Pixie volume loops - a Minecraft puzzle (challenge)

Challenge this week. Solution next week.
The Pixie font by @_KK : / @_kk
The questions:
1) do all starting numbers reach a loop eventually?
2) what are all the loops?
3) out of all the numbers under 100 that reach a loop, which takes the longest to do so?

[@thatchapthere]

Given that there's a finite number of values lower than a given number the only way to avoid looping is divergence. Divergence is impossible since volume increases logarithmically.

[@TheGoodMorty]

New Collatz Conjecture just dropped

[@_KK]

Huh, this reminds me a lot about "Four has four letters" problem, but unlike there here it has a positive feedback loop

[@br0kenPixel]

like two is three, three is five, five is four, four is the magic number

[@troybaxter]

This kind of reminds me of the Collatz Conjecture or the 3n+1 problem. To those that don't know, the conjecture starts with three conditions: 1. You have to choose any positive integer (>0) 2. If number is odd, multiply by 3 and add 1. 3. If the number is even, divide by 2. The conjecture is that any positive integer will ultiamtely reach 1.

[@cerberus0225]

Played around with this a bit, haven't watched the subsequent episodes. One interesting thing I noticed is that the three possible decimal values (.25, .5, .75) can effectively be ignored in order to just focus on the whole portion with one neat trick. For any string containing a decimal value, there exists at least one (and often several) strings without a decimal value that yield the same resulting number. The segment .5 adds a value of 6.75 to whatever string it's appended to, while .25 adds 11.75 and .75 adds 11. So, any number ending in .5 can have that swapped with a final 0 instead and give an equivalent result, while .25 can be swapped with the two-digit sequences 02 or 56 (or their reversals) and .75 can be swapped with 07, 26, 35, or 44, or their reversals. Add on that 6, 8 and 9 all add a value of 6 to the next step, and we can effectively ignore 8s and 9s and swap them for 6s in all sequences. We also don't care about the order of our digits, in a whole number anyway- but hey, we just showed that if we can search for all possible loops and chains while staying in whole numbers with some swapping rules, so that's easy enough. For all possible two-digit values (using only 0, 1, 2, 3, 4, 5, 6, 7, giving 36 combinations), there's only 19 unique possible outcomes as well, further collapsing down our space. We start running into an interesting behavior where our smallest 3-digit values, such as 777 yielding 12.75, are actually smaller than or equal to our largest 2-digit values, such as 00 yielding 13.5 or 06 (or 60) yielding 12.75. This overlap will continue to grow as the length of our whole numbers grows, such that many n-length numbers will start having equivalencies to n-1 length numbers. Another way to think of it is that, for any sequence of n digits (assuming we start with a whole number), our smallest possible output would come from all 7's, while our largest comes from all 0's. All 7s would yield 4.25n while all 0s yields 6.75n. Our output range would have length 2.5n, which means that there would be 10n possible outputs, accounting for decimals. We'd actually have 1 less than this because for any n-length number our second-highest value would be n-1 0s with 1 6, giving 6.75n - 0.5, so we skip one, but that almost doesn't matter. If we look at all n-length numbers with an additional decimal allowed (and not counted toward that length) then our largest possible outcome would be a sequence of 0's with .25 attached, giving 6.75n + 11.75. For any 2

[@josiahtaylor8714]

This reminds me of that Sethbling challenge of the longest running time bomb in a 5x5x1 box in minecraft

[@celetial3287]

just discovered your channel through a mogswamp shoutout! this channel is a hidden gem; i love math and minecraft, and then together is very entertaining :)

[@krloz7493]

you have the mind of a mathematician. keep it up the good work.

[@rubixube4239]

sounds like syracuse problem ! I just discovered your channel and it's amazing, math in minecraft... just amazing

[@alchemicalguns]

reminds me of the collatz conjecture in a way; patterns that define what number you go to next.

[@crumblinggolem6327]

So here are a few points I'm going to be spitballing. 1) By virtue of the smallest possible unit being a quarter, any number you choose will end up as an integer followed by nothing, .25, .5 or .75. I don't know exactly how much but it seems like each digit seems to reach at most 6 blocks. However, to simplify things for now, I'm going to assume a 6 block volume for each digit and keep a 1 block volume for the dot. If we look at numbers, before 100 with at most two decimals, they will fall below 25 since each digit is at most 6 which would give you 4*6+1 for a number like 12.25. In fact, any number that you choose will at most end as n*6+12+1 with n being the number of digits. Since there is a finite amount of numbers with 0, 1, 2 or 3 quarters as a decimal between 0 and 25, you will end up with chains for those numbers. Then, for any number above that, you will go down until you reach a number between 0 and 25 and join one of the possible chains. In a more generalized manner, unless you have a font where each number to the left is more than 10 times as big as the one to the right, you will always fall to an interval in which your 'end' chains will be. And if the volume of your digits is finite, you can only make your chains out of a finite amount of numbers and thus you have a finite amount of chains. 2) All the loops for this font exists between 0 and 25 with branches going out of them for any higher number. The only way I see of knowing all of them is through computing every result for the 100 valid numbers between 0 and 25. 3) Only way I see of finding that is to compute all results from 0 to 25 and the longest one is one longer at most

[@svenskahugo3199]

came down to comment something like this, but it looks like you got most of it. some additions after running a python script (it should be right, could be wrong though): 2) pretty much. fun fact: brute forcing reveals that for the pixie font the longest loops are all length 3 3) for numbers 15.25 -> 22 -> 10 -> 11.25 -> 20.75 -> 22.75 -> 21 -> 9.5 -> 12.75 -> 20.5 -> 18.5 -> 17.25 -> 20.5 -> back to 18.5 First instances of drop lengths (measured by steps until the first repeat) 3: .25 4: .5 7: 1.25 10: 1.5 11: 2 13: 7.75 14: 10002.75 script hung itself after that

[@rawesomeness_]

the challenges sound like the infamous 3n+1

[@1224chrisng]

reminds me of the Four puzzle, where the function is the length of the word in english, so Four has 4 letters which is the only loop.

[@subway_surfers5251]

I used to do that with segmented displays

[@wilh3lmmusic]

Solved it, will send you the solution

[@wilh3lmmusic]

You should’ve gotten it over email

[@keoel3444]

i think you miscounted the zero (5.75 instead of 6.75)

[@wilh3lmmusic]

It is 6.75