Turing Tumble Community

Multiplexer problem


The two gear bits form a register. As usual, the less significant bit (1) is on top, the more significant (2) below. Zero is pointing to the left, one to the right.

If the register shows 1 then put a ball in the left-hand interceptor.
If the register shows 2 then put a ball in the middle interceptor.
If the register shows 3 then put a ball in the right-hand interceptor.
If the register shows 0 then stop with no ball in an interceptor.

In other words the interceptors are labelled left to right 1, 2 and 3 matching the number in the register.

Note that it’s an easier problem if the interceptors are numbered 3, 2, 1.

Solution on request.

I have tried different gear and gear bit combinations until I found a solution.
Here is the link.
Press here to upload the image.
If you want to see it here, I have a screenshot.

That’s excellent. More compact than my solution.