# Cormen Edition 3 Exercise 8.1 Question 4 (Page No. 194)

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Suppose that you are given a sequence of $n$ elements to sort.The input sequence consists of $n/k$ subsequences, each containing $k$ elements.The elements in a given subsequence are all smaller than the elements in the succeeding subsequence and larger than the elements in the preceding subsequence. Thus, all that is needed to sort the whole sequence of length $n$ is to sort the $k$ elements in each of the $n/k$ sequences Show an $\Omega(n\ lg\ k)$ lower bound on the number of comparisons needed to solve this variant of the sorting problem. (Hint: It is not rigorous to simply combine the lower bounds for the individual subsequences.)

## Related questions

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Show that there is no comparison sort whose running time is linear for at least half of the $n!$ inputs of length $n$.What about a fraction of $1/n$ inputs of length $n$? What about a fraction $1/2^n$?
BUCKET-SORT(A) 1 let B[0...n-1] be a new array 2 n = A.length 3 for i – 0 to n – 1 4 make B[i] an empty list 5 for i = 1 to n 6 insert A[i] into list B[nA[i]] 7 for i = 0 to n – 1 8 sort list B[i] with insertion sort 9 concatenate the lists B[0] , B[1] , ….,B[n-1] together in order illustrate the operation of BUCKET-SORT on the array $A=\langle .79,.13,.16,.64,.39,.20,.89,.53,.71,.42\rangle$
Obtain asymptotically tight bounds on $lg\ (n!)$ without using Stirling’s approximation. Instead, evaluate the summation $\sum_{k=1}^{n} lg\ k$.
A probability distribution function $P(x)$ for a random variable $X$ is defined by $P(x) =Pr\{X\leq x\}$.Suppose that we draw a list of $n$ random variables $X_1,X_2,…,X_n$ from a continuous probability distribution function $P$ that is computable in $O(1)$ time. Give an algorithm that sorts these numbers in linear averagecase time.