Great twist on the 3 utilities problem! It was very satisfying, and I barely had enough space (only 1 row and 1 column to spare) to get a balanced solution, would love to see the more efficient solutions out there
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).
I noticed that all the bars go down by a fraction of a box a bit one of the houses reaches it's max, enough that the other three houses are prevented from ever reaching their max if you do a straight connection between one of the sources and houses.
what are the specifics for that? (I understand this is just to make sure all houses have a constant supply, and one of the is't getting like 3X more than the others)
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.
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%.
I would have responded with the above immediately if my account was working.
I have now fixed my account and made a solution that involves a loop containing some accumulators (hypothetically, a box could cycle through all of them twice, but it would be extremely unlikely) and some more that loop on themselves to achieve an even number of outputs. the big loop is random enough that outputs are evenly dispersed. this is probably the easiest solution, granted that you can get the timing right to avoid collision.
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Great twist on the 3 utilities problem! It was very satisfying, and I barely had enough space (only 1 row and 1 column to spare) to get a balanced solution, would love to see the more efficient solutions out there
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).
omy thats crazy HAHA
what does 90% of the max of each box mean?
I noticed that all the bars go down by a fraction of a box a bit one of the houses reaches it's max, enough that the other three houses are prevented from ever reaching their max if you do a straight connection between one of the sources and houses.
what are the specifics for that? (I understand this is just to make sure all houses have a constant supply, and one of the is't getting like 3X more than the others)
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.
I currently have a solution that supplies all 4 houses with all 3 resources, but there is 1 house/per resource that gets too much.
additional clarity on this mechanic would be appreciated.
I adjusted it and got the accepted screen for like one frame. how many accumulators does the official solution use?
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%.
actually, if I let it run, some of them balance out at 80-85%. I simply delayed the excess product that causes it to come out of balance.
how many accumulators are needed in the final solution?
I would have responded with the above immediately if my account was working.
I have now fixed my account and made a solution that involves a loop containing some accumulators (hypothetically, a box could cycle through all of them twice, but it would be extremely unlikely) and some more that loop on themselves to achieve an even number of outputs. the big loop is random enough that outputs are evenly dispersed. this is probably the easiest solution, granted that you can get the timing right to avoid collision.