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

其他回答

令人惊讶的是,没有人尝试使用PHP版本。

这是John Fouhy的Python解决方案的PHP版本。

虽然我从其他人的答案中得到了一些建议,但这主要是抄袭约翰的。

$boggle = "fxie
           amlo
           ewbx
           astu";

$alphabet = str_split(str_replace(array("\n", " ", "\r"), "", strtolower($boggle)));
$rows = array_map('trim', explode("\n", $boggle));
$dictionary = file("C:/dict.txt");
$prefixes = array(''=>'');
$words = array();
$regex = '/[' . implode('', $alphabet) . ']{3,}$/S';
foreach($dictionary as $k=>$value) {
    $value = trim(strtolower($value));
    $length = strlen($value);
    if(preg_match($regex, $value)) {
        for($x = 0; $x < $length; $x++) {
            $letter = substr($value, 0, $x+1);
            if($letter == $value) {
                $words[$value] = 1;
            } else {
                $prefixes[$letter] = 1;
            }
        }
    }
}

$graph = array();
$chardict = array();
$positions = array();
$c = count($rows);
for($i = 0; $i < $c; $i++) {
    $l = strlen($rows[$i]);
    for($j = 0; $j < $l; $j++) {
        $chardict[$i.','.$j] = $rows[$i][$j];
        $children = array();
        $pos = array(-1,0,1);
        foreach($pos as $z) {
            $xCoord = $z + $i;
            if($xCoord < 0 || $xCoord >= count($rows)) {
                continue;
            }
            $len = strlen($rows[0]);
            foreach($pos as $w) {
                $yCoord = $j + $w;
                if(($yCoord < 0 || $yCoord >= $len) || ($z == 0 && $w == 0)) {
                    continue;
                }
                $children[] = array($xCoord, $yCoord);
            }
        }
        $graph['None'][] = array($i, $j);
        $graph[$i.','.$j] = $children;
    }
}

function to_word($chardict, $prefix) {
    $word = array();
    foreach($prefix as $v) {
        $word[] = $chardict[$v[0].','.$v[1]];
    }
    return implode("", $word);
}

function find_words($graph, $chardict, $position, $prefix, $prefixes, &$results, $words) {
    $word = to_word($chardict, $prefix);
    if(!isset($prefixes[$word])) return false;

    if(isset($words[$word])) {
        $results[] = $word;
    }

    foreach($graph[$position] as $child) {
        if(!in_array($child, $prefix)) {
            $newprefix = $prefix;
            $newprefix[] = $child;
            find_words($graph, $chardict, $child[0].','.$child[1], $newprefix, $prefixes, $results, $words);
        }
    }
}

$solution = array();
find_words($graph, $chardict, 'None', array(), $prefixes, $solution);
print_r($solution);

如果你想尝试的话,这里有一个实时链接。虽然在我的本地机器上需要大约2秒,但在我的web服务器上需要大约5秒。无论哪种情况,它都不是很快。尽管如此,它还是很可怕,所以我可以想象时间可以大大缩短。任何关于如何实现这一目标的建议都将不胜感激。PHP缺少元组,这使得坐标处理起来很奇怪,而且我无法理解到底发生了什么,这对我一点帮助都没有。

编辑:一些修复使它在本地少于1秒。

我很快完美地解决了这个问题。我把它放进了一个安卓应用程序。在play store链接中查看视频,看看它是如何运作的。

单词作弊是一个应用程序,“破解”任何矩阵风格的文字游戏。这个应用程序 来帮我在文字混淆器上作弊。它可以用于单词搜索, 沙沙,单词,单词查找器,单词破解,拼字游戏,和更多!

在这里可以看到 https://play.google.com/store/apps/details?id=com.harris.wordcracker

在视频中查看应用程序的操作 https://www.youtube.com/watch?v=DL2974WmNAI

该解决方案还提供了在给定的板中搜索的方向

一件事:

1. Uses trie to save all the word in the english to fasten the search
2. The uses DFS to search the words in Boggle

输出:

Found "pic" directions from (4,0)(p) go  → →
Found "pick" directions from (4,0)(p) go  → → ↑
Found "pickman" directions from (4,0)(p) go  → → ↑ ↑ ↖ ↑
Found "picket" directions from (4,0)(p) go  → → ↑ ↗ ↖
Found "picked" directions from (4,0)(p) go  → → ↑ ↗ ↘
Found "pickle" directions from (4,0)(p) go  → → ↑ ↘ →

代码:

from collections import defaultdict
from nltk.corpus import words
from nltk.corpus import stopwords
from nltk.tokenize import word_tokenize

english_words = words.words()

# If you wan to remove stop words
# stop_words = set(stopwords.words('english'))
# english_words = [w for w in english_words if w not in stop_words]

boggle = [
    ['c', 'n', 't', 's', 's'],
    ['d', 'a', 't', 'i', 'n'],
    ['o', 'o', 'm', 'e', 'l'],
    ['s', 'i', 'k', 'n', 'd'],
    ['p', 'i', 'c', 'l', 'e']
]

# Instead of X and Y co-ordinates
# better to use Row and column
lenc = len(boggle[0])
lenr = len(boggle)

# Initialize trie datastructure
trie_node = {'valid': False, 'next': {}}

# lets get the delta to find all the nighbors
neighbors_delta = [
    (-1,-1, "↖"),
    (-1, 0, "↑"),
    (-1, 1, "↗"),
    (0, -1, "←"),
    (0,  1, "→"),
    (1, -1, "↙"),
    (1,  0, "↓"),
    (1,  1, "↘"),
]


def gen_trie(word, node):
    """udpates the trie datastructure using the given word"""
    if not word:
        return

    if word[0] not in node:
        node[word[0]] = {'valid': len(word) == 1, 'next': {}}

    # recursively build trie
    gen_trie(word[1:], node[word[0]])


def build_trie(words, trie):
    """Builds trie data structure from the list of words given"""
    for word in words:
        gen_trie(word, trie)
    return trie


def get_neighbors(r, c):
    """Returns the neighbors for a given co-ordinates"""
    n = []
    for neigh in neighbors_delta:
        new_r = r + neigh[0]
        new_c = c + neigh[1]

        if (new_r >= lenr) or (new_c >= lenc) or (new_r < 0) or (new_c < 0):
            continue
        n.append((new_r, new_c, neigh[2]))
    return n


def dfs(r, c, visited, trie, now_word, direction):
    """Scan the graph using DFS"""
    if (r, c) in visited:
        return

    letter = boggle[r][c]
    visited.append((r, c))

    if letter in trie:
        now_word += letter

        if trie[letter]['valid']:
            print('Found "{}" {}'.format(now_word, direction))

        neighbors = get_neighbors(r, c)
        for n in neighbors:
            dfs(n[0], n[1], visited[::], trie[letter], now_word, direction + " " + n[2])


def main(trie_node):
    """Initiate the search for words in boggle"""
    trie_node = build_trie(english_words, trie_node)

    # print the board
    print("Given board")
    for i in range(lenr):print (boggle[i])
    print ('\n')

    for r in range(lenr):
        for c in range(lenc):
            letter = boggle[r][c]
            dfs(r, c, [], trie_node, '', 'directions from ({},{})({}) go '.format(r, c, letter))


if __name__ == '__main__':
    main(trie_node)

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}"

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。