Pao*_*ino 374 puzzle algorithm boggle
最近我一直在我的iPhone上玩一款名为Scramble的游戏.有些人可能认为这个游戏是Boggle.基本上,当游戏开始时你得到一个像这样的字母矩阵:
F X I E
A M L O
E W B X
A S T U
游戏的目标是尽可能多地找到可以通过将字母链接在一起形成的单词.你可以从任何字母开始,并且它周围的所有字母都是公平的游戏,然后一旦你继续下一个字母,围绕那个字母的所有字母都是合理的游戏,除了以前使用的任何字母.因此在上面的网格,例如,我能想出的话LOB,TUX,SEA,FAME,等词必须至少有3个字符,并且不超过N×N个字符以上,这将是本场比赛16,但可以在一些实现改变.虽然这个游戏很有趣且令人上瘾,但我显然不是很擅长它而且我想通过制作一个可以给我最好的单词的程序来作弊(单词越长,得分就越多).
样本博格http://www.boggled.org/sample.gif
遗憾的是,我不熟悉算法或效率等等.我第一次尝试使用字典,如这一个(〜2.3MB),并确实试图以配合字典条目组合线性搜索.这需要很长时间才能找到可能的单词,而且由于每轮只有2分钟,所以根本就不够.
我很想知道Stackoverflowers是否可以提供更有效的解决方案.我主要是在寻找使用Big 3 Ps的解决方案:Python,PHP和Perl,尽管Java或C++也很酷,因为速度至关重要.
当前的解决方案:
BOUNTY:
我正在为这个问题增加一笔赏金,作为我向所有投入他们计划的人表示感谢的方式.不幸的是,我只能向你们中的一个人提供接受的答案,所以我将测量7天后谁拥有最快的晃动解算器,并奖励获胜者赏金.
赏金奖励.感谢所有参与的人.
Dar*_*con 141
我的回答与其他人一样,但是我会发布它,因为它看起来比其他Python解决方案更快,从更快地设置字典.(我对John Fouhy的解决方案进行了检查.)设置完成后,解决的时间是噪音.
grid = "fxie amlo ewbx astu".split()
nrows, ncols = len(grid), len(grid[0])
# A dictionary word that could be a solution must use only the grid's
# letters and have length >= 3. (With a case-insensitive match.)
import re
alphabet = ''.join(set(''.join(grid)))
bogglable = re.compile('[' + alphabet + ']{3,}$', re.I).match
words = set(word.rstrip('\n') for word in open('words') if bogglable(word))
prefixes = set(word[:i] for word in words
for i in range(2, len(word)+1))
def solve():
for y, row in enumerate(grid):
for x, letter in enumerate(row):
for result in extending(letter, ((x, y),)):
yield result
def extending(prefix, path):
if prefix in words:
yield (prefix, path)
for (nx, ny) in neighbors(path[-1]):
if (nx, ny) not in path:
prefix1 = prefix + grid[ny][nx]
if prefix1 in prefixes:
for result in extending(prefix1, path + ((nx, ny),)):
yield result
def neighbors((x, y)):
for nx in range(max(0, x-1), min(x+2, ncols)):
for ny in range(max(0, y-1), min(y+2, nrows)):
yield (nx, ny)
样品用法:
# Print a maximal-length word and its path:
print max(solve(), key=lambda (word, path): len(word))
编辑:过滤少于3个字母的单词.
编辑2:我很好奇为什么Kent Fredric的Perl解决方案更快; 事实证明,使用正则表达式匹配而不是一组字符.在Python中做同样的事情会使速度提高一倍.
Ada*_*eld 116
您将获得的最快解决方案可能涉及将您的字典存储在trie中.然后,创建一个三元组队列(x,y,s),其中队列中的每个元素对应于一个单词的前缀s,该单词可以在网格中拼写,结束于位置(x,y).使用N x N个元素(其中N是网格的大小)初始化队列,网格中每个方块的一个元素.然后,算法如下进行:
While the queue is not empty:
Dequeue a triple (x, y, s)
For each square (x', y') with letter c adjacent to (x, y):
If s+c is a word, output s+c
If s+c is a prefix of a word, insert (x', y', s+c) into the queue
如果将字典存储在trie中,则测试s + c是单词还是单词的前缀可以在恒定时间内完成(前提是您还在每个队列数据中保留一些额外的元数据,例如指向当前节点的指针在trie中),因此该算法的运行时间为O(可拼写的单词数).
[编辑]这是我刚编写的Python实现:
#!/usr/bin/python
class TrieNode:
def __init__(self, parent, value):
self.parent = parent
self.children = [None] * 26
self.isWord = False
if parent is not None:
parent.children[ord(value) - 97] = self
def MakeTrie(dictfile):
dict = open(dictfile)
root = TrieNode(None, '')
for word in dict:
curNode = root
for letter in word.lower():
if 97 <= ord(letter) < 123:
nextNode = curNode.children[ord(letter) - 97]
if nextNode is None:
nextNode = TrieNode(curNode, letter)
curNode = nextNode
curNode.isWord = True
return root
def BoggleWords(grid, dict):
rows = len(grid)
cols = len(grid[0])
queue = []
words = []
for y in range(cols):
for x in range(rows):
c = grid[y][x]
node = dict.children[ord(c) - 97]
if node is not None:
queue.append((x, y, c, node))
while queue:
x, y, s, node = queue[0]
del queue[0]
for dx, dy in ((1, 0), (1, -1), (0, -1), (-1, -1), (-1, 0), (-1, 1), (0, 1), (1, 1)):
x2, y2 = x + dx, y + dy
if 0 <= x2 < cols and 0 <= y2 < rows:
s2 = s + grid[y2][x2]
node2 = node.children[ord(grid[y2][x2]) - 97]
if node2 is not None:
if node2.isWord:
words.append(s2)
queue.append((x2, y2, s2, node2))
return words
用法示例:
d = MakeTrie('/usr/share/dict/words')
print(BoggleWords(['fxie','amlo','ewbx','astu'], d))
输出:
['fa','xi','ie','io','el','am','ax','ae','aw','mi','ma','me',' lo','li','oe','ox','em','ea','ea','es','wa','we','wa','bo','bu' ,'as','aw','ae','st','se','sa','tu','ut','fam','fae','imi','eli',' elm','elb','ami','ama','ame','aes','awl','awa','awe','awa','mix','mim','mil' ,'妈','马','马','马','马','记忆','梅','吊球','洛克斯','列伊','利奥','谎言',' lim','oil','olm','ewe','eme','wax','waf','wae','waw','wem','wea','wea','was' ,'waw','wae','bob','blo','bub','但','ast','ase','asa','awl','awa','awe',' awa','aes','swa','swa','sew','sea','sea','saw','tux','tub','tut','twa','twa' ,'tst','utu','fama','fame','ixil','imam','amli','amil','ambo','axil','axe','mimi',' mima','mime','milo','mile','mewl','mese','mesa','lolo','lobo','lima','lime','limb','lile' ,'oime','oleo','olio','oboe','obol','emim','emil','east','ease','wame','wawa','wawa',' weam','west','wese','wast','wase' ,'wawa','wawa','boil','bolo','bole','bobo','blob','bleo','bubo','asem','stub','stut','游泳','半','seme','seam','seax','sasa','sawt','tutu','tuts','twae','twas','twae','ilima' ,'amble','axile','awest','mamie','mambo','maxim','mease','mesem','limax','limes','limbo','limbu',' obole','emesa','embox','awest','swami','famable','mimble','maxima','embolo','embole','wamble','semese','semble' ,'sawbwa','sawbwa']
注意:此程序不输出1个字母的单词,或者根据单词长度进行过滤.这很容易添加,但与问题无关.如果它们可以以多种方式拼写,它还会多次输出一些单词.如果一个给定的单词可以用许多不同的方式拼写(最坏的情况:网格中的每个字母都是相同的(例如'A'),并且在你的字典中有'aaaaaaaaaa'这样的单词),那么运行时间将会非常指数.算法完成后,过滤掉重复项和排序是微不足道的.
Ken*_*ric 39
对于字典加速,您可以做一个通用的转换/过程,以便提前大大减少字典比较.
鉴于上面的网格只包含16个字符,其中一些是重复的,只需过滤掉具有无法获得的字符的条目,就可以大大减少字典中的总键数.
我认为这是明显的优化,但看到没有人这样做,我提到它.
它只是在输入过程中将我从200,000键的字典缩减到仅2,000键.这至少可以减少内存开销,并且由于内存不是无限快,因此肯定会映射到某个地方的速度增加.
我的实现有点头重脚轻,因为我非常重视能够知道每个提取字符串的确切路径,而不仅仅是其中的有效性.
我也有一些适应性,理论上允许带有孔的网格起作用,并且网格具有不同大小的线(假设你得到正确的输入并且它以某种方式排列).
早期过滤器是我应用程序中最重要的瓶颈,正如之前所怀疑的那样,评论该行将其从1.5s膨胀到7.5s.
在执行时,似乎认为所有单个数字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,但至少我从cpan 重用了Tree :: Trie
其中一些部分受到现有实现的启发,其中一些已经考虑过了.
建设性批评及其可以改进的方式受到欢迎(/我注意到他从来没有搜索过CPAN的摇摆解决方案,但这样做更有趣)
更新了新标准
#!/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 ) ;
};
用于比较的Arch /执行信息:
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% ==============================
我使用的正则表达式优化对于多解决字典是无用的,对于多解决方案,你需要一个完整的字典,而不是预先修剪的字典.
然而,那说,一次性解决,它真的很快.(Perl正则表达式在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.
Joh*_*uhy 33
您可以将问题分成两部分:
理想情况下,(2)还应该包括一种测试字符串是否是有效字的前缀的方法 - 这将允许您修剪搜索并节省一大堆时间.
Adam Rosenfield的Trie是(2)的解决方案.它很优雅,可能是你的算法专家所喜欢的,但是对于现代语言和现代计算机,我们可能会有点懒散.此外,正如Kent建议的那样,我们可以通过丢弃网格中不存在字母的单词来减少字典大小.这是一些python:
def make_lookups(grid, fn='dict.txt'):
# Make set of valid characters.
chars = set()
for word in grid:
chars.update(word)
words = set(x.strip() for x in open(fn) if set(x.strip()) <= chars)
prefixes = set()
for w in words:
for i in range(len(w)+1):
prefixes.add(w[:i])
return words, prefixes
哇; 恒定时间前缀测试.加载你链接的字典需要几秒钟,但只有几个:- words <= prefixes) (请注意)
现在,对于第(1)部分,我倾向于用图表来思考.所以我将构建一个类似于下面的字典:
graph = { (x, y):set([(x0,y0), (x1,y1), (x2,y2)]), }
ie graph[(x, y)]是您可以从位置到达的坐标集(x, y).我还将添加一个虚拟节点None,它将连接到所有东西.
建立它有点笨拙,因为有8个可能的位置,你必须做边界检查.这里有一些相应笨拙的python代码:
def make_graph(grid):
root = None
graph = { root:set() }
chardict = { root:'' }
for i, row in enumerate(grid):
for j, char in enumerate(row):
chardict[(i, j)] = char
node = (i, j)
children = set()
graph[node] = children
graph[root].add(node)
add_children(node, children, grid)
return graph, chardict
def add_children(node, children, grid):
x0, y0 = node
for i in [-1,0,1]:
x = x0 + i
if not (0 <= x < len(grid)):
continue
for j in [-1,0,1]:
y = y0 + j
if not (0 <= y < len(grid[0])) or (i == j == 0):
continue
children.add((x,y))
此代码还构建了映射(x,y)到相应字符的字典.这让我可以将一个位置列表转换为一个单词:
def to_word(chardict, pos_list):
return ''.join(chardict[x] for x in pos_list)
最后,我们进行深度优先搜索.基本程序是:
蟒蛇:
def find_words(graph, chardict, position, prefix, results, words, prefixes):
""" Arguments:
graph :: mapping (x,y) to set of reachable positions
chardict :: mapping (x,y) to character
position :: current position (x,y) -- equals prefix[-1]
prefix :: list of positions in current string
results :: set of words found
words :: set of valid words in the dictionary
prefixes :: set of valid words or prefixes thereof
"""
word = to_word(chardict, prefix)
if word not in prefixes:
return
if word in words:
results.add(word)
for child in graph[position]:
if child not in prefix:
find_words(graph, chardict, child, prefix+[child], results, words, prefixes)
运行代码为:
grid = ['fxie', 'amlo', 'ewbx', 'astu']
g, c = make_graph(grid)
w, p = make_lookups(grid)
res = set()
find_words(g, c, None, [], res, w, p)
并检查res以查看答案.这是为您的示例找到的单词列表,按大小排序:
['a', 'b', 'e', 'f', 'i', 'l', 'm', 'o', 's', 't',
'u', 'w', 'x', 'ae', 'am', 'as', 'aw', 'ax', 'bo',
'bu', 'ea', 'el', 'em', 'es', 'fa', 'ie', 'io', 'li',
'lo', 'ma', 'me', 'mi', 'oe', 'ox', 'sa', 'se', 'st',
'tu', 'ut', 'wa', 'we', 'xi', 'aes', 'ame', 'ami',
'ase', 'ast', 'awa', 'awe', 'awl', 'blo', 'but', 'elb',
'elm', 'fae', 'fam', 'lei', 'lie', 'lim', 'lob', 'lox',
'mae', 'maw', 'mew', 'mil', 'mix', 'oil', 'olm', 'saw',
'sea', 'sew', 'swa', 'tub', 'tux', 'twa', 'wae', 'was',
'wax', 'wem', 'ambo', 'amil', 'amli', 'asem', 'axil',
'axle', 'bleo', 'boil', 'bole', 'east', 'fame', 'limb',
'lime', 'mesa', 'mewl', 'mile', 'milo', 'oime', 'sawt',
'seam', 'seax', 'semi', 'stub', 'swam', 'twae', 'twas',
'wame', 'wase', 'wast', 'weam', 'west', 'amble', 'awest',
'axile', 'embox', 'limbo', 'limes', 'swami', 'embole',
'famble', 'semble', 'wamble']
代码(字面上)需要几秒钟来加载字典,但其余部分在我的机器上是即时的.
gin*_*eer 23
我在Java中的尝试.读取文件和构建trie需要大约2秒,解决难题大约需要50毫秒.我使用了问题中链接的词典(它有几个我不知道的单词,如fae,ima)
0 [main] INFO gineer.bogglesolver.util.Util - Reading the dictionary
2234 [main] INFO gineer.bogglesolver.util.Util - Finish reading the dictionary
2234 [main] INFO gineer.bogglesolver.Solver - Found: FAM
2234 [main] INFO gineer.bogglesolver.Solver - Found: FAME
2234 [main] INFO gineer.bogglesolver.Solver - Found: FAMBLE
2234 [main] INFO gineer.bogglesolver.Solver - Found: FAE
2234 [main] INFO gineer.bogglesolver.Solver - Found: IMA
2234 [main] INFO gineer.bogglesolver.Solver - Found: ELI
2234 [main] INFO gineer.bogglesolver.Solver - Found: ELM
2234 [main] INFO gineer.bogglesolver.Solver - Found: ELB
2234 [main] INFO gineer.bogglesolver.Solver - Found: AXIL
2234 [main] INFO gineer.bogglesolver.Solver - Found: AXILE
2234 [main] INFO gineer.bogglesolver.Solver - Found: AXLE
2234 [main] INFO gineer.bogglesolver.Solver - Found: AMI
2234 [main] INFO gineer.bogglesolver.Solver - Found: AMIL
2234 [main] INFO gineer.bogglesolver.Solver - Found: AMLI
2234 [main] INFO gineer.bogglesolver.Solver - Found: AME
2234 [main] INFO gineer.bogglesolver.Solver - Found: AMBLE
2234 [main] INFO gineer.bogglesolver.Solver - Found: AMBO
2250 [main] INFO gineer.bogglesolver.Solver - Found: AES
2250 [main] INFO gineer.bogglesolver.Solver - Found: AWL
2250 [main] INFO gineer.bogglesolver.Solver - Found: AWE
2250 [main] INFO gineer.bogglesolver.Solver - Found: AWEST
2250 [main] INFO gineer.bogglesolver.Solver - Found: AWA
2250 [main] INFO gineer.bogglesolver.Solver - Found: MIX
2250 [main] INFO gineer.bogglesolver.Solver - Found: MIL
2250 [main] INFO gineer.bogglesolver.Solver - Found: MILE
2250 [main] INFO gineer.bogglesolver.Solver - Found: MILO
2250 [main] INFO gineer.bogglesolver.Solver - Found: MAX
2250 [main] INFO gineer.bogglesolver.Solver - Found: MAE
2250 [main] INFO gineer.bogglesolver.Solver - Found: MAW
2250 [main] INFO gineer.bogglesolver.Solver - Found: MEW
2250 [main] INFO gineer.bogglesolver.Solver - Found: MEWL
2250 [main] INFO gineer.bogglesolver.Solver - Found: MES
2250 [main] INFO gineer.bogglesolver.Solver - Found: MESA
2250 [main] INFO gineer.bogglesolver.Solver - Found: MWA
2250 [main] INFO gineer.bogglesolver.Solver - Found: MWA
2250 [main] INFO gineer.bogglesolver.Solver - Found: LIE
2250 [main] INFO gineer.bogglesolver.Solver - Found: LIM
2250 [main] INFO gineer.bogglesolver.Solver - Found: LIMA
2250 [main] INFO gineer.bogglesolver.Solver - Found: LIMAX
2250 [main] INFO gineer.bogglesolver.Solver - Found: LIME
2250 [main] INFO gineer.bogglesolver.Solver - Found: LIMES
2250 [main] INFO gineer.bogglesolver.Solver - Found: LIMB
2250 [main] INFO gineer.bogglesolver.Solver - Found: LIMBO
2250 [main] INFO gineer.bogglesolver.Solver - Found: LIMBU
2250 [main] INFO gineer.bogglesolver.Solver - Found: LEI
2250 [main] INFO gineer.bogglesolver.Solver - Found: LEO
2250 [main] INFO gineer.bogglesolver.Solver - Found: LOB
2250 [main] INFO gineer.bogglesolver.Solver - Found: LOX
2250 [main] INFO gineer.bogglesolver.Solver - Found: OIME
2250 [main] INFO gineer.bogglesolver.Solver - Found: OIL
2250 [main] INFO gineer.bogglesolver.Solver - Found: OLE
2250 [main] INFO gineer.bogglesolver.Solver - Found: OLM
2250 [main] INFO gineer.bogglesolver.Solver - Found: EMIL
2250 [main] INFO gineer.bogglesolver.Solver - Found: EMBOLE
2250 [main] INFO gineer.bogglesolver.Solver - Found: EMBOX
2250 [main] INFO gineer.bogglesolver.Solver - Found: EAST
2250 [main] INFO gineer.bogglesolver.Solver - Found: WAF
2250 [main] INFO gineer.bogglesolver.Solver - Found: WAX
2250 [main] INFO gineer.bogglesolver.Solver - Found: WAME
2250 [main] INFO gineer.bogglesolver.Solver - Found: WAMBLE
2250 [main] INFO gineer.bogglesolver.Solver - Found: WAE
2250 [main] INFO gineer.bogglesolver.Solver - Found: WEA
2250 [main] INFO gineer.bogglesolver.Solver - Found: WEAM
2250 [main] INFO gineer.bogglesolver.Solver - Found: WEM
2250 [main] INFO gineer.bogglesolver.Solver - Found: WEA
2250 [main] INFO gineer.bogglesolver.Solver - Found: WES
2250 [main] INFO gineer.bogglesolver.Solver - Found: WEST
2250 [main] INFO gineer.bogglesolver.Solver - Found: WAE
2250 [main] INFO gineer.bogglesolver.Solver - Found: WAS
2250 [main] INFO gineer.bogglesolver.Solver - Found: WASE
2250 [main] INFO gineer.bogglesolver.Solver - Found: WAST
2250 [main] INFO gineer.bogglesolver.Solver - Found: BLEO
2250 [main] INFO gineer.bogglesolver.Solver - Found: BLO
2250 [main] INFO gineer.bogglesolver.Solver - Found: BOIL
2250 [main] INFO gineer.bogglesolver.Solver - Found: BOLE
2250 [main] INFO gineer.bogglesolver.Solver - Found: BUT
2250 [main] INFO gineer.bogglesolver.Solver - Found: AES
2250 [main] INFO gineer.bogglesolver.Solver - Found: AWA
2250 [main] INFO gineer.bogglesolver.Solver - Found: AWL
2250 [main] INFO gineer.bogglesolver.Solver - Found: AWE
2250 [main] INFO gineer.bogglesolver.Solver - Found: AWEST
2250 [main] INFO gineer.bogglesolver.Solver - Found: ASE
2250 [main] INFO gineer.bogglesolver.Solver - Found: ASEM
2250 [main] INFO gineer.bogglesolver.Solver - Found: AST
2250 [main] INFO gineer.bogglesolver.Solver - Found: SEA
2250 [main] INFO gineer.bogglesolver.Solver - Found: SEAX
2250 [main] INFO gineer.bogglesolver.Solver - Found: SEAM
2250 [main] INFO gineer.bogglesolver.Solver - Found: SEMI
2250 [main] INFO gineer.bogglesolver.Solver - Found: SEMBLE
2250 [main] INFO gineer.bogglesolver.Solver - Found: SEW
2250 [main] INFO gineer.bogglesolver.Solver - Found: SEA
2250 [main] INFO gineer.bogglesolver.Solver - Found: SWA
2250 [main] INFO gineer.bogglesolver.Solver - Found: SWAM
2250 [main] INFO gineer.bogglesolver.Solver - Found: SWAMI
2250 [main] INFO gineer.bogglesolver.Solver - Found: SWA
2250 [main] INFO gineer.bogglesolver.Solver - Found: SAW
2250 [main] INFO gineer.bogglesolver.Solver - Found: SAWT
2250 [main] INFO gineer.bogglesolver.Solver - Found: STU
2250 [main] INFO gineer.bogglesolver.Solver - Found: STUB
2250 [main] INFO gineer.bogglesolver.Solver - Found: TWA
2250 [main] INFO gineer.bogglesolver.Solver - Found: TWAE
2250 [main] INFO gineer.bogglesolver.Solver - Found: TWA
2250 [main] INFO gineer.bogglesolver.Solver - Found: TWAE
2250 [main] INFO gineer.bogglesolver.Solver - Found: TWAS
2250 [main] INFO gineer.bogglesolver.Solver - Found: TUB
2250 [main] INFO gineer.bogglesolver.Solver - Found: TUX
源代码由6个类组成.我将在下面发布它们(如果这不是StackOverflow上的正确做法,请告诉我).
package gineer.bogglesolver;
import org.apache.log4j.BasicConfigurator;
import org.apache.log4j.Logger;
public class Main
{
private final static Logger logger = Logger.getLogger(Main.class);
public static void main(String[] args)
{
BasicConfigurator.configure();
Solver solver = new Solver(4,
"FXIE" +
"AMLO" +
"EWBX" +
"ASTU");
solver.solve();
}
}
package gineer.bogglesolver;
import gineer.bogglesolver.trie.Trie;
import gineer.bogglesolver.util.Constants;
import gineer.bogglesolver.util.Util;
import org.apache.log4j.Logger;
public class Solver
{
private char[] puzzle;
private int maxSize;
private boolean[] used;
private StringBuilder stringSoFar;
private boolean[][] matrix;
private Trie trie;
private final static Logger logger = Logger.getLogger(Solver.class);
public Solver(int size, String puzzle)
{
trie = Util.getTrie(size);
matrix = Util.connectivityMatrix(size);
maxSize = size * size;
stringSoFar = new StringBuilder(maxSize);
used = new boolean[maxSize];
if (puzzle.length() == maxSize)
{
this.puzzle = puzzle.toCharArray();
}
else
{
logger.error("The puzzle size does not match the size specified: " + puzzle.length());
this.puzzle = puzzle.substring(0, maxSize).toCharArray();
}
}
public void solve()
{
for (int i = 0; i < maxSize; i++)
{
traverseAt(i);
}
}
private void traverseAt(int origin)
{
stringSoFar.append(puzzle[origin]);
used[origin] = true;
//Check if we have a valid word
if ((stringSoFar.length() >= Constants.MINIMUM_WORD_LENGTH) && (trie.containKey(stringSoFar.toString())))
{
logger.info("Found: " + stringSoFar.toString());
}
//Find where to go next
for (int destination = 0; destination < maxSize; destination++)
{
if (matrix[origin][destination] && !used[destination] && trie.containPrefix(stringSoFar.toString() + puzzle[destination]))
{
traverseAt(destination);
}
}
used[origin] = false;
stringSoFar.deleteCharAt(stringSoFar.length() - 1);
}
}
package gineer.bogglesolver.trie;
import gineer.bogglesolver.util.Constants;
class Node
{
Node[] children;
boolean isKey;
public Node()
{
isKey = false;
children = new Node[Constants.NUMBER_LETTERS_IN_ALPHABET];
}
public Node(boolean key)
{
isKey = key;
children = new Node[Constants.NUMBER_LETTERS_IN_ALPHABET];
}
/**
Method to insert a string to Node and its children
@param key the string to insert (the string is assumed to be uppercase)
@return true if the node or one of its children is changed, false otherwise
*/
public boolean insert(String key)
{
//If the key is empty, this node is a key
if (key.length() == 0)
{
if (isKey)
return false;
else
{
isKey = true;
return true;
}
}
else
{//otherwise, insert in one of its child
int childNodePosition = key.charAt(0) - Constants.LETTER_A;
if (children[childNodePosition] == null)
{
children[childNodePosition] = new Node();
children[childNodePosition].insert(key.substring(1));
return true;
}
else
{
return children[childNodePosition].insert(key.substring(1));
}
}
}
/**
Returns whether key is a valid prefix for certain key in this trie.
For example: if key "hello" is in this trie, tests with all prefixes "hel", "hell", "hello" return true
@param prefix the prefix to check
@return true if the prefix is valid, false otherwise
*/
public boolean containPrefix(String prefix)
{
//If the prefix is empty, return true
if (prefix.length() == 0)
{
return true;
}
else
{//otherwise, check in one of its child
int childNodePosition = prefix.charAt(0) - Constants.LETTER_A;
return children[childNodePosition] != null && children[childNodePosition].containPrefix(prefix.substring(1));
}
}
/**
Returns whether key is a valid key in this trie.
For example: if key "hello" is in this trie, tests with all prefixes "hel", "hell" return false
@param key the key to check
@return true if the key is valid, false otherwise
*/
public boolean containKey(String key)
{
//If the prefix is empty, return true
if (key.length() == 0)
{
return isKey;
}
else
{//otherwise, check in one of its child
int childNodePosition = key.charAt(0) - Constants.LETTER_A;
return children[childNodePosition] != null && children[childNodePosition].containKey(key.substring(1));
}
}
public boolean isKey()
{
return isKey;
}
public void setKey(boolean key)
{
isKey = key;
}
}
package gineer.bogglesolver.trie;
public class Trie
{
Node root;
public Trie()
{
this.root = new Node();
}
/**
Method to insert a string to Node and its children
@param key the string to insert (the string is assumed to be uppercase)
@return true if the node or one of its children is changed, false otherwise
*/
public boolean insert(String key)
{
return root.insert(key.toUpperCase());
}
/**
Returns whether key is a valid prefix for certain key in this trie.
For example: if key "hello" is in this trie, tests with all prefixes "hel", "hell", "hello" return true
@param prefix the prefix to check
@return true if the prefix is valid, false otherwise
*/
public boolean containPrefix(String prefix)
{
return root.containPrefix(prefix.toUpperCase());
}
/**
Returns whether key is a valid key in this trie.
For example: if key "hello" is in this trie, tests with all prefixes "hel", "hell" return false
@param key the key to check
@return true if the key is valid, false otherwise
*/
public boolean containKey(String key)
{
return root.containKey(key.toUpperCase());
}
}
package gineer.bogglesolver.util;
public class Constants
{
public static final int NUMBER_LETTERS_IN_ALPHABET = 26;
public static final char LETTER_A = 'A';
public static final int MINIMUM_WORD_LENGTH = 3;
public static final int DEFAULT_PUZZLE_SIZE = 4;
}
package gineer.bogglesolver.util;
import gineer.bogglesolver.trie.Trie;
import org.apache.log4j.Logger;
import java.io.File;
import java.io.FileNotFoundException;
import java.util.Scanner;
public class Util
{
private final static Logger logger = Logger.getLogger(Util.class);
private static Trie trie;
private static int size = Constants.DEFAULT_PUZZLE_SIZE;
/**
Returns the trie built from the dictionary. The size is used to eliminate words that are too long.
@param size the size of puzzle. The maximum lenght of words in the returned trie is (size * size)
@return the trie that can be used for puzzle of that size
*/
public static Trie getTrie(int size)
{
if ((trie != null) && size == Util.size)
return trie;
trie = new Trie();
Util.size = size;
logger.info("Reading the dictionary");
final File file = new File("dictionary.txt");
try
{
Scanner scanner = new Scanner(file);
final int maxSize = size * size;
while (scanner.hasNext())
{
String line = scanner.nextLine().replaceAll("[^\\p{Alpha}]", "");
if (line.length() <= maxSize)
trie.insert(line);
}
}
catch (FileNotFoundException e)
{
logger.error("Cannot open file", e);
}
logger.info("Finish reading the dictionary");
return trie;
}
static boolean[] connectivityRow(int x, int y, int size)
{
boolean[] squares = new boolean[size * size];
for (int offsetX = -1; offsetX <= 1; offsetX++)
{
for (int offsetY = -1; offsetY <= 1; offsetY++)
{
final int calX = x + offsetX;
final int calY = y + offsetY;
if ((calX >= 0) && (calX < size) && (calY >= 0) && (calY < size))
squares[calY * size + calX] = true;
}
}
squares[y * size + x] = false;//the current x, y is false
return squares;
}
/**
Returns the matrix of connectivity between two points. Point i can go to point j iff matrix[i][j] is true
Square (x, y) is equivalent to point (size * y + x). For example, square (1,1) is point 5 in a puzzle of size 4
@param size the size of the puzzle
@return the connectivity matrix
*/
public static boolean[][] connectivityMatrix(int size)
{
boolean[][] matrix = new boolean[size * size][];
for (int x = 0; x < size; x++)
{
for (int y = 0; y < size; y++)
{
matrix[y * size + x] = connectivityRow(x, y, size);
}
}
return matrix;
}
}
Mar*_* A. 23
我想你可能会花大部分时间来尝试匹配你的字母网格无法构建的单词.所以,我要做的第一件事就是尝试加快这一步,这应该可以让你在那里大部分时间.
为此,我会将网格重新表达为可能的"移动"表,您可以通过正在查看的字母转换来编制索引.
首先为每个字母分配整个字母表中的数字(A = 0,B = 1,C = 2,......等等).
我们来看这个例子:
h b c d
e e g h
l l k l
m o f p
现在,让我们使用我们所拥有的字母(通常你可能希望每次使用相同的整个字母):
b | c | d | e | f | g | h | k | l | m | o | p
---+---+---+---+---+---+---+---+---+---+----+----
0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11
然后创建一个2D布尔数组,告诉您是否有可用的某个字母转换:
| 0 1 2 3 4 5 6 7 8 9 10 11 <- from letter
| b c d e f g h k l m o p
-----+--------------------------------------
0 b | T T T T
1 c | T T T T T
2 d | T T T
3 e | T T T T T T T
4 f | T T T T
5 g | T T T T T T T
6 h | T T T T T T T
7 k | T T T T T T T
8 l | T T T T T T T T T
9 m | T T
10 o | T T T T
11 p | T T T
^
to letter
现在浏览你的单词列表并将单词转换为过渡:
hello (6, 3, 8, 8, 10):
6 -> 3, 3 -> 8, 8 -> 8, 8 -> 10
然后通过在表格中查找是否允许这些转换来检查:
[6][ 3] : T
[3][ 8] : T
[8][ 8] : T
[8][10] : T
如果他们都被允许,那么可能会找到这个词.
例如,可以在第4次转换(m到e:helMEt)中排除"helmet"一词,因为表中的该条目为false.
并且可以排除仓鼠这个词,因为不允许第一个(h到a)过渡(甚至不存在于你的表中).
现在,对于你没有消除的可能很少的剩余单词,尝试在网格中实际找到它们的方式,就像你现在这样做或者在这里的其他一些答案中所建议的那样.这是为了避免因网格中相同字母之间的跳转而导致误报.例如,表格允许使用"帮助"一词,但网格不允许.
有关此想法的一些进一步的性能改进提示
使用一维数组而不是使用二维数组,只需自己计算第二个字母的索引.因此,不是像上面这样的12x12数组,而是制作一个长度为144的一维数组.如果你总是使用相同的字母表(即标准英文字母表的26x26 = 676x1数组),即使并非所有字母都显示在你的网格中,您可以将索引预先计算到此1D数组中,您需要测试该数组以匹配您的字典单词.例如,上面示例中的'hello'的索引将是
hello (6, 3, 8, 8, 10):
42 (from 6 + 3x12), 99, 104, 128
-> "hello" will be stored as 42, 99, 104, 128 in the dictionary
将想法扩展到3D表格(表示为1D数组),即所有允许的3个字母组合.这样你就可以立即消除更多的单词并将每个单词的数组查找次数减少1:对于'hello',你只需要3个数组查找:hel,ell,llo.顺便说一句,构建此表将非常快,因为网格中只有400个可能的3个字母移动.
预先计算网格中需要包含在表格中的移动索引.对于上面的示例,您需要将以下条目设置为"True":
(0,0) (0,1) -> here: h, b : [6][0]
(0,0) (1,0) -> here: h, e : [6][3]
(0,0) (1,1) -> here: h, e : [6][3]
(0,1) (0,0) -> here: b, h : [0][6]
(0,1) (0,2) -> here: b, c : [0][1]
.
:
我确定如果你使用这种方法,你可以让你的代码疯狂地运行,如果你有预先计算好的字典并且已经加载到内存中.
顺便说一句:另外一件好事,如果你正在建立一个游戏,就是在后台立即运行这些东西.当用户仍在查看应用程序的标题屏幕并让他的手指按下"播放"时,开始生成并解决第一个游戏.然后在用户播放上一个游戏时生成并解决下一个游戏.这应该会给你很多时间来运行你的代码.
(我喜欢这个问题,所以我很可能会在接下来的几天里用Java来实现我的提议,看看它实际上会如何执行......我会在这里发布代码.)
更新:
好的,我今天有一些时间用Java实现了这个想法:
class DictionaryEntry {
public int[] letters;
public int[] triplets;
}
class BoggleSolver {
// Constants
final int ALPHABET_SIZE = 5; // up to 2^5 = 32 letters
final int BOARD_SIZE = 4; // 4x4 board
final int[] moves = {-BOARD_SIZE-1, -BOARD_SIZE, -BOARD_SIZE+1,
-1, +1,
+BOARD_SIZE-1, +BOARD_SIZE, +BOARD_SIZE+1};
// Technically constant (calculated here for flexibility, but should be fixed)
DictionaryEntry[] dictionary; // Processed word list
int maxWordLength = 0;
int[] boardTripletIndices; // List of all 3-letter moves in board coordinates
DictionaryEntry[] buildDictionary(String fileName) throws IOException {
BufferedReader fileReader = new BufferedReader(new FileReader(fileName));
String word = fileReader.readLine();
ArrayList<DictionaryEntry> result = new ArrayList<DictionaryEntry>();
while (word!=null) {
if (word.length()>=3) {
word = word.toUpperCase();
if (word.length()>maxWordLength) maxWordLength = word.length();
DictionaryEntry entry = new DictionaryEntry();
entry.letters = new int[word.length() ];
entry.triplets = new int[word.length()-2];
int i=0;
for (char letter: word.toCharArray()) {
entry.letters[i] = (byte) letter - 65; // Convert ASCII to 0..25
if (i>=2)
entry.triplets[i-2] = (((entry.letters[i-2] << ALPHABET_SIZE) +
entry.letters[i-1]) << ALPHABET_SIZE) +
entry.letters[i];
i++;
}
result.add(entry);
}
word = fileReader.readLine();
}
return result.toArray(new DictionaryEntry[result.size()]);
}
boolean isWrap(int a, int b) { // Checks if move a->b wraps board edge (like 3->4)
return Math.abs(a%BOARD_SIZE-b%BOARD_SIZE)>1;
}
int[] buildTripletIndices() {
ArrayList<Integer> result = new ArrayList<Integer>();
for (int a=0; a<BOARD_SIZE*BOARD_SIZE; a++)
for (int bm: moves) {
int b=a+bm;
if ((b>=0) && (b<board.length) && !isWrap(a, b))
for (int cm: moves) {
int c=b+cm;
if ((c>=0) && (c<board.length) && (c!=a) && !isWrap(b, c)) {
result.add(a);
result.add(b);
result.add(c);
}
}
}
int[] result2 = new int[result.size()];
int i=0;
for (Integer r: result) result2[i++] = r;
return result2;
}
// Variables that depend on the actual game layout
int[] board = new int[BOARD_SIZE*BOARD_SIZE]; // Letters in board
boolean[] possibleTriplets = new boolean[1 << (ALPHABET_SIZE*3)];
DictionaryEntry[] candidateWords;
int candidateCount;
int[] usedBoardPositions;
DictionaryEntry[] foundWords;
int foundCount;
void initializeBoard(String[] letters) {
for (int row=0; row<BOARD_SIZE; row++)
for (int col=0; col<BOARD_SIZE; col++)
board[row*BOARD_SIZE + col] = (byte) letters[row].charAt(col) - 65;
}
void setPossibleTriplets() {
Arrays.fill(possibleTriplets, false); // Reset list
int i=0;
while (i<boardTripletIndices.length) {
int triplet = (((board[boardTripletIndices[i++]] << ALPHABET_SIZE) +
board[boardTripletIndices[i++]]) << ALPHABET_SIZE) +
board[boardTripletIndices[i++]];
possibleTriplets[triplet] = true;
}
}
void checkWordTriplets() {
candidateCount = 0;
for (DictionaryEntry entry: dictionary) {
boolean ok = true;
int len = entry.triplets.length;
for (int t=0; (t<len) && ok; t++)
ok = possibleTriplets[entry.triplets[t]];
if (ok) candidateWords[candidateCount++] = entry;
}
}
void checkWords() { // Can probably be optimized a lot
foundCount = 0;
for (int i=0; i<candidateCount; i++) {
DictionaryEntry candidate = candidateWords[i];
for (int j=0; j<board.length; j++)
if (board[j]==candidate.letters[0]) {
usedBoardPositions[0] = j;
if (checkNextLetters(candidate, 1, j)) {
foundWords[foundCount++] = candidate;
break;
}
}
}
}
boolean checkNextLetters(DictionaryEntry candidate, int letter, int pos) {
if (letter==candidate.letters.length) return true;
int match = candidate.letters[letter];
for (int move: moves) {
int next=pos+move;
if ((next>=0) && (next<board.length) && (board[next]==match) && !isWrap(pos, next)) {
boolean ok = true;
for (int i=0; (i<letter) && ok; i++)
ok = usedBoardPositions[i]!=next;
if (ok) {
usedBoardPositions[letter] = next;
if (checkNextLetters(candidate, letter+1, next)) return true;
}
}
}
return false;
}
// Just some helper functions
String formatTime(long start, long end, long repetitions) {
long time = (end-start)/repetitions;
return time/1000000 + "." + (time/100000) % 10 + "" + (time/10000) % 10 + "ms";
}
String getWord(DictionaryEntry entry) {
char[] result = new char[entry.letters.length];
int i=0;
for (int letter: entry.letters)
result[i++] = (char) (letter+97);
return new String(result);
}
void run() throws IOException {
long start = System.nanoTime();
// The following can be pre-computed and should be replaced by constants
dictionary = buildDictionary("C:/TWL06.txt");
boardTripletIndices = buildTripletIndices();
long precomputed = System.nanoTime();
// The following only needs to run once at the beginning of the program
candidateWords = new DictionaryEntry[dictionary.length]; // WAAAY too generous
foundWords = new DictionaryEntry[dictionary.length]; // WAAAY too generous
usedBoardPositions = new int[maxWordLength];
long initialized = System.nanoTime();
for (int n=1; n<=100; n++) {
// The following needs to run again for every new board
initializeBoard(new String[] {"DGHI",
"KLPS",
"YEUT",
"EORN"});
setPossibleTriplets();
checkWordTriplets();
checkWords();
}
long solved = System.nanoTime();
// Print out result and statistics
System.out.println("Precomputation finished in " + formatTime(start, precomputed, 1)+":");
System.out.println(" Words in the dictionary: "+dictionary.length);
System.out.println(" Longest word: "+maxWordLength+" letters");
System.out.println(" Number of triplet-moves: "+boardTripletIndices.length/3);
System.out.println();
System.out.println("Initialization finished in " + formatTime(precomputed, initialized, 1));
System.out.println();
System.out.println("Board solved in "+formatTime(initialized, solved, 100)+":");
System.out.println(" Number of candidates: "+candidateCount);
System.out.println(" Number of actual words: "+foundCount);
System.out.println();
System.out.println("Words found:");
int w=0;
System.out.print(" ");
for (int i=0; i<foundCount; i++) {
System.out.print(getWord(foundWords[i]));
w++;
if (w==10) {
w=0;
System.out.println(); System.out.print(" ");
} else
if (i<foundCount-1) System.out.print(", ");
}
System.out.println();
}
public static void main(String[] args) throws IOException {
new BoggleSolver().run();
}
}
以下是一些结果:
对于原始问题中发布的图片中的网格(DGHI ...):
Precomputation finished in 239.59ms:
Words in the dictionary: 178590
Longest word: 15 letters
Number of triplet-moves: 408
Initialization finished in 0.22ms
Board solved in 3.70ms:
Number of candidates: 230
Number of actual words: 163
Words found:
eek, eel, eely, eld, elhi, elk, ern, erupt, erupts, euro
eye, eyer, ghi, ghis, glee, gley, glue, gluer, gluey, glut
gluts, hip, hiply, hips, his, hist, kelp, kelps, kep, kepi
kepis, keps, kept, kern, key, kye, lee, lek, lept, leu
ley, lunt, lunts, lure, lush, lust, lustre, lye, nus, nut
nuts, ore, ort, orts, ouph, ouphs, our, oust, out, outre
outs, oyer, pee, per, pert, phi, phis, pis, pish, plus
plush, ply, plyer, psi, pst, pul, pule, puler, pun, punt
punts, pur, pure, puree, purely, pus, push, put, puts, ree
rely, rep, reply, reps, roe, roue, roup, roups, roust, rout
routs, rue, rule, ruly, run, runt, runts, rupee, rush, rust
rut, ruts, ship, shlep, sip, sipe, spue, spun, spur, spurn
spurt, strep, stroy, stun, stupe, sue, suer, sulk, sulker, sulky
sun, sup, supe, super, sure, surely, tree, trek, trey, troupe
troy, true, truly, tule, tun, tup, tups, turn, tush, ups
urn, uts, yeld, yelk, yelp, yelps, yep, yeps, yore, you
your, yourn, yous
对于在原始问题中作为示例发布的字母(FXIE ...)
Precomputation finished in 239.68ms:
Words in the dictionary: 178590
Longest word: 15 letters
Number of triplet-moves: 408
Initialization finished in 0.21ms
Board solved in 3.69ms:
Number of candidates: 87
Number of actual words: 76
Words found:
amble, ambo, ami, amie, asea, awa, awe, awes, awl, axil
axile, axle, boil, bole, box, but, buts, east, elm, emboli
fame, fames, fax, lei, lie, lima, limb, limbo, limbs, lime
limes, lob, lobs, lox, mae, maes, maw, maws, max, maxi
mesa, mew, mewl, mews, mil, mile, milo, mix, oil, ole
sae, saw, sea, seam, semi, sew, stub, swam, swami, tub
tubs, tux, twa, twae, twaes, twas, uts, wae, waes, wamble
wame, wames, was, wast, wax, west
For the following 5x5-grid:
R P R I T
A H H L N
I E T E P
Z R Y S G
O G W E Y
it gives this:
Precomputation finished in 240.39ms:
Words in the dictionary: 178590
Longest word: 15 letters
Number of triplet-moves: 768
Initialization finished in 0.23ms
Board solved in 3.85ms:
Number of candidates: 331
Number of actual words: 240
Words found:
aero, aery, ahi, air, airt, airth, airts, airy, ear, egest
elhi, elint, erg, ergo, ester, eth, ether, eye, eyen, eyer
eyes, eyre, eyrie, gel, gelt, gelts, gen, gent, gentil, gest
geste, get, gets, gey, gor, gore, gory, grey, greyest, greys
gyre, gyri, gyro, hae, haet, haets, hair, hairy, hap, harp
heap, hear, heh, heir, help, helps, hen, hent, hep, her
hero, hes, hest, het, hetero, heth, hets, hey, hie, hilt
hilts, hin, hint, hire, hit, inlet, inlets, ire, leg, leges
legs, lehr, lent, les, lest, let, lethe, lets, ley, leys
lin, line, lines, liney, lint, lit, neg, negs, nest, nester
net, nether, nets, nil, nit, ogre, ore, orgy, ort, orts
pah, pair, par, peg, pegs, peh, pelt, pelter, peltry, pelts
pen, pent, pes, pest, pester, pesty, pet, peter, pets, phi
philter, philtre, phiz, pht, print, pst, rah, rai, rap, raphe
raphes, reap, rear, rei, ret, rete, rets, rhaphe, rhaphes, rhea
ria, rile, riles, riley, rin, rye, ryes, seg, sel, sen
sent, senti, set, sew, spelt, spelter, spent, splent, spline, splint
split, stent, step, stey, stria, striae, sty, stye, tea, tear
teg, tegs, tel, ten, tent, thae, the, their, then, these
thesp, they, thin, thine, thir, thirl, til, tile, tiles, tilt
tilter, tilth, tilts, tin, tine, tines, tirl, trey, treys, trog
try, tye, tyer, tyes, tyre, tyro, west, wester, wry, wryest
wye, wyes, wyte, wytes, yea, yeah, year, yeh, yelp, yelps
yen, yep, yeps, yes, yester, yet, yew, yews, zero, zori
For this I used the TWL06 Tournament Scrabble Word List, since the link in the original question no longer works. This file is 1.85MB, so it's a little bit shorter. And the buildDictionary function throws out all words with less than 3 letters.
Here are a couple of observations about the performance of this:
It's about 10 times slower than the reported performance of Victor Nicollet's OCaml implementation. Whether this is caused by the different algorithm, the shorter dictionary he used, the fact that his code is compiled and mine runs in a Java virtual machine, or the performance of our computers (mine is an Intel Q6600 @ 2.4MHz running WinXP), I don't know. But it's much faster than the results for the other implementations quoted at the end of the original question. So, whether this algorithm is superior to the trie dictionary or not, I don't know at this point.
The table method used in checkWordTriplets() yields a very good approximation to the actual answers. Only 1 in 3-5 words passed by it will fail the checkWords() test (See number of candidates vs. number of actual words above).
Something you can't see above: The checkWordTriplets() function takes about 3.65ms and is therefore fully dominant in the search process. The checkWords() function takes up pretty much the remaining 0.05-0.20 ms.
The execution time of the checkWordTriplets() function depends linearly on the dictionary size and is virtually independent of board size!
The execution time of checkWords() depends on the board size and the number of words not ruled out by checkWordTriplets().
The checkWords() implementation above is the dumbest first version I came up with. It is basically not optimized at all. But compared to checkWordTriplets() it is irrelevant for the total performance of the application, so I didn't worry about it. But, if the board size gets bigger, this function will get slower and slower and will eventually start to matter. Then, it would need to be optimized as well.
One nice thing about this code is its flexibility:
initializeBoard().Ok, but I think by now this post is waaaay long enough. I can definitely answer any questions you might have, but let's move that to the comments.
Pao*_*ino 19
令人惊讶的是,没有人尝试过这个PHP版本.
这是John Fouhy 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指示,但这主要是从约翰那里复制过来的.
$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);
如果您想尝试一下,这是一个实时链接.虽然我的本地机器需要大约2秒,但我的网络服务器需要大约5秒.在任何一种情况下,它都不是很快.尽管如此,它仍然很可怕,所以我可以想象时间可以大大减少.关于如何实现这一点的任何指示将不胜感激.PHP缺乏元组使得坐标很奇怪,我无法理解到底发生了什么并没有帮助.
编辑:一些修复使它在本地不到1秒.
rva*_*her 16
对VB不感兴趣？:)我无法抗拒.我解决这个问题的方式与此处介绍的许多解决方案不同.
我的时代是:
编辑:Web主机服务器上的字典加载时间比我的家用计算机长1到1.5秒.
我不知道服务器上的负载会有多么糟糕.
我在.Net中将我的解决方案写成了一个网页.myvrad.com/boggle
我正在使用原始问题中引用的字典.
字母不会被重复使用.只找到3个字符或更长的字.
我正在使用所有唯一单词前缀和单词的散列表而不是trie.我不知道特里的故事,所以我在那里学到了一些东西.除了完整的单词之外,创建单词前缀列表的想法最终让我的时间减少到可敬的数字.
阅读代码注释以获取更多详细信息.
这是代码:
Imports System.Collections.Generic
Imports System.IO
Partial Class boggle_Default
'Bob Archer, 4/15/2009
'To avoid using a 2 dimensional array in VB I'm not using typical X,Y
'coordinate iteration to find paths.
'
'I have locked the code into a 4 by 4 grid laid out like so:
' abcd
' efgh
' ijkl
' mnop
'
'To find paths the code starts with a letter from a to p then
'explores the paths available around it. If a neighboring letter
'already exists in the path then we don't go there.
'
'Neighboring letters (grid points) are hard coded into
'a Generic.Dictionary below.
'Paths is a list of only valid Paths found.
'If a word prefix or word is not found the path is not
'added and extending that path is terminated.
Dim Paths As New Generic.List(Of String)
'NeighborsOf. The keys are the letters a to p.
'The value is a string of letters representing neighboring letters.
'The string of neighboring letters is split and iterated later.
Dim NeigborsOf As New Generic.Dictionary(Of String, String)
'BoggleLetters. The keys are mapped to the lettered grid of a to p.
'The values are what the user inputs on the page.
Dim BoggleLetters As New Generic.Dictionary(Of String, String)
'Used to store last postition of path. This will be a letter
'from a to p.
Dim LastPositionOfPath As String = ""
'I found a HashTable was by far faster than a Generic.Dictionary
' - about 10 times faster. This stores prefixes of words and words.
'I determined 792773 was the number of words and unique prefixes that
'will be generated from the dictionary file. This is a max number and
'the final hashtable will not have that many.
Dim HashTableOfPrefixesAndWords As New Hashtable(792773)
'Stores words that are found.
Dim FoundWords As New Generic.List(Of String)
'Just to validate what the user enters in the grid.
Dim ErrorFoundWithSubmittedLetters As Boolean = False
Public Sub BuildAndTestPathsAndFindWords(ByVal ThisPath As String)
'Word is the word correlating to the ThisPath parameter.
'This path would be a series of letters from a to p.
Dim Word As String = ""
'The path is iterated through and a word based on the actual
'letters in the Boggle grid is assembled.
For i As Integer = 0 To ThisPath.Length - 1
Word += Me.BoggleLetters(ThisPath.Substring(i, 1))
Next
'If my hashtable of word prefixes and words doesn't contain this Word
'Then this isn't a word and any further extension of ThisPath will not
'yield any words either. So exit sub to terminate exploring this path.
If Not HashTableOfPrefixesAndWords.ContainsKey(Word) Then Exit Sub
'The value of my hashtable is a boolean representing if the key if a word (true) or
'just a prefix (false). If true and at least 3 letters long then yay! word found.
If HashTableOfPrefixesAndWords(Word) AndAlso Word.Length > 2 Then Me.FoundWords.Add(Word)
'If my List of Paths doesn't contain ThisPath then add it.
'Remember only valid paths will make it this far. Paths not found
'in the HashTableOfPrefixesAndWords cause this sub to exit above.
If Not Paths.Contains(ThisPath) Then Paths.Add(ThisPath)
'Examine the last letter of ThisPath. We are looking to extend the path
'to our neighboring letters if any are still available.
LastPositionOfPath = ThisPath.Substring(ThisPath.Length - 1, 1)
'Loop through my list of neighboring letters (representing grid points).
For Each Neighbor As String In Me.NeigborsOf(LastPositionOfPath).ToCharArray()
'If I find a neighboring grid point that I haven't already used
'in ThisPath then extend ThisPath and feed the new path into
'this recursive function. (see recursive.)
If Not ThisPath.Contains(Neighbor) Then Me.BuildAndTestPathsAndFindWords(ThisPath & Neighbor)
Next
End Sub
Protected Sub ButtonBoggle_Click(ByVal sender As Object, ByVal e As System.EventArgs) Handles ButtonBoggle.Click
'User has entered the 16 letters and clicked the go button.
'Set up my Generic.Dictionary of grid points, I'm using letters a to p -
'not an x,y grid system. The values are neighboring points.
NeigborsOf.Add("a", "bfe")
NeigborsOf.Add("b", "cgfea")
NeigborsOf.Add("c", "dhgfb")
NeigborsOf.Add("d", "hgc")
NeigborsOf.Add("e", "abfji")
NeigborsOf.Add("f", "abcgkjie")
NeigborsOf.Add("g", "bcdhlkjf")
NeigborsOf.Add("h", "cdlkg")
NeigborsOf.Add("i", "efjnm")
NeigborsOf.Add("j", "efgkonmi")
NeigborsOf.Add("k", "fghlponj")
NeigborsOf.Add("l", "ghpok")
NeigborsOf.Add("m", "ijn")
NeigborsOf.Add("n", "ijkom")
NeigborsOf.Add("o", "jklpn")
NeigborsOf.Add("p", "klo")
'Retrieve letters the user entered.
BoggleLetters.Add("a", Me.TextBox1.Text.ToLower.Trim())
BoggleLetters.Add("b", Me.TextBox2.Text.ToLower.Trim())
BoggleLetters.Add("c", Me.TextBox3.Text.ToLower.Trim())
BoggleLetters.Add("d", Me.TextBox4.Text.ToLower.Trim())
BoggleLetters.Add("e", Me.TextBox5.Text.ToLower.Trim())
BoggleLetters.Add("f", Me.TextBox6.Text.ToLower.Trim())
BoggleLetters.Add("g", Me.TextBox7.Text.ToLower.Trim())
BoggleLetters.Add("h", Me.TextBox8.Text.ToLower.Trim())
BoggleLetters.Add("i", Me.TextBox9.Text.ToLower.Trim())
BoggleLetters.Add("j", Me.TextBox10.Text.ToLower.Trim())
BoggleLetters.Add("k", Me.TextBox11.Text.ToLower.Trim())
BoggleLetters.Add("l", Me.TextBox12.Text.ToLower.Trim())
BoggleLetters.Add("m", Me.TextBox13.Text.ToLower.Trim())
BoggleLetters.Add("n", Me.TextBox14.Text.ToLower.Trim())
BoggleLetters.Add("o", Me.TextBox15.Text.ToLower.Trim())
BoggleLetters.Add("p", Me.TextBox16.Text.ToLower.Trim())
'Validate user entered something with a length of 1 for all 16 textboxes.
For Each S As String In BoggleLetters.Keys
If BoggleLetters(S).Length <> 1 Then
ErrorFoundWithSubmittedLetters = True
Exit For
End If
Next
'If input is not valid then...
If ErrorFoundWithSubmittedLetters Then
'Present error message.
Else
'Else assume we have 16 letters to work with and start finding words.
Dim SB As New StringBuilder
Dim Time As String = String.Format("{0}:{1}:{2}:{3}", Date.Now.Hour.ToString(), Date.Now.Minute.ToString(), Date.Now.Second.ToString(), Date.Now.Millisecond.ToString())
Dim NumOfLetters As Integer = 0
Dim Word As String = ""
Dim TempWord As String = ""
Dim Letter As String = ""
Dim fr As StreamReader = Nothing
fr = New System.IO.StreamReader(HttpContext.Current.Request.MapPath("~/boggle/dic.txt"))
'First fill my hashtable with word prefixes and words.
'HashTable(PrefixOrWordString, BooleanTrueIfWordFalseIfPrefix)
While fr.Peek <> -1
Word = fr.ReadLine.Trim()
TempWord = ""
For i As Integer = 0 To Word.Length - 1
Letter = Word.Substring(i, 1)
'This optimization helped quite a bit. Words in the dictionary that begin
'with letters that the user did not enter in the grid shouldn't go in my hashtable.
'
'I realize most of the solutions went with a Trie. I'd never heard of that before,
'which is one of the neat things about SO, seeing how others approach challenges
'and learning some best practices.
'
'However, I didn't code a Trie in my solution. I just have a hashtable with
'all words in the dicitonary file and all possible prefixes for those words.
'A Trie might be faster but I'm not coding it now. I'm getting good times with this.
If i = 0 AndAlso Not BoggleLetters.ContainsValue(Letter) Then Continue While
TempWord += Letter
If Not HashTableOfPrefixesAndWords.ContainsKey(TempWord) Then
HashTableOfPrefixesAndWords.Add(TempWord, TempWord = Word)
End If
Next
End While
SB.Append("Number of Word Prefixes and Words in Hashtable: " & HashTableOfPrefixesAndWords.Count.ToString())
SB.Append("<br />")
SB.Append("Loading Dictionary: " & Time & " - " & String.Format("{0}:{1}:{2}:{3}", Date.Now.Hour.ToString(), Date.Now.Minute.ToString(), Date.Now.Second.ToString(), Date.Now.Millisecond.ToString()))
SB.Append("<br />")
Time = String.Format("{0}:{1}:{2}:{3}", Date.Now.Hour.ToString(), Date.Now.Minute.ToString(), Date.Now.Second.ToString(), Date.Now.Millisecond.ToString())
'This starts a path at each point on the grid an builds a path until
'the string of letters correlating to the path is not found in the hashtable
'of word prefixes and words.
Me.BuildAndTestPathsAndFindWords("a")
Me.BuildAndTestPathsAndFindWords("b")
Me.BuildAndTestPathsAndFindWords("c")
Me.BuildAndTestPathsAndFindWords("d")
Me.BuildAndTestPathsAndFindWords("e")
Me.BuildAndTestPathsAndFindWords("f")
Me.BuildAndTestPathsAndFindWords("g")
Me.BuildAndTestPathsAndFindWords("h")
Me.BuildAndTestPathsAndFindWords("i")
Me.BuildAndTestPathsAndFindWords("j")
Me.BuildAndTestPathsAndFindWords("k")
Me.BuildAndTestPathsAndFindWords("l")
Me.BuildAndTestPathsAndFindWords("m")
Me.BuildAndTestPathsAndFindWords("n")
Me.BuildAndTestPathsAndFindWords("o")
Me.BuildAndTestPathsAndFindWords("p")
SB.Append("Finding Words: " & Time & " - " & String.Format("{0}:{1}:{2}:{3}", Date.Now.Hour.ToString(), Date.Now.Minute.ToString(), Date.Now.Second.ToString(), Date.Now.Millisecond.ToString()))
SB.Append("<br />")
SB.Append("Num of words found: " & FoundWords.Count.ToString())
SB.Append("<br />")
SB.Append("<br />")
FoundWords.Sort()
SB.Append(String.Join("<br />", FoundWords.ToArray()))
'Output results.
Me.LiteralBoggleResults.Text = SB.ToString()
Me.PanelBoggleResults.Visible = True
End If
End Sub
End Class
Pau*_*och 11
我一看到问题陈述,就想到了"Trie".但看到其他几个海报都使用了这种方法,我寻找另一种方法只是为了与众不同.唉,Trie方法表现更好.我在我的机器上运行了Kent的Perl解决方案,运行后需要0.31秒才能使用我的字典文件.我自己的perl实现需要0.54秒才能运行.
这是我的方法:
创建转换哈希以模拟合法转换.
迭代所有16 ^ 3可能的三个字母组合.
然后遍历字典中的所有单词.
打印出我找到的单词.
我尝试了3个字母和4个字母的序列,但是4个字母的序列减慢了程序的速度.
在我的代码中,我使用/ usr/share/dict/words作为我的字典.它是MAC OS X和许多Unix系统的标准配置.如果需要,您可以使用其他文件.要破解不同的谜题,只需更改变量@puzzle即可.对于较大的矩阵,这很容易适应.您只需要更改%transitions散列和%legalTransitions散列.
该解决方案的优势在于代码简短,数据结构简单.
这是Perl代码(我知道使用了太多的全局变量):
#!/usr/bin/perl
use Time::HiRes qw{ time };
sub readFile($);
sub findAllPrefixes($);
sub isWordTraceable($);
sub findWordsInPuzzle(@);
my $startTime = time;
# Puzzle to solve
my @puzzle = (
F, X, I, E,
A, M, L, O,
E, W, B, X,
A, S, T, U
);
my $minimumWordLength = 3;
my $maximumPrefixLength = 3; # I tried four and it slowed down.
# Slurp the word list.
my $wordlistFile = "/usr/share/dict/words";
my @words = split(/\n/, uc(readFile($wordlistFile)));
print "Words loaded from word list: " . scalar @words . "\n";
print "Word file load time: " . (time - $startTime) . "\n";
my $postLoad = time;
# Define the legal transitions from one letter position to another.
# Positions are numbered 0-15.
# 0 1 2 3
# 4 5 6 7
# 8 9 10 11
# 12 13 14 15
my %transitions = (
-1 => [0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15],
0 => [1,4,5],
1 => [0,2,4,5,6],
2 => [1,3,5,6,7],
3 => [2,6,7],
4 => [0,1,5,8,9],
5 => [0,1,2,4,6,8,9,10],
6 => [1,2,3,5,7,9,10,11],
7 => [2,3,6,10,11],
8 => [4,5,9,12,13],
9 => [4,5,6,8,10,12,13,14],
10 => [5,6,7,9,11,13,14,15],
11 => [6,7,10,14,15],
12 => [8,9,13],
13 => [8,9,10,12,14],
14 => [9,10,11,13,15],
15 => [10,11,14]
);
# Convert the transition matrix into a hash for easy access.
my %legalTransitions = ();
foreach my $start (keys %transitions) {
my $legalRef = $transitions{$start};
foreach my $stop (@$legalRef) {
my $index = ($start + 1) * (scalar @puzzle) + ($stop + 1);
$legalTransitions{$index} = 1;
}
}
my %prefixesInPuzzle = findAllPrefixes($maximumPrefixLength);
print "Find prefixes time: " . (time - $postLoad) . "\n";
my $postPrefix = time;
my @wordsFoundInPuzzle = findWordsInPuzzle(@words);
print "Find words in puzzle time: " . (time - $postPrefix) . "\n";
print "Unique prefixes found: " . (scalar keys %prefixesInPuzzle) . "\n";
print "Words found (" . (scalar @wordsFoundInPuzzle) . ") :\n " . join("\n ", @wordsFoundInPuzzle) . "\n";
print "Total Elapsed time: " . (time - $startTime) . "\n";
###########################################
sub readFile($) {
my ($filename) = @_;
my $contents;
if (-e $filename) {
# This is magic: it opens and reads a file into a scalar in one line of code.
# See http://www.perl.com/pub/a/2003/11/21/slurp.html
$contents = do { local( @ARGV, $/ ) = $filename ; <> } ;
}
else {
$contents = '';
}
return $contents;
}
# Is it legal to move from the first position to the second? They must be adjacent.
sub isLegalTransition($$) {
my ($pos1,$pos2) = @_;
my $index = ($pos1 + 1) * (scalar @puzzle) + ($pos2 + 1);
return $legalTransitions{$index};
}
# Find all prefixes where $minimumWordLength <= length <= $maxPrefixLength
#
# $maxPrefixLength ... Maximum length of prefix we will store. Three gives best performance.
sub findAllPrefixes($) {
my ($maxPrefixLength) = @_;
my %prefixes = ();
my $puzzleSize = scalar @puzzle;
# Every possible N-letter combination of the letters in the puzzle
# can be represented as an integer, though many of those combinations
# involve illegal transitions, duplicated letters, etc.
# Iterate through all those possibilities and eliminate the illegal ones.
my $maxIndex = $puzzleSize ** $maxPrefixLength;
for (my $i = 0; $i < $maxIndex; $i++) {
my @path;
my $remainder = $i;
my $prevPosition = -1;
my $prefix = '';
my %usedPositions = ();
for (my $prefixLength = 1; $prefixLength <= $maxPrefixLength; $prefixLength++) {
my $position = $remainder % $puzzleSize;
# Is this a valid step?
# a. Is the transition legal (to an adjacent square)?
if (! isLegalTransition($prevPosition, $position)) {
last;
}
# b. Have we repeated a square?
if ($usedPositions{$position}) {
last;
}
else {
$usedPositions{$position} = 1;
}
# Record this prefix if length >= $minimumWordLength.
$prefix .= $puzzle[$position];
if ($prefixLength >= $minimumWordLength) {
$prefixes{$prefix} = 1;
}
push @path, $position;
$remainder -= $position;
$remainder /= $puzzleSize;
$prevPosition = $position;
} # end inner for
} # end outer for
return %prefixes;
}
# Loop through all words in dictionary, looking for ones that are in the puzzle.
sub findWordsInPuzzle(@) {
my @allWords = @_;
my @wordsFound = ();
my $puzzleSize = scalar @puzzle;
WORD: foreach my $word (@allWords) {
my $wordLength = length($word);
if ($wordLength > $puzzleSize || $wordLength < $minimumWordLength) {
# Reject word as too short or too long.
}
elsif ($wordLength <= $maximumPrefixLength ) {
# Word should be in the prefix hash.
if ($prefixesInPuzzle{$word}) {
push @wordsFound, $word;
}
}
else {
# Scan through the word using a window of length $maximumPrefixLength, looking for any strings not in our prefix list.
# If any are found that are not in the list, this word is not possible.
# If no non-matches are found, we have more work to do.
my $limit = $wordLength - $maximumPrefixLength + 1;
for (my $startIndex = 0; $startIndex < $limit; $startIndex ++) {
if (! $prefixesInPuzzle{substr($word, $startIndex, $maximumPrefixLength)}) {
next WORD;
}
}
if (isWordTraceable($word)) {
# Additional test necessary: see if we can form this word by following legal transitions
push @wordsFound, $word;
}
}
}
return @wordsFound;
}
# Is it possible to trace out the word using only legal transitions?
sub isWordTraceable($) {
my $word = shift;
return traverse([split(//, $word)], [-1]); # Start at special square -1, which may transition to any square in the puzzle.
}
# Recursively look for a path through the puzzle that matches the word.
sub traverse($$) {
my ($lettersRef, $pathRef) = @_;
my $index = scalar @$pathRef - 1;
my $position = $pathRef->[$index];
my $letter = $lettersRef->[$index];
my $branchesRef = $transitions{$position};
BRANCH: foreach my $branch (@$branchesRef) {
if ($puzzle[$branch] eq $letter) {
# Have we used this position yet?
foreach my $usedBranch (@$pathRef) {
if ($usedBranch == $branch) {
next BRANCH;
}
}
if (scalar @$lettersRef == $index + 1) {
return 1; # End of word and success.
}
push @$pathRef, $branch;
if (traverse($lettersRef, $pathRef)) {
return 1; # Recursive success.
}
else {
pop @$pathRef;
}
}
}
return 0; # No path found. Failed.
}
我知道我已经很晚了但是我刚才用PHP制作了其中一个- 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个字(133个点)
F.........X..I..............E...............
A......................................M..............................L............................O...............................
E....................W............................B..........................X
A..................S..................................................T.................U....
给出一些关于程序实际执行情况的指示 - 每个字母都是在每个'''开始查看模式的地方.显示它尝试采取的路径.越多'.' 它进一步搜索了.
让我知道你是否想要代码...这是一个可怕的PHP和HTML组合,从来没有意味着看到光明的一天,所以我不敢在这里发布:P
小智 9
我花了3个月的时间研究10个最佳密集5x5 Boggle板问题的解决方案.
该问题现已得到解决,并在5个网页上完整披露.如有疑问,请与我联系.
电路板分析算法使用显式堆栈通过具有直接子信息的定向非循环字图和时间戳跟踪机制伪递归地遍历电路板方块.这可能是世界上最先进的词典数据结构.
该方案在四核上每秒评估大约10,000个非常好的电路板.(9500+分)
父网页:
DeepSearch.c - http://www.pathcom.com/~vadco/deep.html
组件网页:
最佳记分板 - http://www.pathcom.com/~vadco/binary.html
高级词典结构 - http://www.pathcom.com/~vadco/adtdawg.html
电路板分析算法 - http://www.pathcom.com/~vadco/guns.html
并行批处理 - http://www.pathcom.com/~vadco/parallel.html
- 这项全面的工作只会引起最苛刻的人的兴趣.