Syntactic Confectionery Delight PerlMonks

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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 was chipmunk who posted it. I will have to decide what to do with that one. For the bonus round tye is clearly leading.

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:

```sub p{
%p=@_=3..1+pop;@p{2,map\$f*\$_,@_}=\$f=\$_ for@_;grep\$p{\$_},2,@_
}
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:
```sub p{
2,grep{\$f=\$_,@p{map\$f*\$_,3..\$_[0]/\$f}=0if\$p{\$_}}%p=3..1+\$_[0]
}
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...)

In reply to Re (tilly) 1: (Golf): Sieve of Eratosthenes by tilly
in thread (Golf): Sieve of Eratosthenes by tilly

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