Solutions to Exercises 2

LTCCCourse onGraph Theory 2011/12Solutions toExercisesfor Notes21 (⇒) Supposethereexistsapolynomial p(n)ofdegreedsothat|f(n)| ≤ p(n)foralln ≥ 1.To show that this means f(n) = O(n d ) means that we must show that there exist constantsC > 0 and N ≥ 1 so that for all n ≥ N we have |f(n)| ≤ Cn d . Write p(n) = a d n d +a d−1 n d−1 +···+a 1 n+a 0 . Thenweknow a d ̸= 0(since p(n)isofdegreed). Andinfactwemusthave a d > 0,otherwisewewouldhave p(n) −→ −∞ as n → ∞,whichisimpossibleif|f(n)| ≤ p(n) forall n ≥ 1. So we can estimateforall n ≥ 1:|f(n)| ≤ p(n) = a d n d +a d−1 n d−1 +···+a 1 n+a 0≤ a d n d +|a d−1 |n d−1 +···+|a 1 |n+|a 0 |≤ a d n d +|a d−1 |n d +···+|a 1 |n d +|a 0 |n d= (a d +|a d−1 |+···+|a 1 |+|a 0 |)n d .So by setting C = a d +|a d−1 |+···+|a 0 | and N = 1,wefind |f(n)| ≤ Cn d forall n ≥ N.(⇐) Suppose we have f(n) = O(n d ), hence there exist constants C > 0 and N ≥ 1 sothat for all n ≥ N we have |f(n)| ≤ Cn d . So for n ≥ N the polynomial p(n) = Cn dwouldwork. Butthatpolynomialmightnotworkfor n < N. Toovercomethatproblem,setK = max{0,|f(1)|,|f(2)|,...,|f(N −1)|}. (This maximum exists, since we are just takingthemaximumoverafinitesetofnumbers.) Thenweknowthat|f(n)| ≤ K foralln < N. Soby setting p(n) = Cn d +K we are guaranteed |f(n)| ≤ p(n) for all n ≥ 1.2 (a) The flow diagramof the firstTuring Machine on page 5ofthe notes looks likethis:0 : 0,+11 : 1,+1q 0q 1# : #,−1# : #,+1q 2# : #,−1q 41 : #,−10 : #,+10 : #,−1# : .,. 1 : .,.1 : #,+10 : 0,−11 : 1,−1# : .,.0 : 0,+11 : 1,+1# : .,.0 : .,.q Yq NAuthor: Jan van den Heuvel c○London School of Economics, 2012

LTCC Course on Graph Theory Solutions to Exercises for Notes 2 — Page 2To get an ideaof what is happening, let’s try to follow a few steps. Suppose at the start,on the first square the machine reads a “0”. Then this “0” is erased (since # is writtenon the tape), the state becomes q 1 and the head goes one to the right. As long as thereis a “0”or “1”on the tape, the head goes further to the right. Until the head reaches theend of the 0,1-string (reads a blank “#”), then it goes back one step to the left and tostate q 3 . If it then reads a “1”, it goes to the end-state q N . So a string starting with “0”but endingwith “1”willbe rejected.Ifontheotherhandinstateq 3 itreadsa“0”,thenthat“0”iserased,thestatebecomesq 5and the head goes one to the left. As long as there is a “0” or “1” on the tape, the headgoes further to the left. Until the head reaches the left-end of the 0,1-string, then it goesback to the originalstate q 0 and one stepto the left.And then the whole process starts fromthe verybeginning.If in the first step a “1” is read, then something similar to the above will be done, butnowviastates q 2 and q 4 . Andinparticular,ifthefinalbitisa“0”,thenthemachinewillgo to state q N . If the final bit is a “1”, then that “1” is erased. And via q 5 the head goesback to the beginning.So all in all, if you look back over what happens above, then the machine compares thefirst and the last bit of the input. If these are the same, then they are both erased, andthe machine continues with the reduced string. If they are different, then the machineimmediatelyrejectsthestring. Thepositivefinalstageq Y isreachedwhennothingisleftonthe tape.FromthisitfollowsthatthestringacceptedbythisTuringMachineareexactlyso-calledpalindromes: strings thatread fromleftto right and fromrightto leftarethe same.(b) The second Turing Machine on page 5 of the notes is a lot easier to analyse: I’ll leave itto you to check that this accepts wordsthat contain exactly one “1”.3 These questions are actually harder than they look. I’m not going to give full details of theTuring Machine, but some important ideas that can be used. An important issue for bothcases is how to keep track of what part of the string has already been copied or doubled,and what part still needs to be done. Although it seems tempting to use a “counter” toindicatewhatpartneedstobecopied/doublednext,suchacounterisactuallyquitehardtoimplement. So it’s probably much more convenient to use a “marker” to indicate where themachine was. Such amarker can justbe asquare onthe tape with the blank symbol #on it.Of course, that means the machine erases what was on that square before. So somehow themachine needs to store that information. Since there are only two possible original symbols(0 or 1), this can be done using appropriate states. So there should be a set of states just forwhen the machine is dealing with a 0, and a set of states for when the machine is dealingwith 1. (This also happens in the machine for Question 1 above, where states q 1 ,q 3 onlyappear when the machine had read a 0 in state q 0 ; while q 2 ,q 4 are states for the case therewas a1onthe tape whilethe machine was instate q 0 .,)

LTCC Course on Graph Theory Solutions to Exercises for Notes 2 — Page 3So in the descriptions of the two Turing Machines below, we divide the state space Q intothree parts: Q = Q ′ ∪ Q 0 ∪ Q 1 . Here Q ′ contains some general states (including q 0 andso), while Q 0 and Q 1 are sets of states the machine can be in while it is copying a 0, or a 1,respectively.A Apart from the idea to use the blank symbol as a marker, another observation thatmakeslifealoteasierforthisproblemistostartworkingfromthebackofthestring.TounderstandthestepsoftheTuringMachineasgivenbelow,youshouldhaveanideahow the tapeis used. So here ishow thetape wouldlooklikeinthe middleofthe computation of making acopy ofthe string00010001011:··· |#|#|1|0|1|1|#|0|0|0|1|0|0|#|1|0|1|1|#|#| ··· .The underlined square is the starting square, while the black blob is the tape head.So the machine has just erased the symbol on the square to the right of the tapehead. That square was erased, but since it contained originally the symbol 0, themachine entered in one of the states from Q 0 , and the tape head started to move tothe left.The Turing Machine performs steps as follows,all the time using states from Q 0 :– Thetapeheadgoeslefttothebeginningoftheoriginalstring(recognisedbythefirstblank itencounters).– Then it moves back to the beginning of the already copied part (again, recognisedby ablank square).– Onceitencountersthatblanksquare,itwritesa0,andstartsmovingrightagain.– The tape head goes right till the beginning of the original string (recognised byablank).– Then furtherrighttillthe markerof the bititwas workingon (again ablank).– Then it writes a 0 to replace the blank (so that square is back to the originalstate). It also moves one square to the left, and the state becomes one of thenon-specific states from Q ′ .– If it now reads a blank symbol, the copying is done, and the machine halts. If itdoesn’treadablanksymbol,itreplacesthecurrentsymbolbytheblanksymbol;and its state becomes a state from Q 0 or Q 1 , depending if the square containeda0or a1.– And now the process starts as fromthe firststepabove.With the description above, it should be possible to design a full Turing Machine.Youprobablyneedafewinitialstepstogetgoingandtomakesurethatthemachinehalts immediatelyif the empty string is givenas input.B Several of the ideas from A can be used here as well. Here an extra complicationis to make room for the bits that have to be added in between. The best way to dothis is probably by doing it one by one. I.e., every time when making a copy of thenext bit, first more the remainder of the string one to the right (starting at the backagain),and then make the copy inthe freedspace.

LTCC Course on Graph Theory Solutions to Exercises for Notes 2 — Page 4Let’s demonstrate the steps and the main ideas by looking at what the situationcould be in the middle of a computation again. Using the same string 00010001011as example again, a typical situation on the tape in the middle of the computationwillbe something like:··· |#|#|0|0|0|0|0|0|1|1|#|0|0|1|0|1|1|#|#| ··· .So the machine has already doubled the first four bits, and has just erased the fifthbit,andmovedonesteptotheright. Sincethaterasedsymbolwasa0,themachinewillbe insome state from Q 0 .And next the machine willperformthe followingsteps:– It willmove rightto the endof the string (recognisedby ablank square).– Itthenwilldoonestepback to theleft,readthesymbol,eraseit,moveone steptotheright,writethatonesymbolonthatsquare,andonesteptotheleftagain.(These steps arejustmoving one symbol one squareto the right.)– It continues doing the above until it is back to the beginning of the part of theoriginalstring that hasn’t beendoubledyet.– Once it is back at the blank “marker”, there will actually be two blank squaresin a row (since the complete right side of the string has been moved one squarefurther to the right). So it can write two 0s in a row. (The machine “knows”these mustbe 0s since itwas using states from Q 0 inall the steps above.)– The machine moves one square further to the right, and the state becomes oneof the non-specific states from Q ′ .– If it now reads a blank symbol, the doubling is done and the machine halts. If itdoesn’treadablanksymbol,itreplacesthecurrentsymbolbyablank,andgoesinto astate from Q 0 orfrom Q 1 ,according to the symbol that was onthe tape.– And now the process starts as fromthe firststepabove.Again, the final Turing Machine probably needs some extra initial steps to get it allgoing and to make sure it halts if there is nothing on the tape. But apart from that,you shouldbe able to designaTuring Machine that doesallof the above.4 (a) By definition, L 1 is in NP, if there exists a nondeterministic Turing Machine N (1) and apolynomial p (1) (n)suchthatforeveryx ∈ L 1 ,thereexistsasequenceofatmost p (1) (|x|)allowedstepssothatN (1) haltsintheacceptingstateofN (1) . Andasimilarconclusion,using a nondeterministic Turing Machine N (2) and a polynomial p (2) (n), can be drawnfromthe claimthat L 2 isin NP.From this we can easily see that also L 1 ∪ L 2 is in NP. Make a new nondeterministicTuring Machine N that is formed by using disjoint states q (1)ifor N (1) and states q (2)ifor N (2) . Now let q 0 and q Y be two new states. Then form N as follows. Starting atstate q 0 , there is one possible transition to the initial state q (1)0of N (1) and one possibletransitiontotheinitialstateq (2)0ofN (2) . Also,fromtheacceptingstateq (1)Y ofN(1) thereis one transition to the accepting state q Y of N, and the same holds for the acceptingstate of N (2) .

LTCC Course on Graph Theory Solutions to Exercises for Notes 2 — Page 5It’s now straightforward to check that if x ∈ L 1 ∪ L 2 , then the nondeterministic TuringMachine N has a sequence of at most max{p (1) (|x|),p (2) (|x|)} + 2 allowed steps sothat N halts in its accepting state. And if x /∈ L 1 ∪ L 2 , then x /∈ L 1 and x /∈ L 2 , so withinput x, N can never reach the accepting state of the N (1) and the N (2) part of N, andhence such an x willneverbe accepted.(b) We use the set-up from (a) again. And in fact we can use almost the same ideas, exceptnow we construct a nondeterministic Turing Machine N as follows: From the initialstateq 0 thereisonetransitiontotheinitialstateq (1)0ofN (1) ;fromtheacceptingstateq (1)Yof N (1) there is one transition to the initial state q (2)0of N (2) ; and from the acceptingstate q (2)Y of N(2) there isone transitionto the generalaccepting state q Y .Again when we give an input x ∈ L 1 ∩ L 2 to the resulting nondeterministic TuringMachine N, there is a sequence of at most p (1) (|x|)+ p (2) (|x|)+3 allowed transitionsafter which N has reached its accepting state. And for x /∈ L 1 ∩ L 2 , at least one of thetwo parts will preventthatfromhappening.(c) This time we can’t conclude that L 1 \ L 2 is in NP. The reason is that the fact that x ∈L 1 \ L 2 means that x ∈ L 1 and x /∈ L 2 . But we have no good way to deal with thesituation x /∈ L 2 .So in generalwedon’tknow if L 1 \L 2 is inNP ornot.5 Property6states: SupposeL 1 isanNP-completelanguage. IfL 2 isalanguagesothatL 2 isinNPand L 1 ≤ P L 2 ,then L 2 is NP-complete.The fact that La 1 is NP-complete means that L 1 is in NP, and for all L ∈ NP we have thatL ≤ P L 1 . AndtoprovethatL 2 isNP-completeweneedtoprovethatL 2 isinNP,andforallL ∈ NPwehave that L ≤ P L 2 .We are already given that L 2 is in NP, so we are left to prove that L ≤ P L 2 for all L ∈ NP.For this we obviously need to use that L ≤ P L 1 for all L ∈ NP and that L 1 ≤ P L 2 . In fact itfollowsdirectlythat itis enoughto prove the followingproperty:For anythreelanguages L,L ′ ,L ′′ ,if L ≤ P L ′ and L ′ ≤ P L ′′ , then L ≤ P L ′′ .Using the definition of polynomial-time reducibility, L ≤ P L ′ and L ′ ≤ P L ′′ means thatthere exist functions f 1 , f 2 : {0,1} ∗ −→ {0,1} ∗ , Turing Machines M 1 ,M 2 , and polynomialsp 1 ,p 2 so that– forall x ∈ {0,1} ∗ wehave x ∈ L ifand only if f 1 (x) ∈ L ′ ;– forall x ∈ {0,1} ∗ wehave x ∈ L ′ if and only if f 2 (x) ∈ L ′′ ;– forall x ∈ {0,1} ∗ , M 1 computes f 1 (x) inatmost p 1 (|x|) steps;– forall x ∈ {0,1} ∗ , M 2 computes f 2 (x) is atmost p 2 (|x|) steps.Note that since a Turing Machine can only write one bit of output per time step, we alsohave that forall x ∈ {0,1} ∗ the length of f 1 (x) satisfies |f 1 (x)| ≤ p 1 (|x|).Sonowdefineafunction f : {0,1} ∗ −→ {0,1} ∗ by f(x) = f 2 (f 1 (x)). Thenforall x ∈ {0,1} ∗we havethat x ∈ L ifand only if f 1 (x) ∈ L ′ if and only if f(x) = f 2 (f 1 (x)) ∈ L ′′ .

LTCC Course on Graph Theory Solutions to Exercises for Notes 2 — Page 6Moreover,usingaTuringMachinethatfirstperformstheactionsofM 1 andthentheactionsof M 2 will result in a Turing Machine to compute f(x) = f 2 (f 1 (x)) and, for some smallconstant C, uses at most p 2 (p 1 (|x|))+ p 1 (|x|)+C steps to do so. Note that if p 1 (n),p 2 (n)are polynomials,thenso is p 2 (p 1 (n))+ p 1 (n)+C.This proves that L ≤ P L ′′ and completesthe answer.6 Let L ⊆ {0,1} ∗ be a language in P, L ̸= ∅ and L ̸= {0,1} ∗ . In particular this means thatthere are y 1 ,y 2 ∈ {0,1} ∗ so that y 1 ∈ L but y 2 /∈ L. We will show that forall L ′ ∈ Pwe haveL ′ ≤ P L,whichwouldshow that L is P-complete.If L ′ is in P, then there is a Turing Machine M and an integer d so that for all x ∈ {0,1} ∗ wehave: if x ∈ L ′ , then M accepts x in O(|x| d ) steps; while if x /∈ L ′ , then M rejects x afterO(|x| d ) steps.Now define the function f : {0,1} ∗ −→ {0,1} ∗ as follows: f(x) ={y 1 , if x ∈ L ′ ,y 2 , if x /∈ L ′ .Since y 1 ∈ L and y 2 /∈ L, it’s obvious that for all x ∈ {0,1} ∗ we have x ∈ L ′ if and only iff(x) ∈ L. Moreover, using the Turing Machine M from above, for all x ∈ L, f(x) can becomputedinpolynomialtime. (Justgiveasoutput y 1 ifMaccepts x,andasoutput y 2 ifMrejects x.) This shows L ′ ≤ P L and completesthe answer.Someextranotes:Although the above answer is (hopefully) easy to follow, there are some deeper issues ittouches. In particular, the algorithm described assumes we have knowledge of suitable y 1and y 2 . By the formulation of the question, we can be sure suitable y 1 ,y 2 exist. And hencethe algorithmexists.But if we would like to implement the algorithm above, we would need to know explicitlywhat y 1 and y 2 are. And that couldbe aproblem(or alotofwork).So keep this in mind: The definitions of P and so talk about the existence of certain TuringMachines. Itdoesn’trequirethatyou are actuallyable to construct them.7 It is easy to see that there is a cycle in G through u if and only if there is an edge uv and apath from u to v in the graph G ′ formed by removing the edge uv from G. Using SavitchTheorem 11 from Section 2.11 of the notes, it is fairly easy to construct a way to do thischeckinginthesameamountofspaceitrequirestocheckST-CONNECTIVITY(plusalittlebitextrato store the edge uv we arecurrently dealingwith).

LTCC Course on Graph Theory Solutions to Exercises for Notes 2 — Page 7More explicitly, build a 2-tape Turing Machine that does the following. (Here we againassume that the verticesareindicatedby the numbers 1to N.)1. Write the vertices u and v onthe tape and set a = 1.2. Write a onthe tape.3. Readthe input tape to check if ua is anedge.If ua is an edge,continue withstep4below.If ua is notan edgeand a = N, then conclude thatthere is no cycle through u and stop.If ua is notan edgeand a < N, then increase a by oneand gobeck to step2.4. Use Savitch Algorithm to check if there is a path from u to a in G, where we will ignorethe edge ua inthe computation.If there is suchapath, thenconclude that thereis acycle through u and stop.If such apath doesnot existand a = N, then that thereis no cycle through u;stop.If such apath doesno existand a < N, then increase a by one and go beck to step2.Since step 4 can be done using at most O((log(N)) 2 ) memory space, we can easily see thatthe whole procedurecan bedone using atmostO((log(N)) 2 ) memory space.