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Unformatted text preview: CS 202: Data Structures and Discrete Mathematics II Fall 2010 Homework 3 () Solutions 1. Recurrences (60 points) (a) Find an exact solution to the following recurrence: a n = a n 1 + 2 a n 2 + 3 n 2 16 n + 17 . 5 , with a = 1 ,a 1 = . 5 Solution: To solve recurrence a n = a n 1 + 2 a n 2 + 3 n 2 16 n + 17 . 5 • The homogeneous annihilator is E 2 E 2 = ( E 2)( E + 1). • The residue is h 6 n 2 + 38 n 42 i which is annihilated by ( E 1) 3 . • The annihilator is ( E 2)( E + 1)( E 1) 3 , and the generic solution is a n = α 2 n + β ( 1) n + γ + δn + n 2 . • The constants α,β,γ,δ, satisfying the equations a = 1 = α + β + γ a 1 = . 5 = 2 α β + γ + δ + a 2 = 1 = 4 α + β + γ + 2 δ + 4 a 3 = 5 . 5 = 8 α β + γ + 3 δ + 9 a 4 = 6 = 16 α + β + γ + 4 δ + 16 are 1 , 3 4 , 3 4 , 1 2 , 3 2 respectively, • The closed form of the recurrence is a n = 2 n + 3 4 ( 1) n 3 2 n 2 + n 2 3 4 . (b) Let a n be the number of additions performed in evaluating the determinant of an n × n matrix by the cofactor expansion method (expansion by minors) 1 . i. Write a recursive formula for a n . Justify briefly. Solution: a n = n · a n 1 + n 1 a 1 = 0 Explanation: Computing the determinant of an n × n matrix involves computing the de terminants of n other matrices, each of size ( n 1) × ( n 1). Once those are computed we then add them all together. The number of additions performed to compute a determinant of an n × n matrix must be the number of additions to compute the determinant of an 1 See http://mathworld.wolfram.com/DeterminantExpansionbyMinors.html 31 CS 202 Homework 3 ( n 1) × ( n 1) matrix multiplied by n because that’s how many ( n 1) × ( n 1) matrices there and then plus the n 1 additions necessary to sum up all the determinants. The base case is set to n = 1 to represent the situation when the matrix is of size 1 × 1 which is really just a single scalar number requiring no computation work at all. ii. The solution to the recursion is a n = n ! 1. iii. Prove that your formula correctly expresses the number of additions, a n . Solution: We’ll use a Proof by Induction to show that our recurrence relation in part i matches the exact solution in part ii . In other words we’ll show n · a n 1 + n 1 = n ! 1 Base Case: n = 2 . For n = 2, we get: Left Hand Side (LHS) = 2 · 0 + 2 1 = 1 Right Hand Side (RHS) = 2! 1 = 2 1 = 1 Both sides are equal which proves the base case. Inductive Hypothesis: Assume that the recurrence relation is true for all a k where 2 ≤ k < n . Inductive Step: Prove the recurrence relation is also correct for n . Proof: a n = n · a n 1 + n 1 by the definition of a n = n · (( n 1)! 1) + n 1 by the inductive hypothesis = n ! n + n 1 = n ! 1 This proves the recurrence relation from part i is equal to the exact equa tion given in part ii for a n ....
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This note was uploaded on 11/22/2010 for the course CS 202 taught by Professor Staff during the Spring '08 term at Ill. Chicago.
 Spring '08
 STAFF
 Data Structures

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