Java programming example of recursive and non-recursive code with two distributions, two Recursion

Java programming example of recursive and non-recursive code with two distributions, two Recursion

The main content of this article is recursive and non-Recursive Implementation of the binary distribution of Java programming.

Source:

Exercise 27: return (1.1-p) * binomial (N-1, k, p) + p * binomial (N-1, k-1, k-1, p );
How does the recursive formula calculate the number of recursive calls?

Two distributions:

Definition: n Independent discrete probability distributions of the number of successful times k in/non-test. The probability of each successful experiment is p, which is recorded as B (n, p, k ).

Probability formula: P (ε = K) = C (n, k) * p ^ k * (1-p) ^ (n-k)

Where C (n, k) = (n-k )! /(K! * (N-k )!), Recorded as EZ ~ B (n, p), expect: E ε = np, variance: D ε = npq, where q = 1-p.

There is a recursive formula in probability statistics:

This is the source of recursion in the question.

The recursive formula comes from C (n, k) = C (n-1, k) + C (n-1, k-1 ). The actual scenario is to select k from n people. How many combinations are there? Place n people by 1 ~ N is sorted in order. If the k-th individual is not selected, select k from the n-th individual. The k-th individual is selected, you need to select the K-1 from the remaining n-1 individuals.

Recursive Implementation of two item distribution in the book:

Public static double binomial (int N, int k, double p) {COUNT ++; // records the number of recursive calls if (N = 0 & k = 0) {return 1.0;} if (N <0 | k <0) {return 0.0;} return (1.0-p) * binomial (N-1, k, p) + p * binomial (N-1, k-1, p );}

Experiment results:

N k p call times 10 5 0.25 246720 10 0.25 243553830 15 0.25 2440764535

From the results, we can see that the number of calls required for this recursive method is a geometric disaster, and n to 50 is not enough.

Improved Recursive Implementation of two distributions:

Private static long COUNT = 0; private static double [] [] M; private static double binomial (int N, int k, double p) {COUNT ++; if (N = 0 & k = 0) {return 1.0;} if (N <0 | k <0) {return 0.0 ;} if (M [N] [k] =-1) {// Save the calculation result and use the calculated result directly, no need to recursively calculate M [N] [k] = (1.0-p) * binomial (N-1, k, p) + p * binomial (N-1, k-1, p);} return M [N] [k];} public static double Binomial (int N, int k, double p) {M = new double [N + 1] [k + 1]; for (int I = 0; I <= N; I ++) {for (int j = 0; j <= k; j ++) {M [I] [j] =-1 ;}return binomial (N, k, p );}

Experiment results:

N k p call times 10 5 0.25 10120 10 0.25 15 45230 0.25 25 120350 0.25 50 3204100 0.25

The experiment results show that the number of calls is greatly reduced, and the algorithm can be used.

Non-Recursive Implementation of Two-item distribution:

In fact, it is faster to directly calculate the combination number and factorial without recursion.

// Calculate the combination number public static double combination (double N, double k) {double min = k; double max = N-k; double t = 0; double NN = 1; double kk = 1; if (min> max) {t = min; min = max; max = t;} while (N> max) {// you do not need to calculate NN = NN * N; N --;} while (min> 0) {// calculate the factorial kk = kk * min; min --;} return NN/kk ;} // calculate the public static double binomial (int N, int k, double p) {double a = 1; double B = 1; double c = combination (N, k); while (N-k)> 0) {// calculate the (N-k) Power of (1-p) a = a * (1-p); N --;} while (k> 0) {// calculate the k power of p B = B * p; k --;} return c * a * B ;}

Experiment results:

N k p binary distribution value: 10, 5, 0.25 0.05839920043945312520, 10, 0.25 0.00992227527967770650, 25, 0.25 8.44919466990417e-5

Compared with the previous algorithm, the calculation result is correct and the running speed is very fast.

Summary

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