Rather than subtracting 1 from A three times, be could try to subtract 1 and 2. We know that when we add 1 to B, we will summon a red ball. Rather than wasting it to trigger a new blue, we could use it to subtract 2. Blue now only has 2 jobs to do (subtract 1 from A and add 1 from B) so a single bit is sufficient.!
What happens if A isn’t a multiple of 3? I think your solution ends up rounding the answer up rather than discarding the remainder…
Looks like I didn’t include a link to the solution, this should be one with A=7: https://www.lodev.org/jstumble/?board=rle1lee1xfllreer1ereelle0er1errelle0lr0errelie0le
I forgot to mention which lever to start with - it should be the red lever. It will round up if you start with the blue lever, because it wont attempt to subtract 2 until after adding 1, but I think it works if you start red. Does that solve the problem?
Yeah, starting with red works.