Re^2: Challenge: Nearest Palindromic Number

sub build {
    my ($f, $l) = @_;

    my $b = reverse $f;
    substr($b, 0, 1, '') if $l;
    return $f . $b;
}

sub palindrate {
    my $num = shift;

    my $is_odd = length($num) % 2;

    my $first = substr( $num, 0, int(length($num)/2 + .5 ) );

    my $one = build( $first, $is_odd );

    $first += $one < $num ? 1 : -1;
    my $two = build( $first, $is_odd );

    return ((abs($one-$num) < abs($two-$num)) ? $one : $two);
}
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Comment on Re^2: Challenge: Nearest Palindromic Number Download Code

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Re^3: Challenge: Nearest Palindromic Number by tilly (Archbishop) on Feb 03, 2005 at 07:15 UTC
This works perfectly. :-) For those who don't understand it, here's an explanation. Take the first half of the digits and build a palindrome. (Be careful, you need to keep it even/odd in size.) That palindrome is going to be larger or smaller than the original number. Adjust the first half up/down one then build a palindrome on the other side of the original number. One of those two is closest, so take the closer one. To truly verify that it works (I'm just giving the handwaving explanation) you'll find that you actually are taking the nearest palindromes to either side in every case except where adding/subtracting from $first changes the number of digits that you have. But the only case where that happens is in 1 and numbers matching the pattern /^1(0*)(0\|1)$/. In those cases $one winds up being 100...0001 and $two is wildly off. But in those cases, $one is one of the (possibly 2) acceptable answers, so the algorithm is right again.	[reply]

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Re^3: Challenge: Nearest Palindromic Number
by tilly (Archbishop) on Feb 03, 2005 at 07:15 UTC

For those who don't understand it, here's an explanation. Take the first half of the digits and build a palindrome. (Be careful, you need to keep it even/odd in size.) That palindrome is going to be larger or smaller than the original number. Adjust the first half up/down one then build a palindrome on the other side of the original number. One of those two is closest, so take the closer one.

To truly verify that it works (I'm just giving the handwaving explanation) you'll find that you actually are taking the nearest palindromes to either side in every case except where adding/subtracting from $first changes the number of digits that you have. But the only case where that happens is in 1 and numbers matching the pattern /^1(0*)(0|1)$/. In those cases $one winds up being 100...0001 and $two is wildly off. But in those cases, $one is one of the (possibly 2) acceptable answers, so the algorithm is right again.

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