follows trivially from Lemma 12.Theorem 14 Let ϕ be an instance of Positive-NAE-3SAT. If ϕ is connected, then(i). For each truth value assignment that witnesses membership of ϕ in Positive-NAE-3SATthere is a unique up to duality satisfying spin-assignment to T ϕ .(ii). For every pair of duality related satisfying spin-assignment to T ϕ there is exactly one truthvalue assignment that witnesses membership of ϕ in Positive-NAE-3SAT.Proof: Let a 1 , a 2 , . . . , a n be a truth value assignment to the variables x 1 , . . . , x n that is awitness of membership of ϕ in Positive-NAE-3SAT. We claim that there is a unique up toduality satisfying spin-assignment to T ϕ . As usual, let Cl 1 , . . . , Cl m be the clauses of ϕ, letC 1 , . . . , C m denote the associated clause gadgets, and let L 1 , . . . , L n and R k 1, . . . , R kn be thechoice and replicator gadgets associated to variables x 1 , . . . , x n .If a i is True, fix the spins of all nodes of the variable cycle of L i in such a way that the cycleends up being positive and monochromatic (observe that this can be done in a unique way).Otherwise, a i is False, fix the spins of all nodes of the variable cycle of L i in such a waythat the cycle ends up being positive and non-monochromatic (observe that since fundamentaledges of choice gadgets are serious, this can again be done in a unique way). Extend the so far17

partially defined satisfying spin-assignment to the union of choice gadgets (by Proposition 3,such a satisfying assignment extension exists and is unique).Similarly, fix the spins of the nodes of the literal cycles of each clause gadget C j according tothe truth value taken by the associated Boolean formula variable, i.e. if the variable’s value isTrue, make the literal cycle positive and monochromatic, and positive and non-monochromaticotherwise (observe again that such spin-assignments can be done in a unique way). Extend oncemore the so far partially defined spin-assignment to the union of clause gadgets. We claim thatsuch an extension exists and is unique. Indeed, given that a 1 , . . . , a n is a witness of membershipof ϕ in Positive-NAE-3SAT, the variables in each clause Cl j do not take the same truth valueunder the assignment a 1 , . . . , a n . Thus, the aforementioned spin-assignment to the literal cyclesof C j is such that not all literal cycles end up having the same chromaticity. By Proposition 10,each clause gadget has a unique satisfying spin assignment extension, thus establishing our claim.Recall that variable cycles of choice gadgets (respectively, literal cycles of clause gadgets) areidentified with starting cycles (respectively, end cycles) of replicator gadgets. Hence, for alli = 1, . . . , n, the partial spin-assignment thus far defined makes the start and end cycle of thereplicator gadget R k ipositive and monochromatic (respectively, non-monochromatic) if and onlyif a i is True (respectively, False). Given that k i is even and positive for all i = 1, . . . , n, byCorollary 6 (iv) and Proposition 7, there is a unique extension of the previously defined partialspin-assignment to all nodes of replicator gadgets so the resulting spin-assignment is a satisfyingspin-assignment for T ϕ .We now prove the second part of the claimed result. Assume there is a satisfying spinassignments to T ϕ . By Lemma 13 and since ϕ is connected, all literal and variable cyclesof T ϕ have the same sign, say positive. For i = 1, . . . , n, let a i be True if the variablecycle of L i is positive and monochromatic, and False if the variable cycle of L i is positiveand non-monochromatic. We claim that a 1 , . . . , a n is a witness of membership of ϕ inPositive-NAE-3SAT. Indeed, assume x j1 , x j2 , and x j3 are the (not necessarily distinct) variablesappearing in clause Cl j . Let s ∈ {1, 2, 3}. Since the start cycle of the replicator gadgetsR k jsis identified with the variable cycle of the choice gadget Ljs , then by Corollary 6 the endcycles of R k js must be monochromatic if and only if ajs is True. Since the non-capped endcycles of replicator gadgets are identified with the literal cycles of clause gadgets, we have thatthe literal cycle of the clause gadget C j associated to the variable x js is monochromatic if andonly if a js is True. Moreover, all literal cycles of clause gadgets are positive. Since s is asatisfying spin-assignment, by Proposition 9 and Proposition 10, the literal cycles of C j can notall be either monochromatic or non-monochromatic. This implies that a j1 , a j2 , and a j3 are notall equal, as we wanted to establish.Corollary 15 Let ϕ be an instance of Positive-NAE-3SAT. Then, ϕ is satisfiable if and onlyif there is a satisfying spin-assignment to T ϕ .Proof:If ϕ is connected, the result is immediate from Theorem 14. Assume ϕ is not con-18

nected. Then, there are ϕ 1 , . . . , ϕ c instances of Positive-NAE-3SAT such that each ϕ i is connectedand ϕ = ∧ c i=1 ϕ i. Moreover, ϕ belongs to Positive-NAE-3SAT if and only if ϕ 1 , . . . , ϕ cbelong to Positive-NAE-3SAT. By Theorem 14, this is equivalent to saying that ϕ belongs toPositive-NAE-3SAT if and only the union of the surface triangulations T ϕ1 , . . . , T ϕc , i.e. T ϕ ,admits a satisfying spin assignment.The next result allows us to handle, in our reduction, instances of Positive-NAE-3SAT whichare not connected.Lemma 16 Let ϕ be an instance of Positive-NAE-3SAT. Then, there exists a log-space (hence,polynomial time) computable instance ϕ ′ of Positive-NAE-3SAT such that ϕ ′ is connected andthe following equality holds:|{⃗a = (a 1 , . . . , a n ) : ⃗a is a witness of membership of ϕ in Positive-NAE-3SAT}| =1∣∣{ a2⃗′ = (a ′ 1, . . . , a ′ n ′) : a ⃗ ′ is a witness of membership of ϕ ′ in Positive-NAE-3SAT} ∣ .Proof:Assume x 1 , . . . , x n are the variables and C 1 , . . . , C m the clauses of ϕ. Consider twoadditional Boolean variables y and z and define n additional clauses C ′ 1 , . . . , C′ n such that C ′ i =x i ∧y∧z. Let ϕ ′ be the conjunction of C 1 , . . . , C n , C ′ 1 , . . . , C′ n. Clearly, ϕ ′ is a connected instanceof Positive-NAE-3SAT which is log-space computable given ϕ. Note that a membership witnessa 1 , . . . , a s of an instance of Positive-NAE-3SAT can not be such that all a i ’s are equal. Thisimmediately implies that x 1 , . . . , x n is a witness of membership of ϕ in Positive-NAE-3SAT ifand only if x 1 , . . . , x n , y = 0, z = 1 and x 1 , . . . , x n , y = 1, z = 0 are witnesses of membership ofϕ ′ in Positive-NAE-3SAT.Corollary 17 Let ϕ be an instance of Positive-NAE-3SAT.Then, there exists a log-space(hence, polynomial time) computable instance ϕ ′ of Positive-NAE-3SAT such that∣ {s : s is a satisfying spin-assignments to Tϕ ′} ∣ ∣ =4 · |{⃗a = (a 1 , . . . , a n ) : ⃗a is a witness of membership of ϕ in Positive-NAE-3SAT}| .To conclude, note that given an instance ϕ of Positive-NAE-3SAT with n variables and mclauses each of the necessary gadgets can be constructed in O(log |V (T ′ ϕ)|) space, i.e. in log-spacein the size of the encoding of ϕ. The number of choice, block-replicator, and clause gadgets thatneed to be built are n + 2, ∑ ni=1 (2k i− 1) = O(m), and m + n, respectively. Hence, an encodingof a rotation system for T ϕ ′ can be computed in log-space, thence in polynomial time. Then,Corollary 15 and Corollary 17 imply Theorem 1 and Theorem 2, respectively.AcknowledgementsThe authors thank Martin Loebl for his enthusiasm, motivation, and many helpful discussions.19

References[1] R. J. Baxter. Exactly Solved Models in Statistical Mechanics. Dover Publications, 2008.[2] N. Creignou and M. Hermann. Complexity of generalized satisfiability counting problems.Information and Computation, 125:1–12, February 1996.[3] N. Creignou, S. Khanna, and M. Sudan. Complexity Classifications of Boolean ConstraintSatisfaction Problems. SIAM Monographs in Discrete Mathematics and Applications.SIAM, 2001.[4] L. Esperet, F. Kardos, A. King, D. Král, and S. Norine. Exponentially many perfectmatchings in cubic graphs. Advances in Mathematics, 227(4):1646–1664, 2011.[5] A. Galluccio and M. Loebl. On the theory of pfaffian orientations. i. perfect matchings andpermanents. The Electronic Journal of Combinatorics, 6, 1998.[6] A. Jiménez, M. Kiwi, and M. Loebl. Satisfying states of triangulations of a convex n-gon.The Electronic Journal of Combinatorics, 17(1), 2010.[7] M. Loebl. Discrete Mathematics in Statistical Physics. Advanced Lectures in Mathematics.Vieweg and Teubner, 2009.[8] M. Loebl and J. Vondrák. Towards a theory of frustrated degeneracy. Discrete Mathematics,271(1-3):179–193, 2003.[9] B. Mohar and C. Thomassen. Graphs on Surfaces. The Johns Hopkins University Press,Baltimore and London, 2001.[10] T.J. Schaefer. The complexity of satisfiability problem. Proceedings of the 10th AnnualACM Symposium on Theory of Computing, pages 216–226, 1978.[11] J. P. Sethna. Statistical Mechanics: Entropy, Order Parameters and Complexity (OxfordMaster Series in Physics). Oxford University Press, USA, 2006.20

Apéndice DEstado fundamental no degenerado enel modelo de Ising antiferromagnéticoen triangulaciones83

Non-degenerated groundstates in the antiferromagnetic Isingmodel on triangulationsAndrea Jiménez ∗Available online at http://arxiv.org/abs/1110.2507, October 2011AbstractA triangulation is an embedding of a graph into a closed Riemann surface so that eachface boundary is a 3-cycle of the graph. In this work, groundstate degeneracy in the antiferromagneticIsing model on triangulations is studied. We show that for every fixed closedRiemann surface Ω, there are vertex-increasing sequences of triangulations of Ω with a nondegeneratedgroundstate. In particular, we exhibit geometrically frustrated systems with anon-degenerated groundstate.1 IntroductionThe Ising model is one of the most studied models of interacting particles in statistical physics.This model has been strongly linked to the study of discrete mathematics [8, 4]. In this sense,tools and techniques developed in the discrete setting have shown to be very useful to deal withthe solution of problems related to the Ising model and vice versa.Typically, to study the Ising model, particles are located at the vertices of a graph and thetype of interaction between them is determined by the existence and weight of edges in thegraph. In this work, we explore the Ising model where particles and their interaction describetriangulations of closed Riemann surfaces with edge-weight equal to -1.A triangulation of a closed Riemann surface Ω, or simply a triangulation, is an embeddingof a graph in Ω so that each face boundary is a 3-cycle of the graph. Throughout this work, theclosed Riemann surface Ω will be specified just in needed cases.Let’s introduce the main ingredients of the Ising model. Given a triangulation T , a stateof the Ising model on T is a function σ that assigns to each vertex of T a value from the set{+1, -1}. In other words, σ ∈ {+1, -1} |V (T )| , where V (T ) denotes the set of vertices of T . Values+1 and -1 are usually called spins. Then, we also say that a state on T is a spin-assignment onT . For each state σ, the energy or Hamiltonian of the Ising model on a triangulation T is defined∗ Departamento de Ingeniería Matemática, Universidad de Chile & Department of Applied Mathematics,Charles University. Web: www.dim.uchile.cl/∼ajimenez. Gratefully acknowledges the support of Mecesupvia UCH0607 Project, CONICYT and the partial support of the Czech Research Grant MSM 0021620838.1

y H(σ) = - ∑ uv ∈ E(T ) J uvσ u σ v , where E(T ) denotes the set of edges of T and for each uv inE(T ) the parameter J uv is called coupling constant. The main purpose of a coupling constantJ uv is to specify the type of interaction between vertices u and v. In general, this constant mayvary from positive to negative depending on the characteristics of the system to be studied. Itcan also be randomly chosen, like in the case of spin-glasses.The antiferromagnetic variant of the Ising model takes coupling constant equal to -1 forall edges of the graph. Then, given a triangulation T and a state σ, the energy of σ in theantiferromagnetic Ising model is given byH(σ) =∑σ u σ v . (1)uv ∈ E(T )Many mathematical problems naturally arise from the Ising model. One of them is the studyof states that provide the minimum possible energy for the system. Those states are usuallyknown as groundstates. Related to the study of groundstates, is the groundstate degeneracy,which by definition corresponds to the number of different groundstates that a system supports.If the groundstate degeneracy of a system is greater than two (i.e. more than one pair ofgroundstates exist), it is said that the groundstate is degenerated. Otherwise, it is called nondegenerated.The groundstate degeneracy has been vastly studied [9, 7], being of great physicalinterest, because (among others) it determines the entropy of the system, and characterizingentropy’s behaviour helps to understand physical phenomena associated to order and stabilityof the system [13].Let T be a triangulation and σ ∈ {+1, -1} |V (T )| be a state of the antiferromagnetic Isingmodel on T . It is said that uv ∈ E(T ) is frustrated by σ or that σ frustrates uv ∈ E(T ) ifσ u = σ v . Observe that each state on T frustrates at least one edge of each face boundary of thetriangulation since each face boundary is 3-cycle.This feature (every state frustrates at least one edge of each face boundary) is known as geometricalfrustration. The understanding of order and stability of geometrically frustrated systems,is one of the main questions that condensed matter physicists face to explain. It is expectedthat systems which exhibit geometrical frustration lead to highly degenerated groundstates witha non-zero entropy at zero temperature (see [10]). In other words, groundstate degeneracy in ageometrically frustrated system is typically exponentially large as a function of the number ofvertices of the underlying graph. Indeed, groundstate degeneracy of any plane triangulation isexponential in the number of vertices, since groundstate degeneracy of plane triangulations istwice the number of perfect matchings of cubic bridgeless planar graphs (see [3, 6]).Surprisingly, in this work it is shown that there are triangulations of closed Riemann surfaceswith an arbitrary number of vertices and a non-degenerated antiferromagnetic groundstate.More precisely, we establish the next result.Theorem 1 Let Ω be a fixed closed Riemann surface with positive genus (g > 0). Then, forevery n > 0 there is a triangulation T of Ω with n ≤ |V (T )| so that T has a non-degeneratedantiferromagnetic groundstate.2

In particular, when the closed Riemann surface Ω is a torus, we have the following.Theorem 2 For every n > 0 there is a toroidal triangulation T with n ≤ |V (T )| so that T hasa non-degenerated antiferromagnetic groundstate.2 Non-degenerated groundstates in triangulations2.1 PreliminariesThroughout this work, it will be needed to consider triangulations of closed Riemann surfaceswith some removed faces (with holes) so that each hole is circumscribed by a 3-cycle. Triangulationsof this type will be called punctured triangulations (see for example Figure 3(b); trianglesin grey depict holes). In general, every term defined for triangulations is naturally adapted topunctured triangulations. However, there are some facts that hold only for triangulations; theywill be properly specified. We now introduce definitions for both triangulations and puncturedtriangulations.Let T be a (punctured) triangulation. Recall that every spin-assignment on T frustrates atleast one edge of each face boundary of T . Notice that a spin-assignment is a groundstate if ithas the smallest possible number of frustrated edges (see equation 1). A spin-assignment σ onT is said to be satisfying if σ frustrates exactly one edge of each face boundary of T .Let T be a triangulation. Obviously any satisfying spin-assignment on T is a groundstate.The converse is true for plane triangulations [6]. Nevertheless, the equivalence doesnot hold in general because a satisfying spin-assignment doesn’t need to exist. However, notethat when satisfying spin-assignments exist, then every groundstate corresponds to a satisfyingspin-assignment. The situation is more complicated if T is a punctured triangulation, sincedistinct satisfying spin-assignments on T may provide distinct antiferromagnetic energy. InFigure 1 an example is shown.++ −+−++−+−++−+−− +−− +−−++−+−++−(a) H(σ a) = −5(b) H(σ b ) = −6Figure 1: Consider the depicted punctured toroidal triangulation T (triangle in grey representsa hole). In both pictures (a) and (b) a satisfying spin-assignment on T is sketched; σ a and σ brespectively. However, the energy of σ a is greater than energy of σ b .3

Observe that given a (punctured) triangulation T , a spin-assignment σ on T is satisfyingif and only if -σ is also a satisfying spin-assignment on T . We shall refer to this fact as signsymmetry. We will use it in order to reduce the number of cases that need to be analysed in theproofs. Moreover, if T admits exactly two satisfying spin-assignments σ and -σ, we say that Tadmits a unique pair of satisfying spin-assignments.If σ assigns the same spin on all vertices of a subgraph H of T (respectively all elements ofS ⊆ V (T )), we say that H (respectively a subset S) is monochromatic under σ. Similarly, we saythat an edge is monochromatic (respectively non-monochromatic) under σ if σ assigns the same(respectively distinct) spins on both ends of the edge. In other words, an edge is monochromaticunder σ if and only if it is frustrated by σ. Monochromatic and non-monochromatic faces aredefined analogously depending on whether or not its face boundary is either monochromatic ornon-monochromatic. Then, a spin-assignment σ to T is satisfying if and only if every face ofT is non-monochromatic under σ. An edge e in E(T ) will be called serious if and only if e ismonochromatic under every satisfying spin-assignment on T .In what follows, serious edges are depicted as thicker lines and holes of punctured triangulationsare depicted as grey areas.2.2 Non-degenerated groundstates in toroidal triangulationsThis section is devoted to prove Theorem 2 and to discuss some features of toroidal triangulationswith a non-degenerated groundstate. Here, we show a strategy to construct toroidaltriangulations with a non-degenerated groundstate, where a groundstate is a satisfying spinassignment.Nevertheless, we strongly believe there are many other ways to construct them andalso that the class of toroidal triangulations with a non-degenerated groundstate is not small.The proof of Theorem 2 is constructive and it is based on a simple idea. However, it is far frombeing trivial to find the concrete triangulations to make that idea work.Next, we give definitions and an overview of the proof. Let T be a toroidal triangulationthat admits exactly one pair of satisfying spin-assignments (recall that when a satisfying spinassignmentexists, satisfying spin-assignments are identical to groundstates) and σ denote asatisfying spin-assignment on T . Let ˜F be a non-empty set of faces of T . We say that the pair(T, σ) is invariant under removal of ˜F if the punctured triangulation ˜T obtained by removing allfaces contained in the set ˜F , has exactly one pair of satisfying spin-assignments. In particular, ˜σis a satisfying spin-assignment on ˜T if and only ˜σ ∈ {+σ, -σ}. If such a set of faces ˜F exists, wesay that T is a supporting triangulation, ˜F will be called a removable set of faces of T and ˜T willbe referred to as the supporting punctured triangulation associated to T and ˜F ; when T and ˜F areclear (or implicit) from the context we just say that ˜T is a supporting punctured triangulation.Clearly, ˜T is an embedding of a graph in a torus with | ˜F | triangular holes. Observe that theexistence of a removable set of faces is not trivial: one could remove from T a non-empty set offaces in such a way that the obtained triangulation has more satisfying spin-assignments than T .Let T be a supporting triangulation, ˜F be a removable set of faces of T , σ be a satisfyingspin-assignment on T and ˜T be the supporting punctured triangulation associated to T and ˜F .4

The face boundaries of the faces in ˜F (in other words, the 3-cycles circumscribing the holes of ˜T ),will be referred to as the expandable cycles of ˜T . The set of edges contained in the expandablecycles which are monochromatic under σ will be called the fundamental edges of ˜T . Notice thateach expandable cycle contains exactly one fundamental edge, since each expandable cycle isnon-monochromatic under σ. Clearly, fundamental edges of ˜T are serious.Assume that a supporting triangulation T exists. Then, we know that a supporting puncturedtriangulation ˜T associated to T admits exactly one pair of satisfying spin-assignments, sayσ, -σ and all expandable cycles of ˜T are non-monochromatic under σ with its fundamental edgesmonochromatic under σ. In our construction, each hole of ˜T (circumscribed by an expandablecycle) will be covered by a plane triangulation in such a way that the number of vertices of thetoroidal triangulation increases and the number of satisfying spin-assignments keeps constant.To achieve this task, it is needed a plane triangulation ∆ with an arbitrary number of verticesand satisfying the following condition: if {xyz} is the face boundary of the outer face of ∆, thenthere is a unique pair of satisfying spin-assignments on ∆ such that at least one edge from theset of edges {xy, yz, xz} is monochromatic; say xy. We refer to ∆ as an augmenting triangulationand we say that edge xy is a fundamental edge of ∆. An augmenting triangulation alwaysexists, since plane triangulations are duals of cubic bridgeless planar graphs and every perfectmatching of a cubic bridgeless planar graph G corresponds to the set of monochromatic edgesunder a satisfying spin-assignment on G ∗ . Then, an augmenting triangulation is the dual of acubic bridgeless planar graph with a specified edge contained in exactly one perfect matching.Next, we show that supporting triangulations exist. In Subsection 2.2.2 we will describea family of augmenting triangulations. Finally, in Subsection 2.2.3 we formalize the proof ofTheorem 2.2.2.1 Supporting triangulationsThe minimal triangulations of a surface are those that have every edge in a noncontractible3-cycle. A splitting of a vertex v replaces the vertex v by two vertices v 1 and v 2 connected bya new edge v 1 v 2 , and replaces each edge vu incident to v either by the edge v 1 u or by v 2 u. Itis well-known that every triangulation of a given surface Ω, may be generated by a sequence ofvertex-splittings from a minimal triangulation of Ω. In general, the set of minimal triangulationsis finite for every fixed surface [1, 2]. In particular, there are 21 minimal toroidal triangulations(see [11, §5.4]).It is natural and potentially useful to look for supporting triangulations in the set of minimaltoroidal triangulations. On one hand its reduced number of vertices allow to verify the requiredproperties manually, on the other hand, it strongly indicates a possible way to describe thewhole set of toroidal triangulations with a non-degenerated groundstate since every toroidaltriangulation may be generated from a minimal one.We exhibit the subset of minimal triangulations of the torus which have the property ofbeing a supporting triangulation (see Figure 2). However, some minimal triangulations of thetorus which are not supporting can become supporting after some slight modifications, namely,5

Figure 2: In the first row, minimal triangulation of the torus which are supporting. Below eachsupporting triangulation an associated supporting punctured triangulation is depicted (regionsin grey represent the holes generated by the removal of a removable set of faces of each supportingtriangulation).after edge-flippings and vertex-splittings. Nevertheless, this analysis is not the aim of this workand will be studied separately.In the first row of Figure 2, all minimal toroidal triangulations which are supporting triangulationsare depicted (in each picture opposite sides have to be identified). In the second rowof Figure 2 are depicted the supporting punctured triangulations obtained from the removal ofa set of removable faces of each supporting triangulation above. We will formally prove supportabilityonly for one of the triangulation depicted in Figure 2. The same proof ideas can beeasily applied for the remaining cases.Proposition 3 Let T be the toroidal triangulation depicted in Figure 3(a) and ˜F be the subsetof faces of T with face boundary the cycles {uvw} and {ũṽ ˜w}. Then, triangulation T is asupporting triangulation and ˜F is a removable set of faces of T .wwxvxvṽ˜wuṽ˜wuũũ(a) Toroidal triangulation.(b) Punctured toroidal triangulation.Figure 3: The supporting triangulation and punctured triangulation of Proposition 3.6

Proof: Let ˜T be the punctured triangulation obtained by removing from T all faces contained in˜F (see Figure 3(b)). To see that T is a supporting triangulation and ˜F is a removable set of facesof T , it is enough to prove that ˜T has exactly one pair of satisfying spin-assignments and thatboth expandable cycles of ˜T are non-monochromatic under a satisfying spin-assignment on ˜T .To do that, by sign symmetry it suffices to verify that exactly one of the following three initialconfigurations can be extended to a satisfying spin-assignment on ˜T (vertex labels as in Figure3(b)): [1] when the cycle {uv ˜w} is assigned spin ++-, [2] when the cycle {uv ˜w} is assignedspin +-+, [3] when the cycle {uv ˜w} is assigned spin +--. Then, it is necessary to check thatboth expandable cycles are non-monochromatic under such satisfying spin-assignment.In Figures 4(a) and 4(b) the case when uv is non-monochromatic is worked out; a subindex iaccompanying a + or - sign indicates that the spin is forced by the spin-assignments with smallerindices in order for the assignment to be satisfiable — if spins assigned on the vertices of a faceboundary are forced to be all of the same sign, then no satisfying assignment can exist underthe given initial conditions. This establishes that when the cycle {uv ˜w} is assigned +-+ or +--,there is no satisfying spin-assignment extension on ˜T .− 1 + 2+ 2+ 3− 2− 0+ 2 →←+ 1+ 0− 2− 0+ 0− 0− 2− 1+ 0+ 2− 4+ 3− 4 − 4− 4+ 1→←− 2(a) Assignment forced by fixinguv ˜w to +-+.(b) Assignment forced by fixinguv ˜w to +--.Figure 4: Case when uv is non-monochromatic. Forced monochromatic faces are labelledby →←.In Figure 5, the case when edge uv is monochromatic is studied. In this situation, two subcasesarise, depending on whether or not the spin +1 is assigned to vertex x (vertex labels as inFigure 3(b)) — each subcase is dealt in the same way as the previous situation and workedout separately in Figures 5(a) and 5(b). This shows that there exist a unique satisfying spinassignmenton ˜T .Finally, notice that both expandable cycles of ˜T are non-monochromatic under the uniquesatisfying spin-assignment on ˜T (depicted in Figure 5(b)).7

− 1 + 2+ 0 +0 →← + 0+ 3− 4+ 3+ 0− 0 − 0− 2+ 3− 1 − 1− 0 + 0− 1 + 2(a) Sub-case when + is prescribedto vertex x. Initial spinassignmentforces a monochromatictriangle, labelled by→←.+ 1− 0 + 0+ 1+ 3 − 2 −4(b) Sub-case when - is prescribedto vertex x. Initial spinassignmentcan be extendedto a unique satisfying spinassignmenton ˜T .Figure 5: The case when {uv ˜w} is prescribed spin ++-.2.2.2 Augmenting triangulationsWe already mentioned that an augmenting triangulation always exists. In this subsection wejust show an easy way to construct a vertex-increasing family of such triangulations.Let ∆ 0 = {x, y 0 , z} be a plane triangle. For i ≥ 1, let ∆ i be the plane triangulation obtainedby applying the following rule to ∆ i−1 : (1) insert a new vertex y i in the outer face of ∆ i−1 , and(2) connect the new vertex y i to each vertex in the outer face of ∆ i−1 so that the outer face ofthe new plane triangulation has face boundary {xy i z} (See Figure 6). Clearly, the number ofvertices of ∆ n is n + 3. We will see that every triangulation from the collection {∆ j } j>0 is anaugmenting triangulation with fundamental edge xy j . This type of triangulations belongs tothe set of stack triangulations (see [5]) and more families of augmenting triangulations can beeasily found in that set.xzy 0y 1y 2y 3y nFigure 6: Construction of ∆ n .Theorem 4 Let n > 0 and {xy n z} the face boundary of the outer face of ∆ n (see Figure 6).There exist a unique pair of satisfying spin-assignments on ∆ n so that edge xy n is monochro-8

matic.Proof: By sign symmetry, it suffices to prove that if spin ++- is prescribed to {xy n z}, thenthere is a unique satisfying spin-assignment extension on ∆ n . We will proceed by inductionon n. If n = 1 it is trivial to check that the result holds. Let n > 1. If we prescribe spin ++- to{xy n z}, then in order to have a satisfying spin-assignment on ∆ n the vertex y n−1 is forced tohave spin -. It implies that in any extension to a satisfying spin-assignment on ∆ n , the 3-cycle{x, y n−1 , z} has spin +--. Then, by induction hypothesis uniqueness holds.2.2.3 Proof of Theorem 2Let T be a supporting triangulation, ˜F be a set of removable faces of T and ˜T be the supportingpunctured triangulation associated to T and ˜F . Consider | ˜F | = t and let {∆ i } i∈[t] by a collectionof augmenting triangulations. Let C 1 , . . . , C t denote the expandable cycles of ˜TLet T denote the triangulation obtained by gluing together the supporting punctured triangulation˜T and the collection of augmenting triangulations {∆ i } i∈[t] in the following way: takethe collection of augmenting triangulations {∆ i } i∈[t] and for each i ∈ [t], identify the boundaryface of the outer face of ∆ i with the expandable cycle C i of ˜T in such a way that the fundamentaledge the augmenting triangulation coincides with the fundamental edge of C i .It follows from the construction that the toroidal triangulation T has a unique pair of groundstates.This finishes the proof of Theorem 2.2.3 Proof of Theorem 1The proof of Theorem 1 is relatively straightforward from the construction made for provingTheorem 2. Unlike the case of triangulations of the torus, in Theorem 1 the genus of the closedRiemann surface may be arbitrarily large. However, the described supporting punctured triangulationscan perform the task of increasing genus and at the same time keeping all properties ofexistence and uniqueness of satisfying spin-assignments in such a way that the same constructionmade in Subsection 2.2.3 works. Next, we add some new definitions and show how to deal withthe construction for proving Theorem 1.Let T be a supporting triangulation and ˜F be a set of removable faces of T . If ˜F containsat least two faces f 1 , f 2 such that its face boundaries C 1 , C 2 don’t share any vertex, we say thatthe supporting punctured triangulation ˜T associated to T and ˜F is a connector and that C 1and C 2 are its connection cycles. Clearly, a connector exists. Indeed, the supporting puncturedtriangulation depicted in Figure 3(b) is a connector. Properties and names from supportingpunctured triangulations are transferred to connectors.We are ready to describe the construction. Let g be a integer positive number, { ˜T i } i∈[g] bea collection of connectors and t i + 2 be the number of expandable cycles of connector ˜T i foreach i ∈ [g] (clearly, t i ≥ 0 for all i). Moreover, let C 2i−1 , C 2i denote the connection cycles of ˜T ifor each i ∈ [g].9

Let ˜T denote the punctured triangulation of a surface of genus g obtained by gluing togetherthe collection of connectors { ˜T i } i∈[g] in the following way: for each j ∈ [g − 1] identify theconnection cycle C 2j of ˜T j with the connection cycle C 2(j+1)−1 of ˜T j+1 in such a way that thefundamental edge of C 2j coincides with the fundamental edge of C 2(j+1)−1 .It is routine to check that ˜T has exactly one pair of satisfying spin-assignments and thateach hole of ˜T is non-monochromatic under any satisfying state on ˜T .To conclude, the Theorem 1 follow directly from applying the same construction presentedin Subsection 2.2.3, using ˜T instead of a supporting punctured triangulation.3 Final CommentsThe strategy to construct triangulations presented in this work may be (easily) extended toconstruct triangulations of a fixed surface with n vertices and groundstate degeneracy f(n),where f(n) is a function depending on n which can be either constant or polynomial on n; itcan be reached by taking instead of an augmenting triangulation, a plane triangulation with therequired property (which always exists — see for example [5]).We believe that this work leaves many doors open and unanswered questions. First, wethink that it is of particular relevance to find a complete description of triangulations with anon-degenerated groundstate. Also, in this context, we strongly believe that it is of great interestto study the following question: what is the groundstate degeneracy of random triangulationsprovided with the antiferromagnetic Ising model?. This will help to understand the behaviourof geometrically frustrated systems.On the other hand, the problem of spin glasses has attracted considerable attention overrecent years. Both, in solid physics and in statistical physics (for instance see [12]). In the Isingspin glass model, coupling constants are randomly distributed. Usually, each coupling constantis set randomly to either +1 or -1 with equal probability. The case of the antiferromagneticIsing model is the critical case when the coupling constant is set to -1 with probability equalto one. In this context, next goal would be to study the Ising model on toroidal triangulationswith this probability less than one and very close to one.AcknowledgementsThe author thanks Marcos Kiwi and Martin Loebl for their enthusiasm, motivation, and manyhelpful discussions.References[1] D. Barnette and A. Edelson. All orientable 2-manifolds have finitely many minimal triangulations.Israel Journal of Mathematics, 62:90–98, 1988.10

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