R Boggle解算器的优化

Question

Forenote：这是这一个的后续问题.

我在R中编写了一个Boggle游戏解决程序(参见此用于源代码的github页面)，并发现它的性能令人失望。

它是这样运作的..。

# 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字母候选词的可重现性示例

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，这是相当快的。但是13个字母单词的实际路径数是2,644,520。所以找到所有的候选人可能需要一分钟甚至更长的时间。使用doSNOW，我能够平分搜索，大大减少了总时间，但这有一个巨大的缺点:当我使用普通循环时，每当我到达找不到更多单词的时候，我就可以退出/中断。这并不明显(不可能？)与并行过程有关。

所以我的问题是:你能为这个任务想出一个更好的函数/算法吗？一些网站为Boggle游戏提供了几秒钟的解决方案.它们生成所有可能的字母组合并将结果存储在数据库(!)中，否则它们显然使用了更好的算法(可能是编译的语言)来实现这些结果。

有什么想法吗？

Answer 1

使用来自cpp_str_split的Rcpp画廊函数，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方法在我的笔记本上花费了大约80秒。