Just for grins, I decided to code Ackerman's function in Perl.

The definition is really quite simple:

Formal Definition:
* A(0, j)=j+1 for j ≥ 0
* A(i, 0)=A(i-1, 1) for i > 0
* A(i, j)=A(i-1, A(i, j-1)) for i, j > 0

In my typical Brute Force and Ignorance™ method, I coded this:

#!perl

use strict;
use warnings;

use bigint;

my  $value;

my $m;
my $n;

if($#ARGV >= 1){
    $m = shift(@ARGV);
    $n = shift(@ARGV);
    }
elsif($#ARGV == 0){
    $m = shift(@ARGV);
    $n = 0;
    }
else{
    $m = 0;
    $n = 0;
    }

$value = ackermann($m, $n);

print "ackermann( $m, $n ) is $value\n";

sub ackermann {

    (my $m, my $n) = @_;

    if($m == 0){
        return $n+1;
    }
    elsif($m > 0 and $n == 0) {
        return ackermann($m - 1, 1);
    }
    elsif($m > 00 && $n > 0){
        return ackermann($m-1, ackermann($m, $n-1));
    }
}

[download]

When run as ackermann 4 1, I get (many) 'deep recursion' messages.

OK; I know that my implementation is less than brilliant; that's not my question.

This is my question: Is there any way of predicting what conditions are likely to result in deep recursion messages?

added in edit

Thanks to ikegami (I've got to try Memoize) and jdhedden.

emc

e(π√−1) = −1

In reply to The Ackermann Function by swampyankee

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