Mutual recursion

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Task
Mutual recursion
You are encouraged to solve this task according to the task description, using any language you may know.

Two functions are said to be mutually recursive if the first calls the second, and in turn the second calls the first.

Write two mutually recursive functions that compute members of the Hofstadter Female and Male sequences defined as:


\begin{align}
F(0)&=1\ ;\ M(0)=0 \\
F(n)&=n-M(F(n-1)), \quad n>0 \\
M(n)&=n-F(M(n-1)), \quad n>0.
\end{align}


(If a language does not allow for a solution using mutually recursive functions then state this rather than give a solution by other means).

Contents

[edit] ACL2

(mutual-recursion
(defun f (n)
(declare (xargs :mode :program))
(if (zp n)
1
(- n (m (f (1- n))))))
 
(defun m (n)
(declare (xargs :mode :program))
(if (zp n)
0
(- n (f (m (1- n)))))))

[edit] Ada

with Ada.Text_Io; use Ada.Text_Io; 
procedure Mutual_Recursion is
function M(N : Integer) return Integer;
function F(N : Integer) return Integer is
begin
if N = 0 then
return 1;
else
return N - M(F(N - 1));
end if;
end F;
function M(N : Integer) return Integer is
begin
if N = 0 then
return 0;
else
return N - F(M(N-1));
end if;
end M;
begin
for I in 0..19 loop
Put_Line(Integer'Image(F(I)));
end loop;
New_Line;
for I in 0..19 loop
Put_Line(Integer'Image(M(I)));
end loop;
end Mutual_recursion;
Works with: Ada 2012
with Ada.Text_Io; use Ada.Text_Io;
procedure Mutual_Recursion is
function M(N: Natural) return Natural;
function F(N: Natural) return Natural;
 
function M(N: Natural) return Natural is
(if N = 0 then 0 else N – F(M(N–1)));
 
function F(N: Natural) return Natural is
(if N =0 then 1 else N – M(F(N–1)));
begin
for I in 0..19 loop
Put_Line(Integer'Image(F(I)));
end loop;
New_Line;
for I in 0..19 loop
Put_Line(Integer'Image(M(I)));
end loop;
 
end Mutual_recursion;

[edit] Aime

Translation of: C
integer F(integer n);
integer M(integer n);
 
integer F(integer n)
{
integer r;
if (n) {
r = n - M(F(n - 1));
} else {
r = 1;
}
return r;
}
 
integer M(integer n)
{
integer r;
if (n) {
r = n - F(M(n - 1));
} else {
r = 0;
}
return r;
}
 
integer main(void)
{
integer i;
i = 0;
while (i < 20) {
o_winteger(3, F(i));
i += 1;
}
o_byte('\n');
i = 0;
while (i < 20) {
o_winteger(3, M(i));
i += 1;
}
o_byte('\n');
return 0;
}

[edit] ALGOL 68

Translation of: C
Works with: ALGOL 68 version Standard - no extensions to language used
Works with: ALGOL 68G version Any - tested with release mk15-0.8b.fc9.i386
Works with: ELLA ALGOL 68 version Any (with appropriate job cards) - tested with release 1.8.8d.fc9.i386
PROC (INT)INT m; # ONLY required for ELLA ALGOL 68RS - an official subset OF full ALGOL 68 #
 
PROC f = (INT n)INT:
IF n = 0 THEN 1
ELSE n - m(f(n-1)) FI;
 
m := (INT n)INT:
IF n = 0 THEN 0
ELSE n - f(m(n-1)) FI;
 
main:
(
FOR i FROM 0 TO 19 DO
print(whole(f(i),-3))
OD;
new line(stand out);
FOR i FROM 0 TO 19 DO
print(whole(m(i),-3))
OD;
new line(stand out)
)

Output:

  1  1  2  2  3  3  4  5  5  6  6  7  8  8  9  9 10 11 11 12
  0  0  1  2  2  3  4  4  5  6  6  7  7  8  9  9 10 11 11 12

[edit] AutoHotkey

Loop 20
i := A_Index-1, t .= "`n" i "`t " M(i) "`t " F(i)
MsgBox x`tmale`tfemale`n%t%
 
F(n) {
Return n ? n - M(F(n-1)) : 1
}
 
M(n) {
Return n ? n - F(M(n-1)) : 0
}
Translation of: C

This one is an alternative to the above.

main()
Return
 
F(n)
{
If (n == 0)
Return 1
Else
Return n - M(F(n-1))
}
 
M(n)
{
If (n == 0)
Return 0
Else
Return n - F(M(n-1)) ;
}
 
main()
{
i = 0
While, i < 20
{
male .= M(i) . "`n"
female .= F(i) . "`n"
i++
}
MsgBox % "male:`n" . male
MsgBox % "female:`n" . female
}

[edit] AWK

In AWK it is enough that both functions are defined somewhere. It matters not whether the BEGIN block is before or after the function definitions.

function F(n)
{
if ( n == 0 ) return 1;
return n - M(F(n-1))
}
 
function M(n)
{
if ( n == 0 ) return 0;
return n - F(M(n-1))
}
 
BEGIN {
for(i=0; i < 20; i++) {
printf "%3d ", F(i)
}
print ""
for(i=0; i < 20; i++) {
printf "%3d ", M(i)
}
print ""
}

[edit] BASIC

Works with: QBasic
DECLARE FUNCTION f! (n!)
DECLARE FUNCTION m! (n!)
 
FUNCTION f! (n!)
IF n = 0 THEN
f = 1
ELSE
f = m(f(n - 1))
END IF
END FUNCTION
 
FUNCTION m! (n!)
IF n = 0 THEN
m = 0
ELSE
m = f(m(n - 1))
END IF
END FUNCTION

[edit] BBC BASIC

      @% = 3 : REM Column width
PRINT "F sequence:"
FOR i% = 0 TO 20
PRINT FNf(i%) ;
NEXT
PRINT
PRINT "M sequence:"
FOR i% = 0 TO 20
PRINT FNm(i%) ;
NEXT
PRINT
END
 
DEF FNf(n%) IF n% = 0 THEN = 1 ELSE = n% - FNm(FNf(n% - 1))
 
DEF FNm(n%) IF n% = 0 THEN = 0 ELSE = n% - FNf(FNm(n% - 1))

Output:

F sequence:
  1  1  2  2  3  3  4  5  5  6  6  7  8  8  9  9 10 11 11 12 13
M sequence:
  0  0  1  2  2  3  4  4  5  6  6  7  7  8  9  9 10 11 11 12 12

[edit] Bc

define f(n) {
if ( n == 0 ) return(1);
return(n - m(f(n-1)));
}
 
define m(n) {
if ( n == 0 ) return(0);
return(n - f(m(n-1)));
}
Works with: GNU bc
Works with: OpenBSD bc

POSIX bc doesn't have the print statement.

/* GNU bc */
for(i=0; i < 19; i++) {
print f(i); print " ";
}
print "\n";
for(i=0; i < 19; i++) {
print m(i); print " ";
}
print "\n";

[edit] Bracmat

 (F=.!arg:0&1|!arg+-1*M$(F$(!arg+-1)));
(M=.!arg:0&0|!arg+-1*F$(M$(!arg+-1)));
 
-1:?n&whl'(!n+1:~>20:?n&put$(F$!n " "))&put$\n
1 1 2 2 3 3 4 5 5 6 6 7 8 8 9 9 10 11 11 12 13
 
-1:?n&whl'(!n+1:~>20:?n&put$(M$!n " "))&put$\n
0 0 1 2 2 3 4 4 5 6 6 7 7 8 9 9 10 11 11 12 12

[edit] Brat

female = null #yes, this is necessary
 
male = { n |
true? n == 0
{ 0 }
{ n - female male(n - 1) }
}
 
female = { n |
true? n == 0
{ 1 }
{ n - male female(n - 1 ) }
}
 
p 0.to(20).map! { n | female n }
p 0.to(20).map! { n | male n }

[edit] C

To let C see functions that will be used, it is enough to declare them. Normally this is done in a header file; in this example we do it directly in the code. If we do not declare them explicitly, they get an implicit declaration (if implicit declaration matches the use, everything's fine; but it is better however to write an explicit declaration)

#include <stdio.h>
#include <stdlib.h>
 
/* let us declare our functions; indeed here we need
really only M declaration, so that F can "see" it
and the compiler won't complain with a warning */

int F(const int n);
int M(const int n);
 
int F(const int n)
{
return (n == 0) ? 1 : n - M(F(n - 1));
}
 
int M(const int n)
{
return (n == 0) ? 0 : n - F(M(n - 1));
}
 
int main(void)
{
int i;
for (i = 0; i < 20; i++)
printf("%2d ", F(i));
printf("\n");
for (i = 0; i < 20; i++)
printf("%2d ", M(i));
printf("\n");
return EXIT_SUCCESS;
}

[edit] C++

C++ has prior declaration rules similar to those stated above for C, if we would use two functions. Instead here we define M and F as static (class) methods of a class, and specify the bodies inline in the declaration of the class. Inlined methods in the class can still call other methods or access fields in the class, no matter what order they are declared in, without any additional pre-declaration. This is possible because all the possible methods and fields are declared somewhere in the class declaration, which is known the first time the class declaration is parsed.

#include <iostream>
#include <vector>
#include <iterator>
 
class Hofstadter
{
public:
static int F(int n) {
if ( n == 0 ) return 1;
return n - M(F(n-1));
}
static int M(int n) {
if ( n == 0 ) return 0;
return n - F(M(n-1));
}
};
 
using namespace std;
 
int main()
{
int i;
vector<int> ra, rb;
 
for(i=0; i < 20; i++) {
ra.push_back(Hofstadter::F(i));
rb.push_back(Hofstadter::M(i));
}
copy(ra.begin(), ra.end(),
ostream_iterator<int>(cout, " "));
cout << endl;
copy(rb.begin(), rb.end(),
ostream_iterator<int>(cout, " "));
cout << endl;
return 0;
}

The following version shows better what's going on and why we seemingly didn't need pre-declaration (like C) when "encapsulating" the functions as static (class) methods.

This version is equivalent to the above but does not inline the definition of the methods into the definition of the class. Here the method declarations in the class definition serves as the "pre-declaration" for the methods, as in C.

class Hofstadter
{
public:
static int F(int n);
static int M(int n);
};
 
int Hofstadter::F(int n)
{
if ( n == 0 ) return 1;
return n - M(F(n-1));
}
 
int Hofstadter::M(int n)
{
if ( n == 0 ) return 0;
return n - F(M(n-1));
}

[edit] C#

namespace RosettaCode {
class Hofstadter {
static public int F(int n) {
int result = 1;
if (n > 0) {
result = n - M(F(n-1));
}
 
return result;
}
 
static public int M(int n) {
int result = 0;
if (n > 0) {
result = n - F(M(n - 1));
}
 
return result;
}
}
}

[edit] Clojure

(declare F) ; forward reference 
 
(defn M [n]
(if (zero? n)
0
(- n (F (M (dec n))))))
 
(defn F [n]
(if (zero? n)
1
(- n (M (F (dec n))))))

[edit] CoffeeScript

 
F = (n) ->
if n is 0 then 1 else n - M F n - 1
 
M = (n) ->
if n is 0 then 0 else n - F M n - 1
 
console.log [0...20].map F
console.log [0...20].map M
 

output

 
> coffee mutual_recurse.coffee
[ 1, 1, 2, 2, 3, 3, 4, 5, 5, 6, 6, 7, 8, 8, 9, 9, 10, 11, 11, 12 ]
[ 0, 0, 1, 2, 2, 3, 4, 4, 5, 6, 6, 7, 7, 8, 9, 9, 10, 11, 11, 12 ]
 

[edit] Common Lisp

(defun m (n)
(if (zerop n)
0
(- n (f (m (- n 1))))))
 
(defun f (n)
(if (zerop n)
1
(- n (m (f (- n 1))))))

[edit] D

import std.stdio, std.algorithm, std.range;
 
int male(in int n) pure nothrow {
return n ? n - male(n - 1).female : 0;
}
 
int female(in int n) pure nothrow {
return n ? n - female(n - 1).male : 1;
}
 
void main() {
20.iota.map!female.writeln;
20.iota.map!male.writeln;
}
Output:
[1, 1, 2, 2, 3, 3, 4, 5, 5, 6, 6, 7, 8, 8, 9, 9, 10, 11, 11, 12]
[0, 0, 1, 2, 2, 3, 4, 4, 5, 6, 6, 7, 7, 8, 9, 9, 10, 11, 11, 12]

[edit] Déjà Vu

F n:
if n:
- n M F -- n
else:
1
 
M n:
if n:
- n F M -- n
else:
0
 
for i range 0 10:
!.( M i F i )
Output:
0 1 
0 1 
1 2 
2 2 
2 3 
3 3 
4 4 
4 5 
5 5 
6 6 
6 6 

[edit] Dart

int M(int n) => n==0?1:n-F(M(n-1));
int F(int n) => n==0?0:n-M(F(n-1));
 
main() {
String f="",m="";
for(int i=0;i<20;i++) {
m+="${M(i)} ";
f+="${F(i)} ";
}
print("M: $m");
print("F: $f");
}

[edit] Delphi

 
unit Hofstadter;
 
interface
 
type
THofstadterFemaleMaleSequences = class
public
class function F(n: Integer): Integer;
class function M(n: Integer): Integer;
end;
 
implementation
 
class function THofstadterFemaleMaleSequences.F(n: Integer): Integer;
begin
Result:= 1;
if (n > 0) then
Result:= n - M(F(n-1));
end;
 
class function THofstadterFemaleMaleSequences.M(n: Integer): Integer;
begin
Result:= 0;
if (n > 0) then
Result:= n - F(M(n - 1));
end;
 
end.
 

[edit] E

In E, nouns (variable names) always refer to preceding definitions, so to have mutual recursion, either one must be forward-declared or we must use a recursive def construct. Either one of these is syntactic sugar for first binding the noun to an E promise (a reference with an undetermined target), then resolving the promise to the value.

Recursive def:

def [F, M] := [
fn n { if (n <=> 0) { 1 } else { n - M(F(n - 1)) } },
fn n { if (n <=> 0) { 0 } else { n - F(M(n - 1)) } },
]

Forward declaration:

def M
def F(n) { return if (n <=> 0) { 1 } else { n - M(F(n - 1)) } }
bind M(n) { return if (n <=> 0) { 0 } else { n - F(M(n - 1)) } }

def M binds M to a promise, and stashes the resolver for that promise where bind can get to it. When def F... is executed, the function F closes over the promise which is the value of M. bind M... uses the resolver to resolve M to the provided definition. The recursive def operates similarly, except that it constructs promises for every variable on the left side ([F, M]), executes the right side ([fn ..., fn ...]) and collects the values, then resolves each promise to its corresponding value.

But you don't have to worry about that to use it.

[edit] Elixir

 
defmodule MutualRecursion do
def f(0), do: 1
def f(n), do: n - m(f(n - 1))
def m(0), do: 0
def m(n), do: n - f(m(n - 1))
end
 

[edit] Erlang

-module(mutrec).
-export([mutrec/0, f/1, m/1]).
 
f(0) -> 1;
f(N) -> N - m(f(N-1)).
 
m(0) -> 0;
m(N) -> N - f(m(N-1)).
 
mutrec() -> lists:map(fun(X) -> io:format("~w ", [f(X)]) end, lists:seq(0,19)),
io:format("~n", []),
lists:map(fun(X) -> io:format("~w ", [m(X)]) end, lists:seq(0,19)),
io:format("~n", []).

[edit] Euphoria

integer idM, idF
 
function F(integer n)
if n = 0 then
return 1
else
return n - call_func(idM,{F(n-1)})
end if
end function
 
idF = routine_id("F")
 
function M(integer n)
if n = 0 then
return 0
else
return n - call_func(idF,{M(n-1)})
end if
end function
 
idM = routine_id("M")

[edit] F#

let rec f n =
match n with
| 0 -> 1
| _ -> n - (m (f (n-1)))
and m n =
match n with
| 0 -> 0
| _ -> n - (f (m (n-1)))

Like OCaml, the let rec f .. and m ... construct indicates that the functions call themselves (rec) and each other (and).

[edit] Factor

In Factor, if you need a word before it's defined, you have to DEFER: it.

DEFER: F
: M ( n -- n' ) dup 0 = [ dup 1 - M F - ] unless ;
: F ( n -- n' ) dup 0 = [ drop 1 ] [ dup 1 - F M - ] if ;

[edit] FALSE

[$[$1-f;!m;!-1-]?1+]f:
[$[$1-m;!f;!- ]? ]m:
[0[$20\>][\$@$@!." "1+]#%%]t:
f; t;!"
"m; t;!

[edit] Fantom

 
class Main
{
static Int f (Int n)
{
if (n <= 0) // ensure n > 0
return 1
else
return n - m(f(n-1))
}
 
static Int m (Int n)
{
if (n <= 0) // ensure n > 0
return 0
else
return n - f(m(n-1))
}
 
public static Void main ()
{
50.times |Int n| { echo (f(n)) }
}
}
 

[edit] Forth

Forward references required for mutual recursion may be set up using DEFER.

defer m
 
: f ( n -- n )
dup 0= if 1+ exit then
dup 1- recurse m - ;
 
:noname ( n -- n )
dup 0= if exit then
dup 1- recurse f - ;
is m
 
: test ( xt n -- ) cr 0 do i over execute . loop drop ;
 
' m defer@ 20 test \ 0 0 1 2 2 3 4 4 5 6 6 7 7 8 9 9 10 11 11 12
' f 20 test \ 1 1 2 2 3 3 4 5 5 6 6 7 8 8 9 9 10 11 11 12

[edit] Fortran

As long as the code of the two functions is inside the same "block" (module or program) we don't need special care. Otherwise, we should "load" at least the interface of the other function (each module will load mutually the other; of course the compiler won't enter in a infinite loop), e.g. by using a "use" (we do that if M and F function are inside different modules)

Works with: Fortran version 95 and later
module MutualRec
implicit none
contains
pure recursive function m(n) result(r)
integer :: r
integer, intent(in) :: n
if ( n == 0 ) then
r = 0
return
end if
r = n - f(m(n-1))
end function m
 
pure recursive function f(n) result(r)
integer :: r
integer, intent(in) :: n
if ( n == 0 ) then
r = 1
return
end if
r = n - m(f(n-1))
end function f
 
end module

I've added the attribute pure so that we can use them in a forall statement.

program testmutrec
use MutualRec
implicit none
 
integer :: i
integer, dimension(20) :: a = (/ (i, i=0,19) /), b = (/ (i, i=0,19) /)
integer, dimension(20) :: ra, rb
 
forall(i=1:20)
ra(i) = m(a(i))
rb(i) = f(b(i))
end forall
 
write(*,'(20I3)') rb
write(*,'(20I3)') ra
 
end program testmutrec

[edit] Go

It just works. No special pre-declaration is necessary.

package main
import "fmt"
 
func F(n int) int {
if n == 0 { return 1 }
return n - M(F(n-1))
}
 
func M(n int) int {
if n == 0 { return 0 }
return n - F(M(n-1))
}
 
func main() {
for i := 0; i < 20; i++ {
fmt.Printf("%2d ", F(i))
}
fmt.Println()
for i := 0; i < 20; i++ {
fmt.Printf("%2d ", M(i))
}
fmt.Println()
}

[edit] Groovy

Solution:

def f, m  // recursive closures must be declared before they are defined
f = { n -> n == 0 ? 1 : n - m(f(n-1)) }
m = { n -> n == 0 ? 0 : n - f(m(n-1)) }

Test program:

println 'f(0..20): ' + (0..20).collect { f(it) }
println 'm(0..20): ' + (0..20).collect { m(it) }

Output:

f(0..20): [1, 1, 2, 2, 3, 3, 4, 5, 5, 6, 6, 7, 8, 8, 9, 9, 10, 11, 11, 12, 13]
m(0..20): [0, 0, 1, 2, 2, 3, 4, 4, 5, 6, 6, 7, 7, 8, 9, 9, 10, 11, 11, 12, 12]

[edit] Haskell

Haskell's definitions constructs (at the top level, or inside a let or where construct) are always mutually-recursive:

f 0 = 1
f n | n > 0 = n - m (f $ n-1)
 
m 0 = 0
m n | n > 0 = n - f (m $ n-1)
 
main = do
print $ map f [0..19]
print $ map m [0..19]

[edit] Icon and Unicon

procedure main(arglist)
every write(F(!arglist)) # F of all arguments
end
 
procedure F(n)
if integer(n) >= 0 then
return (n = 0, 1) | n - M(F(n-1))
end
 
procedure M(n)
if integer(n) >= 0 then
return (0 = n) | n - F(M(n-1))
end

[edit] J

F =: 1:`(-  M @ $: @ <:) @.* M."0
M =: 0:`(- F @ $: @ <:) @.* M."0

Example use:

   F i. 20
1 1 2 2 3 3 4 5 5 6 6 7 8 8 9 9 10 11 11 12

[edit] Java

Translation of: C
public static int f(final int n)
{
return n == 0 ? 1 : n - m(f(n - 1));
}
 
public static int m(final int n)
{
return n == 0 ? 0 : n - f(m(n - 1));
}
 
public static void main(final String args[])
{
for (int i = 0; i < 20; i++)
System.out.println(f(i));
System.out.println();
for (i = 0; i < 20; i++)
System.out.println(m(i));
}

[edit] JavaScript

function F(n)
{
return n === 0 ? 1 : n - M(F(n - 1));
}
 
function M(n)
{
return n === 0 ? 0 : n - F(M(n - 1));
}
 
var
out = {F: [], M: []},
i;
for (i = 0; i < 20; i++)
{
out.F.push(F(i));
out.M.push(M(i));
}
print(out.F + "\n" + out.M);

outputs:

1,1,2,2,3,3,4,5,5,6,6,7,8,8,9,9,10,11,11,12
0,0,1,2,2,3,4,4,5,6,6,7,7,8,9,9,10,11,11,12

[edit] Julia

F(n) = n < 1 ? one(n) : n - M(F(n - 1))
M(n) = n < 1 ? zero(n) : n - F(M(n - 1))

Output:

julia> [F(i) for i = 0:19], [M(i) for i = 0:19]
([1,1,2,2,3,3,4,5,5,6,6,7,8,8,9,9,10,11,11,12],[0,0,1,2,2,3,4,4,5,6,6,7,7,8,9,9,10,11,11,12])

[edit] Liberty BASIC

 
print "F sequence."
for i = 0 to 20
print f(i);" ";
next
print
print "M sequence."
for i = 0 to 20
print m(i);" ";
next
 
end
 
function f(n)
if n = 0 then
f = 1
else
f = n - m(f(n - 1))
end if
end function
 
function m(n)
if n = 0 then
m = 0
else
m = n - f(m(n - 1))
end if
end function
 
Output:
F sequence.
1 1 2 2 3 3 4 5 5 6 6 7 8 8 9 9 10 11 11 12 13
M sequence.
0 0 1 2 2 3 4 4 5 6 6 7 7 8 9 9 10 11 11 12 12

[edit]

Like Lisp, symbols in Logo are late-bound so no special syntax is required for forward references.

to m :n
if 0 = :n [output 0]
output :n - f m :n-1
end
to f :n
if 0 = :n [output 1]
output :n - m f :n-1
end
 
show cascade 20 [lput m #-1 ?] []
[1 1 2 2 3 3 4 5 5 6 6 7 8 8 9 9 10 11 11 12]
show cascade 20 [lput f #-1 ?] []
[0 0 1 2 2 3 4 4 5 6 6 7 7 8 9 9 10 11 11 12]

[edit] LSL

To test it yourself; rez a box on the ground, and add the following as a New Script.

integer iDEPTH = 100;
integer f(integer n) {
if(n==0) {
return 1;
} else {
return n-m(f(n - 1));
}
}
integer m(integer n) {
if(n==0) {
return 0;
} else {
return n-f(m(n - 1));
}
}
default {
state_entry() {
integer x = 0;
string s = "";
for(x=0 ; x<iDEPTH ; x++) {
s += (string)(f(x))+" ";
}
llOwnerSay(llList2CSV(llParseString2List(s, [" "], [])));
s = "";
for(x=0 ; x<iDEPTH ; x++) {
s += (string)(m(x))+" ";
}
llOwnerSay(llList2CSV(llParseString2List(s, [" "], [])));
}
}

Output:

1, 1, 2, 2, 3, 3, 4, 5, 5, 6, 6, 7, 8, 8, 9, 9, 10, 11, 11, 12, 13, 13, 14, 14, 15, 16, 16, 17, 17, 18, 19, 19, 20, 21, 21, 22, 22, 23, 24, 24, 25, 25, 26, 27, 27, 28, 29, 29, 30, 30, 31, 32, 32, 33, 34, 34, 35, 35, 36, 37, 37, 38, 38, 39, 40, 40, 41, 42, 42, 43, 43, 44, 45, 45, 46, 46, 47, 48, 48, 49, 50, 50, 51, 51, 52, 53, 53, 54, 55, 55, 56, 56, 57, 58, 58, 59, 59, 60, 61, 61
0, 0, 1, 2, 2, 3, 4, 4, 5, 6, 6, 7, 7, 8, 9, 9, 10, 11, 11, 12, 12, 13, 14, 14, 15, 16, 16, 17, 17, 18, 19, 19, 20, 20, 21, 22, 22, 23, 24, 24, 25, 25, 26, 27, 27, 28, 29, 29, 30, 30, 31, 32, 32, 33, 33, 34, 35, 35, 36, 37, 37, 38, 38, 39, 40, 40, 41, 42, 42, 43, 43, 44, 45, 45, 46, 46, 47, 48, 48, 49, 50, 50, 51, 51, 52, 53, 53, 54, 54, 55, 56, 56, 57, 58, 58, 59, 59, 60, 61, 61

[edit] Lua

 
function m(n) return n > 0 and n - f(m(n-1)) or 0 end
function f(n) return n > 0 and n - m(f(n-1)) or 1 end

It is important to note, that if m and f are to be locally scoped functions rather than global, that they would need to be forward declared:

 
local m,n
function m(n) return n > 0 and n - f(m(n-1)) or 0 end
function f(n) return n > 0 and n - m(f(n-1)) or 1 end

[edit] M4

define(`female',`ifelse(0,$1,1,`eval($1 - male(female(decr($1))))')')dnl
define(`male',`ifelse(0,$1,0,`eval($1 - female(male(decr($1))))')')dnl
define(`loop',`ifelse($1,$2,,`$3($1) loop(incr($1),$2,`$3')')')dnl
loop(0,20,`female')
loop(0,20,`male')

[edit] Mathematica

Without caching:

f[0]:=1
m[0]:=0
f[n_]:=n-m[f[n-1]]
m[n_]:=n-f[m[n-1]]

With caching:

f[0]:=1
m[0]:=0
f[n_]:=f[n]=n-m[f[n-1]]
m[n_]:=m[n]=n-f[m[n-1]]

Example finding f(1) to f(30) and m(1) to m(30):

m /@ Range[30] 
f /@ Range[30]

gives back:

{0,1,2,2,3,4,4,5,6,6,7,7,8,9,9,10,11,11,12,12,13,14,14,15,16,16,17,17,18,19}
{1,2,2,3,3,4,5,5,6,6,7,8,8,9,9,10,11,11,12,13,13,14,14,15,16,16,17,17,18,19}

[edit] MATLAB

female.m:

function Fn = female(n)
 
if n == 0
Fn = 1;
return
end
 
Fn = n - male(female(n-1));
end

male.m:

function Mn = male(n)
 
if n == 0
Mn = 0;
return
end
 
Mn = n - female(male(n-1));
end

Sample Output:

>> n = (0:10);
>> arrayfun(@female,n)
 
ans =
 
1 1 2 2 3 3 4 5 5 6 6
 
>> arrayfun(@male,n)
 
ans =
 
0 0 1 2 2 3 4 4 5 6 6

[edit] Maxima

f[0]: 1$
m[0]: 0$
f[n] := n - m[f[n - 1]]$
m[n] := n - f[m[n - 1]]$
 
makelist(f[i], i, 0, 10);
[1, 1, 2, 2, 3, 3, 4, 5, 5, 6, 6]
 
makelist(m[i], i, 0, 10);
[0, 0, 1, 2, 2, 3, 4, 4, 5, 6, 6]
 
remarray(m, f)$
 
f(n) := if n = 0 then 1 else n - m(f(n - 1))$
m(n) := if n = 0 then 0 else n - f(m(n - 1))$
 
makelist(f(i), i, 0, 10);
[1, 1, 2, 2, 3, 3, 4, 5, 5, 6, 6]
 
makelist(m(i), i, 0, 10);
[0, 0, 1, 2, 2, 3, 4, 4, 5, 6, 6]
 
remfunction(f, m)$

[edit] Mercury

 
:- module mutual_recursion.
:- interface.
 
:- import_module io.
:- pred main(io::di, io::uo) is det.
 
:- implementation.
:- import_module int, list.
 
main(!IO) :-
io.write(list.map(f, 0..19), !IO), io.nl(!IO),
io.write(list.map(m, 0..19), !IO), io.nl(!IO).
 
:- func f(int) = int.
 
f(N) = ( if N = 0 then 1 else N - m(f(N - 1)) ).
 
:- func m(int) = int.
 
m(N) = ( if N = 0 then 0 else N - f(m(N - 1)) ).
 

[edit] MMIX

	LOC	Data_Segment
 
GREG @
NL BYTE #a,0
GREG @
buf OCTA 0,0
 
t IS $128
Ja IS $127
 
LOC #1000
 
GREG @
// print 2 digits integer with trailing space to StdOut
// reg $3 contains int to be printed
bp IS $71
0H GREG #0000000000203020
prtInt STO 0B,buf % initialize buffer
LDA bp,buf+7 % points after LSD
% REPEAT
1H SUB bp,bp,1 % move buffer pointer
DIV $3,$3,10 % divmod (x,10)
GET t,rR % get remainder
INCL t,'0' % make char digit
STB t,bp % store digit
PBNZ $3,1B % UNTIL no more digits
LDA $255,bp
TRAP 0,Fputs,StdOut % print integer
GO Ja,Ja,0 % 'return'

// Female function

F GET $1,rJ % save return addr
PBNZ $0,1F % if N != 0 then F N
INCL $0,1 % F 0 = 1
PUT rJ,$1 % restore return addr
POP 1,0 % return 1
1H SUBU $3,$0,1 % N1 = N - 1
PUSHJ $2,F % do F (N - 1)
ADDU $3,$2,0 % place result in arg. reg.
PUSHJ $2,M % do M F ( N - 1)
PUT rJ,$1 % restore ret addr
SUBU $0,$0,$2
POP 1,0 % return N - M F ( N - 1 )

// Male function

M GET $1,rJ
PBNZ $0,1F
PUT rJ,$1
POP 1,0 % return M 0 = 0
1H SUBU $3,$0,1
PUSHJ $2,M
ADDU $3,$2,0
PUSHJ $2,F
PUT rJ,$1
SUBU $0,$0,$2
POP 1,0 $ return N - F M ( N - 1 )

// do a female run

Main SET $1,0 % for (i=0; i<25; i++){
1H ADDU $4,$1,0 %
PUSHJ $3,F % F (i)
GO Ja,prtInt % print F (i)
INCL $1,1
CMP t,$1,25
PBNZ t,1B % }
LDA $255,NL
TRAP 0,Fputs,StdOut
// do a male run
SET $1,0 % for (i=0; i<25; i++){
1H ADDU $4,$1,0 %
PUSHJ $3,M % M (i)
GO Ja,prtInt % print M (i)
INCL $1,1
CMP t,$1,25
PBNZ t,1B % }
LDA $255,NL
TRAP 0,Fputs,StdOut
TRAP 0,Halt,0

Output:

~/MIX/MMIX/Rosetta> mmix mutualrecurs1
1 1 2 2 3 3 4 5 5 6 6 7 8 8 9 9 10 11 11 12 13 13 14 14 15
0 0 1 2 2 3 4 4 5 6 6 7 7 8 9 9 10 11 11 12 12 13 14 14 15

[edit] Nemerle

using System;
using System.Console;
 
module Hofstadter
{
F(n : int) : int
{
|0 => 1
|_ => n - M(F(n - 1))
}
 
M(n : int) : int
{
|0 => 0
|_ => n - F(M(n - 1))
}
 
Main() : void
{
foreach (n in [0 .. 20]) Write("{0} ", F(n));
WriteLine();
foreach (n in [0 .. 20]) Write("{0} ", M(n));
}
}

[edit] Nimrod

proc m(n): int
 
proc f(n): int =
if n == 0: 1
else: n - m(f(n-1))
 
proc m(n): int =
if n == 0: 0
else: n - f(m(n-1))
 
for i in 1 .. 10:
echo f(i)
echo m(i)

[edit] Objective-C

Objective-C has prior declaration rules similar to those stated above for C, for C-like types. In this example we show the use of a two class method; this works since we need an interface block that is like declaration of functions in C code.

#import <Foundation/Foundation.h>
 
@interface Hofstadter : NSObject
+ (int)M: (int)n;
+ (int)F: (int)n;
@end
 
@implementation Hofstadter
+ (int)M: (int)n
{
if ( n == 0 ) return 0;
return n - [self F: [self M: (n-1)]];
}
+ (int)F: (int)n
{
if ( n == 0 ) return 1;
return n - [self M: [self F: (n-1)]];
}
@end
 
int main()
{
int i;
 
for(i=0; i < 20; i++) {
printf("%3d ", [Hofstadter F: i]);
}
printf("\n");
for(i=0; i < 20; i++) {
printf("%3d ", [Hofstadter M: i]);
}
printf("\n");
return 0;
}

[edit] Objeck

Translation of: C
 
class MutualRecursion {
function : Main(args : String[]) ~ Nil {
for(i := 0; i < 20; i+=1;) {
f(i)->PrintLine();
};
"---"->PrintLine();
for (i := 0; i < 20; i+=1;) {
m(i)->PrintLine();
};
}
 
function : f(n : Int) ~ Int {
return n = 0 ? 1 : n - m(f(n - 1));
}
 
function : m(n : Int) ~ Int {
return n = 0 ? 0 : n - f(m(n - 1));
}
}
 

[edit] OCaml

let rec f = function
| 0 -> 1
| n -> n - m(f(n-1))
and m = function
| 0 -> 0
| n -> n - f(m(n-1))
;;

The let rec f ... and m ... construct indicates that the functions call themselves (rec) and each other (and).

[edit] Octave

We don't need to pre-declare or specify in some other way a function that will be defined later; but both must be declared before their use.
(The code is written to handle vectors, as the testing part shows)

function r = F(n)
for i = 1:length(n)
if (n(i) == 0)
r(i) = 1;
else
r(i) = n(i) - M(F(n(i)-1));
endif
endfor
endfunction
 
function r = M(n)
for i = 1:length(n)
if (n(i) == 0)
r(i) = 0;
else
r(i) = n(i) - F(M(n(i)-1));
endif
endfor
endfunction
# testing
ra = F([0:19]);
rb = M([0:19]);
disp(ra);
disp(rb);

[edit] Order

Since Order is powered by the C preprocessor, definitions follow the same rule as CPP macros: they can appear in any order relative to each other as long as all are defined before the ORDER_PP block that calls them.

#include <order/interpreter.h>
 
#define ORDER_PP_DEF_8f \
ORDER_PP_FN(8fn(8N, \
8if(8is_0(8N), \
1, \
8sub(8N, 8m(8f(8dec(8N)))))))

 
#define ORDER_PP_DEF_8m \
ORDER_PP_FN(8fn(8N, \
8if(8is_0(8N), \
0, \
8sub(8N, 8f(8m(8dec(8N)))))))

 
//Test
ORDER_PP(8for_each_in_range(8fn(8N, 8print(8f(8N))), 0, 19))
ORDER_PP(8for_each_in_range(8fn(8N, 8print(8m(8N))), 0, 19))

[edit] Oz

declare
fun {F N}
if N == 0 then 1
elseif N > 0 then N - {M {F N-1}}
end
end
 
fun {M N}
if N == 0 then 0
elseif N > 0 then N - {F {M N-1}}
end
end
in
{Show {Map {List.number 0 9 1} F}}
{Show {Map {List.number 0 9 1} M}}

[edit] PARI/GP

F(n)=if(n,n-M(F(n-1)),1)
M(n)=if(n,n-F(M(n-1)),0)

[edit] Pascal

In Pascal we need to pre-declare functions/procedures; to do so, the forward statement is used.

Program MutualRecursion;
 
{M definition comes after F which uses it}
function M(n : Integer) : Integer; forward;
 
function F(n : Integer) : Integer;
begin
if n = 0 then
F := 1
else
F := n - M(F(n-1));
end;
 
function M(n : Integer) : Integer;
begin
if n = 0 then
M := 0
else
M := n - F(M(n-1));
end;
 
var
i : Integer;
 
begin
for i := 0 to 19 do begin
write(F(i) : 4)
end;
writeln;
for i := 0 to 19 do begin
write(M(i) : 4)
end;
writeln;
end.

[edit] Perl

sub F { my $n = shift; $n ? $n - M(F($n-1)) : 1 }
sub M { my $n = shift; $n ? $n - F(M($n-1)) : 0 }
 
# Usage:
foreach my $sequence (\&F, \&M) {
print join(' ', map $sequence->($_), 0 .. 19), "\n";
}
Output:
1 1 2 2 3 3 4 5 5 6 6 7 8 8 9 9 10 11 11 12 
0 0 1 2 2 3 4 4 5 6 6 7 7 8 9 9 10 11 11 12 

[edit] Perl 6

A direct translation of the definitions of F and M:

multi F(0) { 1 }; multi M(0) { 0 }
multi F(\𝑛) { 𝑛 - M(F(𝑛 - 1)) }
multi M(\𝑛) { 𝑛 - F(M(𝑛 - 1)) }
 
say map &F, ^20;
say map &M, ^20;
Output:
1 1 2 2 3 3 4 5 5 6 6 7 8 8 9 9 10 11 11 12
0 0 1 2 2 3 4 4 5 6 6 7 7 8 9 9 10 11 11 12

[edit] PHP

<?php
function F($n)
{
if ( $n == 0 ) return 1;
return $n - M(F($n-1));
}
 
function M($n)
{
if ( $n == 0) return 0;
return $n - F(M($n-1));
}
 
$ra = array();
$rb = array();
for($i=0; $i < 20; $i++)
{
array_push($ra, F($i));
array_push($rb, M($i));
}
echo implode(" ", $ra) . "\n";
echo implode(" ", $rb) . "\n";
?>

[edit] PicoLisp

(de f (N)
(if (=0 N)
1
(- N (m (f (dec N)))) ) )
 
(de m (N)
(if (=0 N)
0
(- N (f (m (dec N)))) ) )

[edit] PL/I

 
test: procedure options (main);
 
M: procedure (n) returns (fixed) recursive; /* 8/1/2010 */
declare n fixed;
if n <= 0 then return (0);
else return ( n - F(M(n-1)) );
end M;
 
F: procedure (n) returns (fixed) recursive;
declare n fixed;
if n <= 0 then return (1);
else return ( n - M(F(n-1)) );
end F;
 
declare i fixed;
 
do i = 1 to 15;
put skip list ( F(i), M(i) );
end;
end test;
 

[edit] PostScript

 
/female{
/n exch def
n 0 eq
{1}
{
n n 1 sub female male sub
}ifelse
}def
 
/male{
/n exch def
n 0 eq
{0}
{
n n 1 sub male female sub
}ifelse
}def
 
Library: initlib
 
/F {
{
{0 eq} {pop 1} is?
{0 gt} {dup 1 sub F M sub} is?
} cond
}.
 
/M {
{
{0 eq} {pop 0} is?
{0 gt} {dup 1 sub M F sub} is?
} cond
}.
 
 

[edit] PowerShell

function F($n) {
if ($n -eq 0) { return 1 }
return $n - (M (F ($n - 1)))
}
 
function M($n) {
if ($n -eq 0) { return 0 }
return $n - (F (M ($n - 1)))
}

[edit] Prolog

female(0,1).
female(N,F) :- N>0,
N1 is N-1,
female(N1,R),
male(R, R1),
F is N-R1.
 
male(0,0).
male(N,F) :- N>0,
N1 is N-1,
male(N1,R),
female(R, R1),
F is N-R1.
Works with: GNU Prolog
flist(S) :- for(X, 0, S), female(X, R), format('~d ', [R]), fail.
mlist(S) :- for(X, 0, S), male(X, R), format('~d ', [R]), fail.

Testing

| ?- flist(19).
1 1 2 2 3 3 4 5 5 6 6 7 8 8 9 9 10 11 11 12 

no
| ?- mlist(19).
0 0 1 2 2 3 4 4 5 6 6 7 7 8 9 9 10 11 11 12

[edit] Pure

The Pure definitions very closely maps to the mathematical definitions.

F 0 = 1;
M 0 = 0;
F n = n - M(F(n-1)) if n>0;
M n = n - F(M(n-1)) if n>0;
> let females = map F (0..10); females;
[1,1,2,2,3,3,4,5,5,6,6]
> let males = map M (0..10); males;
[0,0,1,2,2,3,4,4,5,6,6]

[edit] PureBasic

Declare M(n)
 
Procedure F(n)
If n = 0
ProcedureReturn 1
ElseIf n > 0
ProcedureReturn n - M(F(n - 1))
EndIf
EndProcedure
 
Procedure M(n)
If n = 0
ProcedureReturn 0
ElseIf n > 0
ProcedureReturn n - F(M(n - 1))
EndIf
EndProcedure
 
Define i
If OpenConsole()
 
For i = 0 To 19
Print(Str(F(i)))
If i = 19
Continue
EndIf
Print(", ")
Next
 
PrintN("")
For i = 0 To 19
Print(Str(M(i)))
If i = 19
Continue
EndIf
Print(", ")
Next
 
Print(#CRLF$ + #CRLF$ + "Press ENTER to exit")
Input()
CloseConsole()
EndIf

Sample output:

1, 1, 2, 2, 3, 3, 4, 5, 5, 6, 6, 7, 8, 8, 9, 9, 10, 11, 11, 12
0, 0, 1, 2, 2, 3, 4, 4, 5, 6, 6, 7, 7, 8, 9, 9, 10, 11, 11, 12

[edit] Python

Works with: Python version 3.0
.
Works with: Python version 2.6

def F(n): return 1 if n == 0 else n - M(F(n-1))
def M(n): return 0 if n == 0 else n - F(M(n-1))
 
print ([ F(n) for n in range(20) ])
print ([ M(n) for n in range(20) ])

Output:

[1, 1, 2, 2, 3, 3, 4, 5, 5, 6, 6, 7, 8, 8, 9, 9, 10, 11, 11, 12]
[0, 0, 1, 2, 2, 3, 4, 4, 5, 6, 6, 7, 7, 8, 9, 9, 10, 11, 11, 12]

In python there is no need to pre-declare M for it to be used in the definition of F. (However M must be defined before F calls it).

[edit] R

F <- function(n) ifelse(n == 0, 1, n - M(F(n-1)))
M <- function(n) ifelse(n == 0, 0, n - F(M(n-1)))
print.table(lapply(0:19, M))
print.table(lapply(0:19, F))

[edit] REBOL

rebol [
Title: "Mutual Recursion"
Date: 2009-12-14
Author: oofoe
URL: http://rosettacode.org/wiki/Mutual_Recursion
References: [http://en.wikipedia.org/wiki/Hofstadter_sequence#Hofstadter_Female_and_Male_sequences]

]
 
f: func [
"Female."
n [integer!] "Value."
] [either 0 = n [1][n - m f n - 1]]
 
m: func [
"Male."
n [integer!] "Value."
] [either 0 = n [0][n - f m n - 1]]
 
fs: [] ms: [] for i 0 19 1 [append fs f i append ms m i]
print ["F:" mold fs crlf "M:" mold ms]

Output:

F: [1 1 2 2 3 3 4 5 5 6 6 7 8 8 9 9 10 11 11 12] 
M: [0 0 1 2 2 3 4 4 5 6 6 7 7 8 9 9 10 11 11 12]

[edit] Racket

#lang racket
(define (F n)
(if (>= 0 n)
1
(- n (M (F (sub1 n))))))
 
(define (M n)
(if (>= 0 n)
0
(- n (F (M (sub1 n))))))

[edit] REXX

[edit] vanilla

This version uses vertical formatting.

/*REXX program shows mutual recursion (via Hofstadter Male & Female seq)*/
parse arg lim .; if lim='' then lim=40; pad=left('',20)
 
do j=0 to lim; jj=Jw(j); ff=F(j); mm=M(j)
say pad 'F('jj") =" Jw(ff) pad 'M('jj") =" Jw(mm)
end /*j*/
exit /*stick a fork in it, we're done.*/
/*─────────────────────────────────────F, M, Jw subroutines────────────*/
F: procedure; parse arg n; if n==0 then return 1; return n-M(F(n-1))
M: procedure; parse arg n; if n==0 then return 0; return n-F(M(n-1))
Jw: return right(arg(1),length(lim)) /*right justifies # for nice look*/

output using the default input of:   40

                     F( 0) =  1                      M( 0) =  0
                     F( 1) =  1                      M( 1) =  0
                     F( 2) =  2                      M( 2) =  1
                     F( 3) =  2                      M( 3) =  2
                     F( 4) =  3                      M( 4) =  2
                     F( 5) =  3                      M( 5) =  3
                     F( 6) =  4                      M( 6) =  4
                     F( 7) =  5                      M( 7) =  4
                     F( 8) =  5                      M( 8) =  5
                     F( 9) =  6                      M( 9) =  6
                     F(10) =  6                      M(10) =  6
                     F(11) =  7                      M(11) =  7
                     F(12) =  8                      M(12) =  7
                     F(13) =  8                      M(13) =  8
                     F(14) =  9                      M(14) =  9
                     F(15) =  9                      M(15) =  9
                     F(16) = 10                      M(16) = 10
                     F(17) = 11                      M(17) = 11
                     F(18) = 11                      M(18) = 11
                     F(19) = 12                      M(19) = 12
                     F(20) = 13                      M(20) = 12
                     F(21) = 13                      M(21) = 13
                     F(22) = 14                      M(22) = 14
                     F(23) = 14                      M(23) = 14
                     F(24) = 15                      M(24) = 15
                     F(25) = 16                      M(25) = 16
                     F(26) = 16                      M(26) = 16
                     F(27) = 17                      M(27) = 17
                     F(28) = 17                      M(28) = 17
                     F(29) = 18                      M(29) = 18
                     F(30) = 19                      M(30) = 19
                     F(31) = 19                      M(31) = 19
                     F(32) = 20                      M(32) = 20
                     F(33) = 21                      M(33) = 20
                     F(34) = 21                      M(34) = 21
                     F(35) = 22                      M(35) = 22
                     F(36) = 22                      M(36) = 22
                     F(37) = 23                      M(37) = 23
                     F(38) = 24                      M(38) = 24
                     F(39) = 24                      M(39) = 24
                     F(40) = 25                      M(40) = 25

[edit] with memoization

This version uses memoization as well as a horizontal output format.

The optimization due to memoization is faster by many orders of magnitude.

/*REXX program shows mutual recursion (via Hofstadter Male & Female seq)*/
parse arg lim .; if lim=='' then lim=99 /*get or assume LIM.*/
hm.=; hm.0=0; hf.=; hf.0=1; Js=; Fs=; Ms=
 
do j=0 to lim; ff=F(j); mm=M(j)
Js=Js jW(j); Fs=Fs jw(ff); Ms=Ms jW(mm)
end /*j*/
say 'Js=' Js
say 'Fs=' Fs
say 'Ms=' Ms
exit /*stick a fork in it, we're done.*/
/*──────────────────────────────────one─liner subroutines──────────────────────────────*/
F: procedure expose hm. hf.; parse arg n; if hf.n=='' then hf.n=n-M(F(n-1)); return hf.n
M: procedure expose hm. hf.; parse arg n; if hm.n=='' then hm.n=n-F(M(n-1)); return hm.n
Jw: return right(arg(1),length(lim)) /*right justifies # for nice look*/

output using the default input of:   99

Js=  0  1  2  3  4  5  6  7  8  9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99
Fs=  1  1  2  2  3  3  4  5  5  6  6  7  8  8  9  9 10 11 11 12 13 13 14 14 15 16 16 17 17 18 19 19 20 21 21 22 22 23 24 24 25 25 26 27 27 28 29 29 30 30 31 32 32 33 34 34 35 35 36 37 37 38 38 39 40 40 41 42 42 43 43 44 45 45 46 46 47 48 48 49 50 50 51 51 52 53 53 54 55 55 56 56 57 58 58 59 59 60 61 61
Ms=  0  0  1  2  2  3  4  4  5  6  6  7  7  8  9  9 10 11 11 12 12 13 14 14 15 16 16 17 17 18 19 19 20 20 21 22 22 23 24 24 25 25 26 27 27 28 29 29 30 30 31 32 32 33 33 34 35 35 36 37 37 38 38 39 40 40 41 42 42 43 43 44 45 45 46 46 47 48 48 49 50 50 51 51 52 53 53 54 54 55 56 56 57 58 58 59 59 60 61 61

[edit] with memoization, specific entry

This version is identical in function to the previous example, but it also can compute and
display a specific request (indicated by a negative number for the argument).

/*REXX program shows mutual recursion (via Hofstadter Male & Female seq)*/
/*If LIM is negative, only show a single result for the abs(lim) entry.*/
 
parse arg lim .; if lim=='' then lim=99; aLim=abs(lim)
parse var lim . hm. hf. Js Fs Ms; hm.0=0; hf.0=1
 
do j=0 to Alim; ff=F(j); mm=M(j)
Js=Js jW(j); Fs=Fs jw(ff); Ms=Ms jW(mm)
end
 
if lim>0 then say 'Js=' Js; else say 'J('aLim")=" word(Js,aLim+1)
if lim>0 then say 'Fs=' Fs; else say 'F('aLim")=" word(Fs,aLim+1)
if lim>0 then say 'Ms=' Ms; else say 'M('aLim")=" word(Ms,aLim+1)
exit /*stick a fork in it, we're done.*/
/*──────────────────────────────────one─liner subroutines──────────────────────────────*/
F: procedure expose hm. hf.; parse arg n; if hf.n=='' then hf.n=n-M(F(n-1)); return hf.n
M: procedure expose hm. hf.; parse arg n; if hm.n=='' then hm.n=n-F(M(n-1)); return hm.n
Jw: return right(arg(1),length(lim)) /*right justifies # for nice look*/

output using the input of:   -70000

J(70000)= 70000
F(70000)= 43262
M(70000)= 43262

output using the input of a ¼ million:   -250000

J(250000)= 250000
F(250000)= 154509
M(250000)= 154509

[edit] Ruby

def F(n)
n == 0 ? 1 : n - M(F(n-1))
end
def M(n)
n == 0 ? 0 : n - F(M(n-1))
end
 
p (Array.new(20) {|n| F(n) })
p (Array.new(20) {|n| M(n) })

Output:

[1, 1, 2, 2, 3, 3, 4, 5, 5, 6, 6, 7, 8, 8, 9, 9, 10, 11, 11, 12]
[0, 0, 1, 2, 2, 3, 4, 4, 5, 6, 6, 7, 7, 8, 9, 9, 10, 11, 11, 12]

In ruby there is no need to pre-declare M for it to be used in the definition of F. (However M must be defined before F calls it).

[edit] Run BASIC

print "F sequence:";
for i = 0 to 20
print f(i);" ";
next i
print :print "M sequence:";
for i = 0 to 20
print m(i);" ";
next i
end
 
function f(n)
f = 1
if n <> 0 then f = n - m(f(n - 1))
end function
 
function m(n)
m = 0
if n <> 0 then m = n - f(m(n - 1))
end function

Output:

F sequence:1 1 2 2 3 3 4 5 5 6 6 7 8 8 9 9 10 11 11 12 13 
M sequence:0 0 1 2 2 3 4 4 5 6 6 7 7 8 9 9 10 11 11 12 12

[edit] Rust

fn f(n: int) -> int {
match n {
0 => 1,
_ => n - m(f(n - 1))
}
}
 
fn m(n: int) -> int {
match n {
0 => 0,
_ => n - f(m(n - 1))
}
}
 
fn main() {
for i in range(0, 20).map(f) {
print!("{} ", i);
}
println!("")
 
for i in range(0, 20).map(m) {
print!("{} ", i);
}
println!("")
}

Output:

1 1 2 2 3 3 4 5 5 6 6 7 8 8 9 9 10 11 11 12
0 0 1 2 2 3 4 4 5 6 6 7 7 8 9 9 10 11 11 12

[edit] Sather

class MAIN is
 
f(n:INT):INT
pre n >= 0
is
if n = 0 then return 1; end;
return n - m(f(n-1));
end;
 
m(n:INT):INT
pre n >= 0
is
if n = 0 then return 0; end;
return n - f(m(n-1));
end;
 
main is
loop i ::= 0.upto!(19);
#OUT + #FMT("%2d ", f(i));
end;
#OUT + "\n";
loop i ::= 0.upto!(19);
#OUT + #FMT("%2d ", m(i));
end;
end;
end;

There's no need to pre-declare someway F or M.

[edit] Scala

def F(n:Int):Int =
if (n == 0) 1 else n - M(F(n-1))
def M(n:Int):Int =
if (n == 0) 0 else n - F(M(n-1))
 
println((0 until 20).map(F).mkString(", "))
println((0 until 20).map(M).mkString(", "))

Output:

1, 1, 2, 2, 3, 3, 4, 5, 5, 6, 6, 7, 8, 8, 9, 9, 10, 11, 11, 12
0, 0, 1, 2, 2, 3, 4, 4, 5, 6, 6, 7, 7, 8, 9, 9, 10, 11, 11, 12

[edit] Scheme

define declarations are automatically mutually recursive:

(define (F n)
(if (= n 0) 1
(- n (M (F (- n 1))))))
 
(define (M n)
(if (= n 0) 0
(- n (F (M (- n 1))))))

If you wanted to use a let-like construct to create local bindings, you would do the following. The define construct above is just a syntactic sugar for the following where the entire rest of the scope is used as the body.

(letrec ((F (lambda (n)
(if (= n 0) 1
(- n (M (F (- n 1)))))))
(M (lambda (n)
(if (= n 0) 0
(- n (F (M (- n 1))))))))
(F 19)) # evaluates to 12

The letrec indicates that the definitions can be recursive, and fact that we placed these two in the same letrec block makes them mutually recursive.

[edit] Seed7

$ include "seed7_05.s7i";
 
const func integer: m (in integer: n) is forward;
 
const func integer: f (in integer: n) is func
result
var integer: res is 0;
begin
if n = 0 then
res := 1;
else
res := n - m(f(n - 1));
end if;
end func;
 
const func integer: m (in integer: n) is func
result
var integer: res is 0;
begin
if n = 0 then
res := 0;
else
res := n - f(m(n - 1));
end if;
end func;
 
const proc: main is func
local
var integer: i is 0;
begin
for i range 0 to 19 do
write(f(i) lpad 3);
end for;
writeln;
for i range 0 to 19 do
write(m(i) lpad 3);
end for;
writeln;
end func;

Output:

  1  1  2  2  3  3  4  5  5  6  6  7  8  8  9  9 10 11 11 12
  0  0  1  2  2  3  4  4  5  6  6  7  7  8  9  9 10 11 11 12

[edit] Smalltalk

Using block closure.

|F M ra rb|
 
F := [ :n |
(n == 0)
ifTrue: [ 1 ]
ifFalse: [ n - (M value: (F value: (n-1))) ]
].
 
M := [ :n |
(n == 0)
ifTrue: [ 0 ]
ifFalse: [ n - (F value: (M value: (n-1))) ]
].
 
ra := OrderedCollection new.
rb := OrderedCollection new.
0 to: 19 do: [ :i |
ra add: (F value: i).
rb add: (M value: i)
].
 
ra displayNl.
rb displayNl.

[edit] SNOBOL4

        define('f(n)') :(f_end)
f f = eq(n,0) 1 :s(return)
f = n - m(f(n - 1)) :(return)
f_end
 
define('m(n)') :(m_end)
m m = eq(n,0) 0 :s(return)
m = n - f(m(n - 1)) :(return)
m_end
 
* # Test and display
L1 s1 = s1 m(i) ' ' ; s2 = s2 f(i) ' '
i = le(i,25) i + 1 :s(L1)
output = 'M: ' s1; output = 'F: ' s2
end

Output:

M: 0 0 1 2 2 3 4 4 5 6 6 7 7 8 9 9 10 11 11 12 12 13 14 14 15 16 16
F: 1 1 2 2 3 3 4 5 5 6 6 7 8 8 9 9 10 11 11 12 13 13 14 14 15 16 16

[edit] SNUSP

The program shown calculates F(3) and demonstrates simple and mutual recursion.

       /======\
F==!/=!\?\+# | />-<-\
| \@\-@/@\===?/<#
| | |
$+++/======|====/
 ! /=/ /+<<-\
| \!/======?\>>=?/<# dup
| \<<+>+>-/
 !  !
\======|====\
| | |
| /===|==\ |
M==!\=!\?\#| | |
\@/-@/@/===?\<#
^ \>-<-/
| ^ ^ ^ ^
| | | | subtract from n
| | | mutual recursion
| | recursion
| n-1
check for zero

[edit] Standard ML

fun f 0 = 1
| f n = n - m (f (n-1))
and m 0 = 0
| m n = n - f (m (n-1))
;

The fun construct creates recursive functions, and the and allows a group of functions to call each other. The above is just a shortcut for the following:

val rec f = fn 0 => 1
| n => n - m (f (n-1))
and m = fn 0 => 0
| n => n - f (m (n-1))
;

which indicates that the functions call themselves (rec) and each other (and).

[edit] Swift

It just works. No special pre-declaration is necessary.

func F(n: Int) -> Int {
return n == 0 ? 1 : n - M(F(n-1))
}
 
func M(n: Int) -> Int {
return n == 0 ? 0 : n - F(M(n-1))
}
 
for i in 0..20 {
print("\(F(i)) ")
}
println()
for i in 0..20 {
print("\(M(i)) ")
}
println()

[edit] Tcl

proc m {n} {
if { $n == 0 } { expr 0; } else {
expr {$n - [f [m [expr {$n-1}] ]]};
}
}
proc f {n} {
if { $n == 0 } { expr 1; } else {
expr {$n - [m [f [expr {$n-1}] ]]};
}
}
 
for {set i 0} {$i < 20} {incr i} {
puts -nonewline [f $i];
puts -nonewline " ";
}
puts ""
for {set i 0} {$i < 20} {incr i} {
puts -nonewline [m $i];
puts -nonewline " ";
}
puts ""

[edit] TI-89 BASIC

Define F(n) = when(n=0, 1, n - M(F(n - 1)))
Define M(n) = when(n=0, 0, n - F(M(n - 1)))

[edit] TXR

@(do
(defun f (n)
(if (>= 0 n)
1
(- n (m (f (- n 1))))))
 
(defun m (n)
(if (>= 0 n)
0
(- n (f (m (- n 1))))))
 
(each ((n (range 0 15)))
(format t "f(~s) = ~s; m(~s) = ~s\n" n (f n) n (m n))))
$ txr mutual-recursion.txr
f(0) = 1; m(0) = 0
f(1) = 1; m(1) = 0
f(2) = 2; m(2) = 1
f(3) = 2; m(3) = 2
f(4) = 3; m(4) = 2
f(5) = 3; m(5) = 3
f(6) = 4; m(6) = 4
f(7) = 5; m(7) = 4
f(8) = 5; m(8) = 5
f(9) = 6; m(9) = 6
f(10) = 6; m(10) = 6
f(11) = 7; m(11) = 7
f(12) = 8; m(12) = 7
f(13) = 8; m(13) = 8
f(14) = 9; m(14) = 9
f(15) = 9; m(15) = 9

[edit] UNIX Shell

Works with: Bourne Again SHell
M()
{
local n
n=$1
if [[ $n -eq 0 ]]; then
echo -n 0
else
echo -n $(( n - $(F $(M $((n-1)) ) ) ))
fi
}
 
F()
{
local n
n=$1
if [[ $n -eq 0 ]]; then
echo -n 1
else
echo -n $(( n - $(M $(F $((n-1)) ) ) ))
fi
}
 
for((i=0; i < 20; i++)); do
F $i
echo -n " "
done
echo
for((i=0; i < 20; i++)); do
M $i
echo -n " "
done
echo

[edit] Ursala

Forward declarations are not an issue in Ursala, which allows any definition to depend on any symbol declared within the same scope. However, cyclic dependences are not accepted unless the programmer explicitly accounts for their semantics. If the recurrence can be solved using a fixed point combinator, the compiler can be directed to use one by the #fix directive as shown, in this case with one of a family of functional fixed point combinators from a library. (There are easier ways to define these functions in Ursala than by mutual recursion, but fixed points are useful for other things as well.)

#import std
#import nat
#import sol
 
#fix general_function_fixer 0
 
F = ~&?\1! difference^/~& M+ F+ predecessor
M = ~&?\0! difference^/~& F+ M+ predecessor

This test program applies both functions to the first twenty natural numbers.

#cast %nLW
 
test = ^(F*,M*) iota 20

output:

(
   <1,1,2,2,3,3,4,5,5,6,6,7,8,8,9,9,10,11,11,12>,
   <0,0,1,2,2,3,4,4,5,6,6,7,7,8,9,9,10,11,11,12>)

[edit] x86 Assembly

Works with: nasm

Since all "labels" (symbols), if not local, can be seen by the whole code in the same source unit, we don't need special care to let the subroutine func_f call func_m. If the function would have been in another source unit, we should have declared it extern (the linker will resolve the symbol), as done for printf.
(It must be linked with the C standard library libc or similar and a startup code; lazyly a gcc mutrec.o works, being mutrec.o produced by e.g. nasm -f elf mutrec.asm)

	global	main
extern printf
 
section .text
 
func_f
mov eax, [esp+4]
cmp eax, 0
jz f_ret
dec eax
push eax
call func_f
mov [esp+0], eax
call func_m
add esp, 4
mov ebx, [esp+4]
sub ebx, eax
mov eax, ebx
ret
f_ret
mov eax, 1
ret
 
func_m
mov eax, [esp+4]
cmp eax, 0
jz m_ret
dec eax
push eax
call func_m
mov [esp+0], eax
call func_f
add esp, 4
mov ebx, [esp+4]
sub ebx, eax
mov eax, ebx
ret
m_ret
xor eax, eax
ret
 
main
mov edx, func_f
call output_res
mov edx, func_m
call output_res
ret
 
output_res
xor ecx, ecx
loop0
push ecx
call edx
 
push edx
 
push eax
push form
call printf
add esp, 8
 
pop edx
pop ecx
 
inc ecx
cmp ecx, 20
jnz loop0
 
push newline
call printf
add esp, 4
 
ret
 
 
section .rodata
form
db '%d ',0
newline
db 10,0
 
end

[edit] XPL0

code    ChOut=8, CrLf=9, IntOut=11;
 
ffunc M; \forward-referenced function declaration
 
func F(N);
int N;
return if N=0 then 1 else N - M(F(N-1));
 
func M(N);
int N;
return if N=0 then 0 else N - F(M(N-1));
 
int I;
[for I:= 0 to 19 do [IntOut(0, F(I)); ChOut(0, ^ )];
CrLf(0);
for I:= 0 to 19 do [IntOut(0, M(I)); ChOut(0, ^ )];
CrLf(0);
]

Output:

1 1 2 2 3 3 4 5 5 6 6 7 8 8 9 9 10 11 11 12 
0 0 1 2 2 3 4 4 5 6 6 7 7 8 9 9 10 11 11 12 

[edit] zkl

zkl doesn't like referencing undefined objects so there is a bit of sleaze here.

fcn f(n,m){if(n==0)return(1); n-m(f(n-1,m),f)}
fcn m(n,f){if(n==0)return(0); n-f(m(n-1,f),m)}
[0..19].apply(f.fp1(m))
[0..19].apply(m.fp1(f))
Output:
L(1,1,2,2,3,3,4,5,5,6,6,7,8,8,9,9,10,11,11,12)
L(0,0,1,2,2,3,4,4,5,6,6,7,7,8,9,9,10,11,11,12)
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