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| author | brxfork <ph.bricout@gmail.com> | 2022-06-23 19:59:25 +0200 |
|---|---|---|
| committer | brxfork <ph.bricout@gmail.com> | 2022-06-23 19:59:25 +0200 |
| commit | e48611ef36ee059b6e8791469f4cd7bbd610ba02 (patch) | |
| tree | acedf0ac989453d2383a7e4eebb1a5a6db55bd87 | |
| parent | dc0541bcb73b41aa3d2960dffaa884226ae035af (diff) | |
| download | perlweeklychallenge-club-e48611ef36ee059b6e8791469f4cd7bbd610ba02.tar.gz perlweeklychallenge-club-e48611ef36ee059b6e8791469f4cd7bbd610ba02.tar.bz2 perlweeklychallenge-club-e48611ef36ee059b6e8791469f4cd7bbd610ba02.zip | |
one-liner with regex for challenge 144
| -rw-r--r-- | challenge-144/brxfork/README | 1 | ||||
| -rwxr-xr-x | challenge-144/brxfork/perl/ch-1.sh | 32 |
2 files changed, 33 insertions, 0 deletions
diff --git a/challenge-144/brxfork/README b/challenge-144/brxfork/README new file mode 100644 index 0000000000..5eb20c6be7 --- /dev/null +++ b/challenge-144/brxfork/README @@ -0,0 +1 @@ +Solution by Philippe Bricout. diff --git a/challenge-144/brxfork/perl/ch-1.sh b/challenge-144/brxfork/perl/ch-1.sh new file mode 100755 index 0000000000..5c2d7672bf --- /dev/null +++ b/challenge-144/brxfork/perl/ch-1.sh @@ -0,0 +1,32 @@ +#!/bin/sh +perl -E 'while (++$n <= 100) {undef @pf ; $_="S" x $n ; m{^(S{2,}?)(??{ $1=~/^(S{2,})\1+$/?"(?!)":"" })\1+$(?{push @pf, length($1)})(?!)};next if (@pf<1 or @pf>2 or ($pf[0]*($pf[1]//$pf[0])!=$n)) ;print "$n is Semiprime as $n = $pf[0] x ",$pf[1]//$pf[0],"\n"}' +# +# Default variable $_ contains several "S" (exactly $n chars). +# +# 1) +# m{^(S{2,}?)(??{ $1=~/^(S{2,})\1+$/?"(?!)":"" })\1+$(?{push @pf, length($1)})(?!)} +# --------- ---- +# +# This finds any factor of $n by capturing a sequence in $1 (factor == length($1) ) +# for example it finds "SSS" in "SSSSSS" (3 is a factor of 6 :) +# +# 2) +# m{^(S{2,}?)(??{ $1=~/^(S{2,})\1+$/?"(?!)":"" })\1+$(?{push @pf, length($1)})(?!)} +# ---------------------------------- +# Here we verified the factor is a prime number +# (??{}) produces a pattern "(?!)" if $1 is not prime => always fail => backtracking +# (??{}) produces a pattern "" if $1 is prime => always ok => continue +# +# 3) +# m{^(S{2,}?)(??{ $1=~/^(S{2,})\1+$/?"(?!)":"" })\1+$(?{push @pf, length($1)})(?!)} +# ------------------------- +# When a prime factor is found, we keep it in @pf +# +# 4) +# m{^(S{2,}?)(??{ $1=~/^(S{2,})\1+$/?"(?!)":"" })\1+$(?{push @pf, length($1)})(?!)} +# ---- +# This pattern always fails. Backtracking permits to find all prime factors. +# +# Result : we keep only numbers with exactly one or two prime factors +# and we verify the product is correct (the square if only one prime factor). +# So we don't keep 8 because 2*2!=8 |