an np-complete example for the subpower membership problem

Transcription

1 an np-complete example for the subpower membership problem Graduate Algebra Seminar, CU Boulder Jeff Shriner December 5, 2017

2 the problem

3 the problem Fix a finite algebra A with fintely many operations smp(a) (Willard, 2007) Instance: A positive integer m and tuples a 1,, a n, b A m Question: Is b a 1,, a n? 2

4 the story so far

5 a picture T AYLOR CM CUBE CP CD

6 a picture T AYLOR CM CUBE CP CD More structure (easier?)

7 a picture T AYLOR CM CUBE CP CD Groups (Multilinear Exp) More structure (easier?)

8 a picture T AYLOR CM CUBE CP CD Groups (Multilinear Exp) NU (eg Lattices) More structure (easier?)

9 a picture T AYLOR CM CUBE CP Groups (Multilinear Exp) CD NU (eg Lattices) Hard: Semigroups P, NP-comp, PSPACE-comp [Steindl, Mayr, Bulatov] Ugly Algebra EXPTIME-comp [Kozik] More structure (easier?)

10 a picture T AYLOR CM CUBE CP CD Groups (Multilinear Exp) NU (eg Lattices) A Hard: Semigroups P, NP-comp, PSPACE-comp [Steindl, Mayr, Bulatov] Ugly Algebra EXPTIME-comp [Kozik] More structure (easier?)

11 a picture T AYLOR CM CUBE CP CD Groups (Multilinear Exp) NU (eg Lattices) A Hard: Semigroups P, NP-comp, PSPACE-comp [Steindl, Mayr, Bulatov] Ugly Algebra EXPTIME-comp [Kozik] More structure (easier?) Conjecture If A has a cube term, then SMP(A) is in P

12 a picture T AYLOR CM CUBE CP CD Groups (Multilinear Exp) Group Exp Exp p-grps Near Rings NU (eg Lattices) A Hard: Semigroups P, NP-comp, PSPACE-comp [Steindl, Mayr, Bulatov] Ugly Algebra EXPTIME-comp [Kozik] Res Small More structure (easier?) Conjecture If A has a cube term, then SMP(A) is in P 4

13 the result and construction

14 strong linear maltsev conditions and cube terms A strong linear Maltsev condition consists of 1 a finite set of operation symbols, H, and 2 a finite set of identities, Σ, over symbols from H such that no identity in Σ involves compositions 6

15 strong linear maltsev conditions and cube terms A strong linear Maltsev condition consists of 1 a finite set of operation symbols, H, and 2 a finite set of identities, Σ, over symbols from H such that no identity in Σ involves compositions We say Σ is consistent if it is satisfied by some non-trivial algebra 6

16 strong linear maltsev conditions and cube terms A strong linear Maltsev condition consists of 1 a finite set of operation symbols, H, and 2 a finite set of identities, Σ, over symbols from H such that no identity in Σ involves compositions We say Σ is consistent if it is satisfied by some non-trivial algebra We say Σ entails φ if every model of Σ is a model of φ 6

17 strong linear maltsev conditions and cube terms A strong linear Maltsev condition consists of 1 a finite set of operation symbols, H, and 2 a finite set of identities, Σ, over symbols from H such that no identity in Σ involves compositions We say Σ is consistent if it is satisfied by some non-trivial algebra We say Σ entails φ if every model of Σ is a model of φ Examples Having an m-cube term for some m 2 y c x 1,, x n y where x 1, x n {x, y} m \ (y,, y) 6

18 strong linear maltsev conditions and cube terms Examples Belonging to a Congruence 3-Distributive variety (CD(3)) d 1 (x, y, x) d 2 (x, y, x) x x d 1 (x, x, y), d 1 (x, y, y) d 2 (x, y, y), d 2 (x, x, y) y Belonging to a Congruence 3-Permutable variety (CP(3)) x p 1 (x, y, y), p 1 (x, x, y) p 2 (x, y, y), p 2 (x, x, y) y 7

19 strong linear maltsev conditions and cube terms Examples Belonging to a Congruence 3-Distributive variety (CD(3)) d 1 (x, y, x) d 2 (x, y, x) x x d 1 (x, x, y), d 1 (x, y, y) d 2 (x, y, y), d 2 (x, x, y) y Belonging to a Congruence 3-Permutable variety (CP(3)) x p 1 (x, y, y), p 1 (x, x, y) p 2 (x, y, y), p 2 (x, x, y) y Remark: Neither of these strong linear Maltsev conditions entail the existence of a cube term 7

20 strong linear maltsev conditions and cube terms Examples Belonging to a Congruence 3-Distributive variety (CD(3)) d 1 (x, y, x) d 2 (x, y, x) x x d 1 (x, x, y), d 1 (x, y, y) d 2 (x, y, y), d 2 (x, x, y) y Belonging to a Congruence 3-Permutable variety (CP(3)) x p 1 (x, y, y), p 1 (x, x, y) p 2 (x, y, y), p 2 (x, x, y) y Remark: Neither of these strong linear Maltsev conditions entail the existence of a cube term Theorem (Kearnes, Szendrei) Σ entails the existence of a cube term if and only if Σ entails that some h H is a cube term 7

21 the result Theorem Let M = (H, Σ) be a consistent strong linear Maltsev condition If Σ does not entail cube identities for any h H, then for any finite algebra A, there exists a finite algebra A M such that A M = Σ and SMP(A M ) is at least as hard as SMP(A) 8

22 the result Theorem Let M = (H, Σ) be a consistent strong linear Maltsev condition If Σ does not entail cube identities for any h H, then for any finite algebra A, there exists a finite algebra A M such that A M = Σ and SMP(A M ) is at least as hard as SMP(A) This is a statement which says we can add more structure to an algebra and the SMP will not get easier! 8

23 the result Theorem Let M = (H, Σ) be a consistent strong linear Maltsev condition If Σ does not entail cube identities for any h H, then for any finite algebra A, there exists a finite algebra A M such that A M = Σ and SMP(A M ) is at least as hard as SMP(A) This is a statement which says we can add more structure to an algebra and the SMP will not get easier! T AYLOR CM CUBE CP CD

24 the result Theorem Let M = (H, Σ) be a consistent strong linear Maltsev condition If Σ does not entail cube identities for any h H, then for any finite algebra A, there exists a finite algebra A M such that A M = Σ and SMP(A M ) is at least as hard as SMP(A) This is a statement which says we can add more structure to an algebra and the SMP will not get easier! T AYLOR CM CUBE CP CD A [Kozik]

25 the result Theorem Let M = (H, Σ) be a consistent strong linear Maltsev condition If Σ does not entail cube identities for any h H, then for any finite algebra A, there exists a finite algebra A M such that A M = Σ and SMP(A M ) is at least as hard as SMP(A) This is a statement which says we can add more structure to an algebra and the SMP will not get easier! T AYLOR CM CUBE CP CD AM A [Kozik] 8

26 construction Let A = A; F and M = (H, Σ) Set A M = A {0}; F H Extend operations in F to A {0} so that 0 is an absorbing element with respect to all f F Interpret operations in H to map to 0 whenever possible and still maintain A M = Σ 9

27 construction Let A = A; F and M = (H, Σ) Set A M = A {0}; F H Extend operations in F to A {0} so that 0 is an absorbing element with respect to all f F Interpret operations in H to map to 0 whenever possible and still maintain A M = Σ Example: CP(3) x p 1 (x, y, y), p 1 (x, x, y) p 2 (x, y, y), p 2 (x, x, y) y For (a 1, a 2, a 3 ) (A {0}) 3, let p 1 (a 1, a 2, a 3 ) = a 1 if a 2 = a 3 p 2 (a 1, a 2, a 3 ) = a 3 if a 1 = a 2 p i (a 1, a 2, a 3 ) = 0 otherwise 9

28 setup To show hardness, we reduce SMP(A) to SMP(A M ): a 1,, a n, b SMP(A) a 1,, a n, b SMP(A M ) 10

29 setup To show hardness, we reduce SMP(A) to SMP(A M ): a 1,, a n, b SMP(A) a 1,, a n, b SMP(A M ) Must show SMP(A) instance is yes if and only if SMP(A M ) instance is yes Easy Show that if b a 1,, a n AM, then there is a term p such that p(a 1,, a n ) = b which is h-free for all h H 10

30 proof sketch If b a 1,, a n AM, let p be a term such that p(a 1,, a n ) = b and has a minimum number of occurrences of operations from H in the term tree of p p

31 proof sketch If b a 1,, a n AM, let p be a term such that p(a 1,, a n ) = b and has a minimum number of occurrences of operations from H in the term tree of p p b a 1,, a n

32 proof sketch If b a 1,, a n AM, let p be a term such that p(a 1,, a n ) = b and has a minimum number of occurrences of operations from H in the term tree of p p b h a 1,, a n

33 proof sketch If b a 1,, a n AM, let p be a term such that p(a 1,, a n ) = b and has a minimum number of occurrences of operations from H in the term tree of p p b h c 1,, c l a 1,, a n 11

34 proof sketch If b a 1,, a n AM, let p be a term such that p(a 1,, a n ) = b and has a minimum number of occurrences of operations from H in the term tree of p p b h 0 in each coord! c 1,, c l a 1,, a n 11

35 proof sketch If b a 1,, a n AM, let p be a term such that p(a 1,, a n ) = b and has a minimum number of occurrences of operations from H in the term tree of p h c1,, cl = no 0 s p b h 0 in each coord! c 1,, c l a 1,, a n 11

36 proof sketch If b a 1,, a n AM, let p be a term such that p(a 1,, a n ) = b and has a minimum number of occurrences of operations from H in the term tree of p h c1,, cl = no 0 s p b h 0 in each coord! c 1,, c l In every row, there is an identity provable from Σ which forces the output to agree with at least one of the arguments a 1,, a n 11

37 proof sketch If b a 1,, a n AM, let p be a term such that p(a 1,, a n ) = b and has a minimum number of occurrences of operations from H in the term tree of p h c1,, cl = no 0 s p b h c i c 1,, c l a 1,, a n In every row, there is an identity provable from Σ which forces the output to agree with at least one of the arguments Since h cannot satisfy cube identities, we can show there is some 1 i l such that h(c 1,, c l ) = c i Replace the subterm with root h by the appropriate proper subterm 11

38 corollaries Corollary 1 (SMP for Congruence Distributive Varieties) 1 If k 2, then SMP(A) P for every finite algebra A in CD(k) 2 If k 3, then there exists a finite algebra A CD(k) such that SMP(A) is EXPTIME-complete 12

39 corollaries Corollary 1 (SMP for Congruence Distributive Varieties) 1 If k 2, then SMP(A) P for every finite algebra A in CD(k) 2 If k 3, then there exists a finite algebra A CD(k) such that SMP(A) is EXPTIME-complete Corollary 2 (SMP for Congruence k-permutable Varieties) For k 3, there exists a finite algebra A CP(k) such that SMP(A) is EXPTIME-complete 12

40 corollaries Corollary 1 (SMP for Congruence Distributive Varieties) 1 If k 2, then SMP(A) P for every finite algebra A in CD(k) 2 If k 3, then there exists a finite algebra A CD(k) such that SMP(A) is EXPTIME-complete Corollary 2 (SMP for Congruence k-permutable Varieties) For k 3, there exists a finite algebra A CP(k) such that SMP(A) is EXPTIME-complete Remark: We can use the semigroup examples which are NP-complete and PSPACE-complete to get finite algebras in the varieties above with SMP which are NP-hard and PSPACE-hard 12

41 corollaries Corollary 1 (SMP for Congruence Distributive Varieties) 1 If k 2, then SMP(A) P for every finite algebra A in CD(k) 2 If k 3, then there exists a finite algebra A CD(k) such that SMP(A) is EXPTIME-complete Corollary 2 (SMP for Congruence k-permutable Varieties) For k 3, there exists a finite algebra A CP(k) such that SMP(A) is EXPTIME-complete Remark: We can use the semigroup examples which are NP-complete and PSPACE-complete to get finite algebras in the varieties above with SMP which are NP-hard and PSPACE-hard What is the complexity of SMP for these expanded semigroups? 12

42 an np-complete example

43 the algebras S = {z, a, 1}; { } SMP(S) is NP-Complete S z a 1 z z z z a z z a 1 z a 1 14

44 the algebras S = {z, a, 1}; { } SMP(S) is NP-Complete S z a 1 z z z z a z z a 1 z a 1 Let M be the the strong linear Maltsev condition for CD(3) and CP(3) 14

45 the algebras S = {z, a, 1}; { } SMP(S) is NP-Complete S z a 1 z z z z a z z a 1 z a 1 Let M be the the strong linear Maltsev condition for CD(3) and CP(3) SM 0 z a z 0 z z z a 0 z z a 1 0 z a 1 14

46 the algebras S = {z, a, 1}; { } SMP(S) is NP-Complete S z a 1 z z z z a z z a 1 z a 1 Let M be the the strong linear Maltsev condition for CD(3) and CP(3) SM 0 z a z 0 z z z a 0 z z a 1 0 z a 1 d S M 1 (x 1, x 2, x 3 ) = d S M 2 (x 1, x 2, x 3 ) = p S M 1 (x 1, x 2, x 3 )= p S M 2 (x 1, x 2, x 3 )= { x1 if x 1 = x 2 or x 1 = x 3 0 otherwise { x3 if x 1 = x 2 or x 1 = x 3 0 otherwise { x1 if x 2 = x 3 0 otherwise { x3 if x 1 = x 2 0 otherwise 14

47 the algebras S = {z, a, 1}; { } SMP(S) is NP-Complete S z a 1 z z z z a z z a 1 z a 1 Let M be the the strong linear Maltsev condition for CD(3) and CP(3) SM 0 z a z 0 z z z a 0 z z a 1 0 z a 1 d S M 1 (x 1, x 2, x 3 ) = d S M 2 (x 1, x 2, x 3 ) = p S M 1 (x 1, x 2, x 3 )= p S M 2 (x 1, x 2, x 3 )= { x1 if x 1 = x 2 or x 1 = x 3 0 otherwise { x3 if x 1 = x 2 or x 1 = x 3 0 otherwise { x1 if x 2 = x 3 0 otherwise { x3 if x 1 = x 2 0 otherwise 14

48 the goal From the general hardness result, we know SMP(S M ) is NP-hard To show that SMP(S M ) is NP-complete, we will show Theorem For every instance a 1,, a n, b {0, z, a, 1} m of SMP(S M ) with a positive answer, there is a term t in the language {, d 1, d 2 } such that the size of t is O(nm 3 ) and t(a 1,, a n ) = b 15

49 proof idea t a 1,, a n = b 1 b r

50 proof idea t a 1,, a n = b 1 b r 0 0 To get a short term t which evaluates to b, we will: (i) For every j [m] \ [r], generate an element c j = b 1 b r 0 (j (Induct on size of subsets of [r])

51 proof idea t a 1,, a n = b 1 b r 0 0 To get a short term t which evaluates to b, we will: (i) For every j [m] \ [r], generate an element c j = b 1 b r 0 (j (Induct on size of subsets of [r]) (ii) For every subset K of [m] \ [r], generate an element t K = b 1 b r 0 K 0 (Induct on size of K) 16

52 generating c j We show that For every j [m] \ [r] and for every nonempty subset L [r], there is a term of size O(n L ) which evaluates to an m-tuple c j such that c j i = b i if i L, and c j j = 0 We proceed by induction on L b i (i Assume the base case is proven: c j = 0 (j 17

53 generating t k We show that For every nonempty subset K [m] \ [r], there is a term of size O(nm 2 K ) which evaluates to an m-tuple t K such that t K i = b i if i [r], and t K i = 0 if i K We proceed by induction on K The base case is proven: t {j} = c j 18

54 thank you!