I have been trying to write a program which prints out a factorial without actually using recursion.
Here is the code
let factorial x =
let mutable n = x
while n > 0 do
let result = n*(n-1)
n <- (n-1)
result
The issue is that when I try to run the code it tells me that the expected result is a unit, whereas the input is clearly an integer, which obviously results in an error. However, I have checked all my variables and the compiler interprets them as integers, so what is the issue
There are several issues with your code here, you must keep in mind indentation in f# is very important, for you want to do your code should be:
let factorial x =
let mutable n = x
let mutable result = 1
while n > 0 do
result <- n * (n-1)
n <- (n - 1)
result
You were declaring the result variable inside the while scope and returning it outside it. Your code however is incorrect, I took the liberty of modify it, here what I did:
let factorial x =
let mutable n = x
let mutable result = 1
while n >= 1 do
result <- result * n
n <- (n - 1)
result
Related
In C, I would solve the problem with a loop. To represent the idea, something like:
void foo(int x){
while(x > 0){
printf("%d", x % 10);
x /= 10;
}
}
With F#, I am unable to make the function return the single values. I tried:
let reverse =
let aux =
fun x ->
x % 10
let rec aux2 =
fun x ->
if x = 0 then 0
else aux2(aux(x / 10))
aux2 n
but it returns always the base case 0.
I cannot get my mind beyond this approach, where the recursion results are maintained with an operation, and cannot be reported (according to may comprehension) individually:
let reverse2 =
let rec aux =
fun x ->
if x = 0 then 0
else (x % 10) + aux (x / 10) // The operation returning the result
aux n
This is a simple exercise I am doing in order to "functionalize" my mind. Hence, I am looking for an approach to this problem not involving library functions.
A for loop that changes the value of mutable variables can be rewritten as a recursive function. You can think of the mutable variables as implicit parameters to the function. So if we have a mutable variable x, we need to instead pass the new state of x explicitly as a function parameter. The closest equivalent to your C function as a recursive F# function is this:
let rec foo x =
if x > 0 then
printf "%d" (x % 10)
foo (x / 10)
This in itself isn't particularly functional because it returns unit and only has side effects. You can collect the result of each loop using another parameter. This is often called an accumulator:
let foo x =
let rec loop x acc =
if x > 0 then
loop (x / 10) (x % 10 :: acc)
else acc
loop x [] |> List.rev
foo 100 // [0; 0; 1]
I made an inner loop function that is actually the recursive one. The outer foo function starts off the inner loop with [] as the accumulator. Items are added to the start of the list during each iteration and the accumulator list is reversed at the end.
You can use another type as the accumulator, e.g. a string, and append to the string instead of adding items to the list.
I'm writing some thing really easy, a program that finds all factors of an int. Here is what I have
let factor n=
let ls=[]
for i=1 to n do
if i % n =0 then ls = i::ls
l
If I do this then it pops an error This expression was expected to have the type unit. But I tried to put an expression that prints something after if..then.., which is suppose to return type unit, but it still gives the same error. I am lost about this. Can someone help please? Thanks
You are trying to make ls into a mutable variable nd assign it with =. While this is possible, by using mutable (1) or ref (2) along with <- or := assignment operators, this is generally discouraged in the functional world.
A possibly more idiomatic implementation of the naive algorithm could be:
let factor n =
let rec factorLoop curr divs =
if curr > n then divs
else
if n % curr = 0
then factorLoop (curr+1) (curr::divs)
else factorLoop (curr+1) divs
factorLoop 1 [] |> List.rev
> factor 12;;
val it : int list = [1; 2; 3; 4; 6; 12]
Here the main function defines an inner factorLoop function that is recursive. Recursion is the way we can avoid many uses of mutable variables in functional languages. The recursive inner function threads along a curr variable that is the current divisor to be tested and a list divs of currently found divisors. The result includes 1 and n. This can be altered respectively by changing the initial value of curr and the terminating condition in the first line of factorLoop.
It is worth noting that it can all be shrunk down to one line by making use of the F# library:
let factor n =
[1..n] |> List.filter (fun x -> n % x = 0)
Here we build a list of values 1..n and feed them to List.filter which applies the given predicate (at the end of the line) to select only divisors on n. If n is large, however, the temp list will grow very large. We can use a lazily evaluated sequence instead, which won't blow the memory usage:
let factor n =
{1..n} |> Seq.filter (fun x -> n % x = 0) |> Seq.toList
Here we filter on a 'lazy' sequence and only convert the (much smaller) sequence of results to a list at the end:
> factor 10000000;;
val it : int list =
[1; 2; 4; 5; 8; 10; 16; 20; 25; 32; ... etc
= is comparison, not assignment. You want either
let factor n =
let mutable ls = []
for i = 1 to n do
if n % i = 0 then ls <- i::ls
ls
or
let factor n =
let ls = ref []
for i = 1 to n do
if n % i = 0 then ls := i::(!ls)
!ls
Note, however, that both of these solutions are highly unidiomatic, as there are equally easy immutable solutions to this problem.
Am trying to count down from 6 to 1 in a non-recursive factorial function and getting a compiler error:
let fact x =
let mutable product = 1
for i in x..-1..1 do
product <- product * i
product
// Error on the 6 - type 'int' does not support the operator '..-'
let answer = fact 6
printfn "%i" answer
I got this idea from near the bottom here
Have changed the function to just count up and it works, but I'm interested to know why this failed. Is there a better way to count down?
Using VS2012 update 3
Have changed the function to just count up and it works, but I'm interested to know why this failed.
Your example fails because .. and - are two different operators, the compiler needs separation between those. Instead of wrapping -1 by parentheses, you could add spaces:
let fact x =
let mutable product = 1
for i in x .. -1 .. 1 do
product <- product * i
product
Is there a better way to count down?
The less well-known for .. downto .. do construct is more appropriate to use here.
let fact x =
let mutable product = 1
for i = x downto 1 do
product <- product * i
product
This works:
let fact x =
let mutable product = 1
for i in x..(-1)..1 do
product <- product * i
product
As does this (as used in the link in the question):
let fact x =
let mutable product = 1
for i in x .. -1 .. 1 do
product <- product * i
product
PS: also note that there are more functional ways to compute a factorial (bad to use mutable variables); the most obvious using recursion:
let rec fact x =
if x > 2
then x * (fact (x - 1))
else x
or a one-liner using lists:
let fact x = [2..x] |> List.reduce (*)
Try to use brackets:
...
for i in x..(-1)..1 do
...
Creating a Parallel.ForEach expression of this form:
let low = max 1 (k-m)
let high = min (k-1) n
let rangesize = (high+1-low)/(PROCS*3)
Parallel.ForEach(Partitioner.Create(low, high+1, rangesize), (fun j ->
let i = k - j
if x.[i-1] = y.[j-1] then
a.[i] <- b.[i-1] + 1
else
a.[i] <- max c.[i] c.[i-1]
)) |> ignore
Causes me to receive the error: No overloads match for method 'ForEach'. However I am using the Parallel.ForEach<TSource> Method (Partitioner<TSource>, Action<TSource>) and it seems right to me. Am I missing something?
Edited: I am trying to obtain the same results as the code below (that does not use a Partitioner):
let low = max 1 (k-m)
let high = min (k-1) n
let rangesize = (high+1-low)/(PROCS*3)
let A = [| low .. high |]
Parallel.ForEach(A, fun (j:int) ->
let i = k - j
if x.[i-1] = y.[j-1] then
a.[i] <- b.[i-1] + 1
else
a.[i] <- max c.[i] c.[i-1]
) |> ignore
Are you sure that you have opened all necessary namespaces, all the values you are using (low, high and PROCS) are defined and that your code does not accidentally redefine some of the names that you're using (like Partitioner)?
I created a very simple F# script with this code and it seems to be working fine (I refactored the code to create a partitioner called p, but that does not affect the behavior):
open System.Threading.Tasks
open System.Collections.Concurrent
let PROCS = 10
let low, high = 0, 100
let p = Partitioner.Create(low, high+1, high+1-low/(PROCS*3))
Parallel.ForEach(p, (fun j ->
printfn "%A" j // Print the desired range (using %A as it is a tuple)
)) |> ignore
It is important that the value j is actually a pair of type int * int, so if the body uses it in a wrong way (e.g. as an int), you will get the error. In that case, you can add a type annotation to j and you would get a more useful error elsewhere:
Parallel.ForEach(p, (fun (j:int * int) ->
printfn "%d" j // Error here, because `j` is used as an int, but it is a pair!
)) |> ignore
This means that if you want to perform something for all j values in the original range, you need to write something like this:
Parallel.ForEach(p, (fun (loJ, hiJ) ->
for j in loJ .. hiJ - 1 do // Iterate over all js in this partition
printfn "%d" j // process the current j
)) |> ignore
Aside, I guess that the last argument to Partitioner.Create should actually be (high+1-low)/(PROCS*3) - you probably want to divide the total number of steps, not just the low value.
I am trying to learn F# so I paid a visit to Project Euler and I am currently working on Problem 3.
The prime factors of 13195 are 5, 7,
13 and 29.
What is the largest prime
factor of the number 600851475143?
Some things to consider:
My first priority is to learn good functional habits.
My second priority is I would like it to be fast and efficient.
Within the following code I have marked the section this question is regarding.
let isPrime(n:int64) =
let rec check(i:int64) =
i > n / 2L or (n % i <> 0L && check(i + 1L))
check(2L)
let greatestPrimeFactor(n:int64) =
let nextPrime(prime:int64):int64 =
seq { for i = prime + 1L to System.Int64.MaxValue do if isPrime(i) then yield i }
|> Seq.skipWhile(fun v -> n % v <> 0L)
|> Seq.hd
let rec findNextPrimeFactor(number:int64, prime:int64):int64 =
if number = 1L then prime else
//************* No variable
(fun p -> findNextPrimeFactor(number / p, p))(nextPrime(prime))
//*************
//************* Variable
let p = nextPrime(prime)
findNextPrimeFactor(number / p, p)
//*************
findNextPrimeFactor(n, 2L)
Update
Based off some of the feedback I have refactored the code to be 10 times faster.
module Problem3
module private Internal =
let execute(number:int64):int64 =
let rec isPrime(value:int64, current:int64) =
current > value / 2L or (value % current <> 0L && isPrime(value, current + 1L))
let rec nextPrime(prime:int64):int64 =
if number % prime = 0L && isPrime(prime, 2L) then prime else nextPrime(prime + 1L)
let rec greatestPrimeFactor(current:int64, prime:int64):int64 =
if current = 1L then prime else nextPrime(prime + 1L) |> fun p -> greatestPrimeFactor(current / p, p)
greatestPrimeFactor(number, 2L)
let execute() = Internal.execute(600851475143L)
Update
I would like to thank everyone for there advice. This latest version is a compilation of all the advice I received.
module Problem3
module private Internal =
let largestPrimeFactor number =
let rec isPrime value current =
current > value / 2L || (value % current <> 0L && isPrime value (current + 1L))
let rec nextPrime value =
if number % value = 0L && isPrime value 2L then value else nextPrime (value + 1L)
let rec find current prime =
match current / prime with
| 1L -> prime
| current -> nextPrime (prime + 1L) |> find current
find number (nextPrime 2L)
let execute() = Internal.largestPrimeFactor 600851475143L
Functional programming becomes easier and more automatic with practice, so don't sweat it if you don't get it absolutely right on the first try.
With that in mind, let's take your sample code:
let rec findNextPrimeFactor(number:int64, prime:int64):int64 =
if number = 1L then prime else
//************* No variable
(fun p -> findNextPrimeFactor(number / p, p))(nextPrime(prime))
//*************
//************* Variable
let p = nextPrime(prime)
findNextPrimeFactor(number / p, p)
//*************
Your no variable version is just weird, don't use it. I like your version with the explicit let binding.
Another way to write it would be:
nextPrime(prime) |> fun p -> findNextPrimeFactor(number / p, p)
Its ok and occasionally useful to write it like this, but still comes across as a little weird. Most of the time, we use |> to curry values without needing to name our variables (in "pointfree" style). Try to anticipate how your function will be used, and if possible, re-write it so you can use it with the pipe operator without explicit declared variables. For example:
let rec findNextPrimeFactor number prime =
match number / prime with
| 1L -> prime
| number' -> nextPrime(prime) |> findNextPrimeFactor number'
No more named args :)
Ok, now that we have that out of the way, let's look at your isPrime function:
let isPrime(n:int64) =
let rec check(i:int64) =
i > n / 2L or (n % i <> 0L && check(i + 1L))
check(2L)
You've probably heard to use recursion instead of loops, and that much is right. But, wherever possible, you should abstract away recursion with folds, maps, or higher order functions. Two reasons for this:
its a little more readable, and
improperly written recursion will result in a stack overflow. For example, your function is not tail recursive, so it'll blow up on large values of n.
I'd rewrite isPrime like this:
let isPrime n = seq { 2L .. n / 2L } |> Seq.exists (fun i -> n % i = 0L) |> not
Most of the time, if you can abstract away your explicit looping, then you're just applying transformations to your input sequence until you get your results:
let maxFactor n =
seq { 2L .. n - 1L } // test inputs
|> Seq.filter isPrime // primes
|> Seq.filter (fun x -> n % x = 0L) // factors
|> Seq.max // result
We don't even have intermediate variables in this version. Coolness!
My second priority is I would like it
to be fast and efficient.
Most of the time, F# is going to be pretty comparable with C# in terms of speed, or it's going to be "fast enough". If you find your code takes a long time to execute, it probably means you're using the wrong data structure or a bad algorithm. For a concrete example, read the comments on this question.
So, the code I've written is "elegant" in the sense that its concise, gives the correct results, and doesn't rely on any trickery. Unfortunately, its not very fast. For start:
it uses trial division to create a sequence of primes, when the Sieve of Eratosthenes would be much faster. [Edit: I wrote a somewhat naive version of this sieve which didn't work for numbers larger than Int32.MaxValue, so I've removed the code.]
read Wikipedia's article on the prime counting function, it'll give you pointers on calculating the first n primes as well as estimating the upper and lower bounds for the nth prime.
[Edit: I included some code with a somewhat naive implementation of a sieve of eratosthenes. It only works for inputs less than int32.MaxValue, so it probably isn't suitable for project euler.]
Concerning "good functional habit" or rather good practice I see three minor things. Using the yield in your sequence is a little harder to read than just filter. Unnecessary type annotations in a type inferred language leads to difficult refactoring and makes the code harder to read. Don't go overboard and try to remove every type annotation though if you're finding it difficult. Lastly making a lambda function which only takes a value to use as a temp variable reduces readability.
As far as personal style goes I prefer more spaces and only using tupled arguments when the data makes sense being grouped together.
I'd write your original code like this.
let isPrime n =
let rec check i =
i > n / 2L || (n % i <> 0L && check (i + 1L))
check 2L
let greatestPrimeFactor n =
let nextPrime prime =
seq {prime + 1L .. System.Int64.MaxValue}
|> Seq.filter isPrime
|> Seq.skipWhile (fun v -> n % v <> 0L)
|> Seq.head
let rec findNextPrimeFactor number prime =
if number = 1L then
prime
else
let p = nextPrime(prime)
findNextPrimeFactor (number / p) p
findNextPrimeFactor n 2L
Your updated code is optimal for your approach. You would have to use a different algorithm like Yin Zhu answer to go faster. I wrote a test to check to see if F# makes the "check" function tail recursive and it does.
the variable p is actually a name binding, not a variable. Using name binding is not a bad style. And it is more readable. The lazy style of nextPrime is good, and it actually prime-test each number only once during the whole program.
My Solution
let problem3 =
let num = 600851475143L
let rec findMax (n:int64) (i:int64) =
if n=i || n<i then
n
elif n%i=0L then
findMax (n/i) i
else
findMax n (i+1L)
findMax num 2L
I basically divides num from 2, 3, 4.. and don't consider any prime numbers. Because if we divides all 2 from num, then we won't be able to divide it by 4,8, etc.
on this number, my solution is quicker:
> greatestPrimeFactor 600851475143L;;
Real: 00:00:01.110, CPU: 00:00:00.702, GC gen0: 1, gen1: 1, gen2: 0
val it : int64 = 6857L
>
Real: 00:00:00.001, CPU: 00:00:00.000, GC gen0: 0, gen1: 0, gen2: 0
val problem3 : int64 = 6857L
I think that the code with the temporary binding is significantly easier to read. It's pretty unusual to create an anonymous function and then immediately apply it to a value as you do in the other case. If you really want to avoid using a temporary value, I think that the most idiomatic way to do that in F# would be to use the (|>) operator to pipe the value into the anonymous function, but I still think that this isn't quite as readable.