F# Power issues which accepts both arguments to be

问题:

I am currently experimenting with F#. The articles found on the internet are helpful, but as a C# programmer, I sometimes run into situations where I thought my solution would help, but it did not or just partially helped.

So my lack of knowledge of F# (and most likely, how the compiler works) is probably the reason why I am totally flabbergasted sometimes.

For example, I wrote a C# program to determine perfect numbers. It uses the known form of Euclids proof, that a perfect number can be formed from a Mersenne Prime 2p−1(2p−1) (where 2p-1 is a prime, and p is denoted as the power of).

Since the help of F# states that '**' can be used to calculate a power, but uses floating points, I tried to create a simple function with a bitshift operator (<<<) (note that I've edit this code for pointing out the need):

 let PowBitShift (y:int32) = 1 <<< y;;

However, when running a test, and looking for performance improvements, I also tried a form which I remember from using Miranda (a functional programming language also), which uses recursion and a pattern matcher to calculate the power. The main benefit is that I can use the variable y as a 64-bit Integer, which is not possible with the standard bitshift operator.

    let rec Pow (x : int64) (y : int64) = 
    match y with
        | 0L -> 1L
        | y -> x * Pow x (y - 1L);;

It turns out that this function is actually faster, but I cannot (yet) understand the reason why. Perhaps it is a less intellectual question, but I am still curious.

The seconds question then would be, that when calculating perfect numbers, you run into the fact that the int64 cannot display the big numbers crossing after finding the 9th perfectnumber (which is formed from the power of 31). I am trying to find out if you can use the BigInteger object (or bigint type) then, but here my knowledge of F# is blocking me a bit. Is it possible to create a powerfunction which accepts both arguments to be bigints?

I currently have this:

let rec PowBigInt (x : bigint) (y : bigint) = 
    match y with
        | bigint.Zero -> 1I
        | y -> x * Pow x (y - 1I);;

But it throws an error that bigint.Zero is not defined. So I am doing something wrong there as well. 0I is not accepted as a replacement, since it gives this error:

Non-primitive numeric literal constants cannot be used in pattern matches because they    
can be mapped to multiple different types through the use of a NumericLiteral module.  
Consider using replacing with a variable, and use 'when <variable> = <constant>' at the 
end of the match clause.    

But a pattern matcher cannot use a 'when' statement. Is there another solution to do this?

Thanks in advance, and please forgive my long post. I am only trying to express my 'challenges' as clear as I can.

回答1:

I failed to understand why you need y to be an int64 or a bigint. According to this link, the biggest known Mersenne number is the one with p = 43112609, where p is indeed inside the range of int.

Having y as an integer, you can use the standard operator pown : ^T -> int -> ^T instead because:

let Pow (x : int64) y = pown x y
let PowBigInt (x: bigint) y = pown x y

Regarding your question of pattern matching bigint, the error message indicates quite clearly that you can use pattern matching via when guards:

let rec PowBigInt x y = 
    match y with
    | _ when y = 0I -> 1I
    | _ -> x * PowBigInt x (y - 1I)

回答2:

I think the easiest way to define PowBigInt is to use if instead of pattern matching:

let rec PowBigInt (x : bigint) (y : bigint) =  
  if y = 0I then 1I   
  else x * PowBigInt x (y - 1I) 

The problem is that bigint.Zero is a static property that returns the value, but patterns can only contain (constant) literals or F# active patterns. They can't directly contain property (or other) calls. However, you can write additional constraints in where clause if you still prefer match:

let rec PowBigInt (x : bigint) (y : bigint) =  
  match y with 
  | y when y = bigint.Zero -> 1I 
  | y -> x * PowBigInt x (y - 1I)

As a side-note, you can probably make the function more efficent using tail-recursion (the idea is that if a function makes recursive call as the last thing, then it can be compiled more efficiently):

let PowBigInt (x : bigint) (y : bigint) =   
  // Recursive helper function that stores the result calculated so far
  // in 'acc' and recursively loops until 'y = 0I'
  let rec PowBigIntHelper (y : bigint) (acc : bigint) =
    if y = 0I then acc 
    else PowBigIntHelper (y - 1I) (x * acc)
  // Start with the given value of 'y' and '1I' as the result so far
  PowBigIntHelper y 1I

Regarding the PowBitShift function - I'm not sure why it is slower, but it definitely doesn't do what you need. Using bit shifting to implement power only works when the base is 2.

回答3:

You don't need to create the Pow function. The (**) operator has an overload for bigint -> int -> bigint. Only the second parameter should be an integer, but I don't think that's a problem for your case. Just try

bigint 10 ** 32 ;;

val it : System.Numerics.BigInteger =
  100000000000000000000000000000000 {IsEven = true;
                                     IsOne = false;
                                     IsPowerOfTwo = false;
                                     IsZero = false;
                                     Sign = 1;}

回答4:

Another option is to inline your function so it works with all numeric types (that support the required operators: (*), (-), get_One, and get_Zero).

let rec inline PowBigInt (x:^a) (y:^a) : ^a =  
  let zero = LanguagePrimitives.GenericZero 
  let one = LanguagePrimitives.GenericOne
  if y = zero then one
  else x * PowBigInt x (y - one) 

let x = PowBigInt 10 32     //int
let y = PowBigInt 10I 32I   //bigint
let z = PowBigInt 10.0 32.0 //float

I'd probably recommend making it tail-recursive, as Tomas suggested.