Mutual recursion
From Rosetta Code
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:
(If a language does not allow for a solution using mutually recursive functions then state this rather than give a solution by other means).
[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;
[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
}
[edit] AWK
In AWK it is enough that both functions are defined before either is called.
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: QuickBasic version 4.5
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 f! (n!)
IF n = 0 THEN
m = 0
ELSE
m = f(m(n - 1))
END IF
END FUNCTION
[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. 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] AutoHotkey
Translation of: C
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] 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>
/* 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(int n);
int M(int n);
int F(int n)
{
if ( n==0 ) return 1;
return n - M(F(n-1));
}
int M(int n)
{
if ( n == 0 ) return 0;
return n - F(M(n-1));
}
int main()
{
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 0;
}
[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] 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
long male (long n)
{
if (n)
return n - female (male (n - 1));
else
return 0;
}
long female (long n)
{
if (n)
return n - male (female (n - 1));
else
return 1;
}
[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] 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] 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] 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 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 seems to be 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
[edit] Icon
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] Unicon
This Icon solution works in Unicon.
[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(int n)
{
if ( n == 0 ) return 1;
return n - m(f(n - 1));
}
public static int m(int n)
{
if ( n == 0 ) return 0;
return n - f(m(n - 1));
}
public static void main(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));
}
out = {F: [], M: []};
for (var 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] Logo
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] 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
[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] 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] 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 <objc/Object.h>
@interface Hofstadter : Object
+ (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] 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] 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] 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
use strict;
sub F
{
my $n = shift;
return 1 if $n==0;
return $n - M(F($n-1));
}
sub M
{
my $n = shift;
return 0 if $n==0;
return $n - F(M($n-1));
}
my @ra = ();
my @rb = ();
foreach my $i (0 .. 19) {
push @ra, F($i);
push @rb, M($i);
}
print join(" ", @ra) . "\n";
print join(" ", @rb) . "\n";
sub F {!$_[0] or $_[0] - M(F($_[0]-1))}
sub M {$_[0] and $_[0] - F(M($_[0]-1))}
for my $f (\&F, \&M)
{print "@{[map $f->($_), 0..20]}\n"}
[edit] Perl 6
Works with: Rakudo version #23 "Lisbon"
A direct translation of the definitions of F and M:
multi F (0) { 1 }; multi M (0) { 0 }
multi F (Int $n where (* > 0)) { $n - M F $n - 1 }
multi M (Int $n where (* > 0)) { $n - F M $n - 1 }
[edit] PHP
<?php
function F($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(" ", $rb) . "\n";
?>
[edit] PicoLisp
(de f (N)
(if (=0 N)
1
(- N (m (f (dec N)))) ) )
(de m (N)
(if (=0 N)
0
(- N (f (m (dec N)))) ) )
[edit] Prolog
female(0,1).
female(N,F) :- N>0,
N1 is N-1,
female(N1,R),
male(R, R1),
F is N-R1.
male(0,0).
male(N,F) :- N>0,
N1 is N-1,
male(N1,R),
female(R, R1),
F is N-R1.
Works with: GNU Prolog
flist(S) :- for(X, 0, S), female(X, R), format('~d ', [R]), fail.
mlist(S) :- for(X, 0, S), male(X, R), format('~d ', [R]), fail.
Testing
| ?- flist(19). 1 1 2 2 3 3 4 5 5 6 6 7 8 8 9 9 10 11 11 12 no | ?- mlist(19). 0 0 1 2 2 3 4 4 5 6 6 7 7 8 9 9 10 11 11 12
[edit] PostScript
/female{
/n exch def
n 0 eq
{1}
{
n n 1 sub female male sub
}ifelse
}def
/male{
/n exch def
n 0 eq
{0}
{
n n 1 sub male female sub
}ifelse
}def
[edit] PowerShell
function F($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] 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] 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] 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] 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] 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] 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
