The Cantor set with Regex

1 0 1
1 0 1 0 0 0 1 0 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

The endpoints of Cantor set iteration n are always multiples of 3^-n (exercise).

So we can represent that iteration using each of the midpoints 3^-n·(j+0.5).

      ('\d' ⎕R '\0 0 \0') '1 0 1'
1 0 1 0 0 0 1 0 1

      ('\d' ⎕R '\0 0 \0')⍣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

The Power (⍣) operator iterates its left operand like in mathematics.