note
tilly
Well both of my solutions have been beaten so I am willing
to post them. But there are still a couple of hours for
people to try to beat the current entries. But right now
I am in a quandry about who is leading in the first. The
best solution so far is 47 characters. It is basically
[MeowChow]'s solution with a trivial fix from [chipmunk].
I feel that it is "really" [MeowChow]'s, but technically
it <i>was</i> [chipmunk] who posted it. I will have to
decide what to do with that one. For the bonus round
[tye] is clearly leading.<P>
What I did differently than everyone else is I used a hash.
My idea was to use a hash and then mark off entries by
assigning to a hash reference. If you assign a range of
numbers starting with an odd, the odds conveniently become
the keys. This unfortunately makes 2 a special case. For
the general problem I snuck 2 past like this in 60
characters:
<code>
sub p{
%p=@_=3..1+pop;@p{2,map$f*$_,@_}=$f=$_ for@_;grep$p{$_},2,@_
}
</code>
For the second problem I just dropped the special logic
for 2 and stuck it at the beginning of the list with this
61 character solution:
<code>
sub p{
2,grep{$f=$_,@p{map$f*$_,3..$_[0]/$f}=0if$p{$_}}%p=3..1+$_[0]
}
</code>
The observant will notice that this is indeed startlingly
similar to [tye]'s solution. If he tried to
reverse-engineer mine, the copy is better than the original!
(Possibly because he did not have my blind spot for using
a hash...)
81695
81695