zetamath reacts LIVE to Simon's solve of Distinction

[@philinahammock]

Simon emphatically says “Zetamath, take a bow” and you don’t take a bow. What sort of reaction video is this?

[@philiplindley7384]

Simon has a bizarre ability to make everything ultimately complicated. But then get through it. He does it over and over again.

[@rosiefay7283]

Simon's explaining things for people who might not immediately understand them. If you didn't need them explaining, then that might make them seem to be more complicated than they actually are.

[@v.sandrone4268]

I believe that Simon has been damaged by setters over the years. If you watch his older videos he would freak out over a puzzle with few numbers now he casually accepts no number puzzles (and recently a 120 size cage...which would have caused old Simon to go nuts)

[@SaarRocks]

I love this format, I would love to see more of these live reaction videos!

[@KateMedland]

That was great to watch. I hope you do more of these!

[@zetamathdoespuzzles]

It was a lot of fun to do, I hope I am lucky enough to get another feature on a day I'm streaming, in which case I certainly will!

[@stevieinselby]

Having solved the puzzle myself in about the same time as Simon did (which is rare) and really enjoyed it, I was laughing like you are at the way Simon massively over-complicates things. I just want to say, thank you for thinking about whether adding extra information will help to make a puzzle _fun_ rather than just whether it is humanable 👍🏻

[@maartenwitpaard1737]

Nice to see your reaction video to Simon's solve.

[@mickwoolley8267]

The break-in by Simon is really hilarious, but turns out brilliantly. It's clear Simon never worked with the pairing up of the 2 cell renbans to cut down combinations. But great fun anyway. Happy to hear how long Zetamath works on building puzzles, as sometimes I take absolutely ages to get anything that I am happy with.

[@jomo87]

43:56 "Okay.... that's very interesting but doesn't seem to do anything..." Excellent poker face at this point, Zetamath :p

[@r.w.emersonii3501]

I'm trying to get a mathematical "handle" on a simple arithmetical puzzle. As an aside, the puzzle can serve as a model for certain principles of quantum theory. Take a grid, 7x7, say, and put a digit in each cell, either randomly or according to a pattern. Let's use a ring pattern: 1 in the center, surrounded by a ring of 2's, surrounded by a ring of 3's, etc.. Let's call this grid the "particle universe", where the numbers indicate the particle "property" or "frequency". The "particle universe" is spatial, in that cells can be called "adjacent" and cell position matters Now stand on one of the cells -- top left corner, say -- and use the digit you are standing as the number of moves you are allowed in any direction. Perform the move, and then repeat until you can go no further. The set of all possible moves has a tree structure, with the initial cell as the trunk. Let's call the set of trees -- one for each cell -- the "wave universe". The "wave universe" is non-spatial: The trees intertwine, and all of the cells on a given tree are "quantum entangled". Questions: What is the maximum depth or height of a given tree? What cells, if any, are not reached by the tree? What are the conditions for a tree to reach every cell of the grid? My problem is that there are just too many possibilities for mathematical analysis to produce interesting results. The situation is hopelessly amorphous. I've tried reducing the number of digits to just one or two, and I've tried looking for tree equivalence classes, but have found nothing earthshaking. Has anyone else analyzed this puzzle?

[@violetasuklevska9074]

25:11 You elude to the given 2 being unnecessary and I got a 24 minute time with this constraint, albeit my original solve with the given 2 was 30-40 minutes so I had some experience going in. Infamously, I will now proceed to write my solution (up to the 'given' 2) in a needlessly long comment that nobody will read. I will also share some thoughts at the end, so here we go: Consider the set of 8 renbans of length 2 (A) and the set of 7 renbans of length 3 (B), because there are exactly 8 & 7 distinct options for a renban in A & B respectively, we must use all available options for these renbans exactly once, we deduce the following statements: -The count of the numbers 1 and 9 in all of A is 1 and for all other numbers in A the count is 2 (*). -The count of the numbers 1 and 9 in all of B is 1, for 2 and 8 the count is 2 and for all other numbers in B the count is 3 (~). R5C5 must be 1 or 9 because of (*), this number also must appear in one cell of A which has to be R1C4, by extension R2C5 is 2 or 8. The 1 or 9 in row 5 that isn't in column 5 is also a part of 2R, because of (*) it can't reappear in A so this 1 or 9, which has to appear somewhere in column 5, must be in R1C5. Since R1C4 and R5C5 are the same digit (1 or 9), this digit must appear in the intersection of box 8 and column 6 which is precisely a renban from B which is now either 123 or 789. The renban containing R1C5 is now either 123 or 789, with the given 8 we can fill the options for R1C6 as 237 and R2C6 as 238. Thus, we have approximately located the positions of the 1 and 9 in A and B and because of (*) and (~) they can't reappear in A and B. Looking at row 3, the 9 can only appear in R3C9 since it can't appear in A or B and the renban of length 4 can't contain an 8 to complement the 9. For the same reasons the 1 in row 3 can only appear on the renban of length 4 which forces that renban to be 1234. Looking at previous deductions, if there's no 8 between R2C5 and R2C6 this would break at least one of the two renbans containing those 2 cells, so there's an 8 between those 2 cells which combines nicely with the given 8 to yield an 8 on the renban in box 3 and because of uniqueness this renban becomes 678. Consider R3C3, this digit can't reappear on it's own renban and also can't reappear in the row, effectively seeing 4 cells in box 2, these 4 cells contain 4, 5 and 6 but also since this renban is in B it also can't contain 1 or 9. Effectively R3C3 can only be 2, 3 or 7 (also considering the given 8). If R1C5 is a 1 we can fill in the 9 and 8 adjacent to it and approximate the 2 and 3 on its renban, the remaining cells in box 2 form a 4567 quadruple. Now considering R3C3, it also can't be a 7 since it 'sees' the quadruple and it can't be a 2 because of its renban. It must be a 3 which forces its renban to have a 4 and 5, which splits the previous quadruple into two dominoes, but the 67 domino breaks the 678 renban in box 3. The assumption is incorrect and therefore R1C5 is not a 1, but rather a 9. We fill in the 1, 2, 7 and 8 in box 2, the 1 in R5C5 and the 1 in box 8 is now on a renban from B which becomes a 123 triple. The reaming cells in box 2 form a 3456 quadruple, which means R3C3 is not a 3 and is either 2 or 7. If R3C3 is a 7 then its renban contains a 56 domino in column 4. Now looking at the renban from B partially inside box 8, the domino that is in box 8 can only contain 4, 7, 8 or 9, but 8 and 9 can't reappear in B because of (~)and 4 and 7 also can't suffice for this renban. The assumption is incorrect and therefore R3C3 is not a 7, but rather a 2. The 2 in R3C3 forces a 2 in R4C2 (since that renban was a 1234) and also a 34 domino onto its own renban. The 3 must appear a third time in B because of (~). Because of the 123 triple in box 8 and the 34 domino in column 4 a three can't go in R3C9 or that renban either. With the 123 triple in column 6 the only remaining spots for the third 3 in B are all in box 6. Now looking at row 5, the 3 can only appear in the first 4 four columns. Because of (*) there's a second 2 in A which must be in row 5 and also must partner up with a 3 on its renban (since it has already partnered up with a 1). The 2 must also be in the first 4 columns of row 5 and with the help of the 2 in R4C2 this new 2 gets placed in R5C4. Finally, the 2 in box 6 is placed in R6C9 since it can't a appear a third time in B because of (~) and the rest is history. Unsurprisingly, it's faster to communicate this verbally than to write it all down. I tried to tidy it as much as I could while condensing it into as few steps as possible, but rarely sparing words. This was not too complicated which I guess was the original point, but at this point this has become something else. Thanks for reading! Have a cookie🍪

[@ellaenchanted2399]

Excited to watch this!

[@JohnRandomness105]

I did the puzzle and watched him do it -- last night? I was surprised at how long it took him to find the 12-34-56-78-9 and 1-23-45-67-89 patterns. I found actually getting the right doubles and triples a lot harder. I"m NOT a Sudoku expert. It took me time disambiguating the 19 and placing the top 8 at the right spot, for example. I think that I finished in a couple hours. 11:45 -- Now that someone mentioned it, I do see a frowning face at the left. 18:40 -- He's completely missing the point, right? I saw and heard your guffaw. 45:00 -- I can't remember if I kept shouting, "The 2, the 2, the 2!" here. I've forgotten so much.

[@57thorns]

18:40 is a classic Simon: If you can solve it the easy way, you can solve it the hard way.

[@TPH250290]

Simon's always had a problem with never following his logic through all the way to the sensible conclusion. As soon as you work out the thing with the two-cell renbans it should be obvious that the central cell is a 1 or a 9. He basically said it and then didn't follow it up.

[@Jonas.Nilsson]

I love this! "oh no he doing SET" 🤣🤣🤣

[@57thorns]

I love that you put in that 2, having to keep track of all the possible pairing of 19 using colouring instead would have been a pain, and it would detract from what I feel is the main theme of the puzzle.

[@zetamathdoespuzzles]

Yes, I have no regrets about the 2. I could have done it with an inequality symbol, but I know a lot of solvers really, really hate random inequality symbols or kropki.

[@davidmiller9485]

Part of Simons "problem" is that he's trying to come up with the wording for what he's both thinking and doing while also doing the logic in his head. As someone who has done public speaking, it's rough trying to get complicated thoughts across for a general audience. (He also tends to miss things because he's in a rush ... :P )

[@iabervon]

The first thought I had when looking at the grid was to ask whether the renbans in row 5 had to be adjacent, and trying to prove that led me to the intended break-in. It was a bit disconcerting to see Simon solve 2/3 of the puzzle and 3/4 of the 2-cell renbans without using the first conjecture I proved. I was relying the whole time on being able to determine at a glance what went with any digit on a particular 2-cell renban, and he just didn't need that.

[@douglasreay1648]

If you allow diagonal arrows, there are 19 valid length 2 arrows with distinct digits (1x3 2x4 2x5 3x6 3x7 4x8 4x9)

[@JohnRandomness105]

24:00 Watching parts again. "This is definitely not the approachable way of doing the logic." Snerk.

[@57thorns]

Tried this a few time earlier buy always messed it up som I did have the breakin "for free", still a 27 minute solve is not that bad.

[@57thorns]

23:50 it does give a few more 19 cells, but going immediately for the disambiguation of 19 in r5c6 feels a lot more rewarding, as it gives digits in the grid. 36:55 I don't think Simon ever realise you need two sets of two cell renbans, the odds low (12 34 56 78) and the even low (23 45 67 89) that has to be on different parts of the cross. At least not until 38:43. Keeping that in mind helps a lot with solving the puzzle. At 43:10 Kuraban mentions that a piece of logic was bypassed in a solve of one of their puzzles. I would not say that Simon "bypassed" the approachable path here. A bypass in my book is that a solve becomes significantly easier using unintended techniques. Say the first Mephistomefel's theorem puzzle where Simon did not get the memo from Mark on the theorem.

[@zetamathdoespuzzles]

I don't think Simon bypassed what was going on. I think he correctly identified it was going to be hard to put many 19s on cells, and then did basically step one and step two of the puzzle together simultaneously!

[@OG29]

I think Simon's glasses somehow limit his ability to scan; perhaps in combination with his laptop.

[@philiplindley7384]

Yes, I've often considered this. His ability for panoramic vision is limited.

[@emdiar6588]

I've said it many times. Simon will stubbornly ignore the implication of a given digit if he can use the square root of -1 to solve Fermat's last theorem to get the same result. It makes me laugh and drives me mad in equal measure.

[@ryanrauber4669]

Yo, if you're just going to pause him when you talk could you make sure your levels are even? I'm trying to watch and every time you talk it's 98,000x louder than Simon and I have to change my volume. Great puzzle BTW. I love your math vids.

[@Roblilley999]

Zygotically Identical Twin is a great Prog Rock Band Name

[@matthewjohnson6360]

I came by your channel from Chameleon's. It's crazy SOMETIME Simon doesn't use proper grammar his says maths instead of math he could just say mathematics. It also gets annoying when he says sorry all the time