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Walterpachl (talk | contribs) (→version 1: why incorrect??) |
(→{{header|Perl 6}}: update to task, add bitwise numeric solution based on XOR) |
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=={{header|Perl 6}}== |
=={{header|Perl 6}}== |
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{{update|Perl 6|It is '00' not '10' for the test.}} |
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.say; |
.say; |
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if /^ (.+) $0+ (.*$) <?{ $0.substr(0,$1.chars) eq $1 }> / { |
if /^ (.+) $0+ (.*$) <?{ $0.substr(0,$1.chars) eq $1 }> / { |
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1 |
1 |
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00 |
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10 |
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0 |
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(no repeat) |
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1 |
1 |
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(no repeat)</pre> |
(no repeat)</pre> |
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Here's a technique that relies on the fact that XORing the shifted binary number should set all the lower bits to 0 if there are repeats. (The cool thing is that shift will automatically throw away the bits on the right that you want thrown away.) This produces the same output as above. |
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<lang perl6>sub repstr(Str $s) { |
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my $bits = :2($s); |
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for reverse 1 .. $s.chars div 2 -> $left { |
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my $right = $s.chars - $left; |
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return $left if $bits +^ ($bits +> $left) == $bits +> $right +< $right; |
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} |
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} |
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for '1001110011 1110111011 0010010010 1010101010 1111111111 0100101101 0100100 101 11 00 1'.words { |
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if repstr $_ -> $rep { |
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say $_; |
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say ' ' x $rep, .substr($rep,$rep), "\n"; |
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} |
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else { |
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say "$_\n (no repeat)\n"; |
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} |
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}</lang> |
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=={{header|Python}}== |
=={{header|Python}}== |
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