证明算法有下界

Question

I'm trying to prove this problem:

if an algorithm exists that can determine if a sorted list of n elements has duplicate elements in it, than the number of comparisons needed has a lower bound of n-1 .

I'm not quite familiar with lower and higher bounds and I seem to confuse it, can someone help me with an easy to understand proof?

Answer 1

The problem statement is not rigorous. It should say "the number of comparisons in the worst case ".

In a sorted array, there are n-1 relations between pairs of successive elements, which are either < or = . If all elements are different, you cannot deduce the outcome of a comparison from that of other comparisons. Hence you cannot avoid an exhaustive search, taking up to n-1 tests.

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By the way, n-1 is also an upper bound on the worst case, as after the exhaustive search you always have the answer.

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In the best case, when the first two elements are equal, you find the answer after exactly 1 comparison. Hence, lower and upper bounds on the best case are both 1 .

Answer 2

If you have the array list = {1, 2, 3, 4, 5} , you have to compare the 1 and the 2, the 2 and the 3, the 3 and the 4 and so on, to determine if it has any duplicates in it.

So the total number of comparisons is 4 and the total number of elements is 5 and thus n-1 comparisons are needed.