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Unformatted text preview: COT 5405  Fall 2010 Homework 6(solution) December 13, 2010 Grading Policy: Please contact TA Chunchun Zhao by email or in his o ce hours for any grading issues. Maximum score is 100 points. Each completed question worth 10 points. Problem 4 , 5 are graded for 25 points each. Partial credit is given if you dont answer a question completely. Please notify the TA if you nd anything wrong in this solution. Problem 1 Ex32.41: Compute the pre x function π for the pattern ababbabbabbabab babb when the alphabet is ∑ = a,b . Solutions: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 P a b a b b a b b a b b a b a b b a b b π [ i ] 1 2 1 2 1 2 1 2 3 4 5 6 7 8 Ex32.45: Give a lineartime algorithm to determine if a text T is a cyclic rotation of another string T ´ . For example, arc and car are cyclic rotations of each other. Solutions: If T.strenglength 6 = T .strenglenth return false; else Let T 1 = TT ( concatenation of two strings T and T ; T 2 = T . This problem is equivalent to following stringmacthing problem: Search for the pattern T 2 in string T 1 , which can be solved by KMP algorithm in linear time. 1 Problem 2 Ex34.43: Professor Jagger proposes to show that SAT ≤ P 3 CNF SAT by using only the truthtable technique in the proof of Theorem 34.10, and not the other steps. That is, the professor proposes to take the boolean formula φ , form a truth table for its variables, derive from the truth table a formula in 3 DNF that is equivalent to ¬ φ , and then negate and apply DeMorgan's laws to produce a 3CNF formula equivalent to φ . Show that this strategy does not yield a polynomialtime reduction. Solutions: Suppose φ has only one clause with n variables, forming the truthtable for this boolean formula will require 2 n rows. Thus the reduction is in time Ω(2 n ) . Ex34.47: Let 2CNFSAT be the set of satis able boolean formulas in CNF with exactly 2 literals per clause. Show that 2 CNF SAT ∈ P . Make your algorithm as e cient as possible. (Hint: Observe that x W y is equivalent to ¬ x → y . Reduce 2CNFSAT to a problem on a directed graph that is e ciently solvable.) Solutions: Let I be an arbitrary instance of 2Sat having n literals { x 1 ,x 2 ,...,x n } . Create a graph G = ( V,E ) represents I , where V = { x 1 ,x 2 ...,x n , ¬ x 1 , ¬ x 2 ,... ¬ x n } , E = { ( x i ,x j )  either ( ¬ x i W x j ) or ( ¬ x j W x i ) in I } . Lemma: I is satis able if and only if there is no x i ∈ V such that there is a path from x i to ¬ x i and from ¬ x i to x i in G . Proof: 1) If there is a path x i to ¬ x i and a path ¬ x i to x i then I is unsatis able: If x is true, then by the transitivity of → , there must be one clause on that path from x i to ¬ x i is false ( T → F ) . I is unsatis able....
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This note was uploaded on 01/15/2012 for the course COT 5405 taught by Professor Ungor during the Fall '08 term at University of Florida.
 Fall '08
 UNGOR
 Algorithms

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