Conjunctive Queries

Three Problems:
CQs and Structural Methods
HOM: The homomorphism problem
BCQ: Boolean conjunctive query evaluation
CSP: Constraint satisfaction problem
Bertinoro International Spring School 2009, 2-13 March 2009
1
Important problems in different areas.
All these problems are hypergraph based.
But actually: HOM = BCQ = CSP
The Homomorphism Problem
Given two relational structures
A ( U , R1,
R
B ( V , S
1
,
S
2
2
,..., R
,...,
S
k
k
)
)
Decide whether there exists a homomorphism h from A to B
h :
U
such
x R
that
V
x,
i
h(
x)
i Si
The Homomorphism Problem
Given two relational structures
A ( U , R1,
R
B ( V , S
1
,
S
2
2
,..., R
,...,
S
k
k
)
)
Decide whether there exists a homomorphism h from A to B
h :
U
such
x R
that
V
x,
i
h(
x)
i Si
Feder & Vardi 93: Relationship to CSP, restrictions on B
Kolaitis & Vardi, 98: Relationship to Query Containment, restrictions on A, B

(well-known)
HOM is NP-complete
Membership: Obvious, guess h.
Hardness: Transformation from 3COL.
(well-known)
HOM is NP-complete
Membership: Obvious, guess h.
Hardness: Transformation from 3COL.
1
3
6
2
4
5
A
1 2
1 3
2 3
3 4
2 5
4 5
3 6
B
red
red
green
green
blue
blue
green
blue
red
blue
red
green
1
3
6
2
4
h
5
A
1 2
1 3
2 3
3 4
2 5
4 5
3 6
h
B
red
red
green
green
blue
blue
green
blue
red
blue
red
green
Graph 3-colourable iff HOM(A,B ) yes-instance.
Graph 3-colourable iff HOM(A,B ) yes-instance.
DATABASE:
Boolean Conjunctive Query
Enrolled Teaches Parent
John Algebra 2003
Robert Logic 2003
Mary DB 2002
Lisa DB 2003
……… ….. …….
McLane Algebra March
Kolaitis Logic May
Lausen DB June
Rahm DB May
……… ….. …….
McLane Lisa
Kolaitis Robert
Rahm Mary
……… …..
QUERY: Is there any teacher having a child
enrolled in his/her course
ans Enrolled(S,C,R) Teaches(P,C,A)
Parent(P,S)
Boolean Conjunctive Query (2)
• Database schema:
– Enrolled (Pers#, Course, Reg-Date)
– Teaches (Pers#, Course, Assigned)
– Parent (Pers1, Pers2)
• Is there any teacher whose child attend some
course
ans Enrolled(S,C’,R) Teaches(P,C,A)
Parent(P,S)

Boolean Conjunctive Queries
The problem BCQ:
Instance: < DB, Q>
Question: Does Q evaluate to true over DB
Conjunctive Query with Output
• Database schema:
– Enrolled (Pers#, Course, Reg-Date)
– Teaches (Pers#, Course, Assigned)
– Parent (Pers1, Pers2)
• Output all students x enrolled in a course
taught by a parent of x
ans(S) Enrolled(S,C,R) Teaches(P,C,A)
Parent(P,S)
ConjunctiveQueriesandRA,SQL
Logical View of CQs
Conjunctive queries correspond to Select-Project-Join queries,and to simple
“SELECT-FROM-WHERE SQL statements with a conjunctive WHERE
clause.
–Enrolled (Pers#, Course, Reg-Date)
–Teaches (Pers#, Course, Assigned)
–Parent (Pers1, Pers2)
Q: Ans(S) Enrolled(S,C,R) Teaches(P,C,A) Parent(P,S)
SELECT Enrolled.Pers#
FROM Enrolled Teaches Parent
WHERE Enrolled Pers# = Parent.Pers2
and Enrolled.Course = Teaches.Course
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and Teaches.Pers# = Parent.Pers1
AconjunctivequeryisaFOqueryoftheform
{r(x) |y (r 1 (x 1 ,y 1 ) r2(x 2 ,y 2 ) … r2(x k ,y k ))}
Where:
x,y aredisjointlistsofvariables,
allvariablesofxoccurfreeintheformulay (r 1 (x 1 ,y 1 ) r2(x 2 ,y 2 ) … r2(x k ,y k ))
eachx i isalistofvariablesfromx,
eachy i iscontainediny
Written:r(x) r 1 (x 1 ,y 1 ) r2(x 2 ,y 2 ) … r2(x k ,y k ).
12

BCQ = HOM
View query Q as a relational structure
Universe: Variables of the query
Relations: sets of query-atoms for
each database relation.
Non-Boolean CQs
Q:r(x) r 1 (x 1 ) r 2 (x 2 )… r m (x m )
The database D is itself a relational structure.
By the definition of the semantics of FO queries, the
Boolean conjunctive query evaluates to true iff there is a
homomorphism h: QD.
R1
R1
DatabaseD
…..
R1
D=QiffHOM(Q,D)
HowdoweformallydefineQ(D)
Vive-versa, every HOM instance can be reformulated as a Boolean conjunctive query. 14
Non-Boolean CQs
head(Q)
body(Q)
r(x) r 1 (x 1 ) r 2 (x 2 )… r m (x m )
h
h homomorphismh h
R1
R1 ….. R1
DatabaseD
Problems Equivalent to BCQ
• Conjunctive Query Containment
Q
( 1
Q2
db
Q1 db)
Q2
( db)
• Query of Tuple Problem
t Q(db)
• Constraint Satisfaction in AI
CSP BCQ
• Clause Subsumption in Theorem Proving
Q(D)={h(head(Q))|h HOM(body(Q),D)}
s.t.
C
D
15

Combinatorial Crossword Puzzle
All known general
solution algorithms
rely on backtracking
Combinatorial Crossword Puzzle
All known general
solution algorithms
rely on backtracking
1h: P A R I S 1v: L I M B O
P A N D A
L I N G O
L A U R A
P E T R A
A N I T A
P A M P A
P E T E R
We are looking for a
homomorphism
f:[1..26][A..Z]
that translates
1H ={}
into 1h, and so on.
and so on
1h: P A R I S 1v: L I M B O
P A N D A
L I N G O
L A U R A
P E T R A
A N I T A
P A M P A
P E T E R
We are looking for a
homomorphism
f:[1..26][A..Z]
that translates
1H ={}
into 1h, and so on.
and so on
ContainmentofCQ’s
• GivenconjunctivequeriesQ1andQ2,wesay
thatQ1iscontainedinQ2,denotedasQ1 Q2,
iff forall databasesD,Q1(D) Q2(D).
• Example:
–Q1:p(X,Y) arc(X,Z),arc(Z,Y)
–Q2:p(X,Y) arc(X,Z),arc(W,Y)
•DBisagraph;Q1producespathsoflength2,
Q2producespairsofnodeswithanarcoutand
in,respectively.
19
WhyCQContainment
Containmenttestsusedforequivalencetests.
Q1=Q2iffQ1 Q2andQ2 Q1
Equivalenceisfundamentaltoqueryoptimization
Answeringqueriesusingviews.
Dataintegration:collectingdatafromdifferentsources
(e.g.onInternet),answeringyourspecificquery need
toknowwhetherthesourcescan(fully,orpartially)
answeryourquery.
20

QueryHomomorphisms
(akaContainmentMappings)
Example
Amappingfromthevariables ofCQQ2to
thevariablesofCQQ1,suchthat:
TheheadofQ2ismappedtotheheadofQ1.
Q1:p(X,Y) r(X,Z),g(Z,Z),r(Z,Y)
Q2:p(A,B) r(A,C),g(C,D),r(D,B)
Eachsubgoal (=bodyatom)ofQ2ismappedto
somesubgoal ofQ1withthesamepredicate.
HomomorphismTheorem:
ThereisahomomorphismfromQ2toQ1if
andonlyifQ1 Q2.
21
Q1looksfor:
X
Q2looksfor:
A
Z
C
Y
D
SinceC=Dispossible,
expectQ1 Q2.
B
22
Example Continued
• WheneverthereisapathfromX toY,it
mustbethatX hasanarcout,andY an
arcin.
• Thus,everyfact(tuple)producedbyQ1is
alsoproducedbyQ2.
•Thatis,Q1 Q2.
Example Continued
Q1: p(X,Y) r(X,Z) , g(Z,Z) , r(Z,Y)
Q2: p(A,B) r(A,C) , g(C,D) , r(D,B)
Homomorphism:h(A)=X;h(B)=Y;h(C)=h(D)=Z.
23
24

ExampleConcluded
Q1: p(X,Y) r(X,Z) , g(Z,Z) , r(Z,Y)
Q2: p(A,B) r(A,C) , g(C,D) , r(D,B)
•NocontainmentmappingfromQ1toQ2.
– g(Z,Z)canonlybemappedtog(C,D).
•Nootherg subgoals inQ2.
–ButthenZmustmaptobothCandD impossible.
• Thus,Q1properlycontainedinQ2.
25
ProofoftheHomomorphismTheorem Hom(Q2,Q1) iff Q1 Q2.
a) hHOM(Q2,Q1) Q1 Q2(thisistheonlyif direction)
LetDbeadatabaseandAQ1(D).
Then gHOM(body(Q1),D),suchthatA=g(head(Q1)).
Q2:head(Q2) body(Q2)
h
h
Q1:head(Q1) body(Q1) A
g
g
AD
Buttheng hisahomomorphisminHOM(body(Q2),D),
andg h(head(Q2))=g(h(head(Q2))=g(head(Q1))=A
ThusA Q2(D).
Insummary,AQ1(D),AQ2(D),thusQ1 Q2.
26
ProofoftheHomomorphismTheorem Hom(Q2,Q1) iff Q1 Q2.
b)Q1 Q2 hHOM(Q2,Q1)(thisistheif direction)
a) hHOM(Q2,Q1) Q1 Q2(thisistheonlyif direction)
LetDbeadatabaseandAQ1(D).
Then gHOM(body(Q1),D),suchthatA=g(head(Q1)).
Q2:head(Q2) body(Q2)
g h
h
h
g h
Q1:head(Q1) body(Q1) A
g
g
AD
Buttheng hisahomomorphisminHOM(body(Q2,D)),
andg h(head(Q2))=g(h(head(Q2))=g(head(Q1))=A
ThusA Q2(D).
Insummary,AQ1(D),AQ2(D),thusQ1 Q2.
27
Q2: p(A,B) r(A,C) , g(C,D) , r(D,B)
Q1: p(X,Y) r(X,Z) , g(Z,Z) , r(Z,Y)
CanonicalDatabaseCan Q1 :
Generateanewconstantf(U)=U foreachvariable U
Can Q1 =f(atoms(body(Q)))={r(X,Z), g(Z,Z), r(Z,Y)}
dom(Can Q1 )={X,Y,Z}
Note:Themappingf:var(Q1) {newconstants}isanisomorphism where
f(atoms(body(Q1)))=Can Q1 and f 1 (Can Q1 )=atoms(body(Q1)
28

Q2: p(A,B) r(A,C) , g(C,D) , r(D,B)
Q2: p(A,B) r(A,C) , g(C,D) , r(D,B)
Q1: p(X,Y) r(X,Z) , g(Z,Z) , r(Z,Y)
Q1: p(X,Y) r(X,Z) , g(Z,Z) , r(Z,Y)
f
f 1 f f 1 f f 1
f f 1
f
f 1 f f 1 f f 1
{r(X,Z) , g(Z,Z) , r(Z,Y)}= Can Q1 :
A=p(X,Z) {r(X,Z) , g(Z,Z) , r(Z,Y)}= Can Q1 :
Sincefmapsbody(Q)intoCan Q1’
AnewfactA=f(head(Q1)) isgenerated
A=f(head(Q1))Q1(Can Q1 ).
SinceQ1 Q2,A Q2(Can Q1 ).
Therefore hHom(Q2,Can Q1 {A}).
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30
Q2: p(A,B) r(A,C) , g(C,D) , r(D,B)
h
h h h
Q1: p(X,Y) r(X,Z) , g(Z,Z) , r(Z,Y)
f f 1 f f 1 f f 1 f f 1
AnotherExample
Q1: p(X,Y) r(X,Y) , g(Y,Z)
Q2: p(A,B) r(A,B) , r(A,C)
A=p(X,Z) {r(X,Z) , g(Z,Z) , r(Z,Y)}= Can Q1 :
Q1looksfor:
X
Y
Z
A=f(head(Q1))Q1(Can Q1 ).
Q2looksfor:
SinceQ1 Q2,A Q2(Can Q1 ).
Therefore hHom(Q2,Can Q1 {A}).
A
B
Thenf 1 h HOM(Q2,Q1)isthedesiredhomomorphism.
C
QED
31
32

Example Continued
Q1: p(X,Y) r(X,Y) , g(Y,Z)
Q2: p(A,B) r(A,B) , r(A,C)
Andnot
everysubgoal
needbea
target.
Noticetwo
subgoals can
maptoone.
ExampleConcluded
Q1: p(X,Y):-r(X,Y) & g(Y,Z)
Q2: p(A,B):-r(A,B) & r(A,C)
•NocontainmentmappingfromQ1toQ2.
– g(Y,Z)cannotmapanywhere,sincethereisnog
subgoal inQ2.
Containmentmapping:m(A)=X;m(B)=m(C)=Y.
33
• Thus,Q1properlycontainedinQ2.
34
LetCan + Q1
Query Containment BCQ
=Can + Q1 f(head(Q1))
LetQ2 + =Q2 head(Q2)(Q2 + isaBoolenquery)
Then,
Q1 Q2iffCan + Q1 =Q2 +
i.e.,iffQ2 + evaluatestotrueover Can + Q1
Therefore,allcomplexityresultsforBCQalsoholdfor
(Boolean)ConjunctiveQuery
evaluation
•{enc(I),enc()|I² }
–if isaconjunctivequery,combined
complexity
– n =|I|,m =||
– n m possibleassignmentsforvars()
– Givenanassignment,thetestingisin
polynomialtime
–InNP
thequerycontainmentproblem.
35
36

ConjunctiveQueryevaluation
CQminimization
THEOREM:BCQisNPComplete(combinedcomplexity)
Proof:FollowsfromtheNPcompletenessofHOM
Exercise: AsanalternativeproofofNPhardness,givea
reductionfrom3SAT.
COROLLARY: ConjunctiveQueryContainmentisNPcomplete
COROLLARY: ConjunctiveQueryEquivalenceisNPcomplete
•AconjunctivequeryQisminimal,ifitcontains
minimalnumberofsubgoals –
(subgoal=bodyatom)
Thismeans:whenewerwedropanatom
fromthebodyofQ,wegetaqueryQ’Q
•ThetaskoftheCQminimization:minimizethe
numberofsubgoals ofagivenCQ.
Note:ThelatterproblemsareundecidableforFO 37
38
Cores
Cores
endomorphism h: {Y b}
BCQ:
Pp(X,Y), p(X,b), p(a,b), p(U,c), p(U,V), q(a,c,d)
B= { p(X,Y), p(X,b), p(a,b), p(U,c), p(U,V), q(a,c,d)}
{ p(X,Y), p(X,b), p(a,b), p(U,c), p(U,V), q(a,c,d) }
Logical meaning
X, Y, U, V:
p(X,Y) & p(X,b) &
p(a,b) & p(U,c) & p(U,V) & q(a,c,d)
39
40

Cores
endomorphism h: {Y b}
B = { p(X,Y), p(X,b), p(a,b), p(U,c), p(U,V), q(a,c,d) }
Cores
endomorphism h: {Y b}
B = { p(X,Y), p(X,b), p(a,b), p(U,c), p(U,V), q(a,c,d) }
{ p(X,Y), p(X,b), p(a,b), p(U,c), p(U,V), q(a,c,d) }
{ p(X,Y), p(X,b), p(a,b), p(U,c), p(U,V), q(a,c,d) }
REDUNDANT!
X,Y p(X,Y) & P(X,b)
X p(X,b)
41
REDUNDANT!
X,Y p(X,Y)
X p(X,b)
42
Cores
endomorphism h: {Y b}
B = { p(X,Y), p(X,b), p(a,b), p(U,c), p(U,V), q(a,c,d) }
Cores
endomorphism h: {Y b}
B = { p(X,Y), p(X,b), p(a,b), p(U,c), p(U,V), q(a,c,d) }
h(B) =
{ p(X,b), p(a,b), p(U,c), p(U,V), q(a,c,d) }
h(B) =
{ p(X,b), p(a,b), p(U,c), p(U,V), q(a,c,d) }
X a
V c
h(B) can be further reduced by endomorphism g: {X a, V c}
43
44

Cores
Cores
endomorphism h: {Y b}
B = { p(X,Y), p(X,b), p(a,b), p(U,c), p(U,V), q(a,c,d) }
B = { p(X,Y), p(X,b), p(a,b), p(U,c), p(U,V), q(a,c,d) }
h(B) =
{ p(X,b), p(a,b), p(U,c), p(U,V), q(a,c,d) }
h(B) =
{ p(X,b), p(a,b), p(U,c), p(U,V), q(a,c,d) }
{ p(a,b), p(U,c), q(a,c,d) }
f(B)=g(h(B))=gh(B)=
{ p(a,b), p(U,c), q(a,c,d) }
h(B) can be further reduced by endomorphism g: {X a, V c}
endomorphism f: {X a, Yb, V c}
45
46
Cores
Cores
B = { p(X,Y), p(X,b), p(a,b), p(U,c), p(U,V), q(a,c,d) }
B = { p(X,Y), p(X,b), p(a,b), p(U,c), p(U,V), q(a,c,d) }
h(B) =
{ p(X,b), p(a,b), p(U,c), p(U,V), q(a,c,d) }
h(B) =
{ p(X,b), p(a,b), p(U,c), p(U,V), q(a,c,d) }
f(B)=g(h(B))=gh(B)= { p(a,b), p(U,c), q(a,c,d) }
no refinement by endomorphisms possible !
endomorphism f: {X a, Yb, V c}
f(B)=g(h(B))=gh(B)=
{ p(a,b), p(U,c), q(a,c,d) }
Core(I)
unique up to variable-renaming!
47
endomorphism f: {X a, Yb, V c}
48

CQminimization
Ans(X,Y,Z) R(X 2 ,Y 1 ,Z),
R(X,Y 1 ,Z 1 ),
R(X 1 ,Y,Z 1 ),
R(X,Y 2 ,Z 2 ),
R(X 2 ,Y 2 ,Z)
Observations
•Thegoal:minimizethenumberofrowsin
thetableau.
Theheadatomcannotbeeliminated!
–Canwejustcheckthesubgoals onebyone(if
onesubgoal cannotberemovedatcontain
moment,coulditberemovedatalater
moment)
– Doesthesequenceoftheeliminationcount
(or,istheminimizedqueryunique)
49
50
Minimizationalgorithm
•Canwejustchecktherowsonebyone
Yes.Takethesubgoal u i ,inaCQQ1,check
whetherthereisahomomorphismh,suchthat
h(Q1) (Q1 u i ).Ifthereexistssucha
homomorphism,thenu i canberemoved.
Otherwise,u i cannotberemovedforever.
BecauseQ1isgettingsmaller.
THEOREM: The Core is Unique up to isomorphism
Isomorphism here means: bijective variable renaming
PROOF: Homework.
•Doesthesequenceoftheeliminationcount(or,
istheminimizedqueryunique)
51
52
homework

CQContainmentvs.
BCQevaluation
• BothareNPcomplete,but:
– thepotentialnumberofmappingsisdifferent:
n m :wheren is
•thesizeofQ1forCQcontainment(small,and
normallydoesnothurt)
•isthesizeofadatabaseforBCQevaluation(couldbe
huge)
• Highlydesirable:IdentifyfragmentsofBCQ,
whoseevaluationproblemistractable(inP).
53
BCQevaluation
• Intuition:theproblemwithcyclesinthequery:
– Considerthequeries:
•Q1:ans e(A,B),e(B,C),e(C,D),e(D,A)
•Q2:ans e(A,B),e(B,C),e(C,D),e(D,E)
•InprocessingQ1,maintainintermediatejoin
resultse’(A,B,C,D) e(D,A)
•InprocessingQ2,allthejoinoperationsareonly
betweentwopredicates(locally)
54
Hypergraph of Crossword Puzzle
Queries, CSPs, and
Hypergraphs
ans Enrolled(S,C,R) Teaches(P,C,A) Parent(P,S)
C
R
A
Theorem:Theevilisinthecycles.
S
P

Acyclic CSP
Acyclic queries or CSPs
ans Enrolled(S,C’,R) Teaches(P,C,A) Parent(P,S)
ans Enrolled(S,C’,R) Teaches(P,C,A) Parent(P,S)
C’
C
C’
C
Parent(P,S)
R
A
R
A
Teaches(P,C,A)
Enrolled(S,C’,R)
S
P
S
P
Join Tree
Join Tree
Definition: AJointreeofaBCQisanorganisationof
itsatomsintoatreeTsuchthatforeachvariableX,
thenodesinwhichXoccursspanaconnected
subtree ofT.(ConnectednessCondition)
Definition: ACQisatreequery iffithasajointree.
Example:Q=R(Y,Z,U)&s(Z,U,W)&d(Y,P)&t(V,Z)
d(Y,P)
r(Y,Z,U)
Acyclicity
GYO-Reduction (Graham, Yu, and Ozsoyoglu):
Input: Hypergraph H
Output: GYO-reduct of H
Apply the following rules as long as possible:
• Eliminate vertices that are contained in at
most one hyperedge (ear vertices)
• Eliminate hyperedges that are empty or
contained in other hyperedges
Definition: H is acyclic iff GYO-reduct(H)=(,).
s(Z,U,W)
t(V,Z)
59
Definition: Q is acyclic iff its hypergraph H(Q) is acyclic.

H
Y
Z
U
P
V
W
Y
Z
rule 1
U
rule 2
Y
Z
U
rule 1
0
7
9
8
6
5
0
9
6
GYO-irreducible
2 3 4
H*= (ø,ø)
GYO reduct
rule 2
1
AternativeProcedure
Definition: Anearedge(short: ear) ofahypergraph (V,E)isa
hyperedge e Esuchthatoneofthefollowingconditionsholds:
Hypergraph ofBCQs
(1)Thereisawitnesse’ E,suchthate’ e andeach
vertexfrome iseitheronlyine orine’,or
(2)ehasnointersectionwithanyotherhyperedge.
(C,D,E)is
anear
Ears
Ears
63
64

GYO’ algorithm– removingears
Algorithm(GYO’ Algorithm)
Input:Hypergraph H =(V,E)
Output: GYO’reductofH
H
P
Y
Z V
U
W
HowdoesGYO’ workhere
DountilEhasnoears:
ChooseanyeareofE.
SetH:=(V’ ,E {e}),whereV’ =vertices(E {e}).
65
Acyclicity Theorems
Lemma: AhypergraphGisaGYOreductof
HiffitisaGYO’reductofH.
THEOREM: Ahypergraphisacycliciff
itsonlyGYO’reductis(,).
THEOREM: Aqueryhasajointreeiff
itisacyclic.
Summary
•TheGYO(andGYO’)algorithmtakeslineartime
•Thehypergraph isacycliciff theoutputofGYO
(orequivalently,GYO’)algorithmisempty
•ABCQisacyclicifitshypergraph isacyclic
•IfaBCQisacyclic,thenthereisajointree,
obtainedfromthesequenceoftheearremoving
oftheGYOalgorithm.
Easyalgorithmsexistfortcheckingwhether
Itremainstoshow: Thequeryevaluationof
acyclicBCQisinpolynomialtime
Aqueryisacyclicandforcomputingajointree. 67
68

Theacyclicqueryevaluation
•ans d(Y,P),t(V,Z),s(Y,Z,U),s(Z,U,W)
•ThepolynomialalgorithmofacyclicBCQ
evaluation:
• Bottomup
•Semijoin
Boolean conjunctive queries
CSPs
The Homomorphism Problem….
….polynomialloy solvable for on instances
having acyclic hypergraphs
69
Semijoin Operation
Employee
Employee Name Employee Salary Department Name
Ackman 5000 Car
Baker 4500 Car
Carson 4800 Toy
Davis 5100 Toy
Temp=
Employee semijoin (DepartmentName=DepartmentName)
Department
Employee
Employee Name Employee Salary Department Name
Ackman 5000 Car
Baker 4500 Car
Department
Department Name Department Budget
Car 450000000
Figure by Dr. James A.
Larson
Emp A=B Dept= attr(Emp) (Emp A=B Dept)
71
n 2 log n

d:
3 8
3 7
5 7
6 7
d(Y,P)
r(Y,Z,U)
r:
3 8 9
9 3 8
8 3 8
3 8 4
3 8 3
8 9 4
9 4 7
d:
3 8
3 7
5 7
6 7
d(Y,P)
r(Y,Z,U)
r:
3 8 9
9 3 8
8 3 8
3 8 4
3 8 3
8 9 4
9 4 7
s:
3 8 9
9 3 8
8 3 8
3 8 4
3 8 3
8 9 4
9 4 7
s(Z,U,W)
t(V,Z)
t:
9 8
9 3
9 5
s:
3 8 9
9 3 8
8 3 8
3 8 4
3 8 3
8 9 4
9 4 7
s(Z,U,W)
t(V,Z)
t:
9 8
9 3
9 5
d:
3 8
3 7
5 7
6 7
d(Y,P)
r(Y,Z,U)
r:
3 8 9
9 3 8
8 3 8
3 8 4
3 8 3
8 9 4
9 4 7
d:
3 8
3 7
5 7
6 7
d(Y,P)
r(Y,Z,U)
r:
3 8 9
9 3 8
8 3 8
3 8 4
3 8 3
8 9 4
9 4 7
s:
3 8 9
9 3 8
8 3 8
3 8 4
3 8 3
8 9 4
9 4 7
s(Z,U,W)
t(V,Z)
t:
9 8
9 3
9 5
s:
3 8 9
9 3 8
8 3 8
3 8 4
3 8 3
8 9 4
9 4 7
s(Z,U,W)
t(V,Z)
t:
9 8
9 3
9 5

d:
3 8
3 7
5 7
6 7
d(Y,P)
r(Y,Z,U)
r:
3 8 9
9 3 8
8 3 8
3 8 4
3 8 3
8 9 4
9 4 7
…
d:
3 8
3 7
5 7
6 7
d(Y,P)
r(Y,Z,U)
r:
3 8 9
9 3 8
8 3 8
3 8 4
3 8 3
8 9 4
9 4 7
…
s:
3 8 9
9 3 8
8 3 8
3 8 4
3 8 3
8 9 4
9 4 7
s(Z,U,W)
t(V,Z)
t:
9 8
9 3
9 5
s:
3 8 9
9 3 8
8 3 8
3 8 4
3 8 3
8 9 4
9 4 7
s(Z,U,W)
t(V,Z)
t:
9 8
9 3
9 5
d:
3 8
3 7
5 7
6 7
d(Y,P)
r(Y,Z,U)
r:
3 8 9
9 3 8
8 3 8
3 8 4
3 8 3
8 9 4
9 4 7
d:
3 8
3 7
5 7
6 7
d(Y,P)
r(Y,Z,U)
r:
3 8 9
9 3 8
8 3 8
3 8 4
3 8 3
8 9 4
9 4 7
s:
3 8 9
9 3 8
8 3 8
3 8 4
3 8 3
8 9 4
9 4 7
s(Z,U,W)
t(V,Z)
t:
9 8
9 3
9 5
s:
3 8 9
9 3 8
8 3 8
3 8 4
3 8 3
8 9 4
9 4 7
s(Z,U,W)
t(V,Z)
t:
9 8
9 3
9 5

s:
3 8 9
9 3 8
8 3 8
3 8 4
3 8 3
8 9 4
9 4 7
d:
3 8
3 7
5 7
6 7
s(Z,U,W)
d(Y,P)
r(Y,Z,U)
r:
t(V,Z)
3 8 9
9 3 8
8 3 8
3 8 4
3 8 3
8 9 4
9 4 7
t:
9 8
9 3
9 5
s:
3 8 9
9 3 8
8 3 8
3 8 4
3 8 3
8 9 4
9 4 7
d:
3 8
3 7
5 7
6 7
s(Z,U,W)
d(Y,P)
r(Y,Z,U)
r:
t(V,Z)
3 8 9
9 3 8
8 3 8
3 8 4
3 8 3
8 9 4
9 4 7
…
t:
9 8
9 3
9 5
d:
3 8
3 7
5 7
6 7
d(Y,P)
r(Y,Z,U)
r:
3 8 9
9 3 8
8 3 8
3 8 4
3 8 3
8 9 4
9 4 7
d:
3 8
3 7
5 7
6 7
d(Y,P)
r(Y,Z,U)
r:
3 8 9
9 3 8
8 3 8
3 8 4
3 8 3
8 9 4
9 4 7
s:
3 8 9
9 3 8
8 3 8
3 8 4
3 8 3
8 9 4
9 4 7
s(Z,U,W)
t(V,Z)
t:
9 8
9 3
9 5
s:
3 8 9
9 3 8
8 3 8
3 8 4
3 8 3
8 9 4
9 4 7
s(Z,U,W)
t(V,Z)
t:
9 8
9 3
9 5

d:
3 8
3 7
5 7
6 7
d(Y,P)
r(Y,Z,U)
r:
3 8 9
9 3 8
8 3 8
3 8 4
3 8 3
8 9 4
9 4 7
d:
3 8
3 7
5 7
6 7
d(Y,P)
r(Y,Z,U)
r:
3 8 9
9 3 8
8 3 8
3 8 4
3 8 3
8 9 4
9 4 7
s:
…
3 8 9
9 3 8
8 3 8
3 8 4
3 8 3
8 9 4
9 4 7
s(Z,U,W)
t(V,Z)
t:
9 8
9 3
9 5
s:
3 8 9
9 3 8
8 3 8
3 8 4
3 8 3
8 9 4
9 4 7
s(Z,U,W)
t(V,Z)
t:
9 8
9 3
9 5
d:
3 8
3 7
5 7
6 7
d(Y,P)
r(Y,Z,U)
r:
3 8 9
9 3 8
8 3 8
3 8 4
3 8 3
8 9 4
9 4 7
…
…
d:
3 8
3 7
5 7
6 7
d(Y,P)
r(Y,Z,U)
r:
3 8 9
9 3 8
8 3 8
3 8 4
3 8 3
8 9 4
9 4 7
s:
3 8 9
9 3 8
8 3 8
3 8 4
3 8 3
8 9 4
9 4 7
s(Z,U,W)
t(V,Z)
t:
9 8
9 3
9 5
s:
3 8 9
9 3 8
8 3 8
3 8 4
3 8 3
8 9 4
9 4 7
s(Z,U,W)
t(V,Z)
t:
9 8
9 3
9 5

…
d:
3 8
3 7
5 7
6 7
d(Y,P)
r(Y,Z,U)
r:
3 8 9
9 3 8
8 3 8
3 8 4
3 8 3
8 9 4
9 4 7
d:
3 8
3 7
5 7
6 7
d(Y,P)
r(Y,Z,U)
r:
3 8 9
9 3 8
8 3 8
3 8 4
3 8 3
8 9 4
9 4 7
s:
3 8 9
9 3 8
8 3 8
3 8 4
3 8 3
8 9 4
9 4 7
s(Z,U,W)
t(V,Z)
t:
9 8
9 3
9 5
s:
3 8 9
9 3 8
8 3 8
3 8 4
3 8 3
8 9 4
9 4 7
s(Z,U,W)
t(V,Z)
t:
9 8
9 3
9 5
s:
3 8 9
9 3 8
8 3 8
3 8 4
3 8 3
8 9 4
9 4 7
d:
3 8
3 7
5 7
6 7
s(Z,U,W)
d(Y,P)
r(Y,Z,U)
r:
t(V,Z)
3 8 9
9 3 8
8 3 8
3 8 4
3 8 3
8 9 4
9 4 7
t:
9 8
9 3
9 5
A solution: Y=3, P=7, Z=8, U=9, W=4, V=9
Computing the result
• The result size can be exponential (even in case of ACQs).
• Even when the result is of polynomial size, it is in general
hard to compute.
• In case of acyclic queries, the result can be computed in time
polynomial in the result size (i.e., in output-polynomial time).
• This will remain true for the subsequent generalizations of ACQs.
• The result of ACQs can be computed by adding a top-down
phase to Yannakakis’ algorithm for ABCQs and by joining the
partial results.

Precise complexity of acyclic Boolean queries
•CanbedonebetterParallelalgorithm
Theorem:AnsweringacyclicBCQs isLOGCFL
complete
AC0 NC1 L NL LOGCFL NC2
Theorem [GLS99]: Acyclic CSP-solvability is LOGCFL-complete.
Answering acyclic BCQs is LOGCFL-complete
LOGCFL: class of problems/languages that are logspace-reducible to a CFL
AC 0 NL
LOGCFL SAC1
AC1
NC
2
NC AC P NP
LOCFL:Classofallproblemsthatare
logspacereducibletoacontextfree
language.
93
Characterization of LOGCFL [Ruzzo80]:
LOGCFL = Class of all problems solvable with a logspace ATM
with polynomial tree-size
ABCQ is in LOGCFL
ABCQ is in LOGCFL
p(x,y)
p(x,y)
p
…
q(x,z)
h(y,v,w)
q(x,z)
…
q
h(y,v,w)
…
h
q(z,t) p(y,v) q(v,w) p(w,u)
q(z,t)
q
p(y,v) q(v,w) p(w,u)
…
p
q
p
…
…
…

p(x,y)
q
q(x,z)
…
ABCQ is in LOGCFL
p
…
h
h(y,v,w)
…
q(z,t)
q
p(y,v) q(v,w) p(w,u)
…
p
q
p
…
…
…
Is this query hard
ans a(
S,
X , X ', C,
F)
b(
S,
Y,
Y ', C',
F')
c(
C,
C',
Z)
d(
X , Z)
e(
Y,
Z)
f ( F,
F',
Z')
g(
X ', Z')
h(
Y ', Z')
j(
J,
X , Y,
X ', Y ') p(
B,
X ', F)
q(
B',
X ', F)
n size of the database
m number of atoms in the query m = 11 !
• Classical methods worst-case complexity: O(n m )
• Despite its apparence, this query is nearly acyclic
It can be evaluated in O(m·n 2· logn)
HardandEasyInstancesand • ThegeneralproblemisNPHard
• Techniquesforidentifyingtractablecases
• Twomainkindoftechniques:
– Basedonthevaluesinthedatabaserelations
e.g., g, [FederandVardi,1993]
[Jeavons,Cohen,andGyssens, 1997]
– Basedonthestructureofthequery
e.g., g, [Freuder,1985],
[DechterandPearl,1989],
[Gyssens,Jeavons,andCohen,1994],
[Gottlob,Leone,andS.,2002]
Nearly Acyclic Queries
Inpractice,manyqueriesareacyclic,
Butmanyothersare“nearlyacyclic”.
Canwemakethisconceptprecise
Canwedevelopefficientalgorithmsfor
Nearlyacyclicqueries
99

ans a(
S,
X , X ', C,
F)
b(
S,
Y,
Y ', C',
F')
c(
C,
C',
Z)
d(
X , Z)
e(
Y,
Z)
f ( F,
F',
Z')
g(
X ', Z')
h(
Y ', Z')
j(
J,
X , Y,
X ', Y ') p(
B,
X ', F)
q(
B',
X ', F)
B B’
X X’ C F
J Z Z’
S
Y Y’ C’ F’
Nearly Acyclic Queries & CSPs
• Bounded Treewidth (tw)
– a measure of the cyclicity of graphs
– for queries: tw(Q) = tw(G(Q))
• For fixed k:
– checking tw(Q) k
– Computing a tree decomposition
linear time
(Bodlaender’96)
Deciding CSP of treewidth k:
It can be evaluated in O(m·n 2· logn)
O(n k log n) [Dechter; Chekuri & Rajaraman’97; Kolaitis & Vardi, 98]
LOGCFL-complete (G.L.S.’98)
Primal graphs of
CSPs/Queries
ans Enrolled(S,C,R) Teaches(P,C,A) Parent(P,S)
S
R
C
A
Hypergraph H(Q)
P
R
S
C
P
A
Primal graph G(Q)
FurtherGraphRepresentations
S
X1
2
S 1
Decther, 92
X 2
S 2
Dual Graph
X 2 S 3
X 1
X 3 X 2
S
S 4 1
X 2 X 4 X 1
S
S 1
3
X 2
S
X 2
5 X S4
3
Seidel, 81 S 3
X 4
S 4
Incidence Graph
X 5
(Hidden variable encoding)

NearlyAcyclicGraphs
y Example: a cyclic graph
q
g
a
b
c
e
d
f
p
k
l
o
h
i
m
j
n
A tree decomposition of width 2
Connectedness condition for h
ah
ah
ahq
hkl
ahq
hkl
hij
abc
hkp
klo
hij
abc
hkp
klo
mno
mno
ag
cef
bcd
ag
cef
bcd

Logical characterization of Treewidth
LogicL
: FO based on ,
( ,
,
disallowed )
k
L FO ,
Basic Querying Logic
TW[k]
k 1
L
L 1
[Kolaitis & Vardi ’98]
: L with at most k
1
vars.
LogicalcharacterizationofTreewidth
A generalization:
NRS- DATALOG - TW[
k
]
FO
(Flum, Frick, and Grohe ‘01)
k1
Game characterization of Treewidth
• A robber and k cops play the game on a graph
• The cops have to capture the robber
• Each cop controls a vertex of the graph
• Each cop, at any time, can fly to any vertex of the
graph
• The robber tries to elude her capture, by running
arbitrarily fast on the vertices of the graph,
but on those vertices controlled by cops
Playing the game
q
b
a
d
g
c
f
e
p
h
l
k
j
i
o
n
m

Playing the game
Playing the game
q
g
a
b
c
e
d
f
q
g
a
b
c
e
d
f
p
k
l
o
h
i
m
j
n
p
k
l
o
h
i
m
j
n
Playing the game
Whenistheevaluationofconjunctive
queriestractable
q
g
p
k
l
a
o
b
c
h
i
m
e
d
j
n
f
Incasetheyarecharacterizedbygraphs
e.g.,throughprimalorGaifmangraphs:
or graphs:
G
:
class of
graphs
Q(
G
)
:
all
queries characterized by graphs inG
G
Q( G
)
tractablet iff
G
unless P = W[1] or other collapses occur
(Grohe, Schwentick, and Segoufin, ’01)
has
bounded
dtreewidth

Hypergraphs vs Graphs (1)
Hypergraphs vs Graphs (2)
C’
R
C
A
R
C’
C
A
C’
R
C
A
R
C’
C
A
S
P
S
P
S
P
S
P
An acyclic hypergraph
Its cyclic primal graph
There are two cliques.
We cannot know where they come from
Drawbacks of treewidth
Acyclic queries may have unbounded TW!
Example:
q p 1 (X 1 , X 2 ,…, X n ,Y 1 ) … p n (X 1 , X 2 ,…, X n ,Y n )
is acyclic, obviously polynomial,
but has treewidth n-1
Acounterexample
• LetQQ 1 betheclassofconjunctivequeriesconsistingofa
conjunctive consisting a
singleatom(withanarbitrarynumberofvariables)
• EvaluatinganyqueryinQ 1 istrivial
• LetGG 1 bethesetoftheprimalgraphsofqueriesinQ in Q 1
• G 1 isthesetofallcliques
• But…Q(G 1 ) Q 1 ,e.g.,Q(G 1 )containsCLIQUE m :
• Thus,ingeneral,classesofqueriescannotbedefined
, fromtheclassesoftheirprimalgraphs
• Forgeneral(nonbinary)queriestheabovetheoremdoes
nothelp
• Boundedtreewidthisnotnecessaryfortractability

Beyond treewidth
Bounded Degree of Cyclicity (Hinges)
[Gyssens & Paredaens ’84, Gyssens, Jeavons, Cohen’94]
Does not generalize bounded treewidth.
Query Decomposition
q p 1 (X 1 , X 2 ,…, Y m ) … p m (X 1 , X 2 ,…, Y n )
Query width = 1 = acyclicity
Bounded Query width
[Chekuri & Rajaraman ’97]
Group together query atoms
(hyperedges) instead of variables
p 1 (X 1 ,..., X n ,Y 1 )
p 2 (X 1 ,..., X n ,Y 2 ) p 3 (X 1 ,..., X n , Y 3 ) p n (X 1 ,..., X 2 , Y n )
• Every atom/hyperarc appears in some node
• Connectedness conditions for variables and atoms
Decomposition of cyclic queries
q s(Y,Z,U) g(X,Y) t(Z,X) s(Z,W,X) t(Y,Z)
Transform a query of bounded width into
an acyclic query over a modified database
q s(Y,Z,U) g(X,Y) t(Z,X) s(Z,W,X) t(Y,Z)
Query width = 2
g(X,Y), t(Y,Z)
g(X,Y), t(Y,Z)
gt(X,Y,Z)
t(Z,X)
s(Y,Z,U)
t(Z,X)
s(Y,Z,U)
t(Z,X)
s(Y,Z,U)
s(Z,W,X)
s(Z,W,X)
s(Z,W,X)
Relations:
Relations:
BCQ is polynomial for queries of bounded
query width, if a query decomposition is given
g t s
t
gt= g t
s

Problems by Chekuri &
Rajaraman ‘97
A negative answer (G.L.S. ’99)
Are the following problems solvable in
polynomial time for fixed k
– Decide whether Q has query width at most k
– Compute a query decomposition of Q of
width k
Theorem:
Proof:
Deciding whether a query has
query width at most k is
NP-complete
Very involved reduction from
EXACT COVERING BY 3-SETS
Important Observation
Important Observation
NP-hardness id due to an overly strong condition
in the definition of query decomposition
But the reuse of p(X,Y,Z) is harmless here:
we can have added an atom p(X,Y,Z’) without changing the query
p(X,Y,Z), q(U,V,Z)
a(X,U,W), b(Y,V,W)
p(X,Y,Z), q(U,V,Z)
a(X,U,W), b(Y,V,W)
Forbidden !
p(X,Y,Z), c(T,W)
p(X,Y,Z’), p(X,Y,Z), c(T,W)
d(X,T)
c(Y,T)
d(X,T)
c(Y,T)

Hypertree Decompositions
Grouping and Reusing Atoms
Query atoms can be used “partially”
as long as the full atom appears
somewhere else
We group atoms
We use p(X,Y,Z) partially
p(X,Y,Z), q(U,V,Z)
a(X,U,W), b(Y,V,W)
p(X,Y,_), c(T,W)
More liberal than query decomposition
d(X,T)
c(Y,T)
Reusing atoms
ans a(
S,
X , X ', C,
F)
b(
S,
Y,
Y ', C',
F')
c(
C,
C',
Z)
d(
X , Z)
e(
Y,
Z)
f ( F,
F',
Z')
g(
X ', Z')
h(
Y ', Z')
j(
J,
X , Y,
X ', Y ') p(
B,
X ', F)
q(
B',
X ', F)
p(X,Y,Z), q(U,V,Z)
a(X,U,W), b(Y,V,W)
j(J,X,Y,X’,Y’)
a(S,X,X’,C,F), b(S,Y,Y’,C’,F’)
We use p(X,Y,Z) partially
p(X,Y,_), c(T,W)
j(_,X,Y,_,_), c(C,C’,Z)
j(_,_,_,X’,Y’), f(F,F’,Z’)
d(X,T)
c(Y,T)
d(X,Z) e(Y,Z) g(X’,Z’), f(F,_,Z’) h(Y’,Z’)
p(B,X’,F)
q(B’,X’,F)
Hypertree of width 2

Generalized Hypertree decomposition
= Hypertree + Connectedness condition
j(_,X,Y,_,_), c(C,C’,Z)
j(J,X,Y,X’,Y’)
a(S,X,X’,C,F), b(S,Y,Y’,C’,F’)
j(_,_,_,X’,Y’), f(F,F’,Z’)
d(X,Z) e(Y,Z) g(X’,Z’), f(F,_,Z’) h(Y’,Z’)
Alternative view and
generalized HT-Decompositions:
GHD = tree decomp + set covering
p(B,X’,F)
q(B’,X’,F)
Tree decomposition of hypergraph
11
H
h8
h1
1 2
h15
h10 h2 h3
12 13 19 4 h7
h9
h6
h11
17 7 5 h4
16
18
14 15 8 9 6
h14 h12 h13
h5 10
3
Tree decomp of G(H)
1,2,3,4,5,6
5,6,7,8,9
1,11,17,19
11,12,17,18,19
3,4,5,6,7,8 12,16,17,18,19
7,9,10
12,15,16,18,19
12,13,14,15,18,19
12
h9
h14
18
11
Hypergraph and the tree
decomposition
h8
h1
1 2
h15
h10 h2 h3
13 19 4
h6 h7
h11
17 7 5 h4
16
14 15 8 9 6
h5
h12 h13 10
3
1,2,3,4,5,6
5,6,7,8,9
1,11,17,19
11,12,17,18,19
3,4,5,6,7,8 12,16,17,18,19
7,9,10
12,15,16,18,19
12,13,14,15,18,19

Hypergraph and the tree
decomposition
Hypergraph and the tree
decomposition
11
h8
h1
1 2
3
1,11,17,19
11
h8
h1
1 2
3
1,11,17,19
12
h9
h14
18
h15
h10 h2 h3
13 19 4
h6 h7
h11
17 7 5 h4
16
14 15 8 9 6
h12
h5
h13 10
1,2,3,4,5,6 11,12,17,18,19
3,4,5,6,7,8 12,16,17,18,19
5,6,7,8,9
7,9,10
12,15,16,18,19
12,13,14,15,18,19
12
h9
h14
18
h15
h10 h2 h3
13 19 4
h6 h7
h11
17 7 5 h4
16
14 15 8 9 6
h12
h5
h13 10
1,2,3,4,5,6 11,12,17,18,19
3,4,5,6,7,8 12,16,17,18,19
5,6,7,8,9
7,9,10
12,15,16,18,19
12,13,14,15,18,19
Hypergraph and the tree
decomposition
Hypergraph and the tree
decomposition
11
h8
h1
1 2
3
1,11,17,19
11
h8
h1
1 2
3
1,11,17,19
12
h9
h14
18
h15
h10 h2 h3
13 19 4
h6 h7
h11
17 7 5 h4
16
14 15 8 9 6
h5
h12 h13 10
1,2,3,4,5,6 11,12,17,18,19
3,4,5,6,7,8 12,16,17,18,19
5,6,7,8,9
7,9,10
12,15,16,18,19
12,13,14,15,18,19
12
h9
h14
18
h15
h10 h2 h3
13 19 4
h6 h7
h11
17 7 5 h4
16
14 15 8 9 6
h5
h12 h13 10
1,2,3,4,5,6 11,12,17,18,19
3,4,5,6,7,8 12,16,17,18,19
5,6,7,8,9
7,9,10
12,15,16,18,19
12,13,14,15,18,19

Hypergraph and the tree
decomposition
Hypergraph and the tree
decomposition
11
h8
h1
1 2
3
1,11,17,19
11
h8
h1
1 2
3
1,11,17,19
12
h9
h14
18
h15
h10 h2 h3
13 19 4
h6 h7
h11
17 7 5 h4
16
14 15 8 9 6
h12
h5
h13 10
1,2,3,4,5,6 11,12,17,18,19
3,4,5,6,7,8 12,16,17,18,19
5,6,7,8,9
7,9,10
12,15,16,18,19
12,13,14,15,18,19
12
h9
h14
18
h15
h10 h2 h3
13 19 4
h6 h7
h11
17 7 5 h4
16
14 15 8 9 6
h12
h5
h13 10
1,2,3,4,5,6 11,12,17,18,19
3,4,5,6,7,8 12,16,17,18,19
5,6,7,8,9
7,9,10
12,15,16,18,19
12,13,14,15,18,19
Hypergraph and the tree
decomposition
Generalized hypertree
decomposition
11
h8
h1
1 2
3
1,11,17,19
11
h8
h1
1 2
3
h8(1,11), h15(1,17,19)
12
h9
h14
18
h15
h10 h2 h3
13 19 4
h6 h7
h11
17 7 5 h4
16
14 15 8 9 6
h5
h12 h13 10
1,2,3,4,5,6 11,12,17,18,19
3,4,5,6,7,8 12,16,17,18,19
5,6,7,8,9
7,9,10
12,15,16,18,19
12,13,14,15,18,19
12
h9
h14
18
h15
h10 h2
13 19 4
h6
h11
17 7
16
14 15 8 9
h12 h13
h5
h3
5
h7
6
10
h4
h1(1,2,3), h2(1,4,5,6)
h2(_,4,5,6), h3(3,4,7,8)
h4(5,7), h5(6,8,9)
h6(7,9,10)
h9(11,12,18), h15(_,17,19)
h10(12,_,19), h14(16,17,18)
h9(_,12,18), h13(15,16,19)
h10(12,13,19), h12(14,15,18)
Generalized hypetree decomposition of width 2

QUESTION:
Hypertree Decomposition
= Generalized HTD +Special Condition
Can we determine in polynomial time
Whether ghw(H) < k for constant k
j(J,X,Y,X’,Y’)
a(S,X,X’,C,F), b(S,Y,Y’,C’,F’)
Each variable
that disappeared
at some vertex v
Theorem: ghw(H) < 4 NP-complete
j(_,X,Y,_,_), c(C,C’,Z)
j(_,_,_,X’,Y’), JXY f(F,F’,Z’)
[G., Miklos, Schwentick et. al. 06]
d(X,Z) e(Y,Z) g(X’,Z’), f(F,_,Z’) h(Y’,Z’)
Does not reappear in
the subtrees rooted
at v
p(B,X’,F)
q(B’,X’,F)
Special Condition
Generalized hypertree
decomposition
j(_,X,Y,_,_), c(C,C’,Z)
j(J,X,Y,X’,Y’)
a(S,X,X’,C,F), b(S,Y,Y’,C’,F’)
j(_,_,_,X’,Y’), JXY f(F,F’,Z’)
d(X,Z) e(Y,Z) g(X’,Z’), f(F,_,Z’) h(Y’,Z’)
Does not appear in
the subtrees rooted
at v
p(B,X’,F)
q(B’,X’,F)
Each variable
that disappeared
at some vertex v
11
h8
h1
1 2
h15
h10 h2 h3
12 13 19 4 h7
h9
h6
h11
17 7 5 h4
16
18
8 9 6
14 15
h14 h12 h13
h5 10
3
Special condition violated
h8(1,11), h15(1,17,19)
h1(1,2,3), h2(1,4,5,6) h9(11,12,18), h15(_,17,19)
h2(_,4,5,6), h3(3,4,7,8) h10(12,_,19), h14(16,17,18)
h4(5,7), h5(6,8,9) h9(_,12,18), h13(15,16,19)
h6(7,9,10)
h10(12,13,19), h12(14,15,18)
Generalized hypetree decomposition of width 2

Hypertree decomposition
Hypertree decomposition
12
h9
h14
18
11
h8
h1
1 2
h15
h10 h2 h3
13 19 4
h6
h7
h11
17 7 5 h4
16
8 9 6
14 15
h5
h12 h13 10
3
h10(12,13,19), h12(14,15,18), h14(16,17,18)
h9(11,12,18), h15(1,17,19)
h2(1,4,5,6), h3(3,4,7,8)
h4(5,7), h5(6,8,9) h1(1,2,3)
h6(7,9,10)
12
h9
h14
18
11
h8
h1
1 2
h15
h10 h2 h3
13 19 4
h6 h7
h11
17 7 5 h4
16
14 15 8 9 6
h5
h12 h13 10
3
h10(12,13,19), h12(14,15,18), h14(16,17,18)
h9(11,12,18), h15(1,17,19)
h2(1,4,5,6), h3(3,4,7,8)
h4(5,7), h5(6,8,9)
h6(7,9,10)
h1(1,2,3)
Hypertee decomposition of width 3
Hypertee decomposition of width 3
Hypertree decomposition
Hypertree decomposition
12
h9
h14
18
11
h8
h1
1 2
h15
h10 h2 h3
13 19 4
h6 h7
h11
17 7 5 h4
16
14 15 8 9 6
h5
h12 h13 10
3
h10(12,13,19), h12(14,15,18), h14(16,17,18)
h9(11,12,18), h15(1,17,19)
h2(1,4,5,6), h3(3,4,7,8)
h4(5,7), h5(6,8,9)
h6(7,9,10)
h1(1,2,3)
12
h9
h14
18
11
h8
h1
1 2
h15
h10 h2 h3
13 19 4
h6 h7
h11
17 7 5 h4
16
14 15 8 9 6
h5
h12 h13 10
3
h10(12,13,19), h12(14,15,18), h14(16,17,18)
h9(11,12,18), h15(1,17,19)
h2(1,4,5,6), h3(3,4,7,8)
h4(5,7), h5(6,8,9)
h6(7,9,10)
h1(1,2,3)
Hypertee decomposition of width 3
Hypertee decomposition of width 3

Hypertree decomposition
Hypertree decomposition
12
h9
h14
18
11
h8
h1
1 2
h15
h10 h2 h3
13 19 4
h6 h7
h11
17 7 5 h4
16
14 15 8 9 6
h5
h12 h13 10
3
h10(12,13,19), h12(14,15,18), h14(16,17,18)
h9(11,12,18), h15(1,17,19)
h2(1,4,5,6), h3(3,4,7,8)
h4(5,7), h5(6,8,9)
h6(7,9,10)
h1(1,2,3)
12
h9
h14
18
11
h8
h1
1 2
h15
h10 h2 h3
13 19 4
h6 h7
h11
17 7 5 h4
16
14 15 8 9 6
h5
h12 h13 10
3
h10(12,13,19), h12(14,15,18), h14(16,17,18)
h9(11,12,18), h15(1,17,19)
h2(1,4,5,6), h3(3,4,7,8)
h4(5,7), h5(6,8,9)
h6(7,9,10)
h1(1,2,3)
Hypertee decomposition of width 3
Hypertee decomposition of width 3
Hypertree decomposition
Positive Results on
Hypertree Decompositions
12
h9
h14
18
11
h8
h1
1 2
h15
h10 h2 h3
13 19 4
h6 h7
h11
17 7 5 h4
16
14 15 8 9 6
h5
h12 h13 10
3
h10(12,13,19), h12(14,15,18), h14(16,17,18)
h9(11,12,18), h15(1,17,19)
h2(1,4,5,6), h3(3,4,7,8)
h4(5,7), h5(6,8,9)
h6(7,9,10)
h1(1,2,3)
Hypertee decomposition of width 3
• For each query Q, hw(Q) qw(Q)
• In some cases, hw(Q) < qw(Q)
• For fixed k, deciding whether
hw(Q) k is in polynomial time (LOGCFL)
• Computing hypertree decompositions is
feasible in polynomial time (for fixed k).
But: FP-intractable wrt k: W[2]-hard.

Evaluating CSP’s having bounded
(gen.) hypertree width
k fixed
Given:
a database db of relations
a CSP Q over db such that hw(Q) k or ghw k
a width k hypertree decomposition of Q
• Deciding whether (Q,db) solvable is in
O(n k+1 log n) and complete for LOGCFL
• Computing Q(db) is feasible in
output-polynomial time
Observation: If H has n vertices, then
HW(H)n/2+1
Does not hold for TW: TW(K n )=n-1
Often HW < TW. H-Decomps are interesting
In case of bounded arity, too.
ComparingDecompositionMethods
We compare these methods according to
their ability in indentifying tractable classes of
CSP instances:
Comparison results
Hypertree Decomposition
• D1 D2 (D2generalizes D1):
0suchthat, k>0,C(D1,k) C(D2,k+)
everyclassofCSPstractableaccordingtoD1istractableaccordingtoD2,
according to is according to D2
with(almost)thesamecost.
• D2
D1 (D2beats D1):thereexistsanintegerksuchthat
exists integer such that
m,C(D2,k) C(D1,m)
thereareclassesofCSPstractableaccordingtoD2butnottractable
accordingtoD1.
• D1

Relationship GHW vs. HW:
Observation:
ghw(H) = hw(H*)
where H* = H {E´| E in edges(H): E´ E}
Exponential!
Q: Are there other
approximations to
GHW
Hypertree Decomposition
Hinge Decomposition
+
Tree Clustering
GHD
Cycle Hypercutset
Approximation Theorem [Adler,G.,Grohe 05] :
Hinge
Decomposition
Tree Clustering
w* treewidth
ghw(H)

Open Question:
Are there approximations
of GHW better than HW
GHD
k
Question:
Are all tractable classes
of CSPs of bounded GHW
GHD
Can we get closer
Explore this area!
Hypertree Decomposition 3k+1 Hypertree Decomposition
Hinge Decomposition
+
Tree Clustering
Cycle Hypercutset
Hinge Decomposition
+
Tree Clustering
Cycle Hypercutset
Hinge
Decomposition
Tree Clustering
w* treewidth
Hinge
Decomposition
Tree Clustering
w* treewidth
Biconnected Components
Cycle Cutset
Biconnected Components
Cycle Cutset
No!
Fractional HT-Decompositions
[Grohe,Marx ’05]
Fractional HD +
GHD
Open Question:
What is the best
“tractable” measure
of hypergraph cyclicity
Explore this area!
GHD
Hypertree Decomposition
[Chen&Dalmau,…]
Hypertree Decomposition
Hinge Decomposition
+
Tree Clustering
Cycle Hypercutset
Hinge Decomposition
+
Tree Clustering
Cycle Hypercutset
Hinge
Decomposition
Tree Clustering
w* treewidth
Hinge
Decomposition
Tree Clustering
w* treewidth
Biconnected Components
Cycle Cutset
Biconnected Components
Cycle Cutset