Simultaneous Multiparty Communication Protocols for Composed Functions
Yassine Hamoudi IRIF , Université Paris Diderot, CNRS MFCS 2018
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Simultaneous Multiparty Communication Protocols for Composed Functions Yassine Hamoudi IRIF , Universit Paris Diderot, CNRS MFCS 2018 Number-On-Forehead model [Chandra, Furst, Lipton83] 2 F : X 1 X k {0,1} Player 1 Player
Yassine Hamoudi IRIF , Université Paris Diderot, CNRS MFCS 2018
Number-On-Forehead model [Chandra, Furst, Lipton’83]
2
F(x1, x2, x3, x4) = ?
x2, x3, x4 x1, x3, x4 x1, x2, x3 x1, x2, x4
Player 1 Player 2 Player 4 Player 3
F : X1 × ⋯ × Xk → {0,1}
Number-On-Forehead model [Chandra, Furst, Lipton’83]
2
xi
F(x1, x2, x3, x4) = ?
x2, x3, x4 x1, x3, x4 x1, x2, x3 x1, x2, x4
Player 1 Player 2 Player 4 Player 3
F : X1 × ⋯ × Xk → {0,1}
N
a n d
n e s s i n t h i s t a l k
Examples
3
An always-O(n) protocol: F is easy / protocol is efficient ⇔ communication cost logO(1)(n) x1, …, xn ∈ {0, 1}n
Examples
3
Equality: x1 = … = xk? Two players: Ω(n) k ≥ 3 players: O(1)
An always-O(n) protocol: F is easy / protocol is efficient ⇔ communication cost logO(1)(n) x1, …, xn ∈ {0, 1}n
Applications of the NOF model
4
Applications of the NOF model
4
polysize constant-depth circuits with AND, OR, NOT, MODm gates
F is hard to compute for k ≥ log(n) players
F is not in ACC0
[Håstad, Goldmann’91] [Babai, Gál, Kimmel, Lokam’04]
Conjecture: MAJORITY ∉ ACC0 Conjecture: NP ⊈ ACC0
(log n)ω(1)
communication cost:
Best lower bounds so far: ˜
Ω ( n 2k)
Applications of the NOF model
4
polysize constant-depth circuits with AND, OR, NOT, MODm gates
F is hard to compute for k ≥ log(n) players
F is not in ACC0
[Håstad, Goldmann’91] [Babai, Gál, Kimmel, Lokam’04]
Conjecture: MAJORITY ∉ ACC0 Conjecture: NP ⊈ ACC0
Log(n) barrier problem Find a function that is hard to compute for log(n) or more players in the Number-On-Forehead model.
(log n)ω(1)
communication cost:
Best lower bounds so far: ˜
Ω ( n 2k)
Applications of the NOF model
4
polysize constant-depth circuits with AND, OR, NOT, MODm gates
F is hard to compute for k ≥ log(n) players
F is not in ACC0
[Håstad, Goldmann’91] [Babai, Gál, Kimmel, Lokam’04]
Conjecture: MAJORITY ∉ ACC0 Conjecture: NP ⊈ ACC0
Log(n) barrier problem Find a function that is hard to compute for log(n) or more players in the Number-On-Forehead model.
(log n)ω(1)
simultaneous
communication cost:
even in the simultaneous NOF model
Best lower bounds so far: ˜
Ω ( n 2k)
Simultaneous Number-On-Forehead model
5
F(x1, x2, x3, x4) = ?
x2, x3, x4 x1, x3, x4 x1, x2, x3 x1, x2, x4
Player 1 Player 2 Player 4 Player 3
One-way communication to a referee, no interactions
Referee
g1 g2 g3 gn
Candidates to break the log(n) barrier: the composed functions
6
x1
1 1
x2
1 1
⋮
…
xk
1 1
n bits k players
Composed function: f ∘ (g1, …, gn) where f : {0, 1}n → {0, 1}
and gj : {0, 1}k → {0, 1} f
g1 g2 g3 gn
Candidates to break the log(n) barrier: the composed functions
6
x1
1 1
x2
1 1
⋮
…
xk
1 1
n bits k players
Composed function: f ∘ (g1, …, gn) where f : {0, 1}n → {0, 1}
and gj : {0, 1}k → {0, 1} f
Examples:
g1 g2 g3 gn
Candidates to break the log(n) barrier: the composed functions
6
x1
1 1
x2
1 1
⋮
…
xk
1 1
n bits k players
Composed function: f ∘ (g1, …, gn) where f : {0, 1}n → {0, 1}
and gj : {0, 1}k → {0, 1} f
[Grolmusz’94] [Babai, Gál, Kimmel, Lokam’04] [Ada, Chattopadhyay, Fawzi, Nguyen’15] Examples:
There is an efficient simultaneous protocol for f ○ (g1,…,gn) when f is symmetric and k ≥ Ω(log n).
Block-composed functions
7
x1
1 1
x2
1 1
⋮
… …
xk
1 1
tn bits t bits
g1 gn f
k players
Block-composed functions
7
x1
1 1
x2
1 1
⋮
… …
xk
1 1
tn bits t bits
g1 gn f The simultaneous communication cost of MAJ ◦ (MAJ,…,MAJ) is (log n)ω(1) for t ≥ √n and k ≥ Ω(log n).
Unknown even for t = 2
Conjecture [Babai et. al.’04]
k players
Our result
8
1 1 1 1 … … 1 1
t bits
g1 f
If t is constant, there is an efficient simultaneous protocol for f ○ (g1,…,gn) when f, g1, …, gn are symmetric and k ≥ Ω(log n).
MAJ ◦ (MAJ,…,MAJ) cannot break the log n barrier for constant t
Theorem:
gn
Our result
8
1 1 1 1 … … 1 1
t bits
g1 f
If t is constant, there is an efficient simultaneous protocol for f ○ (g1,…,gn) when f, g1, …, gn are symmetric and k ≥ Ω(log n).
MAJ ◦ (MAJ,…,MAJ) cannot break the log n barrier for constant t
Theorem:
gn
Block-width t Model Conditions
[Ada, Chattopadhyay, Fawzi, Nguyen’15]
1 simultaneous f symmetric
[Chattopadhyay, Saks’14]
log log n non-simultaneous f symmetric
[Chattopadhyay, Saks’14]
log n non-simultaneous f, g1, …, gn symmetric
Our result
constant simultaneous f, g1, …, gn symmetric
Our result
9
1 1 1 1 … … 1 1
t bits
g1 f
If t is constant, there is an efficient simultaneous protocol for f ○ (g1,…,gn) when f, g1, …, gn are symmetric and k ≥ Ω(log n).
Roadmap (when k = Θ(log n)):
MAJ ◦ (MAJ,…,MAJ) cannot break the log n barrier for constant t
Theorem:
gn
Step 1: equal inner functions
10
x1
1
x2
1 1 1
⋮
… …
xk
1 1
f
g1 gn B1 Bn
Step 1: equal inner functions
10
x1
1
x2
1 1 1
⋮
… …
xk
1 1
f
g1 gn B1 Bn gj
gj(x) = ∑a ca(gj) ⋅ ma(x)
Step 1: equal inner functions
10
x1
1
x2
1 1 1
⋮
… …
xk
1 1
f
g1 gn B1 Bn gj
gj(x) = ∑a ca(gj) ⋅ ma(x)
ca(gj)
Step 1: equal inner functions
10
x1
1
x2
1 1 1
⋮
… …
xk
1 1
f
g1 gn B1 Bn gj
gj(x) = ∑a ca(gj) ⋅ ma(x)
ca(gj)
Step 1: equal inner functions
10
x1
1
x2
1 1 1
⋮
… …
xk
1 1
f
g1 gn B1 Bn gj
gj(x) = ∑a ca(gj) ⋅ ma(x)
ca(gj)
∑a SUM ∘ (ma, …, ma)(Ma) = ∑a ∑j ca(gj) ⋅ ma(Bj) = ∑j gj(Bj)
Step 1: equal inner functions
10
x1
1
x2
1 1 1
⋮
… …
xk
1 1
f
g1 gn B1 Bn gj
gj(x) = ∑a ca(gj) ⋅ ma(x)
ca(gj)
∑a SUM ∘ (ma, …, ma)(Ma) = ∑a ∑j ca(gj) ⋅ ma(Bj) = ∑j gj(Bj)
Enough to compute since f is symmetric. f ∘ (g1, …, gn)
Step 1: equal inner functions
10
x1
1
x2
1 1 1
⋮
… …
xk
1 1
f
g1 gn B1 Bn gj
gj(x) = ∑a ca(gj) ⋅ ma(x)
ca(gj)
∑a SUM ∘ (ma, …, ma)(Ma) = ∑a ∑j ca(gj) ⋅ ma(Bj) = ∑j gj(Bj)
Enough to compute since f is symmetric. f ∘ (g1, …, gn)
mod p
Size ≤ k×n2
for prime p ∈ (n,2n)
→ Knowing y(M) is enough to compute f○(g,…,g) when f and g are symmetric.
11
Step 3: equation solving protocol
[Babai, Gál, Kimmel, Lokam’04] Protocol for t = 1: x1
1 1 1 1 1 1
x2
1 1 1 1
x3
1 1 1
x4
1 1 1
x5
1 1 1 1 1
y0 = 0 y1 = 1 y2 = 1 y3 = 3 y4 = 1 y5 = 1
→ Knowing y(M) is enough to compute f○(g,…,g) when f and g are symmetric.
11
Step 3: equation solving protocol
[Babai, Gál, Kimmel, Lokam’04] Protocol for t = 1: x1
1 1 1 1 1 1
x2
1 1 1 1
x3
1 1 1
x4
1 1 1
x5
1 1 1 1 1
→ This will be the only communication part of our protocol. y0 = 0 y1 = 1 y2 = 1 y3 = 3 y4 = 1 y5 = 1
M1
→ The referee computes y(M) and then f○(g,…,g).
→ Knowing y(M) is enough to compute f○(g,…,g) when f and g are symmetric.
11
Step 3: equation solving protocol
[Babai, Gál, Kimmel, Lokam’04] Protocol for t = 1: x1
1 1 1 1 1 1
x2
1 1 1 1
x3
1 1 1
x4
1 1 1
x5
1 1 1 1 1
→ This will be the only communication part of our protocol.
y0 = 0 y1 = 1 y2 = 1 y3 = 3 y4 = 1 y5 = 1
M1
→ The referee computes y(M) and then f○(g,…,g).
→ Knowing y(M) is enough to compute f○(g,…,g) when f and g are symmetric.
11
Step 3: equation solving protocol
[Babai, Gál, Kimmel, Lokam’04] Protocol for t = 1: x1
1 1 1 1 1 1
x2
1 1 1 1
x3
1 1 1
x4
1 1 1
x5
1 1 1 1 1
→ This will be the only communication part of our protocol.
y0 = 0 y1 = 1 y2 = 1 y3 = 3 y4 = 1 y5 = 1
M1
We generalize this protocol to t > 1, and show that the corresponding equation admits exactly one integral solution when .
k ≥ Ω(log n)
Conclusion
12
→ only general method known: discrepancy method and its variants → [Podolskii, Sherstov’17]: first ω(1) lower bound when k ≥ log n for explicit function Our result: MAJ ◦ (MAJ,…,MAJ) cannot break the log n barrier for any constant t
(in fact, any symmetric f○(g1,…,gn))
Future directions: arXiv: 1710.01969
Step 0: decreasing the number of players to k = Θ(log n)
13
If g : Y1,…,Yk → Y is symmetric then is symmetric. g′ (y1, …, yk′ ) = g(y1, …, yk′ , yk′
+1, …, yk)
Lemma:
any fixed values variables
Step 0: decreasing the number of players to k = Θ(log n)
13
If g : Y1,…,Yk → Y is symmetric then is symmetric. g′ (y1, …, yk′ ) = g(y1, …, yk′ , yk′
+1, …, yk)
Lemma:
x1
1
x2
1 1 1
⋮
… …
xk’
1 1
⋮ xk
1
f
any fixed values variables
g1 gn
Step 0: decreasing the number of players to k = Θ(log n)
13
If g : Y1,…,Yk → Y is symmetric then is symmetric. g′ (y1, …, yk′ ) = g(y1, …, yk′ , yk′
+1, …, yk)
Lemma:
x1
1
x2
1 1 1
⋮
… …
xk’
1 1
⋮ xk
1
f
any fixed values variables
g1 gn x1
1
x2
1 1 1
⋮
… …
xk’
1 1
⋮ xk
1
f
g′
1
g′
n