9.1 Minimum and maximum

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## 9.1 Minimum and maximum

How many comparisons are necessary to determine the minimum of a set of n elements? We can easily obtain an upper bound of n - 1 comparisons: examine each element of the set in turn and keep track of the smallest element seen so far. In the following procedure, we assume that the set resides in array A, where length[A] = n.

    MINIMUM(A)
    1  min ← A[1]
    2  for i ← 2 to length[A]
    3         do if min > A[i]
    4                then min ← A[i]
    5  return min

Finding the maximum can, of course, be accomplished with n - 1 comparisons as well.

Is this the best we can do? Yes, since we can obtain a lower bound of n - 1 comparisons for the problem of determining the minimum. Think of any algorithm that determines the minimum as a tournament among the elements. Each comparison is a match in the tournament in which the smaller of the two elements wins. The key observation is that every element except the winner must lose at least one match. Hence, n - 1 comparisons are necessary to determine the minimum, and the algorithm MINIMUM is optimal with respect to the number of comparisons performed.

#### Simultaneous minimum and maximum

In some applications, we must find both the minimum and the maximum of a set of n elements. For example, a graphics program may need to scale a set of (x, y) data to fit onto a rectangular display screen or other graphical output device. To do so, the program must first determine the minimum and maximum of each coordinate.

It is not difficult to devise an algorithm that can find both the minimum and the maximum of n elements using Θ(n) comparisons, which is asymptotically optimal. Simply find the minimum and maximum independently, using n - 1 comparisons for each, for a total of 2n - 2 comparisons.

In fact, at most 3 ⌊n/2⌋ comparisons are sufficient to find both the minimum and the maximum. The strategy is to maintain the minimum and maximum elements seen thus far. Rather than processing each element of the input by comparing it against the current minimum and maximum, at a cost of 2 comparisons per element, we process elements in pairs. We compare pairs of elements from the input first with each other, and then we compare the smaller to the current minimum and the larger to the current maximum, at a cost of 3 comparisons for every 2 elements.

Setting up initial values for the current minimum and maximum depends on whether n is odd or even. If n is odd, we set both the minimum and maximum to the value of the first element, and then we process the rest of the elements in pairs. If n is even, we perform 1 comparison on the first 2 elements to determine the initial values of the minimum and maximum, and then process the rest of the elements in pairs as in the case for odd n.

Let us analyze the total number of comparisons. If n is odd, then we perform 3 ⌊n/2⌋ comparisons. If n is even, we perform 1 initial comparison followed by 3(n - 2)/2 comparisons, for a total of 3n/2 - 2. Thus, in either case, the total number of comparisons is at most 3 ⌊n/2⌋.

Exercises 9.1-1

Show that the second smallest of n elements can be found with n + ⌈lg n⌉ - 2 comparisons in the worst case. (Hint: Also find the smallest element.)

Exercises 9.1-2: ⋆

Show that ⌈3n/2⌉ - 2 comparisons are necessary in the worst case to find both the maximum and minimum of n numbers. (Hint: Consider how many numbers are potentially either the maximum or minimum, and investigate how a comparison affects these counts.)

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