Відео

Very pretty grid!!! Well done Zeta and chat!

How to stymie Simon: force him to do sudoku. It insults his intelligence to do basic sudoku. It's like asking Mozart to play Guitar Hero.

@50:00 A simpler deduction is that you cannot put any more 5s on the left side of the zipper, so you cannot put any 3s on the right. That determines the V is 23. Not sure what follows after, but it is less look-ahead.

Interesting with 'Outset' how the entire diagonal consists of threes except for two cells which add to three.

Always fun to watch you following my logic only to figure out why it’s horrendously wrong. It’s quite reassuring to watch others hit CTRL+Z. As I’m still not that familiar with renbans, I’d never have seen this as a colouring puzzle, so for me this was educational as well as a fun watch. Cheers my friend!

I hope you're still planning on continuing the maths series..

just thought i'd mention how i figured out the middle german whispers line because it was slightly different to simon's approach. and it has to do with the way i started solving german whispers sudokus which is coloring them (mostly longer ones) in colors which are not "my" low and high variant until i have it figured out but instead randomly selected colors and put them into context when applicable. meaning i had 2 colors on the left side, 2 different ones on the right and after finding the 1/9 pair distilled it down to 2 colors. so i'm mostly thinking in a "low" and "high" way but colored. then i thought about the 2cell segments of 10-line and realized that they both always needed at least 1 low digit. that paired with the low high pair on the outside leaves only 1 low digit for the middle german whispers line which obviously goes into the middle. it's not much different than asking "where do 7/8 aka the high numbers go in the row but since i was already in the mindset of high/low it felt very natural to me edit: oh and i think it goes without saying but this was once again an absolutely wonderful puzzle with some beautiful logic in it!

Absolutely loved this puzzle!

At 1:42 when he is trying to avoid using the knowledge of the 3 in the corner - and ignores the 27 siting next to a known 7!

I made extensive use of a virtual 46 pair in c5 to resolve various things around box 8 and 9 and am surprised to see neither simon nor zeta remark on it, so presumably not intended, cool nonetheless.

I love listening to the logic you employ to set these and it occurred to me that you speak like a sculptor and the logic ‘reveals’ itself to you as you form it. Like the puzzle is hidden in the grid and you bring it to fruition ❤

I am voting for ¨ Differ by at least two¨

At 1:06:55, another approach is that if there is the 343 ten sequence that Simon missed, that would create a known 1, 3 and 6 in column 7 which forces the remaining two cell ten sequence to be 28. This proves that the 2 pencil mark in column 7 of box 3 is indeed correct.

I tend to use the colour cells by religion to see what regions a cell sees - as long as there are less regions then there are colours

Happy birthday!

Hi again, glad you’ve found time for the 6x6. It is also cool to see that many clever people in chat you can rely on to spot the lovely logic and baked-in beauty. The solve path was exactly as intended.

Thank you so much for featuring Yin Yagn. Hope you had a wonderful birthday!

I definitely found this one easier to see with the grid topology effect, which is that each digit must appear in one of: a diagonal pair of corners, a corner and the middle of the three opposite faces, or the middles of all six faces. When you found that there are two repeats on the arrows, each repeat must therefore be both cells of an arrow or both circles (because those are your diagonal pairs of corners), and putting both repeats on the arrow cells doesn't work. Then the diagonal arrow with different digits has to avoid putting a triple of faces somewhere that gets both thermo cells. (The proof from my earlier comment: Each cell is in 1 or 3 regions, and there are 6 regions. The ways to make 6 out of 1s and 3s are: 3+3, 3+1+1+1, and 1+1+1+1+1+1.)

pushing the right buttons is boring Tris - way more fun for Zeta to randomly do correct/opposite pencil marks

I have no idea how anti-knight could work on that grid - but it could be fun to find out

Glad to see Tristan in one of these videos and keeping Zeta in line.

Love that Zeta articulates that there is not 1 on this black dot because it would see a node marked 12. 3 minutes later, forgetting that, pencil marks a 14 option on the black dot. Then finds a different 15 pair in the slice to rule it out. :) It all solves out in the end.

I couldn't tell what would have been useful to know in this puzzle, but I did have some grid thoughts: 12 regions and each cell being in 3 of them means that there are 4 of every digit. If you pick an orientation to be the top and a digit, there's got to be one of that digit in the top, one of that digit in the bottom, and two around the middle. The middle is kind of a zig-zag, alternating digits closer to the top and digits closer to the bottom. Of the regions around the top, a digit in the middle poking up takes care of two of them, while a digit in the middle poking down takes care of one of them. The top itself takes care of two of them, and, since there are five total, there must be one digit poking up and one poking down to get the other three. Considering the top and bottom, (as you noticed thinking of them as opposite corners), the matching digits can't be opposite each other, but I think every other alignment works.

Well done Zeta. Doing logic puzzles while sick is difficult.

Opposite faces are key to this topology.

I was saying what blue is for the longest time after you sorted out the 3x7 and the 4-8 black dot on the second horizontal slice.

Sorry, but what about a new video on zetamath cannel? I really appreciate your work and what you introduce and presenting to us, like a really mean that im very thankful for you, i didn't understand math before like that after seeing your pretty cool videos

WARNING: SPOILERS! Edit: You get there later on 18:10, but don't seem to fully generalize it yet, I simply need to keep watching :) I think that at 6:18 you can place a 1 at layer 4 in the middle. The reasoning is as follows: The 345 and the 34 on the outer edges of layer 2 (from the bottom) both see the center digits of layers 1, 2 and 4. Combine that with them being unique, because they are on the same layer. As you mentioned in the beginning of the video, the 5 center cells are unique. That should mean that they need to exist in layers 3 and 5 in the center, thereby these layers can only contain the digits 345. And as 1 is already eliminated from layer 1, that leaves a 1 in layer 4. I suspect that this thinking could be applied to all layers in this geometry, where the 2 nodes that have 4 connections have to be different and be exactly 2 center cells.

The whole grid has every digit twice. The equator ring has every digit once. Therefore, the top renban triangle plus the bottom whispers together have every digit once... I am just saying that the puzzles are getting difficult enough to start thinking about SETs from time to time ;-)

Are you going to continue your videos on the main channel about analytic continuation and the Riemann Zeta function? They're one of the best videos I've seen on the topic, and it would be a shame to not continue it

The 1 pencilmarked on a thermo tip was only there for a couple minutes and it didn't make any difference.

At 7:10, when you claim that the arrow point sees the other arrow cell, so it can't be 1+1, I am not sure I followed how they see each other. But in looking at the arrow cells, the 135 cell that sees both arrow cells can't be a 1 because it forces the arrow to be a 2+3=5 pair and the 2 would be in the same vertical ring/slice as the corner marked 1, but the 1 was on a white dot so would have repeated a 2 in the vertical slice.

Thank you for remembering to check your regions.

After you've noticed that 4 must be in one of the two white nodes in a straight box, you could notice that those white nodes must be of different high-low polarity. Thus, they can only be from 1234. This includes an end of a thermo, forcing the thermo bulb to be low. Same logic works for the other thermo as well.

The parity thing in four colours that become two is very cool. "By convention, if your last name isn't Anthony, you make evens blue [...]". That callout is hilarious.

Damn, that puzzle was a lot. And that's without the 30 minutes cut from the video.

Geryon - earlier iterations of the puzzle used sum dots in ways that were not emulatable with killer cages. It just so happened that the final version didn't.

Good evening everyone

That was a very nice puzzle.

This one was super fast. Maybe in part because it's the first time the dodecahedron appears.

Around 7:10, you had 3 cells highlighted for the 4 of a certain region. You could have placed the 4 at that point, but I don't think it would have changed the rest of the solve. This varies from person to person but, when you already have multiple of a digit on this grid in particular, it's easier for me to see things if I ask "Which colours still don't have this digit?" and find the intersection of those colours, instead of choosing a region and asking where it can go there.

I came up with a concise proof of how this grid works. There are 6 regions that need each digit. A digit in a corner accounts for 3, while a digit in a center accounts for 1. How do you make 6 out of 3s and 1s? The options are two 3s, one 3 and three 1s, or six 1s. This means that the possibilities for where a digit is are: a opposite pair of corners, one corner and all three centers around the opposite corner, or all of the centers. One consequence of this is that every pair of opposite centers in the puzzle has the same sum, which gives you two digits immediately in this one by subtracting cages.

What's with the red hearts between 6 and 9s?

My favorite part is when Zeta asks if any cell sees all 3 of the remaining cells on the length 4 renban, highlights a cell and says, "I guess there are none that see them all," when that cell sees them all. :)

When you got the 2 in the tip of the 2 cell thermo, you made said that is a 4, and placed 4 in the bulb. I think you just inverted that thermo.

"I bet what I did is I said these 'two are 25' and I hit 25, rather than hitting 16." You did leave 25 pencilmarks, but in the wrong cells. Which sounds like easy mix-up to make on a round puzzle.

This is actually my personal favorite grid in the game. Although the most fun stuff starts happening at level 6 of this grid. I see this grid as a set of 8 intersecting circles placed around the center of the grid. Every circle is a box. Every two circles except for the opposite ones intersect in a single node. Every node is an intersection of two circles, which are either 1 apart for the outer ring of nodes, 2 apart for the middle ring, or 3 apart for the inner ring. BTW, if you have 3 copies of the same digit placed, you can always place the last one.

Good job on catching and correcting the mistake of removing the wrong pencilmark from a cell.

Some of these puzzle where you can see all the dots in one or two orientations, it may actually be easier to not lose things if you let it be flat instead of the constant rotation.

Other people pointed out alternative ways to look at the 4, but I don't think your reasoning is that convoluted. In a regular sudoku, once we placed seven 4s, it's natural to look at the last two placements of 4 together because either of them forces the other. But here there are only three 4s, so you can do that as soon as you place the first 4.