QB3 - Dynamic Programming - Solution

# QB3 - Dynamic Programming - Solution - COMP 271 Design and...

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Unformatted text preview: COMP 271 Design and Analysis of Algorithms 2004 Fall Semester Solutions to Question Bank Number 3 (Selected Problems) 4. (a) Consider the case where wt [ i ] = 1 for all i (the worst case must be at least as bad as this special case). The proof boils down to observing that the recursion tree is a complete binary tree whose height is essentially h = min( n, W ). The number nodes of will be 2 h . More formally, let T ( i, W ) denote the running time of the algorithm for a given pair i and W . We can see that we have the following recurrence (up to constant factors): T ( i, W ) = braceleftBigg 1 if i = 0 or w < T ( i- 1 , W ) + T ( i- 1 , W- 1) otherwise. It is an easy induction proof that T ( i, W ) ≥ 2 min( i,W ) . The basis case i = 0 or W = 0 is trivial. For the induction step we have T ( i, W ) ≥ T ( i- 1 , W ) + T ( i- 1 , W- 1) ≥ 2 min( i- 1 ,W ) + 2 min( i- 1 ,W- 1) ≥ 2 · 2 min( i- 1 ,W- 1) = 2 · 2 min( i,W )- 1 = 2 · 2 min( i,W ) / 2 = 2 min( i,W ) . (b) The problem with the recursive version is that it recomputes many of the same function values over and over again. Again assume that wt [ i ] = 1 for all i . Let R ( i, W ) be a shorthand for the call with parameters i and W . R ( i, W ) calls R ( i- 1 , W ) and R ( i- 1 , W- 1). Both of these call R ( i- 2 , W- 1). As you trace the algorithm deeper, you will see that the same procedure is invoked over and over again. The dynamic programming version avoids this duplication, since once a value has been computed for a given i and W , this effort is never repeated. 5. Let X = < x 1 , . . ., x n > be the given sequence of n numbers. We need to find the longest increasing subsequence in X ....
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## This note was uploaded on 10/18/2009 for the course COMP 271 taught by Professor Arya during the Spring '07 term at HKUST.

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QB3 - Dynamic Programming - Solution - COMP 271 Design and...

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