Seminar on Communication Complexity

Rotem ZachNovember 1st, 2009

Quick Overview• A rectangle in X × Y is a subset R X × Y such that R = ⊆

A × B for some A X and B Y.⊆ ⊆• A rectangle R ⊆ X × Y is called

f-monochromatic if f is fixed on R.• Every protocol f partitions X × Y into f-monochromatic

rectangles.• If every partition of X × Y into

f-monochromatic rectangles requires at least t rectangles, then D(f) ≥ ceil(log2t). These are the 0-rectangles and the 1-rectangles.

Lower Bound• No function has a single “hard” input. Why?– Alice and Bob can each send a bit saying “we have

the half of the hard input”.• A lower bound D(m) gives a set of inputs for

which any protocol that computes f must use at least m bits of communication

Material • Finding lower bounds for the communication

complexity of problems:– Fooling Sets– Weights and Distributions– Rank Lower Bound

Fooling Set (Definition)Let f: X × Y → {0,1} . A set S X × Y is called a ⊆fooling set (for f) if there exists a value z {0,1} such that:∈

• For every (x,y) S, f(x,y) = z∈• For every two pairs (x1,y1) ≠ (x2,y2) in S,

either f(x1,y2)≠z or f(x2,y1)≠z 0 0 0 11 1 0 10 1 0 11 0 1 0

Example• EQ: 0 ≤ x , y < 22

– EQ(x,y) = 1 iff x = y

• How many mono-rectangles are there?• Which fooling set can you find?• S = {(00,00),(01,01),(10,10),(11,11)}

EQ 00 01 10 11

00 1 0 0 0

01 0 1 0 0

10 0 0 1 0

11 0 0 0 1

Lemma 1If f has a fooling S set of size t, then D(f) ≥ log2tProof:

• Claim: No x,y S can be in the same rectangle∈– Assume R contains (x1,y1) ≠ (x2,y2) (both belong to S)

– From the definition, f(x1,y2) ≠ z or f(x2,y1) ≠ z– Thus, R is not monochromatic.

• Therefore, at least t monochromatic rectangles are need to cover f

Return

Exercise 1• GT: 0 ≤ x , y < 2n

– GT(x,y) = 1 iff x > y• Examples:– GT(5,6) = 0– GT(6,5) = 1– GT(6,6) = 0

• Prove D(GT) ≥ n

x\y 1 … … 2n

1 0 0 … 0

… 1 0 … …

… … … 0 0

2n 1 … 1 0

Solution 1• S = {(x,x) | 0 ≤ x < 2n}• For every (x,x) S, GT(x,x) = 0∈• For every (x,x) ≠ (y,y) S,∈

GT(x,y) = 1 or GT(y,x) = 1 ⇒ S is a fooling set

• |S| = 2n

• From lemma 1 , D(GT) ≥ log2(2n) = n

x\y 1 … … 2n

1 0 0 … 0

… 1 0 … …

… … … 0 0

2n 1 … 1 0

We can do better!• What did we actually prove?• We proved: number of 0-rectangles is at least 2n

– Because every member of fooling set is in a 0-rectangle, and we have at least 2n different members

• But, we have at least one 1-rectangle• Thus D(GT) ≥ log2(2n + 1) = n + 1• And because trivial upper bound is n+1:

D(GT) = n + 1

Material • Finding lower bounds for the communication

complexity of problems:– Fooling Sets ✓– Weights and Distributions– Rank Lower Bound

Weighting Rectangles• Idea - proving that if every monochromatic

rectangle is “small”, there must be many of them.

• Weight:– µ: X × Y → [0,1] – Sum of µ(x,y) for every (x,y) X × Y = 1∈

• We choose the weights!

Weighted Fooling Sets• Our choice of weights:– Given fooling set S of size t– µ(x,y) = 0 for (x,y) S∉– µ(x,y) = 1/t for (x,y) S∈– If we can prove that S is a fooling set, only one

weighted point can be in every rectangle. Thus µ(R) ≤ 1/t.

Weighted Fooling Sets (Example)

GTx\y 1 … … 2n

1 0 0 … 0

… 1 0 … …

… … … 0 0

2n 1 … 1 0

x\y 1 … … 2n

1 12n

0 … 0

… 0 12n

… …

… … … 12n

0

2n 0 … 0 12n

Weights

Probability Distribution• Formalization of weights• Definition:

µ: 2F → [0,1]µ (F) = 1A ∩ B = µ (A B) = µ (A) + µ (B)∅⇒ ∪

• Proposition: Let µ be a probability distribution of X × Y. If every monochromatic rectangle R has measure µ(R) ≤ δ, then D(f) ≥ log2 (1/ δ)

Exercise 2• DISJ: x, y {1,2,…,n}⊆– DISJ(x,y) = 1 iff x ∩ y = ∅

• Examples:– DISJ({1,2,3},{3,4}) = 0– DISJ({3,7,9},{1,2}) = 1

• Prove D(DISJ) ≥ n

A E

Solution 2• Define P = 2({1,2,3,…,n})

• The weights: For every A,E P: µ(A,Ā) = 1/2∈ n

(A ≠ Ē) µ(A,E) = 0• Claim: Every rectangle contains at most one

pair (A,Ā)– We will prove DISJ(A,Ā) ≠ DISJ(A, Ē). Why is this

sufficient?

Solution 2 (continued)• Claim: Every rectangle contains at most one

pair (A,Ā)– For every A ≠ E P, without loss of generality, ∈

exists x A and x E.∈ ∉ ⇒ A ∩ Ē ≠ ∅

⇒ DISJ(A,Ē) = 0– DISJ(A,Ā) = 1 and DISJ(A, Ē) = 0 ⇒

Every rectangle contains at most one pair (A,Ā), has size ≤ 1/2n

Solution 2 (final)• Definition: For every A,E P: µ(A,Ā) = 1/2∈ n

(A ≠ Ē) µ(A,E) = 0• We proved: Every rectangle contains at most

one pair (A,Ā) ⇒every rectangle has size ≤ 1/2n

• From the proposition, D(f) ≥ log2 [1/ (1/2n)] = n

• (S = {(A,Ā) | A P} is a fooling set)∈

Material • Finding lower bounds for the communication

complexity of problems:– Fooling Sets ✓– Weights and Distributions ✓– Rank Lower Bound

Rank Lower Bound• Algebraic tool to calculating lower bound• Create matrix M of size |X| * |Y|– Indexed by elements of X and Y– M(x,y) = f(x,y)

Lemma 2• For any function f: X × Y → {0,1},

D(f) ≥ log2 rank(M)

Proof• For each leaf k where output is 1:– Mk(x,y) = 1 iff (x,y) R∈ k where Rk is all the inputs

leading to k– Claim: If f(x,y) = 1 there exists a single matrix

where Mk(x,y) = 1 • Each input leads to exactly one leaf ⇒

for each input (x,y), there is exactly one k such that Mk (x,y) = 1

– Thus, M = ∑ Mk where k goes over all leaves

D(f) ≥ log2 rank(M)

R1 …

… R2f =

1 0

0 0M1 =0 0

0 1M2 =

Proof (continued)• M = ∑ Mk where k goes over all leaves

• Because Mk has a single rectangle, rank(Mk)=1• Remember rank(A + B) ≤ rank(A) + rank(B)– We conclude, rank(M) ≤ ∑ rank(Mk)– rank(M) ≤ |Leaves of output 1| ≤ |Leaves|

= |mono-rectangles| ⇒ D(f) ≥ log2 rank(M)

D(f) ≥ log2 rank(M)

Improvement• We actually found a bound for the number of 1 rectangles, 0

rectangles are symmetric– Look at not(f), instead of f, and exact same proof

• Define J = (ji,k) s.t. j∀ i,k = 1• M = J – Mnot M = J + (–M⇒ not) ⇒

rank(M) ≤ rank(Mnot) + 1 ⇒ rank(Mnot) ≥ rank(M) - 1

• D(f) = log2 (|Rectangles|) ≥ log2(rank(M) + rank(Mnot)) ≥ log2(2*rank(M) - 1)

Example 3• MAJ: x,y {0,1}∈ 2

– MAJ(x,y) = 1 iff |0| < |1| (x and y combined)– If |0| = |1|, we return 0

• Examples:– MAJ(00,10) = 0– MAJ(01,11) = 1– MAJ(10,01) = 0

• Find D(MAJ)

Solution 300 01 10 11

00 0 0 0 0

01 0 0 0 1

10 0 0 0 1

11 0 1 1 1• rank(M) = 2• D(MAJ) ≥ log2(2*2-1) = 2• By previous means we could find better lower

bound!

M =

Questions?

Food for Thought• Is every partition a legal protocol? • Suppose Alice starts:

There are rectangles that aren’tcontained in a any (non-trivial) partition.

• Same with Bob

y y’ y’’

x 1 0 0

x’ 1 1 1

x’’ 0 0 1

NO!