Alice’s tree
Bob’s tree
x y
Figure 11.2: Protocol for disjointness of subtrees
Example 11.2.2. Given is a graph G with n points. Alice knows a vertex set SA spanning a complete subgraph and Bob knows an independent vertex set SB in the graph. They want to decide whether the two subgraphs have a common vertex.
If Alice wants to give the complete information to Bob about the vertex set known to her thenlogM bits would be needed, whereM is the number of complete subgraphs.
This can be, however, even 2n/2, i.e., (in the worst case) Alice must useΩ(n)bits. The situation is similar with Bob.
The following protocol is significantly more economical. Alice checks whether the setSA has a vertex with degree at mostn/2−1. If there is one, then she sends to Bob a 1 and then the name of such a vertex v. Then both of them know that Alice’s set consists only of v and some of its neighbors, i.e., they reduced the problem to a graph with n/2vertices.
If every node ofSA has degree larger thann/2−1then Alice sends to Bob only a 0.
Then Bob checks whether the setSB has a vertex with degree larger thann/2−1. If it has then it sends Alice a 1 and the name of such a node w. Similarly to the foregoing, after this both of them will know that besides w, the setSB can contain only vertices that are not neighbors of w, and they thus again succeeded in reducing the problem to a graph with at most (n+ 1)/2 vertices.
Finally, if every vertex of SB has degree at most n/2−1, Bob sends a 0 to Alice.
After this, they know that their sets are disjoint.
The above turn uses at most O(logn) bits and since it decreases the number of vertices of the graph to half, it will be repeated at most logn times. Therefore, the complete protocol is onlyO((logn)2). More careful computation shows that the number of used bits is at most dlogne(2 +dlogne)/2.
11.3 Non-deterministic communication complexity
As with algorithms, the nondeterministic version plays an important role also with protocols. This can be defined—in a fashion somewhat analogous to the notion of
11. Chapter: Communication complexity 191
“witness”, or “testimony”—in the following way. We want that for every input of Alice and Bob for which the answer is 1, a “superior being” can reveal a short 0-1 sequence convincing both Alice and Bob that the answer is indeed 1. They do not have to believe the revelation of the “superior being” but if they signal anything at all this can only be whether on their part they accept the proof or not. This non-deterministic protocol consists therefore of certain possible “revelations” z1, . . . , zn ∈ {0,1}t all of which are acceptable for certain inputs of Alice and Bob. For a given pair of inputs, there is an zi acceptable for both of them if and only if for this pair of inputs, the answer to the communication problem is 1. The parametert, the length of the zi’s is the complexity of the protocol. Finally, thenondeterministic communication complexityof matrixC is the minimum complexity of all non-deterministic protocols applicable to it; we denote this byκN D(C)
Example 11.3.1. In Example 11.1.1, if the superior being wants to prove that the two strings are different it is enough for her to declare: “Alice’s i-th bit is 0 while Bob’s is not.” This is—apart from the textual part, which belongs to the protocol—
only dlogne+ 1 bits, i.e., much less than the complexity of the optimal deterministic protocol.
We remark that even the superior being cannot give a proof that two words are equal in fewer thann bits, as we will see right away.
Example 11.3.2. Suppose that Alice and Bob know a convex polygon each in the plane, and they want to decide whether the two polygons have a common point.
If the superior being wants to convince the players that their polygons are not disjoint she can do this by revealing a common point. Both players can check that the revealed point indeed belongs to their polygon.
We can notice that in this example, the superior being can also easily prove the negative answer: if the two polygons are disjoint then it is enough to reveal a straight line such that Alice’s polygon is on its left side, Bob’s polygon is on its right side. (We do not discuss here the exact number of bits in the inputs and the revelations.)
Let z be a possible revelation of the superior being and let Hz be the set of all possible pairs (i, j) for which z “convinces” the players that cij = 1. We note that if (i1, j1) ∈ Hz and (i2, j2) ∈ Hz then (i1, j2) and (i2, j1) also belong to Hz: since (i1, j1) ∈ Hz, Alice, possessing i1, accepts the revelation z; since (i2, j2) ∈ Hz, Bob, possessing j2, accepts the revelation z; thus, when they have (i1, j2) both accept z, hence(i1, j2)∈Hz.
We can therefore also consider Hz as a submatrix of C consisting of all 1’s. The submatrices belonging to the possible revelations of a nondeterministic protocol cover the 1’s of the matrixC since the protocol must apply to all inputs with answer 1 (it is possible that a matrix element belongs to several such submatrices). The 1’s of C can therefore be covered with at most2κN D(C) all-1 submatrices.
Conversely, if the 1’s of the matrix C can be covered with 2t all-1 submatrices then it is easy to give a non-deterministic protocol of complexity t: the superior being reveals only the number of the submatrix covering the given input pair. Both players verify whether their respective input is a row or column of the revealed submatrix. If yes then they can be convinced that the corresponding matrix element is 1. We have
192 11.3. Non-deterministic communication complexity thus proved the following statement:
Lemma 11.3.1. κN D(C) is the smallest natural number t for which the 1’s of the matrix C can be covered with 2t all-1 submatrices.
In the negation of Example 11.3.1, the matrix C is the 2n×2n identity matrix.
Obviously, only the 1×1 submatrices of this are all-1, the covering of the 1’s requires therefore2n such submatrices. Thus, the non-deterministic complexity of this problem is also n.
Letκ(C) =s. ThenC can be decomposed into2s submatrices half of which are all-0 and half are all-1. According to Lemma 11.3.1, the nondeterministic communication complexity of C is therefore at most s−1. Hence
κN D(C)≤κ(C)−1.
Example 11.3.1 shows that there can be a big difference between the two quantities.
LetC denote the matrix obtained from C by changing all 1’s to 0 and all 0’s to 1.
Obviously, κ(C) =κ(C). Example 11.3.1 also shows thatκN D(C)and κN D(C)can be very different. On the basis of the previous remarks, we have
max{1 +κN D(C), 1 +κN D(C)} ≤κ(C).
The following important theorem shows that here, already, the difference between the two sides of the inequality cannot be too great.
Theorem 11.3.2 (Aho-Ullman-Yannakakis).
κ(C)≤(κN D(C) + 2)·(κN D(C) + 2).
We will prove a sharper inequality. In case of an arbitrary 0-1 matrix C, let %(C) denote the largest numbertfor whichChas at×tsubmatrix in which—after a suitable rearrangement of the rows and columns—there are all 1’s in the main diagonal and all 0’s everywhere above the main diagonal. Obviously,
%(C)≤rk(C), and Lemma 11.3.1 implies
log%(C)≤κN D(C).
The following inequality therefore implies Theorem 11.3.2.
Theorem 11.3.3.
κ(C)≤2 + log%(C)(2 +κN D(C)) unless C is the all zero matrix.
Proof. We use induction on%(C). If%(C) = 1then the protocol is trivial. Let%(C)>1 andp=κN D(C). Then the 0’s of the matrixCcan be covered with2pall-0 submatrices, say,M1, . . . , M2p. We want to give a protocol that decides the communication problem
11. Chapter: Communication complexity 193 with at most (p + 2) log%(C) bits. The protocol fixes the submatrices Mi, this is therefore known to the players.
For every submatrix Mi, let us consider the matrix Ai formed by the rows of C intersecting Mi and the matrix Bi formed by the columns of C intersecting Mi. The basis of the protocol is the following, very easily verifiable, statement.
Claim 11.3.4.
%(Ai) +%(Bi)≤%(C).
Now, we can describe the following protocol.
Alice checks whether there is an index i for which Mi intersects her row and for which %(Ai) ≤ 12%(C). If yes, then she sends “1” and the index i to Bob, with this the first phase of the protocol has ended. If not, then she sends “0”. Now, Bob checks whether there is an index i for which Mi intersects his column and %(Bi)≤ 12%(C). If yes, then he sends a “1” and the indexito Alice. Else he sends “0”. Now the first phase has ended in any case.
If either Alice or Bob find a suitable index in the first phase, then by the commu-nication of at most p+ 2 bits they have restricted the problem to a matrix C0 (=Ai orBi) for which %(C0)≤ 12%(C). Hence the theorem follows by induction.
If both players sent “0” in the first phase then they can finish the protocol: the answer is “1”. Indeed, if there was a 0 in the intersection of Alice’s row and Bob’s column then this would belong to some submatrixMi. However, for these submatrices, we have on the one hand
%(Ai)> 1 2%(C)
(since they did not suit Alice), on the other hand
%(Bi)> 1 2%(C)
since they did not suit Bob. But this contradicts the above claim.
It is interesting to formulate another corollary of the above theorem (compare it with Lemma 11.1.1):
Corollary 11.3.5.
κ(C)≤2 +¡
logrk(C)¢¡
κN D(C) + 2) unless C is the all zero matrix.
To show the power of Theorems 11.3.2 and 11.3.3 consider the examples treated in Section 11.2. If C is the matrix corresponding to Example 11.2.1 (in which 1 means that the subtrees are disjoint) thenκN D(C)≤ dlogne(it is sufficient to name a common vertex). It is also easy to obtain thatκN D(C) ≤1 +dlog(n−1)e (if the subtrees are disjoint then it is sufficient to name an edge of the path connecting them, together with telling that after deleting it, which component will contain TA and which one TB). It can also be shown that the rank ofC is2n. Therefore, whichever of Theorem 11.3.2 or
194 11.4. Randomized protocols