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我正在根据这个算法使用回溯实现填字游戏生成器:这是我的伪代码:

>  solve(words,grid):    if words is empty:
>        if grid.isValudSol():
>           return grid
>        else:
>           return None    for each word in words:
>        possibleSol <- grid.fillFirst(word)
>        ret <- solve(words\{word},possibleSol)
>        if (ret != None):
>           return ret    return None

这是我当前的代码(C#):

Crossword.Crossword solveCrossword(List<String> words, Crossword.Crossword board){
        if (words.Count == 0)
        {
            return board;
        }
        //Create local copy of board
        Crossword.Crossword localBoard = new Crossword.Crossword(board);

        foreach (var word in words)
        {
            localBoard = new Crossword.Crossword(board);
            int positions = localBoard.getPositionsNumber(word);
            if(positions == 0){
                return null;
            }
            for (int i = 0; i < positions; ++i)
            {
                localBoard = new Crossword.Crossword(board);
                int addResult = localBoard.AddWord(word, i);        //possible solution is now in localBoard  
                if (addResult != -1)
                {
                    List<String> myWords = new List<String>();
                    for (int k = 0; k < words.Count; ++k)
                    {
                        if (words[k] != word)
                            myWords.Add(words[k]);
                    }
                    Crossword.Crossword ret = solveCrossword(myWords, localBoard);
                    if (ret != null)
                        return ret;
                }
                else
                {
                    return null;
                }
            }               

        }
        return null;
    }

我的问题是,对于网格 NxN,例如 6X6,我发现对于大于 N 的字数,我的算法找不到解决方案 - 10 秒或更长时间后;/,对于小于或等于 N 的字数,它可以正常工作,提前感谢任何帮助!

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1 回答 1

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也许,您可以使用此算法生成填字游戏。我认为这个问题可以。

根据这个算法,我认为问题出在强板 N 中。你的下一个单词可以与最后一个单词相交成负坐标,因为填字游戏不仅可以增长 left->right & top->bottom=>你会得到例外。尝试在每个单词放置后重新计算您的棋盘网格,并在每次重新计算后纠正棋盘大小。并且要小心 NxN 网格:从单词列表生成填字游戏的可能性很低=>您可以获得 IndexOutOfRangeException。

此外,您必须检查 place 事件:如果下一个单词没有与任何已使用的单词正确相交,它可以与 next->next word、next->next->next 等相交。在您的解决方案中,没有交集的单词可能导致 StackOverflow 异常或无限循环。

我的链接中的算法首先提供了排序单词列表。但我试图自己解决你的问题。代码如下所示。我的算法在链接中没有使用排序作为示例,但它使用 2 个递归而不是 1 个。此外,我没有包括网格重新计算,因为坐标在我的和您的解决方案中可能具有不同的含义。

    public class Crossword
{
    List<string> words = new List<string>();//for words
    List<string> nomatchWords = new List<string>();//for words those have no match with others
    List<string> usedWords = new List<string>();//for words we already used
    char[,] renderResult;//your result as grid of words. You can draw it in WindowsFormsApp, WPF, etc.

    //a grid.
    //3 values of int[]:
    //[0]==horizontal/vertical word(1/0)
    //[1]==beginning (number of the first letter for horizontal axis)
    //[2]==beginning(number of the first letter for vertical axis)
    List<int[]> Grid = new List<int[]>();


    int HorizontalSize=17;//size of renderResult. I left it as a default for this example. Its calculation depends on your algoritm for rendering
    int VerticalSize=17;

    public Crossword()
    {
        FillWords(this.words);//fill the list of words
    }

    //just for example; it can be reading from file etc.
    void FillWords(List<string> words)
    {
        words.Add("alphabet");
          words.Add("gramin");//may intersect the first at 'a'
          words.Add("asturias");//may intersect the first or second at 'a'
          words.Add("fitness");//may intersect all words
          words.Add("shit");//? too difficult to calculate where it can place in the crossword. Lets do it by the algorithm
          words.Add("programmer");//? you can watch it in debug how my recursion works
    }

    /// <summary>
    /// the main method. Here we render all words in your words-list.
    /// </summary>
    /// <param name="crossword"></param>
    public void Generate(Crossword crossword, DataGridView dgvr)
    {
        List<int[]> letterPosition=new List<int[]>();//can be empty because words may unmatch
        //place EACH word in the crossword
        while (words.Count != 0)//if words.Count=0, all words are placed (but there's also can be words with no match)
        {
            string currentWord = words[0];
            Grid.Add(new int[3] { 1, 0, 0 });//new grid position for this word
            if (usedWords.Count == 0)//this is the first word
            {
                usedWords.Add(currentWord);
                crossword.Place(currentWord, Grid[0], letterPosition, 0);//place it as a new word
                words.RemoveAt(0);
            }
            else//not the first word=> place the new word in position relative to any of used words
                SolveCrossword(currentWord, letterPosition, 0);
            PrintCrossword(????);//just for my app, do your logic here
        }
        //the step with nomatch words. Words that can be intersected with the others words properly will appear in crossword solution. Completely nomatch will not appear.
        foreach(string currentWord in nomatchWords)
            SolveCrossword(currentWord,letterPosition,0);//process NomatchWords like words
        //HOW TO CHECK: try to place as word-list any list of words
    }


    /// <summary>
    /// solution of crossword
    /// </summary>
    /// <param name="currentWord">the word that we try to place</param>
    /// <param name="letterPosition">positions of letter where current word can intersect with the other</param>
    /// <param name="intersectWord">the number of intersect word</param>
    void SolveCrossword(string currentWord, List<int[]>letterPosition, int intersectWord)
    {
        if (usedWords.Count - intersectWord > 0)//we must check the intersection with all this words=>SolveCrossword checks the intersection from the last used word to first
            //the other order will cause generation from the last words=>crossword will look like a tree instead of sharp
        {
            try
            {
                if (MatchWords(usedWords[intersectWord], currentWord, letterPosition))//no matches for its word with last word
                {//usedWords.Count - 1 - 
                    int positionCounter = 0;
                    if (this.Place(currentWord, Grid[intersectWord], letterPosition, positionCounter) == false)//no places for this word related to last placed word=> it's temporarily unmatched
                    {
                        letterPosition.Clear();
                        SolveCrossword(currentWord, letterPosition, ++positionCounter);//we're here because there's no valid intersections between current words=> lets seek the others
                    }
                    if (words.Count != 0)
                        words.RemoveAt(0);//backtrack will end when we'll find the correct position of word
                    return;
                }
                else
                {
                    letterPosition.Clear();
                    SolveCrossword(currentWord, letterPosition, intersectWord + 1);
                }


            //if we're here, the intersection was not found=>it is nomatch word. It can be correct or completely nomatch word but we'll figure it out at last step
            nomatchWords.Add(currentWord);
            words.RemoveAt(0);
            }
             catch//we've got an exception if unchecked positions number->0
            {
                letterPosition.Clear();
                SolveCrossword(currentWord, letterPosition, intersectWord + 1);
            }
        }
    }

    //how to find matches. Doesn't matter, you can use your own method to solve this
    static bool MatchWords(string newword, string matchedword, List<int[]> letterPosition)
    {
        for (int i = 0; i < newword.Length; i++)
            for (int j = 0; j < matchedword.Length; j++)
                if (newword[i] == matchedword[j])
                    letterPosition.Add(new int[2]{i,j});//we need ALL matches, not only FIRST!<= it may cause problems with size of crossword and bug count of unmatched words
        if (letterPosition.Count != 0)
            return true;
        else
            return false;
    }

    /// <summary>
    /// This is how to place words into crossword grid
    /// </summary>
    /// <param name="newword">our word</param>
    /// <param name="letterPosition">Letter Positions list</param>
    /// <param name="el">the number of element of LetterPositions. If current intersection is not valid, we move to next</param>
    /// <returns>true if sucsess</returns>
    bool Place(string newword, int[] intersectedWordGrid, List<int[]> letterPosition, int el)//
    {
        //placing newword related to lastword (setting up the last Grid element)
        //try
        //{
            if (Grid.Count > 1)
            {
                if (intersectedWordGrid[0] == 1)//last word is horizontal=>we need vertical
                {
                    Grid[Grid.Count - 1][0] = 0;
                    Grid[Grid.Count - 1][1] = intersectedWordGrid[1] - letterPosition[el][1];
                    Grid[Grid.Count - 1][2] = intersectedWordGrid[2] + letterPosition[el][0];
                }
                else//vertical=>horizontal
                {
                    Grid[Grid.Count - 1][0] = 1;
                    Grid[Grid.Count - 1][1] = intersectedWordGrid[1] + letterPosition[el][0];
                    Grid[Grid.Count - 1][2] = intersectedWordGrid[2] - letterPosition[el][1];
                }
                usedWords.Add(newword);
            }
        //TODO: recalculate rendering grid for crossword (if you use another mechanic you're welcome)
        //I use default boundaries of size 17x17
        //try to render: if crossword with these Grid positions of words is valid=>good
        if (Render())//renderResult now contains the current solution of crossword
        {
            letterPosition.Clear();
            return true;
        }
        else//crossword is not valid with this grid
                Place(newword, intersectedWordGrid, letterPosition, ++el);

        return true;
    }

    //rendering your crossword
    bool Render()
    {
        if (renderResult == null)
            renderResult = new char[HorizontalSize, VerticalSize];            
        int[] renderword = Grid[Grid.Count - 1];//for each Grid coordinates
        {
            if (usedWords.Count != 0)
            {
                string currentword = usedWords[usedWords.Count-1];
                //usedWords.RemoveAt(0);
                int i=0, j = 0;//need i & j because changing grid coordinates may cause mistakes in other words positions
                try
                {
                    //rresultTempCopy = renderRes;
                    if (CheckWordsInSolution(currentword, renderResult, renderword))
                        for (int k = 0; k < currentword.Length; k++)//word were checked and it's valid=>write it to board
                        {
                            renderResult[HorizontalSize / 2 + renderword[1] + i, VerticalSize / 2 + renderword[2] + j] = currentword[k];
                            if (renderword[0] == 1)
                                j++;
                            else
                                i++;
                        }
                    else
                    {
                        usedWords.RemoveAt(usedWords.Count - 1);
                        return false;
                    }
                }
                catch
                {
                    usedWords.RemoveAt(usedWords.Count - 1);
                    return false;
                }
            }
        }
        //we're here=>no mistakes, this is a valid result
        return true;//sucsess
    }

    bool CheckWordsInSolution(string currentword, char[,] solutionTable, int[] wordCoordinates)
    {
        int i = 0, j = 0;
        for (int k = 0; k < currentword.Length; k++)
            //validation
            if ((renderResult[HorizontalSize / 2 + wordCoordinates[1] + i, VerticalSize / 2 + wordCoordinates[2] + j] == '\0' ||//must be empty place for each letter
                    renderResult[HorizontalSize / 2 + wordCoordinates[1] + i, VerticalSize / 2 + wordCoordinates[2] + j] == currentword[k])//or the same letter
                //we can use other rules, it depends on your type of crossword
                //&& ((i == 0 && renderResult[HorizontalSize / 2 + wordCoordinates[1] + i - 1, VerticalSize / 2 + wordCoordinates[2] + j] == '\0') == false ||//before the beginning of word must be empty place
                //    (j == 0 && renderResult[HorizontalSize / 2 + wordCoordinates[1] + i, VerticalSize / 2 + wordCoordinates[2] + j - 1] == '\0') == false)
                //&& ((i == k && renderResult[HorizontalSize / 2 + wordCoordinates[1] + i + 1, VerticalSize / 2 + wordCoordinates[2] + j] == '\0') == false ||//after the end of word must be empty place
                //    (j == k && renderResult[HorizontalSize / 2 + wordCoordinates[1] + i, VerticalSize / 2 + wordCoordinates[2] + j + 1] == '\0') == false)
                )
                {
                if (wordCoordinates[0] == 1)
                    j++;
                else
                    i++;
            }   
            else
                return false;
        return true;
    }

    void PrintCrossword(???)
    {
        //draw your crossword here
    }
}
于 2017-02-15T09:46:49.960 回答