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Rep-string: Difference between revisions
→{{header|Perl 6}}: use double-struck digits for more compact notation
(→version 1: remove incorrect incorrect marking :)) |
(→{{header|Perl 6}}: use double-struck digits for more compact notation) |
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=={{header|Perl 6}}==
<lang perl6>for <1001110011 1110111011 0010010010 1010101010 1111111111 0100101101 0100100 101 11 00 1> {
if /^ (.+) $0+: (.*$) <?{ $0.substr(0,$1.chars) eq $1 }> / {▼
my $rep = $0.chars;
▲ if /^ (.+) $0+ (.*$) <?{ $0.substr(0,$1.chars) eq $1 }> / {
say .substr(0,$rep), .substr($rep,$rep).trans('01' => '𝟘𝟙'), .substr($rep*2);
}
else {
}
}</lang>
{{out}}
<pre>10011𝟙𝟘𝟘𝟙𝟙
1110𝟙𝟙𝟙𝟘11
001𝟘𝟘𝟙0010
1010𝟙𝟘𝟙𝟘10
11111𝟙𝟙𝟙𝟙𝟙
010𝟘𝟙𝟘0
1𝟙
0𝟘
▲ (no repeat)
▲ (no repeat)
▲ (no repeat)</pre>
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.
<lang perl6>sub repstr(Str $s) {
my $bits = :2($s);
for reverse 1 .. $s.chars div 2 -> $left {
}
}
Line 187 ⟶ 166:
for '1001110011 1110111011 0010010010 1010101010 1111111111 0100101101 0100100 101 11 00 1'.words {
if repstr $_ -> $rep {
say .substr(0,$rep), .substr($rep,$rep).trans('01' => '𝟘𝟙'), .substr($rep*2);
}
else {
}
}</lang>
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