A Minimum/Maximum Puzzle

We solve a puzzle: make as many numbers as possible by putting the integers from 1 to 8 once each into min{ max{ min{ _ , _ }, min{ _ , _ } }, max{ min{ _ , _ }, min{ _ , _ } } }.
00:00 Intro
00:24 Understanding the problem
00:55 Considering 8, 1, and 7
01:44 2 and 3
03:11 4 and 5
04:19 6
05:52 Extensions

[@Umbra451]

I took the liberty to write up a Python program that runs through all permutations of the digits 1-8 and checks which numbers end up "winning" in this case. 3 wins the most often with about 17000 permutations, then 2, then 4, and finally 5. Swapping the order of operations (max first, then min, then max) changes the order to 6, 7, 5, 4. I tried to run the program with 16 digits, but my laptop ran out of memory. :D

[@Aaron-cs3xl]

Interesting to note the symmetry there when you flip it from min max min to max min max. 3 and 6 are both two away from the ends (1+2 or 8-2 respectively.) And so on throughout. To figure out maybe not the frequency but perhaps the viable integers you could easily calculate what the lowest viable integer and highest viable integer are and determine all integers between those two are also valid.

[@DrBarker]

That's really interesting - it would be fun to generate some graphs/bar charts showing the number of permutations where each number wins for bigger numbers too!

[@DrBarker]

For the symmetry of swapping max and min, there's a nice 1:1 duality between our orderings where you swap 1 with 8, 2 with 7, 3 with 6, and 4 with 5. It's helpful to notice that max{a,b} = 9 - min{9 - a, 9 - b} and similarly min{a,b} = 9 - max{9 - a, 9 - b}. We can use this to show that min{ max{ min{a,b}, min{c,d} }, max{ min{e,f}, min{g,h} } } = 9 - max{ min{ max{9-a,9-b}, max{9-c,9-d} }, min{ max{9-e,9-f}, max{9-g,9-h} } }, so each ordering where x wins with one choice of max/min corresponds to another ordering where 9 - x wins with the opposite choice of min/max.

[@EnteiFire4]

If feel like there's a smart, recursive way to go on about it. The 3-level tournament doesn't just apply for numbers 1 to 8: it applies for any set of 8 distinct numbers. A 4-level tournament is just two 3-level tournaments, with another max round. So you could use that knowledge to generate the list of possible results and maybe even their number of occurrence. Something else to think about: you can greatly reduce your search if you ignore symmetric tournaments: if you swap the order of two brackets, the result is still the same. So if you can find a way to only generate the unique tournaments, that could help a ton. To give you an idea, the 3-level tournament has 40,320 permutations, but creates only 315 unique tournaments. The 4-level tournament has a bit less than 21 trillion permutations, but only around 640 millions unique ones.

[@Umbra451]

@@EnteiFire4 True! I reckon the 3-level problem could functionally be reduced to 315 equivalence classes with 2^7 members each by only considering the ones where the numbers are in order

[@heartache5742]

i would love to watch videos about more different elaborate number tournaments

[@DrBarker]

This is a great idea! It would be fun to have all sorts of different weird rules for which number wins, to make it unpredictable!

[@hypebeastuchiha9229]

I like these types of videos Type of maths problem I loved doing as a kid

[@adiaphoros6842]

This is the min-max algorithm with 3 layers. The easiest solution that came to my mind is putting the permutations of 1 through 8 on the leaves and using alpha-beta pruning to get the possible roots, therefore the possible numbers. This brute force could be a fun exercise for alpha-beta pruning for maybe 2 hours.

[@emanuellandeholm5657]

This reminds me of Hasse diagrams from my 1992 Discrete Mathematics course...

[@kobethebeefinmathworld953]

That's elegant

[@3c3k]

For 2, 3, 4 and 5 you should have gone bottom to top to check

[@astherphoenix9648]

thanks for the video. where can i find problems like this to solve?

[@DrBarker]

I don't actually know, but you could maybe try computer science textbooks for similar styles of problem. You may have to just come up with your own problems to explore if you want something closely-related to this puzzle. This can be a more fun way of exploring something further anyway!

[@MrRyanroberson1]

Let's see... wuth each min, the bigfest number is removed. With each max, the smallest. At the most superficial level, therefore, the possible numbers are a subset of 2 to 6. Brute force: min max min 1 5 min 3 7 max min 2 6 min 4 8 = min max 1 3 max 2 4 = 3, and min max min 1 2 min 6 8 max min 3 4 min 5 7 = min max 1 6 max 3 5 = 5. Seems the bounds are one smaller than i initially guessed, but 4 should also be attainable, probably

[@unflexian]

i think that only 2,3,4,5 are possible? the scale here is small enough that i could manually figure out each number, using logical observations i made along the way to rule out or prove different combinations. if this was, say, two or more layers deeper i would have had to get more clever.

[@Aaron-cs3xl]

Yep, if you figure out the smallest number and the largest number, it's easy to determine that all numbers between must also be valid since their scenarios are less limited than the extremes and can "win" in many different scenarios.

[@DrBarker]

I'm sure there must be an elegant, rigorous argument to show that if a and b are possible, then so are all of the numbers in between a and b. Maybe if we change an ordering where a wins, one pair-swap at a time, until it becomes an ordering where b wins, it may have to pass through orderings where numbers in between win? I'm not sure if this actually works, but an argument like this would be cool.

[@Noah-cu8du]

p̾r̾o̾m̾o̾s̾m̾ 🌹

[@ready1fire1aim1]

We're 4D like quaternion math. (4D circle, not 4D sphere as 4D sphere is a contradiction; lit. 5D) I keep hearing theories like "simulation", "holographic" or back to Leibniz' "contingent" universe. Those theories all match up nicely with the i, j, k in quaternions. Quaternion MATHEMATICS a complex number of the form w + xi + yj + zk, where w, x, y, z are real numbers and i, j, k are imaginary units that satisfy certain conditions. RARE/biblical a set of four parts, things or persons. (dimensions, I think)

[@comma_thingy]

"not 4D sphere as 4D sphere is a contradiction; lit. 5D" What does this even mean? The sphere S4 is 4 dimensional, even though it can be embedded in 5D Euclidean space, this does not make it 5D

[@ready1fire1aim1]

String theory has a 4D sphere. This is a contradiction. 4D circle ✅ 4D sphere 🚫 contra. 5D circle 🚫 contra. 5D sphere ✅ Here's my 1D-9D: 3 sets of 3 dimensions; 1D, 2D, 3D are spatial 4D, 5D, 6D are temporal 7D, 8D, 9D are spectral 1D, 4D, 7D line/length/continuous 2D, 5D, 8D width/breadth/emission 3D, 6D, 9D height/depth/absorption Anyone concur? Thanks.

[@angelmendez-rivera351]

There is nothing contradictory about a 4D sphere.

[@ready1fire1aim1]

@@angelmendez-rivera351 Theory of Everything solution (short version): Swap from Newton "real/necessary" universe over to Leibniz "contingent/not-necessary" universe as our fundamental blueprint of the universe. This includes Leibniz calculus vs Newton calculus. Anywhere Leibniz and Newton thought different. All of it. Full swap. Gottfried Leibniz "contingent/not-necessary" universe just lacked 2022 quantum physics verbiage (just match up definitions) and Hamilton's 4D quaternion algebra (created 200 years after Leibniz died). Lastly, the first number is NOT 1. It's 0. First four numbers are 0, 1, 2, 3 ✅. First four dimensions are 0D, 1D, 2D, 3D ✅. Ask someone to start counting. I bet they start with 1.

[@angelmendez-rivera351]

@@ready1fire1aim1 That was a whole lot of gibberish.