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

我已经在c#中使用DFA算法解决了这个问题。你可以查看我的代码

https://github.com/attilabicsko/wordshuffler/

除了在矩阵中查找单词外,我的算法还保存单词的实际路径,所以在设计单词查找游戏时,你可以检查在实际路径上是否有单词。

其他回答

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

源代码

下面是使用NLTK工具包中的预定义单词的解决方案 NLTK有NLTK。语料库包,我们有一个叫做单词的包,它包含超过20万个英语单词,你可以简单地把它们都用到你的程序中。

一旦创建你的矩阵转换成一个字符数组,并执行这段代码

import nltk
from nltk.corpus import words
from collections import Counter

def possibleWords(input, charSet):
    for word in input:
        dict = Counter(word)
        flag = 1
        for key in dict.keys():
            if key not in charSet:
                flag = 0
        if flag == 1 and len(word)>5: #its depends if you want only length more than 5 use this otherwise remove that one. 
            print(word)


nltk.download('words')
word_list = words.words()
# prints 236736
print(len(word_list))
charSet = ['h', 'e', 'l', 'o', 'n', 'v', 't']
possibleWords(word_list, charSet)

输出:

eleven
eleventh
elevon
entente
entone
ethene
ethenol
evolve
evolvent
hellhole
helvell
hooven
letten
looten
nettle
nonene
nonent
nonlevel
notelet
novelet
novelette
novene
teenet
teethe
teevee
telethon
tellee
tenent
tentlet
theelol
toetoe
tonlet
toothlet
tootle
tottle
vellon
velvet
velveteen
venene
vennel
venthole
voeten
volent
volvelle
volvent
voteen

我希望你能得到它。

我的答案和这里的其他答案一样,但我把它贴出来是因为它看起来比其他Python解决方案快一些,因为设置字典更快。(我对比了John Fouhy的解决方案。)设置后,解决的时间在噪声中下降。

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)

示例用法:

# Print a maximal-length word and its path:
print max(solve(), key=lambda (word, path): len(word))

编辑:过滤掉长度小于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;它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,速度大约会翻倍。

你的搜索算法是否会随着搜索的继续而不断减少单词列表?

例如,在上面的搜索中,你的单词只能以13个字母开头(有效地减少了一半的开头字母)。

当你添加更多的字母排列时,它会进一步减少可用的单词集,减少必要的搜索。

我会从这里开始。

I wrote my solver in C++. I implemented a custom tree structure. I'm not sure it can be considered a trie but it's similar. Each node has 26 branches, 1 for each letter of the alphabet. I traverse the branches of the boggle board in parallel with the branches of my dictionary. If the branch does not exist in the dictionary, I stop searching it on the Boggle board. I convert all the letters on the board to ints. So 'A' = 0. Since it's just arrays, lookup is always O(1). Each node stores if it completes a word and how many words exist in its children. The tree is pruned as words are found to reduce repeatedly searching for the same words. I believe pruning is also O(1).

CPU: Pentium SU2700 1.3GHz 内存:3 gb

在< 1秒内加载178,590个单词的字典。 在4秒内解决100x100 Boggle (Boggle .txt)。约44000字。 解决4x4 Boggle游戏的速度太快,无法提供有意义的基准。:)

快速Boggle求解GitHub回购