Prime number check - recursion

I'm having some issues with my prime number checker in F#. It doesn't seem to give the right results so I'm guessing I've screwed up the logic somewhere but I can't figure out where. The implementation is a simple brute forcing one so the logic isn't complicated and I've implemented similiar solutions using for loops in imperative languages before.
let rec isPrime iterator (n : int) =
match iterator with
| 1 -> isPrime (iterator + 1) n
| a when a = n -> isPrime (iterator + 1) n
| _ -> match n % iterator = 0 with
| true -> false
| false -> isPrime (iterator + 1) n

As you already figured out in the comments, the problem is that the function should terminate and say true when the iterator reaches n. You can actually make it faster just by iterating up to square root of n or at least n/2 because by the time you reach n/2, you know it will be a prime.
This kind of logic seems to be easier to write using if rather than match - although you can easily fix it by fixing the case in match, I'd probably write something like:
let rec isPrime iterator (n : int) =
if iterator = n / 2 then true
elif iterator = 1 then isPrime (iterator + 1) n
elif n % iterator = 0 then false
else isPrime (iterator + 1) n
Also, you might not want to expose the iterator parameter to the user - you can write the code using a nested function which calls the loop starting with iterator = 2 (and then you don't need the iterator = 1 case at all):
let isPrime (n : int) =
let rec loop iterator =
if iterator = n/2 then true
elif n % iterator = 0 then false
else loop (iterator + 1)
loop 2

Related

how do I count the amount of times a (recursive) function executes itself in ocaml?

needing some help (if possible) in how to count the amount of times a recursive function executes itself.
I don't know how to make some sort of counter in OCaml.
Thanks!
Let's consider a very simple recursive function (not Schroder as I don't want to do homework for you) to calculate Fibonacci numbers.
let rec fib n =
match n with
| 0 | 1 -> 1
| _ when n > 0 -> fib (n - 2) + fib (n - 1)
| _ -> raise (Invalid_argument "Negative values not supported")
Now, if we want to know how many times it's been passed in, we can have it take a call number and return a tuple with that call number updated.
To get each updated call count and pass it along, we explicitly call fib in let bindings. Each time c shadows its previous binding, as we don't need that information.
let rec fib n c =
match n with
| 0 | 1 -> (1, c + 1)
| _ when n > 0 ->
let n', c = fib (n - 1) (c + 1) in
let n'', c = fib (n - 2) (c + 1) in
(n' + n'', c)
| _ -> raise (Invalid_argument "Negative values not supported")
And we can shadow that to not have to explicitly pass 0 on the first call.
let fib n = fib n 0
Now:
utop # fib 5;;
- : int * int = (8, 22)
The same pattern can be applied to the Schroder function you're trying to write.
You can create a reference in any higher scope like so
let counter = ref 0 in
let rec f ... =
counter := !counter + 1;
... (* Function body *)
If the higher scope happens to be the module scope (or file top-level scope) you should omit the in
You can return a tuple (x,y) where y you increment by one for each recursive call. It can be useful if your doing for example a Schroder sequence ;)

imperative programing to functional programming using recursion

I'm very new to functional programming. I'm struggling using recursion instead of for loop. Here's what I have so far.
let max_factor n =
let rec loop k =
if k >= n then []
else
begin
if k < n && n % k = 0 then
k :: loop(k+1)
end
my plan is to insert the ones into a list and then find the largest from the list. But I have a feeling I'm doing it wrong. With functional programming, is it always like "going around" or am I just a bad at this? is my approach way off? Can someone please guide me how I should approach this simple problem...
The equivalent to your java code would be
let max_factor (n : int) : int =
let rec loop i =
if i < 2 then 1
else if n mod i = 0 then i
else loop (i-1)
in loop (n / 2);; (* you don't want to start at n, which would trivially divide n *)

Unexpected output type

I am doing practice with F#. I am trying to create a simple program capable to find me out a couple of prime numbers that, summed together, equal a natural number input. It is the Goldbach conjecture. A single couple of primes will be enough. We will assume the input to be a even number.
I first created a function to check if a number is prime:
let rec isPrime (x: int) (i: int) :bool =
match x % i with
| _ when float i > sqrt (float x) -> true
| 0 -> false
| _ -> isPrime x (i + 1)
Then, I am trying to develop a function that (a) looks for prime numbers, (b) compare their sum with the input 'z' and (c) returns a tuple when it finds the two numbers. The function should not be correct yet, but I would get the reason behind this problem:
let rec sumPrime (z: int) (j: int) (k: int) :int * int =
match isPrime j, isPrime k with
| 0, 0 when j + k > z -> (0, 0)
| 0, 0 -> sumPrime (j + 1) (k + 1)
| _, 0 -> sumPrime j (k + 1)
| 0, _ -> sumPrime (j + 1) k
| _, _ -> if j + k < z then
sumPrime (j + 1) k
elif j + k = z then
(j, k)
The problem: even if I specified that the output should be a tuple :int * int the compiler protests, claiming that the expected output should be of type bool. When in trouble, I usually refer to F# for fun and profit, that i love, but this time I cannot find out the problem. Any suggestion is greatly appreciated.
Your code has three problems that I've spotted:
Your isPrime returns a bool (as you've specified), but your match expression in sumPrime is matching against integers (in F#, the Boolean value false is not the same as the integer value 0). Your match expression should look like:
match isPrime j, isPrime k with
| false, false when j + k > z -> (0, 0)
| false, false -> ...
| true, false -> ...
| false, true -> ...
| true, true -> ...
You have an if...elif expression in your true, true case, but there's no final else. By default, the final else of an if expression returns (), the unit type. So once you fix your first problem, you'll find that F# is complaining about a type mismatch between int * int and unit. You'll need to add an else condition to your final match case to say what to do if j + k > z.
You are repeatedly calling your sumPrime function, which takes three parameters, with just two parameters. That is perfectly legal in F#, since it's a curried language: calling sumPrime with two parameters produces the type int -> int * int: a function that takes a single int and returns a tuple of ints. But that's not what you're actually trying to do. Make sure you specify a value for z in all your recursive calls.
With those three changes, you should probably see your compiler errors go away.

For Loop Over a Recursive Call Ocaml

I'm working on an implementation of prime decomposition in OCaml. I am not a functional programmer; Below is my code. The prime decomposition happens recursively in the prime_part function. primes is the list of primes from 0 to num. The goal here being that I could type prime_part into the OCaml interpreter and have it spit out when n = 20, k = 1.
2 + 3 + 7
5 + 7
I adapted is_prime and all_primes from an OCaml tutorial. all_primes will need to be called to generate a list of primes up to b prior to prime_part being called.
(* adapted from http://www.ocaml.org/learn/tutorials/99problems.html *)
let is_prime n =
let n = abs n in
let rec is_not_divisor d =
d * d > n || (n mod d <> 0 && is_not_divisor (d+1)) in
n <> 1 && is_not_divisor 2;;
let rec all_primes a b =
if a > b then [] else
let rest = all_primes (a + 1) b in
if is_prime a then a :: rest else rest;;
let f elem =
Printf.printf "%d + " elem
let rec prime_part n k lst primes =
let h elem =
if elem > k then
append_item lst elem;
prime_part (n-elem) elem lst primes in
if n == 0 then begin
List.iter f lst;
Printf.printf "\n";
()
end
else
if n <= k then
()
else
List.iter h primes;
();;
let main num =
prime_part num 1 [] (all_primes 2 num)
I'm largely confused with the reclusive nature with the for loop. I see that List.ittr is the OCaml way, but I lose access to my variables if I define another function for List.ittr. I need access to those variables to recursively call prime_part. What is a better way of doing this?
I can articulate in Ruby what I'm trying to accomplish with OCaml. n = any number, k = 1, lst = [], primes = a list of prime number 0 to n
def prime_part_constructive(n, k, lst, primes)
if n == 0
print(lst.join(' + '))
puts()
end
if n <= k
return
end
primes.each{ |i|
next if i <= k
prime_part_constructive(n - i, i, lst+[i], primes)
}
end
Here are a few comments on your code.
You can define nested functions in OCaml. Nested functions have access to all previously defined names. So you can use List.iter without losing access to your local variables.
I don't see any reason that your function prime_part_constructive returns an integer value. It would be more idiomatic in OCaml for it to return the value (), known as "unit". This is the value returned by functions that are called for their side effects (such as printing values).
The notation a.(i) is for accessing arrays, not lists. Lists and arrays are not the same in OCaml. If you replace your for with List.iter you won't have to worry about this.
To concatenate two lists, use the # operator. The notation lst.concat doesn't make sense in OCaml.
Update
Here's how it looks to have a nested function. This made up function takes a number n and a list of ints, then writes out the value of each element of the list multiplied by n.
let write_mults n lst =
let write1 m = Printf.printf " %d" (m * n) in
List.iter write1 lst
The write1 function is a nested function. Note that it has access to the value of n.
Update 2
Here's what I got when I wrote up the function:
let prime_part n primes =
let rec go residue k lst accum =
if residue < 0 then
accum
else if residue = 0 then
lst :: accum
else
let f a p =
if p <= k then a
else go (residue - p) p (p :: lst) a
in
List.fold_left f accum primes
in
go n 1 [] []
It works for your example:
val prime_part : int -> int list -> int list list = <fun>
# prime_part 12 [2;3;5;7;11];;
- : int list list = [[7; 5]; [7; 3; 2]]
Note that this function returns the list of partitions. This is much more useful (and functional) than writing them out (IMHO).

Reversing an int in OCaml

I'm teaching myself OCaml, and the main resources I'm using for practice are some problem sets Cornell has made available from their 3110 class. One of the problems is to write a function to reverse an int (i.e: 1234 -> 4321, -1234 -> -4321, 2 -> 2, -10 -> -1 etc).
I have a working solution, but I'm concerned that it isn't exactly idiomatic OCaml:
let rev_int (i : int) : int =
let rec power cnt value =
if value / 10 = 0 then cnt
else power (10 * cnt) (value/10) in
let rec aux pow temp value =
if value <> 0 then aux (pow/10) (temp + (value mod 10 * pow)) (value / 10)
else temp in
aux (power 1 i) 0 i
It works properly in all cases as far as I can tell, but it just seems seriously "un-OCaml" to me, particularly because I'm running through the length of the int twice with two inner-functions. So I'm just wondering whether there's a more "OCaml" way to do this.
I would say, that the following is idiomatic enough.
(* [rev x] returns such value [y] that its decimal representation
is a reverse of decimal representation of [x], e.g.,
[rev 12345 = 54321] *)
let rev n =
let rec loop acc n =
if n = 0 then acc
else loop (acc * 10 + n mod 10) (n / 10) in
loop 0 n
But as Jeffrey said in a comment, your solution is quite idiomatic, although not the nicest one.
Btw, my own style, would be to write like this:
let rev n =
let rec loop acc = function
| 0 -> acc
| n -> loop (acc * 10 + n mod 10) (n / 10) in
loop 0 n
As I prefer pattern matching to if/then/else. But this is a matter of mine personal taste.
I can propose you some way of doing it:
let decompose_int i =
let r = i / 10 in
i - (r * 10) , r
This function allows me to decompose the integer as if I had a list.
For instance 1234 is decomposed into 4 and 123.
Then we reverse it.
let rec rev_int i = match decompose_int i with
| x , 0 -> 10 , x
| h , t ->
let (m,r) = rev_int t in
(10 * m, h * m + r)
The idea here is to return 10, 100, 1000... and so on to know where to place the last digit.
What I wanted to do here is to treat them as I would treat lists, decompose_int being a List.hd and List.tl equivalent.

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