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生成纵横字谜的算法
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Stack Overflow用户
提问于 2009-06-03 04:52:45
回答 5查看 95.1K关注 0票数 129

给定一个单词列表,您将如何将它们排列到纵横填字游戏网格中?

它不需要像一个对称的“适当的”字谜游戏或任何类似的东西:基本上只需要输出每个单词的起始位置和方向。

EN

回答 5

Stack Overflow用户

发布于 2009-06-20 15:07:59

为什么不从随机概率的方法开始呢?从一个单词开始,然后重复选择一个随机单词,并尝试在不打破大小等限制的情况下使其适合当前的拼图状态。如果你失败了,就重新开始。

你会惊讶于像这样的蒙特卡洛方法是多么的有效。

票数 9
EN

Stack Overflow用户

发布于 2014-03-08 01:04:32

下面是一些基于尼克夫的答案和布莱恩的JavaScript代码的Python代码。只是在js中发布它,以防其他人需要它。

代码语言:javascript
复制
function board(cols, rows) { //instantiator object for making gameboards
this.cols = cols;
this.rows = rows;
var activeWordList = []; //keeps array of words actually placed in board
var acrossCount = 0;
var downCount = 0;

var grid = new Array(cols); //create 2 dimensional array for letter grid
for (var i = 0; i < rows; i++) {
    grid[i] = new Array(rows);
}

for (var x = 0; x < cols; x++) {
    for (var y = 0; y < rows; y++) {
        grid[x][y] = {};
        grid[x][y].targetChar = EMPTYCHAR; //target character, hidden
        grid[x][y].indexDisplay = ''; //used to display index number of word start
        grid[x][y].value = '-'; //actual current letter shown on board
    }
}

function suggestCoords(word) { //search for potential cross placement locations
    var c = '';
    coordCount = [];
    coordCount = 0;
    for (i = 0; i < word.length; i++) { //cycle through each character of the word
        for (x = 0; x < GRID_HEIGHT; x++) {
            for (y = 0; y < GRID_WIDTH; y++) {
                c = word[i];
                if (grid[x][y].targetChar == c) { //check for letter match in cell
                    if (x - i + 1> 0 && x - i + word.length-1 < GRID_HEIGHT) { //would fit vertically?
                        coordList[coordCount] = {};
                        coordList[coordCount].x = x - i;
                        coordList[coordCount].y = y;
                        coordList[coordCount].score = 0;
                        coordList[coordCount].vertical = true;
                        coordCount++;
                    }

                    if (y - i + 1 > 0 && y - i + word.length-1 < GRID_WIDTH) { //would fit horizontally?
                        coordList[coordCount] = {};
                        coordList[coordCount].x = x;
                        coordList[coordCount].y = y - i;
                        coordList[coordCount].score = 0;
                        coordList[coordCount].vertical = false;
                        coordCount++;
                    }
                }
            }
        }
    }
}

function checkFitScore(word, x, y, vertical) {
    var fitScore = 1; //default is 1, 2+ has crosses, 0 is invalid due to collision

    if (vertical) { //vertical checking
        for (i = 0; i < word.length; i++) {
            if (i == 0 && x > 0) { //check for empty space preceeding first character of word if not on edge
                if (grid[x - 1][y].targetChar != EMPTYCHAR) { //adjacent letter collision
                    fitScore = 0;
                    break;
                }
            } else if (i == word.length && x < GRID_HEIGHT) { //check for empty space after last character of word if not on edge
                 if (grid[x+i+1][y].targetChar != EMPTYCHAR) { //adjacent letter collision
                    fitScore = 0;
                    break;
                }
            }
            if (x + i < GRID_HEIGHT) {
                if (grid[x + i][y].targetChar == word[i]) { //letter match - aka cross point
                    fitScore += 1;
                } else if (grid[x + i][y].targetChar != EMPTYCHAR) { //letter doesn't match and it isn't empty so there is a collision
                    fitScore = 0;
                    break;
                } else { //verify that there aren't letters on either side of placement if it isn't a crosspoint
                    if (y < GRID_WIDTH - 1) { //check right side if it isn't on the edge
                        if (grid[x + i][y + 1].targetChar != EMPTYCHAR) { //adjacent letter collision
                            fitScore = 0;
                            break;
                        }
                    }
                    if (y > 0) { //check left side if it isn't on the edge
                        if (grid[x + i][y - 1].targetChar != EMPTYCHAR) { //adjacent letter collision
                            fitScore = 0;
                            break;
                        }
                    }
                }
            }

        }

    } else { //horizontal checking
        for (i = 0; i < word.length; i++) {
            if (i == 0 && y > 0) { //check for empty space preceeding first character of word if not on edge
                if (grid[x][y-1].targetChar != EMPTYCHAR) { //adjacent letter collision
                    fitScore = 0;
                    break;
                }
            } else if (i == word.length - 1 && y + i < GRID_WIDTH -1) { //check for empty space after last character of word if not on edge
                if (grid[x][y + i + 1].targetChar != EMPTYCHAR) { //adjacent letter collision
                    fitScore = 0;
                    break;
                }
            }
            if (y + i < GRID_WIDTH) {
                if (grid[x][y + i].targetChar == word[i]) { //letter match - aka cross point
                    fitScore += 1;
                } else if (grid[x][y + i].targetChar != EMPTYCHAR) { //letter doesn't match and it isn't empty so there is a collision
                    fitScore = 0;
                    break;
                } else { //verify that there aren't letters on either side of placement if it isn't a crosspoint
                    if (x < GRID_HEIGHT) { //check top side if it isn't on the edge
                        if (grid[x + 1][y + i].targetChar != EMPTYCHAR) { //adjacent letter collision
                            fitScore = 0;
                            break;
                        }
                    }
                    if (x > 0) { //check bottom side if it isn't on the edge
                        if (grid[x - 1][y + i].targetChar != EMPTYCHAR) { //adjacent letter collision
                            fitScore = 0;
                            break;
                        }
                    }
                }
            }

        }
    }

    return fitScore;
}

function placeWord(word, clue, x, y, vertical) { //places a new active word on the board

    var wordPlaced = false;

    if (vertical) {
        if (word.length + x < GRID_HEIGHT) {
            for (i = 0; i < word.length; i++) {
                grid[x + i][y].targetChar = word[i];
            }
            wordPlaced = true;
        }
    } else {
        if (word.length + y < GRID_WIDTH) {
            for (i = 0; i < word.length; i++) {
                grid[x][y + i].targetChar = word[i];
            }
            wordPlaced = true;
        }
    }

    if (wordPlaced) {
        var currentIndex = activeWordList.length;
        activeWordList[currentIndex] = {};
        activeWordList[currentIndex].word = word;
        activeWordList[currentIndex].clue = clue;
        activeWordList[currentIndex].x = x;
        activeWordList[currentIndex].y = y;
        activeWordList[currentIndex].vertical = vertical;

        if (activeWordList[currentIndex].vertical) {
            downCount++;
            activeWordList[currentIndex].number = downCount;
        } else {
            acrossCount++;
            activeWordList[currentIndex].number = acrossCount;
        }
    }

}

function isActiveWord(word) {
    if (activeWordList.length > 0) {
        for (var w = 0; w < activeWordList.length; w++) {
            if (word == activeWordList[w].word) {
                //console.log(word + ' in activeWordList');
                return true;
            }
        }
    }
    return false;
}

this.displayGrid = function displayGrid() {

    var rowStr = "";
    for (var x = 0; x < cols; x++) {

        for (var y = 0; y < rows; y++) {
            rowStr += "<td>" + grid[x][y].targetChar + "</td>";
        }
        $('#tempTable').append("<tr>" + rowStr + "</tr>");
        rowStr = "";

    }
    console.log('across ' + acrossCount);
    console.log('down ' + downCount);
}

//for each word in the source array we test where it can fit on the board and then test those locations for validity against other already placed words
this.generateBoard = function generateBoard(seed = 0) {

    var bestScoreIndex = 0;
    var top = 0;
    var fitScore = 0;
    var startTime;

    //manually place the longest word horizontally at 0,0, try others if the generated board is too weak
    placeWord(wordArray[seed].word, wordArray[seed].displayWord, wordArray[seed].clue, 0, 0, false);

    //attempt to fill the rest of the board 
    for (var iy = 0; iy < FIT_ATTEMPTS; iy++) { //usually 2 times is enough for max fill potential
        for (var ix = 1; ix < wordArray.length; ix++) {
            if (!isActiveWord(wordArray[ix].word)) { //only add if not already in the active word list
                topScore = 0;
                bestScoreIndex = 0;

                suggestCoords(wordArray[ix].word); //fills coordList and coordCount
                coordList = shuffleArray(coordList); //adds some randomization

                if (coordList[0]) {
                    for (c = 0; c < coordList.length; c++) { //get the best fit score from the list of possible valid coordinates
                        fitScore = checkFitScore(wordArray[ix].word, coordList[c].x, coordList[c].y, coordList[c].vertical);
                        if (fitScore > topScore) {
                            topScore = fitScore;
                            bestScoreIndex = c;
                        }
                    }
                }

                if (topScore > 1) { //only place a word if it has a fitscore of 2 or higher

                    placeWord(wordArray[ix].word, wordArray[ix].clue, coordList[bestScoreIndex].x, coordList[bestScoreIndex].y, coordList[bestScoreIndex].vertical);
                }
            }

        }
    }
    if(activeWordList.length < wordArray.length/2) { //regenerate board if if less than half the words were placed
        seed++;
        generateBoard(seed);
    }
}
}
function seedBoard() {
    gameboard = new board(GRID_WIDTH, GRID_HEIGHT);
    gameboard.generateBoard();
    gameboard.displayGrid();
}
票数 7
EN

Stack Overflow用户

发布于 2009-06-03 05:26:02

我会生成两个数字:长度和拼写分数。假设拼写分数低意味着更容易加入(低分数=大量常见字母)。按长度降序和拼写分数升序对列表进行排序。

接下来,从列表中往下看。如果单词没有与现有单词交叉(分别根据每个单词的长度和拼写分数检查每个单词),则将其放入队列中,并检查下一个单词。

冲洗并重复,这应该会生成一个纵横字谜。

当然,我非常确定这是O(n!)它不能保证为你完成纵横字谜,但也许有人可以改进它。

票数 5
EN
页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接:

https://stackoverflow.com/questions/943113

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