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Recall that a function f is symmetric if f (x) only depends on the Hamming weight |x| of its input, so permuting the input does not change the value of the function.

2.6. Symmetric Functions 43

2.6.1 Tight bounds

A symmetric f is fully described by giving a vector (f0, f1, . . . , fN) ∈ {0, 1}N +1, where fk is the value of f (x) for |x| = k. Because of this and Lemma 2.2.2, there is a close relationship between polynomials that represent symmetric functions and single-variate polynomials that assume values 0 or 1 on{0, 1, . . . , N}. Using this relationship, von zur Gathen and Roche [79] proved deg(f ) = (1− o(1))N for all symmetric f :

2.6.1. Theorem (von zur Gathen & Roche [79]). If f is non-constant and symmetric, then deg(f ) = N − O(N0.548). If, furthermore, N + 1 is prime, then deg(f ) = N .

In fact, von zur Gathen and Roche conjecture that deg(f ) = N − O(1) for every symmetric f . The biggest gap they found is deg(f ) = N − 3 for some specific f and N . Since D(f )≥ deg(f) and QE(f )≥ deg(f)/2 (Theorem 2.4.2), the above degree lower bounds give strong lower bounds on D(f ) and QE(f ).

For the case of approximate degrees of symmetric f , Paturi [136] gave the following tight characterization. Define Γ(f ) = min{|2k − N + 1| : fk 6= fk+1}.

Informally, this quantity measures the length of the interval around Hamming weight N/2 where fk is constant. The following theorem of Paturi [136] implies a strong lower bound on Q2(f ) for all symmetric functions.

2.6.2. Theorem (Paturi [136]). If f is non-constant and symmetric, then we have gdeg(f )∈ Θ(p

N (N − Γ(f))).

We can prove a matching upper bound using the result about quantum count-ing from Section 1.7:

2.6.3. Theorem (BBCMW [17]). If f is non-constant and symmetric, then we have Q2(f )∈ Θ(p

N (N− Γ(f))).

Proof. We will sketch a strategy that computes f with bounded error prob-ability ≤ 1/3. Let fk = f (x) for x with |x| = k. First note that since Γ(f) = min{|2k −N +1| : fk6= fk+1 and 0≤ k ≤ N −1}, fkmust be identically 0 or 1 for k∈ {(N −Γ(f))/2, . . . , (N +Γ(f)−2)/2}. Consider some x with |x| = t. In order to be able to compute f (x), it is sufficient to know t exactly if t < (N − Γ(f))/2 or t > (N + Γ(f )− 2)/2, or to know that (N − Γ(f))/2 ≤ t ≤ (N + Γ(f) − 2)/2 otherwise.

Run the quantum counting algorithm for Θ(p

(N − Γ(f))N) steps to count the number of 1s in x. If t is in one of the two tails (t < (N − Γ(f))/2 or t > (N + Γ(f )− 2)/2), then with high probability the algorithm gives us an exact count of t. If (N−Γ(f))/2 ≤ t ≤ (N +Γ(f)−2)/2, then with high probability the counting algorithm returns some ˜t which is in this interval. Thus with bounded

error probability we have obtained sufficient information to compute ft = f (x), using only O(p

N (N − Γ(f))) queries. Repeating this procedure some constant number of times, we can limit the probability of error to at most 1/3. 2 In contrast to Theorem 2.6.3, it can be shown that a randomized classical strategy needs Θ(N ) queries to compute any non-constant symmetric f with bounded-error, the reason being that we can reduce ORN/2 to f (which is easy to see). Also, together with the bound Q0(ORN) = N that we prove below (Proposition 2.6.5), this reduction gives QE(f ), Q0(f )∈ Θ(N). To summarize, we have the following tight characterizations of the various decision tree complexities of all symmetric f :

2.6.4. Theorem. If f is non-constant and symmetric, then

• D(f) = (1 − o(1))N

• R0(f ), R2(f )∈ Θ(N)

• QE(f ), Q0(f ) ∈ Θ(N)

• Q2(f )∈ Θ(p

N (N − Γ(f)))

2.6.2 OR

Before continuing with monotone functions, we will take a closer look at three important symmetric functions: OR, PARITY, and MAJORITY.

First we will consider the OR-function, which is related to database search.

Grover’s search algorithm can find an index i such that xi = 1 with high prob-ability of success in O(√

N ) queries (if there is such an i). This implies that we can also compute the OR-function with high success probability in O(√

N ):

let Grover’s algorithm generate an index i, and return xi. Since bs(ORN) = N , Theorem 2.5.7 gives us a lower bound of 14

N on computing ORN with bounded error probability,2 so we have Q2(ORN) ∈ Θ(√

N ), where classically we require R2(ORN)∈ Θ(N) queries. Now suppose we want to get rid of the probability of error: can we compute ORN exactly or with zero-error using O(√

N ) queries? If not, can quantum computation give us at least some advantage over the classical deterministic case? Both questions have a negative answer:

2.6.5. Proposition (BBCMW [17]). Q0(ORN) = N .

2This Ω(

N ) lower bound on ORN is well known [25, 83], and is given in a tighter form in [30, 167], but the way we obtained it here is rather different from existing proofs. Many more bounds for OR and search will be proven in Chapter 3.

2.6. Symmetric Functions 45 Proof. Consider a quantum algorithm that computes ORN with zero-error using T = Q0(ORN) queries. By Lemma 2.4.1, there are complex-valued polynomials pi of degree at most T , such that the final state of the algorithm on input x is

X

i∈{0,1}m

αi(x)|ii.

Let B be the set of all basis states ending in 10 (i.e., where the output is the answer 0). Then for every i ∈ B we have pi(x) = 0 if x 6= ~0, otherwise the probability of getting the incorrect answer 0 on input x would be non-zero. On the other hand, there must be at least one j ∈ B such that pj(~0) 6= 0, since the probability of getting the correct output 0 on x = ~0 must be non-zero. Let p(x) be the real part of 1− pj(x)/pj(~0). This polynomial p has degree at most T and represents ORN. But then p has degree at least deg(ORN) = N , so T ≥ N. 2

2.6.3 PARITY

Secondly we consider PARITY. Using the Deutsch-Jozsa algorithm for n = 1, we can determine the parity of two variables using only 1 query. The parity of an N -bit input x is the parity of N/2 such pairs, so QE(PARITYN) ≤ ⌈N/2⌉.

A matching lower bound for bounded-error quantum algorithms follows from the next lemma, which is essentially due to Minsky and Papert:

2.6.6. Lemma (Minsky & Papert [122]). gdeg(PARITYN) = N .

Proof. Let f be PARITY on N variables. Let p be a polynomial of degree gdeg(f ) that approximates f . Since p approximates f , its symmetrization psym also approximates f . By Lemma 2.2.2, there is a polynomial q, of degree at most gdeg(f ), such that q(|x|) = psym(x) for all inputs. Thus we must have

|f(x) − q(|x|)| ≤ 1/3, so q(0) ≤ 1/3, q(1) ≥ 2/3, . . . , q(N − 1) ≥ 2/3, q(N) ≤ 1/3 (assuming N even). We see that the polynomial q(x)− 1/2 must have at least N zeroes, hence q has degree at least N and gdeg(f )≥ N. 2 Thus we having the following optimal result for parity, which was proven inde-pendently at around the same time as our result by Farhi, Goldstone, Gutmann, and Sipser.

2.6.7. Corollary (BBCMW [17]; FGGS [71]).

QE(PARITYN) = Q0(PARITYN) = Q2(PARITYN) =⌈N/2⌉.

2.6.4 MAJORITY

Thirdly we consider MAJORITY. Since the OR of N variables can be reduced to MAJORITY on 2N − 1 variables (if we set the first N − 1 variables to 1, then the MAJORITY of all variables equals the OR of the last N variables) and ORN requires N queries to be computed exactly or with zero-error, we have QE(MAJN)≥ Q0(MAJN)≥ (N + 1)/2. It is known that QE(MAJN)≤ N + 1 − w(N ), where w(N ) is the number of 1s in the binary expansion of N . This was first noted by Hayes, Kutin and Van Melkebeek [87]. It also follows immediately from classical results [146, 5] that show that an item with the majority value can be identified classically deterministically with N − w(N) comparisons between bits (a comparison between two input bits is the parity of the two bits, which can be computed with 1 quantum query). One further query to this item suffices to determine the majority value. For N satisfying w(N ) ≥ 2 we thus have QE(MAJN)≤ N − w(N) + 1 < N = D(MAJN).

For the zero-error case, Van Melkebeek, Hayes and Kutin give an algorithm that works in roughly 23N queries, which is still slightly worse than the best known lower bound Q0(MAJN)≥ (N + 1)/2. For the bounded-error case, we can apply Theorem 2.6.3: Γ(MAJN) = 1, so we need Q2(MAJN)∈ Θ(N) queries. The best upper bound we have here is N/2 +√

N , which follows from [60].