Puzzle 60 solution - 37 parts

[jcross]

I independently arrived at something very similar to Samuel’s 40-part solution, but with the gear bits in a mirror-image orientation, which seems to make things a little more compact:

https://www.lodev.org/jstumble/?board=00llllr0r0eerllleexlleeagagafxareelrlrelexelieilie0

The blue lever is used to start, and the output will always be a red ball. There appears to be a condition where a ball could fall off the left-most gear bit, but in practice this won’t happen because a red ball always winds up in one of the interrupters before a blue ball can come through with the gear bits pointing right.

[rmsgrey]

I’ll see your 37 parts and raise you an always output blue (matching the illustration): start with the red lever.

https://www.lodev.org/jstumble/?board=00llllr0r0eerllreexlelerreleebgbgbeelbxeelrrreieilie1_8_8

[RingTheBell]

Nice! I came up with an almost identical solution, but it looks like you saved a red gear:

jstumble
Tumble Together

Red starts. Registers B and A count down alternately by triggering each other through the one-shot in the middle of the gear bit assembly. If A underflows first, it was less than B (because B was decremented first) and the underflow ball is routed to the left interceptor. If B underflows first, its underflow ball reroutes A’s underflow path from the left interceptor to the middle one, and A’s non-underflow path to the right interceptor. Thus, if A underflows on the next ball, the two registers were equal and that ball lands in the middle incerceptor. If A does not underflow on the next ball, A was greater than B and that ball lands in the right interceptor.