Communication Complexity D 2 F : cost of the most efficient - - PowerPoint PPT Presentation

Communication Complexity

Yassine Hamoudi June 20, 2016

Carnegie Mellon University

Two player model F : {0, 1}n × {0, 1}n → {0, 1}

Alice Bob x ∈ {0, 1}n y ∈ {0, 1}n

0, 1 channel

F(x, y) =? F(x, y) =? Number of bits communicated?

• D2 F : cost of the most efficient deterministic protocol
• R2 F : cost of the most efficient randomized protocol with error 1 3

1

Two player model F : {0, 1}n × {0, 1}n → {0, 1}

Alice Bob x ∈ {0, 1}n y ∈ {0, 1}n

0, 1 channel

F(x, y) =? F(x, y) =? Number of bits communicated?

• D2(F) : cost of the most efficient deterministic protocol
• R2(F) : cost of the most efficient randomized protocol with error 1/3

1

Two player simultaneous model

Alice Bob Referee x ∈ {0, 1}n y ∈ {0, 1}n F(x, y) =? Simultaneous communication complexity: D||

2 (F) and R|| 2 (F) 2

Number On the Forehead model

x1 x2 x3 x4 F(x1, . . . , xk) = ? NOF model:

• Player i does not see xi. Communicate by broadcasting
• Communication cost: Dk(F), Rk(F), D||

k (F) and R|| k (F) 3

Applications of Communication Complexity

Circuit complexity [HG91, BT94] Ramsey theory [CFL83] Branching programs [CFL83] Proof complexity [BPS07] Quasirandom graphs [CT93] Streaming algorithms [AMS96] Property testing [BBM12] Game theory [CS04, NS06] Data structures [MNSW95]

4

Contents

The log n barrier and composed functions Decision tree complexity and log-rank conjecture Conclusion

5

The log n barrier and composed functions

ACC0 and the log n barrier

The log n barrier: Find a function F such that D||

k (F) = ω(polylog n) when k = polylog n.

Motivations:

• ACC0 = functions computable by polysize constant-depth circuits made
• f AND, OR, NOT and MODm gates
• NEXP

ACC0 [Wil14]

• Conjecture: NP

ACC0 F breaks the log n barrier

HG91

F ACC0

6

ACC0 and the log n barrier

The log n barrier: Find a function F such that D||

k (F) = ω(polylog n) when k = polylog n.

Motivations:

• ACC0 = functions computable by polysize constant-depth circuits made
• f AND, OR, NOT and MODm gates
• NEXP ⊈ ACC0 [Wil14]
• Conjecture: NP ⊈ ACC0

F breaks the log n barrier

HG91

F ACC0

6

ACC0 and the log n barrier

The log n barrier: Find a function F such that D||

k (F) = ω(polylog n) when k = polylog n.

Motivations:

• ACC0 = functions computable by polysize constant-depth circuits made
• f AND, OR, NOT and MODm gates
• NEXP ⊈ ACC0 [Wil14]
• Conjecture: NP ⊈ ACC0

F breaks the log n barrier

[HG91]

= = = ⇒ F / ∈ ACC0

6

Composed functions

· · · . . .

x1,1 x1,2 x1,3 x2,1 x2,2 x2,3 xk,1 xk,2 xk,3 x1,n x2,n xk,n Player 1 (x1) Player 2 (x2) Player k (xk) n k

Given f 0 1 n t 0 1 and g 0 1 t k 0 1 : f g x1 xk

7

Composed functions

· · · · · · . . .

x1,1 x1,t x2,1 x2,t xk,1 xk,t x1,tn x2,tn xk,tn Player 1 (x1) Player 2 (x2) Player k (xk) t k g g f

Given f : {0, 1}n/t → {0, 1} and g : {0, 1}t·k → {0, 1}: f ◦ g(x1, . . . , xk)

7

Prior work

Symmetric function = invariant under any permutation of the input When k polylog n:

• Dk Sym

AND1 log2 n [Gro94]

• Dk Sym

Sym1 log3 n [BGKL04]

• Dk Sym

Any1 log3 n [ACFN15]

• Dk Sym

Anyt polylogn for t log log n [CS14] Our result:

• D||

k Sym

Symt polylogn for constant t

8

Prior work

Symmetric function = invariant under any permutation of the input When k = polylog n:

• Dk(Sym ◦ AND1) = O

( log2 n ) [Gro94]

• D||

k (Sym ◦ Sym1)

= O ( log3 n ) [BGKL04]

• D||

k (Sym ◦ Any1)

= O ( log3 n ) [ACFN15]

• Dk(Sym ◦ Anyt)

= O (polylogn) for t ≤ log log n [CS14] Our result:

• D||

k Sym

Symt polylogn for constant t

8

Prior work

Symmetric function = invariant under any permutation of the input When k = polylog n:

• Dk(Sym ◦ AND1) = O

( log2 n ) [Gro94]

• D||

k (Sym ◦ Sym1)

= O ( log3 n ) [BGKL04]

• D||

k (Sym ◦ Any1)

= O ( log3 n ) [ACFN15]

• Dk(Sym ◦ Anyt)

= O (polylogn) for t ≤ log log n [CS14] Our result:

• D||

k (Sym ◦ Symt) = O (polylogn) for constant t 8

Proof sketch

Symmetric f and g with t = 2:

g g g f

· · ·

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

yi1 i2 i3 # columns with exactly i1 occurrences of 1, i2 of 2 and i3 of 3 Recovering the yi1 i2 i3’s is enough since f and g are symmetric

9

Proof sketch

Symmetric f and g with t = 2:

g g g f

· · ·

1 2 2 3 2 1 1 1 3 1 1 1 2 3 2 1 2 1 2 1 1 2 1 2 3 3 1 1 2

yi1 i2 i3 # columns with exactly i1 occurrences of 1, i2 of 2 and i3 of 3 Recovering the yi1 i2 i3’s is enough since f and g are symmetric

9

Proof sketch

Symmetric f and g with t = 2:

g g g f

· · ·

1 2 2 3 2 1 1 1 3 1 1 1 2 3 2 1 2 1 2 1 1 2 1 2 3 3 1 1 2

yi1,i2,i3 = # columns with exactly i1 occurrences of 1, i2 of 2 and i3 of 3 Recovering the yi1 i2 i3’s is enough since f and g are symmetric

9

Proof sketch

Symmetric f and g with t = 2:

g g g f

· · ·

1 2 2 3 2 1 1 1 3 1 1 1 2 3 2 1 2 1 2 1 1 2 1 2 3 3 1 1 2

yi1,i2,i3 = # columns with exactly i1 occurrences of 1, i2 of 2 and i3 of 3 → y0,0,0 = 1 Recovering the yi1 i2 i3’s is enough since f and g are symmetric

9

Proof sketch

Symmetric f and g with t = 2:

g g g f

· · ·

1 2 2 3 2 1 1 1 3 1 1 1 2 3 2 1 2 1 2 1 1 2 1 2 3 3 1 1 2

yi1,i2,i3 = # columns with exactly i1 occurrences of 1, i2 of 2 and i3 of 3 → y0,0,0 = 1, y1,0,0 = 2 Recovering the yi1 i2 i3’s is enough since f and g are symmetric

9

Proof sketch

Symmetric f and g with t = 2:

g g g f

· · ·

1 2 2 3 2 1 1 1 3 1 1 1 2 3 2 1 2 1 2 1 1 2 1 2 3 3 1 1 2

yi1,i2,i3 = # columns with exactly i1 occurrences of 1, i2 of 2 and i3 of 3 → y0,0,0 = 1, y1,0,0 = 2, y0,4,1 = 0, . . . Recovering the yi1 i2 i3’s is enough since f and g are symmetric

9

Proof sketch

Symmetric f and g with t = 2:

g g g f

· · ·

1 2 2 3 2 1 1 1 3 1 1 1 2 3 2 1 2 1 2 1 1 2 1 2 3 3 1 1 2

yi1,i2,i3 = # columns with exactly i1 occurrences of 1, i2 of 2 and i3 of 3 → y0,0,0 = 1, y1,0,0 = 2, y0,4,1 = 0, . . . Recovering the yi1,i2,i3’s is enough since f and g are symmetric

9

Proof sketch

1 2 2 3 2 1 1 1 3 1 1 1 2 3 2 1 2 1 2 1 1 2 1 2 3 3 1 1 2

• Player 1 sends to the referee:

a1

i1,i2,i3 = # columns he sees with i1 occurrences of 1, i2 of 2 and i3 of 3

→ a1

0,0,0 = 2, a1 1,0,0 = 1, a1 2,1,1 = 1, . . .

• Players 2 to 5 do the same

10

Proof sketch

1 2 2 3 2 1 1 1 3 1 1 1 2 3 2 1 2 1 2 1 1 2 1 2 3 3 1 1 2

• Player 1 sends to the referee:

a1

i1,i2,i3 = # columns he sees with i1 occurrences of 1, i2 of 2 and i3 of 3

→ a1

0,0,0 = 2, a1 1,0,0 = 1, a1 2,1,1 = 1, . . .

• Players 2 to 5 do the same

10

Proof sketch

The referee computes: bi1,i2,i3 = a1

i1,i2,i3 + · · · + a5 i1,i2,i3

It verifies: yi1 i2 i3 yi1 i2 i3 n k i1 i2 i3 yi1 i2 i3 i1 1 yi1

1 i2 i3

i2 1 yi1 i2

1 i3

i3 1 yi1 i2 i3

1

bi1 i2 i3 Theorem If k 52t log n then it admits exactly one integral solution. the referee recovers the yi1 i2 i3’s and computes the output

11

Proof sketch

The referee computes: bi1,i2,i3 = a1

i1,i2,i3 + · · · + a5 i1,i2,i3

It verifies:                    yi1,i2,i3 ≥ 0 yi1 i2 i3 n k i1 i2 i3 yi1 i2 i3 i1 1 yi1

1 i2 i3

i2 1 yi1 i2

1 i3

i3 1 yi1 i2 i3

1

bi1 i2 i3 Theorem If k 52t log n then it admits exactly one integral solution. the referee recovers the yi1 i2 i3’s and computes the output

11

Proof sketch

The referee computes: bi1,i2,i3 = a1

i1,i2,i3 + · · · + a5 i1,i2,i3

It verifies:                    yi1,i2,i3 ≥ 0 ∑ yi1,i2,i3 = n k i1 i2 i3 yi1 i2 i3 i1 1 yi1

1 i2 i3

i2 1 yi1 i2

1 i3

i3 1 yi1 i2 i3

1

bi1 i2 i3 Theorem If k 52t log n then it admits exactly one integral solution. the referee recovers the yi1 i2 i3’s and computes the output

11

Proof sketch

The referee computes: bi1,i2,i3 = a1

i1,i2,i3 + · · · + a5 i1,i2,i3

It verifies:                    yi1,i2,i3 ≥ 0 ∑ yi1,i2,i3 = n (k − (i1 + i2 + i3))yi1,i2,i3 + (i1 + 1)yi1+1,i2,i3 +(i2 + 1)yi1,i2+1,i3 + (i3 + 1)yi1,i2,i3+1 = bi1,i2,i3 Theorem If k 52t log n then it admits exactly one integral solution. the referee recovers the yi1 i2 i3’s and computes the output

11

Proof sketch

The referee computes: bi1,i2,i3 = a1

i1,i2,i3 + · · · + a5 i1,i2,i3

It verifies:                    yi1,i2,i3 ≥ 0 ∑ yi1,i2,i3 = n (k − (i1 + i2 + i3))yi1,i2,i3 + (i1 + 1)yi1+1,i2,i3 +(i2 + 1)yi1,i2+1,i3 + (i3 + 1)yi1,i2,i3+1 = bi1,i2,i3 Theorem If k ≥ 52t log n then it admits exactly one integral solution. → the referee recovers the yi1,i2,i3’s and computes the output

11

Decision tree complexity and log-rank conjecture

Log-rank conjecture

F : {0, 1}n × {0, 1}n → {0, 1} Proposition ([MS82]) Let MF ∈ {0, 1}n×n be the communication matrix: MF(x, y) = F(x, y). log rank MF ≤ D2(F) Conjecture For some absolute constant c: log rank MF D2 F logc rank MF

12

Log-rank conjecture

F : {0, 1}n × {0, 1}n → {0, 1} Proposition ([MS82]) Let MF ∈ {0, 1}n×n be the communication matrix: MF(x, y) = F(x, y). log rank MF ≤ D2(F) Conjecture For some absolute constant c: log rank MF ≤ D2(F) ≤ logc rank MF

12

XOR and AND functions

• A function F : {0, 1}n × {0, 1}n → {0, 1} is an XOR function if:

F(x, y) = f(x ⊕ y) for some f : {0, 1}n → {0, 1}

• A function F

0 1 n 0 1 n 0 1 is an AND function if: F x y f x y Examples: Equality x y NOR x y , Hammingd x y GAPd x y , Disjointness x y NOR x y , InnerProduct x y MOD2 x y , etc. Interests:

• For XOR functions: rank MF

mon f [BC99]

• For AND functions: rank MF

mon f [BdW01]

• Connections with Decision Tree complexity

13

XOR and AND functions

• A function F : {0, 1}n × {0, 1}n → {0, 1} is an XOR function if:

F(x, y) = f(x ⊕ y) for some f 0 1 n 0 1

• A function F : {0, 1}n × {0, 1}n → {0, 1} is an AND function if:

F(x, y) = f(x ∧ y) Examples: Equality x y NOR x y , Hammingd x y GAPd x y , Disjointness x y NOR x y , InnerProduct x y MOD2 x y , etc. Interests:

• For XOR functions: rank MF

mon f [BC99]

• For AND functions: rank MF

mon f [BdW01]

• Connections with Decision Tree complexity

13

XOR and AND functions

• A function F : {0, 1}n × {0, 1}n → {0, 1} is an XOR function if:

F(x, y) = f(x ⊕ y) for some f 0 1 n 0 1

• A function F : {0, 1}n × {0, 1}n → {0, 1} is an AND function if:

F(x, y) = f(x ∧ y) Examples: Equality(x, y) = NOR(x ⊕ y), Hammingd(x, y) = GAPd(x ⊕ y), Disjointness(x, y) = NOR(x ∧ y), InnerProduct(x, y) = MOD2(x ∧ y), etc. Interests:

• For XOR functions: rank MF

mon f [BC99]

• For AND functions: rank MF

mon f [BdW01]

• Connections with Decision Tree complexity

13

XOR and AND functions

• A function F : {0, 1}n × {0, 1}n → {0, 1} is an XOR function if:

F(x, y) = f(x ⊕ y) for some f 0 1 n 0 1

• A function F : {0, 1}n × {0, 1}n → {0, 1} is an AND function if:

F(x, y) = f(x ∧ y) Examples: Equality(x, y) = NOR(x ⊕ y), Hammingd(x, y) = GAPd(x ⊕ y), Disjointness(x, y) = NOR(x ∧ y), InnerProduct(x, y) = MOD2(x ∧ y), etc. Interests:

• For XOR functions: rank MF = mon f

[BC99]

• For AND functions: rank MF = mon⋆ f [BdW01]
• Connections with Decision Tree complexity

13

Decision tree complexity

A decision tree is an ordered tree where each internal node is labeled with a query, and each leaf is labeled with 0 or 1. x3 x2 x1 x1 x2 1 1 1

14

Decision tree complexity

A decision tree is an ordered tree where each internal node is labeled with a query, and each leaf is labeled with 0 or 1. x3 x2 x1 x1 x2 1 1 1 Input: x1x2x3 = 011 on a regular decision tree DT(f), RDT(f) and QDT(f)

14

Decision tree complexity

A decision tree is an ordered tree where each internal node is labeled with a query, and each leaf is labeled with 0 or 1. x1 ⊕ x2 ⊕ x3 x2 x1 ⊕ x2 x2 ⊕ x3 x2 1 1 1 Input: x1x2x3 = 011 on a parity decision tree DT⊕(f), RDT⊕(f) and QDT⊕(f)

14

Decision tree complexity

A decision tree is an ordered tree where each internal node is labeled with a query, and each leaf is labeled with 0 or 1. x2 ∧ x3 x1 ∧ x3 x1 x2 x1 ∧ x3 1 1 1 Input: x1x2x3 = 011 on a conjunctive decision tree DT∧(f), RDT∧(f) and QDT∧(f)

14

Connections

Proposition ([ZS10]) For any XOR function F(x, y) = f(x ⊕ y): D2(F) ≤ 2 · DT⊕(f) For any AND function F(x, y) = f(x ∧ y): D2(F) ≤ 2 · DT∧(f) Conjecture

• Communication and Decision Tree complexities are polynomially related
• Log-rank conjecture for decision trees:
• XOR function: log mon f

D2 F 2 DT f logc mon f

• AND function: log mon

f D2 F 2 DT f logc mon f

15

Connections

Proposition ([ZS10]) For any XOR function F(x, y) = f(x ⊕ y): D2(F) ≤ 2 · DT⊕(f) For any AND function F(x, y) = f(x ∧ y): D2(F) ≤ 2 · DT∧(f) Conjecture

• Communication and Decision Tree complexities are polynomially related
• Log-rank conjecture for decision trees:
• XOR function: log mon f

D2 F 2 DT f logc mon f

• AND function: log mon

f D2 F 2 DT f logc mon f

15

Connections

Proposition ([ZS10]) For any XOR function F(x, y) = f(x ⊕ y): D2(F) ≤ 2 · DT⊕(f) For any AND function F(x, y) = f(x ∧ y): D2(F) ≤ 2 · DT∧(f) Conjecture

• Communication and Decision Tree complexities are polynomially related
• Log-rank conjecture for decision trees:
• XOR function: log mon(f) ≤ D2(F) ≤ 2 · DT⊕(f) ≤ logc mon(f)
• AND function: log mon⋆(f) ≤ D2(F) ≤ 2 · DT∧(f) ≤ logc mon⋆(f)

15

Symmetric XOR and AND functions

Communication complexity1 of (nontrivial) XOR and AND functions, for symmetric f: XOR functions AND functions Deterministic Θ (n) Θ ( (n − t(f)) ( 1 + log

n n−t(f)

)) Randomized Θ (r(f)) Θ† ( (n − t(f)) ( 1 + log

n n−t(f)

)) Quantum Θ (r(f)) Θ⋆ (√ n · ℓ0(f) + ℓ1(f) )

1[ZS09, BdW01, Raz03]

16

Symmetric functions

Decision tree complexities2 of (nontrivial) symmetric functions: Regular Parity Conjunctive Deterministic Θ (n) Θ (n) Θ ( (n − t(f)) ( 1 + log

n n−t(f)

)) Randomized Θ (n) Θ (r(f)) Θ† ( (n − t(f)) ( 1 + log

n n−t(f)

)) Quantum Θ (√ n · ℓ(f) ) Θ (r(f)) Θ⋆ (√ n · ℓ0(f) + ℓ1(f) ) Result: Communication and Decision Tree complexities are polynomially related for symmetric functions.

2[ZS09, BdW01, Raz03, BBC+01]

17

Symmetric functions

Decision tree complexities2 of (nontrivial) symmetric functions: Regular Parity Conjunctive Deterministic Θ (n) Θ (n) Θ ( (n − t(f)) ( 1 + log

n n−t(f)

)) Randomized Θ (n) Θ (r(f)) Θ† ( (n − t(f)) ( 1 + log

n n−t(f)

)) Quantum Θ (√ n · ℓ(f) ) Θ (r(f)) Θ⋆ (√ n · ℓ0(f) + ℓ1(f) ) Result: Communication and Decision Tree complexities are polynomially related for symmetric functions.

2[ZS09, BdW01, Raz03, BBC+01]

17

Conclusion

Our contributions:

• first efficient simultaneous protocol for Sym ◦ Symt
• full characterization of the decision tree complexities of symmetric

functions

• efficient construction for Ramsey numbers over Fn

p

Future work:

• other protocols for larger families of composed functions
• breaking the log n barrier
• log-rank conjecture for XOR and AND functions (using decision tree

complexity?)

18

Our contributions:

• first efficient simultaneous protocol for Sym ◦ Symt
• full characterization of the decision tree complexities of symmetric

functions

• efficient construction for Ramsey numbers over Fn

p

Future work:

• other protocols for larger families of composed functions
• breaking the log n barrier
• log-rank conjecture for XOR and AND functions (using decision tree

complexity?)

18

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25

Equality function

Equality(x1, . . . , xk) = 1 ⇔ x1 = · · · = xk D2(Equality) = Ω(n)

• log-rank method

R||

2 (Equality) = O (1)

• Alice and Bob test x · r = y · r mod 2 for two random r ∈ {0, 1}n

D||

k (Equality) = O (1) when k > 2

• Player 1 checks x2 = · · · = xk
• Player 2 checks x1 = x3 = · · · = xk

26

ACC0 and Sym+

Sym+(s, k) = depth-2 circuits whose top gate is a symmetric gate of fan-in s, and each bottom gate is an AND gate of fan-in k SYM AND AND · · ·

s k k

• ACC0 ⊂ SYM+(2polylog n, polylog n) [Yao90, BT94]
• f is computed by a SYM+(s, k − 1) circuit ⇒ for any partition of the input

between k players, there is a protocol of cost O (k log s) computing f

27

Symmetric XOR and AND functions

F(x, y) = f(x ⊕ y) is symmetric iff f is symmetric f x depends only on x . Hence f n 0 1

• t f

min p f p 1 f p

• 0 f

min p n 2 f i f n 2 for i p n 2

• 1 f

min p n 2 f i f n 2 for i n 2 n p

• f

min p f i f i 1 for i p n p 1

• r f

min p f i f i 2 for i p n p 2

28

Symmetric XOR and AND functions

F(x, y) = f(x ⊕ y) is symmetric iff f is symmetric → f(x) depends only on |x|. Hence f : {0, . . . , n} → {0, 1}

• t f

min p f p 1 f p

• 0 f

min p n 2 f i f n 2 for i p n 2

• 1 f

min p n 2 f i f n 2 for i n 2 n p

• f

min p f i f i 1 for i p n p 1

• r f

min p f i f i 2 for i p n p 2

28

Symmetric XOR and AND functions

F(x, y) = f(x ⊕ y) is symmetric iff f is symmetric → f(x) depends only on |x|. Hence f : {0, . . . , n} → {0, 1}

• t(f)

= min{p : f(p − 1) ̸= f(p)}

• ℓ0(f) = min{p ≤ n/2 : f(i) = f(n/2) for i ∈ [p, n/2]}
• ℓ1(f) = min{p ≤ n/2 : f(i) = f(n/2) for i ∈ [n/2, n − p]}
• ℓ(f) = min{p : f(i) = f(i + 1) for i ∈ [p, n − p − 1]}
• r(f)

= min{p : f(i) = f(i + 2) for i ∈ [p, n − p − 2]}

28

Ramsey numbers and EvalG

EvalG

For any Abelian group G and x1, . . . , xk ∈ G: EvalG(x1, . . . , xk) = 1 ⇔ x1 + · · · + xk = 0 Communication complexity:

• Rk EvalG

1 since x1 xk x1 x2 xk

• Dk EvalG

connections to Ramsey theory

29

EvalG

For any Abelian group G and x1, . . . , xk ∈ G: EvalG(x1, . . . , xk) = 1 ⇔ x1 + · · · + xk = 0 Communication complexity:

• R||

k (EvalG) = O (1) since

x1 + · · · + xk = 0 ⇔ x1 = −(x2 · · · + xk)

• Dk(EvalG) → connections to Ramsey theory

29

Ramsey numbers

k-dimensional corner in Gk: (x1, x2, . . . , xk), (x1 + λ, x2, . . . , xk), (x1, x2 + λ, . . . , xk), . . . , (x1, x2, . . . , xk + λ) Ramsey numbers:

• ck G = min # of colors to avoid monochromatic k-dim corner in Gk
• rk G = size of largest subset of Gk without any k-dim corner

Chandra, Furst and Lipton [CFL83]: log ck G Dk

1 EvalG

k log ck G

30

Ramsey numbers

k-dimensional corner in Gk: (x1, x2, . . . , xk), (x1 + λ, x2, . . . , xk), (x1, x2 + λ, . . . , xk), . . . , (x1, x2, . . . , xk + λ) Ramsey numbers:

• c∠

k (G) = min # of colors to avoid monochromatic k-dim corner in Gk

• r∠

k (G) = size of largest subset of Gk without any k-dim corner

Chandra, Furst and Lipton [CFL83]: log ck G Dk

1 EvalG

k log ck G

30

Ramsey numbers

k-dimensional corner in Gk: (x1, x2, . . . , xk), (x1 + λ, x2, . . . , xk), (x1, x2 + λ, . . . , xk), . . . , (x1, x2, . . . , xk + λ) Ramsey numbers:

• c∠

k (G) = min # of colors to avoid monochromatic k-dim corner in Gk

• r∠

k (G) = size of largest subset of Gk without any k-dim corner

Chandra, Furst and Lipton [CFL83]: log(c∠

k (G)) ≤ Dk+1(EvalG) ≤ k + log(c∠ k (G)) 30

Connections

Chandra, Furst and Lipton [CFL83]: log(c∠

k (G)) ≤ Dk+1(EvalG) ≤ k + log(c∠ k (G)) 31

Ramsey numbers and EvalFn

p

Motivations for G = Fn

p:

• the proofs are easier and cleaner
• they can be adapted to any other group [Gre05]
• EvalFn

p ∈ Sym ◦ Symp

Prior work:

• D3 Eval n

p

1 [LM07]

• ck

n 2

2n 2k

2nk

1

[ACFN15]

• an explicit large corner free set over

n 2 [ACFN15]

• ck

n p

2

p log2 n p p log n when k

1 p log 3n [CS14] Our result:

• the first explicit large corner-free set over

n p, of size pnk Ck2 pk

k2

32

Ramsey numbers and EvalFn

p

Motivations for G = Fn

p:

• the proofs are easier and cleaner
• they can be adapted to any other group [Gre05]
• EvalFn

p ∈ Sym ◦ Symp

Prior work:

• D3(EvalFn

p) = ω(1) [LM07]

• c∠

k (Fn 2) ≤ O

( 2n/2k−2nk+1) [ACFN15]

• an explicit large corner free set over Fn

2 [ACFN15]

• c∠

k (Fn p) ≤ 2O(p log2 n)pO(p log n) when k > 1 + p log(3n) [CS14]

Our result:

• the first explicit large corner-free set over

n p, of size pnk Ck2 pk

k2

32

Ramsey numbers and EvalFn

p

Motivations for G = Fn

p:

• the proofs are easier and cleaner
• they can be adapted to any other group [Gre05]
• EvalFn

p ∈ Sym ◦ Symp

Prior work:

• D3(EvalFn

p) = ω(1) [LM07]

• c∠

k (Fn 2) ≤ O

( 2n/2k−2nk+1) [ACFN15]

• an explicit large corner free set over Fn

2 [ACFN15]

• c∠

k (Fn p) ≤ 2O(p log2 n)pO(p log n) when k > 1 + p log(3n) [CS14]

Our result:

• the first explicit large corner-free set over Fn

p, of size pnk Ck2 pk+k2 32

Results

Our contribution: the first explicit large corner-free set in Fn

p

Definitions:

• M

n p k is seen as a k

n matrix over

p

• d c cj

Hamming distance between columns c and cj

• ni c M

number of columns at distance i to c in M For any c

k p, Nk

0 and N0 Nk

1

0 such that

k i 0 Ni

n: Sk

c

M

n p k

i k ni c M Ni is a corner-free set. If k

log n log 1

1 p 1

and Ni

k i p 1 i pk

n then Sk

c pnk Ck2 pk

k2

33

Results

Our contribution: the first explicit large corner-free set in Fn

p

Definitions:

• M ∈ (Fn

p)k is seen as a k × n matrix over Fp

• d(c, cj) = Hamming distance between columns c and cj
• ni,c(M) = number of columns at distance i to c in M

For any c

k p, Nk

0 and N0 Nk

1

0 such that

k i 0 Ni

n: Sk

c

M

n p k

i k ni c M Ni is a corner-free set. If k

log n log 1

1 p 1

and Ni

k i p 1 i pk

n then Sk

c pnk Ck2 pk

k2

33

Results

Our contribution: the first explicit large corner-free set in Fn

p

Definitions:

• M ∈ (Fn

p)k is seen as a k × n matrix over Fp

• d(c, cj) = Hamming distance between columns c and cj
• ni,c(M) = number of columns at distance i to c in M

For any c ∈ Fk

p, Nk = 0 and N0, . . . , Nk−1 ≥ 0 such that ∑k i=0 Ni = n:

Sk

c = {M ∈ (Fn p)k : ∀i ∈ {0, . . . , k}, ni,c(M) = Ni}

is a corner-free set. If k

log n log 1

1 p 1

and Ni

k i p 1 i pk

n then Sk

c pnk Ck2 pk

k2

33

Results

Our contribution: the first explicit large corner-free set in Fn

p

Definitions:

• M ∈ (Fn

p)k is seen as a k × n matrix over Fp

• d(c, cj) = Hamming distance between columns c and cj
• ni,c(M) = number of columns at distance i to c in M

For any c ∈ Fk

p, Nk = 0 and N0, . . . , Nk−1 ≥ 0 such that ∑k i=0 Ni = n:

Sk

c = {M ∈ (Fn p)k : ∀i ∈ {0, . . . , k}, ni,c(M) = Ni}

is a corner-free set. If k ≥ ⌈

log n log ( 1+

1 p−1

)

⌉ and Ni = ⌊(k

i

) (p−1)i

pk

n ⌋ then |Sk

c| ≥ pnk Ck2 pk+k2 33

The log n barrier and composed functions

Composed functions

Given f : {0, 1}n → {0, 1} and − → g = (g1, . . . , gn) where gi : {0, 1}k → {0, 1}: f ◦ − → g (x1, . . . , xk) = f(. . . , gi(x1,i, . . . , xk,i), . . . ) · · · . . .

x1,1 x1,2 x1,3 x2,1 x2,2 x2,3 xk,1 xk,2 xk,3 x1,n x2,n xk,n Player 1 (x1) Player 2 (x2) Player k (xk) n k g1 g2 g3 gn f

34

Composed functions

Definitions:

• f ◦ g if g1 = · · · = gn
• Symmetric = invariant under any permutation of the input
• Any ◦ −

− → Any, Any ◦ Any, Sym ◦ − − → Any, Sym ◦ Sym... Motivations:

• very simple structure
• most of the important functions: GIP

MOD2 AND Sym Sym, MAJ MAJ Sym Sym, DISJ NOR AND Sym Sym

• major open problems still unknown for composed functions

35

Composed functions

Definitions:

• f ◦ g if g1 = · · · = gn
• Symmetric = invariant under any permutation of the input
• Any ◦ −

− → Any, Any ◦ Any, Sym ◦ − − → Any, Sym ◦ Sym... Motivations:

• very simple structure
• most of the important functions: GIP = MOD2 ◦ AND ∈ Sym ◦ Sym,

MAJ ◦ MAJ ∈ Sym ◦ Sym, DISJ = NOR ◦ AND ∈ Sym ◦ Sym

• major open problems still unknown for composed functions

35

Prior work

Conjecture ([BKL95]): MAJ ◦ MAJ breaks the log n barrier When k log n :

• Dk f

g log2 n for f g Sym AND [Gro94]

• Dk f

g log3 n for f g Sym Comp [BGKL04]

• Dk f

g log3 n for f g Sym Any [ACFN15] none of the functions in Sym Any can break the log n barrier

36

Prior work

Conjecture ([BKL95]): MAJ ◦ MAJ breaks the log n barrier When k = Ω(log n):

• Dk(f ◦ g) = O

( log2 n ) for f ◦ g ∈ Sym ◦ AND [Gro94]

• D||

k (f ◦ g) = O

( log3 n ) for f ◦ g ∈ Sym ◦ Comp [BGKL04]

• D||

k (f ◦ −

→ g ) = O ( log3 n ) for f ◦ − → g ∈ Sym ◦ − − → Any [ACFN15] none of the functions in Sym Any can break the log n barrier

36

Prior work

Conjecture ([BKL95]): MAJ ◦ MAJ breaks the log n barrier When k = Ω(log n):

• Dk(f ◦ g) = O

( log2 n ) for f ◦ g ∈ Sym ◦ AND [Gro94]

• D||

k (f ◦ g) = O

( log3 n ) for f ◦ g ∈ Sym ◦ Comp [BGKL04]

• D||

k (f ◦ −

→ g ) = O ( log3 n ) for f ◦ − → g ∈ Sym ◦ − − → Any [ACFN15] → none of the functions in Sym ◦ − − → Any can break the log n barrier

36

Composed functions of block-width t

· · · · · · . . .

x1,1 x1,t x2,1 x2,t xk,1 xk,t x1,tn x2,tn xk,tn Player 1 (x1) Player 2 (x2) Player k (xk) t · n k g1 gn f

• MAJt

0 1 k t 0 1

• Conjecture : MAJ

MAJ

n breaks the barrier 37

Composed functions of block-width t

· · · · · · . . .

x1,1 x1,t x2,1 x2,t xk,1 xk,t x1,tn x2,tn xk,tn Player 1 (x1) Player 2 (x2) Player k (xk) t · n k g1 gn f

• MAJt : {0, 1}k·t → {0, 1}
• Conjecture : MAJ ◦ MAJ√n breaks the barrier

37

Composed functions of block-width t {0, 1}t ∼ F2t

· · · · · · . . .

x1,1 x1,t x2,1 x2,t xk,1 xk,t x1,tn x2,tn xk,tn Player 1 (x1) Player 2 (x2) Player k (xk) t · n k g1 gn f

Given f 0 1 n 0 1 and g g1 gn where gi

k p

0 1 : f g x1 xk f gi x1 i xk i Any Anyp Any Anyp Sym Anyp

38

Composed functions of block-width t

F2t

· · · . . .

x1,1 x1,2 x1,3 x2,1 x2,2 x2,3 xk,1 xk,2 xk,3 x1,n x2,n xk,n Player 1 (x1) Player 2 (x2) Player k (xk) n k g1 g2 g3 gn f

Given f 0 1 n 0 1 and g g1 gn where gi

k p

0 1 : f g x1 xk f gi x1 i xk i Any Anyp Any Anyp Sym Anyp

38

Composed functions of block-width t

F2t

· · · . . .

x1,1 x1,2 x1,3 x2,1 x2,2 x2,3 xk,1 xk,2 xk,3 x1,n x2,n xk,n Player 1 (x1) Player 2 (x2) Player k (xk) n k g1 g2 g3 gn f

Given f : {0, 1}n → {0, 1} and − → g = (g1, . . . , gn) where gi : Fk

p → {0, 1}:

f ◦ − → g (x1, . . . , xk) = f(. . . , gi(x1,i, . . . , xk,i), . . . ) → Any ◦ − − → Anyp, Any ◦ Anyp, Sym ◦ Anyp, . . .

38

Prior work

Conjecture: MAJ ◦ MAJ√

log n breaks the log n barrier

When k polylog n :

• Dk f

g log3 n for f g Sym Any2 [ACFN15]

• Dk f

g polylogn for f g Sym Anyp and p polylog n [CS14] New results for constant p:

• Dk f

g polylogn for f g Sym Symp (k polylog n)

• Dk f

g polylogn for f g Sym Compp (k polylog n)

• MAJ

MAJt cannot break the barrier for constant t

39

Prior work

Conjecture: MAJ ◦ MAJ√

log n breaks the log n barrier

When k = Ω(polylog n):

• D||

k (f ◦ −

→ g ) = O ( log3 n ) for f ◦ − → g ∈ Sym ◦ − − → Any2 [ACFN15]

• Dk(f ◦ g) = O (polylogn) for f ◦ g ∈ Sym ◦ −

− → Anyp and p ≤ polylog n [CS14] New results for constant p:

• Dk f

g polylogn for f g Sym Symp (k polylog n)

• Dk f

g polylogn for f g Sym Compp (k polylog n)

• MAJ

MAJt cannot break the barrier for constant t

39

Prior work

Conjecture: MAJ ◦ MAJ√

log n breaks the log n barrier

When k = Ω(polylog n):

• D||

k (f ◦ −

→ g ) = O ( log3 n ) for f ◦ − → g ∈ Sym ◦ − − → Any2 [ACFN15]

• Dk(f ◦ g) = O (polylogn) for f ◦ g ∈ Sym ◦ −

− → Anyp and p ≤ polylog n [CS14] New results for constant p:

• D||

k (f ◦ g) = O (polylogn) for f ◦ g ∈ Sym ◦ Symp (k = polylog n)

• D||

k (f ◦ g) = O (polylogn) for f ◦ g ∈ Sym ◦ Compp (k ≥ polylog n)

• MAJ ◦ MAJt cannot break the barrier for constant t

39

Proof sketch

Symmetric f and g over F3:

g g g f

· · ·

1 2 2 3 2 1 1 1 3 1 1 1 2 3 2 1 2 1 2 1 1 2 1 2 3 3 1 1 2

yi,j = # columns with i one’s and j two’s Recovering the yi j’s is enough since f and g are symmetric

40

Proof sketch

Symmetric f and g over F3:

g g g f

· · ·

1 2 2 3 2 1 1 1 3 1 1 1 2 3 2 1 2 1 2 1 1 2 1 2 3 3 1 1 2

yi,j = # columns with i one’s and j two’s → y0,0 = 1 Recovering the yi j’s is enough since f and g are symmetric

40

Proof sketch

Symmetric f and g over F3:

g g g f

· · ·

1 2 2 3 2 1 1 1 3 1 1 1 2 3 2 1 2 1 2 1 1 2 1 2 3 3 1 1 2

yi,j = # columns with i one’s and j two’s → y0,0 = 1, y1,0 = 2 Recovering the yi j’s is enough since f and g are symmetric

40

Proof sketch

Symmetric f and g over F3:

g g g f

· · ·

1 2 2 3 2 1 1 1 3 1 1 1 2 3 2 1 2 1 2 1 1 2 1 2 3 3 1 1 2

yi,j = # columns with i one’s and j two’s → y0,0 = 1, y1,0 = 2, y1,1 = 1 Recovering the yi j’s is enough since f and g are symmetric

40

Proof sketch

Symmetric f and g over F3:

g g g f

· · ·

1 2 2 3 2 1 1 1 3 1 1 1 2 3 2 1 2 1 2 1 1 2 1 2 3 3 1 1 2

yi,j = # columns with i one’s and j two’s → y0,0 = 1, y1,0 = 2, y1,1 = 1, . . . Recovering the yi,j’s is enough since f and g are symmetric

40

Proof sketch

1 2 2 3 2 1 1 1 3 1 1 1 2 3 2 1 2 1 2 1 1 2 1 2 3 3 1 1 2

• Player 1 sends to the referee:

a1

i,j = # columns she sees with i one’s and j two’s

→ a1

0,0 = 2, a1 1,0 = 1, a1 1,1 = 3, . . .

• Players 2 to 5 do the same

41

Proof sketch

1 2 2 3 2 1 1 1 3 1 1 1 2 3 2 1 2 1 2 1 1 2 1 2 3 3 1 1 2

• Player 1 sends to the referee:

a1

i,j = # columns she sees with i one’s and j two’s

→ a1

0,0 = 2, a1 1,0 = 1, a1 1,1 = 3, . . .

• Players 2 to 5 do the same

41

Proof sketch

1 2 2 3 2 1 1 1 3 1 1 1 2 3 2 1 2 1 2 1 1 2 1 2 3 3 1 1 2

The referee computes: bi,j = a1

i,j + · · · + a5 i,j

Note that: bi j k i j yi j i 1 yi

1 j

j 1 yi j

1 42

Proof sketch

1 2 2 3 2 1 1 1 3 1 1 1 2 3 2 1 2 1 2 1 1 2 1 2 3 3 1 1 2

The referee computes: bi,j = a1

i,j + · · · + a5 i,j

Note that:

• b0,0 =

bi j k i j yi j i 1 yi

1 j

j 1 yi j

1 42

Proof sketch

1 2 2 2 2 1 1 1 1 1 1 2 1 2 1 2 1 2 1 1 2 1 2 2 1 1 1 2

The referee computes: bi,j = a1

i,j + · · · + a5 i,j

Note that:

• b0,0 = 5y0,0

bi j k i j yi j i 1 yi

1 j

j 1 yi j

1 42

Proof sketch

1 2 2 2 2 1 1 1 1 1 1 2 1 2 1 2 1 2 1 1 2 1 2 2 1 1 1 2

The referee computes: bi,j = a1

i,j + · · · + a5 i,j

Note that:

• b0,0 = 5y0,0 + y1,0

bi j k i j yi j i 1 yi

1 j

j 1 yi j

1 42

Proof sketch

1 2 2 3 2 1 1 1 3 1 1 1 2 3 2 1 2 1 2 1 1 2 1 2 3 3 1 1 2

The referee computes: bi,j = a1

i,j + · · · + a5 i,j

Note that:

• b0,0 = 5y0,0 + y1,0 + y0,1

bi j k i j yi j i 1 yi

1 j

j 1 yi j

1 42

Proof sketch

1 2 2 2 2 1 1 1 1 1 1 2 1 2 1 2 1 2 1 1 2 1 2 2 1 1 1 2

The referee computes: bi,j = a1

i,j + · · · + a5 i,j

Note that:

• b0,0 = 5y0,0 + y1,0 + y0,1
• b1,0 = 4y1,0

bi j k i j yi j i 1 yi

1 j

j 1 yi j

1 42

Proof sketch

1 2 2 2 2 1 1 1 1 1 1 2 1 2 1 2 1 2 1 1 2 1 2 2 1 1 1 2

The referee computes: bi,j = a1

i,j + · · · + a5 i,j

Note that:

• b0,0 = 5y0,0 + y1,0 + y0,1
• b1,0 = 4y1,0 + y1,1

bi j k i j yi j i 1 yi

1 j

j 1 yi j

1 42

Proof sketch

1 2 2 3 2 1 1 1 3 1 1 1 2 3 2 1 2 1 2 1 1 2 1 2 3 3 1 1 2

The referee computes: bi,j = a1

i,j + · · · + a5 i,j

Note that:

• b0,0 = 5y0,0 + y1,0 + y0,1
• b1,0 = 4y1,0 + y1,1 + 2y2,0

bi j k i j yi j i 1 yi

1 j

j 1 yi j

1 42

Proof sketch

1 2 2 3 2 1 1 1 3 1 1 1 2 3 2 1 2 1 2 1 1 2 1 2 3 3 1 1 2

The referee computes: bi,j = a1

i,j + · · · + a5 i,j

Note that:

• b0,0 = 5y0,0 + y1,0 + y0,1
• b1,0 = 4y1,0 + y1,1 + 2y2,0
• . . .

bi,j = (k − (i + j))yi,j + (i + 1)yi+1,j + (j + 1)yi,j+1

42

Proof sketch

Let (bi1,...,ip)0≤i1+···+ip≤k−1 be integers. Consider the system of equations:        (k − (i1 + · · · + ip))yi1,...,ip +

p

∑

j=1

(ij + 1)yi1,...,ij−1,ij+1,ij+1,...,ip = bi1,...,ip 0 ≤ i1 + · · · + ip ≤ k − 1 Assume further that yi1,...,ip ≥ 0, 0 ≤ i1 + · · · + ip ≤ k and ∑

i1+···+ip≤k

yi1,...,ip ≤ n Theorem If k 1 5p log n then it admits at most one integral solution. the referee recovers the yi j’s and computes the output

43

Proof sketch

Let (bi1,...,ip)0≤i1+···+ip≤k−1 be integers. Consider the system of equations:        (k − (i1 + · · · + ip))yi1,...,ip +

p

∑

j=1

(ij + 1)yi1,...,ij−1,ij+1,ij+1,...,ip = bi1,...,ip 0 ≤ i1 + · · · + ip ≤ k − 1 Assume further that yi1,...,ip ≥ 0, 0 ≤ i1 + · · · + ip ≤ k and ∑

i1+···+ip≤k

yi1,...,ip ≤ n Theorem If k > 1 + 5p log n then it admits at most one integral solution. the referee recovers the yi j’s and computes the output

43

Proof sketch

Let (bi1,...,ip)0≤i1+···+ip≤k−1 be integers. Consider the system of equations:        (k − (i1 + · · · + ip))yi1,...,ip +

p

∑

j=1

(ij + 1)yi1,...,ij−1,ij+1,ij+1,...,ip = bi1,...,ip 0 ≤ i1 + · · · + ip ≤ k − 1 Assume further that yi1,...,ip ≥ 0, 0 ≤ i1 + · · · + ip ≤ k and ∑

i1+···+ip≤k

yi1,...,ip ≤ n Theorem If k > 1 + 5p log n then it admits at most one integral solution. → the referee recovers the yi,j’s and computes the output

43

Proof sketch

Conclusion:

• [BGKL04] proved the uniqueness for p = 2
• we generalized to any p
• sending all the aℓ

i,j has cost O (k(k + p) log n) → not efficient is

k = ω(polylog n) (compressibility) Future work:

• remove the compressibility condition
• handle larger p

44

Proof sketch

Conclusion:

• [BGKL04] proved the uniqueness for p = 2
• we generalized to any p
• sending all the aℓ

i,j has cost O (k(k + p) log n) → not efficient is

k = ω(polylog n) (compressibility) Future work:

• remove the compressibility condition
• handle larger p

44