Reusing computations

Jordi CortadellaDepartment of Computer Science

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Reusing computations• There are certain algorithms that require a

repeated sequence of similar computations. For example:

• Strategy: try to reuse the work done from previous computations as much as possible.

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Calculating

can be calculated using the following series:

where ( odd)

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Calculating

Since an infinite sum cannot be computed, it may often be sufficient to compute an approximation with a finite sum number of terms.

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Calculating : naïve approach// Pre: nterms > 0// Returns an estimation of using nterms terms// of the series.

double Pi(int nterms) { double sum = 1; // Approximation of /2

// Inv: sum is an approximation of /2 with k terms, // term is the k-th term of the series. for (int k = 1; k < nterms; ++k) {

term = factorial(k)/double_factorial(2k + 1); sum = sum + term; } return 2sum;

}

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It should be a real division

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Calculating : reusing computations

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Calculating // Pre: nterms > 0// Returns an estimation of using nterms terms// of the series.

double Pi(int nterms) { double sum = 1; // Approximation of /2 double term = 1; // Current term of the sum

// Inv: sum is an approximation of /2 with k terms, // term is the k-th term of the series. for (int k = 1; k < nterms; ++k) {

term = termk/(2.0k + 1.0); sum = sum + term; } return 2sum;

}

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Calculating • = 3.14159265358979323846264338327950288...• The series approximation:

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nterms Pi(nterms)1 2.0000005 3.098413

10 3.14057815 3.14156620 3.14159225 3.141593

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Taylor and McLaurin series• Many functions can be approximated by using

Taylor or McLaurin series, e.g.:

• Example: sin x

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Calculating sin x• McLaurin series:

• It is a periodic function (period is 2)

• Convergence improves as x gets closer to zero

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Calculating sin x• Reducing the computation to the (-2,2)

interval:

• Incremental computation of terms:

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Calculating sin x#include <cmath>

// Returns an approximation of sin x.double sin_approx(double x) { int k = int(x/(2M_PI)); x = x – 2kM_PI; // reduce to the (-2,2) interval double term = x; double x2 = xx; int d = 1; double sum = term;

while (abs(term) >= 1e-8) { term = -termx2/((d+1)(d+2)); sum = sum + term; d = d + 2; }

return sum;}

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CombinationsCalculating

– Naïve method: 2 multiplications and 1 division(potential overflow with )

– Recursion:

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Combinations// Pre: n k 0// Returns combinations(n k)int combinations(int n, int k) { int c = 1; for (; k > 0; --k) { c = cn/k; --n; } return c;}

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Problem: integer divisioncn/k may not be integer

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Combinations: recursive solution

// Pre: n k 0// Returns combinations(n k)int combinations(int n, int k) { if (k == 0) return 1; return ncombinations(n – 1, k – 1)/k;}

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Always aninteger division!

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Conclusions

• Try to avoid the brute force to perform complex computations.

• Reuse previous computations whenever possible.

• Use your knowledge about the domain of the problem to minimize the number of operations.

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