算法导论（CLRS, 2nd） 个人答案 Ch5.1

5.1-1:

Since we can compare between any two elements, we know the total order on the ranks of the candidates.

5.1-2:

RANDOM(a,b)

%range = b - a;

%d = the number of binary bits %range has

 

do

for(%i = 1 to %d):

%i th bit of integer %res = rand(0,1);

while(%res > %range);

return (a + %res);

 

Each time of while loop need O(log(b-a)) time, and the expectation of that while loop is constant. Therefore, it takes time O(log(b-a))

 

5.1-3:

UNBIASED-RANDOM()

do:

a = random(0,1);

b = random(0,1);

while(a==b);

return a;

 

The possibility for (1,0) is p(1-p), and the same for (0,1).

Thus, the possibility for the termination of while loop is 2p(1-p);

the expectation is 1/(2p(1-p))

 

Thus the expected running time is O(1/(2p(1-p)))

 

 

 

I've spent some time thinking about how to find this expectation:

 

Suppose the possibility for event e happen is p, then the exception is:

转载于:https://www.cnblogs.com/flyfy1/archive/2011/02/16/1955685.html