最近我一直在iPhone上玩一款名为《Scramble》的游戏。有些人可能知道这个游戏叫拼字游戏。从本质上讲,当游戏开始时,你会得到一个字母矩阵:

F X I E
A M L O
E W B X
A S T U

The goal of the game is to find as many words as you can that can be formed by chaining letters together. You can start with any letter, and all the letters that surround it are fair game, and then once you move on to the next letter, all the letters that surround that letter are fair game, except for any previously used letters. So in the grid above, for example, I could come up with the words LOB, TUX, SEA, FAME, etc. Words must be at least 3 characters, and no more than NxN characters, which would be 16 in this game but can vary in some implementations. While this game is fun and addictive, I am apparently not very good at it and I wanted to cheat a little bit by making a program that would give me the best possible words (the longer the word the more points you get).

(来源:boggled.org)

不幸的是,我不太擅长算法或它们的效率等等。我的第一次尝试使用一个像这样的字典(约2.3MB),并进行线性搜索,试图匹配字典条目的组合。这需要花费很长时间来找到可能的单词,因为你每轮只有2分钟的时间,这是不够的。

我很有兴趣看看是否有任何Stackoverflowers可以提出更有效的解决方案。我主要是在寻找使用三大p的解决方案:Python、PHP和Perl,尽管任何使用Java或c++的东西也很酷,因为速度是至关重要的。

目前的解决方案:

Adam Rosenfield, Python, ~20岁 John Fouhy, Python, ~3秒 Kent Fredric, Perl, ~1s Darius Bacon, Python, ~1s rvarcher, VB。净,~ 1 s Paolo Bergantino, PHP(实时链接),~5s(本地~2s)


当前回答

我知道我已经非常晚了,但是我之前用PHP做了一个——只是为了好玩……

http://www.lostsockdesign.com.au/sandbox/boggle/index.php?letters=fxieamloewbxastu 在0.90108秒内找到75个单词(133分)

F……X . .我 .............. E ............... 一个 ...................................... 米 .............................. L ............................ O ............................... E .................... W ............................ B .......................... X 一个 .................. 年代 .................................................. T ................. U…

给出了一些程序实际在做什么的指示-每个字母是它开始查看模式的地方,而每个'。这显示了中国试图走的一条道路。越多越好。“它搜索得越远。

如果你想要密码,请告诉我…这是一个可怕的PHP和HTML的混合体,从来没有想过要看到阳光,所以我不敢在这里张贴:P

其他回答

我不得不对一个完整的解决方案进行更多的思考,但作为一种方便的优化,我想知道是否值得根据字典中的所有单词预先计算一个图表和三字母组合(2字母和3字母组合)的频率表,并使用它来确定搜索的优先级。我会选择单词的首字母。因此,如果你的字典包含“印度”、“水”、“极端”和“非凡”这些词,那么你预先计算的表可能是:

'IN': 1
'WA': 1
'EX': 2

然后按照共性的顺序(首先是EX,然后是WA/ in)搜索这些图表

我建议根据单词做一个字母树。这棵树将由字母结构组成,像这样:

letter: char
isWord: boolean

然后构建树,每个深度添加一个新字母。换句话说,第一层是字母表;然后从这些树中,会有另外26个条目,以此类推,直到你把所有的单词都拼出来。坚持这个解析树,它将使所有可能的答案更快地查找。

使用这个解析过的树,您可以非常快速地找到解决方案。下面是伪代码:

BEGIN: 
    For each letter:
        if the struct representing it on the current depth has isWord == true, enter it as an answer.
        Cycle through all its neighbors; if there is a child of the current node corresponding to the letter, recursively call BEGIN on it.

这可以通过一些动态编程来加快。例如,在你的样本中,两个“A”都在一个“E”和一个“W”旁边,这(从它们击中它们的点来看)是相同的。我没有足够的时间来详细说明这个代码,但我想你们可以理解。

此外,我相信你会找到其他解决方案,如果你谷歌“Boggle solver”。

这是我想出的解决填字游戏的办法。我想这是最“python”的做事方式:

from itertools import combinations
from itertools import izip
from math import fabs

def isAllowedStep(current,step,length,doubleLength):
            # for step == length -1 not to be 0 => trivial solutions are not allowed
    return length > 1 and \
           current + step < doubleLength and current - step > 0 and \
           ( step == 1 or step == -1 or step <= length+1 or step >= length - 1)

def getPairwiseList(someList):
    iterableList = iter(someList)
    return izip(iterableList, iterableList)

def isCombinationAllowed(combination,length,doubleLength):

    for (first,second) in  getPairwiseList(combination):
        _, firstCoordinate = first
        _, secondCoordinate = second
        if not isAllowedStep(firstCoordinate, fabs(secondCoordinate-firstCoordinate),length,doubleLength):
            return False
    return True

def extractSolution(combinations):
    return ["".join([x[0] for x in combinationTuple]) for combinationTuple in combinations]


length = 4
text = tuple("".join("fxie amlo ewbx astu".split()))
textIndices = tuple(range(len(text)))
coordinates = zip(text,textIndices)

validCombinations = [combination for combination in combinations(coordinates,length) if isCombinationAllowed(combination,length,length*length)]
solution = extractSolution(validCombinations)

我善意地建议你不要将这部分用于所有可能的匹配,但它实际上提供了一种检查你生成的单词是否真的构成有效单词的可能性:

import mechanize
def checkWord(word):
    url = "https://en.oxforddictionaries.com/search?filter=dictionary&query="+word
    br = mechanize.Browser()
    br.set_handle_robots(False)
    response = br.open(url)
    text = response.read()
    return "no exact matches"  not in text.lower()

print [valid for valid in solution[:10] if checkWord(valid)]

我知道我在派对上迟到了,但我已经实现了,作为编码练习,在几种编程语言(c++, Java, Go, c#, Python, Ruby, JavaScript, Julia, Lua, PHP, Perl)中使用了一个填字器,我认为有人可能会对这些感兴趣,所以我在这里留下了链接: https://github.com/AmokHuginnsson/boggle-solvers

我也用Java解决了这个问题。我的实现有269行,非常容易使用。首先,您需要创建Boggler类的一个新实例,然后用网格作为参数调用solve函数。在我的电脑上加载5万个单词的字典大约需要100毫秒,它在大约10-20毫秒内找到单词。找到的单词存储在一个数组列表中,即foundWords。

import java.io.BufferedReader;
import java.io.File;
import java.io.FileInputStream;
import java.io.FileNotFoundException;
import java.io.IOException;
import java.io.InputStreamReader;
import java.net.URISyntaxException;
import java.net.URL;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Comparator;

public class Boggler {
    private ArrayList<String> words = new ArrayList<String>();      
    private ArrayList<String> roundWords = new ArrayList<String>(); 
    private ArrayList<Word> foundWords = new ArrayList<Word>();     
    private char[][] letterGrid = new char[4][4];                   
    private String letters;                                         

    public Boggler() throws FileNotFoundException, IOException, URISyntaxException {
        long startTime = System.currentTimeMillis();

        URL path = GUI.class.getResource("words.txt");
        BufferedReader br = new BufferedReader(new InputStreamReader(new FileInputStream(new File(path.toURI()).getAbsolutePath()), "iso-8859-1"));
        String line;
        while((line = br.readLine()) != null) {
            if(line.length() < 3 || line.length() > 10) {
                continue;
            }

            this.words.add(line);
        }
    }

    public ArrayList<Word> getWords() {
        return this.foundWords;
    }

    public void solve(String letters) {
        this.letters = "";
        this.foundWords = new ArrayList<Word>();

        for(int i = 0; i < letters.length(); i++) {
            if(!this.letters.contains(letters.substring(i, i + 1))) {
                this.letters += letters.substring(i, i + 1);
            }
        }

        for(int i = 0; i < 4; i++) {
            for(int j = 0; j < 4; j++) {
                this.letterGrid[i][j] = letters.charAt(i * 4 + j);
            }
        }

        System.out.println(Arrays.deepToString(this.letterGrid));               

        this.roundWords = new ArrayList<String>();      
        String pattern = "[" + this.letters + "]+";     

        for(int i = 0; i < this.words.size(); i++) {

            if(this.words.get(i).matches(pattern)) {
                this.roundWords.add(this.words.get(i));
            }
        }

        for(int i = 0; i < this.roundWords.size(); i++) {
            Word word = checkForWord(this.roundWords.get(i));

            if(word != null) {
                System.out.println(word);
                this.foundWords.add(word);
            }
        }       
    }

    private Word checkForWord(String word) {
        char initial = word.charAt(0);
        ArrayList<LetterCoord> startPoints = new ArrayList<LetterCoord>();

        int x = 0;  
        int y = 0;
        for(char[] row: this.letterGrid) {
            x = 0;

            for(char letter: row) {
                if(initial == letter) {
                    startPoints.add(new LetterCoord(x, y));
                }

                x++;
            }

            y++;
        }

        ArrayList<LetterCoord> letterCoords = null;
        for(int initialTry = 0; initialTry < startPoints.size(); initialTry++) {
            letterCoords = new ArrayList<LetterCoord>();    

            x = startPoints.get(initialTry).getX(); 
            y = startPoints.get(initialTry).getY();

            LetterCoord initialCoord = new LetterCoord(x, y);
            letterCoords.add(initialCoord);

            letterLoop: for(int letterIndex = 1; letterIndex < word.length(); letterIndex++) {
                LetterCoord lastCoord = letterCoords.get(letterCoords.size() - 1);  
                char currentChar = word.charAt(letterIndex);                        

                ArrayList<LetterCoord> letterLocations = getNeighbours(currentChar, lastCoord.getX(), lastCoord.getY());

                if(letterLocations == null) {
                    return null;    
                }       

                for(int foundIndex = 0; foundIndex < letterLocations.size(); foundIndex++) {
                    if(letterIndex != word.length() - 1 && true == false) {
                        char nextChar = word.charAt(letterIndex + 1);
                        int lastX = letterCoords.get(letterCoords.size() - 1).getX();
                        int lastY = letterCoords.get(letterCoords.size() - 1).getY();

                        ArrayList<LetterCoord> possibleIndex = getNeighbours(nextChar, lastX, lastY);
                        if(possibleIndex != null) {
                            if(!letterCoords.contains(letterLocations.get(foundIndex))) {
                                letterCoords.add(letterLocations.get(foundIndex));
                            }
                            continue letterLoop;
                        } else {
                            return null;
                        }
                    } else {
                        if(!letterCoords.contains(letterLocations.get(foundIndex))) {
                            letterCoords.add(letterLocations.get(foundIndex));

                            continue letterLoop;
                        }
                    }
                }
            }

            if(letterCoords != null) {
                if(letterCoords.size() == word.length()) {
                    Word w = new Word(word);
                    w.addList(letterCoords);
                    return w;
                } else {
                    return null;
                }
            }
        }

        if(letterCoords != null) {
            Word foundWord = new Word(word);
            foundWord.addList(letterCoords);

            return foundWord;
        }

        return null;
    }

    public ArrayList<LetterCoord> getNeighbours(char letterToSearch, int x, int y) {
        ArrayList<LetterCoord> neighbours = new ArrayList<LetterCoord>();

        for(int _y = y - 1; _y <= y + 1; _y++) {
            for(int _x = x - 1; _x <= x + 1; _x++) {
                if(_x < 0 || _y < 0 || (_x == x && _y == y) || _y > 3 || _x > 3) {
                    continue;
                }

                if(this.letterGrid[_y][_x] == letterToSearch && !neighbours.contains(new LetterCoord(_x, _y))) {
                    neighbours.add(new LetterCoord(_x, _y));
                }
            }
        }

        if(neighbours.isEmpty()) {
            return null;
        } else {
            return neighbours;
        }
    }
}

class Word {
    private String word;    
    private ArrayList<LetterCoord> letterCoords = new ArrayList<LetterCoord>();

    public Word(String word) {
        this.word = word;
    }

    public boolean addCoords(int x, int y) {
        LetterCoord lc = new LetterCoord(x, y);

        if(!this.letterCoords.contains(lc)) {
            this.letterCoords.add(lc);

            return true;
        }

        return false;
    }

    public void addList(ArrayList<LetterCoord> letterCoords) {
        this.letterCoords = letterCoords;
    } 

    @Override
    public String toString() {
        String outputString = this.word + " ";
        for(int i = 0; i < letterCoords.size(); i++) {
            outputString += "(" + letterCoords.get(i).getX() + ", " + letterCoords.get(i).getY() + ") ";
        }

        return outputString;
    }

    public String getWord() {
        return this.word;
    }

    public ArrayList<LetterCoord> getList() {
        return this.letterCoords;
    }
}

class LetterCoord extends ArrayList {
    private int x;          
    private int y;          

    public LetterCoord(int x, int y) {
        this.x = x;
        this.y = y;
    }

    public int getX() {
        return this.x;
    }

    public int getY() {
        return this.y;
    }

    @Override
    public boolean equals(Object o) {
        if(!(o instanceof LetterCoord)) {
            return false;
        }

        LetterCoord lc = (LetterCoord) o;

        if(this.x == lc.getX() &&
                this.y == lc.getY()) {
            return true;
        }

        return false;
    }

    @Override
    public int hashCode() {
        int hash = 7;
        hash = 29 * hash + this.x;
        hash = 24 * hash + this.y;
        return hash;
    }
}