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

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)

其他回答

我意识到这个问题的时间来了又去了,但由于我自己正在研究一个求解器,并在谷歌搜索时偶然发现了这个,我想我应该发布一个参考,因为它似乎与其他一些问题有点不同。

我选择在游戏棋盘上使用平面数组,并从棋盘上的每个字母进行递归搜索,从有效邻居遍历到有效邻居,如果索引中的有效前缀是当前字母列表,则扩展搜索。而遍历当前单词的概念是进入板的索引列表,而不是组成单词的字母。在检查索引时,将索引转换为字母并完成检查。

索引是一个蛮力字典,有点像trie,但允许对索引进行python查询。如果单词'cat'和'cater'在列表中,你会在字典中看到:

   d = { 'c': ['cat','cater'],
     'ca': ['cat','cater'],
     'cat': ['cat','cater'],
     'cate': ['cater'],
     'cater': ['cater'],
   }

因此,如果current_word是'ca',您就知道它是一个有效的前缀,因为'ca'在d中返回True(因此继续遍历板)。如果current_word是'cat',那么你知道它是一个有效的单词,因为它是一个有效的前缀,并且d['cat']中的'cat'也返回True。

如果感觉这允许一些可读的代码,似乎不是太慢。像其他人一样,这个系统的费用是读取/构建索引。解这个板子相当麻烦。

代码在http://gist.github.com/268079。它是故意垂直和幼稚的,有很多明确的有效性检查,因为我想理解问题,而不是用一堆魔法或晦涩难懂的东西把它弄得乱七八糟。

下面是我的java实现:https://github.com/zouzhile/interview/blob/master/src/com/interview/algorithms/tree/BoggleSolver.java

Trie构建耗时0小时0分1秒532毫秒 单词搜索花了0小时0分0秒92毫秒

eel eeler eely eer eke eker eld eleut elk ell 
elle epee epihippus ere erept err error erupt eurus eye 
eyer eyey hip hipe hiper hippish hipple hippus his hish 
hiss hist hler hsi ihi iphis isis issue issuer ist 
isurus kee keek keeker keel keeler keep keeper keld kele 
kelek kelep kelk kell kelly kelp kelper kep kepi kept 
ker kerel kern keup keuper key kyl kyle lee leek 
leeky leep leer lek leo leper leptus lepus ler leu 
ley lleu lue lull luller lulu lunn lunt lunule luo 
lupe lupis lupulus lupus lur lure lurer lush lushly lust 
lustrous lut lye nul null nun nupe nurture nurturer nut 
oer ore ort ouphish our oust out outpeep outpeer outpipe 
outpull outpush output outre outrun outrush outspell outspue outspurn outspurt 
outstrut outstunt outsulk outturn outusure oyer pee peek peel peele 
peeler peeoy peep peeper peepeye peer pele peleus pell peller 
pelu pep peplus pepper pepperer pepsis per pern pert pertussis 
peru perule perun peul phi pip pipe piper pipi pipistrel 
pipistrelle pipistrellus pipper pish piss pist plup plus plush ply 
plyer psi pst puerer pul pule puler pulk pull puller 
pulley pullus pulp pulper pulu puly pun punt pup puppis 
pur pure puree purely purer purr purre purree purrel purrer 
puru purupuru pus push puss pustule put putt puture ree 
reek reeker reeky reel reeler reeper rel rely reoutput rep 
repel repeller repipe reply repp reps reree rereel rerun reuel 
roe roer roey roue rouelle roun roup rouper roust rout 
roy rue ruelle ruer rule ruler rull ruller run runt 
rupee rupert rupture ruru rus rush russ rust rustre rut 
shi shih ship shipper shish shlu sip sipe siper sipper 
sis sish sisi siss sissu sist sistrurus speel speer spelk 
spell speller splurt spun spur spurn spurrer spurt sput ssi 
ssu stre stree streek streel streeler streep streke streperous strepsis 
strey stroup stroy stroyer strue strunt strut stu stue stull 
stuller stun stunt stupe stupeous stupp sturnus sturt stuss stut 
sue suer suerre suld sulk sulker sulky sull sully sulu 
sun sunn sunt sunup sup supe super superoutput supper supple 
supplely supply sur sure surely surrey sus susi susu susurr 
susurrous susurrus sutu suture suu tree treey trek trekker trey 
troupe trouper trout troy true truer trull truller truly trun 
trush truss trust tshi tst tsun tsutsutsi tue tule tulle 
tulu tun tunu tup tupek tupi tur turn turnup turr 
turus tush tussis tussur tut tuts tutu tutulus ule ull 
uller ulu ululu unreel unrule unruly unrun unrust untrue untruly 
untruss untrust unturn unurn upper upperer uppish uppishly uppull uppush 
upspurt upsun upsup uptree uptruss upturn ure urn uro uru 
urus urushi ush ust usun usure usurer utu yee yeel 
yeld yelk yell yeller yelp yelper yeo yep yer yere 
yern yoe yor yore you youl youp your yourn yoy 

注意: 我在这个线程的开头使用了字典和字符矩阵。代码在我的MacBookPro上运行,下面是关于这台机器的一些信息。

型号:MacBook Pro 型号标识符:MacBookPro8,1 处理器名称:Intel Core i5 处理器速度:2.3 GHz 处理器数量:1 总核数:2 L2缓存(每核):256kb L3 Cache: 3mb 内存:4gb 引导ROM版本:MBP81.0047.B0E SMC版本(系统):1.68f96

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

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

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

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

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

代码:

var fs = require('fs')

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

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

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

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

    return false
}

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

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

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

    return word
}

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

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

    var nodes = []

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

    return new Board(nodes, dict)
}

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

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

            path.push(node)

            this.dfs_search(path)
        }
    }
}

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

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

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

    var neighbours = []

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

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

    return neighbours
}

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

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

    var neighbours = this.get_unsearched_neighbours(path)

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

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

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

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

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

    this.dict_array = dict_array.sort()
}

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

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

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

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

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

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

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

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

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

var dict = new Dict()

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

只是为了好玩,我在bash中实现了一个。 它不是超级快,但很合理。

http://dev.xkyle.com/bashboggle/