I wrote a simple recursive fibonacci program that works fine without an assert statement, but when I add an assert statement, even with various permutations of parentheses, spaces, and double semicolons, I keep getting a syntax error during compilation.
Working function:
let rec fib n =
if n = 1
then 1
else
n*(fib (n-1))
Not working:
let rec fib n =
assert (n>=0)
if n = 1
then 1
else
n*(fib (n-1))
Any thoughts appreciated.
Thanks.
You have two expressions: assert (n >= 0) and if ... then ... else .... If you want the two expressions to be evaluated in sequence (which you do), you need to separate them with a semicolon:
let rec fib n =
assert (n >= 0);
if n = 1 then 1
else n * fib (n - 1);;
val fib : int -> int = <fun>
# fib 3;;
- : int = 6
# fib (-3);;
Exception: Assert_failure ("//toplevel//", 2, 4).
Extra spaces do not affect the semantics of OCaml programs. There are also no statements per se in OCaml - everything is an expression. To evaluate two expressions in a sequence you may use the semicolon, e.g.,
print_endline "Hello";
assert (1 > 2);
print_endline "World";
You can also use let .. in .. to chain expressions, especially, if you need to the expression values, e.g.,
let x = 1 + 2 in
let y = 3 + 4 in
Format.printf "%d + %d = %d\n" x y (x + y)
Going back to your example, it should be
let rec fib n =
assert (n >= 0);
if n = 1 then 1
else n * fib (n - 1)
P.S. The double semicolons are not really a part of the language but a special input sequence to be used in the interactive toplevel.
Related
I am trying to write a simple function in OCaml
let rec pell (i: int) =
(if i <= 2 then i (*if given n is less tahn 2 then return 2, else return previous n-1 th term and n-2 nd term recursively*)
else if i>2 then
2 * pell i - 1 + pell i - 2
else failwith "unimplemented" (*else fail with unimplemented message*)
);;
Write an infinite precision version of the pell function from before
pell2 0 = []
pell2 1 = [1]
pell2 7 = [9; 6; 1]
pell2 50 = [2; 2; 5; 3; 5; 1; 4; 2; 9; 2; 4; 6; 2; 5; 7; 6; 6; 8; 4]
I have written below code for this:
let rec pell2 i =
(if i <= 2 then
[] -> i;
else if i=0 then [];
else if i>2 then (*finding pell number and using sum function to
output list with infinite precision...*)
[] -> pell2 i-1 + pell2 i-2;
else failwith "unimplemented"
);;
but still has some syntax errors. Can someone help me with this please.
if i <= 2 then
[] -> i
In snippets like this, the -> is invalid. It looks like you might be mixing pattern matching with match ... with ... and if/else up.
Also, you're first checking if i is less than or equal to 2, but then you have an else to test for i being equal to zero. The first check means the second is never going to happen.
First, let's look at the examples for the output of pell2. We see that pell2 has a single integer parameter, and returns a list of integers. So, we know that the function we want to create has the following type signature:
pell2: int -> int list
Fixing (some but not all of) the syntax errors and trying to maintain your logic,
let rec pell2 i =
if i=0 then []
else if i <= 2 then i
else if i>2 then pell2 i-1 + pell2 i-2
Note that I removed the semicolons at the end of each expression since OCaml's use of a semicolon in its syntax is specifically for dealing with expressions that evaluate to unit (). See ivg's excellent explanation on this. The major flaw with this code is that it does not type check. We see that we conditionally return a list, and otherwise return an int. Notice how above we defined that pell2 should return an int list. So, we can begin fixing this by wrapping our int results in a list:
let rec pell2 n =
if n = 0 then []
else if n <= 2 then [n]
else ... something that will return the Pell number as a list ...
As you have already written, the else branch can be written using recursive calls to the pell2 function. However, we can't write it as you did previously, because pell2 evaluates to a list, and the binary operator + only works on two integers. So, we will have to define our own way of summing lists. Calling this sum_lists, we are left with the following code:
We can now fully define our function pell2:
let rec pell2 n =
if n = 0 then []
else if n <= 2 then [n]
else (* Pell(n) = (2 * Pell(n-1)) + Pell(n-2) *)
let half_of_first_term = pell2 n-1 in
let first_term = sum_lists half_of_first_term half_of_first_term in
let second_term = pell2 n-2 in
sum_lists first_term second_term
So, all that is left is to define sum_lists, so that we are properly summing together two lists of the same format as the return type of pell2. The signature for sum_lists would be
sum_lists: int list -> int list -> int list
I'll give a basic outline of the implementation, but will leave the rest for you to figure out, as this is the main crux of the assignment problem.
let sum_lists lst1 lst2 =
let rec sum_lists_helper lst1 lst2 carry =
match lst1, lst2 with
| [], [] -> if carry = 1 then [1] else []
| h::t, []
| [], h::t -> ...
| h1::t1, h2::t2 -> ...
in
sum_lists_helper lst1 lst2 0
I am new in sml. I tried to convert int to int list. For example, assume that there is an input 1234, then output is a list like [1,2,3,4]. And my question is, how can I type nested functions in sml? let in end? There is my code.
fun digit (a : int): int =
let
fun size (a) = if a < 0 then nil
else x = Int.toString x then digit s = size(x)
fun insert (num, nil) = [num]
| insert (num,xs) = x :: insert ()
fun convert (a, s) = if s < 0 then nil
else insert (a / (10*(s - 1)), xs)
then convert(a - (10*(s - 1), s - 1)
in
end
Nested functions are just one way to split up your workload in multiple, smaller parts. Another option is non-nested library functions. The main differences are that functions that aren't nested don't inherit its parent's variable scope, so they can only work with their own input, and functions that are nested aren't available anywhere else and can't be re-used. Let's say you're giving this problem a first stab:
fun digit_meh n = if n < 10 then [n] else n mod 10 :: digit_meh (n div 10)
And you realize it isn't doing exactly as you want:
- digit_meh 1234;
> val it = [4, 3, 2, 1] : int list
You could remove the most significant digit first, but the calculation isn't as trivial as n mod 10, since it depends on the number of digits.
You could generate this list and then reverse it:
fun digit n = rev (digit_meh n)
But the function digit_meh isn't particularly useful outside of this function, so it could be hidden using local-in-end or let-in-end:
local
fun digit_meh n = if n < 10 then [n] else n mod 10 :: digit_meh (n div 10)
in
val digit = rev o digit_meh
end
fun digit n =
let fun meh n = if n < 10 then [n] else n mod 10 :: meh (n div 10)
in rev (meh n) end
Do notice that the function meh's copy of n shadows digit's copy of n.
For clarity you could also name the variables differently.
Or you could look at how rev is doing its thing and do that. It basically treats its input as a stack and puts the top element in a new stack recursively so that the top becomes the bottom, much like StackOverflow's logo would look like if it jumped out and landed upside down like a slinky spring:
fun rev L =
let fun rev_stack [] result = result
| rev_stack (x::xs) result = rev_stack xs (x::result)
in rev_stack L [] end
Because the result is accumulated in an additional argument, and rev should only take a single argument, nesting a function with an extra accumulating argument is a really useful trick.
You can mimic this behavior, too:
fun digit N =
let fun digit_stack n result =
if n < 10
then n::result
else digit_stack (n div 10) (n mod 10::result)
in f N [] end
This way, we continue to treat the least significant digit first, but we put it in the stack result which means it ends up at the bottom / end. So we don't need to call rev and save that iteration of the list.
In practice, you don't have to hide helper functions using either local-in-end or let-in-end; while it can be useful in the case of let-in-end to inherit a parent function's scope, it is not necessary to hide your functions once you start using modules with opaque signatures (the :> operator):
signature DIGIT =
sig
val digit : int -> int list
end
structure Digit :> DIGIT =
struct
fun digit_stack n result =
if n < 10
then n::result
else digit_stack (n div 10) (n mod 10::result)
fun digit n = digit_stack n []
end
As this is entered into a REPL, only the relevant function is available outside of the module:
> structure Digit : {val digit : int -> int list}
signature DIGIT = {val digit : int -> int list}
- Digit.digit 1234;
> val it = [1, 2, 3, 4] : int list
fun aFunctionCallingF2F3 someVariables =
let
<define some functions and local variables here>
fun F2 ...
fun F3 ...
val v1 ...
val v2 ...
in
<Make use of the functions/variables you defined above and `someVariables`>
end
For example,
fun areaCirle r:real =
let fun square x:real = x*x
val pi = 3.14
in
pi * square r
end
Or define functions you need to call beforehand if they are not mutually recursive. If they are mutually recursive, you can look up the keyword and.
fun F2 ...
fun F3 ...
fun aFunctionCallingF2F3 = <make use of F2 F3 directly>
For example,
fun square x:real = x * x
fun areaCircle r = square r * 3.14
Note that you cannot do
fun areaCircle r = square r * 3.14
fun square x:real = x * x
square needs to be defined before areaCircle.
This F# code is an attempt to solve Project Euler problem #58:
let inc = function
| n -> n + 1
let is_prime = function
| 2 -> true
| n when n < 2 || n%2=0-> false
| n ->
[3..2..(int (sqrt (float n)))]
|> List.tryFind (fun i -> n%i=0)
|> Option.isNone
let spir = Seq.initInfinite (fun i ->
let n = i%4
let a = 2 * (i/4 + 1)
(a*n) + a + (a-1)*(a-1))
let rec accum se p n =
match se with
| x when p*10 < n && p <> 0 -> 2*(n/4) + 1
| x when is_prime (Seq.head x) -> accum (Seq.tail x) (inc p) (inc n)
| x -> accum (Seq.tail x) p (inc n)
| _ -> 0
printfn "%d" (accum spir 0 1)
I do not know the running time of this program because I refused to wait for it to finish. Instead, I wrote this code imperatively in C++:
#include "stdafx.h"
#include "math.h"
#include <iostream>
using namespace std;
int is_prime(int n)
{
if (n % 2 == 0) return 0;
for (int i = 3; i <= sqrt(n); i+=2)
{
if (n%i == 0)
{
return 0;
}
}
return 1;
}
int spir(int i)
{
int n = i % 4;
int a = 2 * (i / 4 + 1);
return (a*n) + a + ((a - 1)*(a - 1));
}
int main()
{
int n = 1, p = 0, i = 0;
cout << "start" << endl;
while (p*10 >= n || p == 0)
{
p += is_prime(spir(i));
n++; i++;
}
cout << 2*(i/4) + 1;
return 0;
}
The above code runs in less than 2 seconds and gets the correct answer.
What is making the F# code run so slowly? Even after using some of the profiling tools mentioned in an old Stackoverflow post, I still cannot figure out what expensive operations are happening.
Edit #1
With rmunn's post, I was able to come up with a different implementation that gets the answer in a little under 30 seconds:
let inc = function
| n -> n + 1
let is_prime = function
| 2 -> true
| n when n < 2 || n%2=0-> false
| n ->
[3..2..(int (sqrt (float n)))]
|> List.tryFind (fun i -> n%i=0)
|> Option.isNone
let spir2 =
List.unfold (fun state ->
let p = fst state
let i = snd state
let n = i%4
let a = 2 * (i/4 + 1)
let diag = (a*n) + a + (a-1)*(a-1)
if p*10 < (i+1) && p <> 0 then
printfn "%d" (2*((i+1)/4) + 1)
None
elif is_prime diag then
Some(diag, (inc p, inc i))
else Some(diag, (p, inc i))) (0, 0)
Edit #2
With FuleSnabel's informative post, his is_prime function makes the above code run in under a tenth of a second, making it faster than the C++ code:
let inc = function
| n -> n + 1
let is_prime = function
| 1 -> false
| 2 -> true
| v when v % 2 = 0 -> false
| v ->
let stop = v |> float |> sqrt |> int
let rec loop vv =
if vv <= stop then
if (v % vv) <> 0 then
loop (vv + 2)
else
false
else
true
loop 3
let spir2 =
List.unfold (fun state ->
let p = fst state
let i = snd state
let n = i%4
let a = 2 * (i/4 + 1)
let diag = (a*n) + a + (a-1)*(a-1)
if p*10 < (i+1) && p <> 0 then
printfn "%d" (2*((i+1)/4) + 1)
None
elif i <> 3 && is_prime diag then
Some(diag, (inc p, inc i))
else Some(diag, (p, inc i))) (0, 0)
There is no Seq.tail function in the core F# library (UPDATE: Yes there is, see comments), so I assume you're using the Seq.tail function from FSharpx.Collections. If you're using a different implementation of Seq.tail, it's probably similar -- and it's almost certainly the cause of your problems, because it's not O(1) like you think it is. Getting the tail of a List is O(1) because of how List is implemented (as a series of cons cells). But getting the tail of a Seq ends up creating a brand new Seq from the original enumerable, discarding one item from it, and returning the rest of its items. When you go through your accum loop a second time, you call Seq.tail on that "skip 1 then return" seq. So now you have a Seq which I'll call S2, which asks S1 for an IEnumerable, skips the first item of S1, and returns the rest of it. S1, when asked for its first item, asks S0 (the original Seq) for an enumerable, skips its first item, then returns the rest of it. So for S2 to skip two items, it had to create two seqs. Now on your next run through when you ask for the Seq.tail of S2, you create S3 that asks S2 for an IEnumerable, which asks S1 for an IEnumerable, which asks S0 for an IEnumerable... and so on. This is effectively O(N^2), when you thought you were writing an O(N) operation.
I'm afraid I don't have time right now to figure out a solution for you; using List.tail won't help since you need an infinite sequence. But perhaps just knowing about the Seq.tail gotcha is enough to get you started, so I'll post this answer now even though it's not complete.
If you need more help, comment on this answer and I'll come back to it when I have time -- but that might not be for several days, so hopefully others will also answer your question.
Writing performant F# is very possible but requires some knowledge of patterns that have high relative CPU cost in a tight loop. I recommend using tools like ILSpy to find hidden overhead.
For instance one could imagine F# exands this expression into an effective for loop:
[3..2..(int (sqrt (float n)))]
|> List.tryFind (fun i -> n%i=0)
|> Option.isNone
However it currently doesn't. Instead it creates a List that spans the range using intrinsic operators and passes that to List.tryFind. This is expensive when compared to the actual work we like to do (the modulus operation). ILSpy decompiles the code above into something like this:
public static bool is_prime(int _arg1)
{
switch (_arg1)
{
case 2:
return true;
default:
return _arg1 >= 2 && _arg1 % 2 != 0 && ListModule.TryFind<int>(new Program.Original.is_prime#10(_arg1), SeqModule.ToList<int>(Operators.CreateSequence<int>(Operators.OperatorIntrinsics.RangeInt32(3, 2, (int)Math.Sqrt((double)_arg1))))) == null;
}
}
These operators aren't as performant as they could be (AFAIK this is currently being improved) but no matter how effecient allocating a List and then search it won't beat a for loop.
This means the is_prime is not as effective as it could be. Instead one could do something like this:
let is_prime = function
| 1 -> false
| 2 -> true
| v when v % 2 = 0 -> false
| v ->
let stop = v |> float |> sqrt |> int
let rec loop vv =
if vv <= stop then
(v % vv) <> 0 && loop (vv + 2)
else
true
loop 3
This version of is_prime relies on tail call optimization in F# to expand the loop into an efficient for loop (you can see this using ILSpy). ILSpy decompile the loop into something like this:
while (vv <= stop)
{
if (_arg1 % vv == 0)
{
return false;
}
int arg_13_0 = _arg1;
int arg_11_0 = stop;
vv += 2;
stop = arg_11_0;
_arg1 = arg_13_0;
}
This loop doesn't allocate memory and is just a rather efficient loop. One see some non-sensical assignments but hopefully the JIT:er eliminate those. I am sure is_prime can be improved even further.
When using Seq in performant code one have to keep in mind it's lazy and it doesn't use memoization by default (see Seq.cache). Therefore one might easily end up doing the same work over and over again (see #rmunn answer).
In addition Seq isn't especially effective because of how IEnumerable/IEnumerator are designed. Better options are for instance Nessos Streams (available on nuget).
In case you are interested I did a quick implementation that relies on a simple Push Stream which seems decently performant:
// Receiver<'T> is a callback that receives a value.
// Returns true if it wants more values, false otherwise.
type Receiver<'T> = 'T -> bool
// Stream<'T> is function that accepts a Receiver<'T>
// This means Stream<'T> is a push stream (as opposed to Seq that uses pull)
type Stream<'T> = Receiver<'T> -> unit
// is_prime returns true if the input is prime, false otherwise
let is_prime = function
| 1 -> false
| 2 -> true
| v when v % 2 = 0 -> false
| v ->
let stop = v |> float |> sqrt |> int
let rec loop vv =
if vv <= stop then
(v % vv) <> 0 && loop (vv + 2)
else
true
loop 3
// tryFind looks for the first value in the input stream for f v = true.
// If found tryFind returns Some v, None otherwise
let tryFind f (s : Stream<'T>) : 'T option =
let res = ref None
s (fun v -> if f v then res := Some v; false else true)
!res
// diagonals generates a tuple stream of all diagonal values
// The first value is the side length, the second value is the diagonal value
let diagonals : Stream<int*int> =
fun r ->
let rec loop side v =
let step = side - 1
if r (side, v + 1*step) && r (side, v + 2*step) && r (side, v + 3*step) && r (side, v + 4*step) then
loop (side + 2) (v + 4*step)
if r (1, 1) then loop 3 1
// ratio computes the streaming ratio for f v = true
let ratio f (s : Stream<'T>) : Stream<float*'T> =
fun r ->
let inc r = r := !r + 1.
let acc = ref 0.
let count = ref 0.
s (fun v -> (inc count; if f v then inc acc); r (!acc/(!count), v))
let result =
diagonals
|> ratio (snd >> is_prime)
|> tryFind (fun (r, (_, v)) -> v > 1 && r < 0.1)
I am trying to make a program to compute the length of the Collatz sequence on all numbers from 1 to 100. Basically if I have an odd number I have to multiply it by 3 and add 1(n*3+1), and if I have a even number I need to divide it by 2(n/2) and then keep doing it untill it gets to 1 and in the end print out the count of times the number was either divided by 2 or multiplyed by 3 and added 1.
Here is what i have so far:
let stevec = ref 0;
let n = ref 1;
for i = 1 to 100 do
n := i;
while !n != 1 do
if (n mod 2 = 0) then
stevec := !stevec + 1;
n := !n / 2;
if (n mod 2 = 1) then
stevec := !stevec + 1;
n := 3 * !n + 1;
done
print_int (stevec);
done;;
After I run the code I get a syntax error and the print_int get underlined so i guess there is a problem with that but I'm not even sure about that.
There are several problems with your code, so let's take a look at it.
let stevec = ref 0;
let n = ref 1;
You shouldn't write that kind of code, as ; is an expression separator (and you are using it here as a declaration separator).
The right approach depends on wether you want your declaration to be local or toplevel.
(* local declaration *)
let stevec = ref 0 in
let n = ref 1 in
(* toplevel declaration *)
let stevec = ref 0;;
let n = ref 1;;
Then you typed while !n != 1 do. This shouldn't be used as you do physical inequality between your integers, whereas you want structural equality. Well, it will work too because of OCaml's behavior on integers but good practice requires you to use <> instead of !=.
Now let's look at your loop's body:
if (!n mod 2 = 0) then
stevec := !stevec + 1;
n := !n / 2;
if (!n mod 2 = 1) then
stevec := !stevec + 1;
n := 3 * !n + 1;
Notice the absence of any fi or closing bracket? That's because in OCaml, only the next expression after the then is executed. And precedence over ; doesn't go as you want it to. You can use parens or the more explicit begin ... end construction. To prove that the begin ... end works, I replaced your second test by a else statement.
if (!n mod 2 = 0) then
begin
stevec := !stevec + 1;
n := !n / 2;
end
else
begin
stevec := !stevec + 1;
n := 3 * !n + 1;
end
Finally while ... done being itself an expression, you should put a ; at the end of it.
And that's how you remove the errors from your code.
Yet...
This is clearly not the "right way" to do it in OCaml. The main perk of FP is its closeness to maths and you are here trying to define a mathematical function. So let's do this in a functionnal way:
let is_even x = (x mod 2) = 0;;
let rec collatz counter n =
if n = 1
then counter
else collatz (counter+1) (if is_even n then n/2 else 3*n+1);;
let () =
for i = 1 to 100 do
print_int (collatz 0 i);
print_newline ();
done;;
Doesn't that look nicer? Feel free to ask for any clarification of course.
I'm writing a code that can find the median of a list, and I cannot use rec and should use List.fold_left/right. I wrote the following code, which should work.
It finds the length of the list, if it's an odd number like 5, then it sets len1, len2 to 2, 3, if it's an even number like 6, then it sets len1, len2 to 2, 3.
Then for each member in the list I match the number of those elements that are less than it.
However, the following pattern matching always math lessNum elmt to len1 - can someone tell me why it is so?
let median (lst : int list) : float option =
let len = List.length lst in
if lst = [] then None
else
let len1, len2 = (len - 1) / 2, (len + 1) / 2 in
let lessNum a =
List.length (List.find_all (fun n -> n < a) lst) in
let answer = List.fold_left (fun accm elmt ->
match (lessNum elmt) with
| len1 -> accm + elmt
| len2 -> failwith "len2"
| _ -> failwith "other"
) 0 lst in
if len mod 2 = 0
then Some ((float_of_int answer) /. 2.0)
else Some (float_of_int answer)
An identifier appearing in a pattern always matches, and binds the corresponding value to the identifier. Any current value of the identifier doesn't matter at all: the pattern causes a new binding, i.e., it gives a new value to the identifier (just inside the match).
# let a = 3;;
val a : int = 3
# match 5 with a -> a;;
- : int = 5
# a;;
- : int = 3
#
So, your match statement isn't doing what you think it is. You'll probably have to use an if for that part of your code.
Update
Here's how to use an association list to approximate the function f in your followup question:
let f x = List.assoc x [(pat1, ans1); (pat2, ans2)]
This will raise a Not_found exception if x is not equal to pat1 or pat2.
(I think your Python code is missing return.)