在 OCaml 中使用 memoization 解决序列比对无法得到正确答案

Question

我使用 Hashtbl 存储(int*string*string)，这是问题的答案。但是，当我运行时，答案不正确。我认为它跳过了一些值并且没有计算它们。

这是我的代码：

let solve_memo (x:string) (y:string) : (int*string*string)  =
    let x_len = String.length x in
    let y_len = String.length y in
    let tbl_size = x_len * y_len in
    let tbl = Hashtbl.create tbl_size in
    let case1 xi yi =
        if x.[xi] = y.[yi] then 1 else -1
    in
    let choose_max (pt1,strxi,stryi) (pt2,strx2,stry2) (pt3,strx3,stry3) =
        if pt1>pt2 then
            if pt1>pt3 then (pt1,strxi,stryi)
            else (pt3,strx3,stry3)
        else 
            if pt2>pt3 then (pt2,strx2,stry2) 
            else (pt3,strx3,stry3)
    in
    let rec aux xi yi (pt,strx,stry) =
    if (xi = x_len && yi = y_len) then (pt,strx,stry)
    else
    if xi = x_len then 
        (
            pt + (-2 * (y_len - yi)),
            String.concat "" [strx; String.make (y_len - yi) ' '],
            String.concat "" [stry; String.sub y (yi) (y_len - yi)]
        )
    else if yi = y_len then
        (pt + (-2 * (x_len - xi)),
        String.concat "" [strx; String.sub x (xi) (x_len - xi)],
        String.concat "" [stry; String.make (x_len - xi) ' ']
        )
    else 
        try
        Hashtbl.find tbl (xi,yi)
        with _ ->
        let r1 = aux (xi+1) (yi+1) (pt+(case1 xi yi), String.concat "" [strx; Char.escaped x.[xi]] , String.concat "" [stry; Char.escaped y.[yi]]) in
        let r2 = aux (xi+1) yi ((pt-2), String.concat "" [strx; Char.escaped x.[xi]] , String.concat "" [stry; " "]) in
        let r3 = aux xi (yi+1) ((pt-2), String.concat "" [strx; " "] , String.concat "" [stry; Char.escaped y.[yi]]) in
        let r = choose_max r1 r2 r3 in
        let _ = Hashtbl.add tbl (xi,yi) r in
        r
in aux 0 0 (0,"","")
;;

当我打印在 hashtbl 中找到的值时，它会重复 2-3 个可能在迭代开始时创建的值。 谢谢。

添加：

这是原始的简单求解函数，我想添加一个 Hashtbl 以使其更快，但它返回错误的结果

let solve (x:string) (y:string) : (int*string*string)  =
let x_len = String.length x in
let y_len = String.length y in
let rec aux xi yi (pt,strx,stry) =
let case1 xi yi =
    if x.[xi] = y.[yi] then 1 else -1
    in
    let choose_max (pt1,strxi,stryi) (pt2,strx2,stry2) (pt3,strx3,stry3) =
        if pt1>pt2 then
            if pt1>pt3 then (pt1,strxi,stryi)
            else (pt3,strx3,stry3)
        else 
            if pt2>pt3 then (pt2,strx2,stry2) 
            else (pt3,strx3,stry3)
    in
    if xi = x_len then 
        (
            pt + (-2 * (y_len - yi)),
            String.concat "" [strx; String.make (y_len - yi) ' '],
            String.concat "" [stry; String.sub y (yi) (y_len - yi)]
        )
    else if yi = y_len then
        (pt + (-2 * (String.length x - xi)),
        String.concat "" [strx; String.sub x (xi) (x_len - xi)],
        String.concat "" [stry; String.make (x_len - xi) ' ']
        )
    else
        let r1 = aux (xi+1) (yi+1) (pt+(case1 xi yi), String.concat "" [strx; Char.escaped x.[xi]] , String.concat "" [stry; Char.escaped y.[yi]]) in
        let r2 = aux (xi+1) yi ((pt-2), String.concat "" [strx; Char.escaped x.[xi]] , String.concat "" [stry; " "]) in
        let r3 = aux xi (yi+1) ((pt-2), String.concat "" [strx; " "] , String.concat "" [stry; Char.escaped y.[yi]]) in
        choose_max r1 r2 r3
in aux 0 0 (0,"","")
;;

Answer 1

我认为您无法轻松修改此函数以获取 hashtbl 进行记忆。这是因为您从三个不同的方向（n0..n2，在下图中）将累加器 (pt,strx,stry) 传递到单元格 (xi,yi)（累加器表示子序列的对齐）。在与序列的其余部分对齐之前，您尚未确定这三个中的哪一个是最佳的 (r0..r2)。

通过添加 memoization，您只存储了一次此类调用的结果，而不是最佳结果。因此，您的结果取决于先前的单元格（来自n0..n2 的累加器）以及对序列其余部分的最佳对齐的重新生成r0..r2。正因为如此，您无法在不详尽的情况下记住结果。要使记忆化工作，您需要明确的子问题，而这些子问题在这里没有。