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前注：这是这个问题的后续问题。

我在 R 中编写了一个 Boggle Game Solver（参见这个github 页面的源代码），但发现它的性能令人失望。

这是它的工作原理...

# Say we have the following set of letters
bog.letters <- c("t", "e", "n", "s", "d", "a", "i", "o",
                 "l", "e", "r", "o", "c", "f", "i", "e")

# We get the list of paths (permutations) from a pre-existing list
paths <- paths.by.length[[6]] # 6th element corresponds to 8-element "paths"
dim(paths) # [1] 183472      8

# The following function is the key here, 
# mapping the 183,472 combinations to the 16 letters
candidates <- apply(X = paths, MARGIN = 1, FUN = function(x) paste(bog.letters[x], collapse=""))

# The only remaining thing is to intersect the candidate words 
# with the actual words from our dictionary
dict.words <- dict.fr$mot[dict.fr$taille == 8]
valid.words <- intersect(candidates, dict.words)

13个字母单词候选的可重复示例

bog.letters <- c("t", "e", "n", "s", "d", "a", "i", "o", "l", "e", "r", "o", "c", "f", "i", "e")
n.letters <- 13
paths <- structure(list(V1 = c(1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
  1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1), V2 = c(2,
  2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2,
  2, 2, 2, 2, 2, 2, 2, 2), V3 = c(3, 3, 3, 3, 3, 3, 3, 3, 3, 3,
  3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3),
  V4 = c(4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4,
  4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4), V5 = c(7, 7, 7, 7,
  7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7,
  7, 7, 7, 7, 7, 7, 7), V6 = c(6, 6, 6, 6, 6, 6, 6, 6, 6, 6,
  6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6,
  6), V7 = c(5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5,
  5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5), V8 = c(9, 9, 9,
  9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9,
  9, 9, 9, 9, 9, 9, 9, 9), V9 = c(10, 10, 10, 10, 10, 10, 10,
  10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10,
  10, 10, 10, 10, 10, 10, 10, 10), V10 = c(11, 11, 11, 11,
  11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 13, 13, 13, 13,
  13, 13, 13, 14, 14, 14, 14, 14, 14, 14, 14), V11 = c(8, 8,
  12, 12, 12, 14, 14, 15, 15, 15, 15, 16, 16, 16, 16, 14, 14,
  14, 14, 14, 14, 14, 11, 11, 11, 11, 11, 11, 11, 11), V12 = c(12,
  12, 15, 15, 16, 15, 15, 12, 12, 14, 16, 12, 12, 15, 15, 11,
  11, 11, 11, 15, 15, 15, 8, 12, 12, 12, 15, 15, 16, 16), V13 = c(15,
  16, 14, 16, 15, 12, 16, 8, 16, 13, 12, 8, 15, 12, 14, 8,
  12, 15, 16, 11, 12, 16, 12, 8, 15, 16, 12, 16, 12, 15)), .Names = c("V1",
  "V2", "V3", "V4", "V5", "V6", "V7", "V8", "V9", "V10", "V11",
  "V12", "V13"), row.names = c(NA, 30L), class = "data.frame")

candidates <- apply(X = paths, MARGIN = 1, FUN = function(x) paste(bog.letters[x], collapse=""))

对于这么小的路径列表，这非常快。但是 13 个字母单词的实际路径数是 2,644,520。因此，找到所有候选人可能需要一分钟甚至更长时间。使用 doSNOW，我可以并行化搜索，从而大大减少总时间，但这样做有一个巨大的缺点：当使用正常循环时，只要我到达没有更多单词的点，我就可以退出/中断成立。这与并行过程无关（不可能？）。

所以我的问题是：你能想出一个更好的函数/算法来完成这项任务吗？一些网站在几秒钟内就为 Boggle 游戏提供了解决方案......要么他们生成所有可能的字母组合并将结果存储在数据库中（！），要么他们显然使用更好的算法（可能是编译语言）来实现这些结果。

有任何想法吗？

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1 回答 1

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使用Rcpp Gallerycpp_str_split中的函数，2644520 条路径的运行时间现在减少到 3 秒。

library(stringi)
paths <- data.frame(matrix(sample(1:16, 13*2644520, TRUE), ncol=13))
a1 <- stri_c(bog.letters[t(as.matrix(paths))], collapse="")
candidates <- cpp_str_split(a1, 13)[[1]]

对于 2644520 条路径，apply我的笔记本上的方法大约需要 80 秒。

于 2015-02-28T14:52:58.277 回答