在matlab中解决一个单词搜索难题

Question

发布我的全部代码，希望有人可以帮助我调试这个废话。真的希望尽快得出结论，因为我已经玩这个太久了。

我这里有一个函数提取，它传递了一个整数网格，代表我的 wordsearch 函数中的字母（即 A=1，Z=26）。给定一个方向和一个由整数的行向量表示的目标单词，它应该遍历网格以找到第一个字母存在的位置，并从这里向所有方向移动以获取单词的长度并提取单词，例如如果我们正在寻找 [14, 5, 9, 2] 并且 14 首先定位在 (4,4) 我们应该最终定位在 (4,8)。

然后在搜索功能中比较该单词，如果它与目标单词匹配，则会在实际单词搜索的图像上从第一个字母到最后一个字母画一条线。

我知道我的 if 和 for 循环在某些地方是关闭的，但我发现很难更正我的代码以使其正常工作。帮助！特别是我遇到困难的一件事是控制流量，以便如果在检查包含第一个字母的正方形的所有方向之后，评估该字母的下一个实例。最好在哪里执行此操作？

代码有很多错误，可以通过几个指针告诉我它需要更改或清理的地方。

%//A function to find a word in a grid.
function test = extract(grid, direction, target) 

%//switch through different cases that allow us to move to any adjacent cell to the current    
switch upper(direction)
    case 1 
        rowdir = -1; 
        coldir = 0; 
    case 2 
        rowdir = -1; 
        coldir = 1; 
    case 3 
        rowdir = 0; 
        coldir = 1; 
    case 4 
        rowdir = 1; 
        coldir = 1; 
    case 5 
        rowdir = 1; 
        coldir = 0; 
    case 6 
        rowdir = 1; 
        coldir = -1; 
    case 7 
        rowdir = 0; 
        coldir = -1; 
    case 8 
        rowdir = -1; 
        coldir = -1;  
end

[i, j] = size(grid); 
len = length(target); 
[row,column] = find(target(1)==grid); %//find the letter of the word we are looking for in grid
%//row and column of last letter having moved in a particular direction
rowN = row + (len-1) * rowdir; 
colN = column + (len-1) * coldir; 

%//trying to say here to only move in a particular direction if we don't go out of bounds. 
%//not sure I've succeeded
if (rowN > 1) | (rowN < i) | (colN > 1) | (colN < j) 
     testword = []; %empty array created
    for index = 1:len 
        index_1 = index-1;
        %//on loop get the letter in adjacent cell for direction we have moved
        word = grid(row + (index_1 * rowdir), column + (index_1 * coldir)); 
        testword{index} = word; %//letters are appended to create word for which we compare.
        %//get co-ords of start letter. change to pixel  co-ordinates so we can evaluate on image
        wordstart = [(row*30)-15, (column*30)-15 ]; 
        wordend = [((row + (len-1 * rowdir))*30)-15, ((column + (len-1 * coldir))*30)-15];
    end 
else
    word = '';
end 

  x1 = wordstart(1);
  x2 = wordend(1);
  y1 = wordstart(2);
  y2 = wordend(2);

  test = [ word , [x1,x2] , [y1,y2]]; %//only way I could think of to get all of these as outputs

end

---

%//test is the image we want to evaluate on
%//words is the list of words

function trial1 = wordsearch(test, words)

imagesc(test);
colormap(gray);
hold on;

grid = %//grid is a 15x15 matrix

[row, column] = size(grid); 

for iword = 1 : length(words) 
    target = char(words(iword)) - 'a' + 1;
    for i = 1:row 
        for j = 1:column 
            for direction_num = 1:8 %//for each direction
                direction = directions(direction_num, :);
                testword = extract(grid,direction,target); 
                if testword(1)==target %//if word we have extracted equals the target word

%//draw_line function takes x co-ordinates and y co-ordinates and plots line.
                    draw_line(testword(2),testword(3),testword(4),testword(5)); 
                end
            end
        end
    end
end

hold off;
end

---

@丹

我的提取函数现在看起来像：

[i, j] = size(grid);
len = length(target); 
[row,column] = find(target(1)==grid);
for ii = 1:length(row)
    start_row = row(ii);
    start_column = column(ii); 
    rowN = start_row + len-1 * rowdir; 
    colN = start_column + len-1 * coldir; 
    if (rowN > 1) || (rowN < i) || (colN > 1) || (colN < j) 
        testword = [];
        for index = 1:len
        index_1 = index-1;
        word = grid(start_row + (index_1 * rowdir), start_column + (index_1 * coldir));
        testword{index} = word;
        wordstart = [(start_row*30)-15, (start_column*30)-15 ];
        wordend = [((start_row + (len-1 * rowdir))*30)-15, ((start_column + (len-1 * coldir))*30)-15];
        end 
    else

    end
end 

如果先前在那个特定方向上让你超出界限，我会用什么作为 else 语句来检查这个词？

Answer 1

for iword = 1 : length(words) 
    target = char(words(iword)) - 'a' + 1;
    for i = 1:row 
        for j = 1:column 
            for direction_num = 1:8 %//for each direction

对于每个单词，您都在遍历网格中的每个元素（即 i 和 j 循环），但实际上您从未使用过这些 i 或 j 值。所以这两个循环什么都不做！这是因为您似乎在 extract 函数中完成了所有这些操作。所以放弃这两个循环，它们浪费了大量的时间。

在extract 函数中，有一行[row,column] = find...。这将找到 ALL 可能的起点。所以你实际上需要在某个地方循环遍历其中的任何一个。因此，我建议不要使用您的 if (rowN > 1) | (rowN < i) | (colN > 1) | (colN < j)，而是建议：

for ii = 1:length(row)
    start_row = row(ii);
    start_column = column(ii); 
    %// And now re-use your code, but swap out all your row for start_row and your column for start_column
    .
    .
    .
end

这是遍历每个可能的起始字母的循环。