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

首先,阅读c#语言设计师如何解决一个相关问题: http://blogs.msdn.com/ericlippert/archive/2009/02/04/a-nasality-talisman-for-the-sultana-analyst.aspx。

像他一样,您可以从字典开始,并通过从字母排序的字母数组到可以根据这些字母拼写的单词列表创建字典来规范化单词。

接下来,开始从黑板上创建可能的单词并查找它们。我怀疑这将让你走得很远,但肯定有更多的技巧可以加快速度。

其他回答

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

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

对于字典加速,有一个通用的转换/过程可以大大减少提前的字典比较。

鉴于上面的网格只包含16个字符,其中一些字符是重复的,您可以通过简单地过滤掉具有不可获取字符的条目来大大减少字典中的总键数。

我认为这是明显的优化,但看到没有人这么做,我就提出来了。

在输入过程中,它将我的字典从20万个键减少到只有2000个键。这至少减少了内存开销,并且这肯定会映射到某个地方的速度增加,因为内存不是无限快的。

Perl实现

我的实现有点头重脚轻,因为我重视能够知道每个提取的字符串的确切路径,而不仅仅是其中的有效性。

我也有一些适应在那里,理论上允许一个网格中有洞的功能,网格有不同大小的线(假设你得到了正确的输入,它以某种方式对齐)。

早期筛选器是我的应用程序中最重要的瓶颈,正如之前怀疑的那样,注释掉了一行从1.5s膨胀到7.5s的代码。

在执行时,它似乎认为所有的个位数都在他们自己的有效单词上,但我很确定这是由于字典文件的工作方式。

它有点臃肿,但至少我重用了cpan中的Tree::Trie

其中有些部分是受到现有实现的启发,有些是我已经想到的。

建设性的批评和改进的方法欢迎(/我注意到他从来没有在CPAN上搜索过一个拼字游戏解决器,但这更有趣)

新标准更新

#!/usr/bin/perl 

use strict;
use warnings;

{

  # this package manages a given path through the grid.
  # Its an array of matrix-nodes in-order with
  # Convenience functions for pretty-printing the paths
  # and for extending paths as new paths.

  # Usage:
  # my $p = Prefix->new(path=>[ $startnode ]);
  # my $c = $p->child( $extensionNode );
  # print $c->current_word ;

  package Prefix;
  use Moose;

  has path => (
      isa     => 'ArrayRef[MatrixNode]',
      is      => 'rw',
      default => sub { [] },
  );
  has current_word => (
      isa        => 'Str',
      is         => 'rw',
      lazy_build => 1,
  );

  # Create a clone of this object
  # with a longer path

  # $o->child( $successive-node-on-graph );

  sub child {
      my $self    = shift;
      my $newNode = shift;
      my $f       = Prefix->new();

      # Have to do this manually or other recorded paths get modified
      push @{ $f->{path} }, @{ $self->{path} }, $newNode;
      return $f;
  }

  # Traverses $o->path left-to-right to get the string it represents.

  sub _build_current_word {
      my $self = shift;
      return join q{}, map { $_->{value} } @{ $self->{path} };
  }

  # Returns  the rightmost node on this path

  sub tail {
      my $self = shift;
      return $self->{path}->[-1];
  }

  # pretty-format $o->path

  sub pp_path {
      my $self = shift;
      my @path =
        map { '[' . $_->{x_position} . ',' . $_->{y_position} . ']' }
        @{ $self->{path} };
      return "[" . join( ",", @path ) . "]";
  }

  # pretty-format $o
  sub pp {
      my $self = shift;
      return $self->current_word . ' => ' . $self->pp_path;
  }

  __PACKAGE__->meta->make_immutable;
}

{

  # Basic package for tracking node data
  # without having to look on the grid.
  # I could have just used an array or a hash, but that got ugly.

# Once the matrix is up and running it doesn't really care so much about rows/columns,
# Its just a sea of points and each point has adjacent points.
# Relative positioning is only really useful to map it back to userspace

  package MatrixNode;
  use Moose;

  has x_position => ( isa => 'Int', is => 'rw', required => 1 );
  has y_position => ( isa => 'Int', is => 'rw', required => 1 );
  has value      => ( isa => 'Str', is => 'rw', required => 1 );
  has siblings   => (
      isa     => 'ArrayRef[MatrixNode]',
      is      => 'rw',
      default => sub { [] }
  );

# Its not implicitly uni-directional joins. It would be more effient in therory
# to make the link go both ways at the same time, but thats too hard to program around.
# and besides, this isn't slow enough to bother caring about.

  sub add_sibling {
      my $self    = shift;
      my $sibling = shift;
      push @{ $self->siblings }, $sibling;
  }

  # Convenience method to derive a path starting at this node

  sub to_path {
      my $self = shift;
      return Prefix->new( path => [$self] );
  }
  __PACKAGE__->meta->make_immutable;

}

{

  package Matrix;
  use Moose;

  has rows => (
      isa     => 'ArrayRef',
      is      => 'rw',
      default => sub { [] },
  );

  has regex => (
      isa        => 'Regexp',
      is         => 'rw',
      lazy_build => 1,
  );

  has cells => (
      isa        => 'ArrayRef',
      is         => 'rw',
      lazy_build => 1,
  );

  sub add_row {
      my $self = shift;
      push @{ $self->rows }, [@_];
  }

  # Most of these functions from here down are just builder functions,
  # or utilities to help build things.
  # Some just broken out to make it easier for me to process.
  # All thats really useful is add_row
  # The rest will generally be computed, stored, and ready to go
  # from ->cells by the time either ->cells or ->regex are called.

  # traverse all cells and make a regex that covers them.
  sub _build_regex {
      my $self  = shift;
      my $chars = q{};
      for my $cell ( @{ $self->cells } ) {
          $chars .= $cell->value();
      }
      $chars = "[^$chars]";
      return qr/$chars/i;
  }

  # convert a plain cell ( ie: [x][y] = 0 )
  # to an intelligent cell ie: [x][y] = object( x, y )
  # we only really keep them in this format temporarily
  # so we can go through and tie in neighbouring information.
  # after the neigbouring is done, the grid should be considered inoperative.

  sub _convert {
      my $self = shift;
      my $x    = shift;
      my $y    = shift;
      my $v    = $self->_read( $x, $y );
      my $n    = MatrixNode->new(
          x_position => $x,
          y_position => $y,
          value      => $v,
      );
      $self->_write( $x, $y, $n );
      return $n;
  }

# go through the rows/collums presently available and freeze them into objects.

  sub _build_cells {
      my $self = shift;
      my @out  = ();
      my @rows = @{ $self->{rows} };
      for my $x ( 0 .. $#rows ) {
          next unless defined $self->{rows}->[$x];
          my @col = @{ $self->{rows}->[$x] };
          for my $y ( 0 .. $#col ) {
              next unless defined $self->{rows}->[$x]->[$y];
              push @out, $self->_convert( $x, $y );
          }
      }
      for my $c (@out) {
          for my $n ( $self->_neighbours( $c->x_position, $c->y_position ) ) {
              $c->add_sibling( $self->{rows}->[ $n->[0] ]->[ $n->[1] ] );
          }
      }
      return \@out;
  }

  # given x,y , return array of points that refer to valid neighbours.
  sub _neighbours {
      my $self = shift;
      my $x    = shift;
      my $y    = shift;
      my @out  = ();
      for my $sx ( -1, 0, 1 ) {
          next if $sx + $x < 0;
          next if not defined $self->{rows}->[ $sx + $x ];
          for my $sy ( -1, 0, 1 ) {
              next if $sx == 0 && $sy == 0;
              next if $sy + $y < 0;
              next if not defined $self->{rows}->[ $sx + $x ]->[ $sy + $y ];
              push @out, [ $sx + $x, $sy + $y ];
          }
      }
      return @out;
  }

  sub _has_row {
      my $self = shift;
      my $x    = shift;
      return defined $self->{rows}->[$x];
  }

  sub _has_cell {
      my $self = shift;
      my $x    = shift;
      my $y    = shift;
      return defined $self->{rows}->[$x]->[$y];
  }

  sub _read {
      my $self = shift;
      my $x    = shift;
      my $y    = shift;
      return $self->{rows}->[$x]->[$y];
  }

  sub _write {
      my $self = shift;
      my $x    = shift;
      my $y    = shift;
      my $v    = shift;
      $self->{rows}->[$x]->[$y] = $v;
      return $v;
  }

  __PACKAGE__->meta->make_immutable;
}

use Tree::Trie;

sub readDict {
  my $fn = shift;
  my $re = shift;
  my $d  = Tree::Trie->new();

  # Dictionary Loading
  open my $fh, '<', $fn;
  while ( my $line = <$fh> ) {
      chomp($line);

 # Commenting the next line makes it go from 1.5 seconds to 7.5 seconds. EPIC.
      next if $line =~ $re;    # Early Filter
      $d->add( uc($line) );
  }
  return $d;
}

sub traverseGraph {
  my $d     = shift;
  my $m     = shift;
  my $min   = shift;
  my $max   = shift;
  my @words = ();

  # Inject all grid nodes into the processing queue.

  my @queue =
    grep { $d->lookup( $_->current_word ) }
    map  { $_->to_path } @{ $m->cells };

  while (@queue) {
      my $item = shift @queue;

      # put the dictionary into "exact match" mode.

      $d->deepsearch('exact');

      my $cword = $item->current_word;
      my $l     = length($cword);

      if ( $l >= $min && $d->lookup($cword) ) {
          push @words,
            $item;    # push current path into "words" if it exactly matches.
      }
      next if $l > $max;

      # put the dictionary into "is-a-prefix" mode.
      $d->deepsearch('boolean');

    siblingloop: foreach my $sibling ( @{ $item->tail->siblings } ) {
          foreach my $visited ( @{ $item->{path} } ) {
              next siblingloop if $sibling == $visited;
          }

          # given path y , iterate for all its end points
          my $subpath = $item->child($sibling);

          # create a new path for each end-point
          if ( $d->lookup( $subpath->current_word ) ) {

             # if the new path is a prefix, add it to the bottom of the queue.
              push @queue, $subpath;
          }
      }
  }
  return \@words;
}

sub setup_predetermined { 
  my $m = shift; 
  my $gameNo = shift;
  if( $gameNo == 0 ){
      $m->add_row(qw( F X I E ));
      $m->add_row(qw( A M L O ));
      $m->add_row(qw( E W B X ));
      $m->add_row(qw( A S T U ));
      return $m;
  }
  if( $gameNo == 1 ){
      $m->add_row(qw( D G H I ));
      $m->add_row(qw( K L P S ));
      $m->add_row(qw( Y E U T ));
      $m->add_row(qw( E O R N ));
      return $m;
  }
}
sub setup_random { 
  my $m = shift; 
  my $seed = shift;
  srand $seed;
  my @letters = 'A' .. 'Z' ; 
  for( 1 .. 4 ){ 
      my @r = ();
      for( 1 .. 4 ){
          push @r , $letters[int(rand(25))];
      }
      $m->add_row( @r );
  }
}

# Here is where the real work starts.

my $m = Matrix->new();
setup_predetermined( $m, 0 );
#setup_random( $m, 5 );

my $d = readDict( 'dict.txt', $m->regex );
my $c = scalar @{ $m->cells };    # get the max, as per spec

print join ",\n", map { $_->pp } @{
  traverseGraph( $d, $m, 3, $c ) ;
};

Arch/执行信息进行比较:

model name      : Intel(R) Core(TM)2 Duo CPU     T9300  @ 2.50GHz
cache size      : 6144 KB
Memory usage summary: heap total: 77057577, heap peak: 11446200, stack peak: 26448
       total calls   total memory   failed calls
 malloc|     947212       68763684              0
realloc|      11191        1045641              0  (nomove:9063, dec:4731, free:0)
 calloc|     121001        7248252              0
   free|     973159       65854762

Histogram for block sizes:
  0-15         392633  36% ==================================================
 16-31          43530   4% =====
 32-47          50048   4% ======
 48-63          70701   6% =========
 64-79          18831   1% ==
 80-95          19271   1% ==
 96-111        238398  22% ==============================
112-127          3007  <1% 
128-143        236727  21% ==============================

关于正则表达式优化的更多嘟囔

我使用的正则表达式优化对于多解字典是无用的,而对于多解字典,您将需要一个完整的字典,而不是一个预先修整过的字典。

然而,也就是说,对于一次性解决,它真的很快。(Perl正则表达式是在C!:))

以下是一些不同的代码添加:

sub readDict_nofilter {
  my $fn = shift;
  my $re = shift;
  my $d  = Tree::Trie->new();

  # Dictionary Loading
  open my $fh, '<', $fn;
  while ( my $line = <$fh> ) {
      chomp($line);
      $d->add( uc($line) );
  }
  return $d;
}

sub benchmark_io { 
  use Benchmark qw( cmpthese :hireswallclock );
   # generate a random 16 character string 
   # to simulate there being an input grid. 
  my $regexen = sub { 
      my @letters = 'A' .. 'Z' ; 
      my @lo = ();
      for( 1..16 ){ 
          push @lo , $_ ; 
      }
      my $c  = join '', @lo;
      $c = "[^$c]";
      return qr/$c/i;
  };
  cmpthese( 200 , { 
      filtered => sub { 
          readDict('dict.txt', $regexen->() );
      }, 
      unfiltered => sub {
          readDict_nofilter('dict.txt');
      }
  });
}
           s/iter unfiltered   filtered
unfiltered   8.16         --       -94%
filtered    0.464      1658%         --

Ps: 8.16 * 200 = 27分钟。

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;
    }
}