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找到找到 k 对的时间复杂度的下限

[英]Finding lower bound on time complexity of finding k pairs

原文 2016-10-07 02:33:00 6 1 algorithm/ asymptotic-complexity

We are given n bolts and n nuts of different sizes, where each bolt exactly matches one nut.我们有 n 个不同尺寸的螺栓和 n 个螺母，其中每个螺栓与一个螺母完全匹配。 Our goal is to find the matching nut for each bolt.我们的目标是为每个螺栓找到匹配的螺母。 The nuts and bolts are too similar to compare directly;螺母和螺栓太相似无法直接比较； however, we can test whether any nut is too big, too small, or the same size as any bolt.但是，我们可以测试任何螺母是否太大、太小或与任何螺栓尺寸相同。

Prove that in the worst case Ω(n + k log n) nut-bolt comparisons are required to find k matching pairs.证明在最坏的情况下需要进行 Ω(n + k log n) 螺母螺栓比较才能找到 k 个匹配对。

I'm thoroughly stumped on how to do this, I figure a 3-ary decision tree with for this would need n^k nodes, thus giving klog(n) for the height of the tree, but I can't figure out where the +n comes from.我对如何做到这一点感到非常困惑，我认为一个三元决策树需要 n^k 个节点，从而为树的高度提供 klog(n)，但我不知道在哪里+n 来自。

1 个解决方案

Here is a very loose answer that might form the basis of a proper proof:这是一个非常松散的答案，可能构成正确证明的基础：

Suppose that you have to ask me (your adversary) every time you want to test a new nut or a new bolt and I hand you a new nut or a new bolt.假设您每次要测试新螺母或新螺栓时都必须问我（您的对手），而我递给您一个新螺母或新螺栓。 Before you start I sort the nuts and bolts into matching pairs, then divide the pairs into two heaps, each of size n/2.在开始之前，我将螺母和螺栓分类成匹配的对，然后将这些对分成两堆，每堆大小为 n/2。 For the first n/2 steps I hand you a nut from the first heap, or a bolt from the second heap, depending on what you ask for.对于前 n/2 个步骤，我会从第一个堆中给您一个螺母，或者从第二个堆中取一个螺栓，具体取决于您的要求。 So I can always delay you from finding any matches until you have made at least n/2 matching attempts, because until I run out of one of the heaps, you will never get a nut and bolt that match.因此，我总是可以延迟您找到任何匹配项，直到您至少进行了 n/2 次匹配尝试，因为在我用完其中一个匹配项之前，您将永远无法获得匹配项。

Your life is not made any harder if I label the bolts in order of size, because you can always choose to ignore this information.如果我按尺寸顺序标记螺栓，您的生活不会变得更难，因为您总是可以选择忽略这些信息。 Now if you can find k matches in time less than k log n, for all possible k and n >= k, then you can solve the problem of sorting n numbers using only comparisons, where the numbers are known to be the set {1,2,3...n}.现在，如果您可以在小于 k log n 的时间内找到 k 个匹配项，对于所有可能的 k 并且 n >= k，那么您可以解决仅使用比较对 n 个数字进行排序的问题，其中已知数字为集合 {1 ,2,3...n}。 In fact, even if you have a magic method that works only for eg k<=3 and all n, you can still do this low-comparison sort by repeatedly finding 3 matches between the set of (unlabelled) nuts that remain and the set of (labelled) bolts.事实上，即使你有一个只对 k<=3 和所有 n 有效的魔法方法，你仍然可以通过在剩余的（未标记的）坚果集和剩余的坚果集之间重复找到 3 个匹配来进行这种低比较排序（标记的）螺栓。 So if you can find matches with fewer than k log n comparison, you can sort numbers known to be {1,2...n} with fewer than n log n comparisons - but the usual information-theoretic lower bound on sorting numbers still holds here.因此，如果您可以找到少于 k log n 比较的匹配项，您可以使用少于 n log n 比较对已知为 {1,2...n} 的数字进行排序 - 但排序数字的常用信息理论下界仍然在这里举行。 So you need at least k log n comparisons.所以你至少需要 k log n 比较。

So now we have an lower bound of max(n/2, k log n).所以现在我们有一个下界 max(n/2, k log n)。 We don't care about factors, so let's have max(n, k log n).我们不关心因素，所以让我们有 max(n, k log n)。 But (a + b) / 2 <= max(a, b) <= a + b for a,b >= 0 so again neglecting factors we can turn this into n + k log n.但是 (a + b) / 2 <= max(a, b) <= a + b for a,b >= 0 所以再次忽略因素我们可以把它变成 n + k log n。