Um... the way I learned it, "0-8" represents 9 digits. (1-8 would be 8 digits, as would 0-7.) Is there some digit between 0 and 8 that you intend to leave out, and if so, which one?
Anyway, if you had said "permutations of 0-8 with each digit occurring once, using all 9 digits", then I would understand that you are looking only for strings of nine digit characters, such that all nine characters are distinct, and none of them is the digit "9":
Update: I suspect that the regex used as the last stage in my conditional is a fairly expensive operation; for strings that actually meet the criteria (are not rejected), it has to do 8+7+6+...+1 (total of 36) character comparisons to finish. (There should be some sort of "O(...n...)" expression for that, but it escapes me.) So, it would most likely be better to use a split/hash solution, as suggested by others, especially if you'll be handling large quantities of input with a relatively high "success" rate. Something like this:#!/usr/bin/perl use strict; while (<DATA>) { chomp; if ( length() == 9 and not ( /[^0-8]/ or /(.).*\1/ )) { print "$_\n"; } else { warn "rejected input: $_\n"; } } __DATA__ 01234 123456780 1234567890 223456781 345678012 234567890 a12345678 012345678 456782011 456782013
while (<DATA>) { chomp; if ( length() == 9 and not /[^0-8]/ ) { my %c = map { $_ => undef } split //; if ( keys %c == 9 ) { print "$_\n"; next; } } warn "rejected input: $_\n"; }
In reply to Re^3: RegEx Question
by graff
in thread RegEx Question
by yoda54
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