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copied!<p>Okay, here's a solution to the recursive factorial problem based on Tomas's solution to the waiting problem. I'm still not 100% satisfied with it; I feel like the inner method should return some kind of continuation tuple or something; maintaining the running total seems somehow like "cheating", but anyway:</p> <pre><code>open System.Threading open System let decr x = Thread.Sleep 1000 x - 1 let product x y = Thread.Sleep 1000 x * y let fact (n:int) :Async<int> = let rec fact2 (x:int) (xMinus1Op:Async<int>) (runningProduct:Async<int>) :Async<int> = async { if x = 0 then return! runningProduct else let! xMinus1 = xMinus1Op let! xMinus2Op = async {return decr xMinus1} |> Async.StartChild let! prod = runningProduct let! runningProduct = async {return product x prod} |> Async.StartChild // start both threads to execute simultaneously and feed them forward. return! fact2 xMinus1 xMinus2Op runningProduct } fact2 n (async{return decr n}) (async{return 1}) let start = Environment.TickCount let result = fact 10 |> Async.RunSynchronously printfn "%A" <| result printfn "%A" <| Environment.TickCount - start //runs in 11 seconds, not 20. </code></pre> <p><strong>Edit: maybe it's more straightforward using tasks:</strong></p> <pre><code>let fact (n:int) :int = let rec fact2 (x:int) (xMinus1:int) (runningProduct:int) :int = if x = 0 then runningProduct else let xMinus2Op = new Task<int>(fun () -> decr xMinus1) let runningProductOp = new Task<int>(fun () -> product x runningProduct) xMinus2Op.Start() runningProductOp.Start() let xMinus2 = xMinus2Op.Result let runningProduct = runningProductOp.Result fact2 xMinus1 xMinus2 runningProduct fact2 n (decr n) (1) </code></pre> <p>No workflows necessary, just plain imperative code; it might even translate into C# easily.</p>

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