Re: (Golf) Dependency List Prioritization

[AidanLee]

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}

[AidanLee]

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}

[Juerd]

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)

U28geW91IGNhbiBhbGwgcm90MTMgY
W5kIHBhY2soKS4gQnV0IGRvIHlvdS
ByZWNvZ25pc2UgQmFzZTY0IHdoZW4
geW91IHNlZSBpdD8gIC0tIEp1ZXJk

[tilly]

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

[Sidhekin]

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)},"

[tilly]

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     
}

UPDATE
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'],
);