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Interesting solution. That is clever! It is cool to see what different players come up with to solve this.

To answer your question, it is possible to solve using just 3 accumulators.

Congrats on solving it and glad you enjoyed it! There are some very efficient solutions but they require careful timing. (The most space efficient solution fits within the bounding box formed by the utilities and houses).

Noted about the clarity. Will potentially amend the instructions. Congrats on solving it! If you got the accepted screen for a frame and your system was balanced, letting it continue to run for a few extra cycles will allow it to get to the point where small timing differences in the number of boxes received don't cause any bar to dip below 90%.

Once one house receives 10 boxes of some type, the % filled for the bars are normalized by dividing the total received (for the corresponding house and box type) by the maximum received (over all houses and box types).

Ex. if the first house receives 20 A boxes (and this is the max over all houses and boxes), then all box types must be received at least 18 times (90% of 20) at all houses to count as a win.