Question

着色问题：

您好，我试图实现一个bool函数，当颜色可以扩展到某个国家/地区时返回true，否则返回false但是我在使用集合时遇到问题，因为我们无法对它们进行模式匹配。

我的代码：

type Country = string;;
type Chart = Set<Country*Country>;;
type Colour = Set<Country>;;
type Colouring = Set<Colour>;;

(* This is how you tell that two countries are neghbours.  It requires a chart.*)
let areNeighbours ct1 ct2 chart =
  Set.contains (ct1,ct2) chart || Set.contains (ct2,ct1) chart;;
(* val areNeighbours :
  ct1:'a -> ct2:'a -> chart:Set<'a * 'a> -> bool when 'a : comparison
  *)

(* The colour col can be extended by the country ct when they are no neighbours
according to chart.*)

let canBeExtBy (col:Colouring) (ct:Country) (chart:Chart) = col |> Set.fold (fun x -> (if (areNeighbours x ct chart) then true else false))

预期结果：根据图表中定义的邻域，我们需要检查ct是否是col中任何国家/地区的邻居（假设col中有国家/地区）。所以，如果

chart = set
    [("Andorra", "Benin"); ("Andorra", "Canada"); ("Andorra", "Denmark");
     ("Benin", "Canada"); ("Benin", "Denmark"); ("Canada", "Denmark");
     ("Estonia", "Canada"); ("Estonia", "Denmark"); ("Estonia", "Finland");
     ...]

并且

col = set 
    ["Andorra"]

当canBeExt安道尔与这些国家/地区共享边界时，ct = "Benin" or "Canada" or "Denmark" etc...应该返回false，因此它们不能与Andora颜色相同。

显然我在canBeExtBy中有一个类型错误，因为我试图返回一个bool并且它期待＆＃39; a：着色。我不知道如何实施它。

感谢您的帮助！

Answer 1

这个怎么样？

type Country      = string
type Neighbours   = Set<Country*Country>
type SharesColour = Set<Country>

let areNeighbours (ns : Neighbours) (ct1 : Country) (ct2 : Country) : bool =
  Set.contains (ct1,ct2) ns || Set.contains (ct2,ct1) ns

let canShareColour (ns : Neighbours) (ct : Country) (s : SharesColour) : bool =
  s |> Seq.exists (areNeighbours ns ct) |> not

let neighbours : Neighbours = 
  set [| 
    ("Andorra", "Benin")  ; ("Andorra", "Canada") ; ("Andorra", "Denmark");
    ("Benin"  , "Canada") ; ("Benin"  , "Denmark"); ("Canada" , "Denmark");
    ("Estonia", "Canada") ; ("Estonia", "Denmark"); ("Estonia", "Finland");
  |]

let sharesColour : SharesColour =
  set [|
    "Andorra"
  |]

[<EntryPoint>]
let main argv =
  printfn "%A" <| canShareColour neighbours "Estonia" sharesColour
  printfn "%A" <| canShareColour neighbours "Benin"   sharesColour
  0

将名称更改为对我更有意义的内容。你可能不同意。

Answer 2

要检查集合中的任何元素是否满足给定条件：fold不是exists的作业，而是forall或let fornone f = Set.forall (f >> not) let doesnotexist f = Set.exists f >> not的适当否定。

之一

fold

会做，如FuleSnabel的回答所示。但是，当然可以从let fornone f = Set.fold (fun s -> f >> not >> (&&) s) true let doesnotexist f = Set.fold (fun s -> f >> (||) s) false >> not构建这些函数，尽管除了作为currying，函数组合和pointfree风格的例证之外，没有人会这样做。

background

Set.Fold F＃

2 个答案: