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II. Group-Theoretic Approach Chris Umans Caltech Based on joint - - PowerPoint PPT Presentation

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II. Group-Theoretic Approach Chris Umans Caltech Based on joint work with Noga Alon, Henry Cohn, Bobby Kleinberg, Amir Shpilka, Balazs Szegedy Modern Applications of Representation Theory, IMA, Chicago July 2014 Introduction A X B = C

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SLIDE 1
  • II. Group-Theoretic Approach

Chris Umans

Caltech

Based on joint work with Noga Alon, Henry Cohn, Bobby Kleinberg, Amir Shpilka, Balazs Szegedy

Modern Applications of Representation Theory, IMA, Chicago July 2014

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SLIDE 2

Introduction

  • Standard method: O(n3) operations
  • Strassen (1969): O(n2.81) operations

X = A B C

July 31, 2014 2

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Introduction

  • Standard method: O(n3) operations
  • Strassen (1969): O(n2.81) operations

X = A B C

The exponent of matrix multiplication: smallest number ω such that for all ε>0 O(nω + ε) operations suffice

July 31, 2014 3

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History

  • Standard algorithm

ω ≤ 3

  • Strassen (1969)

ω < 2.81

  • Pan (1978)

ω < 2.79

  • Bini; Bini et al. (1979)

ω < 2.78

  • Schönhage (1981)

ω < 2.55

  • Pan; Romani; Coppersmith

+ Winograd (1981-1982) ω < 2.50

  • Strassen (1987)

ω < 2.48

  • Coppersmith + Winograd (1987)

ω < 2.375

  • Stothers (2010)

ω < 2.3737

  • Williams (2011)

ω < 2.3729

  • Le Gall (2014)

ω < 2.37286

July 31, 2014 4

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SLIDE 5

Outline

Lecture I:

– crash course on main ideas from Strassen 1969 through Le Gall 2014 – conjectures implying ! = 2

This lecture and next one:

  • II. group-theoretic approach
  • III. extending to coherent configurations

July 31, 2014 5

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SLIDE 6

A different approach

  • So far...

– bound border rank of small tensor (by hand) – asymptotic bound from high tensor powers

  • Disadvantages

– limited universe of “starting” tensors – high tensor powers hard to analyze

  • Next: matrix multiplication via groups

July 31, 2014 6

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July 31, 2014 7

The Group Algebra

  • Given a group G
  • The group algebra C[G] has elements

Σg agg

with multiplication

(Σgagg)(Σhbhh) = Σf (Σgh = f agbh)f

Also think of elements as vectors a with |G| entries

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SLIDE 8

July 31, 2014 8

The Group Algebra

C[G] ' (Cd1×d1) × (Cd2×d2) × … × (Cdk×dk)

  • d1, d2, …, dk are character degrees of G
  • two facts:

– Σdi

2 = |G|

– all character degrees are 1 for abelian groups

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SLIDE 9

Main idea: multiply in Fourier domain

C[G] ' (Cd1×d1) × (Cd2×d2) × … × (Cdk×dk)

DFT a DFT b = = = × DFT-1 = a*b

convolution (with respect to G)

becomes

block-diagonal matrix multiplication

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SLIDE 10

Matrix Multiplication

  • Two input matrices: A=(aij), B=(bkl)
  • “embed” A → A∈C[G], B → B∈C[G]

DFT

A

DFT

B

= = = × DFT-1 = A*B

read off C = AB from A*B

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SLIDE 11

July 31, 2014 11

Can this work?

  • All depends on choice of group G
  • need G to permit an embedding

A → A∈C[G], B → B∈C[G] so that we can read off entries of AB from A*B.

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SLIDE 12

July 31, 2014 12

The embedding:

Subgroups X, Y, Z of G satisfy the triple product property if for all x∈ X , y∈ Y , z∈ Z : xyz = 1 iff x = y = z = 1.

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July 31, 2014 13

The embedding:

Subsets X, Y, Z of G satisfy the triple product property if for all x∈Q(X), y∈ Q(Y), z∈ Q(Z): xyz = 1 iff x = y = z = 1. A = Σax1,y1(x1y1

  • 1)

B = Σby2,z2(y2z2

  • 1)

Claim: (AB)x3,z3 = coeff. on (x3z3

  • 1) in A*B.

Q(S) = {s-1t: s, t ∈ S}

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SLIDE 14

July 31, 2014 14

The embedding:

Subsets X, Y, Z of G satisfy the triple product property if for all x∈Q(X), y∈ Q(Y), z∈ Q(Z): xyz = 1 iff x = y = z = 1. A = Σax1,y1(x1y1

  • 1)

B = Σby2,z2(y2z2

  • 1)

Claim: (AB)x3,z3 = coeff. on (x3z3

  • 1) in A*B.

(x1y1

  • 1)(y2z2
  • 1) = x3z3
  • 1 ) x3
  • 1x1y1
  • 1y2z2
  • 1z3=1

Q(S) = {s-1t: s, t ∈ S}

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SLIDE 15

July 31, 2014 15

How many multiplications?

Fact: method to multiply k × k matrices using m multiplications proves ! ≤ logkm

  • we use m ≤ Σdi

3 mults

  • really m = Σdi! mults
  • at least m ≥ Σdi

2 = |G| mults

First Challenge: embed k × k matrix multiplication in group of size ¼ k2

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July 31, 2014 16

The embedding

  • simple pigeonhole argument:

– embedding in an abelian group requires group to have size k3

First Challenge: embed k × k matrix multiplication in group of size ¼ k2

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July 31, 2014 17

The triangle construction

need X, Y, Z in Sn all with size ≈ |Sn|1/2 Theorem: can embed k × k matrix multiplication in symmetric group of size k2 + o(1)

n objects

  • subgroup X
  • subgroup Y
  • subgroup Z
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SLIDE 18

July 31, 2014 18

The triangle construction

– X moves points within rows – Y moves points within columns – Z moves points within diagonals – want: xyz = 1 ⇒ x = y = z = 1

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July 31, 2014 19

Character degrees

  • We have described a reduction from k × k
  • mat. mult. to block-diagonal mat. mult.

Theorem: in group G with character degrees d1, d2, d3,…, we obtain: kω · ∑i di

ω

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July 31, 2014 20

Potential barrier

Can use this framework to prove ω < 3 if and only if can find X, Y, Z subsets of G satisfying the triple product property, and |X||Y||Z| > ∑di

3.

“beating the sum of the cubes”

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July 31, 2014 21

Recall: the triangle construction

unfortunately, dmax > |X| (= |Y| = |Z|) Theorem: can embed k × k matrix multiplication in symmetric group of size k2 + o(1)

n objects

  • subgroup X
  • subgroup Y
  • subgroup Z
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SLIDE 22

What should we be aiming for?

  • If X, Y, Z µ G satisfy T.P.P. and

– (|X|¢|Y|¢|Z|)1/3 = k ¸ |G|1/2 – o(1) – dmax · |G|1/2 – ²

then ! = 2 Theorem: in group G supporting k x k matrix multiplication with character degrees d1, d2, d3,…, we obtain: kω · ∑i di

ω

∑i di

! ·

dmax

! – 2|G|

July 31, 2014 22

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SLIDE 23

Constructions in linear groups

  • Good candidate family:

SL(n, q) for fixed dimension n because dmax · |G|1/2 - ²n

  • a non-trivial construction (i.e., k3 > |G|):

X = { } Y = { } Z = { }

1 x 0 1 1 0 y 1 1+z z

  • z

1-z 1 x 0 1 1 0 y 1 1+xy x y 1

= x

July 31, 2014 23

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SLIDE 24

Constructions in linear groups

  • Good candidate family:

SL(n, q) for fixed dimension n because dmax · |G|1/2 - ²n

  • best we know, in SL(2, q) for q = p2 :

X = { } Y = { } Z = { }

1 x 0 1 1 0 y 1 z w w z

July 31, 2014 24

1 x 0 1 1 0 y 1 1+xy x y 1

= x

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Constructions in linear groups

  • Good candidate family:

SL(n, q) for fixed dimension n because dmax · |G|1/2 - ²n

  • best we know, in SL(2, q) for q = p2 :

X = { } Y = { } Z = { } – (|X|¢|Y|¢|Z|)1/3 = |G|18/7 – o(1)

1 x 0 1 1 0 y 1 z w w z

July 31, 2014 25

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SLIDE 26

Constructions in linear groups

  • Good candidate family:

SL(n, q) for fixed dimension n

  • In SL(n, R) these three subgroups satisfy

the triple product property:

– upper-triangular with ones on the diagonal – lower-triangular with ones on the diagonal – the special orthogonal group SO(n, R)

and dim. of each is ½ dim. of G as n ! 1

July 31, 2014 26

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July 31, 2014 27

an example yielding ω < 3

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July 31, 2014 28

Wreath product groups

  • A abelian group
  • G semidirect product of

(Aw)N and SN (symmetric group)

6 9 8 0 7 4 8 3 0 0 6 2

à w ! N rows

3 7 3 8 4 5 1 3 2 0 9 5

+

9 6 1 8 1 9 9 6 2 0 5 7

=

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SLIDE 29

July 31, 2014 29

Wreath product groups

  • A abelian group
  • G semidirect product of

(Aw)N and SN (symmetric group)

6 9 8 0 7 4 8 3 0 0 6 2

π =

0 6 2 6 9 8 0 7 4 8 3 0

π

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July 31, 2014 30

Beating the sum of the cubes

Three subsets of (A3)2 semidirect S2:

π

X =

X 0 0 0 X’ 0 π

Y =

0 Y 0 0 0 Y’ π

Z =

0 0 Z Z’ 0 0

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July 31, 2014 31

Beating the sum of the cubes

X 0 0 0 X 0 - π 0 Y 0 0 0 Y - ρ 0 0 Z Z 0 0 - τ

+ + =

0 0 0 0 0 0 X 0 0 0 X 0 0 Y 0 0 0 Y 0 0 Z Z 0 0

Q(X) Q(Y) Q(Z)

  • Group is (A3)2 semidirect S2
  • π, ρ, τ 2 S2 either “flip” or “no flip”
  • must be even number of flips
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July 31, 2014 32

Beating the sum of the cubes

X 0 0 0 X 0 - 0 Y 0 0 0 Y - 0 0 Z Z 0 0 -

+ + =

0 0 0 0 0 0 X 0 0 0 X 0 0 Y 0 0 0 Y 0 0 Z Z 0 0

Q(X) Q(Y) Q(Z)

  • Group is (A3)2 semidirect S2
  • π, ρ, τ 2 S2 either “flip” or “no flip”
  • must be even number of flips

– CASE 1: π = ρ = τ = “no flip”

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July 31, 2014 33

Beating the sum of the cubes

X 0 0 0 X 0 - 0 Y 0 0 0 Y - ρ 0 0 Z Z 0 0 -

+ + =

0 0 0 0 0 0 X 0 0 0 X 0 0 Y 0 0 0 Y 0 0 Z Z 0 0

Q(X) Q(Y) Q(Z)

  • Group is (A3)2 semidirect S2
  • π, ρ, τ 2 S2 either “flip” or “no flip”
  • must be even number of flips

– CASE 2: π = ρ = “flip”; τ = “no flip”

π

0 X 0 X 0 0

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SLIDE 34

July 31, 2014 34 0 X 0 X 0 0

Beating the sum of the cubes

X 0 0 0 X 0 - 0 Y 0 0 0 Y - ρ 0 0 Z Z 0 0 -

+ + =

0 0 0 0 0 0 0 Y 0 0 0 Y 0 0 Z Z 0 0

Q(X) Q(Y) Q(Z)

  • Group is (A3)2 semidirect S2
  • π, ρ, τ 2 S2 either “flip” or “no flip”
  • must be even number of flips

– CASE 2: π = ρ = “flip”; τ = “no flip” – contradiction.

π

0 0 Y 0 Y 0 -

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SLIDE 35

July 31, 2014 35

Beating the sum of the cubes

  • |X||Y||Z| = 8(|A|-1)6
  • ∑i di

3 · dmax∑i di 2 = 2|G| = 4|A|6

G semidirect product of (A3)2 and S2

X = { π : π 2 S2} Y = { π : π 2 S2} Z = { π : π 2 S2}

>

X X’ Y Y’ Z Z’

|A| = 17 yields ω < 2.908…

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SLIDE 36

July 31, 2014 36

generalizing the construction via Uniquely Solvable Puzzles

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July 31, 2014 37

π

X X 0 0 0 0 X 0 X 0 0 0 X 0 0 0 0 0 0 X X 0 0 0 0 X 0 0 0 0 0 0 X 0 0 0

X =

π

0 0 Y Y Y 0 0 Y 0 Y 0 Y 0 Y Y Y 0 0 Y 0 0 0 Y Y Y 0 Y 0 Y 0 Y Y 0 0 0 Y

Y =

π

0 0 0 0 0 Z 0 0 0 0 Z 0 0 0 0 0 Z Z 0 0 0 Z 0 0 0 0 0 Z 0 Z 0 0 0 Z Z 0

Z =

Three subsets of (Aw)N semidirect SN:

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July 31, 2014 38

π

X X 0 0 0 0 X 0 X 0 0 0 X 0 0 0 0 0 0 X X 0 0 0 0 X 0 0 0 0 0 0 X 0 0 0

X =

π

0 0 Y Y Y 0 0 Y 0 Y 0 Y 0 Y Y Y 0 0 Y 0 0 0 Y Y Y 0 Y 0 Y 0 Y Y 0 0 0 Y

Y =

π

0 0 0 0 0 Z 0 0 0 0 Z 0 0 0 0 0 Z Z 0 0 0 Z 0 0 0 0 0 Z 0 Z 0 0 0 Z Z 0

Z = Want: partition of table that ensures Triple Product Property holds.

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July 31, 2014 39

Uniquely Solvable Puzzle (USP)

X X Y Y Y Z X Y X Y Z Y X Y Y Y Z Z Y X X Z Y Y Y X Y Z Y Z Y Y X Z Z Y X X X X X X X X X Y Y Y Y Y Y Y Y Y Y Y Y Y Y Y Y Y y Z Z Z Z Z Z Z Z Z

à w ! N rows

  • USP: unique way to assemble “puzzle pieces”

without overlap (implies no duplicate pieces)

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July 31, 2014 40

Uniquely Solvable Puzzle (USP)

  • Goal: maximize N as function
  • f w
  • No duplicate pieces implies:

N · (w choose w/3)

  • Coppersmith/Winograd: USPs with

N = (w choose w/3)1 - o(1) exist

X X Y Y Y Z X Y X Y Z Y X Y Y Y Z Z Y X X Z Y Y Y X Y Z Y Z Y Y X Z Z Y

à w ! N rows

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July 31, 2014 41

Uniquely Solvable Puzzle (USP)

Strong USP: every unintended way of assembling pieces produces

  • verlap of 2 or 3 symbols in some cell
  • verlap of exactly 2 symbols in some cell

X X Y Y Y Z X Y X Y Z Y X Y Y Y Z Z Y X X Z Y Y Y X Y Z Y Z Y Y X Z Z Y

à w ! N rows

Conjecture: Strong USPs with N = (w choose w/3)1 - o(1) rows exist.

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July 31, 2014 42

a different generalization via the Two Families Conjecture

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July 31, 2014 43

Three subsets of (H3)N semidirect SN:

π

X =

A1 B1 A2 B2 A3 B3 A4 B4 ฀ ฀ ฀ An Bn

π

Y =

A1 B1 A2 B2 A3 B3 A4 B4 ฀ ฀ ฀ An Bn

π

Z =

B1 A1 B2 A2 B3 A3 B4 A4 ฀ ฀ ฀ Bn An

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July 31, 2014 44

Sufficient condition for Triple Product Property

  • subsets A1, A2, …, An, B1, B2, …, Bn of

Abelian group H

1.|Ai + Bi| = |Ai|¢|Bi|

  • 2. (Ai + Bi) Å (Aj + Bk) = ;

if j ≠ k A1 A2 ฀ An

all distinct all distinct all distinct

B1 B2 Bn ฀ ≠

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July 31, 2014 45

Example

  • H = Ak for Abelian group A

– A1 = (≠ 0, ≠ 0, ≠ 0, 0, 0, 0) – B1 = ( 0, 0, 0, ≠ 0, ≠ 0, ≠ 0) – A2 = (≠ 0, ≠ 0, 0, ≠ 0, 0, 0) – B2 = ( 0, 0, ≠ 0, 0, ≠ 0, ≠ 0) – … all equipartitions of k coords

  • |Ai + Bi| = |Ai|¢|Bi| for all i
  • Elements of (Ai + Bi) have nonzero in every coordinate
  • Elements of (Aj + Bk) have zero in some coordinate

all distinct all distinct all distinct

≠

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July 31, 2014 46

Parameters

  • Optimal:

– n = |H|1/2 – o(1) – |Ai| = |Bi| = |H|1/2 – o(1) – yields ω = 2

  • Best construction we know (on previous slide):

– n = |H|0.3868… – |Ai| = |Bi| = |H|0.4491… – yields ω · 2.48…

Conjecture: exists

  • ptimal construction

all distinct all distinct all distinct

≠

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Outline

Lecture I:

– crash course on main ideas from Strassen 1969 through Le Gall 2014 – conjectures implying ! = 2

Lecture II:

– group-theoretic approach

Tomorrow: extending to coherent configurations

July 31, 2014 47