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)
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}"
import java.util.HashSet;
import java.util.Set;
/**
* @author Sujeet Kumar (mrsujeet@gmail.com) It prints out all strings that can
* be formed by moving left, right, up, down, or diagonally and exist in
* a given dictionary , without repeating any cell. Assumes words are
* comprised of lower case letters. Currently prints words as many times
* as they appear, not just once. *
*/
public class BoggleGame
{
/* A sample 4X4 board/2D matrix */
private static char[][] board = { { 's', 'a', 's', 'g' },
{ 'a', 'u', 't', 'h' },
{ 'r', 't', 'j', 'e' },
{ 'k', 'a', 'h', 'e' }
};
/* A sample dictionary which contains unique collection of words */
private static Set<String> dictionary = new HashSet<String>();
private static boolean[][] visited = new boolean[board.length][board[0].length];
public static void main(String[] arg) {
dictionary.add("sujeet");
dictionary.add("sarthak");
findWords();
}
// show all words, starting from each possible starting place
private static void findWords() {
for (int i = 0; i < board.length; i++) {
for (int j = 0; j < board[i].length; j++) {
StringBuffer buffer = new StringBuffer();
dfs(i, j, buffer);
}
}
}
// run depth first search starting at cell (i, j)
private static void dfs(int i, int j, StringBuffer buffer) {
/*
* base case: just return in recursive call when index goes out of the
* size of matrix dimension
*/
if (i < 0 || j < 0 || i > board.length - 1 || j > board[i].length - 1) {
return;
}
/*
* base case: to return in recursive call when given cell is already
* visited in a given string of word
*/
if (visited[i][j] == true) { // can't visit a cell more than once
return;
}
// not to allow a cell to reuse
visited[i][j] = true;
// combining cell character with other visited cells characters to form
// word a potential word which may exist in dictionary
buffer.append(board[i][j]);
// found a word in dictionary. Print it.
if (dictionary.contains(buffer.toString())) {
System.out.println(buffer);
}
/*
* consider all neighbors.For a given cell considering all adjacent
* cells in horizontal, vertical and diagonal direction
*/
for (int k = i - 1; k <= i + 1; k++) {
for (int l = j - 1; l <= j + 1; l++) {
dfs(k, l, buffer);
}
}
buffer.deleteCharAt(buffer.length() - 1);
visited[i][j] = false;
}
}
