Basis Collapse in Holographic Algorithms Over All Domain Sizes

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1 Basis Collapse in Holographic Algorithms Over All Domain Sizes Sitan Chen Harvard College March 29, 2016

2 Introduction Matchgates/Holographic Algorithms: A Crash Course Basis Size and Domain Size Collapse Theorems Setup Overview of Proof Group Property Simulation Rank Rigidity Matchgate Identities Implies Cluster Existence Base Case Inductive Step Epilogue k 2 K? Next Steps Acknowledgments

3 Holographic algorithms reduce counting problems into the problem of counting perfect matchings in a graph G = (V, E). Perfect matching: M E for which every v V belongs to exactly one edge e M [Valiant 79]: Counting perfect matchings in arbitrary graphs is #P-complete. [Fisher-Temperley 1961, Kasteleyn 1961]: Counting perfect matchings in planar graphs is in P.

4 More generally, if every edge e of G has some weight w(e), define ( ) PerfMatch(G) = w(e). Theorem (FKT algorithm) perfect matchings M e M If G is a planar weighted graph, PerfMatch(G) can be computed in polynomial time. Idea. For an arbitrary graph G with adjacency matrix A, the Pfaffian ( ) Pf(A) = sgn(m) w(e) perfect matchings M e M satisfies Pf(A) 2 = det(a). For planar graphs, can flip the signs of some entries of A to make Pf and PerfMatch agree.

5 (x 1 x 2 x 3 ) (x 1 x 3 x 4 ) (x 5 x 6 x 7 ) (x 4 x 5 x 6 )

6 (x 1 x 2 x 3 ) (x 1 x 3 x 4 ) (x 5 x 6 x 7 ) (x 4 x 5 x 6 )

7 Imagine: each vertex v on the left propagates signals along its outgoing edges indicating whether v is assigned 1 (green) or 0 (black).

8 Each satisfying assignment corresponds to a collection of signals satisfying two constraints: Consistency: If x i is a vertex on the left, the two signals x i generates must be the same Satisfaction: If C j is a vertex on the right, at least one of the three signals it receives must be

9 Goal: encode these bit vectors using the matching properties of graphs Definition A matchgate is a weighted graph G with designated subsets of its vertices called external nodes X. We say that it is of arity X. Definition The standard signature G of matchgate G of arity n is a vector of dimension 2 n with entries indexed by bitstrings of length n. For Z X corresponding to bitstring α, Γ α = PerfMatch(Γ\Z).

12 We want planar matchgates G and R whose standard signatures respectively match the vectors encoding the consistency and satisfaction constraints: Consistency: If x i is a vertex on the left, the two signals x i generates must be the same Satisfaction: If C j is a vertex on the right, at least one of the three signals it receives must be

13 (x 1 x 2 x 3 ) (x 1 x 3 x 4 ) (x 5 x 6 x 7 ) (x 4 x 5 x 6 )

14 (x 1 x 2 x 3 ) (x 1 x 3 x 4 ) (x 5 x 6 x 7 ) (x 4 x 5 x 6 )

15 (x 1 x 2 x 3 ) (x 1 x 3 x 4 ) (x 5 x 6 x 7 ) (x 4 x 5 x 6 )

16 Unfortunately, no recognizer has standard signature (0, 1, 1, 1, 1, 1, 1, 1): Observation (Parity Condition) Because a graph with an odd number of vertices has no perfect matchings, given any matchgate G, the indices of the nonzero entries in its standard signature must have the same parity.

17 The saving grace: rewrite number of perfect matchings of matchgrid Ω as an inner product and apply a change of basis. Suppose there are w wires in Ω, generators G 1,...G g, and R 1,..., R r recognizers, then g r PerfMatch(Ω) = y G i x i R j j = G, R, z {0,1} w, z=x 1 x r y 1 y g i=1 j=1 where G = i G i and R = i R i with the order of tensoring specified by the wires. Regard G as an element in X = C 2w and R as an element in X : PerfMatch(Ω) is the result of applying dual vector R to G, which is independent of the choice of basis for X.

18 Definition Given a 2 2 basis matrix M, the signature with respect to M of a generator G of arity n is the vector G satisfying G = M n G. The signature with respect to M of a recognizer R of arity n is the vector R satisfying R = RM n.

19 Suffices to find a basis M of matchgates G and R whose signatures with respect to M match the vectors encoding the consistency and satisfaction constraints. Over C and F 2, this still cannot be done. [Valiant 06, Cai-Lu 07]: Over F 7, take M = G = (3, 0, 0, 5), and R = (0, 3, 3, 0, 3, 0, 0, 5). ( ) 1 3, 6 5

21 Introduction Matchgates/Holographic Algorithms: A Crash Course Basis Size and Domain Size Collapse Theorems Setup Overview of Proof Group Property Simulation Rank Rigidity Matchgate Identities Implies Cluster Existence Base Case Inductive Step Epilogue k 2 K? Next Steps Acknowledgments

22 The number of different values that objects in a counting problem can take on is called the domain size. Domain size 2: Boolean satisfying assignments Vertex covers Perfect matchings Ice problems Domain size k k-colorings

23 Over domain size k: Arity-n signatures are now vectors of dimension k n. M now has width k because G = M n G R = RM n.

26 Domain size 2: encode True/False by presence/absence of one external node Domain size k: encode colors {1,..., k} by removal of some subset of a group of l external nodes Arities are now multiples of l External nodes grouped into blocks of l, with wires connecting matchgates blockwise. If Γ has n blocks, Γ has 2 ln entries. M has height 2 l because We call l the basis size. G = M n G R = RM n.

27 Introduction Matchgates/Holographic Algorithms: A Crash Course Basis Size and Domain Size Collapse Theorems Setup Overview of Proof Group Property Simulation Rank Rigidity Matchgate Identities Implies Cluster Existence Base Case Inductive Step Epilogue k 2 K? Next Steps Acknowledgments

28 We will regard standard signatures as matrices: Definition For standard signature G of generator G, the t-th matrix form G(t) (1 t n) is the 2 l 2 (n 1)l matrix of entries of G where the rows are indexed by α t {0, 1} l and the columns are indexed by α 1 α t 1 α t+1 α n {0, 1} (n 1)l.

29 We will also regard signatures as matrices: Definition For signature G of generator G, the t-th matrix form G(t) (1 t n) is the k k n 1 matrix of entries of G where the rows are indexed by α t [k] and the columns are indexed by α 1 α t 1 α t+1 α n [k] n 1. Note: we will denote row indices by superscripts and column indices by subscripts.

30 Definition A generator G is full rank if there exists t for which rank(g(t)) = k. It turns out we may assume that rank(m) = k. But we know So if G is of full rank, G(t) = MG(t)(M T ) (n 1). rank(g(t)) = k.

31 Key to understanding the ultimate capabilities of holographic algorithms for solving counting problems over a given domain size: Question Given k, what is the smallest l for which any holographic algorithm over domain size k with a full-rank matchgate can be simulated by one with basis size l? domain size basis size Cai-Lu

32 Key to understanding the ultimate capabilities of holographic algorithms for solving counting problems over a given domain size: Question Given k, what is the smallest l for which any holographic algorithm over domain size k with a full-rank matchgate can be simulated by one with basis size l? domain size basis size Cai-Lu Cai-Fu

33 Key to understanding the ultimate capabilities of holographic algorithms for solving counting problems over a given domain size: Question Given k, what is the smallest l for which any holographic algorithm over domain size k with a full-rank matchgate can be simulated by one with basis size l? domain size basis size Cai-Lu Cai-Fu C 15, Xia 15 k log 2 k

34 Introduction Matchgates/Holographic Algorithms: A Crash Course Basis Size and Domain Size Collapse Theorems Setup Overview of Proof Group Property Simulation Rank Rigidity Matchgate Identities Implies Cluster Existence Base Case Inductive Step Epilogue k 2 K? Next Steps Acknowledgments

35 Definition Z {0, 1} n is a cluster if there exists s {0, 1} n and positions p 1,..., ( p m [n] such that each member of Z is of the form ) s j J e p j for some J {p 1,..., p m }, where e pj is the bitstring consisting of zeroes everywhere except position p j. We write Z as s + {e p1,..., e pm } (s only unique up to the bits outside of positions p 1,..., p m ). e.g. {000, 001, 100, 101} is a cluster denoted {e 1, e 3 }.

36 For now, assume k = 2 K. Steps of proof: 1. Cluster existence: Any standard signature of rank at least 2 K contains a cluster of 2 K linearly independent rows 2. Group property: Inverses of standard signatures are also standard signatures 3. Simulation: Use 1 and 2 to simulate with a basis of size K

37 For now, assume k = 2 K. Steps of proof: 1. Cluster existence: Any standard signature of rank at least 2 K contains a cluster of 2 K linearly independent rows (hardest part of the proof ) 2. Group property: Inverses of standard signatures are also standard signatures 3. Simulation: Use 1 and 2 to simulate with a basis of size K

38 For now, assume k = 2 K. Steps of proof: 1. Cluster existence: Any standard signature of rank at least 2 K contains a cluster of 2 K linearly independent rows (hardest part of the proof ) 2. Group property: Inverses of standard signatures are also standard signatures (adapted from Li/Xia result in matchgate character theory) 3. Simulation: Use 1 and 2 to simulate with a basis of size K

39 For now, assume k = 2 K. Steps of proof: 1. Cluster existence: Any standard signature of rank at least 2 K contains a cluster of 2 K linearly independent rows (hardest part of the proof ) 2. Group property: Inverses of standard signatures are also standard signatures (adapted from Li/Xia result in matchgate character theory) 3. Simulation: Use 1 and 2 to simulate with a basis of size K (technique due to Cai/Fu)

40 Introduction Matchgates/Holographic Algorithms: A Crash Course Basis Size and Domain Size Collapse Theorems Setup Overview of Proof Group Property Simulation Rank Rigidity Matchgate Identities Implies Cluster Existence Base Case Inductive Step Epilogue k 2 K? Next Steps Acknowledgments

41 Lemma (Group Property) Full-rank 2 K 2 (n 1)K standard signatures G(t) have right inverses (under matrix multiplication) that are also standard signatures.

43 Introduction Matchgates/Holographic Algorithms: A Crash Course Basis Size and Domain Size Collapse Theorems Setup Overview of Proof Group Property Simulation Rank Rigidity Matchgate Identities Implies Cluster Existence Base Case Inductive Step Epilogue k 2 K? Next Steps Acknowledgments

44 We use the approach introduced by Cai-Fu for simulation given cluster existence and group property have been proven. Take any holographic algorithm over domain size k = 2 K. 1. By cluster existence, can pick out a generator G with full-rank signature and find a cluster Z = s + {e p1,..., e pk } of 2 K linearly independent rows. Suppose WLOG s = 0 l. 2. Let M Z denote the submatrix of M with rows indexed by Z. This will be the basis of size log k = K we use for the simulation.

45 G Z = (M Z ) n G G tc Z = (M Z ) (t 1) M (M Z ) (n t) G

46 Modifying generators is easy: G Z i = (M Z ) n i G i has signature G i with respect to new basis M Z Modifying recognizers is more subtle. Can write ( ) R j = R j (M/M Z ) m i (M Z ) m j. Is R j (M/M Z ) m i a valid recognizer standard signature?

47 3. Define T = M(M Z ) By construction, G tc Z (t) = T G Z (t). 5. By group property, G Z (t) has a right-inverse, so right-multiply by this on both sides to conclude that T is a standard signature.

48 Over the new basis M Z : 6. Replace each recognizer R i with R i T m i 7. Replace each generator G j with G Z j. 8. These new matchgates have the same signatures as the originals, but over a basis of size K, so we re done.

49 Introduction Matchgates/Holographic Algorithms: A Crash Course Basis Size and Domain Size Collapse Theorems Setup Overview of Proof Group Property Simulation Rank Rigidity Matchgate Identities Implies Cluster Existence Base Case Inductive Step Epilogue k 2 K? Next Steps Acknowledgments

50 The key ingredient: Theorem (Rank Rigidity) The rank of any standard signature Γ (in matrix form) is always a power of two.

51 Introduction Matchgates/Holographic Algorithms: A Crash Course Basis Size and Domain Size Collapse Theorems Setup Overview of Proof Group Property Simulation Rank Rigidity Matchgate Identities Implies Cluster Existence Base Case Inductive Step Epilogue k 2 K? Next Steps Acknowledgments

52 Our methods are primarily algebraic and rely on the characterization of the set of all standard signatures as the variety cut out by a certain collection of quadratic relations: Theorem (Matchgate Identities) A 2 l 2 (n 1)l matrix Γ is the t-th matrix form of the standard signature of some generator matchgate iff for all ζ, η {0, 1} (n 1)l and σ, τ {0, 1} l, the following matchgate identity (MGI) holds. Let ζ η = e q1 e qd and σ τ = e p1 e pd, where q 1 < < q d and p 1 < < p d. Then if d is even, d ( 1) i+1 Γ (σ ep 1 ep i ) ζ Γ (τ ep 1 ep i ) d η = ± ( 1) j+1 Γ (σ ep 1 ) i=1 j=1 (ζ e qj ) Γ(τ ep 1 ) (η e qj ).

53 d ( 1) i+1 Γ (σ ep 1 ep i ) ζ Γ (τ ep 1 ep i ) d η = ± ( 1) j+1 Γ (σ ep 1 ) i=1 j=1 (ζ e qj ) Γ(τ ep 1 ) (η e qj ). Definition A 2 l 2 m matrix M is a pseudo-signature if for all σ, τ for which wt(σ τ) is even, its entries satisfy the corresponding MGI up to a factor of ±1 on the right-hand side. E.g. (matrix-form) standard signatures, clusters of rows, and their transposes are all pseudo-signatures.

54 The matchgate identities allow us to deduce key linear algebraic relationships between the rows of any pseudo-signature Γ. Example Suppose that d = 2, σ = 0000, τ = 0011, ζ = 1100, η = Then the MGIs become Γ Γ Γ Γ = ± ( Γ Γ Γ Γ )

55 Example (Cont d) Rows Γ 1100 and Γ 1111 are linearly dependent if Γ 1101 and Γ 1110 are linearly dependent. Similarly, rows Γ 0000 and Γ 1111 are linearly dependent if Γ 0001 and Γ 1110 Γ 0010 and Γ 1101 Γ 0100 and Γ 1011 Γ 1000 and Γ 0111 are linearly dependent

56 Lemma Let σ, τ {0, 1} l be such that σ τ = 2d j=1 e p i. If row Γ (σ ep i ) is linearly dependent with row Γ (τ ep i ) for all 1 i 2d, then row Γ σ is linearly dependent with row Γ τ. Coordinate-free interpretation: linear relations among wedges of rows of even parity yield linear relations among wedges of rows of odd parity.

57 Definition For V a vector space with basis {e j }, the second exterior power of V, denoted Λ 2 V, is the vector space given by quotienting V V by the relation v w w v for all v, w V. We denote the image of v w under this quotient map by v w. Λ 2 V has basis {e i e j } i<j. Explicitly, if v = v i e i and w = w i e i, then v w = (v i w j v j w i )e i e j = v i w i i<j i<j v j w j e i e j. In particular, v and w are linearly dependent iff v w = 0.

58 By the MGIs, linear relations among wedges of rows of even parity yield linear relations among wedges of rows of odd parity. Example implies that Γ 0000 Γ 1111 = 0 Γ 0001 Γ 1110 Γ 0010 Γ Γ 0100 Γ 1011 Γ 1000 Γ 0111 = 0

59 By the MGIs, linear relations among wedges of rows of even parity yield linear relations among wedges of rows of odd parity. Example implies that µ (Γ 0000 Γ 1111 ) + ν (Γ 0011 Γ 1100 ) = 0 µ (Γ 0001 Γ 1110 Γ 0010 Γ Γ 0100 Γ 1011 Γ 1000 Γ 0111 )± ν (Γ 0010 Γ 1101 Γ 0001 Γ Γ 0111 Γ 1000 Γ 1011 Γ 0100 ) = 0

60 Introduction Matchgates/Holographic Algorithms: A Crash Course Basis Size and Domain Size Collapse Theorems Setup Overview of Proof Group Property Simulation Rank Rigidity Matchgate Identities Implies Cluster Existence Base Case Inductive Step Epilogue k 2 K? Next Steps Acknowledgments

61 Claim Rank rigidity implies cluster existence. Proof. Suppose Γ is a 2 l 2 m pseudo-signature of rank 2 K. Can assume Γ has no proper clusters of rows with the same rank as Γ. Otherwise, if there were such a cluster Z = σ + {e q1,..., e ql }, replace Γ by Γ Z and l by l, and ignore bits in positions outside of q 1,..., q l.

62 l K+1 {}}{ K 1 {}}{

63 l K+1=1 {}}{ K 1 {}}{

64 l K+1 {}}{ K 1 {}}{

65 l K+1 {}}{ K 1 {}}{

66 l K+1 {}}{ K 1 {}}{

67 l K+1 {}}{ K 1 {}}{

68 l K+1 {}}{ K 1 {}}{

69 l K+1 {}}{ K 1 {}}{

70 l K+1 {}}{ K 1 {}}{

71 l K+1 {}}{ K 1 {}}{ All other rows must be zero. By the MGIs, a contradiction. Γ 0z 0 K 1 Γ 1z 0 K 1 = 0,

72 Introduction Matchgates/Holographic Algorithms: A Crash Course Basis Size and Domain Size Collapse Theorems Setup Overview of Proof Group Property Simulation Rank Rigidity Matchgate Identities Implies Cluster Existence Base Case Inductive Step Epilogue k 2 K? Next Steps Acknowledgments

73 Theorem If Γ is a 2 K+1 2 m pseudo-signature with rank at least 2 K + 1, then rank(γ) = 2 K+1. Sketch. Inductively, we know that Γ contains a cluster Z of 2 K linearly independent rows, say 0 K+1 {e 2,..., e K+1 }. Because rank(γ) 2 K + 1, there exists a row outside the linear span of Z.

74 Even columns: Odd columns:

75 Even columns: Odd columns: Suppose Γ 1001 lay in the span of the red rows, so Γ 1001 Γ 0000 = a σ (Γ σ Γ 0000). LHS: Γ 1000 Γ 0001 RHS: Γ 0 Γ 0 σ red, even

76 Even columns: Odd columns: Suppose Γ 1010 lay in the span of the red rows, so Γ 1010 Γ 0000 = a σ (Γ σ Γ 0000). σ red, even LHS: Γ 1000 Γ 0010 RHS: Γ 0 Γ 0, Γ 1000 Γ 0001

77 Even columns: Odd columns: Suppose Γ 1100 lay in the span of the red rows, so Γ 1100 Γ 0000 = a σ (Γ σ Γ 0000). σ red, even LHS: Γ 1000 Γ 0100 RHS: Γ 0 Γ 0, Γ 1000 Γ 0001, Γ 1000 Γ 0010

78 Even columns: Odd columns: Suppose Γ 1011 lay in the span of the red rows, so Γ 1011 Γ 0001 = a σ (Γ σ Γ 0001). σ red, odd LHS: Γ 1001 Γ 0011 RHS: Γ 0 Γ 0, Γ 0000 Γ 1001

79 Even columns: Odd columns: Suppose Γ 1101 lay in the span of the red rows, so Γ 1011 Γ 0001 = a σ (Γ σ Γ 0001). σ red, odd LHS: Γ 1001 Γ 0101 RHS: Γ 0 Γ 0, Γ 0000 Γ 1001, Γ 1001 Γ 0011

80 Even columns: Odd columns: Suppose Γ 1110 lay in the span of the red rows, so Γ 1110 Γ 0001 = a σ (Γ σ Γ 0001). σ red, odd LHS: Γ 1111 Γ 0000, Γ 1100 Γ 0011, Γ 1010 Γ 0101, Γ 0110 Γ 1001 RHS: Γ 0 Γ 0, Γ 0000 Γ 1001, Γ 1001 Γ 0011, Γ 1001 Γ 0101

81 Even columns: Odd columns: Suppose Γ 1111 lay in the span of the red rows, so Γ 1111 Γ 0000 = a σ (Γ σ Γ 0000). σ red, even LHS: Γ 1110 Γ 0001, Γ 1101 Γ 0010, Γ 1011 Γ 0100, Γ 0111 Γ 1000 RHS: Γ 0 Γ 0, Γ 1000 Γ 0001, Γ 1000 Γ 0010, Γ 1000 Γ 0100

82 Even columns: Odd columns:

83 Introduction Matchgates/Holographic Algorithms: A Crash Course Basis Size and Domain Size Collapse Theorems Setup Overview of Proof Group Property Simulation Rank Rigidity Matchgate Identities Implies Cluster Existence Base Case Inductive Step Epilogue k 2 K? Next Steps Acknowledgments

84 Theorem Suppose l > K + 1. If Γ is a 2 l 2 m pseudo-signature of rank 2 K + 1, then there exists a cluster Z {0, 1} l for which Γ Z is also of rank 2 K + 1. Suppose to the contrary. Inductively we know Γ has a cluster Z of 2 K linearly independent rows, say 0 l + {e p1,..., e pk }.

85 l K K {}}{{}}{

86 l K K {}}{{}}{

87 l K K {}}{{}}{

88 l K K {}}{{}}{

89 l K K {}}{{}}{

90 To show Γ 0l and Γ 1l K 0 l are linearly dependent, by MGIs it s enough to show: Lemma Γ 0l e j = 0 for all j p 1,..., p K. Proof. For i {p 1,..., p K } and j {p 1,..., p K }, define: T i : all rows u for which u i = 0 T j i : all rows u for which u i = u j = 0 Z i : Z T i Note that Z i T j i T i. So inductively, proper cluster T j i has rank a power of two, either 2 K 1 or 2 K.

91 Proof (Cont d). If rank(t j i ) = 2K 1, then span(t j i ) = span(z i). This is true for all i {p 1,..., p K }, so Γ 0l e j K i=1span(z i ) = span({γ 0l }). If rank(t j i ) = 2K, then span(t i ) = span(t j i ) span(z), a contradiction.

92 Introduction Matchgates/Holographic Algorithms: A Crash Course Basis Size and Domain Size Collapse Theorems Setup Overview of Proof Group Property Simulation Rank Rigidity Matchgate Identities Implies Cluster Existence Base Case Inductive Step Epilogue k 2 K? Next Steps Acknowledgments

93 Theorem (Fu/Yang 13) Suppose a basis collapse theorem holds on domain size 2. Then if a holographic algorithm uses a 2 l k basis of rank 2, then the same collapse theorem holds for this holographic algorithm.

94 Theorem Suppose a basis collapse theorem holds on domain size r. Then if a holographic algorithm uses a 2 l k basis of rank r, then the same collapse theorem holds for this holographic algorithm.

95 By rank rigidity, G(t) must have rank a power of two. If G is a full-rank signature, by G(t) = MG(t)(M T ) (n 1), we know M must have rank a power of two. So if k 2 K, we re done inductively by the collapse theorem for domain size 2 K, where K = log 2 k.

96 Introduction Matchgates/Holographic Algorithms: A Crash Course Basis Size and Domain Size Collapse Theorems Setup Overview of Proof Group Property Simulation Rank Rigidity Matchgate Identities Implies Cluster Existence Base Case Inductive Step Epilogue k 2 K? Next Steps Acknowledgments

97 Work out the case where no full-rank matchgate exists Use the collapse theorem to initiate a study of holographic algorithms over higher domains

98 Thank you to: Professor Leslie Valiant (Harvard University) Professor Jin-Yi Cai (University of Wisconsin) Harvard Herchel-Smith Research Fellowship Simons Institute for Computing

Some Results on Matchgates and Holographic Algorithms. Jin-Yi Cai Vinay Choudhary University of Wisconsin, Madison. NSF CCR and CCR

Some Results on Matchgates and Holographic Algorithms Jin-Yi Cai Vinay Choudhary University of Wisconsin, Madison NSF CCR-0208013 and CCR-0511679. 1 #P Counting problems: #SAT: How many satisfying assignments