算法导论习题[Exercises 32.1-3 ]

Suppose that pattern P and text T are randomly chosen strings of length m and n, respectively, from the d-ary alphabet Σ_d = {0, 1, . . . , d - 1}, where d ≥ 2. Show that the expected number of character-to-character comparisons made by the implicit loop in line 4 of the naive algorithm is

over all executions of this loop. (Assume that the naive algorithm stops comparing characters for a given shift once a mismatch is found or the entire pattern is matched.) Thus, for randomly chosen strings, the naive algorithm is quite efficient.

证明：

令P为单个字符比较匹配的概率，1-P为失配的概率,P=d^-1

比较次数 概率 比较次数 * 概率

1 1 - P 1 - P

2 P（1 - P） 2P - 2P²

3 P² (1 - P) 3P - 3P³

…

m-1 P^m-2(1 - P) (m-1)P^m-2 - (m-1) P^m-1

m P^m-1(1 - P)+P^m-1P (m)P^m-1 - (m) P^m+ (m)P^m

E =∑(比较次数 * 概率) = 1+P+P²+…..P^m-1 = (1-P^m)/1-P = (1- d ^–m )/( 1- d^-1)

令f(x) =(1- x ^–m )/( 1- x^-1),对f（x）求导可知，在m>1,x>=2时导数为负，则f(x)在x>=2严格减函数，所以f(x)<=f(2)<=2.证毕。

由此可知，对于随机的字符串，朴素的字符串比较还是有效的。