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Unformatted text preview: 1 CMPS 101 Algorithms and Abstract Data Types Recurrence Relations Iteration Method Recall the following example from the induction handout. ≥ + = = 2 1 ) 2 / ( 1 ) ( n n T n n T We begin by illustrating a solution technique called Iteration , which consists of repeatedly substituting the recurrence into itself until a pattern emerges. ) 2 / ( 1 ) ( n T n T + = ) 2 / ( 2 ) 2 / 2 / ( 1 1 2 n T n T + = + + = ) 2 / ( 3 ) 2 / 2 / ( 1 2 3 2 n T n T + = + + = M ) 2 / ( k n T k + = This process must terminate when the recursion depth k is at its maximum, i.e. when 1 2 / = k n . To solve this equation for k in terms of n , we use the inequality definition of the floor function. 2 2 / 1 < ≤ k n ∴ 1 2 2 + < ≤ k k n ∴ 1 ) lg( + < ≤ k n k ∴ ) lg( n k = Thus for the recursion depth ) lg( n k = we have ) 1 ( ) 2 / ( = = T n T k , and hence the solution to the above recurrence is ) lg( ) ( n n T = . It follows that )) (log( ) ( n n T Θ = . Exercise Check directly that ) lg( ) ( n n T = is the solution to the above recurrence relation, i.e. check that ) 1 ( = T , and for any 2 ≥ n , that ) 2 / ( 1 ) ( n T n T + = . Exercise Use this same technique to show that the recurrence ≥ + = = 2 1 ) 2 / ( 1 ) ( n n S n n S has solution ) lg( ) ( n n S = , and hence also )) (log( ) ( n n S Θ = . 2 Comparing the solutions to the preceding examples, we see that replacing floor by ceiling has no affect on the asymptotic solution )) (log( n Θ , while the exact solutions are different: ) lg( n vs. ) lg( n . We can change other details in a recurrence without changing the asymptotic solution. The following recurrence satisfies )) (log( ) ( n n T Θ = for any values of the constants c, d , and n ....
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 Spring '09
 AgoreBack
 Algorithms, 2 K, 1 K, Recurrence relation, 0 k, Master Theorem, 2 2 K

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