Question

我正在尝试创建一个找到n * n二进制[0,1]矩阵的最大可能行列式的程序。我目前的代码存在的问题是，在4x4之后，这个数字对于“长”而来说太长了。坚持。下面的代码是我用来生成矩阵的代码。

for (int j = 0; j < Math.pow(2, Math.pow(dim, 2)); j++) { //for each possible matrix
            int[][] matrix = new int[dim][dim];//make a 2d i*i matrix
            long t = Long.valueOf(Long.toBinaryString(j)); //line 58

            //format the string so its length = # of elements in the matrix
            String format = "%0" + (dim * dim) + "d";
            String s = String.format(format, t);

            //fill matrix
            int counter = 0;
            for (int k = 0; k < dim; k++) {//for each row
                for (int l = 0; l < dim; l++) {//for each column
                    matrix[k][l] = s.charAt(counter) - 48; //the -48 is for ascii conversions
                    counter++;
                }// end l
            }// end k -- matrix is filled

当我超过4 * 4时我得到的错误：

     Exception in thread "main" java.lang.NumberFormatException: For input string: "10000000000000000000"
at java.lang.NumberFormatException.forInputString(NumberFormatException.java:65)
at java.lang.Long.parseLong(Long.java:444)
at java.lang.Long.valueOf(Long.java:540)
at determinants.Determinants.findMaxBinDet(Determinants.java:58)
at determinants.Determinants.main(Determinants.java:38)

我还能做些什么来将数字解析成二进制文件？

Answer 1

您希望在单元格中创建0和1的所有组合。一种更简单的方法是

for (long j = 0, limit = 1<<(dim*dim); j < limit; j++) { //for each possible matrix
    int[][] matrix = new int[dim][dim];//make a 2d i*i matrix

    //fill matrix
    for (int k = 0, counter = 0; k < dim; k++) {//for each row
        for (int l = 0; l < dim; l++, counter++) {//for each column
            matrix[k][l] = (j >>> counter) & 1;
        }
    }

这最多只能用于7x7矩阵，但由于生成所有8x8组合需要花费超过一生的时间，因此需要采用另一种方法。

将大数字解析为二进制

1 个答案: