1 Introduction
In this note we study the problem in which given two discrete functions and (, ) we have to determine if there is a collision, i.e., inputs such that . In contrast to the problem, where the input is a single function and we have to determine if is injective, is nontrivial even when . This is the setting we focus on.
Both and have wide applications as useful subroutines in more complex algorithms [5, 12] and as a means of lower bounding complexity [10, 1].
and were first tackled by Buhrman et al. in 2000 [8] where they gave an algorithm and lower bound. In 2003 Ambainis, introducing a novel technique of quantum walks, improved the upper bound to in the query model [4]. It was soon realized that a similar approach works for [9, 13, 15]. Meanwhile Aaronson and Shi showed a lower bound that holds if the range [2]. Eventually Ambainis showed that the bound holds even if [3]. The same lower bound has since been reproved using the adversary method [14]. Until now, only the bound based on reduction of searching was known for with [8].
We consider quantum query complexity of where the input functions are given as a list of their values in black box. Let denote the bounded error quantum query complexity of . For a short overview of black box model refer to Buhrman and de Wolf’s survey [7]. Let denote . Let be defined as
2 Results
Theorem 1.
For all , we have
Proof.
Let , be the inputs of the function. We denote .
Consider the following algorithm parametrized by .

Select a random sample of size and query the variables .
Denote by the set containing their values. Do a Grover search for an element such that . If found, output 1. 
Select a random sample of size and query the variables .
Denote by the set containing their values. Do a Grover search for an element such that . If found, output 1. 
Run algorithm (with the value of specified below) with the following oracle:

To get : do a pseudorandom permutation on using seed and using Grover’s minimum search return the first value such that .

To get : do a pseudorandom permutation on using seed and using Grover’s minimum search return the first value such that .

Let , be the sets containing the indices of the variables which have values not seen in the steps 1 and 1’. We denote .
Let us calculate the probability that after step 1 there exists an unseen value
which is represented in at least variables, i.e., . Consider an arbitrary value such that . For , let be the event that . . Let . Then . Using Chernoff inequality (see e.g. [11]),The probability that there exists such is at most . Therefore, with probability after step , every value is represented in the input less than times. The same reasoning can be applied to step and the set . Therefore, with probability both and are at most .
Similarly, we show that with probability each appears as the first element from in at least one of the permutations of the oracle in step 2. Let be the event that appears in the th permutation as the first element from . . Let . . . . The same argument works for . Therefore, if there is a collision, it will be found by the algorithm with probability .
We also show that with probability , in all permutations the first element from appears no further than in position (and similarly for ). We denote by the event that in the th permutation in the th position is an element from . . We denote . . . . Therefore, the Grover’s minimum search will use at most queries.
The steps 1 and 1’ use queries to obtain the random sample, and queries to check if there is a colliding element on the other side of the input. The oracle in step 2 uses queries to obtain one value of or .
Therefore the total complexity of the algorithm is
By using the algorithm in step 2,
and the total complexity is minimized by setting . However, we can do better than that. Notice that the algorithm might not be the best choice for solving in step 2.
Let denote the regular algorithm. For , let denote a version of algorithm from Theorem 1 that in step 2 calls . Then we show that for all and all ,
where .
The proof is by induction on . For , we trivially have that . For the inductive step, consider the analysis of our algorithm. Let us set . First, notice that is nondecreasing in and for all . Thus for all , we have , hence and . Second, since the coefficient of is the function is above for , establishing . This confirms that is a valid choice of .
It remains to show that the complexity of step 2 does not exceed . By the inductive assumption and analysis of the algorithm, the complexity (up to logarithmic factors) of the second step is to the power of . Finally, we have to show that
By expanding and with a slight rearrangement, we obtain
We can further rearrange the required inequality by bringing to right hand side and everything else to the other. Then we get
After simplification we obtain , which is true.
Since and , the result follows. ∎
3 Lower Bound
We show a quantum query complexity lower bound for .
Theorem 2.
For all , we have .
Proof.
Let be the partial function defined as
Consider the function . One can straightforwardly reduce to by setting
and
4 Open Problems
Can we show that ? In particular, our algorithm struggles with instances where there are singletons only two (or none) of which are matching and the remaining variables are evenly distributed with copies each, such that none are matching. Thus our algorithm then either has to waste time sampling all the highfrequency decoy values or have most variables not sampled by step 2. If this lower bound held, it would imply a better lower bound for evaluating constant depth formulas and Boolean matrix product verification [10, Theorem 5].
References
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