java - 如何为 nxn 矩阵测试 1000 组数据，每组 20 次，直到 n=2^i？

Question

我正在做一个测试不同矩阵乘法方法的项目。目标是测试 1000 组数据，每组 20 次，n x n 矩阵最多 n = 2^i，在我的情况下，我将达到 n = 256。
我的矩阵乘法方法一切正常，但我无法让它测试正确的次数。
下面是我的主要方法，我试图在其中测试此功能的输出正确性。

    public static void main(String[] args)
    {
        final int SETS = 1;
        final int TIMES = 2;
        final int POWER = 1;
        long timeStart, timeEnd;
        long totalTimeClassical = 0;
        long totalTimeDivideConquer = 0;
        long totalTimeStrassen = 0;
        long avgTimeClassical = 0;
        long avgTimeDivideConquer = 0;
        long avgTimeStrassen = 0;
        int[][] a, b;
        int n;

        for (int i=1; i<=POWER; i++)
        {
            n = (int) Math.pow(2, i);

            for (int j=1; j<=TIMES; j++)
            {
                for (int k=1; k<=SETS; k++)
                {
                    a = buildMatrix(n);
                    b = buildMatrix(n); 

                    timeStart = System.nanoTime();
                    classicalMatrixMult(a, b, n);
                    timeEnd = System.nanoTime();
                    totalTimeClassical += timeEnd - timeStart;

                    timeStart = System.nanoTime();
                    divideConquerMatrixMult(a, b, n);
                    timeEnd = System.nanoTime();
                    totalTimeDivideConquer += timeEnd - timeStart;

                    timeStart = System.nanoTime();
                    strassenMatrixMult(a, b);
                    timeEnd = System.nanoTime();
                    totalTimeStrassen += timeEnd - timeStart;
                    
                    outputMatrix(a,n);
                    outputMatrix(b,n);
                    System.out.println("         -----------");
                    
                }

                totalTimeClassical = totalTimeClassical / TIMES;
                totalTimeDivideConquer = totalTimeDivideConquer / TIMES;
                totalTimeStrassen = totalTimeStrassen / TIMES;
            }

            avgTimeClassical = totalTimeClassical / SETS;
            avgTimeDivideConquer = totalTimeDivideConquer / SETS;
            avgTimeStrassen = totalTimeStrassen / SETS;

            System.out.print("For " + n + "x" + n + ":");
            System.out.println("\nClassical Matrix multiplication took "
                + avgTimeClassical + " nanoseconds to run." 
                +"\nDivide and Conquer Matrix multiplcation took "
                + avgTimeDivideConquer + " nanoseconds to run."
                +"\nStrassen's Matrix Multiplication took "
                + avgTimeStrassen + " nanoseconds to run."
                + "\n");
        }
    }

以上是我的主要方法，我目前将它设置为打印输入矩阵，以便我可以查看它是否使用正确数量的数据集运行了正确的次数。使用我插入的当前值，我希望它测试 1 组 2 x 2 矩阵，2 次。相反，它测试 2 个 2 x 2 矩阵的集合，因为打印出的值不同。

         9        -9
        -4         4

        -2         0
         3         8

         -----------
        -5        10
        -1         1

        -2        -8
        -5        -5

         -----------
For 2x2:
Classical Matrix multiplication took 2514 nanoseconds to run.
Divide and Conquer Matrix multiplcation took 10575 nanoseconds to run.
Strassen's Matrix Multiplication took 1356 nanoseconds to run.

我试图重新排列循环，但最终破坏了代码。我如何格式化它以便它测试一组数据指定的次数？

Answer 1

看起来您已经错误地订购了矩阵创建和时间循环。
首先，您需要为“SETS”值的数量创建矩阵数量。
然后在该循​​环内，您需要运行“TIMES”数量的操作。

for (int i=1; i<=POWER; i++)
{
     n = (int) Math.pow(2, i);

    for (int j=1; j<=SETS; j++)
    {
         a = buildMatrix(n);
         b = buildMatrix(n); 
            
         for (int k=1; k<=TIMES; k++)
         {
            // operations code
         }
     }
 }