The complexity of Sur-Hom(C 4 )

Transcription

1 The complexity of Sur-Hom(C 4 ) Barnaby Martin and Danïel Paulusma Algorithms and Complexity in Durham, 7th Slovenian International Conference on Graph Theory Bled 11 25th June, 2011

2 My introduction to the problem. For some finite and relational signature σ, we consider only finite σ-structures A, B etc. A homomorphism from A to B is a function h : A B such that, for all R σ of arity i, if R(a 1,..., a i ) A then R(h(a 1 ),..., h(a i )) B. The homomorphism problem Hom(B) takes as input some A and asks whether there is a homomorphism from A to the fixed template B (denoted A B)? Hom(B) is also known as CSP(B); a constraint satisfaction problem.

3 Hom(B) is always in NP and it is conjectured to possess dichotomy between P and NP-complete. However, quite a lot is known about Hom(H), for instance, when H is a symmetric digraph (graph possibly with loops). The following is Hell and Nešetřil (1990), augmented with Reingold (2005). Theorem (Hell and Nešetřil 1990) If H has a self-loop or is bipartite, Hom(H) is in L. Otherwise Hom(H) is NP-complete.

4 The most important problem for me ancestrally is list homomorphism. The problem List-Hom(B) takes as input some A together with, for each a A, lists l a B and asks whether there is a homomorphism h from A to B s.t. h(a) l a. Much is known about the complexity of List-Hom(B). classification for (irreflexive) graphs graphs by Feder, Hell and Huang (1999). classification for reflexive graphs graphs by Feder and Hell (1998). full classification by Bulatov (2003).

5 Since, trivially, Hom(H) L List-Hom(H), in order to classify List-Hom(H) for graphs it is necessary only to classify List-Hom(H) for bipartite H. Theorem (Feder, Hell and Huang 1999) Let H be a bipartite graph. If the complement of H is a circular arc graph then List-Hom(H) is in P. Otherwise it is NP-complete. The smallest bipartite graph that is not the complement of a circular arc graph is C 6...

6 Theorem (Feder and Hell 1998) Let H be a reflexive graph. If H is an interval graph then List-Hom(H) is in P. Otherwise it is NP-complete. The smallest reflexive graph s.t. List-Hom(H) is NP-complete turns out to be C 4...

7 We introduce two further important problems. The retraction problem Ret(B) takes as input some A, with B an induced substructure of A, and asks whether there is a homomorphism h : A B s.t. h is the identity on B. The problem Ret(B) is logspace equivalent with the problem Hom(B c ), where B c is B expanded with all constants.

8 The compaction problem Comp(H) takes as input some G and asks whether there is a surjective homomorphism h : G H s.t. h is edge-surjective modulo self-loops. 1 The definition of compaction is horrible! Blame the graph-theorists! Except for the self-loops it is quite similar to the modern notion of homomorphic image. 1 Formally: b 1, b 2 H s.t. b 1 b 2 and E(b 1, b 2) H, a 1, a 2 G s.t. E(a 1, a 2) G and h(a 1) = b 1 and h(a 2) = b 2.

9 Examples. There is a surjective homomorphism from the 3-path P 3 to the reflexive 4-cycle C 4 ; there is no compaction There is a surjective homomorphism from the 4-cycle C 4 to the reflexive 4-cycle C 4 ; this is also a compaction

10 It is not too hard to see Sur-Hom(B) Tur P Comp(B) Tur P Ret(B) L List-Hom(B). Sketch proof. For Sur-Hom(B) to Comp(B) consider all element-preimages of B in the input and enforce B on these. For Comp(B) to Ret(B) consider all edge-preimages of B that satisfy edge -surjectivity and identify elements as necessary and enforce B on them. The reduction from Ret(B) to List-Hom(B) is trivial. In fact we have already proved that Ret(C 6 ) and Ret(C 4 ) are NP-complete (using the same reductions as before).

11 Our understanding of these problems now appears to be Hom(C 6 ) L Sur-Hom(C 6 ) Tur P Comp(C 6 ) Tur P Ret(C 6 ) L List-Hom(C 6 ) P?? NP c NP c Hom(C 4 ) L Sur-Hom(C 4 ) Tur P Theorem (Vikas 2004) Comp(C 2k ) is NP-complete for k 3. Theorem (Vikas 2003) Comp(C k ) is NP-complete for k 4. Comp(C 4 ) Tur P Ret(C 4 ) L List-Hom(C 4 )

12 Comp(C 6 ) and Sur-Hom(C 6 ) coincide on inputs of diameter 4 Vikas s construction is of diameter 5. A similar story with C 4... Comp(C 4 ) and Sur-Hom(C 4 ) coincide on inputs of diameter 2 Vikas s construction is of diameter 3.

13 Another introduction to the problem. The Brazilian school. The recent history of Sur-Hom(C 4 ) probably begins with a paper in 2005 by Dantas, de Figueiredo, Gravier and Klein: Finding H-partitions efficiently. It appears as the only one in a family of problems whose complexity is unknown. 2 In this world, the problem is known as 2K 2 -partition: to partition the vertices of a graph into four nonempty classes A, B, C and D such that every vertex in A is adjacent to every vertex in B and every vertex in C is adjacent to every vertex in D. 2K 2 -partition is logspace-equivalent to Sur-Hom(C 4 ) under the reduction G complement(g). 2 Essentially this coincides with the fact that the complexity of Sur-Hom(H) is easy to classify for all H of size 4 except H := C 4.

14 Our Brazilian friends have written extensively on this problem (and its list variant) including. Theorem (Dantas, Maffray and Silva 2010) 2K 2 -partition is in P for K 4 -free graphs diamond-free graphs planar graphs graphs of bounded treewidth claw-free graphs (C 5, P 5 )-free graphs graphs with few P 4 s

15 In a recent paper of de Figueiredo, in press at Discrete Applied Mathematics, The P versus NP-complete dichotomy of some challenging problems in graph theory, such is the importance of this open problem, that the complexity class(!) 2K 2-hard is introduced.

16 In Europe, meanwhile, Sur-Hom(C 4 ) has recently appeared as disconnected cut. Theorem (Fleischner, Mujuni, Paulusma and Szeider 2009) Disconnected cut is in P for graphs of diameter 2 graphs of bounded maximum degree graphs not locally connected triangle-free graphs graphs with a dominating edge Theorem (Ito, Kaminski, Paulusma and Thilikos 2009) Disconnected cut is in P for apex-minor-free graphs. connected chordal graphs

17 Working around the problem Sur-Hom(C 4 ) my colleagues in ACiD proved instead: Theorem (Golovach, Paulusma and Song 2011) Let T be a partially reflexive tree. Then, if the vertices in T with a self-loop induce a subtree of T, Sur-Hom(T) is in P. Otherwise, it is NP-complete. I also mention the following result. Theorem (Creignou, Khanna and Sudan 2001) If all relations of B are from one among Horn, dual Horn, bijunctive or affine, Sur-Hom(B) is in P. Otherwise, it is NP-complete.

18 Sur-Hom(C 4 ) appears to be very close to being tractable. Any input of diameter 3 is a trivial yes-instance. If Ret(C 4 ) is in P for inputs of diameter 2, it follows that so is Sur-Hom(C 4 ). Now, Ret(C 4 ) on inputs of diameter 2 really is close to be encodeable as a 2-SAT (the standard method to prove that list homomorphism is tractable).

19 Sur-Hom(C 4 ) appears to be very close to being tractable. Any input of diameter 3 is a trivial yes-instance. If Ret(C 4 ) is in P for inputs of diameter 2, it follows that so is Sur-Hom(C 4 ). Now, Ret(C 4 ) on inputs of diameter 2 really is close to be encodeable as a 2-SAT (the standard method to prove that list homomorphism is tractable). Theorem (M. 2011) Ret(C 4 ) is NP-complete on inputs of diameter 2 (with a dominating non-edge).

20 p p a p b p a p bp q c q d q q dq h0 cq h 1 h 3 h 2 Figure: Possible evaluations of... Figure: H-gadget (q; red) and V-gadget (p; blue).

21 p p 3 1 a p 3 0 b p 2 1 q 3 1 c q 3 2 d q 0 1 q cq a p dq h0 bp h 3 h 1 h 2 Figure: Clause gadget for ( p q)... Figure: i.e. {(3, 1), (1, 3), (3, 3)}.

22 p p 3 1 a p 3 0 b p 2 1 a p bp q 3 1 c q 3 2 d q 0 1 q dq h0 cq h 1 h 3 h 2 Figure: Clause gadget for (p q)... Figure: i.e. {(1, 1), (3, 1), (1, 3)}.

23 In order to satisfy the diameter constraints, we join all a p (in different gadgets) in a big clique. We do likewise for b p, c q and d q. If we were interested in NLogspace-hardness, we would be done, by reduction from 2-SAT. The same reduction works for Ret(C 4 ) on inputs of diameter 2 as for Sur-Hom(C 4 ).

24 p p a p b p a p bp q c q d q Figure: Gadget for... q dq cq h 0 h 3 h 1 h 2 Figure: {(0, 3), (3, 3), (3, 1), (1, 1)}.

25 p p a p b p a p bp q c q d q Figure: Gadget for... q dq cq h 0 h 3 h 1 h 2 Figure: {(3, 3), (3, 1), (1, 0), (1, 1)}.

26 I now claim we are done, as D := ({3, 1, 0} : {(3, 1), (1, 3), (3, 3)}, {(1, 1), (3, 1), (1, 3)}, {(0, 3), (3, 3), (3, 1), (1, 1)}, {(3, 3), (3, 1), (1, 0), (1, 1)}) is such that Hom(D) is NP-complete, by: Theorem (Bulatov, Jeavons and Krokhin 2005) Let B be a finite core with A B, A = 2, a subset that as a unary relation belongs to B. If for each f Pol(B), f A is not majority, semilattice or Mal tsev, then Hom(B) is NP-complete.

27 Unfortunately, this reduction for Ret(C 4 ), on diameter 2 with dominating non-edge, does not immediately work for Sur-Hom(C 4 ): the degenerate cases no longer vanish. Instead we have to borrow most, but not quite all, of the construction of Vikas. x vv v v h 0 h 3 u v u v y v y v wv w v h 1 h 2