CS580 assignment #2 (Due: Wed. Feb. 10)

Problem 1 (10 points): Exercise 33.4-5.

Problem 2 (10 points): Problem 9-1.

Problem 3 (10 points): Exercise 9.3-1.

Problem 4 (10 points): Exercise 8.2-4.

Problem 5 (10 points): Exercise 8.3-4.

Problem 6 (10 points): Using adversary argument, prove that any comparison sort algorithm requires Omega(n lg n) comparisons to sort an array of n integers in worst case.

Problem 7 (10 points): We saw in class that the SELECT algorithm can find the k th largest element in O(n) worst-case time. Now, using adversary argument, prove that n - 1 + min(k - 1, n - k) is a lower bound on the number of comparisons needed to find the k th largest element in an n-element array.

Problem 8 (15 points): Consider an n-element array A[1..n]:

(1) Give an O(n)-time algorithm to determine if any element of A appears at least n/2 times in A. If such an element exists, your algorithm should return its value; else it should return NULL.

(2) Give an O(nk)-time algorithm to determine if any element of A appears at least n/k times in A. If such an element exists, your algorithm should return its value; else it should return NULL.

Problem 9 (15 points):

Let be a bit string of length 5, i.e. for . Consider the problem of determining whether b contains three consecutive zeros, i.e. whether or not b contains the substring 111. We restrict our attention to those algorithms whose only allowable operation is to peek at a bit. Obviously 5 peeks are sufficient. A decision tree argument provides the (useless) fact that at least one peek is necessary.

(1) Use an adversary argument to show that 4 peeks are necessary in general.

(2) Design an algorithm which solves the problem using only 4 peeks in worst case. Express your algorithm as a decision tree.