matlab中高效准确的矩阵搜索方法

时间:2013-10-03 17:28:21

标签: matlab matrix

我正在刷掉我的MATLAB技能,这些技能在很长一段时间内都没有使用过。为了做到这一点,我一直在项目eueler做谜题。好吧,我有点难以理解这一点,因为当我闯入零件时,一切似乎运行正常,但我最大的数字显然是不正确的。

无论如何,这是我的代码         %此脚本将采用网格并找到4个数字的最大乘积         %up,down,left right和diagonal

    %create the grid
    grid = [08 02 22 97 38 15 00 40 00 75 04 05 07 78 52 12 50 77 91 08;
    49 49 99 40 17 81 18 57 60 87 17 40 98 43 69 48 04 56 62 00;
    81 49 31 73 55 79 14 29 93 71 40 67 53 88 30 03 49 13 36 65;
    52 70 95 23 04 60 11 42 69 24 68 56 01 32 56 71 37 02 36 91;
    22 31 16 71 51 67 63 89 41 92 36 54 22 40 40 28 66 33 13 80;
    24 47 32 60 99 03 45 02 44 75 33 53 78 36 84 20 35 17 12 50;
    32 98 81 28 64 23 67 10 26 38 40 67 59 54 70 66 18 38 64 70;
    67 26 20 68 02 62 12 20 95 63 94 39 63 08 40 91 66 49 94 21;
    24 55 58 05 66 73 99 26 97 17 78 78 96 83 14 88 34 89 63 72;
    21 36 23 09 75 00 76 44 20 45 35 14 00 61 33 97 34 31 33 95;
    78 17 53 28 22 75 31 67 15 94 03 80 04 62 16 14 09 53 56 92;
    16 39 05 42 96 35 31 47 55 58 88 24 00 17 54 24 36 29 85 57;
    86 56 00 48 35 71 89 07 05 44 44 37 44 60 21 58 51 54 17 58;
    19 80 81 68 05 94 47 69 28 73 92 13 86 52 17 77 04 89 55 40;
    04 52 08 83 97 35 99 16 07 97 57 32 16 26 26 79 33 27 98 66;
    88 36 68 87 57 62 20 72 03 46 33 67 46 55 12 32 63 93 53 69;
    04 42 16 73 38 25 39 11 24 94 72 18 08 46 29 32 40 62 76 36;
    20 69 36 41 72 30 23 88 34 62 99 69 82 67 59 85 74 04 36 16;
    20 73 35 29 78 31 90 01 74 31 49 71 48 86 81 16 23 57 05 54;
    01 70 54 71 83 51 54 69 16 92 33 48 61 43 52 01 89 19 67 48;];

    %grid(row, column)

    %find how many rows and columns there are
    [rowNum, columnNum] = size(grid);

    %Current greatest product of four consecutive
    greatest = 0;

    %test left right

    %iterate through all rows
    for i = 1:rowNum
        %iterate through all columns except the last tthree
        for x = 1:columnNum-3
            %test the consecutive numbers starting with the current colum
            %iteration
            current = 1;
            for test = x:x+3
                current = current * grid(i, test);
                if greatest < current
                    greatest = current;
                end
            end
        end
    end

    %iterate through all columns
    for i = 1:columnNum
        %iterate throw all rows except the last three
        for x = 1:rowNum-3
            %test the consecutive numbers starting with the current colum
            %iteration
            current = 1;
            for test = x:x+3
                current = current * grid(test, i);
                if greatest < current
                    greatest = current;
                end
            end
        end
    end

    for i = 1:columnNum-3
        %iterate throw all column except the last two
        for x = 1:rowNum-3
            %test the consecutive numbers starting with the current colum
            %iteration
            current = 1;

            %count adds the number of iterations to the row number in grid
            %forcing it to move in a diagonal line
            count = 0;

            %test consecutively from x to three to the right
            for test = x:x+3

                %make new current product
                current = current * grid(i+count, test);

                %add to count to shift down one row next iteration
                count = count + 1;

                %check for greatest
                if greatest < current
                    greatest = current;
                end
            end
        end
    end
    disp(greatest)

我认为问题在于搜索对角线的嵌套for循环,但我不确定。 我把它分解成碎片(自己测试每个循环,并在搜索它们时输出每个元素)。我很确定水平和垂直搜索方法正常工作。

只是想要一双新鲜的眼睛看这个,谢谢!

此外,我确信有一种更有效的方法可以做到这一点,所以如果有人对更好的算法有一些想法,我很乐意听到它!

0 个答案:

没有答案