问题
What is the most elegant way to implement dynamic programming algorithms that solve problems with overlapping subproblems? In imperative programming one would usually create an array indexed (at least in one dimension) by the size of the problem, and then the algorithm would start from the simplest problems and work towards more complicated once, using the results already computed.
The simplest example I can think of is computing the Nth Fibonacci number:
int Fibonacci(int N)
{
var F = new int[N+1];
F[0]=1;
F[1]=1;
for(int i=2; i<=N; i++)
{
F[i]=F[i-1]+F[i-2];
}
return F[N];
}
I know you can implement the same thing in F#, but I am looking for a nice functional solution (which is O(N) as well obviously).
回答1:
One technique that is quite useful for dynamic programming is called memoization. For more details, see for example blog post by Don Syme or introduction by Matthew Podwysocki.
The idea is that you write (a naive) recursive function and then add cache that stores previous results. This lets you write the function in a usual functional style, but get the performance of algorithm implemented using dynamic programming.
For example, a naive (inefficient) function for calculating Fibonacci number looks like this:
let rec fibs n =
if n < 1 then 1 else
(fibs (n - 1)) + (fibs (n - 2))
This is inefficient, because when you call fibs 3
, it will call fibs 1
three times (and many more times if you call, for example, fibs 6
). The idea behind memoization is that we write a cache that stores the result of fib 1
and fib 2
, and so on, so repeated calls will just pick the pre-calculated value from the cache.
A generic function that does the memoization can be written like this:
open System.Collections.Generic
let memoize(f) =
// Create (mutable) cache that is used for storing results of
// for function arguments that were already calculated.
let cache = new Dictionary<_, _>()
(fun x ->
// The returned function first performs a cache lookup
let succ, v = cache.TryGetValue(x)
if succ then v else
// If value was not found, calculate & cache it
let v = f(x)
cache.Add(x, v)
v)
To write more efficient Fibonacci function, we can now call memoize
and give it the function that performs the calculation as an argument:
let rec fibs = memoize (fun n ->
if n < 1 then 1 else
(fibs (n - 1)) + (fibs (n - 2)))
Note that this is a recursive value - the body of the function calls the memoized fibs
function.
回答2:
Tomas's answer is a good general approach. In more specific circumstances, there may be other techniques that work well - for example, in your Fibonacci case you really only need a finite amount of state (the previous 2 numbers), not all of the previously calculated values. Therefore you can do something like this:
let fibs = Seq.unfold (fun (i,j) -> Some(i,(j,i+j))) (1,1)
let fib n = Seq.nth n fibs
You could also do this more directly (without using Seq.unfold
):
let fib =
let rec loop i j = function
| 0 -> i
| n -> loop j (i+j) (n-1)
loop 1 1
回答3:
let fibs =
(1I,1I)
|> Seq.unfold (fun (n0, n1) -> Some (n0 , (n1, n0 + n1)))
|> Seq.cache
来源:https://stackoverflow.com/questions/7985632/dynamic-programming-in-f