Print a Cantor Set

~.abs.3\?'_'*\{.3%..,' '*\++}*](0>2*(%n*

With thanks to hammar for 42->40.

My best attempt yet at a more number-theoretic approach is sadly much longer:

~.abs:^3\?,{3^)?+3base(;1+1?.'_'*^@-)' '*+}%zip\0>2*(%n*

~.abs 3\?:^,{6^*+3base.1+1?.('_'*@,@-' '*+}%zip\0>2*(%n*

and I suspect that the length of base and zip will make it impossible to catch up.

n=input()
d=n>0 or-1
for i in range(n*d+1)[::d]:
 s='_'*3**i
 while i<n*d:s+=len(s)*' '+s;i+=1
 print s

Older algorithm topped out at 113 characters

r=input()
u='_'
l=[u]
for _ in abs(r)*u:o=len(l[0]);l=[s+o*' '+s for s in l]+[u*o*3]
print'\n'.join(l[::r>0 or-1])

Ruby (97)

Based on Steven Rumbalski's python version:

n,r=$*[0].to_i,[?_]
n.abs.times{z=r[0].size;r=r.map{|s|s+' '*z+s}+[?_*z*3]}
puts n<0?r:r.reverse

Build lines from parts:

c=->x,n{n<1??_*x :(z=c[s=x/3,n-1])+' '*s+z}
r=(0..m=(n=$*[0].to_i).abs).map{|i|c[3**m,i]}
puts n<0?r.reverse: r

Start with one line, make holes in it:

r=[?_*3**a=(n=$*[0].to_i).abs]
a.times{|c|r<<r[-1].gsub((x=?_*o=3**(a-c-1))*3,x+' '*o+x)}
puts n<0?r.reverse: r