重构算法作为计算表达式？

Question

此问题包含对尚未完成的人的剧透problem 61 of Project Euler。我为这个必要的问题写了一个答案，所以我开始做一个更通用、更实用的答案。我成功了，但现在正试图弄清楚如何将其重构为或使用计算表达式，并且非常困惑。该问题在下面详细描述，但要点是您正在尝试建立一个数字链，当按顺序排列时，它会在所有相邻对中显示一个属性。链的每个候选者都来自不同的数字池，这意味着蛮力算法必须很聪明，以避免需要搜索所有可能的排列。

我对包含计算表达式的猜测是将搜索算法以某种方式变成一个单子，它继续添加到解决方案中或将空列表转储出去。但我不完全确定。

(*
Triangle, square, pentagonal, hexagonal, heptagonal, and octagonal numbers are
all figurate (polygonal) numbers and are generated by the following formulae:

Triangle        P3,n=n(n+1)/2       1, 3, 6, 10, 15, ...
Square          P4,n=n2             1, 4, 9, 16, 25, ...
Pentagonal      P5,n=n(3n−1)/2      1, 5, 12, 22, 35, ...
Hexagonal       P6,n=n(2n−1)        1, 6, 15, 28, 45, ...
Heptagonal      P7,n=n(5n−3)/2      1, 7, 18, 34, 55, ...
Octagonal       P8,n=n(3n−2)        1, 8, 21, 40, 65, ...

The ordered set of three 4-digit numbers: 8128, 2882, 8281, has three 
interesting properties.

The set is cyclic, in that the last two digits of each number is the first two 
digits of the next number (including the last number with the first).
Each polygonal type: triangle (P3,127=8128), square (P4,91=8281), and pentagonal 
(P5,44=2882), is represented by a different number in the set.
This is the only set of 4-digit numbers with this property.

Find the sum of the only ordered set of six cyclic 4-digit numbers for which 
each polygonal type: triangle, square, pentagonal, hexagonal, heptagonal, and 
octagonal, is represented by a different number in the set.
*)


let rec distribute e = function
    | [] -> [[e]]
    | x::xs' as xs -> (e::xs)::[for xs in distribute e xs' -> x::xs]

// Return a list of all permutations of a list
let rec permute = function
    | [] -> [[]]
    | e::xs -> List.collect (distribute e) (permute xs)

// Return a list rotated until it's minimum element is the head
let canonicalCyclicPermutation (permutationList : 'a list) = 
    let min = Seq.min permutationList
    let rec loop ourList = 
        match ourList with
        | head :: tail when head = min -> ourList
        | head :: tail -> loop (tail @ [head])
    loop permutationList

// Return a list of all permutations of a list that is rotationally/cylically unique
let permutateCycUniq seedList = 
    permute seedList
    |> List.distinctBy canonicalCyclicPermutation

// Generate a sequence of all s-gonal numbers
let polygonalGenerator s = 
    Seq.initInfinite (fun idx -> ((pown (idx+1) 2) * (s-2) - (idx+1)*(s-4))/2)

// Generate a sequence of s-gonal numbers relevant for our exercise
let polygonalCandidates s = 
    s
    |> polygonalGenerator
    |> Seq.skipWhile (fun x -> x <= 999)
    |> Seq.takeWhile (fun x -> x <= 9999)
    |> Seq.cache

// Create the polygonal numbers as a list not seq
let polygonalCandidatesL s = 
    polygonalCandidates s
    |> Seq.toList

// Returns true if the last digits of first input are first digits in last input
let sharesDigits xxvv vvyy = 
    (xxvv / 100) = (vvyy % 100)

// Returns true if a sequence is cyclical
let isCyclical intSeq = 
    (Seq.append intSeq (Seq.take 1 intSeq))
    |> Seq.pairwise 
    |> Seq.fold (fun acc (num1,num2) -> acc && (sharesDigits num1 num2)) true

// Returns an empty list if the candidate number does not share digits
// with the list head, otherwise returns the list with the candidate at the head
let addCandidateToSolution (solution : int list) (number : int) =
    match solution with
    | (head::tail) when sharesDigits number head -> number::head::tail
    | _ -> []

// Returns a sequence of all valid solutions generated by trying to add
// a sequence of candidates to all solutions in a sequence
let addCandidatesToSolution (solutions : int list seq) (candidates : int seq) =
    Seq.collect (fun solution -> 
                 Seq.map (fun candidate -> 
                          addCandidateToSolution solution candidate)
                          candidates
                |> Seq.filter (not << List.isEmpty)) 
              solutions

// Given a list of side lengths, we return a sequence of cyclical solutions
// from the polygonal number families in the order they appear in the list
let generateSolutionsFromPolygonalFamilies (seedList : int list) = 
    let solutionSeeds = 
        seedList 
        |> List.head
        |> polygonalCandidates
        |> Seq.map (fun x -> [x])

    let solutions = 
        Seq.fold (fun acc elem -> (addCandidatesToSolution acc elem)) 
                 solutionSeeds 
                 ((List.tail seedList) |> List.map polygonalCandidatesL)
        |> Seq.filter isCyclical
    solutions

// Find all cyclical sequences from a list of polygonal number families
let FindSolutionsFromFamilies intList = 
    intList
    |> permutateCycUniq
    |> Seq.collect generateSolutionsFromPolygonalFamilies
    |> Seq.toList

// Given in the problem
let sampleAnswer = FindSolutionsFromFamilies [3;4;5]

// The set of answers that answers the problem
#time
let problemAnswer = FindSolutionsFromFamilies [3 .. 8]
#time // 0.09s wow!

Answer 1

虽然一开始持怀疑态度，但我不得不承认，这个问题背后的想法非常合理，而实际的实现似乎非常难以捉摸。由于需要为给定的数据结构提供 member Bind : ma:'a list * f:('a -> 'b list) -> 'b list 的等效单子签名，因此很自然地坚持 F# 列表并使用其相应的高阶函数 List.collect。

type ListBuilder () =
    member __.Return x = [x]
    member __.Bind(ma, f) = List.collect f ma

let myList a b = ListBuilder () {
    let! x = a
    let! y = b
    return x, y } 

myList [1..2] [3..4]    // [(1, 3); (1, 4); (2, 3); (2, 4)]

这个最小但光滑的笛卡尔积并没有让我们走得太远。我们需要使执行链有条件，需要一个额外的成员，Zero。显然，固定数量是这种方法的主要缺点。

type MyListBuilder () =
    member __.Zero _ = []
    member __.Return x = [x]
    member __.Bind(ma, f) = List.collect f ma

let myListXY cmp a b c = MyListBuilder () {
    let! r = a
    let! s = b
    if cmp r s then 
        let! t = c
        if cmp s t then 
            if cmp t r then 
                return r, s, t } 

let T n k = if n < 2 then 0 else ((n - 2) * k * k - (n - 4) * k) / 2

let figurate s min max =
    Seq.initInfinite ((+) 1)
    |> Seq.map (T s)
    |> Seq.takeWhile (fun n -> n < max)
    |> Seq.filter (fun n -> n >= min)
    |> Seq.toList

myListXY (fun x y -> x % 100 = y / 100)
    (figurate 3 1000 10000)
    (figurate 5 1000 10000)
    (figurate 4 1000 10000) // [(8128, 2882, 8281)]