In this paper we deal with the problem of finding the smallest and the largest elements of a totally ordered set of size $n$ using pairwise comparisons if $k$ of the comparisons might be erroneous where $k$ is a fixed constant. We prove that at least $(k+1.5)n+\Theta(k)$ comparisons are needed in the worst case thus disproving the conjecture that $(k+1+\epsilon)n$ comparisons are enough.
Search problems with lies have been studied in many different settings (see surveys Deppe [2] and Pelc [5]). In this paper we deal with the model when a fixed number, k, of the answers may be false, which we call lies. There are also several models depending on what kind of questions are allowed as well, the most famous being the Rényi-Ulam game. In this paper we deal with the case when we are given n different elements and we can use pairwise comparisons to decide which element is bigger from the two.
The problem of finding the maximum (or the minimum) element with k lies was first solved by Ravikumar et al. [8]. They have shown that (k + 1)n -1 comparisons are necessary and sufficient. The topic of this paper is finding the maximum and the minimum. If all answers have to be correct then the minimum number of comparisons needed is ⌈ 3n 2 ⌉ -2 (see [6]). Aigner in [1] proved that (k + Θ( √ k))n + Θ(k) comparisons are always sufficient * . It was proved by Gerbner et al. [3] that if k = 1, then 87n 32 + Θ(1) comparisons are necessary and sufficient. We also made the conjecture that for general k, there is an algorithm using only (k + 1 + c k )n comparisons where c k tends to 0 as k tends to infinity. Hoffmann et al. [4] showed that (k + 1 + C)n + O(k 3 ) comparisons are sufficient for some absolute constant C (whose value is less than 10 but no attempts to optimize it were made yet). Until now the best lower bound on c k was Ω((1 + √ 2) -k ) by Aigner [1]. The main result of this paper is the following theorem.
Theorem 1 At least ⌈(k + 1.5)(n -1) -0.5⌉ = (k+1.5)n+Θ(k) comparisons are needed in the worst case to find the largest and the smallest element if there might be k erroneous answers. This bound is tight for k = 0 (see Theorem 4) but not for k = 1 as shown in [3] and using a slightly more involved argument than the one presented here it is easy to see that the bound can be simply improved for any k ≥ 1. The reason why the theorem is presented in this “weak” form is that it already disproves the conjecture and the argument is simply, yet gives a perfectly matching bound for k = 0. To find a stronger version would involve a thorough case analysis, similar to the one in [3] and improving the constant a bit is uninteresting at the moment. It would be more interesting to study the behavior of c k in future works. Now we know that 1.5 ≤ c k ≤ C ∼ 10. But is c k monotonously increasing as k grows? This would imply, of course, the existence of a limit, which is likely to exist.
The rest of the paper is organized as follows. In Section 2 we develop a method to increase the lower bound by k for many search problems and give proofs using it for some known results. In Section 3 we prove our main result, Theorem 1.
In this section, as a warm-up, we prove a very general result that holds for all search problems and generally gives an additional constant to the lower bounds that are proved using a consistent adversary.
Claim 2 Suppose we have a search problem where we want to determine the value of some function f using (not necessarily yes-no) questions from a family of allowed questions. The answers are given by an adversary who can lie at most k times. Suppose that we have already asked some questions and the answers we got are consistent, i.e. it is possible that none of them is a lie. If we do not yet know the value of f, then we need at least k + 1 further questions to determine it. This claim has an immediate, quite weak corollary.
Corollary 3 If there is a search problem as in Claim 2 with a non-trivial f, then we need at least 2k + 1 questions to determine f. Although it is not too standard, we first give a proof of the Corollary, as it is a simplified version of the proof of the Claim.
Proof. Take two possible elements of the universe, x and y, for which f(x) = f(y). The adversary can answer the first k questions according to x and the next k questions according to y, thus after 2k questions both are still possible.
Proof of Claim 2. Suppose we have already asked some consistent questions, i.e. there is an x such that they are all true for x. However, if we do not yet know f, there is a y for which at most k of these questions would be false, such that f(x) = f(y). We can answer the next k questions according to y.
To show the power of this simple claim, let us prove the following theorem.
Theorem 4 (Ravikumar et al. [8]) To find the maximum among n elements using comparisons of which k might be incorrect, we need (k + 1)n -1 comparisons in the worst case.
Proof. The upper bound follows from using any tournament scheme and comparing any two elements until one of them is bigger than the other k + 1 times. This is (k + 1)(n -1) plus the possible k lies that might prolong our search.
To prove the lower bound, answer the first (k + 1)(n -1) -1 questions consistently. Now we have an element that was always bigger, and another that was the smaller one at most k times, thus the conditions of Claim 2 ar
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