NO NEWS ON MATRIX MULTIPLICATION. Manuel Kauers Institute for Algebra JKU
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1 NO NEWS ON MATRIX MULTIPLICATION Manuel Kauers Institute for Algebra JKU
2 ( ) ( ) ( ) a1,1 a 1,2 b1,1 b 1,2 c1,1 c = 1,2 a 2,1 a 2,2 b 2,1 b 2,2 c 2,1 c 2,2 c 1,1 = a 1,1 b 1,1 + a 1,2 b 2,1 c 1,2 = a 1,1 b 1,2 + a 1,2 b 2,2 c 2,1 = a 2,1 b 1,1 + a 2,2 b 2,1 c 2,2 = a 2,1 b 1,2 + a 2,2 b 2,2 1
3 ( ) ( ) ( ) a1,1 a 1,2 b1,1 b 1,2 c1,1 c = 1,2 a 2,1 a 2,2 b 2,1 b 2,2 c 2,1 c 2,2 c 1,1 = M 1 + M 4 M 5 + M 7 c 1,2 = M 3 + M 5 c 2,1 = M 2 + M 4 c 2,2 = M 1 M 2 + M 3 + M 6 1
4 ... where ( ) ( ) ( ) a1,1 a 1,2 b1,1 b 1,2 c1,1 c = 1,2 a 2,1 a 2,2 b 2,1 b 2,2 c 2,1 c 2,2 M 1 = (a 1,1 + a 2,2 ) (b 1,1 + b 2,2 ) M 2 = (a 2,1 + a 2,2 ) b 1,1 M 3 = a 1,1 (b 1,2 b 2,2 ) M 4 = a 2,2 (b 2,1 b 1,1 ) M 5 = (a 1,1 + a 1,2 ) b 2,2 M 6 = (a 2,1 a 1,1 ) (b 1,1 + b 1,2 ) M 7 = (a 1,2 a 2,2 ) (b 2,1 + b 2,2 ) 1
5 ( ) ( ) ( ) a1,1 a 1,2 b1,1 b 1,2 c1,1 c = 1,2 a 2,1 a 2,2 b 2,1 b 2,2 c 2,1 c 2,2 This scheme needs 7 multiplications instead of 8. Recursive application allows to multiply n n matrices with O(n log 2 7 ) operations in the ground ring. Let ω be the smallest number so that n n matrices can be multiplied using O(n ω ) operations in the ground domain. Then 2 ω < 3. What is the exact value? 1
6 Strassen 1969: ω log
7 Strassen 1969: ω log Pan 1978: ω Bini et al : ω Schönhage 1981: ω Romani 1982: ω Coppersmith/Winograd 1981: ω Strassen 1986: ω Coppersmith/Winograd 1990: ω
8 Strassen 1969: ω log Pan 1978: ω Bini et al : ω Schönhage 1981: ω Romani 1982: ω Coppersmith/Winograd 1981: ω Strassen 1986: ω Coppersmith/Winograd 1990: ω Stothers 2010: ω Williams 2011: ω Le Gall 2014 : ω
9 Only Strassen s algorithm beats the classical algorithm for reasonable problem sizes. 3
10 Only Strassen s algorithm beats the classical algorithm for reasonable problem sizes. Want: a matrix multiplication algorithm that beats Strassen s algorithm for matrices of moderate size. 3
11 Only Strassen s algorithm beats the classical algorithm for reasonable problem sizes. Want: a matrix multiplication algorithm that beats Strassen s algorithm for matrices of moderate size. Idea: instead of dividing the matrices into 2 2-block matrices, divide them into 3 3-block matrices. 3
12 Only Strassen s algorithm beats the classical algorithm for reasonable problem sizes. Want: a matrix multiplication algorithm that beats Strassen s algorithm for matrices of moderate size. Idea: instead of dividing the matrices into 2 2-block matrices, divide them into 3 3-block matrices. Question: What s the minimal number of multiplications needed to multiply two 3 3 matrices? 3
13 Only Strassen s algorithm beats the classical algorithm for reasonable problem sizes. Want: a matrix multiplication algorithm that beats Strassen s algorithm for matrices of moderate size. Idea: instead of dividing the matrices into 2 2-block matrices, divide them into 3 3-block matrices. Question: What s the minimal number of multiplications needed to multiply two 3 3 matrices? Answer: Nobody knows. 3
14 Question: What s the minimal number of multiplications needed to multiply two 3 3 matrices? naive algorithm: 27 padd with zeros, use Strassen twice, cleanup: 25 best known upper bound: 23 (Laderman 1976) best known lower bound: 19 (Bläser 2003) maximal number of multiplications allowed if we want to beat Strassen: 21 (because log 3 21 < log 2 7 < log 3 22). 4
15 Using SAT solvers, Courtois et al. discovered a new scheme with 23 multiplication in 2011, but none with 22. Can we do better? 5
16 Using SAT solvers, Courtois et al. discovered a new scheme with 23 multiplication in 2011, but none with 22. Can we do better? Idea: Find a better encoding Exploit symmetries of the problem Use more powerful SAT solvers Let it run on bigger computers Wait for a solution with fewer multiplications 5
17 Using SAT solvers, Courtois et al. discovered a new scheme with 23 multiplication in 2011, but none with 22. Can we do better? Idea: Find a better encoding Exploit symmetries of the problem Use more powerful SAT solvers Biere s Treengeling Let it run on bigger computers SG Altix UV 1000 Wait for a solution with fewer multiplications 5
18 Using SAT solvers, Courtois et al. discovered a new scheme with 23 multiplication in 2011, but none with 22. Can we do better? Idea: Find a better encoding Exploit symmetries of the problem Use more powerful SAT solvers Biere s Treengeling Let it run on bigger computers SG Altix UV 1000 Wait for a solution with fewer multiplications We are still waiting. 5
19 Using SAT solvers, Courtois et al. discovered a new scheme with 23 multiplication in 2011, but none with 22. Can we do better? Idea: Find a better encoding Exploit symmetries of the problem Use more powerful SAT solvers Biere s Treengeling Let it run on bigger computers SG Altix UV 1000 Wait for a solution with fewer multiplications We are still waiting. (Possible other approach: use QBF instead of SAT.) 5
20 How to encode the search for a matrix multiplication scheme as a SAT problem? 6
21 How to encode the search for a matrix multiplication scheme as a SAT problem? Make an ansatz M 1 = (α (1) 1,1 a 1,1 + α (1) 1,2 a 1,2 + )(β (1) 1,1 b 1,1 + ) M 2 = (α (2) 1,1 a 1,1 + α (2) 1,2 a 1,2 + )(β (2) 1,1 b 1,1 + ). c 1,1 = γ (1) 1,1 M 1 + γ (2) 1,1 M
22 How to encode the search for a matrix multiplication scheme as a SAT problem? Make an ansatz M 1 = (α (1) 1,1 a 1,1 + α (1) 1,2 a 1,2 + )(β (1) 1,1 b 1,1 + ) M 2 = (α (2) 1,1 a 1,1 + α (2) 1,2 a 1,2 + )(β (2) 1,1 b 1,1 + ). c 1,1 = γ (1) 1,1 M 1 + γ (2) 1,1 M 2 +. Set c i,j = k a i,kb k,j for all i, j and compare coefficients. 6
23 How to encode the search for a matrix multiplication scheme as a SAT problem? This gives the Brent equations (e.g., for 3 3 with 21 multiplications) 21 i, j, k, l, m, n {1, 2, 3} : α (q) i,j β(q) k,l γ(q) m,n = δ j,kδ i,m δ l,n q=1 6
24 How to encode the search for a matrix multiplication scheme as a SAT problem? This gives the Brent equations (e.g., for 3 3 with 21 multiplications) 21 i, j, k, l, m, n {1, 2, 3} : α (q) i,j β(q) k,l γ(q) m,n = δ j,kδ i,m δ l,n q=1 3 6 = 729 cubic equations = 567 variables 6
25 How to encode the search for a matrix multiplication scheme as a SAT problem? This gives the Brent equations (e.g., for 3 3 with 21 multiplications) 21 i, j, k, l, m, n {1, 2, 3} : α (q) i,j β(q) k,l γ(q) m,n = δ j,kδ i,m δ l,n q=1 3 6 = 729 cubic equations = 567 variables Laderman claims that he solved this system by hand, but he doesn t say exactly how. 6
26 How to encode the search for a matrix multiplication scheme as a SAT problem? This gives the Brent equations (e.g., for 3 3 with 21 multiplications) 21 i, j, k, l, m, n {1, 2, 3} : α (q) i,j β(q) k,l γ(q) m,n = δ j,kδ i,m δ l,n Solve this system in Z 2. q=1 6
27 How to encode the search for a matrix multiplication scheme as a SAT problem? This gives the Brent equations (e.g., for 3 3 with 21 multiplications) 21 i, j, k, l, m, n {1, 2, 3} : α (q) i,j β(q) k,l γ(q) m,n = δ j,kδ i,m δ l,n Solve this system in Z 2. q=1 Reading α (q) i,j, β(q) k,l, γ(q) m,n as boolean variables and + as XOR, the problem becomes a SAT problem. 6
28 Problem: SAT solvers don t like XOR. They want CNF as input. 7
29 Problem: SAT solvers don t like XOR. They want CNF as input. a + b = 1 (ā b) (a b) 7
30 Problem: SAT solvers don t like XOR. They want CNF as input. a + b = 1 (ā b) (a b) a + b + c = 1 (ā b c) (ā c b) ( b c a) (a b c) 7
31 Problem: SAT solvers don t like XOR. They want CNF as input. a + b = 1 (ā b) (a b) a + b + c = 1 (ā b c) (ā c b) ( b c a) (a b c) a + b + c + d = 1 (ā b c d) (ā b c d) (ā c b d) (ā d b c) ( b c a d) ( b d a c) ( c d a b) (a b c d). 7
32 Problem: SAT solvers don t like XOR. They want CNF as input. a + b = 1 (ā b) (a b) a + b + c = 1 (ā b c) (ā c b) ( b c a) (a b c) a + b + c + d = 1 (ā b c d) (ā b c d) (ā c b d) (ā d b c) ( b c a d) ( b d a c) ( c d a b) (a b c d). Expanding 21-term sum into CNF like this gives a million clauses. 7
33 SAT people avoid this explosion by assigning new variables ( Tseitin variables ) to subexpressions before converting to CNF: 8
34 SAT people avoid this explosion by assigning new variables ( Tseitin variables ) to subexpressions before converting to CNF: a + b + c + d + e + f + g + h + i = 0 8
35 SAT people avoid this explosion by assigning new variables ( Tseitin variables ) to subexpressions before converting to CNF: a + b + c + d + e + f + g + h + i = 0 a + b + c = T 1 d + e + f = T 2 g + h + i = T 3 T 1 + T 2 + T 3 = 0 8
36 SAT people avoid this explosion by assigning new variables ( Tseitin variables ) to subexpressions before converting to CNF: a + b + c + d + e + f + g + h + i = 0 a + b + c = T 1 d + e + f = T 2 g + h + i = T 3 T 1 + T 2 + T 3 = 0 CNF CNF CNF CNF 8
37 SAT people avoid this explosion by assigning new variables ( Tseitin variables ) to subexpressions before converting to CNF: a + b + c + d + e + f + g + h + i = 0 a + b + c = T 1 d + e + f = T 2 g + h + i = T 3 T 1 + T 2 + T 3 = 0 CNF CNF CNF CNF This decreases the number (and length) of clauses at the cost of increasing the number of variables. 8
38 Breaking all the sums of length 21 into chunks of size 3 terms each, we end up with a CNF with variables and clauses. 9
39 Breaking all the sums of length 21 into chunks of size 3 terms each, we end up with a CNF with variables and clauses. Next, we enrich this formula with some additional information that the solver might find helpful. 9
40 Breaking all the sums of length 21 into chunks of size 3 terms each, we end up with a CNF with variables and clauses. Next, we enrich this formula with some additional information that the solver might find helpful. For example: For each i, j, at least one of α (1) i,j,..., α(q) i,j must be nonzero. Likewise for β and γ. 9
41 Breaking all the sums of length 21 into chunks of size 3 terms each, we end up with a CNF with variables and clauses. Next, we enrich this formula with some additional information that the solver might find helpful. For example: For each i, j, at least one of α (1) i,j,..., α(q) i,j must be nonzero. Likewise for β and γ. For each q, at least one of the α (q) i,j must be nonzero. Likewise for β and γ. 9
42 We should also exploit symmetries. 10
43 We should also exploit symmetries. 10
44 We should also exploit symmetries. 10
45 We should also exploit symmetries. 10
46 We should also exploit symmetries. 10
47 We should also exploit symmetries. 10
48 We should also exploit symmetries. 10
49 We should also exploit symmetries. 10
50 We should also exploit symmetries. 10
51 We should also exploit symmetries. 10
52 We should also exploit symmetries. 10
53 We should also exploit symmetries. 10
54 We should also exploit symmetries. 10
55 We should also exploit symmetries. 10
56 We should also exploit symmetries. 10
57 We should also exploit symmetries. 10
58 We should also exploit symmetries. 10
59 We should also exploit symmetries. 10
60 We should also exploit symmetries. 10
61 We should also exploit symmetries. 10
62 We should also exploit symmetries. 10
63 We should also exploit symmetries. 10
64 We should also exploit symmetries. 10
65 We should also exploit symmetries. 10
66 We should also exploit symmetries. 10
67 We should also exploit symmetries. 10
68 We should also exploit symmetries. 10
69 We should also exploit symmetries. 10
70 We should also exploit symmetries. 10
71 We should also exploit symmetries. 10
72 We should also exploit symmetries. 10
73 We should also exploit symmetries. 10
74 We should also exploit symmetries. 10
75 We should also exploit symmetries. 10
76 We should also exploit symmetries. 10
77 We should also exploit symmetries. 10
78 We should also exploit symmetries. 10
79 We should also exploit symmetries. 10
80 We should also exploit symmetries. 10
81 We should also exploit symmetries. 10
82 We should also exploit symmetries. 10
83 We should also exploit symmetries. 10
84 We should also exploit symmetries. 10
85 We should also exploit symmetries. 10
86 We should also exploit symmetries. 10
87 We should also exploit symmetries. 10
88 We should also exploit symmetries. 10
89 We should also exploit symmetries. 10
90 We should also exploit symmetries. 10
91 We should also exploit symmetries. 10
92 We should also exploit symmetries. 10
93 We should also exploit symmetries. 10
94 We should also exploit symmetries. 10
95 We should also exploit symmetries. 10
96 We should also exploit symmetries. 10
97 We should also exploit symmetries. 10
98 We should also exploit symmetries. 10
99 We should also exploit symmetries. 10
100 We should also exploit symmetries. 10
101 We should also exploit symmetries. 10
102 We should also exploit symmetries. 10
103 We should also exploit symmetries. 10
104 We should also exploit symmetries. It suffices to inspect one configuration per orbit. 10
105 Matrix multiplication AB = C enjoys several symmetries: 11
106 Matrix multiplication AB = C enjoys several symmetries: AUU 1 B = C for every invertible U 11
107 Matrix multiplication AB = C enjoys several symmetries: AUU 1 B = C for every invertible U VAB = VC for every invertible V 11
108 Matrix multiplication AB = C enjoys several symmetries: AUU 1 B = C for every invertible U VAB = VC for every invertible V ABW = CW for every invertible W 11
109 Matrix multiplication AB = C enjoys several symmetries: AUU 1 B = C for every invertible U VAB = VC for every invertible V ABW = CW for every invertible W B A = C 11
110 Matrix multiplication AB = C enjoys several symmetries: AUU 1 B = C for every invertible U VAB = VC for every invertible V ABW = CW for every invertible W B A = C and one more that is a little more subtle 11
111 Matrix multiplication AB = C enjoys several symmetries: AUU 1 B = C for every invertible U VAB = VC for every invertible V ABW = CW for every invertible W B A = C and one more that is a little more subtle Altogether, the symmetry group is S 3 GL(n) 3. 11
112 Matrix multiplication AB = C enjoys several symmetries: AUU 1 B = C for every invertible U VAB = VC for every invertible V ABW = CW for every invertible W B A = C and one more that is a little more subtle Altogether, the symmetry group is S 3 GL(n) 3. We have worked out a small set of clauses that has exactly one solution in each orbit under this group action. 11
113 Simpler example: For the group action GL(3) 2 Z Z 3 3 2, (U, V) A := U A V 1 we have four orbits. 12
114 Simpler example: For the group action GL(3) 2 Z Z 3 3 2, (U, V) A := U A V 1 we have four orbits. A choice of orbit representatives is , 0 0 0, 0 1 0,
115 Simpler example: For the group action GL(3) 2 Z Z 3 3 2, (U, V) A := U A V 1 we have four orbits. A choice of orbit representatives is , 0 0 0, 0 1 0, not needed 12
116 Simpler example: For the group action GL(3) 2 Z Z 3 3 2, (U, V) A := U A V 1 we have four orbits. A choice of orbit representatives is , 0 0 0, 0 1 0, not needed Stabilizer H 1 GL(3) 2 H 1 = 576 Stabilizer H 2 GL(3) 2 H 2 = 96 Stabilizer H 3 GL(3) 2 H 3 =
117 Next, for each i = 1, 2, 3 consider the group action ( Hi GL(3) ) Z Z 3 3 2, (U, V, W) B := V B W 1. 13
118 Next, for each i = 1, 2, 3 consider the group action ( Hi GL(3) ) Z Z 3 3 2, (U, V, W) B := V B W 1. For instance for i = 2 we get 6 orbits. 13
119 Next, for each i = 1, 2, 3 consider the group action ( Hi GL(3) ) Z Z 3 3 2, (U, V, W) B := V B W 1. For instance for i = 2 we get 6 orbits. A list of representatives is , 0 0 0, 0 0 0, , 0 0 0,
120 Next, for each i = 1, 2, 3 consider the group action ( Hi GL(3) ) Z Z 3 3 2, (U, V, W) B := V B W 1. For instance for i = 2 we get 6 orbits. A list of representatives is , 0 0 0, 0 0 0, , 0 0 0, Similarly, we have 6 orbits for i = 1 and 4 orbits for i = 3. 13
121 Altogether, we have found nontrivial orbits for the combined group action GL(3) 3 (Z ) 2 (Z ) 2, (U, V, W) (A, B) := (U A V 1, V B W 1 ). 14
122 Altogether, we have found nontrivial orbits for the combined group action GL(3) 3 (Z ) 2 (Z ) 2, (U, V, W) (A, B) := (U A V 1, V B W 1 ). For each representative, determine the stabilizer H GL(3) 3 and find the orbits of the action of H on C Z
123 Altogether, we have found nontrivial orbits for the combined group action GL(3) 3 (Z ) 2 (Z ) 2, (U, V, W) (A, B) := (U A V 1, V B W 1 ). For each representative, determine the stabilizer H GL(3) 3 and find the orbits of the action of H on C Z Merge orbits that are equivalent under the action of S 3. 14
124 Altogether, we have found nontrivial orbits for the combined group action GL(3) 3 (Z ) 2 (Z ) 2, (U, V, W) (A, B) := (U A V 1, V B W 1 ). For each representative, determine the stabilizer H GL(3) 3 and find the orbits of the action of H on C Z Merge orbits that are equivalent under the action of S 3. Finally, instead of (2 9 1) 3 = matrix triples, we have to consider only 94 orbit representatives. 14
125 Actually there are some much more obvious and much more effective symmetries: 21 i, j, k, l, m, n {1, 2, 3} : α (q) i,j β(q) k,l γ(q) m,n = δ j,kδ i,m δ l,n q=1 15
126 Actually there are some much more obvious and much more effective symmetries: 21 i, j, k, l, m, n {1, 2, 3} : α (q) i,j β(q) k,l γ(q) m,n = δ j,kδ i,m δ l,n q=1 We can change the order of summation in 21! possible ways. 15
127 Actually there are some much more obvious and much more effective symmetries: 21 i, j, k, l, m, n {1, 2, 3} : α (q) i,j β(q) k,l γ(q) m,n = δ j,kδ i,m δ l,n q=1 We can change the order of summation in 21! possible ways. We impose a lexicographic order on the summands in order to break this symmetry. 15
128 At the end of the day, we have an enriched CNF with variables and clauses. 16
129 At the end of the day, we have an enriched CNF with variables and clauses. Unfortunately, we haven t been able to solve it within some 10 years of CPU time. 16
130 At the end of the day, we have an enriched CNF with variables and clauses. Unfortunately, we haven t been able to solve it within some 10 years of CPU time. In contrast, finding Strassen s original algorithm for 2 2 matrices is a matter of a few seconds only. 16
131 At the end of the day, we have an enriched CNF with variables and clauses. Unfortunately, we haven t been able to solve it within some 10 years of CPU time. In contrast, finding Strassen s original algorithm for 2 2 matrices is a matter of a few seconds only. The optimal algorithms for the rectangular cases (2 2)(2 3) and (2 3)(3 3) are found within a few minutes. 16
132 At the end of the day, we have an enriched CNF with variables and clauses. Unfortunately, we haven t been able to solve it within some 10 years of CPU time. In contrast, finding Strassen s original algorithm for 2 2 matrices is a matter of a few seconds only. The optimal algorithms for the rectangular cases (2 2)(2 3) and (2 3)(3 3) are found within a few minutes. We will keep trying to find some improvement for 3 3. For the time being, we have no news on matrix multiplication. 16
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