Question

(25pts) You are given two sorted lists of size m and n. Give an O(log m log n) time algorithm for computing the k-th smallest element in the union of the two lists Note that the only way you can access these values is through queries to the databases. Ina single query, you can specify a value k to one of the two databases, and the chosen database will return the k-th smallest value that it contains. Since queries are expensive, you would like to compute the median using as few queries as possible. Give an algorithm that finds the median value using at most O(log m +log n) queries.

Answer 1

Algorithm to compute the k th smallest element in the union of two lists: function kelement(S[1,... ,pl.[1...q].k) return t[k] return s[k] if k<(p+q2: return kelement(s[1 p/21.41...,q].k) else return kelement(s( 1 , . . . , p],t(Q2)+1, . . . ,aj, k-q/2) else if k<(p+q2: return kelement(s[1, Річ1, q/2],k) else return kelement(s[(p/2)+1,...pl,t1,..q,k-q/2)

Explanation .kelement" is the function which accepts the array s[1,..,pl. t[1...ql,and k as the input parameters and compute the kth smallest element in the union of two list. Initially, it checks whether the value of p is equal to 0 and returns the value t[k] . It checks whether the value of q is equal to 0 and returns the value s[k] . Then it checks whether the value of s[p/2] is greater than t[q/2]. o It checks whether the value of k is less than (p+q)/2. . If yes, then it returns the value returned by the function kelement which accepts the array s[1, ..p/2j, 4[1,..q].k) as the input parameters. -Otherwise, t returns the value returned by the tunction kelement wnich accepts the aray st pl.t[(q/2)+1,...,q],K-q/2 as the input parameters o Otherwise, it checks whether the value of k is less than (p+q)/2. . If yes, then it returns the value returned by the function kelement which accepts the array s[1, ....pl.1..q/2],k as the input parameters .Otherwise, it returns the value returned by the function kelement which accepts the array SI(p/2)+. .pl.,..,q],k-q/2 as the input parameters.

The algorithm first checks whether the middle value of first array is greater than second array If yes, it checks whether the value of k is less than the middle value of the combined array o If yes, only the first half of the first array and entire second array is processed by calling the same function kelement. o Otherwise, second half of the second array and entire first array is processed by calling the same function kelement. Otherwise, it checks whether the value of k is less than the middle value of the combined array. o If yes, only the first half of the second array and entire first array is processed by calling the same function kelement. o Otherwise, second half of the first array and entire first array is processed by calling the same function kelement. Thus, a constant time is required in each recursive call to halves the array with m elements and n elements. Hence, it takes log m + log n times. Therefore, the algorithm takes a running time of O(log m+ logn)" times.