{0, 1}n by replacing xi by xixi in x, and y′ ∈ {0, 1}n by replacing yi by yiyi in y.
It is easy to see that EQn/2(x, y) = DISJn(x′, y′) so from the previous proposition we obtain:
7.4.11. Corollary (Buhrman & de Wolf [48]). If ε≥ 2−n/2, then we have Qccε(DISJn)∈ Ω(log(n/ε)).
In particular, both equality and disjointness require Ω(n) qubits of communi-cation if we want the error probability to be exponentially small.
7.5 Non-Deterministic Complexity
Above we showed a lower bound on Qcc∗E(f ) in terms of log rank(f ), but were not able to prove the desired general upper bound on Qcc∗E(f ) in terms of log rank(f ).
In this section we will prove that such a result holds for the non-deterministic case: the non-deterministic quantum communication complexity NQcc(f ) equals log nrank(f ) up to a factor of 2, where nrank(f ) is the non-deterministic analogue of rank(f ), to be defined below.
7.5.1 Some definitions
Consider some communication complexity problem f :{0, 1}n× {0, 1}n→ {0, 1}.
A non-deterministic protocol for f is a protocol whose acceptance probability (=
the probability of outputting 1) on an input (x, y) is positive iff f (x, y) = 1. For a discussion of this choice of definition and a comparison with other potential definitions, we refer to Section 5.6. We use Ncc(f ) and NQcc(f ) for the cost of optimal classical and quantum non-deterministic protocols for f , respectively.
It is well known that the classical non-deterministic complexity Ncc(f ) is closely related to the minimal size of a 1-cover for f , defined as follows. A rectangle is a subset R = S × T ⊆ X × Y . Such an R is a 1-rectangle (for f ) if f (x, y) = 1 for all (x, y) ∈ R. A 1-cover for f is a set of 1-rectangles whose union contains all (x, y) ∈ {0, 1}n × {0, 1}n for which f (x, y) = 1. We use Cov1(f ) to denote the minimal size (i.e., minimal number of rectangles) of a 1-cover for f . Similarly we define 0-rectangles, 0-covers, and Cov0(f ). Now it is easy to prove that Ncc(f ) =⌈log Cov1(f )⌉ (see e.g. [109, Section 2.1]), so the classical non-deterministic communication complexity is completely determined by the combinatorial notion of 1-covers.
Below we show that the quantum non-deterministic communication complex-ity is almost completely determined by the algebraic notion of non-deterministic rank, defined as follows. Recall that the communication matrix Mf of f is the 2n× 2n Boolean matrix whose x, y entry is f (x, y) and that rank(f ) denotes the rank of Mf over the reals. A real 2n× 2n matrix M is called a non-deterministic
communication matrix for f if it has the property that M (x, y)6= 0 iff f(x, y) = 1.
Thus M is any matrix obtainable by replacing 1-entries in Mf by non-zero reals.
Let the non-deterministic rank of f , denoted nrank(f ), be the minimum rank over all non-deterministic matrices M for f . Without loss of generality we can assume all M -entries are in [−1, 1], because we can divide by maxx,y|M[x, y]|
without changing the rank of M .
7.5.2 Equality to non-deterministic rank
Here we characterize the non-deterministic quantum communication complexity NQcc(f ) in terms of the non-deterministic rank nrank(f ):
7.5.1. Theorem (de Wolf [160]). log nrank(f )
2 ≤ NQcc(f) ≤ ⌈log nrank(f)⌉.
Proof. Consider an NQcc(f )-qubit non-deterministic quantum protocol for f . Using Lemma 7.2.1 in the same way as in Theorems 7.2.2 and 7.4.4, its acceptance probabilities P (x, y) form a matrix of rank≤ 22NQcc(f ). It is easy to see that this is a non-deterministic matrix for f , hence nrank(f ) ≤ 22NQcc(f ) and the first inequality follows.
For the upper bound, let r = nrank(f ) and M be a rank-r non-deterministic matrix for f . Let MT = U ΣV be the singular value decomposition of MT (see Appendix A.3), so U and V are unitary, and Σ is a diagonal matrix whose first r diagonal entries are positive real numbers and whose other diagonal entries are 0.
Below we describe a 1-round non-deterministic protocol for f , using⌈log r⌉ qubits.
First Alice prepares the vector |φxi = cxΣV|xi, where cx > 0 is a normalizing real number that depends on x. Because only the first r diagonal entries of Σ are non-zero, only the first r amplitudes of |φxi are non-zero, so |φxi can be compressed into⌈log r⌉ qubits. Alice sends these qubits to Bob. Bob then applies U to|φxi and measures the resulting state. If he observes |yi then he outputs 1 and otherwise he outputs 0. The acceptance probability of this protocol is
P (x, y) = |hy|U|φxi|2
= c2x|hy|UΣV |xi|2
= c2x|MT(y, x)|2
= c2x|M(x, y)|2.
Since M (x, y) is non-zero iff f (x, y) = 1, P (x, y) will be positive iff f (x, y) = 1.
Thus we have a non-deterministic protocol for f with ⌈log r⌉ qubits. 2 In sum: classically we have Ncc(f ) = ⌈log Cov1(f )⌉ and quantumly we have NQcc(f )≈ log nrank(f).
7.5. Non-Deterministic Complexity 135
7.5.3 Exponential quantum-classical separation
We now give an f with an exponential gap between the classical complexity Ncc(f ) and the quantum complexity NQcc(f ). For n > 1, define f by
f (x, y) = 1 iff |x ∧ y| 6= 1.
We first show that the quantum complexity NQcc(f ) is low:
7.5.2. Theorem (de Wolf [160]). NQcc(f )≤ ⌈log(n + 1)⌉ for the above f.
Proof. By Theorem 7.5.1, it suffices to prove nrank(f )≤ n + 1. Let Mi be the Boolean matrix whose (x, y)-entry is 1 if xi = yi = 1, and whose (x, y)-entry is 0 otherwise. Notice that Mi has rank 1. Now define a 2n× 2n matrix M by
M (x, y) = P
iMi(x, y)− 1
n− 1 .
Note that M (x, y) is non-zero iff the Hamming weight of x∧ y is different from 1, hence M is a non-deterministic matrix for f . Because M is the sum of n + 1
rank-1 matrices, it has rank at most n + 1. 2
Now we show that the classical Ncc(f ) is high (both for f and its complement):
7.5.3. Theorem (de Wolf [160]). Ncc(f )∈ Ω(n) and Ncc(f) ≥ n − 1 for the above f .
Proof. Let R1, . . . , Rk be a minimal 1-cover for f . We use the following result from [109, Example 3.22 and Section 4.6], which is essentially due to Razborov [139].
There exist sets A, B⊆ {0, 1}n×{0, 1}nand a probability distribution µ :{0, 1}n× {0, 1}n→ [0, 1] such that all (x, y) ∈ A have |x ∧ y| = 0, all (x, y)∈ B have |x ∧ y| = 1, µ(A) = 3/4, and there are constants α, δ > 0 (independent of n) such that for all rectangles R, µ(R∩ B) ≥ α· µ(R ∩ A) − 2−δn.
Since the Riare 1-rectangles, they cannot contain elements from B. Hence µ(Ri∩ B) = 0 and µ(Ri∩ A) ≤ 2−δn/α. But since all elements of A are covered by the Ri we have
3
4 = µ(A) = µ Ã k
[
i=1
(Ri∩ A)
!
≤ Xk
i=1
µ(Ri∩ A) ≤ k ·2−δn α . Therefore Ncc(f ) =⌈log k⌉ ≥ δn + log(3α/4).
For the lower bound on Ncc(f ), consider the set S = {(x, y) | x1 = y1 = 1, xi = yi for i > 1}. This S contains 2n−1 elements, all of which are 1-inputs for f . Note that if (x, y) and (x′, y′) are two elements from S then |x ∧ y′| > 1 or |x′ ∧ y| > 1, so a 1-rectangle for f can contain at most one element of S.
This shows that a minimal 1-cover for f requires at least 2n−1 rectangles and
Ncc(f )≥ n − 1. 2
Another quantum-classical separation was obtained earlier by Massar et al. [120]:
7.5.4. Theorem (MBCC [120]). For the non-equality problem on n bits, we have NQcc(NEn) = 1 versus Ncc(NEn) = log n.
Proof. Ncc(NEn) = log n is well known (see [109, Example 2.5]). Below we give the [120]-protocol for NEn.
Viewing her input x as a number ∈ [0, 2n− 1], Alice rotates a |0i-qubit over an angle xπ/2n, obtaining a qubit cos(xπ/2n)|0i + sin(xπ/2n)|1i which she sends to Bob. Bob rotates the qubit back over an angle yπ/2n, obtaining cos((x− y)π/2n)|0i + sin((x − y)π/2n)|1i. Bob now measures the qubit and outputs the observed bit. If x = y then sin((x− y)π/2n) = 0, so Bob always outputs 0. If x6= y then sin((x − y)π/2n)6= 0, so Bob will output 1 with probability > 0. 2 Note that nrank(EQn) = 2n, since every non-deterministic matrix for equal-ity will be a diagonal 2n × 2n matrix with non-zero diagonal entries. Thus NQcc(EQn)≥ (log nrank(EQn))/2 = n/2, which contrasts sharply with the non-deterministic quantum complexity NQcc(NEn) = 1 of its complement.