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Re: (Golf) Dependency List Prioritization

by AidanLee (Chaplain)
on Mar 14, 2002 at 18:32 UTC ( #151777=note: print w/replies, xml ) Need Help??

in reply to (Golf) Dependency List Prioritization

Thanks for making such an accessible golf problem (at least it feels so to me). here's 116 (my first golf):

sub f{my%h=@_;for(values %h){for(@$_){$h{$_}=[]if!$h{$_}}} print join(', ',sort{(grep /^$b$/,@{$h{$a}})?1:-1}keys%h)}
update: cool, i can rip some stuff out if i assume the test harness that tadman provided. 97:
sub f{%h=@_;for(values %h){for(@$_){$h{$_}=[]if!$h{$_}}} sort{(grep /^$b$/,@{$h{$a}})?1:-1}keys%h}

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Re: Re: (Golf) Dependency List Prioritization
by AidanLee (Chaplain) on Mar 15, 2002 at 14:16 UTC

    I think i'm getting the hang of this. Here's 90:

    sub f{%h=@_;for(values %h){for(@$_){$h{$_}||=[]}} sort{(grep /^$b$/,@{$h{$a}})?1:-1}keys%h}
      Nice idea.

      A few minor nits though. First is that it is customary to only count the body of the sub, so you actually did slightly better than you thought! A second minor issue is that your grep is not quite right - the string "fooey\n" will match /^fooey$/ so you don't really have an equality test.

      Further than that, there are some mechanical tricks that can be used to improve scores. Such as using an inline for loop, using map to avoid double loops, and the like. Using those I can take your answer to 75 fairly easily:

      sub f{ #23456789_123456789_123456789_123456789_123456789_123456789_123456789_ +12345 %h=@_;$h{$_}||=[]for map@$_,values%h;sort{-1+2*grep$_ eq$b,@{$h{$a}}}k +eys%h }

      Oops. I didn't notice the lack of transitivity in your code. This is a fatal flaw as you see with the following data set.

      my @list = ( a => ['b'], b => ['c'], c => ['d'], d => ['e'], );

      Assuming the test harness provided, I doubt anyone can beat this:

      sub f{d,c,b,a,f,e}

      Try this for a test harness:

      my @list = ( a => [ 'b', 'c', 'e' ], b => [ 'd' ], c => [ 'b', 'd' ], f => [ 'a' ], ); print join (',', map { "'$_'" } f(@list)),"\n";

      Your code produces 'd','b','c','f','e','a', which does not reflect the fact that 'f' depends on 'a'.

      So, I am afraid your solution holds only for a limited set of test harnesses -- and is certainly not the shortest among such solutions.

      The Sidhekin
      print "Just another Perl ${\(trickster and hacker)},"

      I hope you don't mind me removing a few bytes:

      sub f{for(values(%h=@_)){$h{$_}||=[]for@$_}sort{(grep/^$b$/,@{$h{$a}}) +||-1}keys%h}
      (Note: I didn't test)


        If you test, you will find it doesn't compile.

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