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  • 【算法】乱序检查

    乱序检查。通过实验检查表中的乱序代码是否能够产生预期的效果。编写一个程序

    ShuffleTest,接受命令行参数M和N,将大小为M的数组打乱N次且在每次打乱之前都将数组

    重新初始化为a[i] = i。打印一个MxM的表格,对于所有的列j,行i表示的是i在打乱后

    落到j的位置的次数。数组中的所有元素的值都应该接近于N/M。

    乱序代码:(StdRandom来自http://algs4.cs.princeton.edu/code/stdlib.jar

        /**
         * Rearrange the elements of a double array in random order.
         */
        public static void shuffle(double[] a) {
            int N = a.length;
            for (int i = 0; i < N; i++) {
                int r = i + uniform(N-i);     // between i and N-1
                double temp = a[i];
                a[i] = a[r];
                a[r] = temp;
            }
        }

    实验代码:

    package com.beyond.algs4.experiment;
    
    import com.beyond.algs4.lib.StdIn;
    import com.beyond.algs4.lib.StdOut;
    import com.beyond.algs4.lib.StdRandom;
    
    public class ShuffleTest {
    
        public static class RandomArray {
            private int M = 0;
            private int[] a = null;
    
            public int[] getA() {
                return a;
            }
    
            public void setA(int[] a) {
                this.a = a;
            }
    
            public RandomArray(int M) {
                this.M = M;
                this.a = new int[M];
                for (int i = 0; i < M; i++) {
                    a[i] = i;
                }
            }
        }
        
        public static class RandomChecker {
            private double[][] matrix = null;
            public double[][] getMatrix() {
                return matrix;
            }
    
    
            private int M = 0;
            private int N = 0;
            
            public RandomChecker(int M, int N) {
                this.M = M;
                this.N = N;
                matrix = new double[M][];
                for (int i = 0; i < matrix.length; i++) {
                    matrix[i] = new double[M];
                }
            }
            
            public void execute() {
                for (int n = 0; n < this.N; n++) {
                    RandomArray c = new RandomArray(M);
                    int[] a = c.getA();
                    StdRandom.shuffle(a);
                    for (int k = 0; k < a.length; k++) {
                        int i = a[k];
                        int j = k;
                        matrix[i][j] += 1;
                    }
                    
                }
            }
            
            public void print() {
                for (int i = 0; i < M; i++) {
                    for (int j = 0; j < M; j++) {
                        StdOut.print(matrix[i][j] + "	");
                    }
                    StdOut.print("
    ");
                }
            }
        }
        
        public static class BadRandomChecker {
            private double[][] matrix = null;
            public double[][] getMatrix() {
                return matrix;
            }
    
    
            private int M = 0;
            private int N = 0;
            
            public BadRandomChecker(int M, int N) {
                this.M = M;
                this.N = N;
                matrix = new double[M][];
                for (int i = 0; i < matrix.length; i++) {
                    matrix[i] = new double[M];
                }
            }
            
            public void execute() {
                for (int n = 0; n < this.N; n++) {
                    RandomArray c = new RandomArray(M);
                    int[] a = c.getA();
                    shuffle(a);
                    for (int k = 0; k < a.length; k++) {
                        int i = a[k];
                        int j = k;
                        matrix[i][j] += 1;
                    }
                    
                }
            }
            
            public void print() {
                for (int i = 0; i < M; i++) {
                    for (int j = 0; j < M; j++) {
                        StdOut.print(matrix[i][j] + "	");
                    }
                    StdOut.print("
    ");
                }
            }
            
            public static void shuffle(int[] a) {
                int N = a.length;
                for (int i = 0; i < N; i++) {
                    int r = StdRandom.uniform(N-i);     // between 0 and N-1
                    int temp = a[i];
                    a[i] = a[r];
                    a[r] = temp;
                }
            }
        }
    
        public static void main(String[] args) {
            StdOut.println("Please input M: ");
            int M = StdIn.readInt();
            StdOut.println("Please input N: ");
            int N = StdIn.readInt();
            StdOut.println("Random Check: ");
            RandomChecker rc = new RandomChecker(M, N);
            rc.execute();
            rc.print();
            
            StdOut.println("Bad Random: ");
            BadRandomChecker br = new BadRandomChecker(M, N);
            br.execute();
            br.print();
        }
    }

    实验结果:

     Please input M:
    10
    Please input N:
    1000

    ij 1 2 3 4 5 6 7 8 9 10
    1 108 106 95 102 98 85 104 101 112 89
    2 102 102 96 99 113 84 88 105 102 109
    3 101 99 84 92 103 104 96 118 100 103
    4 107 104 112 88 96 104 101 98 84 106
    5 76 104 111 102 97 113 101 93 110 93
    6 107 103 91 121 96 123 86 103 87 83
    7 98 103 94 98 94 109 116 89 98 101
    8 97 79 91 102 94 97 105 99 107 129
    9 98 104 102 116 93 107 105 97 90 88
    10 106 96 124 80 116 74 98 97 110 99

    Please input M:
    5
    Please input N:
    100

    ij 1 2 3 4 5
    1 13 17 22 20 28
    2 27 15 25 19 14
    3 15 23 22 23 17
    4 19 26 18 17 20
    5 26 19 13 21 21

    结果分析:
    当M<N时,数组中的所有元素的值都应该接近于N/M。但当M>N时,无法得出结论。

    糟糕的打乱。假设在我们的乱序代码中你选择的是一个0到N-1而非i到N-1之间的随机整数。

    证明得到的结果并非均匀的分布在N!种可能性之间。用乱序检查的测试检验这个版本。

    实验结果:

    Please input M:
    10
    Please input N:
    1000

    ij 1 2 3 4 5 6 7 8 9 10
    1 95 101 131 89 103 83 114 96 93 95
    2 0 107 112 100 116 128 118 111 100 108
    3 0 62 98 119 134 118 146 136 132 55
    4 0 54 100 118 136 180 136 139 95 42
    5 0 47 82 154 146 188 187 97 72 27
    6 0 32 75 161 228 110 147 136 64 47
    7 0 54 139 213 43 77 60 211 151 52
    8 0 148 252 23 48 66 50 23 260 130
    9 0 395 6 15 32 33 31 38 17 433
    10 905 0 5 8 14 17 11 13 16 11

    Please input M:
    5
    Please input N:
    100

    ij 1 2 3 4 5
    1 18 18 23 17 24
    2 0 22 29 32 17
    3 0 15 31 40 14
    4 0 44 10 8 38
    5 82 1 7 3 7

    参考资料:

    算法 第四版  谢路云 译 Algorithms Fourth Edition [美] Robert Sedgewick, Kevin Wayne著

    http://algs4.cs.princeton.edu/home/

    源码下载链接:

     http://pan.baidu.com/s/1pJ0xicF

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  • 原文地址:https://www.cnblogs.com/richaaaard/p/4576969.html
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