1. Shengyu Zhang The Chinese University of Hong Kong
On the Power of
Randomness in Communication
Shengyu Zhang
The Chinese University of Hong Kong

2. Communication complexity
[Yao79] Two parties, Alice and Bob, jointly compute a function f(x,y) with x known only to Alice and y only to Bob.
Communication complexity: how many bits are needed to be exchanged? D(f) x y
Alice
Bob
f(x,y)
f(x,y)

3. Communication complexity: other models
One-way: Alice sends a message to Bob D1(f)
SMP (Simultaneous Message Passing): They both send a message to a Referee D∥(f)
Relation: D(f) ≤ D1(f) ≤ D∥(f) y
Alice
Bob
f(x,y) x y
Alice
Bob
Ref
f(x,y)

4. Applications of CC
Though defined in an information theoretical setting, it turned out to provide lower bounds to many computational models.
Data structures, circuit complexity, streaming algorithms, decision tree complexity, VLSI, algorithmic game theory, optimization, pseudo-randomness…

5. Rank lower bound
Lower bounds in communication complexity are crucial (for those applications).
Communication Matrix Mf = [f(x,y)](x,y).
A c-bit protocol partitions
the matrix into 2c
rectangles (submatrices).
All entries in a rectangle
are the same
either all 0 or all 1.

6. Rank lower bound So, where Mi is the submatrix in rectangle Ri.
rank(Mf) = rank(∑i Mi) ≤ ∑i rank(Mi) ≤ 2c.
[Thm] (logrank l.b.)
D(f) ≥ log2 rank(Mf).
[Open] (logrank Conj.) D(f) = poly(log2 rank(Mf)) M f = P 2 c i 1

7. Equality EQ (Equality): MEQ = [EQ(x,y)](x,y) = IN, where N = 2n.
D(EQ) ≥ log2 rank(MEQ) = log2 rank(IN) = n. E Q ( x ; y ) = 1 i f 6 I N = 2 6 4 1 . 3 7 5

8. Communication complexity: Randomized
Randomized communication complexity: min # bits exchanged s.t. output is correct w.p
Private-coin: r1 and r2 are independent. --- Rpriv(f)
Public-coin: r1 and r2 are the same Rpub(f) r r2 x y
Alice
Bob

9. Communication complexity: other models
One-way:
Private-coin: Rpriv(f)
Public-coin: Rpub(f)
SMP r y
Alice
Bob x r1 r r2 y
Alice
Bob
Ref

10. Randomized protocol for EQ
Recall that we’ve proved that D(EQ) ≥ n.
[RY] R1,priv(EQ) = O(log n).
Fix some p in [n2, 2n2]. ∀a0…an-1∊{0,1}n, define
a(t) = a0 + a1t + … + an-1tn-1 mod p
Communication: 2log2 p = O(log n). x y
Alice
Bob
Protocol:
t ∊R {0, …, p-1}
t, x(t) O u t p ( 1 i f x ) = y 6

11. Correctness Recall: p∊[n2,2n2], a(t) = a0 + a1t + … + an-1tn-1 mod px
Alice
Bob y
Protocol:
t ∊R {0, …, p-1}
t, x(t) O u t p ( 1 i f x ) = y 6
Recall: p∊[n2,2n2], a(t) = a0 + a1t + … + an-1tn-1 mod p
Case x = y: x(t) = y(t), ∀t. So Bob outputs 1.
Case x  y: x(t) and y(t) are different polynomials of degree n-1.
[Thm] Any polynomial of degree d has ≤d roots.
A random t∊{0, …, p-1} is one of these roots w.p. (n-1)/p < 1/n.
So Bob outputs 1 w.p. < 1/n.

12. Public coin [Thm] R∥,pub(EQ) = O(1).
Complexity: 1 from Alice and 1 from Bob.
Case x = y: x·r = y·r, ∀r.
Case x  y: x·r = y·r ⇔ (x⊕y)·r = 0. So Bob outputs 1 w.p. ½.
[Fact] If z  00…0, then x y r 2 R f ; 1 g n
Alice
Bob
Protocol: x r y r
Ref O u t p ( 1 i f x r = y 6
x·r = x1r1 + … + xnrn mod 2
Repeat the protocol k times (with fresh r) decreases the error prob. to 2-k. z r = ( w . p 1 2

13. Approximate rank Given a matrix M,
rankε(M) = min {rank(M’): |Mij - M’ij| ≤ ε}.
Rpub(f) = Ω(log rankε(Mf))
Rpub(EQ) = O(1)
⇒ IN can be made to O(1) rank
by perturbing each entry by 0.01. I N = 2 6 4 1 . 3 7 5

14. Newman’s result We’ve seen the difference between Rpub and Rpriv.
[Newman91] Rpriv(f) ≤ Rpub(f) + O(log n).
Let’s see whether we have time for the proof at the end of the talk.

15. Hamming Distance HD(x,y) = |x⊕y| = |{i: xi  yi}|.
[Yao03] R||,pub(Hamd) = O(d2)
[GKdW04] R||,pub(Hamd) = O(d log n)
[HSZZ06] R||,pub(Hamd) = O(d log d)
Also shows Rpub(Hamd) = Ω(d). H a m d ( x ; y ) = 1 i f D
> Q

16. [HSZZ06] R||,pub(Hamd) = O(d log d)
Assume that HD(x,y) ≤ m = O(d2).
The case of distinguishing HD(x,y) ≤ d and HD(x,y) > m is easy.

17. [HSZZ06] R||,pub(Hamd) = O(d log d)
Assume that HD(x,y) ≤ m = O(d2).
Complexity: O(d log(d2)) = O(d log d) x
Alice
Bob y
Protocol: R a n d o m l y p r t i [ ] b c k s B ( 1 ) ; : . a j = P r i t y ( x B ) b j = P a r i t y ( B ) ( 8 j 2 [ m ] )
GKdW04-Protocol for Hamd(a,b)
Ref

18. [HSZZ06] R||,pub(Hamd) = O(d log d)
Assume that HD(x,y) ≤ m = O(d2).
[Fact] Whp, each block contains ≤1 i s.t. xi  yi.
Thus HD(xB(j),yB(j)) = Parity(xB(j)⊕yB(j))
= Parity(xB(j))⊕Parity(yB(j))
Hamd(x,y) = Hamd(a,b). x
Alice
Bob y
Protocol: R a n d o m l y p r t i [ ] b c k s B ( 1 ) ; : . a j = P r i t y ( x B ) b j = P a r i t y ( B ) ( 8 j 2 [ m ] )
GKdW04-Protocol for Hamd(a,b)
Ref

19. XOR functions EQ and Hamd are XOR functions f(x⊕y)
EQ: f = OR
Hamd: f = threshold(d).
XOR functions are an important class of functions for communication complexity
Relation to decision tree complexity
Position in strongly balanced composed functions
What can we say about CC of XOR fn’s?

20. Symmetric XOR functions
f is symmetric:
f(x) only depends on |x|.
i.e. f(x) = f(π(x)), ∀π∊Sn.
i.e. f(x) = S(|x|).
Let r = r0 + r1, where r0, r1 ≤ n/2 are the min integers s.t. S(k)=S(k+2), ∀k∊[r0, n−r1).
[SZ09] For symmetric fn’s,
n/2 n r0 r1 …
n/2 n r0 r1 … R ( f x y ) = ~ r

21. R(f(x⊕y)) = Õ(r) Alice Bob
Protocol:
Use Hamr0 and Hamr1 to see
|x⊕y| ∊ [0,r0), [r0,n-r1), or [n-r1,n]?
If |x⊕y|∊[r0,n-r1): Alice sends Parity(x).
If |x⊕y|∊[0,r0) or [n-r1,n], Alice and Bob use binary
search and Hamd to find |x⊕y|. Output S(|x⊕y|).
Complexity: O(r log r) (log2r)O(r log r) = Õ(r).
Correctness:
If |x⊕y|∊[0,r0) or [n-r1,n], Alice and Bob finds |x⊕y| exactly.
If |x⊕y|∊[r0,n-r1), then f(x) depends only on Parity(x⊕y), by def of r. So Alice sending Parity(x) is enough.

22. Questions for you [SZ09] Question: Close the gap for R1(f(x⊕y))!
One-way: no binary search possible.
Question: Close the gap for R1(f(x⊕y))!
Question: Or even pin down R||,pub(f(x⊕y))?
Question: What can we say about general XOR fn’s? R ( f x y ) = ~ r , R 1 ; p u b ( f x y ) = ~ O r 2

23. Thanks

24. Proof of R p r i v + ( f ) u b O l o g n 1
ε-error pub-coin protocol → (ε+δ)-error pub-coin protocol using O(log n + logδ-1) random bits.
Actually, Ǝr1, …, rt s.t. ∀(x,y), Pri∊[t] [P(x,y,ri)f(x,y)] < ε+δ (*)
Why exist? Choose them randomly! Prr1…rt[(*)] < exp(-δ2t) // Chernoff bound
< 2-2n, when t=O(n/δ2).
Then Prr1…rt[Ǝ(x,y) s.t. (*)] < 1, i.e. Ǝr1, …, rt s.t. ∀(x,y), (*) happens.

25. Thanks Again