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A certificate for an input $x$ is a subset of bits $S \subseteq \{1,...,n\}$ such that for all inputs $y$, $(\forall i \in S \quad y_i = x_i) \rightarrow f(y) = f(x)$. Then $C_x(f)$ is the minimum size of a certificate for input $x$ and the certificate complexity $C(f) = \max_x C_x(f)$. Thus, certificate complexity can be seen as a form of query complexity for nondeterministic machines: guess the smallest certificate for $x$ and then verify with $C_x(f)$ many queries. This is used as an intermediate complexity measure when proving relationships between deterministic query complexity ($D(f)$) and quantum query complexity ($Q_2(f)$).

Is it known, or believed that for a total functions $f$, $C(f) \leq Q_2(f)$? Are there any total functions $f$, where it is know that $C(f) \in o(Q_2(f))$ (or vice-versa)? And just for fun: does this question correspond in some formal way to the questions associated NP vs. BQP?

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    $\begingroup$ well, query complexity can only separate classes in a relativized world. And we know that $NP\nsubseteq BQP$ relative to an oracle because of the lower bound on search. You cannot separate NP from BQP using decision trees or certificate complexity, these are weaker models of computation. $\endgroup$ – Marcos Villagra Mar 8 '11 at 22:28
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In the paper Negative weights make adversaries stronger, Lee, Spalek, and Hoyer give a function for which $ADV^{±}(f)= \Omega( ({C_0(f)C_1(f)})^{0.549})$, where $ADV^±$ denotes the general adversary lower bound. This lower bound has since been proved to characterize quantum query complexity.

This function is obtained by composing a function on a few bits bits with itself $d$ times. You get in the end $C_0(f)=C_1(f)=3^d$ and thus $C(f) \in o(Q_2(f))$.

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There are total functions with $Q_2 = o(C)$ and total functions with $C = o(Q_2)$. Marc gave one direction.

For the other direction, the OR function on $n$ inputs works. The certificate complexity is $n$ and the quantum query complexity is $\sqrt{n}$, by Grover's search algorithm. However, this example really only highlights that the correct quantity to look at is the geometric mean of the true and false certificate complexities, $\sqrt{C_0 C_1}$, and not the certificate complexity $C = \max\{C_0, C_1\}$.

A better example is the function OR($n/2$ inputs, AND($n/2$ inputs)). The 1 and 0 certificate complexities are $n/2$ and $n/2+1$ but the adversary bounds and therefore $Q_2$ are $\sqrt{n}$. (Any read-once AND-OR formula of size $n$ has $\sqrt{n}$ adversary bounds.)

To my knowledge, for functions with a transitive symmetry group on the input bits, e.g., symmetrical under cyclic permutations, no examples are known with $ADV^{\pm} < \sqrt{C_0 C_1}$. (On the other hand, the original adversary bound satisfies $ADV \leq \sqrt{C_0 C_1}$ for every total function; see, e.g., [Spalek and Szegedy: "All quantum adversary bounds are equivalent" Theorem 3.2].)

Finally, if the co-domain of the total function is non-boolean, the most natural quantity to consider is $\max_{i \neq j} \sqrt{C_i C_j}$. See the above Spalek and Szegedy theorem for details, but it should not be surprising. Note that a quantity like $(\prod_{i=1}^{m} C_i)^{1/m}$ is not well behaved when you extend the function's codomain. There are good questions here.

Another reference you might find interesting is [Aaronson, quant-ph/0210020]. Aaronson defines and relates randomized and quantum certificate complexities in a natural manner.

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  • $\begingroup$ does the geometric mean continue to make sense when we look at functions to an arbitrary set S of size m > 2? Would we then just have $(C_0C_1...C_{m - 1})^{1/m}$ or something more subtle? $\endgroup$ – Artem Kaznatcheev Mar 11 '11 at 12:05

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