# L04 - Dynamic Programming An efficient way to implement some

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CS0250, Set 4 Dynamic Programming An efficient way to implement some divide-and-conquer algorithms

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CS0250, Set 4 Problem with divide-and-conquer Usually, a direct implementation of a divide-and-conquer algorithm is very inefficient. Example: Computing the Fibonacci numbers: F(0) = 1; F(1)=1; F(n)=F(n-1)+F(n-2); The first few Fibonacci numbers : 1,1,2,3,5,8,13,21,34,55,89,144,233,377,610,…, // fib.cpp int F(int n) { int m1,m2; if (n < 2) return 1; m1 = F(n-1); m2 = F(n-2); return (m1+m2); } It takes a long time to compute F(n) even for small n, such as n=50. Why?
CS0250, Set 4 The execution of F(7)

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CS0250, Set 4 The execution of F(7) We have made a lot of redundant computations. How to avoid them?
CS0250, Set 4 Idea for improvement Memorization: Store F(i) somewhere after we have computed its value. Afterward, we don’t need to re-compute F(i); we can retrieve its value from our memory. int v[100]; int F( int n) { if (v[n] <0 ) v[n] = F(n-1)+F(n-2); return v[n]; } int main() { int i, n; for (i =0; i<100; i++) v[i] = -1; v[0]=1; v[1]=1; cin >> n; cout << F(n) << end; return 0; }

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CS0250, Set 4 Look at the execution of F(7) -1 -1 -1 -1 -1 -1 1 1 v[0] v[1] v[2] v[3] v[4] v[5] v[6] v[7]
CS0250, Set 4 Look at the execution of F(7) -1 -1 -1 -1 -1 -1 1 1 v[0] v[1] v[2] v[3] v[4] v[5] v[6] v[7]

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CS0250, Set 4 Look at the execution of F(7) -1 -1 -1 -1 -1 2 1 1 v[0] v[1] v[2] v[3] v[4] v[5] v[6] v[7]
CS0250, Set 4 Look at the execution of F(7) -1 -1 -1 -1 -1 2 1 1 v[0] v[1] v[2] v[3] v[4] v[5] v[6] v[7]

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CS0250, Set 4 Look at the execution of F(7) -1 -1 -1 -1 3 2 1 1 v[0] v[1] v[2] v[3] v[4] v[5] v[6] v[7]
CS0250, Set 4 Look at the execution of F(7) -1 -1 -1 5 3 2 1 1 v[0] v[1] v[2] v[3] v[4] v[5] v[6] v[7]

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CS0250, Set 4 Look at the execution of F(7) -1 -1 -1 5 3 2 1 1 v[0] v[1] v[2] v[3] v[4] v[5] v[6] v[7]
CS0250, Set 4 Look at the execution of F(7) -1 -1 8 5 3 2 1 1 v[0] v[1] v[2] v[3] v[4] v[5] v[6] v[7]

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CS0250, Set 4 Look at the execution of F(7) -1 -1 8 5 3 2 1 1 v[0] v[1] v[2] v[3] v[4] v[5] v[6] v[7]
CS0250, Set 4 Look at the execution of F(7) -1 1 3 8 5 3 2 1 1 v[0] v[1] v[2] v[3] v[4] v[5] v[6] v[7]

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CS0250, Set 4 Look at the execution of F(7) -1 1 3 8 5 3 2 1 1 v[0] v[1] v[2] v[3] v[4] v[5] v[6] v[7]
CS0250, Set 4 Look at the execution of F(7) 2 1 1 3 8 5 3 2 1 1 v[0] v[1] v[2] v[3] v[4] v[5] v[6] v[7]

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CS0250, Set 4 Further (minor) improvement Observation Our algorithm makes many function calls, and each wastes time in parameters passing, dynamic linking,… In the computation of F(n), we need F(n-1), F(n-2). In general, to compute F(i), we need F(i-1) and F(i-2). Idea: Compute the values in bottom-up fashion. That is, compute F(2) (we already know F(0)=F(1)=1), then F(3), then F(4)… This new implementation saves lots of overhead . int F(int n) { int i, A[1000]; A[0]=A[1]=1; for (i=2; i <=n; i++) A[i]=A[i-1]+A[i-2]; }
CS0250, Set 4 Summary on the methodology Write down a formula that relates a solution of a problem with those of subproblems. E.g. F(n) = F(n-1) + F(n-2). Index the subproblems so that they can be stored and retrieved easily in a table (i.e., array) Fill the table in some bottom-up manner; start filling the solution of the smallest problem. (This ensures that when we solve a particular subproblem, the solutions of all the smaller subproblems that it depends are available.) For historical reasons, we call such methodology Dynamic Programming . In the late 40’s (when computers were rare), programming refers to the

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## This note was uploaded on 07/05/2008 for the course CS CSIS0250 taught by Professor Dr.hing-fungting during the Summer '08 term at HKU.

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L04 - Dynamic Programming An efficient way to implement some

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