Thank you for this wonderfully smooth feature - I think quite a few people who have attempted this so far struggled with the bit you mentioned in your closing remarks, but you sailed through it very nicely indeed! I didn't put too much thought into the title, only that it should relate to the TOroidAL nature of the grid... There'a a bit of a funny story behind the publication of this one. I was watching Scojo's Impostor stream and my pair of puzzles (one by me, one by my impostor) was about to come up, when @patrickgass787 aka ViKingPrime (one of the kindest and funniest members of the community) playfully mocked the complexity of my rulesets by predicting that they would include ludicrous nonsense such as toroidal little killers... I had this puzzle ready to publish the following morning, but after his comment, I published it there and then and sent him the link. He really does know me well!! Thanks so much to those who recommended the puzzle, I'm feeling very loved by the sudoku community at the moment! I'll hopefully see as many of you as possible in Stratford next month!! :)
Could I request an edit to the wording of the rules on svens website if that's possible? To me, the current writing of the rules specifying that there must be a "continuous" run of "cellS" makes me think that you could alternate colours along a diagonal until a certain point, and only start counting once you have two or more of the same colour in a row.
You've gone and done it! Ever since my first Yin Yang construction, I'd vaguely had the idea of Yin Yang on a torus, precisely because it messes with the checkerboard and border lemmas. I just hadn't taken it any further than just a thought. Bravo!
But it can't be a torus: top meets bottom AND left meets right. There's no hole. It also can't be spherical, because: say we start with left meeting right, then the only way to seal the ends would be top to top, bottom to bottom. It's probably a shape that can't exist in a 3 dimensional space. I think it's most probably a 4d hyper-torus.
Actually that is how a torus works. Imagine rolling this grid into a cylinder, with the top connected to the bottom, then connecting the ends of the cylinder into a torus, connecting the left side to the right.
@@Tahgtahv it did for mine though - I had 3 of the corners in one colour and realized that meant the 4th one had to be the other colour as they form a 2x2
Absolutely stunning little puzzle from the marvelous mind of the PedallingPianist. I'm afraid the 3 in the corner song must be rescinded though - there are no corners on a torus 😜
True that. Simon's "bipolarish" explanations and solves: he goes into extreme detail with the rules, and seems to have troubles understanding/explaining simple things....but regularly spots & solves extremely difficult logic. Slow to accelerate...but hard to keep up with him once he gets going.
@@leestoddart7014 That’s true, his brain most likely sees patterns differently than someone like I see them. I especially thank that if someone only really does very difficult puzzles, they get used to having to see more out of the box patterns instead of the usual easy ones. In that way, one no longer notices the simple patterns and solutions. Of course, this is only speculation, but I think it makes sense.
I found it helpful to color the squares outside the grid to keep track of what colors were virtually next to the edges due to the toroidal. Then it was easier to spot potential islands and 4x4s, particularly at the corners.
The infinity is new. You've solved two other sudokus that uses the torus rule: 1. Poisoned Bagel by Bismuth 4 years ago 2. Mozzarella by Perladel almost 2 years ago
@pairot01 haha they probably checked the video index (link in the description) but I'd be impressed if someone just liked the toroidal rule enough to carry that in their back pocket
@@pairot01 I remember there were two videos before. The oldest one I remember the video were called "the 98% sudoku". The other one I search "cracking the cryptic peradel" and got the name of the other one.
I think there is some form of checker-board avoidance on a toroidal grid: namely there can be at most two checkerboard 2x2s (which may be overlapping as in this puzzle). Proving that requires a bit of topology: In this proof I will use the colors red and blue. Suppose to the contrary there was a grid with at least 3 checkerboard 2x2s. Then choose one such 2x2 and consider the orthogonal paths connecting the diagonal squares. Cutting the torus along those paths, one obtains (up to homeomorphism) a disk, whose boundary has two opposing stretches of red and two opposing stretches of blue. Now in the interior of this disk, there are 2 more checkerboards. For each of these checkerboards, the red square have to connect. But they cannot connect to one-another, as that would isolate the blues (as in non-toroidal ying yang). Hence, they all connect to the boundary. But whichever way one does this, it will create isolated regions of blue, which yields a contradiction. Hence, there can be at most two checkerboard 2x2s.
My immediate thought was that there would be only one checkerboard, but that was proven wrong by this puzzle. In fact, Simon colored a valid toroidal with no checkerboards. Thank you for the thoughts on three!
This is one of my all time favourite puzzles and really could only have been done by ThePedallingPianist (and that is a challenge to all you imposters out there!)
There could be a Klein bottle grid if the orientation was flipped in the perpendicular axis upon crossing an edge. I have no idea how to make this into a puzzle, but it is possible!
Rules: 05:16 Let's Get Cracking: 09:47 Simon's time: 26m07s Puzzle Solved: 35:54 What about this video's Top Tier Simarkisms?! Maverick: 2x (08:03, 26:03) The Secret: 2x (12:22, 12:22) Three In the Corner: 1x (35:29) And how about this video's Simarkisms?! Weird: 7x (01:27, 08:35, 09:02, 15:46, 16:03, 16:07, 16:58) Hang On: 5x (09:13, 16:18, 19:40, 24:55, 34:26) Goodness: 4x (02:13, 10:32, 36:54, 36:57) Sorry: 4x (21:22, 21:51, 22:04, 29:48) By Sudoku: 3x (34:17, 34:46) Checkerboard: 3x (11:17, 12:16, 13:02) Lunacy: 3x (35:51, 35:59, 35:59) Ah: 3x (17:59, 26:20, 35:40) What on Earth: 2x (18:07, 36:14) Nonsense: 2x (17:24, 24:44) Lovely: 2x (36:45, 36:45) Ridiculous: 2x (30:17, 30:17) Cake!: 2x (05:01, 05:02) Useless: 1x (31:09) Clever: 1x (36:20) In the Spotlight: 1x (35:30) Beautiful: 1x (32:36) Brilliant: 1x (35:13) Bonkers: 1x (03:19) Approachable: 1x (01:52) Surely: 1x (09:51) Obviously: 1x (22:22) That's Huge: 1x (33:57) Most popular number(>9), digit and colour this video: Ten (13 mentions) One, Two (34 mentions) Green (55 mentions) Antithesis Battles: High (2) - Low (0) Even (5) - Odd (1) Outside (2) - Inside (0) Column (6) - Row (5) FAQ: Q1: You missed something! A1: That could very well be the case! Human speech can be hard to understand for computers like me! Point out the ones that I missed and maybe I'll learn! Q2: Can you do this for another channel? A2: I've been thinking about that and wrote some code to make that possible. Let me know which channel you think would be a good fit!
toroid yin yang was interesting. I appreciated being able to work with the outside grid, copying my colors over to the opposite side helped me keep track of successful connections.
I think there's a pretty important ambiguity in the rules: Simon solves this assuming that each of the diagonal clues also signal the 'start' of their colour strip. For example in the very first deduction he makes that proves that the infinity diagonal and 22 diagonal are different colours he uses the fact that then 10 diagonal needs at least 2 cells of the same colour, however that only follows if you assume that the second cell in the 10 diagonal needs to be r3c2. The rules don't explicitly rule out the possibility that the strip of r5c6 and r4 c1 is a valid possibility for that 10 clue. The rules only state that the clue applies to "the first continuous run on their diagonal" and if the clue sits in the middle of that run, so to speak, that would still be the first run the clue sees. If these kind of 'mid-run clues' are possible then I'm almost certain that the puzzle is impossible to solve so I think they do need to be said to be invalid within the rules text.
I agree it would be better to remove the ambiguity. However, I don't think it's impossible to solve if you allow mid-run clues. Because the 22 shares a diagonal with one of the 5 clues, those clues cannot be mid-run clues. It takes some doing, but I've pretty well convinced myself that whichever color the 22 diagonal is, you can't make either the 10 or 24 a mid-run clue without the coloring breaking down with either a 2x2 block or an isolated group somewhere.
If the clue sits in the middle of a run, it can't possibly be the first continuous run on the diagonal. The first cell starts (and may end) the first continuous run on the diagonal!
@@themorebeer3072 I don't think its obvious in any way that the first cell starts the run - after all the fact that its the first cell is just an artifact of how we've chosen to place the grid, which shouldn't matter since the grip is representing a torus which doesn't have any edges
@@alexhawco2970 I place an arrow pointing at a grid. What is the first cell along that arrow? Sure it's an artifact of how we place the grid. It's also an artifact of how we placed the arrow. That's still pretty obvious however.
19:16 for me. Very nice puzzle. Mostly a matter of not assuming too much. Eg about checker boards or how many cells can make up a 5 clue. Keep yourself grounded and it is fairly straightforward.
I discovered this on Logic Masters a few days ago and in the process of solving it got really tied up on the topic about checker boards, so I took to a whiteboard and I *think* I've convinced myself that for a toroidal yin-yang puzzle there can be at most two checker boards in the grid. If you imagine an actual doughnut shape in 3D, one checker board would represent a green line going around the doughnut and a purple one looping through the hole, while the other checker board would be a point where a green line loops through the hole and a purple one wraps around the doughnut. You can do that much without green lines intersecting purple ones, but a third point on the torus with nontrivial loops of each colour cannot be placed. I didn't take the time to consider whether there's some issue taking that logic based on a continuous smooth surface and applying it to a grid of cells, but I think it holds. You can see at the end of Simon's solve for example, the purple cells form a loop horizontally across the centre of the grid, but also vertically along the left edge - two loops that the green cells can still manage to navigate. A third loop would be one too many though.
A mathematical way of approaching this (which I thought of but didn't quite use) is to relate the number of allowable checkerboard patterns to the Euler Characteristic of the surface. But here the logic is that you can't have more than two (there was another puzzle on the channel with a hole in the middle of a big grid where the same logic applied). It will help if someone creates a Klein Bottle puzzle, for example.
I had serious new ruleset struggles with this one - I never attempted a torodial yin/yang before - and I kept getting confused about whether edge cells could get out or not. Took me about 45 minutes.
Hi~ First time commenting here, hope it will help. This puzzle reminds me of the one I've solved a couple weeks ago called "Twin Galaxies in Curved Space". It has similar topology and yin-yang rules, with some insane setup that leads to a unique solution. More importantly, it's a 9x9 puzzle which brings more fun ;) It would be amazing to see you guys feature that if it's possible. It only has 14 solves right now, and I truly believe it deserves more.
Stunning and funny simultaneously as a challenge (if not technically hard, still a challenge to wrap your head around it) and some hilariousness with the infinity diagonal. Took me about 22 minutes to solve.
you didn't use it, but in my playthrought, the fact that the cases right outside the 6x6 could be colored to continue the pattern really helped me a lot
Brilliant puzzle, its not often that a completely new set of rules can produce a challenging but approachable puzzle for me. Took me 65 minutes but enjoyed the challenge.
That was a great new twist on an old rule set. I wish Simon had realized you can mirror the cells on the outside of the grid. That makes it easier to visualize.
Fantastic puzzle took a couple of tries to get it right. The successful solve took 33 minutes while being somewhat distracted. Those distractions are probably what caused the earlier attempt to fail.
Simon and Mark, if you ever plan to do a performance during a festival in the Netherlands, probably a lot of people like me (who aren't going to make it to Stratford) will try to be first in the queue to buy a ticket ;-)
One 'secret' of a toroidal yin-yang puzzle that I have just noticed (and no doubt other people have so sorry if I'm stating the obvious) is if all four corners were the same colour, they would form a 2x2 via the torus. So at least one corner must be different from the rest.
Alto as is the musical term (in italian) for high, perhaps? Infinity is pretty high. Also, if you pick only one side for connecting the grid to a cylinder, it may represent a wax cylinder for a phonograph, if not a music-box cylinder.
12:30 The other secret is still valid but void. The perimiter can still have only one change of colour but what use can you make of it when there is no such thing as perimiter?
Finished in about 20 minutes, only i hadnt as i stupidly msde the 5 diagonal only 2 cells long and then found a valid solution that worked. Then when i realised it wasnt the solution back-pedaled and then finished properly in 24 minutes :)
28:51 STOP! Once you made R5C4 green, you've now forced your 22 clue starting at R1C6. 6+6+6+4 = 22 6+5+6+5 = 22 Those are the only options you have. And if it's 6+6+6+4 then the only cell that can hold the 4 is in box 4. So R1C6 and R2C1 are from [56] and R3C2 / R4C3 are from [456] But remember the 10 clue? It can only be [46] And if R4C1 were a 6, all options for the 22 clue would be impossible. So now the 22 clue IS 6+5+6+5 and R3C2 is a 6 and R4C3 is a 5 and R4C1 is a 4. Then, as R2C1 and R1C6 are effectively a [56] pair looking at R1C3 and R2C4, which is along the 24 clue, and those cells can only be as high as a 4. And R4C5 will see a 6. So it can only be as high as a 5. The highest the first two cells on the 24 clue can be is double 4, adding to 8. The highest the 3 remaining cells on the 24 clue can be is 17 (5+6+5 in that exact order for sudoku reasons) so you only have one degree of freedom left on the 24 clue which at most could be 8 + 17 = 25.That degree of freedom can only exist in one of the first three cells as the last 2 would otherwise have one digit only as high as 3 and you would lose 2 degrees of freedom. Point is, the last two cells on the 24 clue are from [56] and the order is known and resolves all [56] pairs you have so far. I have no idea how i just intuitively knew that limiting the 22 clue to stop where it does would resolve a lot of the puzzle but that is what i was seeing.
16:06 had to restart a couple times, because I assumed maybe there was only 1 checkboard allowed. Instead I avoided using checkboard. Coloring the extra cells on the border helped with spotting potential 2x2s, but the connection issues were hard to spot
This is, for me I think, the LEAST approachable sudoku I've attempted on this channel. It's pretty, but I had to be hand-held for pretty much the entire thing.
I would like to suggest 'dominoid', 'triominoid' etc for toroidal dominoes. Except triominoid could be shortened to 'trioid' and tetrominoid to 'tetroid' for absolutely no reason other than that I think it feels quite satisfying.
Building a Klein bottle in a sudoku would actually be quite easy. All you do is take a normal toroidal rule and "flip" one side upside down (so that r1c1 is "orthogonal" to r9c9, r2c1 to r8c9, r3c1 to r7c9, and so on). For a Mobius strip you do the same thing except instead of having a normal toroidal edge you have one edge that's, well, just an edge (reusing the prior example, r1c5 isn't adjacent to anything in column 9).
I found this puzzle extremely hard - or I should say, easy to make a mistake. I ended up in situation where there’s no solution that works and had to restart twice. With third attempt though it took me about 15min to color it and putting in digits was just a minute. It was a good one but I as much as I love ying yang, I prefer to look at a grid as a square 😅
30:51, was hard to see if I was in a dead end or not when the regions looped around and around and around and around. And not being able to use the "rule" that you couldn't have the two colors in the opposite corners of a 2x2 region.
Nice concept... But think "grid is toroidal " should be the second thing stated behind sudoku... Trying to grasp orthogonally connected before noting toriodal reallly was a bugger. Who would have thought the easiest thing to grasp was the inf.
Not sure if I'm looking at the right point, but for a long time the green region occupying r5c2 could take r6c1 and loop round to r6c6 toroidally to connect to the other side.
so while working on this puzzle...a silly pointless thought crossed my head.... how many designers who are fans of this show start off their puzzles wth a 3 in the corner, just to get simon to sing
@@RichSmith77 but it takes at least one to prevent the infinity loop… so first one from edge has to be green I think. Dunno… I am not good at this. I just like watching.
@@betazep I think that's the trap I fell into during my solve. It could have been one purple, one green, four purples from the direction of the 5 clue, with a single 5 in purple. (I too coloured r6c5 green, at first, because I also overlooked this possibility on my first solve.)
The mathematical part of my mind is troubled by the designation of the grid as being a torus The rules of the puzzle are symmetric as to going off the top and coming back in at the bottom or going off the left and coming back in at the right and vice versa in each case. So the grid has a 4-fold symmetry with respect to these transitions. But a torus seen from the side only has a 2-fold symmetry. Going left-right entails going around the equator of the torus, whereas going top-bottom entails diving through the hole in the torus. To me, this clash of symmetries represents a horrid mental dissonance.
Colour scheme today was lit! I would like to know the opinion of colour blind people, maybe this can become a new standard since orange and blue always give an odd/even vibe
I was wondering if that's true in general, but I took it over to a 9x9 grid where there's a little more room to stretch, and it was pretty easy to make 2 checkerboards. However I do think 2 is probably the max. Something about how each color can connect across the north-south edge once, and they can also both connect across the east-west edge once, but then those connections prevent any further ones, so each color can only connect once each on each edge, which can then only accommodate circumventing the normal no-checkerboard rule twice. But.. that's hardly a proof. We'll have to wait for some insane person to make poor Simon prove it generally 🙃
You could call it undefined or divergent, but you'd probably need to explain it in the rules (to distinguish from infinity and negative infinity, unless you want to allow the ambiguity). Also calling it undefined sounds a bit like you just aren't given it as a clue. Interesting idea, I had a quick look on the Wikipedia pages for divergent series and limits, but didn't find any better terminology.
@@caspianmaclean8122I don't see why it wouldn't be zero, if one "lap" sums to zero. N x 0 is zero for all N, so surely in the limit, as N goes to infinity, the sum is still zero, no?
@@RichSmith77 that would only work for a whole number of laps. It might have a sum of 0 after 9N cells (a whole number of laps), but a sum of -4 after 9N+1 cells, if the first cell in each lap is a -4. If the sum keeps changing between 0 and -4 as you include more cells, the sum series generally wouldn't count as converging to a number. There's special series sum definitions you could use to get an answer anyway, but not necessarily 0.
Thank you for this wonderfully smooth feature - I think quite a few people who have attempted this so far struggled with the bit you mentioned in your closing remarks, but you sailed through it very nicely indeed! I didn't put too much thought into the title, only that it should relate to the TOroidAL nature of the grid...
There'a a bit of a funny story behind the publication of this one. I was watching Scojo's Impostor stream and my pair of puzzles (one by me, one by my impostor) was about to come up, when @patrickgass787 aka ViKingPrime (one of the kindest and funniest members of the community) playfully mocked the complexity of my rulesets by predicting that they would include ludicrous nonsense such as toroidal little killers... I had this puzzle ready to publish the following morning, but after his comment, I published it there and then and sent him the link. He really does know me well!!
Thanks so much to those who recommended the puzzle, I'm feeling very loved by the sudoku community at the moment! I'll hopefully see as many of you as possible in Stratford next month!! :)
I've never experienced a bigger mic drop in my entire life, well played, that one
you’re a genius, lovely puzzle once again!
Ah, toirodALTOroidal...
Could I request an edit to the wording of the rules on svens website if that's possible?
To me, the current writing of the rules specifying that there must be a "continuous" run of "cellS" makes me think that you could alternate colours along a diagonal until a certain point, and only start counting once you have two or more of the same colour in a row.
You've gone and done it! Ever since my first Yin Yang construction, I'd vaguely had the idea of Yin Yang on a torus, precisely because it messes with the checkerboard and border lemmas. I just hadn't taken it any further than just a thought. Bravo!
There's no 3 in the corner in this puzzle because a torus has no corners
But it can't be a torus: top meets bottom AND left meets right. There's no hole. It also can't be spherical, because: say we start with left meeting right, then the only way to seal the ends would be top to top, bottom to bottom. It's probably a shape that can't exist in a 3 dimensional space. I think it's most probably a 4d hyper-torus.
Actually that is how a torus works. Imagine rolling this grid into a cylinder, with the top connected to the bottom, then connecting the ends of the cylinder into a torus, connecting the left side to the right.
@@alanclarke4646Woah. If it hadn't lost its religion already... 🤯
@@alanclarke4646 Top meeting bottom and left meeting right simultaneously is what makes a shape toroidal.
@@alanclarke4646 No, there is no hole in a torus's surface, you're on the torus, not looking at it
The sudden realization that he needs to avoid a 2X2 across the torus boundary felt nice.
While that is true, I don't remember it ever coming up in the solve?
@@TahgtahvI don't remember him using it either, and I was waiting for him to use it to colour r5c6 for 10 minutes from around 18:35.
It came up in my solve path rows 1 and 6 columns 4 and 5 somehow I picked up on it right away.
@@Tahgtahv Simon, as usual, found another way to get the same result. He never used what I realized.
@@Tahgtahv it did for mine though - I had 3 of the corners in one colour and realized that meant the 4th one had to be the other colour as they form a 2x2
Absolutely stunning little puzzle from the marvelous mind of the PedallingPianist. I'm afraid the 3 in the corner song must be rescinded though - there are no corners on a torus 😜
The confetti has spoken. There can be no doubt.
I think Simon would have found this a lot easier if he just coloured the squares outside of the Sudoku, so that he can see the pattern more easily.
What I did
Surely that’s the reason for making them fillable “cells”. Was completely obvious to me…
True that.
Simon's "bipolarish" explanations and solves: he goes into extreme detail with the rules, and seems to have troubles understanding/explaining simple things....but regularly spots & solves extremely difficult logic.
Slow to accelerate...but hard to keep up with him once he gets going.
Exactly - it amazes me how that simplification can be invisible to Simon when the most tricky logic just springs out to him as obvious.
@@leestoddart7014 That’s true, his brain most likely sees patterns differently than someone like I see them. I especially thank that if someone only really does very difficult puzzles, they get used to having to see more out of the box patterns instead of the usual easy ones. In that way, one no longer notices the simple patterns and solutions. Of course, this is only speculation, but I think it makes sense.
I thought I would give this a go.
A toroid- no corners to show.
And when 3 came along
I launched into the song
And sang "That's 3 in the... OH NO!"
I found it helpful to color the squares outside the grid to keep track of what colors were virtually next to the edges due to the toroidal. Then it was easier to spot potential islands and 4x4s, particularly at the corners.
Yeah, I think that was the author's intent
The infinity is new.
You've solved two other sudokus that uses the torus rule:
1. Poisoned Bagel by Bismuth 4 years ago
2. Mozzarella by Perladel almost 2 years ago
Did you pull those from memory?
@pairot01 haha they probably checked the video index (link in the description) but I'd be impressed if someone just liked the toroidal rule enough to carry that in their back pocket
@@pairot01 I remember there were two videos before.
The oldest one I remember the video were called "the 98% sudoku".
The other one I search "cracking the cryptic peradel" and got the name of the other one.
@@prodigis. If something is really good you have a higher chance of remembering it
I love ♾️. That’s hilarious.
I think there is some form of checker-board avoidance on a toroidal grid: namely there can be at most two checkerboard 2x2s (which may be overlapping as in this puzzle).
Proving that requires a bit of topology:
In this proof I will use the colors red and blue. Suppose to the contrary there was a grid with at least 3 checkerboard 2x2s. Then choose one such 2x2 and consider the orthogonal paths connecting the diagonal squares. Cutting the torus along those paths, one obtains (up to homeomorphism) a disk, whose boundary has two opposing stretches of red and two opposing stretches of blue. Now in the interior of this disk, there are 2 more checkerboards. For each of these checkerboards, the red square have to connect. But they cannot connect to one-another, as that would isolate the blues (as in non-toroidal ying yang). Hence, they all connect to the boundary. But whichever way one does this, it will create isolated regions of blue, which yields a contradiction. Hence, there can be at most two checkerboard 2x2s.
That's a very interesting point and proof, definitely use it for next time, thank you!
My immediate thought was that there would be only one checkerboard, but that was proven wrong by this puzzle. In fact, Simon colored a valid toroidal with no checkerboards. Thank you for the thoughts on three!
Wow, just wow. This is an incredible puzzle. Thanks so much for setting it, Peddling Pianist!
This is one of my all time favourite puzzles and really could only have been done by ThePedallingPianist (and that is a challenge to all you imposters out there!)
Your invented word "triominic" is not to be sneezed at (especially as it's an old drug used to to treat hay fever).
The word he "wanted" to use was actually, "TRIOMINOIC"....!
There could be a Klein bottle grid if the orientation was flipped in the perpendicular axis upon crossing an edge. I have no idea how to make this into a puzzle, but it is possible!
And a Moebius strip grid if that's only true for one pair of edges and the other pair consists of true edges.
What a brilliant puzzle! Another amazing puzzle from the The PedallingPianist.
Rules: 05:16
Let's Get Cracking: 09:47
Simon's time: 26m07s
Puzzle Solved: 35:54
What about this video's Top Tier Simarkisms?!
Maverick: 2x (08:03, 26:03)
The Secret: 2x (12:22, 12:22)
Three In the Corner: 1x (35:29)
And how about this video's Simarkisms?!
Weird: 7x (01:27, 08:35, 09:02, 15:46, 16:03, 16:07, 16:58)
Hang On: 5x (09:13, 16:18, 19:40, 24:55, 34:26)
Goodness: 4x (02:13, 10:32, 36:54, 36:57)
Sorry: 4x (21:22, 21:51, 22:04, 29:48)
By Sudoku: 3x (34:17, 34:46)
Checkerboard: 3x (11:17, 12:16, 13:02)
Lunacy: 3x (35:51, 35:59, 35:59)
Ah: 3x (17:59, 26:20, 35:40)
What on Earth: 2x (18:07, 36:14)
Nonsense: 2x (17:24, 24:44)
Lovely: 2x (36:45, 36:45)
Ridiculous: 2x (30:17, 30:17)
Cake!: 2x (05:01, 05:02)
Useless: 1x (31:09)
Clever: 1x (36:20)
In the Spotlight: 1x (35:30)
Beautiful: 1x (32:36)
Brilliant: 1x (35:13)
Bonkers: 1x (03:19)
Approachable: 1x (01:52)
Surely: 1x (09:51)
Obviously: 1x (22:22)
That's Huge: 1x (33:57)
Most popular number(>9), digit and colour this video:
Ten (13 mentions)
One, Two (34 mentions)
Green (55 mentions)
Antithesis Battles:
High (2) - Low (0)
Even (5) - Odd (1)
Outside (2) - Inside (0)
Column (6) - Row (5)
FAQ:
Q1: You missed something!
A1: That could very well be the case! Human speech can be hard to understand for computers like me! Point out the ones that I missed and maybe I'll learn!
Q2: Can you do this for another channel?
A2: I've been thinking about that and wrote some code to make that possible. Let me know which channel you think would be a good fit!
25:07 - "TRIOMINOIC" - That's the word you're looking for.
23:17 finish. A fun puzzle, but definitely pushes you, having to picture the wraparound. An excellent offering!
22:34 for me. Wow that was confusing. Great puzzle anyways!!
This puzzle has so original and fun logic!!!
You can color the cells outside the grid to represent what the connected cell across the grid is it helps to see the regions.
“Do come along, we’d love to meet you.”
Exactly what AI might say…
Took me over 40 minutes and over multiple sessions ... felt like sudoku's answer to non-Euclidian geometry
Insane puzzle!
Another video with the amazing Simon!
We had at least one torus here before... Well done, both to the setter and the solver.
Besides, the 4 colour theorem doesn't work at a torus as you need 7.
toroid yin yang was interesting. I appreciated being able to work with the outside grid, copying my colors over to the opposite side helped me keep track of successful connections.
So many incredible 6x6 lately, with new ideas, smart difficulty, and incredible craftsmanship. The Sudoku scene is in very good health !
I saw a torus and I knew immediately it was PedallingPianist lmao
I think there's a pretty important ambiguity in the rules: Simon solves this assuming that each of the diagonal clues also signal the 'start' of their colour strip. For example in the very first deduction he makes that proves that the infinity diagonal and 22 diagonal are different colours he uses the fact that then 10 diagonal needs at least 2 cells of the same colour, however that only follows if you assume that the second cell in the 10 diagonal needs to be r3c2. The rules don't explicitly rule out the possibility that the strip of r5c6 and r4 c1 is a valid possibility for that 10 clue. The rules only state that the clue applies to "the first continuous run on their diagonal" and if the clue sits in the middle of that run, so to speak, that would still be the first run the clue sees. If these kind of 'mid-run clues' are possible then I'm almost certain that the puzzle is impossible to solve so I think they do need to be said to be invalid within the rules text.
I agree it would be better to remove the ambiguity. However, I don't think it's impossible to solve if you allow mid-run clues. Because the 22 shares a diagonal with one of the 5 clues, those clues cannot be mid-run clues. It takes some doing, but I've pretty well convinced myself that whichever color the 22 diagonal is, you can't make either the 10 or 24 a mid-run clue without the coloring breaking down with either a 2x2 block or an isolated group somewhere.
If the clue sits in the middle of a run, it can't possibly be the first continuous run on the diagonal. The first cell starts (and may end) the first continuous run on the diagonal!
@@themorebeer3072 I don't think its obvious in any way that the first cell starts the run - after all the fact that its the first cell is just an artifact of how we've chosen to place the grid, which shouldn't matter since the grip is representing a torus which doesn't have any edges
@@alexhawco2970 I place an arrow pointing at a grid. What is the first cell along that arrow? Sure it's an artifact of how we place the grid. It's also an artifact of how we placed the arrow. That's still pretty obvious however.
I agree, this threw me off. It's hard to define a single cell as "a continuous run".
Clever path building that collapses quickly (not a complaint -- I think the path building is the point) once you get the path.
19:16 for me. Very nice puzzle. Mostly a matter of not assuming too much. Eg about checker boards or how many cells can make up a 5 clue. Keep yourself grounded and it is fairly straightforward.
40:20 I'm definitely stoked I managed to solve this awesome puzzle!
I discovered this on Logic Masters a few days ago and in the process of solving it got really tied up on the topic about checker boards, so I took to a whiteboard and I *think* I've convinced myself that for a toroidal yin-yang puzzle there can be at most two checker boards in the grid. If you imagine an actual doughnut shape in 3D, one checker board would represent a green line going around the doughnut and a purple one looping through the hole, while the other checker board would be a point where a green line loops through the hole and a purple one wraps around the doughnut. You can do that much without green lines intersecting purple ones, but a third point on the torus with nontrivial loops of each colour cannot be placed.
I didn't take the time to consider whether there's some issue taking that logic based on a continuous smooth surface and applying it to a grid of cells, but I think it holds. You can see at the end of Simon's solve for example, the purple cells form a loop horizontally across the centre of the grid, but also vertically along the left edge - two loops that the green cells can still manage to navigate. A third loop would be one too many though.
Had to give it a go - simply outstanding
Lucky the "usual" Yin Yang example still works on the torus ...
A mathematical way of approaching this (which I thought of but didn't quite use) is to relate the number of allowable checkerboard patterns to the Euler Characteristic of the surface. But here the logic is that you can't have more than two (there was another puzzle on the channel with a hole in the middle of a big grid where the same logic applied). It will help if someone creates a Klein Bottle puzzle, for example.
Astonishing again, just beautiful, the puzzle and the solve. Thank you
I had serious new ruleset struggles with this one - I never attempted a torodial yin/yang before - and I kept getting confused about whether edge cells could get out or not. Took me about 45 minutes.
Absolutely brilliant. Loved it!!
Thank you thank you THANK YOU for the colorable external squares! Would love to have this ability in most puzzles with outside-the-grid clues
i wish i would've tried this one myself before watching the video. this was brilliant
I know I could never have solved this having no experience with Toroidal, would have for sure needed that thoroughly explained to me beforehand.
Hi~ First time commenting here, hope it will help.
This puzzle reminds me of the one I've solved a couple weeks ago called "Twin Galaxies in Curved Space". It has similar topology and yin-yang rules, with some insane setup that leads to a unique solution. More importantly, it's a 9x9 puzzle which brings more fun ;)
It would be amazing to see you guys feature that if it's possible. It only has 14 solves right now, and I truly believe it deserves more.
Very bizarrely unique combination of rules with this new toroidal idea! Worked so well together
The numbers and the first half of the shading was straightforward, but I took a while to complete the shading
30:23 Completely ridiculous but completely forced. hats off to PP
23:41 would love to see more puzzles with this torus feature
Stunning and funny simultaneously as a challenge (if not technically hard, still a challenge to wrap your head around it) and some hilariousness with the infinity diagonal. Took me about 22 minutes to solve.
A small hint: copying the rows and columns outside the 6x6 help with seeing many colouring limitation.
One can argue, that the 3 is not in a corner, since it's on a torus.
Next up, a Möbius puzzle. After you cross an edge, 6's become e's, with a value of ~2.718.
A bunch of new rules for me, very proud that I could complete it in 51:37
How about a Dyson Sphere puzzle now? 😂
Also could someone please start counting the infinty diagonal now and let me know when your done 😜
52:17 - That was gorgeous, though the colouring took some working out. Once that was done, the rest was easy!
you didn't use it, but in my playthrought, the fact that the cases right outside the 6x6 could be colored to continue the pattern really helped me a lot
I appreciate that the rules say to shade the grid in two colors, which is what Simon would do anyway.
Brilliant puzzle, its not often that a completely new set of rules can produce a challenging but approachable puzzle for me. Took me 65 minutes but enjoyed the challenge.
According to Claude AI the Ramanujan sum of that infinite diagonal is -8 (the diagonal sums to 16, so it's -16/2).
That was a great new twist on an old rule set. I wish Simon had realized you can mirror the cells on the outside of the grid. That makes it easier to visualize.
Fantastic puzzle took a couple of tries to get it right. The successful solve took 33 minutes while being somewhat distracted. Those distractions are probably what caused the earlier attempt to fail.
Simon and Mark, if you ever plan to do a performance during a festival in the Netherlands, probably a lot of people like me (who aren't going to make it to Stratford) will try to be first in the queue to buy a ticket ;-)
One 'secret' of a toroidal yin-yang puzzle that I have just noticed (and no doubt other people have so sorry if I'm stating the obvious) is if all four corners were the same colour, they would form a 2x2 via the torus. So at least one corner must be different from the rest.
Alto as is the musical term (in italian) for high, perhaps?
Infinity is pretty high.
Also, if you pick only one side for connecting the grid to a cylinder, it may represent a wax cylinder for a phonograph, if not a music-box cylinder.
12:30 The other secret is still valid but void. The perimiter can still have only one change of colour but what use can you make of it when there is no such thing as perimiter?
Finished in about 20 minutes, only i hadnt as i stupidly msde the 5 diagonal only 2 cells long and then found a valid solution that worked. Then when i realised it wasnt the solution back-pedaled and then finished properly in 24 minutes :)
I used that R5 C6 had to be green because of the 2x2 rule. Same for that corner. All corners can not be purple.
28:51 STOP!
Once you made R5C4 green, you've now forced your 22 clue starting at R1C6.
6+6+6+4 = 22
6+5+6+5 = 22
Those are the only options you have.
And if it's 6+6+6+4 then the only cell that can hold the 4 is in box 4.
So R1C6 and R2C1 are from [56] and R3C2 / R4C3 are from [456]
But remember the 10 clue? It can only be [46] And if R4C1 were a 6, all options for the 22 clue would be impossible.
So now the 22 clue IS 6+5+6+5 and R3C2 is a 6 and R4C3 is a 5 and R4C1 is a 4.
Then, as R2C1 and R1C6 are effectively a [56] pair looking at R1C3 and R2C4, which is along the 24 clue, and those cells can only be as high as a 4.
And R4C5 will see a 6. So it can only be as high as a 5.
The highest the first two cells on the 24 clue can be is double 4, adding to 8.
The highest the 3 remaining cells on the 24 clue can be is 17 (5+6+5 in that exact order for sudoku reasons) so you only have one degree of freedom left on the 24 clue which at most could be 8 + 17 = 25.That degree of freedom can only exist in one of the first three cells as the last 2 would otherwise have one digit only as high as 3 and you would lose 2 degrees of freedom.
Point is, the last two cells on the 24 clue are from [56] and the order is known and resolves all [56] pairs you have so far.
I have no idea how i just intuitively knew that limiting the 22 clue to stop where it does would resolve a lot of the puzzle but that is what i was seeing.
>Purple and green
Thanks doc.
That five die pattern is called a quincunx.
I think it would have helped him to see the toroidal rule if he colored the 'ghost' squares outside the puzzle.
30:21 for me, what an excellent puzzle
Really cool idea. 38:18
16:06 had to restart a couple times, because I assumed maybe there was only 1 checkboard allowed. Instead I avoided using checkboard.
Coloring the extra cells on the border helped with spotting potential 2x2s, but the connection issues were hard to spot
This is, for me I think, the LEAST approachable sudoku I've attempted on this channel. It's pretty, but I had to be hand-held for pretty much the entire thing.
Incredible hulk colouring!
I would like to suggest 'dominoid', 'triominoid' etc for toroidal dominoes. Except triominoid could be shortened to 'trioid' and tetrominoid to 'tetroid' for absolutely no reason other than that I think it feels quite satisfying.
25:18 for me. it took me too long to get used to to the connected cells rule.
ALTO (halt, in español) appears on many STOP signs in Central America.
All but one of these diagonals must stop, to avoid having infinite sums.
Sorry guys. The shape not a doughnut, it is clearly a coffee cup!
This one hurt my brain, but i got it. Fun puzzle
Finished in 29:47. It was weird doing a yin yang puzzle where checkerboards were allowed!
The first thing I saw was the checkerboard with the 10 and infinity, so that floored me right off the bat.
Perhaps a triomino on a toroid should be called a triominoid.
I noticed that I could not avoid a checkerboard between the infinity and ten clues and gave up; this is why we watch, to learn.
Building a Klein bottle in a sudoku would actually be quite easy. All you do is take a normal toroidal rule and "flip" one side upside down (so that r1c1 is "orthogonal" to r9c9, r2c1 to r8c9, r3c1 to r7c9, and so on). For a Mobius strip you do the same thing except instead of having a normal toroidal edge you have one edge that's, well, just an edge (reusing the prior example, r1c5 isn't adjacent to anything in column 9).
It is a Klein bottle. Isn't it?
I found this puzzle extremely hard - or I should say, easy to make a mistake. I ended up in situation where there’s no solution that works and had to restart twice. With third attempt though it took me about 15min to color it and putting in digits was just a minute. It was a good one but I as much as I love ying yang, I prefer to look at a grid as a square 😅
30:51, was hard to see if I was in a dead end or not when the regions looped around and around and around and around. And not being able to use the "rule" that you couldn't have the two colors in the opposite corners of a 2x2 region.
Nice concept... But think "grid is toroidal " should be the second thing stated behind sudoku...
Trying to grasp orthogonally connected before noting toriodal reallly was a bugger.
Who would have thought the easiest thing to grasp was the inf.
When Simon got stuck, wasn't it enough to notice that c5r5 has to be green, to connect both green regions?
Not sure if I'm looking at the right point, but for a long time the green region occupying r5c2 could take r6c1 and loop round to r6c6 toroidally to connect to the other side.
so while working on this puzzle...a silly pointless thought crossed my head.... how many designers who are fans of this show start off their puzzles wth a 3 in the corner, just to get simon to sing
Took almost twice as long but I got there in the end.
He is ignoring that the 5 needs one green on purpose. 😂
?
Both 5s take more than one green.
@@RichSmith77 but it takes at least one to prevent the infinity loop… so first one from edge has to be green I think. Dunno… I am not good at this. I just like watching.
@@betazep I think that's the trap I fell into during my solve. It could have been one purple, one green, four purples from the direction of the 5 clue, with a single 5 in purple.
(I too coloured r6c5 green, at first, because I also overlooked this possibility on my first solve.)
@@RichSmith77 ah yes it could be a purple 5… good catch
40.32 for me, but I had to watch the video for some help twice.
The mathematical part of my mind is troubled by the designation of the grid as being a torus
The rules of the puzzle are symmetric as to going off the top and coming back in at the bottom or going off the left and coming back in at the right and vice versa in each case. So the grid has a 4-fold symmetry with respect to these transitions. But a torus seen from the side only has a 2-fold symmetry.
Going left-right entails going around the equator of the torus, whereas going top-bottom entails diving through the hole in the torus.
To me, this clash of symmetries represents a horrid mental dissonance.
21:27 does the 2x2 rule extend through the Torus it must right? if so, column 6, row 5 is green, it can't be purple. Don't think it'll help much.
Colour scheme today was lit! I would like to know the opinion of colour blind people, maybe this can become a new standard since orange and blue always give an odd/even vibe
This puzzle doesn’t convert to SudokuPad
20:17 - It's a toroiomino?
Misunderstood the rules and broke it 7min in then restarted. Total time 32:00
Titi's Time: 37:33
Was it possible to have more then 1 checkerboard pattern in the grid?
I was wondering if that's true in general, but I took it over to a 9x9 grid where there's a little more room to stretch, and it was pretty easy to make 2 checkerboards. However I do think 2 is probably the max. Something about how each color can connect across the north-south edge once, and they can also both connect across the east-west edge once, but then those connections prevent any further ones, so each color can only connect once each on each edge, which can then only accommodate circumventing the normal no-checkerboard rule twice.
But.. that's hardly a proof. We'll have to wait for some insane person to make poor Simon prove it generally 🙃
What if you had an infinitely long diagonal, but all the numbers added to zero (such as using -4 to +4 in a 9×9)? What would the outside clue be then?
You could call it undefined or divergent, but you'd probably need to explain it in the rules (to distinguish from infinity and negative infinity, unless you want to allow the ambiguity). Also calling it undefined sounds a bit like you just aren't given it as a clue.
Interesting idea, I had a quick look on the Wikipedia pages for divergent series and limits, but didn't find any better terminology.
@@caspianmaclean8122I don't see why it wouldn't be zero, if one "lap" sums to zero. N x 0 is zero for all N, so surely in the limit, as N goes to infinity, the sum is still zero, no?
@@RichSmith77 that would only work for a whole number of laps. It might have a sum of 0 after 9N cells (a whole number of laps), but a sum of -4 after 9N+1 cells, if the first cell in each lap is a -4. If the sum keeps changing between 0 and -4 as you include more cells, the sum series generally wouldn't count as converging to a number. There's special series sum definitions you could use to get an answer anyway, but not necessarily 0.
25:12 for me.
32:31 for me