Re: The 10**21 Problem (Part I)

You can easily gain one character on the Perl one:

sub u {IXCMVLD=~$_;"1E@-"%9995}
sub i {XCMVLD=~$_;"1E@+"%9995}
print $_, v9, u, v9, i, $/ for qw(M D C L X V I);
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Output:

M       1000    1000
D       500     500
C       100     100
L       50      50
X       10      10
V       5       5
I       1       1
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Re^2: The 10**21 Problem (Part I)
by eyepopslikeamosquito (Archbishop) on May 30, 2014 at 23:49 UTC

Thanks Grimy.

By the way, the PHP md5 solution works in Perl too since these two languages have essentially the same bitwise string operators:

use Digest::MD5 qw(md5_hex);
sub r {uppp&md5_hex$_.PQcUv}
print "$_: ", 0+r(), "\n" for (qw(I V X L C D M));
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I've updated the Perl summary of shortest solutions in the root node like so:

VLD=~$_*5+IXCM=~$_."E@-"    # Perl
10**(7&5045e8/ord)%2857     # Perl (64-bit)
IXCMVLD=~$_;"1E@-"%9995     # Perl
XCMVLD=~$_;"1E@+"%9995      # Perl (Grimy improvement)
uppp&md5_hex$_.PQcUv        # Perl (needs Digest::MD5 module)
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As you can see, if Perl had a md5 built-in function, this would be the shortest Perl solution -- even with a ridiculously long md5_hex function name!

Update: for full details on how I happened to find this md5 solution, see Re^2: The golf course looks great, my swing feels good, I like my chances (Part III).

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