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 *)
Related
I've written the this radix-2 FFT with the goal of making it functionally idiomatic without sacrificing too much performance:
let reverse x bits =
let rec reverse' x bits y =
match bits with
| 0 -> y
| _ -> ((y <<< 1) ||| (x &&& 1))
|> reverse' (x >>> 1) (bits - 1)
reverse' x bits 0
let radix2 (vector: Complex[]) (direction: int) =
let z = vector.Length
let depth = floor(Math.Log(double z, 2.0)) |> int
if (1 <<< depth) <> z then failwith "Vector length is not a power of 2"
// Complex roots of unity; "twiddle factors"
let unity: Complex[] =
let xpn = float direction * Math.PI / double z
Array.Parallel.init<Complex> (z/2) (fun i ->
Complex.FromPolarCoordinates(1.0, (float i) * xpn))
// Permutes elements of input vector via bit-reversal permutation
let pvec = Array.Parallel.init z (fun i -> vector.[reverse i depth])
let outerLoop (vec: Complex[]) =
let rec recLoop size =
if size <= z then
let mid, step = size / 2, z / size
let rec inrecLoop i =
if i < z then
let rec bottomLoop idx k =
if idx < i + mid then
let temp = vec.[idx + mid] * unity.[k]
vec.[idx + mid] <- (vec.[idx] - temp)
vec.[idx] <- (vec.[idx] + temp)
bottomLoop (idx + 1) (k + step)
bottomLoop i 0
inrecLoop (i + size)
inrecLoop 0
recLoop (size * 2)
recLoop 2
vec
outerLoop pvec
The outerLoop segment is the biggest nested tail-recursive mess I have ever written. I replicated the algorithm in the Wikipedia article for the Cooley-Tukey algorithm, but the only functional constructs I could think to implement using higher-order functions result in massive hits to both performance and memory efficiency. Are there other solutions that would yield the same results without resulting in massive slow-downs, while still being idiomatic?
I'm not an expert on how the algorithm works, so there might be a nice functional implementation, but it is worth noting that using a localised mutation is perfectly idiomatic in F#.
Your radix2 function is functional from the outside - it takes vector array as an input, never mutates it, creates a new array pvec which it then initializes (using some mutation along the way) and then returns it. This is a similar pattern to what built-in functions like Array.map use (which initializes a new array, mutates it and then returns it). This is often a sensible way of doing things, because some algorithms are better written using mutation.
In this case, it's perfectly reasonable to also use local mutable variables and loops. Doing that will make your code more readable compared to the tail-recursive version. I have not tested this, but my naive translation of your outerLoop function would just be to use three nested loops - something like this:
let mutable size = 2
while size <= z do
let mid, step = size / 2, z / size
let mutable i = 0
while i < z do
for j in 0 .. mid - 1 do
let idx, k = i + j, step * j
let temp = pvec.[idx + mid] * unity.[k]
pvec.[idx + mid] <- (pvec.[idx] - temp)
pvec.[idx] <- (pvec.[idx] + temp)
i <- i + size
size <- size * 2
This might not be exactly right (I did this just be refactoring your code), but I think it's actually more idiomatic than using complex nested tail-recursive functions in this case.
i am trying to implement a recursive function which takes a float and returns a list of ints representing the continued fraction representation of the float (https://en.wikipedia.org/wiki/Continued_fraction) In general i think i understand how the algorithm is supposed to work. its fairly simply. What i have so far is this:
let rec float2cfrac (x : float) : int list =
let q = int x
let r = x - (float q)
if r = 0.0 then
[]
else
q :: (float2cfrac (1.0 / r ))
the problem is with the base case obviously. It seems the value r never does reduce to 0.0 instead the algorithm keeps on returning values which are the likes of 0.0.....[number]. I am just not sure how to perform the comparison. How exactly should i go about it. The algorithm the function is based on says the base case is 0, so i naturally interpret this as 0.0. I dont see any other way. Also, do note that this is for an assignment where i am explicitly asked to implement the algorithm recursively. Does anyone have some guidance for me? It would be much appreciated
It seems the value r never does reduce to 0.0 instead the algorithm keeps on returning values which are the likes of 0.0.....[number].
This is a classic issue with floating point comparisons. You need to use some epsilon tolerance value for comparisons, because r will never reach exactly 0.0:
let epsilon = 0.0000000001
let rec float2cfrac (x : float) : int list =
let q = int x
let r = x - (float q)
if r < epsilon then
[]
else
q :: (float2cfrac (1.0 / r))
> float2cfrac 4.23
val it : int list = [4; 4; 2; 1]
See this MSDN documentation for more.
You could define a helper function for this:
let withinTolerance (x: float) (y: float) e =
System.Math.Abs(x - y) < e
Also note your original solution isn't tail-recursive, so it consumes stack as it recurses and could overflow the stack. You could refactor it such that a float can be unfolded without recursion:
let float2cfrac (x: float) =
let q = int x
let r = x - (float q)
if withinTolerance r 0.0 epsilon then None
else Some (q, (1.0 / r))
4.23 |> Seq.unfold float2cfrac // seq [4; 4; 2; 1]
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.
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
I'm trying to write a function that accepts an int n and returns a list that runs down from n to 0.
This is what I have
let rec downFrom n =
let m = n+1 in
if m = 0 then
[]
else
(m-1) :: downFrom (m - 1);;
The function compiles ok but when I test it with any int it gives me the error
Stack overflow during evaluation (looping recursion?).
I know it's the local varible that gets in the way but I don't know another way to declare it. Thank you!!!
First, the real thing wrong with your program is that you have an infinite loop. Why, because your inductive base case is 0, but you always stay at n! This is because you recurse on m - 1 which is really n + 1 - 1
I'm surprised as to why this compiles, because it doesn't include the rec keyword, which is necessary on recursive functions. To avoid stack overflows in OCaml, you generally switch to a tail recursive style, such as follows:
let downFrom n =
let rec h n acc =
if n = 0 then List.rev acc else h (n-1) (n::acc)
in
h n []
Someone suggested the following edit:
let downFrom n =
let rec h m acc =
if m > n then acc else h (m + 1) (m::acc)
in
h 0 [];
This saves a call to List.rev, I agree.
The key with recursion is that the recursive call has to be a smaller version of the problem. Your recursive call doesn't create a smaller version of the problem. It just repeats the same problem.
You can try with a filtering parameter
syntax:
let f = function
p1 -> expr1
| p2 -> expr2
| p3 -> ...;;
let rec n_to_one =function
0->[]
|n->n::n_to_one (n-1);;
# n_to_one 3;;
- : int list = [3; 2; 1]