And the permalink:
The idea is basically the same as that for my puzzle 20 solution, but at first I wasn’t sure this was valid here. I had another look at the puzzle book though and it doesn’t put any constraints on which or how many balls end up in the bottom tray. So depending on the input we might end up with zero, one or two red balls (plus the initial blue one) at the bottom when intercepting a red ball.
The reason that happens here is that we now have three cases where the blue ball crosses over, so we’re not guaranteed that the bits are in the “true” position afterwards… they might just end up in another “false” position. But since the bits always cycle through all four states, they will eventually (after at most two more red balls) end up in the “true” position and send a red ball to the interceptor.