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Asked: 2026-06-01T06:36:23+00:00

In Algorithm Design Manual , there is such an excise 4-26 Consider the problem

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In Algorithm Design Manual, there is such an excise

4-26 Consider the problem of sorting a sequence of n 0’s and 1’s using
comparisons. For each comparison of two values x and y, the algorithm
learns which of x < y, x = y, or x > y holds.

(a) Give an algorithm to sort in n − 1 comparisons in the worst case.
Show that your algorithm is optimal.

(b) Give an algorithm to sort in 2n/3 comparisons in the average case
(assuming each of the n inputs is 0 or 1 with equal probability). Show
that your algorithm is optimal.

For (a), I think it is fairly easy. I can choose a[n-1] as pivot, then do something like in quicksort partition, scan 0 to n – 2, find the middle point where left side is all 0 and right side is all 1, this take n – 1 comparisons.

But for (b), I can’t get a clue. It says “each of the n inputs is 0 or 1 with equal probability”, so I guess I can assume the numbers of 0 and 1 equal? But how can I get a result which is related to 1/3? divide the whole array into 3 groups?

Thanks

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1 Answer

  1. Editorial Team

    “0 or 1 with equal probability” is the condition for “average” case. Other cases may have worse timing.

    Hint 1: 2/3 = 1/2 + 1/8 + 1/32 + 1/128 + …

    Hint 2: Consider the sequence as a sequence of pairs and compare the items in each pair. Half will return equal; half will not. Of the half that are unequal you know which item in the pair is 0 and which is 1, so those need no more comparisons.

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