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performance - 在F#: performance中解决背包问题

转载 作者:行者123 更新时间:2023-12-03 11:21:28 25 4
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我找到了一篇文章:
Solving the 0-1 knapsack problem using continuation-passing style with memoization in F#

关于在F#中实现的背包问题。当我学习这种语言时,我发现这真的很有趣,并尝试对其进行一些研究。这是我编写的代码:

open System
open System.IO
open System.Collections.Generic

let parseToTuple (line : string) =
let parsedLine = line.Split(' ') |> Array.filter(not << String.IsNullOrWhiteSpace) |> Array.map Int32.Parse
(parsedLine.[0], parsedLine.[1])

let memoize f =
let cache = Dictionary<_, _>()
fun x ->
if cache.ContainsKey(x)
then cache.[x]
else
let res = f x
cache.[x] <- res
res

type Item =
{
Value : int
Size : int
}

type ContinuationBuilder() =
member b.Bind(x, f) = fun k -> x (fun x -> f x k)
member b.Return x = fun k -> k x
member b.ReturnFrom x = x

let cont = ContinuationBuilder()

let set1 =
[
(4, 11)
(8, 4)
(10, 5)
(15, 8)
(4, 3)
]

let set2 =
[
(50, 341045); (1906, 4912); (41516, 99732); (23527, 56554); (559, 1818); (45136, 108372); (2625, 6750); (492, 1484)
(1086, 3072); (5516, 13532); (4875, 12050); (7570, 18440); (4436, 10972); (620, 1940); (50897, 122094); (2129, 5558)
(4265, 10630); (706, 2112); (2721, 6942); (16494, 39888); (29688, 71276); (3383, 8466); (2181, 5662); (96601, 231302)
(1795, 4690); (7512, 18324); (1242, 3384); (2889, 7278); (2133, 5566); (103, 706); (4446, 10992); (11326, 27552)
(3024, 7548); (217, 934); (13269, 32038); (281, 1062); (77174, 184848); (952, 2604); (15572, 37644); (566, 1832)
(4103, 10306); (313, 1126); (14393, 34886); (1313, 3526); (348, 1196); (419, 1338); (246, 992); (445, 1390)
(23552, 56804); (23552, 56804); (67, 634)
]

[<EntryPoint>]
let main args =
// prepare list of items from a file args.[0]
let header, items = set1
|> function
| h::t -> h, t
| _ -> raise (Exception("Wrong data format"))

let N, K = header
printfn "N = %d, K = %d" N K
let items = List.map (fun x -> {Value = fst x ; Size = snd x}) items |> Array.ofList

let rec combinations =
let innerSolver key =
cont
{
match key with
| (i, k) when i = 0 || k = 0 -> return 0
| (i, k) when items.[i-1].Size > k -> return! combinations (i-1, k)
| (i, k) -> let item = items.[i-1]
let! v1 = combinations (i-1, k)
let! beforeItem = combinations (i-1, k-item.Size)
let v2 = beforeItem + item.Value
return max v1 v2
}
memoize innerSolver

let res = combinations (N, K) id
printfn "%d" res
0

但是,此实现的问题在于它的速度很慢(实际上,我无法解决50个项目且容量约为300000的问题,而我在不到1秒的时间内就用C#的天真的实现解决了这个问题)。

你能告诉我我是否在某个地方犯了错误吗?也许实现是正确的,这仅仅是解决此问题的低效方法。

最佳答案

通过在FSI中运行此代码:

open System
open System.Diagnostics
open System.Collections.Generic

let time f =
System.GC.Collect()
let sw = Stopwatch.StartNew()
let r = f()
sw.Stop()
printfn "Took: %f" sw.Elapsed.TotalMilliseconds
r

let mutable cacheHits = 0
let mutable cacheMisses = 0

let memoize f =
let cache = Dictionary<_, _>()
fun x ->
match cache.TryGetValue(x) with
| (true, v) ->
cacheHits <- cacheHits + 1
//printfn "Hit for %A - Result is %A" x v
v
| _ ->
cacheMisses <- cacheMisses + 1
//printfn "Miss for %A" x
let res = f x
cache.[x] <- res
res

type Item = { Value : int; Size : int }

type ContinuationBuilder() =
member b.Bind(x, f) = fun k -> x (fun x -> f x k)
member b.Return x = fun k -> k x
member b.ReturnFrom x = x

let cont = ContinuationBuilder()

let genItems n =
[| for i = 1 to n do
let size = i % 5
let value = (size * i)
yield { Value = value; Size = size }
|]

let N, K = (80, 400)
printfn "N = %d, K = %d" N K

let items = genItems N

//let rec combinations_cont =
// memoize (
// fun key ->
// cont {
// match key with
// | (0, _) | (_, 0) -> return 0
// | (i, k) when items.[i-1].Size > k -> return! combinations_cont (i - 1, k)
// | (i, k) -> let item = items.[i-1]
// let! v1 = combinations_cont (i-1, k)
// let! beforeItem = combinations_cont (i-1, k - item.Size)
// let v2 = beforeItem + item.Value
// return max v1 v2
// }
// )
//
//
//cacheHits <- 0
//cacheMisses <- 0

//let res = time(fun () -> combinations_cont (N, K) id)
//printfn "Answer: %d" res
//printfn "Memo hits: %d" cacheHits
//printfn "Memo misses: %d" cacheMisses
//printfn ""

let rec combinations_plain =
memoize (
fun key ->
match key with
| (i, k) when i = 0 || k = 0 -> 0
| (i, k) when items.[i-1].Size > k -> combinations_plain (i-1, k)
| (i, k) -> let item = items.[i-1]
let v1 = combinations_plain (i-1, k)
let beforeItem = combinations_plain (i-1, k-item.Size)
let v2 = beforeItem + item.Value
max v1 v2
)

cacheHits <- 0
cacheMisses <- 0

printfn "combinations_plain"
let res2 = time (fun () -> combinations_plain (N, K))
printfn "Answer: %d" res2
printfn "Memo hits: %d" cacheHits
printfn "Memo misses: %d" cacheMisses
printfn ""

let recursivelyMemoize f =
let cache = Dictionary<_, _>()
let rec memoizeAux x =
match cache.TryGetValue(x) with
| (true, v) ->
cacheHits <- cacheHits + 1
//printfn "Hit for %A - Result is %A" x v
v
| _ ->
cacheMisses <- cacheMisses + 1
//printfn "Miss for %A" x
let res = f memoizeAux x
cache.[x] <- res
res
memoizeAux

let combinations_plain2 =
let combinations_plain2Aux combinations_plain2Aux key =
match key with
| (i, k) when i = 0 || k = 0 -> 0
| (i, k) when items.[i-1].Size > k -> combinations_plain2Aux (i-1, k)
| (i, k) -> let item = items.[i-1]
let v1 = combinations_plain2Aux (i-1, k)
let beforeItem = combinations_plain2Aux (i-1, k-item.Size)
let v2 = beforeItem + item.Value
max v1 v2
let memoized = recursivelyMemoize combinations_plain2Aux
fun x -> memoized x

cacheHits <- 0
cacheMisses <- 0

printfn "combinations_plain2"
let res3 = time (fun () -> combinations_plain2 (N, K))
printfn "Answer: %d" res3
printfn "Memo hits: %d" cacheHits
printfn "Memo misses: %d" cacheMisses
printfn ""

let recursivelyMemoizeCont f =
let cache = Dictionary HashIdentity.Structural
let rec memoizeAux x k =
match cache.TryGetValue(x) with
| (true, v) ->
cacheHits <- cacheHits + 1
//printfn "Hit for %A - Result is %A" x v
k v
| _ ->
cacheMisses <- cacheMisses + 1
//printfn "Miss for %A" x
f memoizeAux x (fun y ->
cache.[x] <- y
k y)
memoizeAux

let combinations_cont2 =
let combinations_cont2Aux combinations_cont2Aux key =
cont {
match key with
| (0, _) | (_, 0) -> return 0
| (i, k) when items.[i-1].Size > k -> return! combinations_cont2Aux (i - 1, k)
| (i, k) -> let item = items.[i-1]
let! v1 = combinations_cont2Aux (i-1, k)
let! beforeItem = combinations_cont2Aux (i-1, k - item.Size)
let v2 = beforeItem + item.Value
return max v1 v2
}
let memoized = recursivelyMemoizeCont combinations_cont2Aux
fun x -> memoized x id

cacheHits <- 0
cacheMisses <- 0

printfn "combinations_cont2"
let res4 = time (fun () -> combinations_cont2 (N, K))
printfn "Answer: %d" res4
printfn "Memo hits: %d" cacheHits
printfn "Memo misses: %d" cacheMisses
printfn ""

我得到这些结果:
N = 80, K = 400
combinations_plain
Took: 7.191000
Answer: 6480
Memo hits: 6231
Memo misses: 6552

combinations_plain2
Took: 6.310800
Answer: 6480
Memo hits: 6231
Memo misses: 6552

combinations_cont2
Took: 17.021200
Answer: 6480
Memo hits: 6231
Memo misses: 6552
  • combinations_plain来自latkin的答案。
  • combinations_plain2显式公开递归备注步骤。
  • combinations_cont2将递归内存功能改编为内存连续结果的功能。
  • combinations_cont2的工作原理是在将延续结果传递给实际延续之前将其截获。随后在同一键上的调用提供了一个延续,并且这个延续被提供给我们最初截获的答案。

  • 这表明我们能够:
  • 使用延续传递样式进行内存。
  • 实现与原始备忘录版本相似的(ish)性能特征。

  • 我希望这可以使事情变得简单。抱歉,我的博客代码段不完整(我认为最近重新格式化时可能会丢失它)。

    关于performance - 在F#: performance中解决背包问题,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/17445855/

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