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I've just started learning F# very recently. I have a function which counts the coefficients of the linear equation: y = ax + b, based on coordinates of two points P1(x1, y1), P2(x1, y2). The function looks like this:
module LinearFit
let generate(x1 : double, y1 : double, x2 : double, y2 : double) =
let w = x1 * 1.0 - x2 * 1.0
let wa = y1 * 1.0 - y2 * 1.0
let wb = x1 * y2 - x2 * y1
printfn "w: %g" w
printfn "wa: %g" wa
printfn "wb: %g" wb
let a = wa/w
let b = wb/w
printfn "a: %g" a
printfn "b: %g" b
printfn "%g %g" a b
(a, b)
I'm trying to somehow return founded coefficients as a tuple result and then assign the result to the new variables so later I can use the result to do some other operations. The trivial thing, for now, would be just displayed a result like:
The generated function is y = 2.5x - 6.5
So far I was trying to do sth like this
open System
let main() =
printf "Linear fit"
(a: double, b: double) <- LinearFit.generate(5.0, 6.0, 7.0, 11.0)
printfn "The generated functi..."
main()
Console.ReadKey() |> ignore
This is only a concept as I'm not even able to compile the project as im getting errors:
"Unexpected symbol ',' in expression"
"Unexpected symbol ')' in binding."
I tried to find some similar approach to C#...
For now what I want to achieve is just to assing the result of generate function to some variables. In C# it would look just like
public (double a, double b) Generate(some params here)
{
// some logic here
return (a, b);
}
(var a, var b) = Generate(...);
Any ideas?
You're making several syntactic mistakes.
First, the arrow-left operator <- is destructive update. It takes a mutable variable on the right and an expression on the left, and pushes the value of the expression into the variable. For example:
let mutable x = 5
x <- 42
In your example, neither a nor b are mutable variables that exist by the time you're trying to use the <- operator. Plus, the operator expects a single mutable variable, not a pattern.
Second, the way to declare new variables in F# is with let. It is roughly equivalent to var in C#, except you can declare multiple variables at once by putting them in a pattern. For example:
let x = 42
let pair = (1, 5)
let a, b = pair
Here, on the last line, I'm declaring two variables a and b by destructuring the pair.
In your example, you're trying to introduce the two new variables a and b without a let keyword. This is not allowed.
So, putting all of the above together, this is the right way to do what you're trying to do:
let main() =
printf "Linear fit"
let a, b = LinearFit.generate(5.0, 6.0, 7.0, 11.0)
printfn "The generated functi..."
P.S. Your question betrays a misunderstanding of some pretty basic principles of F# syntax. Because of this, I would recommend that you read through tutorials, examples, and other articles on F# to familiarize yourself with the syntax before attempting to venture farther.
I am working with an API that represents points like float * float.
These are inconvenient to do arithmetic on:
let a = (3.0, 4.0)
let b = (2.0, 1.0)
let c = (fst a + fst b, snd a + snd b)
I would like to write:
let c = a + b
I can do this if I define my own type:
type Vector2 =
{
X : float;
Y : float;
}
with
static member (+) (a : Vector2, b : Vector2) =
{ X = a.X + b.X; Y = a.Y + b.Y }
But then I need to convert for the API I am using:
let c = a + b
let cAsTuple = (c.X, c.Y)
Alternatively, I could create a free function:
let add (ax, ay) (bx, by) =
(ax + bx, ay + by)
let c = a |> add b
But this is not quite as nice as true infix operators.
Does F# allow me to define custom operators for tuples?
If you are willing to use a different operator like (+.) you can do this:
let inline (+.) (a,b) (c,d) = (a + c, b + d)
it works with ints, floats, strings:
( 4 , 3 ) +. ( 3 , 2 ) // (7, 5)
( 4., 3.) +. ( 3., 2.) // (7.0, 5.0)
("4", "3") +. ("3", "2") // ("43", "32")
TL;DR; #AMieres answer is the real one, this should rather be a comment but comments are length limited and code formatting is not nice ¯_(ツ)_/¯
There is work in progress to make operator extensions become reality: Issue, RFC, PR Once this is done, the following might finally work:
open System
open System.Runtime.CompilerServices
[<Extension>]
type TupleExtensions() =
[<Extension>]
static member inline (+) ((x1, y1), (x2, y2)) = (x1 + x2, y1 + y2)
// or
type Tuple<'T1, 'T2> with
// warning FS1215: Extension members cannot provide operator overloads.
// Consider defining the operator as part of the type definition instead.
static member inline (+) ((x1, y1), (x2, y2)) = (x1 + x2, y1 + y2)
// and then
let t1 = (1., 2.)
let t2 = (42., 3.141)
TupleExtensions.(+) (t1, t2) // (43.0, 5.141)
// error FS0001: Expecting a type supporting the operator '+' but given a tuple type
t1 + t2
I am interested in learning an elegant way to use currying in a functional programming language to numerically evaluate multiple integrals. My language of choice is F#.
If I want to integrate f(x,y,z)=8xyz on the region [0,1]x[0,1]x[0,1] I start by writing down a triple integral of the differential form 8xyz dx dy dz. In some sense, this is a function of three ordered arguments: a (float -> float -> float -> float).
I take the first integral and the problem reduces to the double integral of 4xy dx dy on [0,1]x[0,1]. Conceptually, we have curried the function to become a (float -> float -> float).
After the second integral I am left to take the integral of 2x dx, a (float -> float), on the unit interval.
After three integrals I am left with the result, the number 1.0.
Ignoring optimizations of the numeric integration, how could I succinctly execute this? I would like to write something like:
let diffForm = (fun x y z -> 8 * x * y * z)
let result =
diffForm
|> Integrate 0.0 1.0
|> Integrate 0.0 1.0
|> Integrate 0.0 1.0
Is this doable, if perhaps impractical? I like the idea of how closely this would capture what is going on mathematically.
I like the idea of how closely this would capture what is going on mathematically.
I'm afraid your premise is false: The pipe operator threads a value through a chain of functions and is closely related to function composition. Integrating over an n-dimensional domain however is analogous to n nested loops, i.e. in your case something like
for x in x_grid_nodes do
for y in y_grid_nodes do
for z in z_grid_nodes do
integral <- integral + ... // details depend on integration scheme
You cannot easily map that to a chain of three independet calls to some Integrate function and thus the composition integrate x1 x2 >> integrate y1 y2 >> integrate z1 z2 is actually not what you do when you integrate f. That is why Tomas' solution—if I understood it correctly (and I am not sure about that...)—essentially evaluates your function on an implicitly defined 3D grid and passes that to the integration function. I suspect that is as close as you can get to your original question.
You did not ask for it, but if you do want to evaluate a n-dimensional integral in practice, look into Monte Carlo integration, which avoids another problem commonly known as the "curse of dimensionality", i.e. that fact that the number of required sample points grows exponentially with n with classic integration schemes.
Update
You can implement iterated integration, but not with a single integrate function, because the type of the function to be integrated is different for each step of the integration (i.e. each step turns an n-ary function to an (n - 1)-ary one):
let f = fun x y z -> 8.0 * x * y * z
// numerically integrate f on [x1, x2]
let trapRule f x1 x2 = (x2 - x1) * (f x1 + f x2) / 2.0
// uniform step size for simplicity
let h = 0.1
// integrate an unary function f on a given discrete grid
let integrate grid f =
let mutable integral = 0.0
for x1, x2 in Seq.zip grid (Seq.skip 1 grid) do
integral <- integral + trapRule f x1 x2
integral
// integrate a 3-ary function f with respect to its last argument
let integrate3 lower upper f =
let grid = seq { lower .. h .. upper }
fun x y -> integrate grid (f x y)
// integrate a 2-ary function f with respect to its last argument
let integrate2 lower upper f =
let grid = seq { lower .. h .. upper }
fun x -> integrate grid (f x)
// integrate an unary function f on [lower, upper]
let integrate1 lower upper f =
integrate (seq { lower .. h .. upper }) f
With your example function f
f |> integrate3 0.0 1.0 |> integrate2 0.0 1.0 |> integrate1 0.0 1.0
yields 1.0.
I'm not entirely sure how you would implement this in a normal way, so this might not fully solve the problem, but here are some ideas.
To do the numerical integration, you'll (I think?) need to call the original function diffForm at various points as specified by the Integrate calls in the pipeline - but you actually need to call it at a product of the ranges - so if I wanted to call it only at the borders, I would still need to call it 2x2x2 times to cover all possible combinations (diffForm 0 0 0, diffForm 0 0 1, diffForm 0 1 0 etc.) and then do some calcualtion on the 8 results you get.
The following sample (at least) shows how to write similar code that calls the specified function with all combinations of the argument values that you specify.
The idea is to use continuations which can be called multiple times (and so when we get a function, we can call it repeatedly at multiple different points).
// Our original function
let diffForm x y z = 8.0 * x * y * z
// At the first step, we just pass the function to a continuation 'k' (once)
let diffFormK k = k diffForm
// This function takes a function that returns function via a continuation
// (like diffFormK) and it fixes the first argument of the function
// to 'lo' and 'hi' and calls its own continuation with both options
let range lo hi func k =
// When called for the first time, 'f' will be your 'diffForm'
// and here we call it twice with 'lo' and 'hi' and pass the
// two results (float -> float -> float) to the next in the pipeline
func (fun f -> k (f lo))
func (fun f -> k (f hi))
// At the end, we end up with a function that takes a continuation
// and it calls the continuation with all combinations of results
// (This is where you need to do something tricky to aggregate the results :-))
let integrate result =
result (printfn "%f")
// Now, we pass our function to 'range' for every argument and
// then pass the result to 'integrate' which just prints all results
let result =
diffFormK
|> range 0.0 1.0
|> range 0.0 1.0
|> range 0.0 1.0
|> integrate
This might be pretty confusing (because continuations take a lot of time to get used to), but perhaps you (or someone else here?) can find a way to turn this first attempt into a real numerical integration :-)
I'm learning f# and I've got a pretty trivial problem that doesn't seem to make sense. I'm working on Project Euler problem 2 and I've got this:
let fib (x : BigInteger) (y : BigInteger) (max : BigInteger) =
let added = x + y
if added > max then y
else fib y (x + y) max
I've got the error at the recursive fib call:
Value or constructor 'fib' is not defined
And I'm not sure why. Any help?
Because fib is recursive function, it has to start with let rec.
In F#, if you want to write a recursive function, you have to use the rec keyword:
let rec fib (x : BigInteger) (y : BigInteger) (max : BigInteger) =
let added = x + y
if added > max then y
else fib y (x + y) max
That's because in F# under normal circumstances, you can only use identifiers declared before the current code, unlike in C#.
Talking of Project Euler Problem 2, you may consider instead of recursion going with Seq.unfold, which is very idiomatic and gives you all Fibonacci numbers at once:
let fibs = Seq.unfold (fun (current, next) ->
Some(current, (next, current + next))) (1,2)
Now fibs represents lazy sequence of Fibonacci numbers :
>fibs;;
val it : seq<int> = seq[1; 2; 3; 5; ...]
And to make it of BigInteger just substitute (1,2) by (1I,2I), although the solution allows you to stay within ordinary integers.
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.
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)
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.
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;}
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.