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Figure 23 Each problem of size n is divided into a subproblems of size n/b
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As k goes from 0 (the root) to log b n (the leaves), these numbers form a geometric series with ratio a/bd Finding the sum of such a series in big-O notation is easy (Exercise 02), and comes down to three cases 1 The ratio is less than 1 Then the series is decreasing, and its sum is just given by its rst term, O(n d ) 2 The ratio is greater than 1 The series is increasing and its sum is given by its last term, O(n logb a ): nd a bd
logb n
= nd
alogb n (blogb n )d
= alogb n = a(loga n)(logb a) = nlogb a
3 The ratio is exactly 1 In this case all O(log n) terms of the series are equal to O(n d ) These cases translate directly into the three contingencies in the theorem statement 53
Binary search
The ultimate divide-and-conquer algorithm is, of course, binary search: to nd a key k in a large le containing keys z[0, 1, , n 1] in sorted order, we rst compare k with z[n/2], and depending on the result we recurse either on the rst half of the le, z[0, , n/2 1], or on the second half, z[n/2, , n 1] The recurrence now is T (n) = T ( n/2 ) + O(1), which is the case a = 1, b = 2, d = 0 Plugging into our master theorem we get the familiar solution: a running time of just O(log n)
23 Mergesort
The problem of sorting a list of numbers lends itself immediately to a divide-and-conquer strategy: split the list into two halves, recursively sort each half, and then merge the two sorted sublists function mergesort(a[1 n]) Input: An array of numbers a[1 n] Output: A sorted version of this array if n > 1: return merge(mergesort(a[1 n/2 ]), mergesort(a[ n/2 + 1 n])) else: return a The correctness of this algorithm is self-evident, as long as a correct merge subroutine is speci ed If we are given two sorted arrays x[1 k] and y[1 l], how do we ef ciently merge them into a single sorted array z[1 k + l] Well, the very rst element of z is either x[1] or y[1], whichever is smaller The rest of z[ ] can then be constructed recursively function merge(x[1 k], y[1 l]) if k = 0: return y[1 l] if l = 0: return x[1 k] if x[1] y[1]: return x[1] merge(x[2 k], y[1 l]) else: return y[1] merge(x[1 k], y[2 l]) Here denotes concatenation This merge procedure does a constant amount of work per recursive call (provided the required array space is allocated in advance), for a total running time of O(k + l) Thus merge s are linear, and the overall time taken by mergesort is T (n) = 2T (n/2) + O(n), or O(n log n) Looking back at the mergesort algorithm, we see that all the real work is done in merging, which doesn t start until the recursion gets down to singleton arrays The singletons are 54
Figure 24 The sequence of merge operations in mergesort
Input: 10 2
7 13 1
2 2 10 2 3
5 3 5 10 5
13 7 13 1 6
1 1 7 13 6
7 10 13
merged in pairs, to yield arrays with two elements Then pairs of these 2-tuples are merged, producing 4-tuples, and so on Figure 24 shows an example This viewpoint also suggests how mergesort might be made iterative At any given moment, there is a set of active arrays initially, the singletons which are merged in pairs to give the next batch of active arrays These arrays can be organized in a queue, and processed by repeatedly removing two arrays from the front of the queue, merging them, and putting the result at the end of the queue In the following pseudocode, the primitive operation inject adds an element to the end of the queue while eject removes and returns the element at the front of the queue
function iterative-mergesort(a[1 n]) Input: elements a1 , a2 , , an to be sorted Q = [ ] (empty queue) for i = 1 to n: inject(Q, [ai ]) while |Q| > 1: inject(Q, merge(eject(Q), eject(Q))) return eject(Q)
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