在阅读这个问题时，我想知道为什么没有人会"简单地"迭代boggle网格上所有可能的路径，并让单词trys跟随，然后如果单词trie中没有匹配项，则取消该路径。在一个4乘4的网格上不可能有那么多路径，对吧？有多少条路？所以我开始用F#编写一个路径计数器函数。结果产生了另一页上没有人说的结果：网格上的路径比我想象的要多得多(实际上，路径比单词集中的单词多)。
虽然所有这些都是我问题的背景，但我最终得到的代码运行相当慢，我发现我无法对代码的几个方面给出好的答案。因此，在这里，首先是代码，然后在下面，你会发现我认为值得解释的要点。。。
let moves n state square =
let allSquares = [0..n*n-1] |> Set.ofList
let right = Set.difference allSquares (Set.ofList [n-1..n..n*n])
let left = Set.difference allSquares (Set.ofList [0..n..n*n-1])
let up = Set.difference allSquares (Set.ofList [0..n-1])
let down = Set.difference allSquares (Set.ofList [n*n-n..n*n-1])
let downRight = Set.intersect right down
let downLeft = Set.intersect left down
let upRight = Set.intersect right up
let upLeft = Set.intersect left up
let appendIfInSet se v res =
if Set.contains square se then res @ v else res
[]
|> appendIfInSet right [square + 1]
|> appendIfInSet left [square - 1]
|> appendIfInSet up [square - n]
|> appendIfInSet down [square + n]
|> appendIfInSet downRight [square + n + 1]
|> appendIfInSet downLeft [square + n - 1]
|> appendIfInSet upRight [square - n + 1]
|> appendIfInSet upLeft [square - n - 1]
|> List.choose (fun s -> if ((uint64 1 <<< s) &&& state) = 0UL then Some s else None )
let block state square =
state ||| (uint64 1 <<< square)
let countAllPaths n lmin lmax =
let mov = moves n // line 30
let rec count l state sq c =
let state' = block state sq
let m = mov state' sq
match l with
| x when x <= lmax && x >= lmin ->
List.fold (fun acc s -> count (l+1) state' s acc) (c+1) m
| x when x < lmin ->
List.fold (fun acc s -> count (l+1) state' s acc) (c) m
| _ ->
c
List.fold (fun acc s -> count 0 (block 0UL s) s acc) 0 [0..n*n-1]
[<EntryPoint>]
let main args =
printfn "%d: %A" (Array.length args) args
if 3 = Array.length args then
let n = int args.[0]
let lmin = int args.[1]
let lmax = int args.[2]
printfn "%d %d %d -> %d" n lmin lmax (countAllPaths n lmin lmax)
else
printfn "usage: wordgames.exe n lmin lmax"
0
在第30行中，我用第一个参数创建了
moves函数，希望代码优化能从中受益。也许优化了我在move中创建的9个集，它们只是n的一个函数。毕竟，它们不需要一次又一次地生成，对吧？另一方面，我不会真的打赌它真的会发生
所以，问题#1是：我如何以尽可能少的代码膨胀的方式强制执行此优化？(当然，我可以创建一个有9个成员的类型，然后为每个可能的n创建一个该类型的数组，然后对预先计算的集合进行类似查找表的使用，但在我看来，这将是代码膨胀)。许多资料表明，平行褶皱被认为是至关重要的。如何创建计数函数的并行版本(在多个内核上运行)？
有人知道如何加快速度吗？也许是修剪或记忆等？
起初，当我为n=4 lmin=3 lmax=8运行函数时，我认为它需要很长时间，因为我在fsi中运行了它。但后来我用-O编译了代码，它仍然花了大约相同的时间。。。
更新
在等待你们的输入时，我做了代码膨胀的手动优化版本(运行速度更快)，然后找到了一种在多个核心上运行的方法
总的来说，这两个变化产生了大约30倍的速度。这里是我想出的(膨胀的)版本(仍在寻找避免膨胀的方法)：
let squareSet n =
let allSquares = [0..n*n-1] |> Set.ofList
let right = Set.difference allSquares (Set.ofList [n-1..n..n*n])
let left = Set.difference allSquares (Set.ofList [0..n..n*n-1])
let up = Set.difference allSquares (Set.ofList [0..n-1])
let down = Set.difference allSquares (Set.ofList [n*n-n..n*n-1])
let downRight = Set.intersect right down
let downLeft = Set.intersect left down
let upRight = Set.intersect right up
let upLeft = Set.intersect left up
[|right;left;up;down;upRight;upLeft;downRight;downLeft|]
let RIGHT,LEFT,UP,DOWN = 0,1,2,3
let UPRIGHT,UPLEFT,DOWNRIGHT,DOWNLEFT = 4,5,6,7
let squareSets =
[|Set.empty;Set.empty;Set.empty;Set.empty;Set.empty;Set.empty;Set.empty;Set.empty;|]
::
[ for i in 1..8 do
yield squareSet i
]
|> Array.ofList
let moves n state square =
let appendIfInSet se v res =
if Set.contains square se then res @ v else res
[]
|> appendIfInSet squareSets.[n].[RIGHT] [square + 1]
|> appendIfInSet squareSets.[n].[LEFT] [square - 1]
|> appendIfInSet squareSets.[n].[UP] [square - n]
|> appendIfInSet squareSets.[n].[DOWN] [square + n]
|> appendIfInSet squareSets.[n].[DOWNRIGHT] [square + n + 1]
|> appendIfInSet squareSets.[n].[DOWNLEFT] [square + n - 1]
|> appendIfInSet squareSets.[n].[UPRIGHT] [square - n + 1]
|> appendIfInSet squareSets.[n].[UPLEFT] [square - n - 1]
|> List.choose (fun s -> if ((uint64 1 <<< s) &&& state) = 0UL then Some s else None )
let block state square =
state ||| (uint64 1 <<< square)
let countAllPaths n lmin lmax =
let mov = moves n
let rec count l state sq c =
let state' = block state sq
let m = mov state' sq
match l with
| x when x <= lmax && x >= lmin ->
List.fold (fun acc s -> count (l+1) state' s acc) (c+1) m
| x when x < lmin ->
List.fold (fun acc s -> count (l+1) state' s acc) (c) m
| _ ->
c
//List.fold (fun acc s -> count 0 (block 0UL s) s acc) 0 [0..n*n-1]
[0..n*n-1]
|> Array.ofList
|> Array.Parallel.map (fun start -> count 0 (block 0UL start) start 0)
|> Array.sum
[<EntryPoint>]
let main args =
printfn "%d: %A" (Array.length args) args
if 3 = Array.length args then
let n = int args.[0]
let lmin = int args.[1]
let lmax = int args.[2]
printfn "%d %d %d -> %d" n lmin lmax (countAllPaths n lmin lmax)
else
printfn "usage: wordgames.exe n lmin lmax"
0
关于集合生成的非优化。发布在问题更新中的第二个版本显示，事实就是这样(编译器没有优化)，它大大加快了速度。最终版本(发布在下面的答案中)进一步采用了这种方法，并进一步加快了路径计数(以及难题的解决)。
结合在多个核心上的并行执行，对于n=4 lmin=3 lmax=8的情况，最初非常慢(可能是30秒)的版本可以加速到只有100毫秒左右。
对于n=6类问题，在我的机器上，并行和手动调优的实现在大约60ms内解决了一个难题。这比路径计数更快是有道理的，因为单词列表探测(使用了一本约有80000个单词的词典)以及@GuyCoder指出的动态编程方法使谜题的解决方案比(蛮力)路径计数更不复杂。
经验教训
如果涉及到代码优化，f#编译器似乎并不太神秘和神奇。如果真的需要性能，手动调整是值得的。
在这种情况下，将单线程递归搜索函数转换为并行(并发)函数并不困难。
代码的最终版本
编制单位：
fsc--优化+--尾调用+wordgames.fs
(Microsoft(R)F#编译器14.0.23413.0版)
let wordListPath = @"E:temp12dicts-6.0.2International3of6all.txt"
let acceptableWord (s : string) : bool =
let s' = s.Trim()
if s'.Length > 2
then
if System.Char.IsLower(s'.[0]) && System.Char.IsLetter(s'.[0]) then true
else false
else
false
let words =
System.IO.File.ReadAllLines(wordListPath)
|> Array.filter acceptableWord
let squareSet n =
let allSquares = [0..n*n-1] |> Set.ofList
let right = Set.difference allSquares (Set.ofList [n-1..n..n*n])
let left = Set.difference allSquares (Set.ofList [0..n..n*n-1])
let up = Set.difference allSquares (Set.ofList [0..n-1])
let down = Set.difference allSquares (Set.ofList [n*n-n..n*n-1])
let downRight = Set.intersect right down
let downLeft = Set.intersect left down
let upRight = Set.intersect right up
let upLeft = Set.intersect left up
[|right;left;up;down;upRight;upLeft;downRight;downLeft|]
let RIGHT,LEFT,UP,DOWN = 0,1,2,3
let UPRIGHT,UPLEFT,DOWNRIGHT,DOWNLEFT = 4,5,6,7
let squareSets =
[|Set.empty;Set.empty;Set.empty;Set.empty;Set.empty;Set.empty;Set.empty;Set.empty;|]
::
[ for i in 1..8 do
yield squareSet i
]
|> Array.ofList
let movesFromSquare n square =
let appendIfInSet se v res =
if Set.contains square se then v :: res else res
[]
|> appendIfInSet squareSets.[n].[RIGHT] (square + 1)
|> appendIfInSet squareSets.[n].[LEFT] (square - 1)
|> appendIfInSet squareSets.[n].[UP] (square - n)
|> appendIfInSet squareSets.[n].[DOWN] (square + n)
|> appendIfInSet squareSets.[n].[DOWNRIGHT] (square + n + 1)
|> appendIfInSet squareSets.[n].[DOWNLEFT] (square + n - 1)
|> appendIfInSet squareSets.[n].[UPRIGHT] (square - n + 1)
|> appendIfInSet squareSets.[n].[UPLEFT] (square - n - 1)
let lutMovesN n =
Array.init n (fun i -> if i > 0 then Array.init (n*n-1) (fun j -> movesFromSquare i j) else Array.empty)
let lutMoves =
lutMovesN 8
let moves n state square =
let appendIfInSet se v res =
if Set.contains square se then v :: res else res
lutMoves.[n].[square]
|> List.filter (fun s -> ((uint64 1 <<< s) &&& state) = 0UL)
let block state square =
state ||| (uint64 1 <<< square)
let countAllPaths n lmin lmax =
let mov = moves n
let rec count l state sq c =
let state' = block state sq
let m = mov state' sq
match l with
| x when x <= lmax && x >= lmin ->
List.fold (fun acc s -> count (l+1) state' s acc) (c+1) m
| x when x < lmin ->
List.fold (fun acc s -> count (l+1) state' s acc) (c) m
| _ ->
c
//List.fold (fun acc s -> count 0 (block 0UL s) s acc) 0 [0..n*n-1]
[|0..n*n-1|]
|> Array.Parallel.map (fun start -> count 0 (block 0UL start) start 0)
|> Array.sum
//printfn "%d " (words |> Array.distinct |> Array.length)
let usage() =
printfn "usage: wordgames.exe [--gen n count problemPath | --count n lmin lmax | --solve problemPath ]"
let rng = System.Random()
let genProblem n (sb : System.Text.StringBuilder) =
let a = Array.init (n*n) (fun _ -> char (rng.Next(26) + int 'a'))
sb.Append(a) |> ignore
sb.AppendLine()
let genProblems nproblems n (sb : System.Text.StringBuilder) : System.Text.StringBuilder =
for i in 1..nproblems do
genProblem n sb |> ignore
sb
let solve n (board : System.String) =
let ba = board.ToCharArray()
let testWord (w : string) : bool =
let testChar k sq = (ba.[sq] = w.[k])
let rec testSquare state k sq =
match k with
| 0 -> testChar 0 sq
| x ->
if testChar x sq
then
let state' = block state x
moves n state' x
|> List.exists (testSquare state' (x-1))
else
false
[0..n*n-1]
|> List.exists (testSquare 0UL (String.length w - 1))
words
|> Array.splitInto 32
|> Array.Parallel.map (Array.filter testWord)
|> Array.concat
[<EntryPoint>]
let main args =
printfn "%d: %A" (Array.length args) args
let nargs = Array.length args
let sw = System.Diagnostics.Stopwatch()
match nargs with
| x when x >= 2 ->
match args.[0] with
| "--gen" ->
if nargs = 4
then
let n = int args.[1]
let nproblems = int args.[2]
let outpath = args.[3]
let problems = genProblems nproblems n (System.Text.StringBuilder())
System.IO.File.WriteAllText (outpath,problems.ToString())
0
else
usage()
0
| "--count" ->
if nargs = 4
then
let n = int args.[1]
let lmin = int args.[2]
let lmax = int args.[3]
sw.Start()
let count = countAllPaths n lmin lmax
sw.Stop()
printfn "%d %d %d -> %d (took: %d)" n lmin lmax count (sw.ElapsedMilliseconds)
0
else
usage ()
0
| "--solve" ->
if nargs = 2
then
let problems = System.IO.File.ReadAllLines(args.[1])
problems
|> Array.iter
(fun (p : string) ->
let n = int (sqrt (float (String.length p)))
sw.Reset()
sw.Start()
let found = solve n p
sw.Stop()
printfn "%sn%An%dms" p found (sw.ElapsedMilliseconds)
)
0
else
usage ()
0
| _ ->
usage ()
0
| _ ->
usage ()
0