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

    package ProblemSolving;

import java.util.HashSet;
import java.util.Set;

/**
 * Given a 2-dimensional array of characters and a
 * dictionary in which a word can be searched in O(1) time.
 * Need to print all the words from array which are present
 * in dictionary. Word can be formed in any direction but
 * has to end at any edge of array.
 * (Need not worry much about the dictionary)
 */
public class DictionaryWord {
    private static char[][] matrix = new char[][]{
            {'a', 'f', 'h', 'u', 'n'},
            {'e', 't', 'a', 'i', 'r'},
            {'a', 'e', 'g', 'g', 'o'},
            {'t', 'r', 'm', 'l', 'p'}
    };
    private static int dim_x = matrix.length;
    private static int dim_y = matrix[matrix.length -1].length;
    private static Set<String> wordSet = new HashSet<String>();

    public static void main(String[] args) {
        //dictionary
        wordSet.add("after");
        wordSet.add("hate");
        wordSet.add("hair");
        wordSet.add("air");
        wordSet.add("eat");
        wordSet.add("tea");

        for (int x = 0; x < dim_x; x++) {
            for (int y = 0; y < dim_y; y++) {
                checkAndPrint(matrix[x][y] + "");
                int[][] visitedMap = new int[dim_x][dim_y];
                visitedMap[x][y] = 1;
                recursion(matrix[x][y] + "", visitedMap, x, y);
            }
        }
    }

    private static void checkAndPrint(String word) {
        if (wordSet.contains(word)) {
            System.out.println(word);
        }
    }

    private static void recursion(String word, int[][] visitedMap, int x, int y) {
        for (int i = Math.max(x - 1, 0); i < Math.min(x + 2, dim_x); i++) {
            for (int j = Math.max(y - 1, 0); j < Math.min(y + 2, dim_y); j++) {
                if (visitedMap[i][j] == 1) {
                    continue;
                } else {
                    int[][] newVisitedMap = new int[dim_x][dim_y];
                    for (int p = 0; p < dim_x; p++) {
                        for (int q = 0; q < dim_y; q++) {
                           newVisitedMap[p][q] = visitedMap[p][q];
                        }
                    }
                    newVisitedMap[i][j] = 1;
                    checkAndPrint(word + matrix[i][j]);
                    recursion(word + matrix[i][j], newVisitedMap, i, j);
                }
            }
        }
    }

}

其他回答

一个Node.JS JavaScript解决方案。在不到一秒钟的时间内计算所有100个独特的单词,其中包括阅读字典文件(MBA 2012)。

Output: ["FAM","TUX","TUB","FAE","ELI","ELM","ELB","TWA","TWA","SAW","AMI","SWA","SWA","AME","SEA","SEW","AES","AWL","AWE","SEA","AWA","MIX","MIL","AST","ASE","MAX","MAE","MAW","MEW","AWE","MES","AWL","LIE","LIM","AWA","AES","BUT","BLO","WAS","WAE","WEA","LEI","LEO","LOB","LOX","WEM","OIL","OLM","WEA","WAE","WAX","WAF","MILO","EAST","WAME","TWAS","TWAE","EMIL","WEAM","OIME","AXIL","WEST","TWAE","LIMB","WASE","WAST","BLEO","STUB","BOIL","BOLE","LIME","SAWT","LIMA","MESA","MEWL","AXLE","FAME","ASEM","MILE","AMIL","SEAX","SEAM","SEMI","SWAM","AMBO","AMLI","AXILE","AMBLE","SWAMI","AWEST","AWEST","LIMAX","LIMES","LIMBU","LIMBO","EMBOX","SEMBLE","EMBOLE","WAMBLE","FAMBLE"]

代码:

var fs = require('fs')

var Node = function(value, row, col) {
    this.value = value
    this.row = row
    this.col = col
}

var Path = function() {
    this.nodes = []
}

Path.prototype.push = function(node) {
    this.nodes.push(node)
    return this
}

Path.prototype.contains = function(node) {
    for (var i = 0, ii = this.nodes.length; i < ii; i++) {
        if (this.nodes[i] === node) {
            return true
        }
    }

    return false
}

Path.prototype.clone = function() {
    var path = new Path()
    path.nodes = this.nodes.slice(0)
    return path
}

Path.prototype.to_word = function() {
    var word = ''

    for (var i = 0, ii = this.nodes.length; i < ii; ++i) {
        word += this.nodes[i].value
    }

    return word
}

var Board = function(nodes, dict) {
    // Expects n x m array.
    this.nodes = nodes
    this.words = []
    this.row_count = nodes.length
    this.col_count = nodes[0].length
    this.dict = dict
}

Board.from_raw = function(board, dict) {
    var ROW_COUNT = board.length
      , COL_COUNT = board[0].length

    var nodes = []

    // Replace board with Nodes
    for (var i = 0, ii = ROW_COUNT; i < ii; ++i) {
        nodes.push([])
        for (var j = 0, jj = COL_COUNT; j < jj; ++j) {
            nodes[i].push(new Node(board[i][j], i, j))
        }
    }

    return new Board(nodes, dict)
}

Board.prototype.toString = function() {
    return JSON.stringify(this.nodes)
}

Board.prototype.update_potential_words = function(dict) {
    for (var i = 0, ii = this.row_count; i < ii; ++i) {
        for (var j = 0, jj = this.col_count; j < jj; ++j) {
            var node = this.nodes[i][j]
              , path = new Path()

            path.push(node)

            this.dfs_search(path)
        }
    }
}

Board.prototype.on_board = function(row, col) {
    return 0 <= row && row < this.row_count && 0 <= col && col < this.col_count
}

Board.prototype.get_unsearched_neighbours = function(path) {
    var last_node = path.nodes[path.nodes.length - 1]

    var offsets = [
        [-1, -1], [-1,  0], [-1, +1]
      , [ 0, -1],           [ 0, +1]
      , [+1, -1], [+1,  0], [+1, +1]
    ]

    var neighbours = []

    for (var i = 0, ii = offsets.length; i < ii; ++i) {
        var offset = offsets[i]
        if (this.on_board(last_node.row + offset[0], last_node.col + offset[1])) {

            var potential_node = this.nodes[last_node.row + offset[0]][last_node.col + offset[1]]
            if (!path.contains(potential_node)) {
                // Create a new path if on board and we haven't visited this node yet.
                neighbours.push(potential_node)
            }
        }
    }

    return neighbours
}

Board.prototype.dfs_search = function(path) {
    var path_word = path.to_word()

    if (this.dict.contains_exact(path_word) && path_word.length >= 3) {
        this.words.push(path_word)
    }

    var neighbours = this.get_unsearched_neighbours(path)

    for (var i = 0, ii = neighbours.length; i < ii; ++i) {
        var neighbour = neighbours[i]
        var new_path = path.clone()
        new_path.push(neighbour)

        if (this.dict.contains_prefix(new_path.to_word())) {
            this.dfs_search(new_path)
        }
    }
}

var Dict = function() {
    this.dict_array = []

    var dict_data = fs.readFileSync('./web2', 'utf8')
    var dict_array = dict_data.split('\n')

    for (var i = 0, ii = dict_array.length; i < ii; ++i) {
        dict_array[i] = dict_array[i].toUpperCase()
    }

    this.dict_array = dict_array.sort()
}

Dict.prototype.contains_prefix = function(prefix) {
    // Binary search
    return this.search_prefix(prefix, 0, this.dict_array.length)
}

Dict.prototype.contains_exact = function(exact) {
    // Binary search
    return this.search_exact(exact, 0, this.dict_array.length)
}

Dict.prototype.search_prefix = function(prefix, start, end) {
    if (start >= end) {
        // If no more place to search, return no matter what.
        return this.dict_array[start].indexOf(prefix) > -1
    }

    var middle = Math.floor((start + end)/2)

    if (this.dict_array[middle].indexOf(prefix) > -1) {
        // If we prefix exists, return true.
        return true
    } else {
        // Recurse
        if (prefix <= this.dict_array[middle]) {
            return this.search_prefix(prefix, start, middle - 1)
        } else {
            return this.search_prefix(prefix, middle + 1, end)
        }
    }
}

Dict.prototype.search_exact = function(exact, start, end) {
    if (start >= end) {
        // If no more place to search, return no matter what.
        return this.dict_array[start] === exact
    }

    var middle = Math.floor((start + end)/2)

    if (this.dict_array[middle] === exact) {
        // If we prefix exists, return true.
        return true
    } else {
        // Recurse
        if (exact <= this.dict_array[middle]) {
            return this.search_exact(exact, start, middle - 1)
        } else {
            return this.search_exact(exact, middle + 1, end)
        }
    }
}

var board = [
    ['F', 'X', 'I', 'E']
  , ['A', 'M', 'L', 'O']
  , ['E', 'W', 'B', 'X']
  , ['A', 'S', 'T', 'U']
]

var dict = new Dict()

var b = Board.from_raw(board, dict)
b.update_potential_words()
console.log(JSON.stringify(b.words.sort(function(a, b) {
    return a.length - b.length
})))

如何简单的排序和使用字典中的二进制搜索?

在0.35秒内返回整个列表,并可以进一步优化(例如删除含有未使用字母的单词等)。

from bisect import bisect_left

f = open("dict.txt")
D.extend([line.strip() for line in f.readlines()])
D = sorted(D)

def neibs(M,x,y):
    n = len(M)
    for i in xrange(-1,2):
        for j in xrange(-1,2):
            if (i == 0 and j == 0) or (x + i < 0 or x + i >= n or y + j < 0 or y + j >= n):
                continue
            yield (x + i, y + j)

def findWords(M,D,x,y,prefix):
    prefix = prefix + M[x][y]

    # find word in dict by binary search
    found = bisect_left(D,prefix)

    # if found then yield
    if D[found] == prefix: 
        yield prefix

    # if what we found is not even a prefix then return
    # (there is no point in going further)
    if len(D[found]) < len(prefix) or D[found][:len(prefix)] != prefix:
        return

    # recourse
    for neib in neibs(M,x,y):
        for word in findWords(M,D,neib[0], neib[1], prefix):
            yield word

def solve(M,D):
    # check each starting point
    for x in xrange(0,len(M)):
        for y in xrange(0,len(M)):
            for word in findWords(M,D,x,y,""):
                yield word

grid = "fxie amlo ewbx astu".split()
print [x for x in solve(grid,D)]

我用c语言解决了这个问题。在我的机器上运行大约需要48毫秒(其中98%的时间花在从磁盘加载字典和创建trie上)。字典是/usr/share/dict/american-english,有62886个单词。

源代码

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”。

给定一个有N行M列的Boggle板,让我们假设如下:

N*M基本上大于可能单词的数量 N*M基本上大于可能的最长单词

在这些假设下,该解的复杂度为O(N*M)。

我认为比较这个示例板的运行时间在很多方面都没有重点,但是为了完整性,在我的现代MacBook Pro上,这个解决方案在0.2秒内完成。

这个解决方案将为语料库中的每个单词找到所有可能的路径。

#!/usr/bin/env ruby
# Example usage: ./boggle-solver --board "fxie amlo ewbx astu"

autoload :Matrix, 'matrix'
autoload :OptionParser, 'optparse'

DEFAULT_CORPUS_PATH = '/usr/share/dict/words'.freeze

# Functions

def filter_corpus(matrix, corpus, min_word_length)
  board_char_counts = Hash.new(0)
  matrix.each { |c| board_char_counts[c] += 1 }

  max_word_length = matrix.row_count * matrix.column_count
  boggleable_regex = /^[#{board_char_counts.keys.reduce(:+)}]{#{min_word_length},#{max_word_length}}$/
  corpus.select{ |w| w.match boggleable_regex }.select do |w|
    word_char_counts = Hash.new(0)
    w.each_char { |c| word_char_counts[c] += 1 }
    word_char_counts.all? { |c, count| board_char_counts[c] >= count }
  end
end

def neighbors(point, matrix)
  i, j = point
  ([i-1, 0].max .. [i+1, matrix.row_count-1].min).inject([]) do |r, new_i|
    ([j-1, 0].max .. [j+1, matrix.column_count-1].min).inject(r) do |r, new_j|
      neighbor = [new_i, new_j]
      neighbor.eql?(point) ? r : r << neighbor
    end
  end
end

def expand_path(path, word, matrix)
  return [path] if path.length == word.length

  next_char = word[path.length]
  viable_neighbors = neighbors(path[-1], matrix).select do |point|
    !path.include?(point) && matrix.element(*point).eql?(next_char)
  end

  viable_neighbors.inject([]) do |result, point|
    result + expand_path(path.dup << point, word, matrix)
  end
end

def find_paths(word, matrix)
  result = []
  matrix.each_with_index do |c, i, j|
    result += expand_path([[i, j]], word, matrix) if c.eql?(word[0])
  end
  result
end

def solve(matrix, corpus, min_word_length: 3)
  boggleable_corpus = filter_corpus(matrix, corpus, min_word_length)
  boggleable_corpus.inject({}) do |result, w|
    paths = find_paths(w, matrix)
    result[w] = paths unless paths.empty?
    result
  end
end

# Script

options = { corpus_path: DEFAULT_CORPUS_PATH }
option_parser = OptionParser.new do |opts|
  opts.banner = 'Usage: boggle-solver --board <value> [--corpus <value>]'

  opts.on('--board BOARD', String, 'The board (e.g. "fxi aml ewb ast")') do |b|
    options[:board] = b
  end

  opts.on('--corpus CORPUS_PATH', String, 'Corpus file path') do |c|
    options[:corpus_path] = c
  end

  opts.on_tail('-h', '--help', 'Shows usage') do
    STDOUT.puts opts
    exit
  end
end
option_parser.parse!

unless options[:board]
  STDERR.puts option_parser
  exit false
end

unless File.file? options[:corpus_path]
  STDERR.puts "No corpus exists - #{options[:corpus_path]}"
  exit false
end

rows = options[:board].downcase.scan(/\S+/).map{ |row| row.scan(/./) }

raw_corpus = File.readlines(options[:corpus_path])
corpus = raw_corpus.map{ |w| w.downcase.rstrip }.uniq.sort

solution = solve(Matrix.rows(rows), corpus)
solution.each_pair do |w, paths|
  STDOUT.puts w
  paths.each do |path|
    STDOUT.puts "\t" + path.map{ |point| point.inspect }.join(', ')
  end
end
STDOUT.puts "TOTAL: #{solution.count}"