Running Time of Randomized Binary Search

问题

Consider the following silly randomized variant of binary search. You are given a sorted array A of n integers and the integer v that you are searching for is chosen uniformly at random from A. Then, instead of comparing v with the value in the middle of the array, the randomized binary search variant chooses a random number r from 1 to n and it compares v with A[r]. Depending on whether v is larger or smaller, this process is repeated recursively on the left sub-array or the right sub-array, until the location of v is found. Prove a tight bound on the expected running time of this algorithm.

Here is what I got for the T(n)

T(n) = T(n-r) + T(r) + Θ(1)

However, I have no clue how to get a tight bound.

回答1:

Your formulation of T(n) is not completely correct. Actually,

Lets try to look over all the cases. When we reduce the problem size by partitioning the array across any random point, the reduced sub-problem will have any size from 1 to n with uniform probability. Hence with probability 1/n, search space becomes r. So expected running time becomes

T(n) = sum ( T(r)*Pr(search space becomes r) ) + O(1) = sum ( T(r) )/n + O(1)

Which gives,

T(n) = average(T(r)) + O(1)

Let expected time complexity of random binary sort be T(n).

T(n) = [ T(1)+T(2)+...+T(n)]/n + 1
n*T(n) = T(1)+T(2)+...+T(n) + n
(n-1)*T(n-1) = T(1)+T(2)+...+T(n-1) + n-1       [substituiting n by n-1]
n*T(n) - (n-1)*T(n-1) = T(n) + 1
(n-1)*T(n) - (n-1)*T(n-1) =  1
(n-1)*T(n) = (n-1)*T(n-1) + 1
T(n) = 1/(n-1) + T(n-1)
T(n) = 1/(n-1) + 1/(n-2) + T(n-2)               [ T(n-1) = T(n-2) + 1/(n-2) ]
...
T(n) = 1 + 1/2 + 1/3 + ... + 1/(n-1) = H(n-1) < H(n) = O(log n) 
[ H(n) = reciprocal sum of first n natural numbers   ]

so, T(n) = O(log n)

Harmonic number

bound of H(n)

来源：https://stackoverflow.com/questions/40034475/running-time-of-randomized-binary-search