How to handle negative power in F#? - f#

I'm trying to build the nth power function in F#. (Yes, there's already Math.Pow in .Net). Here is my attempt:
let rec nthPower x n =
match n with
| 0 -> 1
| _ -> x * (nthPower x (n-1))
This works fine when n >= 0; however, I don't know how to handle the negative case: when n < 0.
Question:
How to handle the negative case? (n<0)
Is this recursive algorithm efficient? or are there any efficient ways in F#?

You can implement it like this:
let rec nthPower x n =
match n with
| 0 -> 1m
| t when t < 0 -> 1m / (nthPower x -n)
| _ -> decimal x * (nthPower x (n - 1));;
The t when t < 0 allows the pattern matching to match a range of values. I would say that the RHS of this line is self-explanatory, but let me know if it's unclear.
Regarding question #2, I don't think there's anything particularly inefficient about this approach and there's probably not a much simpler way to do it. I'm not sure what the most efficient approach is, but hopefully some mathematicians can chime in.
Edit: I have found an approach that is more efficient for exponents > ~10. It uses memoization and divide-and-conquer to compute the result in O(log n) time instead of O(n):
let rec nthPower x n =
match n with
| 0 -> 1.0
| 1 -> double x
| t when t < 0 -> 1.0 / (nthPower x -n)
| _ ->
let p = nthPower x (n / 2)
p * p * nthPower x (n % 2)

Related

F# integer comparison

Given a list int from -273 to 5526, I want to print the closest integer to zero. In case of you have equality (n and -n) we should take n.
let temps = // this contains => 1 -2 -8 4 5
let (|Greater|_|) a b = if a > b then Some() else None
let (|Smaller|_|) a b = if a < b then Some() else None
let compareTemperatures a b =
let distanceA = abs a
let distanceB = abs b
match distanceA with
| Greater distanceB -> b
| Smaller distanceB -> a
| _ -> abs a
printfn "%i" (temps |> Seq.reduce compareTemperatures)
And that returns -8 instead of 1. It seems correct to me and I can't find the bug but I'm new to F# so I might have make a mistake anywhere and can't see it :(
Thanks in advance
I think you got the comparison the wrong way round - when you write:
match distanceA with
| Greater distanceB -> b
| Smaller distanceB -> a
Then distanceA gets passed as the second parameter to Greater and so you are returning b (in the first case) in case when b is further away from zero. The following will make it work:
match distanceA with
| Greater distanceB -> a
| Smaller distanceB -> b
That said, using active patterns for this just makes the code unnecessarily complicated (and makes it easy to introduce bugs like this one). The following does the same thing and it is easy to understand and also a lot simpler:
let compareTemperatures a b =
if abs a > abs b then b else a
temps |> Seq.reduce compareTemperatures
I think the lesson here is that pattern matching works really well for things like algebraic data types and option values, but it is not that useful for simple numerical comparisons where if works fine!
So you want to compare the values, using first the absolute value, then the sign. Here's my one-liner:
temps |> Seq.minBy (fun x -> abs x, -sign x)
Test case (prints 2):
let li = [-2; 2; -2; 3; -5]
li |> Seq.minBy (fun x -> abs x, -sign x) |> printfn "%d"
This should have been a comment to Thomas' answer, but I'm too stupid for the mobile app...
How about if abs (2*a-1) > abs (2*b) then b else a?

F#: Not understanding match .. with

I'm messing around with F# and Fable, and trying to test my understanding. To do so, I tried creating a function to calculate e given a certain number of iterations. What I've come up with is
let eCalc n =
let rec internalECalc ifact sum count =
match count = n with
| true -> sum
| _ -> internalECalc (ifact / (float count)) (sum + ifact) (count+1)
internalECalc 1.0 0.0 1
Which works fine, returning 2.7182818284590455 when called with
eCalc 20
However, if I try using, what I think is, the more correct form
let eCalc n =
let rec internalECalc ifact sum count =
match count with
| n -> sum
| _ -> internalECalc (ifact / (float count)) (sum + ifact) (count+1)
internalECalc 1.0 0.0 1
I get a warning "[WARNING] This rule will never be matched (L5,10-L5,11)", and returned value of 0. (and the same thing happens if I swap 'n' and 'count' in the match statement). Is there a reason I can't use 'n' in the match statement? Is there a way around this so I can use 'n'?
Thanks
When you use a name in a match statement, you're not checking it against the value assigned to that variable the way you think you are. You are instead assigning that name. I.e.,
match someInt with
| n -> printfn "%d" n
will print the value of someInt. It's the equivalent of let n = someInt; printfn "%d" n.
What you wanted to do was use a when clause; inside a when clause, you're not pattern-matching, but doing a "standard" if check. So what you wanted was:
let eCalc n =
let rec internalECalc ifact sum count =
match count with
| cnt when cnt = n -> sum
| _ -> internalECalc (ifact / (float count)) (sum + ifact) (count+1)
internalECalc 1.0 0.0 1
Does that make sense, or do you need me to go into more detail?
P.S. In a case like this one where your match function looks like "x when (boolean condition involving x) -> case 1 | _ -> case 2", it's quite a bit more readable to use a simple if expression:
let eCalc n =
let rec internalECalc ifact sum count =
if count = n then
sum
else
internalECalc (ifact / (float count)) (sum + ifact) (count+1)
internalECalc 1.0 0.0 1

Why do I have the feeling my F# code could be more concise [closed]

Closed. This question does not meet Stack Overflow guidelines. It is not currently accepting answers.
This question does not appear to be about programming within the scope defined in the help center.
Closed 7 years ago.
Improve this question
I'm an experienced C# developer trying to teach myself F#. I spent a day or 3 reading throught the F# wikibook trying to get to know the syntax and F# fundamentals.
As an exercise I'm trying to go through the Project Euler problems to get a better feeling of the syntax and working with the language.
I've just solved problem 5. But I'm not so happy about the hoops I had to jump through to get a data structure that represents my solution.
I've used this blogpost to get to the algorithm for solving the problem.
I was wondering if anyone could give me some pointers as to how this code could be improved? My guess is that the inherent immutability of F# values is causing me to have to perform a lot of steps to get the exact data I want...
This is my code:
let main argv =
//calculates the prime factors of a number
let findPrimeFactors x =
let primes = [|2I;3I;5I;7I;11I;13I;17I;19I|]
let rec loop acc counter = function
| x when x = 1I -> failwith "A PRIME IS BY DEFINITION GREATER THAN 1"
| x when primes |> Array.contains x -> x :: acc
| x when counter = primes.Length -> failwith "MY LIST OF KNOWN PRIMES IS NOT BIG ENOUGH"
| x when x%primes.[counter]=0I-> loop (primes.[counter]::acc) (counter) (x/primes.[counter])
| x -> loop acc (counter + 1) x
let primeFactor = loop [] 0 x |> List.rev
primeFactor
//calculates the prime factors for each of the numbers between 2 and n
//then, for each of the prime factorizations it tries to find the highest power for each occurring prime factor
let findPrimeFactorsPowers n =
//builds a map of all the prime factor powers for all prime factorizations
let rec addCounterFactorPowers factorPowers = function
| counter when counter = n -> factorPowers
| (counter : int) -> addCounterFactorPowers ((findPrimeFactors (counter|>bigint) |> List.countBy (fun x-> x)) # factorPowers) (counter + 1)
let allFactorPowers = addCounterFactorPowers [] 2
//group all the powers per prime factor
let groupedFactorPowers = allFactorPowers |> List.groupBy (fun (factor, power) -> factor)
//get the highest power per prime factor
let maxFactorPowers = groupedFactorPowers |> List.map (fun (key, powers) -> (key, powers |> List.map (fun (factor, power) -> power) |> List.max))
//return the result
maxFactorPowers
let n = 20;
let primeFactorSet = findPrimeFactorsPowers n
printfn "%A" primeFactorSet
let smallestNumberDivisableByAllNumbersBelown = (primeFactorSet |> List.fold (fun state (factor, power) -> state * pown factor power) 1I)
printfn "Result %A" smallestNumberDivisableByAllNumbersBelown
System.Console.ReadKey(true)|>ignore
0 // return an integer exit code
There are many direct simplifications you can apply to your code, but I don't think that's the best approach.
This is the way I would solve that problem in F#:
let rec tryDivide n m =
if m = 1 then true
else
if n % m = 0 then tryDivide n (m-1)
else false
let rec findIt i m =
if tryDivide i m then i
else findIt (i+1) m
findIt 1 20
It's a bit slower than yours because it doesn't use hardcoded primes, they're calculated on the fly, but still it takes less than 2 secs on my computer to get the right answer.
Note that I'm not using any list-like data structure and no need to rely on big integers in this specific problem.
UPDATE
Here's a better approach based on Kvb's proposed solution:
let rec gcd x y = if y = 0 then abs x else gcd y (x % y)
let lcm a b =
match (a, b) with
| (_, 0) | (0, _) -> 0
| (x, y) -> abs ((x / (gcd x y)) * y)
Seq.fold lcm 1 {2..20}

F# infinite stream of armstrong numbers

I'm trying to create an infinite Stream in F# that contains armstrong numbers. An armstrong number is one whose cubes of its digits add up to the number. For example, 153 is an armstrong number because 1^3 + 5^3 + 3^3 = 153. so far, I have created several functions to help me do so. They are:
type 'a stream = Cons of 'a * (unit -> 'a stream);;
let rec upfrom n = Cons (n, fun() -> upfrom (n+1));;
let rec toIntArray = function
| 0 -> []
| n -> n % 10 :: toIntArray (n / 10);;
 
let rec makearmstrong = function
| [] -> 0
| y::ys -> (y * y * y) + makearmstrong ys;;
let checkarmstrong n = n = makearmstrong(toIntArray n);;
let rec take n (Cons(x,xsf)) =
match n with
| 0 -> []
| _ -> x :: take (n-1)(xsf());;
let rec filter p (Cons (x, xsf)) =
if p x then Cons (x, fun() -> filter p (xsf()))
else filter p (xsf());;
And finally:
let armstrongs = filter (fun n -> checkarmstrong n)(upfrom 1);;
Now, when I do take 4 armstrongs;;, (or any number less than 4) this works perfectly and gives me [1;153;370;371] but if I do take 5 armstrongs;;nothing happens, it seems like the program freezes.
I believe the issue is that there are no numbers after 407 that are the sums of their cubes (see http://oeis.org/A046197), but when your code evaluates the equivalent of take 1 (Cons(407, filter checkarmstrong (upfrom 408))) it's going to force the evaluation of the tail and filter will recurse forever, never finding a matching next element. Also note that your definition of Armstrong numbers differs from, say, Wikipedia's, which states that the power the digits are raised to should be the number of digits in the number.

Usage about Pattern matching

I thought these two function were the same, but it seems that I was wrong.
I define two function f and g in this way:
let rec f n k =
match k with
|_ when (k < 0) || (k > n) -> 0
|_ when k = n -> 100
|_ -> (f n (k+1)) + 1
let rec g n k =
match k with
|_ when (k < 0) || (k > n) -> 0
| n -> 100
|_ -> (g n (k+1)) + 1
let x = f 10 5
let y = g 10 5
The results are:
val x : int = 105
val y : int = 100
Could anyone tell me what's the difference between these two functions?
EDIT
Why does it work here?
let f x =
match x with
| 1 -> 100
| 2 -> 200
|_ -> -1
List.map f [-1..3]
and we get
val f : x:int -> int
val it : int list = [-1; -1; 100; 200; -1]
The difference is that
match k with
...
when k = n -> 100
is a case that matches when some particular condition is true (k = n). The n used in the condition refers to the n that is bound as the function parameter. On the other hand
match k with
...
n -> 100
is a case that only needs to match k against a pattern variable n, which can always succeed. The n in the pattern isn't the same n as the n passed into the function.
For comparison, try the code
let rec g n k =
match k with
|_ when (k < 0) || (k > n) -> 0
| n -> n
|_ -> (g n (k+1)) + 1
and you should see that when you get to the second case, the value returned is the value of the pattern variable n, which has been bound to the value of k.
This behavior is described in the Variable Patterns section of the MSDN F# Language Reference, Pattern Matching:
Variable Patterns
The variable pattern assigns the value being matched to a variable
name, which is then available for use in the execution expression to
the right of the -> symbol. A variable pattern alone matches any
input, but variable patterns often appear within other patterns,
therefore enabling more complex structures such as tuples and arrays
to be decomposed into variables. The following example demonstrates a
variable pattern within a tuple pattern.
let function1 x =
match x with
| (var1, var2) when var1 > var2 -> printfn "%d is greater than %d" var1 var2
| (var1, var2) when var1 < var2 -> printfn "%d is less than %d" var1 var2
| (var1, var2) -> printfn "%d equals %d" var1 var2
function1 (1,2)
function1 (2, 1)
function1 (0, 0)
The use of when is described in more depth in Match Expressions.
The first function is ok, it calls recursively itself n-k times and returns 100 when matches with the conditional where k = n. So, it returns all the calls adding 1 n-k times. with your example, with n=10 and k=5 it is ok the result had been 105.
The problem is the second function. I tested here. See I changed the pattern n->100 to z->100 and it still matches there and never calls itself recursively. So, it always returns 100 if it does not fail in the first conditional. I think F# don't allow that kind of match so it is better to put a conditional to get what you want.

Resources