Proof Assistants for Graph Non-isomorphism

Transcription

1 Proof Assistants for Graph Non-isomorphism Arjeh M. Cohen 8 January 2007 second lecture of Three aspects of exact computation a tutorial at Mathematics: Algorithms and Proofs (MAP) Leiden, January 8 12, /75

2 Outline 1. Introduction 2. Permutation groups 3. Complexity 4. Graph isomorphism algorithms 5. Proof assistant demo Based on work with Scott Murray and Jan Willem Knopper 2/75

3 1. Introduction Usually algorithms give answers without proofs. To what extent can we expect answers with proofs? Such answers could certify correctness serve as a first step towards formal proofs 3/75

4 Focus on graph non-isomorphism Motivated by universality of graph isomorphism use of group theory 4/75

5 Graph A graph G consists of a finite set of vertices V = V (G) and a set of edges E = E(G) whose elements are subsets of V of size two. Here V = {1, 2, 3, 4} and E = {{1, 2}, {1, 4}, {2, 3}, {3, 4}}. 5/75

6 Isomorphism Let G 0 and G 1 be graphs. An isomorphism σ : G 0 G 1 is a bijective map V (G 0 ) V (G 1 ) preserving edges. 6/75

7 Automorphism An automorphism is an isomorphism from a graph to itself. 7/75

8 Isomorphism automorphism Let G and G be two connected graphs. Take v V (G) and v V (G ). There is an isomorphism G G mapping v to v iff there is an automorphism of order 2 on G v v G interchanging v and v. G G v v 8/75

9 Graph isomorphism G 0 and G 1 are graphs. If G 0 = G1 then the specification of an isomorphism G 0 G 1 suffices for the isomorphism proof. If G 0 = G1 then often a simple invariant will work, but not always. 9/75

10 Two graphs that are not isomorphic The first graph has 4 edges, while the second graph has 5 edges. 10/75

11 Invariants Invariants can tell why two graphs are non-isomorphic. Examples: number of vertices (of given degree) number of edges (and distance multiplicities) eigenvalues and Smith normal form of the adjacency matrix Do not separate all isomorphism classes. 11/75

12 Invariants that always work Lexicographically least adjacency matrix McKay s canonical labelling Hard to compute in general. Non-isomorphism proof strategy: build up from cheaper invariants towards McKay s canonical labelling 12/75

13 2. Permutation groups G is a permutation group. G = A, where A = {a 1, a 2,..., a k } consists of permutations of the points {1, 2,..., n}. A word in A is an expression of the form a e 1 i 1 a e 2 i 2 a e m im where the indices i j are in the range 1,..., k and the exponents e j are integers. 13/75

14 Example The Mathieu group on 11 points, M 11, has generating set A = {a 1, a 2 }, where a 1 = (1, 10)(2, 8)(3, 11)(5, 7), a 2 = (1, 4, 7, 6)(2, 11, 10, 9). 14/75

15 Group membership proving by use of GAP Query input A permutation g. A list A of permutations The fact g A. GAP input G := Group(A); IsIn_Proof(g,G); GAP output A word in A that is equal to g. Query output As G = A, the group G consists of those elements that can be expressed as a word in A. In particular g = IsIn_Proof(g, G), and so belongs to G. 15/75

16 Example a1 = (1,10)(2,8)(3,11)(5,7); a2 = (1,4,7,6)(2,11,10,9) G = Group([a1,a2]) g = (1,3,8,2,11,10)(4,9)(5,7,6) Query: prove that g is in G. Answer: The element g can be written as follows as a word in the generators of G: g = a1*a2^3*a1*a2, and so belongs to G. QED 16/75

17 Example a1 = (1,10)(2,8)(3,11)(5,7); a2 = (1,4,7,6)(2,11,10,9) G = Group([a1,a2]) g = (1,2)(3,4)(5,6,7,8) Query: prove that g is in G. Answer: Oops, the element does not belong to the group 17/75

18 Example a1 = (1,10)(2,8)(3,11)(5,7); a2 = (1,4,7,6)(2,11,10,9) G = Group([a1,a2]) g = (1,2)(3,4)(5,6,7,8) Query: prove that g is in G. Answer: Oops, the element does not belong to the group How is that proved? 18/75

19 Subgroup Suppose that H = B. How to prove that H is a subgroup of G? H is a subgroup of G iff each element of B is contained in G. 19/75

20 Orbit The orbit of x under G is xg = {xg : g G}. If G = a 1, a 2,..., a k and X is an orbit of G, then on vertex set X has labeled edges y the orbit graph i z where ya i = z. 20/75

21 # 12 Example M 11 = a 1 = (1, 10)(2, 8)(3, 11)(5, 7), a 2 = (1, 4, 7, 6)(2, 11, 10, 9). Interrupted lines are labeled by 1 and solid lines by /75

22 Orbit proving by use of GAP Query Input A list A of permutations generating G. A point x. GAP input G := Group(A); Orbit_Proof(G,x); GAP output A set of points X and a list B of words in A indexed by X. The set X is just the orbit xg. The word in B corresponding to y X maps x to y. 22/75

23 Stabilizers The stabilizer subgroup in G of x is G x = {g G : xg = x}. 23/75

24 Stabilizers The stabilizer subgroup in G of x is G x = {g G : xg = x}. A generating set for G x? 24/75

25 Orbit Lemma If y xg, then {g G : xg = y} is a coset of G x. In particular, xg = G / G x. A one-to-one correspondence between the orbit and the cosets. 25/75

26 Schreier tree For every y xg, choose t(y) G with the property that xt(y) = y. By the Orbit Lemma the t(y) form coset reps for G x in G. Efficient construction: the Schreier tree rooted at x: a subgraph T of the orbit graph that is a tree with root x (when we view the edges as being undirected). For every y X, there is a unique minimal path in T from x to y. Then t(y) is the product of the labels of this path. 26/75

27 12 Example M 11 = a 1 = (1, 10)(2, 8)(3, 11)(5, 7), a 2 = (1, 4, 7, 6)(2, 11, 10, 9). A Schreier tree T rooted at 1 is shown /75

28 12 Example For M 11, the values of t written as words in a 1, a 2 are as follows. y t(y) 1 a 1 a 2 2 a 1 a 3 2a 1 a 2 a 2 2a 1 a 3 2 a 2 2 a 1 a 2 2a 1 a 1 a 2 a 1 a 1 a /75

29 The Schreier tree leads to coset reps U for G x in G and a transversal t : G U sending an element g of G to the representative of G x g and satisfying xt(g) = xg and t(g) = t(hg) whenever h G x. 29/75

30 Implementation We store this tree in a linearized form, using Backpointer ω : X X {0} ω(z) = { y if y is adjacent to z and on the minimal path from z to x 0 if z = x Schreier vector v : X { m,..., 1, 0, 1,..., m} i i if y = ω(z) and y z is in T v(z) = i i if y = ω(z) and y z is in T 0 if z = x 30/75

31 Construction of the transversal t For g G, there is a minimal path x = x 0, x 1,..., x m = xg in the Schreier tree T from x to xg. Write b j = a v(xj ) if v(x j ) > 0 and b j = a 1 v(x j ) if v(x j ) < 0. Then t(g) = b 1 b m. 31/75

32 12 Example a 1 = (1, 10)(2, 8)(3, 11)(5, 7) and a 2 = (1, 4, 7, 6)(2, 11, 10, 9). The linearized version of the Schreier tree rooted at 1, is v ω /75

33 12 a 1 = (1, 10)(2, 8)(3, 11)(5, 7) and a 2 = (1, 4, 7, 6)(2, 11, 10, 9) Choosing the root to be 5, we find v ω /75

34 Schreier data proving Query Input A list A = [a 1,..., a k ] of permutations generating G. A point x. GAP input G := Group(A); SchreierData(G,x); GAP output A triple [X, v, ω] of integer sequences consisting of the orbit, the Schreier vector, and the backpointer. 34/75

35 Stabilizer generators Use the coset reps to compute a generating set of G x. Schreier s lemma Suppose G = A, and H G. If U is a set of coset representatives for H in G, and the function t : G U maps an element g of G to the representative of Hg, then a generating set for H is given by { ua t(ua) 1 : u U, a A }. Interpretation: loops in the graph created by the edges left out to create T from G 35/75

36 Stabilizer subgroup identification by use of GAP Query Input G = A and a point x. GAP input G := Group(A); H := Stabilizer(G,x); IsStabiliser_Proof(G,x,H); GAP output A proof that H is a subgroup of G. Schreier data for G wrt x A sequence of quadruples (y, i, g, h) consisting of y X, an index i, the Schreier generator t(y)a i t(t(y)a i ) 1 written as a word g in A, and as a word h in the generators of H. 36/75

37 Base A base for G is a finite sequence B = [x 1,..., x k ] of distinct points in {1, 2,..., n} such that G x1,x 2,...,x k = 1. Hence, the only element of G which fixes all of the points x 1, x 2,..., x k is the identity. Write G (i) = G x1,x 2,...,x i. Then we have a chain of stabilizers G = G (0) G (1) G (k 1) G (k) = 1. We often require the additional property G (i) G (i+1). A base is easy to construct. 37/75

38 Example M 11 = a 1 = (1, 10)(2, 8)(3, 11)(5, 7), a 2 = (1, 4, 7, 6)(2, 11, 10, 9) has a base [1, 2, 3, 4]. The Schreier trees and backpointers for the stabilizer chain are given in the following table v ω v ω v ω v ω /75

39 Base proving by use of GAP Query Input G = A. GAP input G := Group(A); B := BaseOfGroup(G); IsBase_Proof(G,B); GAP output A base B = [x 1,..., x k ] and sets A i with A i 1 xi = A i Query output From IsBase_Proof(G,B) we read off, for each i, a sequence A i of permutations; a proof that A i is the stabilizer of x i in A i 1 ; the determination of the A i 1 -orbit of x i. Check A k = 1. Now B is a base with stabilizer chain { A i }. 39/75

40 Nonmembership Given a base B = [x 1,..., x k ] of G, take G (i) = G x1,...,x i 1 and t(g (i) ) Schreier elements for G (i) rooted at x i 1. Sifting: if g G, then g = u k u k 1 u 1 u 0 where each u i t(g (i) ) is uniquely determined by g. If g G, then sifting fails because at some stage x i h i 1 x i G (i 1), and so h i 1 G (i 1). This gives a proof of nonmembership. 40/75

41 Base proving by use of GAP Query Input g G = A. GAP input G := Group(A); IsNotIn_Proof(g,G) GAP output A base [x 1,..., x k ] and generators for G (i) = G x1,...,x i, the Schreier trees for G (i 1) rooted at x i, and sets A i with G (i) = A i. A sequence of permutations [h 0, h 1,..., h j 1 ] (where j k) such that h i G (i) and x i h i 1 h 1 h 0 = x i g for i = 1,..., j 1 and x j h j 1 h 1 h 0 x j G (j 1). 41/75

42 Order lemma If B = [x 1,..., x k ] is a base for G, then G = k i=1 x ig (i 1). 42/75

43 Example The Mathieu group M 11 on 11 points has order 1 M 11 2 M (1) 11 3 M (2) 11 4 M (3) 11 = = /75

44 Order proving by use of GAP Query Input A list A of permutations generating G. GAP input G := Group(A); Order_Proof(G); GAP output A base [x 1,..., x k ], the corresponding stabilizer chain G (i), and the orbit sizes x i G (i 1). Query output The base and stabilizer chain proof are as before. By the Order Lemma, G is the product of the orbit sizes x i G (i 1). 44/75

45 Permutation groups, conclusion All basic permutation group operations membership orbit Schreier transversal stabilizer base non-membership order are in NP and can be supplied with efficient proofs. 45/75

46 3. Complexity graph non-isomorphism is neither known nor believed to be in NP quantum complexity of graph isomorphism is unknown 46/75

47 Complexity, positive graph isomorphism has time complexity O(exp(n 1/2+o(1) )) Babai and Luks, 1983, compare with n! = O(exp(n log n)) graph non-isomorphism has subexponential size proofs unless the polynomial-time hierarchy collapses Klivans and Van Melkebeek, 2002 subexponential: O(exp(n ɛ )) for every ɛ > 0 there is a zero-knowledge proof of graph non-isomorphism 47/75

48 Arthur-Merlin proof A simplication of the zero-knowledge proof. Arthur is verifier, Merlin is prover. Given two graphs G 0 and G 1 on n vertices. Iterate the following procedure Arthur chooses a random bit α Arthur chooses a random permutation σ of {1,..., n} Arthur constructs H = σ(g α ) Arthur sends H to Merlin Merlin provides β such that H = G β Merlin sends β to Arthur Arthur checks α = β and keeps score 48/75

49 Complexity theorem IP = Interactive Proof using a probabilistic interactive protocol, with unbounded Merlin and polynomial time Arthur Theorem (Shamir 1992) IP = PSPACE. 49/75

50 Graph isomorphism algorithms Merlin s methods 1. Integer programming 2. Weisfeiler-Leman methods 3. bounded degree: polynomial time [Babai, Luks] 4. fast implementation: nauty [McKay] 50/75

51 3.1. Integer programming (Derksen) A G adjacency matrix of G G 0 = G1 iff there exists X O n ({0, 1}) with XA G0 = A G1 X Gives relations on entries of X of degree at most 2 Use Gröbner bases methods to solve for X Efficiency? 51/75

52 3.2. Weisfeiler-Leman methods Label vertices with their degrees Iterate: label vertices with multiset of the labels of their neighbors Between iterations: replace the labels by their order numbers in the lex order of all occurring labels (will be called a partition) Terminate when no new distinctions occur This labels at least 1 c n log(n)/ log log(n) of all graphs on n vertices. Fails on regular graphs. 52/75

53 Higher-dimensional Weisfeiler-Leman method dimension 2: Replace vertices by pairs First label pair by presence of an edge (or equality) between its vertices (isomorphism type) Iterate by computing for pair p the n element multiset of pairs of colors p {v} \ {u} for u p and v arbitrary. Terminate when no new distinctions occur Similarly for higher dimensions. 53/75

54 Higher-dimensional Weisfeiler-Leman method implemented in STABCOL by Babel, Baumann, Lüdecke, Tinhofer at TU München implemented in STABIL by Babel, Chuvaeva, Klin, Pasechnik (1997) 1-dimensional version used in nauty [McKay] 54/75

55 3.3. Luks algorithm Fix G and G. For v, v Luks checks for an isomorphism σ with v = σ(v ). If for all v no such σ exists, then there are no isomorphisms. G G v v 55/75

56 Luks algorithm, cont d The algorithm constructs the automorphism group inductively. In the induction step, families are used: A family of type (2, 0) (left) and one of type (2, 3) (right). 56/75

57 Luks algorithm, cont d The algorithm by Luks inductively constructs the automorphisms. Initialization: A 0 = {(v, v )} Induction step: consider the vertices at distance i and i + 1 of {v, v }. restrict A i to the interior and the vertices at distance i look at the connections to the vertices at distance i + 1 group the edges in families and by their family s type calculate the automorphisms on the boundary that stabilize the sets of families extend automorphisms on the boundary to the graph so far add interior and boundary automorphisms 57/75

58 Luks example A 0 = (1, 5) A 1 = (1, 5)(2, 6)(4, 8), (2, 4), (6, 8) The reduction of A 1 to the boundary is (2, 6)(4, 8), (2, 4), (6, 8). The families are ({2, 4}, {3}), ({6, 8}, {7}) and ({6, 8}, ). A 2 = (2, 4), (6, 8) 58/75

59 Luks algorithm, conclusion The algorithm by Luks is implemented in PAGN with proof production. The permutation group routines discussed above are used. 59/75

60 3.4. Nauty by McKay fast canonical label or automorphism works with building a tree of labelings restricts the computation by Weisfeiler-Leman kind of labeling backtracks using automorphisms 60/75

61 Nauty: partitions A partition is an ordered vertex labeling The partition is here [ ], with WHITE<BLACK. 61/75

62 Nauty: partitions A partition every vertex of which has its own label is called complete. A pair of two complete partitions gives a map on the vertices (smallest label to smallest label, etc.) Check for each complete partition in the tree whether the map to the first complete partition gives an automorphism. Refine partition by Weisfeiler-Leman. 62/75

63 McKay example The graph with starting partition [ ]. 63/75

64 McKay example, The partition [ ] is refined to [ ]. 64/75

65 McKay example, The partition [ ] is refined to [ ]. 65/75

66 McKay example, The partition [ ] is refined to [ ]. 66/75

67 McKay example, The partition [ ] is refined to [ ]. Separation of 1 and 5 suffices for non-isomorphism part /75

68 McKay example, Now case distinction is used: from [ ] we get both [ ] and [ ]. 68/75

69 McKay example, 7a Continuing with the case [ ] gives: Now case distinction is used: from [ ] we get both [ ] and [ ]. Comparing with the first complete partition [ ] gives the maps () and (6, 8) (an automorphism). 69/75

70 McKay example, 7b Continuing with the case [ ] gives: Since we know the automorphism (6, 8) we can choose here: a choice from [ ] is [ ]. Comparing with the first complete partition [ ] gives the map (2, 4) (automorphism). 70/75

71 McKay example, 8 [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] 71/75

72 3.5. Nauty by McKay conclusion Automated proofs: Spot where automorphisms interchanging G 0 and G 1 fail to exist. 72/75

73 4. Proof Assistant for Graph Non-isomorphism 73/75

74 5. Conclusion Merlin s proof construction more or less understood Merlin s automated presentation needs work Gröbner bases methods to be investigated 74/75

75 Coming Friday Rewriting rules for algebras related to Knot theory 75/75