The Cantor set with Outer!

      ({⍵×1 0 1}⍤0) 0 1
0 0 0
1 0 1

The rank mapping uses the same array every iteration. What if we pulled it out?

      0 1 ({⍺×⍵}⍤0 1) 1 0 1
0 0 0
1 0 1

      0 1 ∘.× 1 0 1
0 0 0
1 0 1

      { , ⍵ ∘.× 1 0 1 }⍣3⊢ 1
1 0 1 0 0 0 1 0 1 0 0 0 0 0 0 0 0 0 1 0 1 0 0 0 1 0 1