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Question on Recursion

by prasadbabu (Prior)
on Jan 09, 2009 at 10:49 UTC ( #735151=perlquestion: print w/ replies, xml ) Need Help??
prasadbabu has asked for the wisdom of the Perl Monks concerning the following question:

Hi Monks,

Today I wrote a small script for my friend to check the sum of the digits recursively (till single digit of length), numbers range from 1500 to 2300. For example:

Number 1818 1818 = 1+8+1+8 = 36 step 1 (2 digits output) 36 = 3 + 6 = 9 step 2 (single digit output) result: 9 (final count of the digits )

I wrote the below code and found the final total is printing continuously. I expected only 9 but it is printing both 9 and 18 continuously. I solved the issue later (case 2). But I want to know the reason why it is printing continuously 9 and 18. Because of that I am not getting final output '1818 == 9' in the main program.

I 'super searched' in perlmonks and found following links. subroutine recursion question,Recursion problem, Recursion. But I was not able to find correct answer. Even I googled but I didn't get correct answer. Even I flushed using $| but not able to get expected output. Where am I going wrong? Could someone explain where I am making mistake?

use strict; use warnings; my $total1; for my $num (1818..1818){ my $total1 = &spl($num); print "$num == 9\n" if ($total1 == 9); } ################# not working as i expected ####### case 1 sub spl{ my ($num1) = @_; my $total = 0; $total = $total + $_ for (split '', $num1); print "total: $total\n"; my $len = length ($total); print "length: $len\n"; &spl($total) if ($len != 1); print "final: $total\n"; return $total; } ########## working perfectly ######### case 2 #sub spl{ # #my ($num1) =@_; # #my $total = 0; # #$total = $total + $_ for (split '', $num1); # #my $len = length ($total); # #return $total if ($len == 1); # #&spl($total) if ($len != 1); # #} output: ------- total: 18 length: 2 total: 9 length: 1 final: 9 final: 18 Expected output: ---------------- total: 18 length: 2 total: 9 length: 1 final: 9 1818 == 9

Thanks in advance

Prasad

updated: added the line, 'Because of that I am not getting final output '1818 == 9' in the main program.'

Comment on Question on Recursion
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Re: Question on Recursion
by ikegami (Pope) on Jan 09, 2009 at 11:06 UTC
    You're ignoring the result of the recursive call.
    &spl($total) if ($len != 1);
    should be
    $total = &spl($total) if ($len != 1);

    Now, what we have here what is called tail-end recursion. The recursion is the last thing in the function. That means you can use a loop instead.

    sub spl { my ($num1) = @_; while (1) { my $total = 0; $total = $total + $_ for (split '', $num1); my $len = length ($total); if ($len == 1) { return $total; } $num1 = $total; } }

    After some cleaning up, we get:

    sub spl { my ($n) = @_; while (length($n) > 1) { my $total = 0; $total = $total + $_ for split '', $n; $n = $total; } return $n; }

    or even

    use List::Util qw( sum ); sub spl { my ($n) = @_; while (length($n) > 1) { $n = sum split '', $n; } return $n; }

    Update: Added last snippet.

      FWIW: Yet another snippet, slightly different approach:
      >perl -wMstrict -le "sub T { my $n = 0; $n += $_ for @_; return $n <= 9 ? $n : T(split '', $n); } printf qq{%5s -> %d \n}, $_, T($_) for @ARGV " 1818 1819 1918 1500 5001 51 2300 0230 32 1 9 0 00 00000 10000 00001 1818 -> 9 1819 -> 1 1918 -> 1 1500 -> 6 5001 -> 6 51 -> 6 2300 -> 5 0230 -> 5 32 -> 5 1 -> 1 9 -> 9 0 -> 0 00 -> 0 00000 -> 0 10000 -> 1 00001 -> 1
Re: Question on Recursion
by moritz (Cardinal) on Jan 09, 2009 at 11:07 UTC
    You get two lines with
    final<c> because that print is unconditional. You could write it that +way instead: <c> sub spl { my ($num1) = @_; my $total = 0; $total = $total + $_ for (split '', $num1); if (length($total) == 1) { print "Final: $total\n"; return $total; } else { return spl($total) } }
Re: Question on Recursion
by JavaFan (Canon) on Jan 09, 2009 at 11:52 UTC
    Now that you have found the recursion problem, I'd write it without recursion. Or looping.
    print "$num == 9\n" unless $num % 9;
    will do as well.
      JavaFan:

      Not quite ... it'll only print the 9s! I'd suggest:

      print "num=$num, result=", ($num ? substr("912345678", $num%9, 1) : 0) +, "\n";
      ...roboticus
        Well, considering the OP has:
        print "$num == 9\n" if ($total1 == 9);
        only printing the 9s was done on purpose, as that's what the OP is doing.

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