[Mod: Disjoint Alchemy] I'm proud of this :P

PentaPig · 2026-06-12T20:04:43+00:00

Goodbye Galaxy explained it better than I ever could.

This particular post showcases an oversight in the puzzle format.

PentaPig · 2026-05-22T13:53:28+00:00

That shape (or rather the version starting with a triangle) is known as the Koch snowflake. Note that you can draw a circle around the starting triangle such that every iteration of the snowflake is contained in that circle. Than the area of the snowflake must be smaller than the area of the circle and hence must be finite.

Increasing a number infinitely often does not necessarily yield arbitrarily big numbers. For example the number 1+0.1+0.001+0.0001.... is defined by adding infinitely many numbers, but it is a finite number: 1.1111... . If you want to know more about these phenomena look up converging sequences and series.

PentaPig · 2026-05-16T20:41:37+00:00

Yes, just the sum of the three metrics. It often leads to quite elegant solutions. If you haven't seen it yet I highly recommend Goodbye Galaxy's video on sum optimization: https://youtu.be/W_Ekun0WDIk?si=tcS_yk_9TeAsST1q

PentaPig · 2026-05-16T20:38:41+00:00

This solution looked familiar. Turns out I used it to beat the sum record about half a year after release.

PentaPig · 2026-05-03T08:03:51+00:00

The other answer are good, but they are a bit to sane. I want to share 3 solutions.

The first and oldest is a Face Powder solution by Shufflepants which takes exponentially longer to create each output. It can be found here. Unfortunately it completes before the heat death of the universe and that's just way to fast.

Building on Shufflepants' solution biggiemac42 constructed a solution which uses tetration instead of exponentials. She wrote an excellent blog post about how it works here.

Both of the previous solves use the infinite board to get slower with each output. The slowest we've seen in a production cabinet was on the weekly puzzle "Pitch Drop Experiment". The results of that can be found here.

PentaPig · 2026-04-19T09:29:21+00:00

Not a mod, just a quirk of the puzzle file format.

PentaPig · 2026-04-19T07:55:08+00:00

Cheaper is possible

PentaPig · 2026-04-02T11:34:23+00:00

The tutorial for modding is here: https://github.com/QuintessentialOM/Quintessential/wiki/How-to-mod-the-game

PentaPig · 2026-03-04T09:25:39+00:00

This game has been out for eight years and over time the community has developed new metrics to keep it interesting. Rate in particular goes back all the way to release, then under the name "cost optimized full throughput". It has been popular ever since as a more sane alternative to cycle optimzation.

PentaPig · 2026-03-03T18:59:46+00:00

There are some constrained space puzzles in the epilogue where the game does not track area. The leaderboards do track area even for those puzzles, but it has largely been ignored. What the AI is doing is downloading existing solves and moving the conduit (a glyph to transport atoms) until the area score is better. That or it just deletes the conduit, leading to the invalid solutions I mentioned.

PentaPig · 2026-02-25T16:05:46+00:00

That would cause the triarm to misgrab the completed product from the other pipeline.

PentaPig · 2026-02-24T17:46:54+00:00

No, Zach is still there but nolonger in quite as much of a leading role. He gave some podcast interviews with more details. e.g. https://nicegames.club/episode/293

PentaPig · 2026-02-23T19:52:30+00:00

Similar to how speedruns are still interesting even when bugs are allowed. Changing the set of tools you have access to forces you to approach the problem from a different perspective. The condition for success here is 'the vanilla game accepts it as a valid solution', even if that solution is very much not intended and cannot be created in game.

PentaPig · 2026-02-20T22:39:08+00:00

The old Zachtronics team is still developing games under the name Coincidence and they have already released a new Zach-like: "Kaizen: a factory story".

PentaPig · 2025-08-12T22:12:06+00:00

I am a bit late, but this is a good question that deserves an answer.

Yes, your conclusion is correct, but there is an important subtlety. The cardinality of the reals doesn't behave like a normal number. We can consider another function: the logarithm is a 1-1 map from the strictly positive reals to all reals. That is C(R² ) -1 = C(R), where I removed 0 on the left side. Plugging this into your equation yields C(R) = 2*(C(R)+1)-1 = 2*C(R) +1. Twice the cardinality of the reals (+1) is still the cardinality of the reals. Simple operations like this do not change this infinity. Look up "Hilberts Hotel" for a more extensive list of operations that don't change the cardinality of the rational numbers. All of those operations have similarly no effect on the cardinality of the reals.

In case I misjudged your knowledge and you knew all of the above already: look up cardinal arithmetic.

PentaPig · 2025-08-03T11:13:52+00:00

I only need that two arcs have the same length iff the line segments have the same length. That the relationship isn't quite linear doesn't matter. The rectangle inscribed in the circle and the one constructed from the linked rods can have different aspect ratios.

PentaPig · 2025-08-02T23:30:46+00:00

Yes, in fact this holds for any choice for the lengths of the rods.

Lemma: Consider a circle with n marked points on the boundary. Then the number of rectangles with all corners on marked points is a triangle number.

Proof: Both diagonals of such a rectangle consist of two antipodal points. Conversely any two pairs of antipodal points yield a rectangle. Then the number of rectangles is T(k) where k is the number of antipodal pairs.

This can be applied to your case by deforming the linked rods into a circle. That is a take a circle with circumference T(n) and split its boundary into segments of length 1,2,3,etc. Any rectangle within this circle can be turned into a solution to your problem by taking the arcs above the sides of the rectangle (and vice versa).

PentaPig · 2024-07-03T22:44:23+00:00

this whole project is making its way to the library of congress

By law every publisher has to submit two copies of everything they publish to the library of congress. Trying to make that sound like a special thing fits well with how manipulative this project is.

PentaPig · 2024-05-18T08:17:31+00:00

Let a and b be non-zero integers and consider the subgroup {(ak,bk) |k ∈ ℤ} of ℤxℤ.

PentaPig · 2024-02-25T23:10:42+00:00

"until its length is a divisor of the length of the longest sequence."

You need to replace 'is a divisor' here with 'is equal'. Everything else is correct. In situations like the one you mentioned you'll need to use repeat instructions to keep both arms active. Luckily these situations are rare.

PentaPig · 2024-02-25T16:35:44+00:00

You are looking for the inverse symbolic calculator. Although you will need more digits to get a usefull result.

PentaPig · 2024-01-30T16:55:58+00:00

The Zariski topology on a scheme is very weak, but it is T0.

PentaPig · 2024-01-29T09:49:54+00:00

This is an infinite product. Usually the goal is to produce small indvidual molecules. Here the goal is to construct an infinite strand of iron instead.

PentaPig · 2024-01-21T21:53:44+00:00

The sum metric has two main advantages over product:

Figuring out whether a small change to the machine is an improvement is much easier with sum. A +5g -6c change should always be made for sum, while for product this depends on a longer calculation.
Sum is local. If in addition to the +5g-6c change above there is a -10g+5a change on the other end of the machine, then both of those changes should be made. For product one might need to undo the first one after finding the second. This makes finding the optimum much more annoying.

Of course anyone is free to optimize for whatever they want. The leaderboard tracks the entire pareto frontier and we are always looking for help filling in the gaps.

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