The Phistomefel Meta Conundrum 

Simon’s ability to accidentally hover the mouse over the place he needs to look next never ceases to amaze me

Never ever apologize for taking single digit hours to solve a Phistomefel, Simon.

Only thing that gets me more excited than a CTC video, is a CTC video with Phistomefel in the title

Please never apologize for "taking our time" with long videos. It's always an absolute joy to watch you untangle the logic of these things, and the longer they are, the more time we get to spend enjoying that. :)

My favourite part of Phistomefel puzzles is, just over halfway through watching a video of a puzzle I couldn't even start, thinking "what a fool Simon is, he didn't immediately place that 2 in the top corner!"

Simon: "Do have a go." Me, seeing Phistomefel in the title: *insert Captain American "don't think I will" gif here*

Simon, don't apologize for taking our time. A, because we can see the length of the video and know it is long. B, because it is a Phistomefel puzzle, and every second of watching you crack one of his masterpieces is pure gold.

"What is Cracking the Cryptic for if not to showcase puzzles like this?" says Simon. And I say, "Exactly right." I enjoyed watching every moment of this, and as so often happens in these long videos of Simon's, I was so surprised to notice that over an hour had gone by, and I had not noticed the passage of time. I am glad you took back your apology, Simon, for the video being so long - believe me, if I did not want to watch a long video, I would not watch this. Rather, when I see that it is a long video, and I recognize the iconography representing Phistomefel in the thumbnail, I am happy about it. Thank you so much for this and all the great, elegant, beautiful logic this channel showcases. And also for your delightful delight in the beauty of all of this.

Rules: 02:46 Let's Get Cracking: 05:57 Simon's time: 1h14m41s Puzzle Solved: 1:20:38 What about this video's Top Tier Simarkisms?! Phistomefel: 9x (00:22, 00:29, 00:34, 00:39, 02:45, 06:14, 23:07, 1:18:33, 1:22:16) The Secret: 4x (08:24, 08:26, 08:28, 08:33) Cooking with Gas: 1x (1:02:46) And how about this video's Simarkisms?! By Sudoku: 17x (42:14, 48:10, 49:41, 51:55, 53:49, 53:52, 54:01, 55:27, 1:00:23, 1:02:41, 1:03:06, 1:04:49, 1:17:31, 1:17:34, 1:17:50, 1:18:06, 1:19:27) Ah: 15x (13:26, 14:08, 17:15, 17:39, 18:55, 34:44, 43:27, 43:27, 48:58, 59:16, 1:00:44, 1:02:21, 1:05:21, 1:12:13, 1:13:09, 1:13:47) In Fact: 10x (11:12, 11:27, 25:57, 36:52, 49:41, 1:07:01, 1:14:54, 1:15:39, 1:18:06) Hang On: 9x (06:31, 16:03, 17:39, 18:55, 42:54, 45:19, 45:19, 51:39, 57:28) Sorry: 8x (03:39, 30:46, 38:18, 42:30, 1:18:22, 1:18:29, 1:18:29, 1:20:05) Beautiful: 6x (38:44, 38:47, 38:48, 47:51, 54:44, 1:02:27) What on Earth: 5x (14:35, 21:46, 47:01, 1:07:42, 1:20:56) Break the Puzzle: 5x (15:26, 15:37, 47:01, 53:32, 1:10:09) Obviously: 5x (25:41, 30:41, 37:08, 56:35, 1:15:20) Clever: 4x (59:34, 59:39, 1:02:23, 1:21:20) Brilliant: 4x (1:21:22, 1:22:12, 1:22:12, 1:22:14) Gorgeous: 4x (49:53, 49:56, 1:13:58, 1:14:18) Wow: 4x (17:30, 36:08, 1:20:41, 1:20:42) The Answer is: 3x (43:48, 52:25, 1:19:08) Lovely: 3x (55:10, 1:00:45, 1:21:40) Bother: 2x (1:03:19, 1:15:58) Nonsense: 2x (22:21, 42:30) Progress: 2x (1:20:58, 1:21:01) Fabulous: 2x (00:52, 00:57) Good Grief: 1x (38:50) Goodness: 1x (1:20:49) Naughty: 1x (03:44) Fascinating: 1x (59:20) Elegant: 1x (21:21) Of All Things: 1x (1:15:33) Masterpiece: 1x (48:31) Surely: 1x (38:10) I've Got It!: 1x (38:44) Propitious: 1x (07:40) We Can Do Better Than That: 1x (45:57) What Does This Mean?: 1x (33:55) Have a Think: 1x (06:03) Thingy Thing: 1x (1:09:18) Most popular number(>9), digit and colour this video: Forty (28 mentions) One (136 mentions) Orange (22 mentions) Antithesis Battles: High (2) - Low (1) Even (7) - Odd (0) Column (11) - Row (10) 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!

So many things to say about this, where to begin! First off, that’s absolutely incredible from Phistomefel. I love the concept of the puzzle and I could see greedy cages becoming a more common variant going forward. The break-in is so cool, but really my favorite thing was the series of deductions after that. There’s just so much *joy* to be found in this one. Don’t get me wrong, it’s nowhere near easy, but the deductions feel more graceful than we normally get from Phistomefel. He has such devilish connotations about his name, but this one felt far more angelic. As for you, Simon, an absolutely brilliant solve. I would’ve never attempted anything like this. You crushed it out of the park. The length of the video feels like it has a lot less to do with you being stumped as it did with you having to do a lot of deductions, and I feel like you did those rather quickly. This video *deserved* to be this long and kept me thoroughly entertained from start to finish. You’re absolutely right - puzzles like this are what CTC exists for. Also, nice cryptic clue on Twitter!!

I solved Phistomefel's newest sudoku on LMD yesterday and felt very pleased with myself. Stared at this one for 10 minutes before giving up completely and deciding to just enjoy the video instead lol

I'm confident Phistomefel also admires Simon, a kind of distance relationship between two great minds p.s. I just wish Simon had used shades for those 2/3 cells and left full colours for cages

I just love Phistomefel puzzles and I love watching Simon solve them. This is definitely what the channel is all about as he said.

I want to start by saying there is absoluetely no way I could do this puzzle. That being said I figured out the shape and location of the 40 cage waaaay faster than Simon did. I couldn't do the rest of if but I'm pretty proud I figured that one out!

I've started to realise that Simon feels like a genius on maths, in that he can flip digits in his head so amazingly where I'd be at paper and pencil. But the number of times I see him work out something in a really complicated way when there is a digit looking at it, and that I juggled the pieces in the central box and concluded that it only worked as an hourglass shape makes me think my spatial visualisation may be quite good. Nowhere near Mark and Simon level puzzling, but one tiny bit if the job being good at gives me warm fuzzies. Thanks for another delightful long video, Simon!

61:02 for me. Another Phistomefel masterpiece. It took me like 15 minutes to understand how the 40 cage would work, but from there I didn’t feel too stuck at any point. Awesome ruleset and awesome puzzle!

Despite all my rage I'm still just a digit in a cage

Regarding the opening logic: you can quickly determine that a box can contain 1 cell from the 40 cage, and that the arrangement of 7 cells that actually happened is also acceptable. The question then becomes whether any other number of cells in a single box is possible, which then bleeds into the question of whether the 40 cage can span multiple boxes. It's fairly easy to see that 2 and 3 are impossible, as there is no arrangement (of the very small number of arrangements available) that doesn't leave at least 3 other cells in the box on the border of the cage. However, I believe 4 and 5 are more difficult to show. Fortunately, these can be proven simply by looking at the single box in question, and you don't even need to take into account the other boxes or which box it is or whether it's a corner box or whatever. All you need is that if the box has >2 cells that border the cage, the cage is illegal. Each cell in a box borders at least two other cells in that same box. Thus, if any cell is added to the 40 cage inside that box, it must border 2 other cells in that box at minimum. We know that 3 cells are impossible, as it must border at least 3 cells in that cage. If we make that 4 cells, we need to select one of the cells not currently in the cage and add it to the currently illegal arrangement. In doing so, we would need to reduce the border cells in the box by 1. The newly added cell must have already been on the border of the current arrangement, so we would remove that border cell. However, the newly added cell always must border at least 2 other cells in that cage. Thus, in order to reduce the same-cage border count by 1, both of those cells the new cell borders must have already been in the cage. This 2 border cells situation occurs with the newly added cell being in the corner, therefor the 2 cells that must already be in the cage are the middle-outer cells bordering that corner. Then, the only possible way that those two cells could have been in a 3 arrangement was if the center cell was also in the cage. This 3 arrangement bordered 5 cells before adding the 4th in the cage, so although this addition reduces the border count by 1, it's not enough. Thus 4 cells in a 40 cage in a single box is impossible. The proof for 5 being impossible is nearly identical. Given that we have 4 cells in the cage in the box, we know we have more than 2 cells in that box that border the cage, by the above proof. Then, we must add a 5th cell to eliminate 1 border cell. This is once again only possible by adding a corner cell, and that corner cell must be surrounded on both in-box sides by cells already in the 4 arrangement. This demands that the center cell in the box is in the cage, and we already know that just these 3 cells borders 5 cells, and wherever the 4th cell was placed could not have eliminated the necessary 2 border cells for the 5th cell's reduction of 1 to bring the total in-box border cell count down to the necessary 2. Thus 5 cells in the 40 cage in a box is impossible. 6 cells is actually possible. This would require that there are 3 cells in the box that are not in the cage, and at least one of these cells is not bordering the cage. This is possible if the three non-cage cells are an L on the corner of the box, as the cornermost cell doesn't border the cage. However, this is the only possible arrangement (or 4 arrangements, 1 for each corner). We know this because the corner cell is the only cell that borders only 2 other cells in the box, which both must be not in the cage. Then, including the corner cell, that's 3 cells not in the cage, which is our budget spent. Looking outside the box, however, the arrangement becomes difficult, though not impossible. The cells in the cage in the box span all 3 cells of two of the box's edges. If either of those edges are borders to another box, then that other box will have 3 cells that border the cage. So the only way to make 6 cells work would be to put it in a corner box, with both of the fully in-cage edges of the box towards the border of the grid. This is the only way to have 6 cells in a 40 cage in a single box, with the 7th cell in the cage being legally pickable from one of the two bordering boxes.

I think one of the most impressive things about watching Simon solve is how easily he can force his brain to adjust to different rulesets. I would constantly be forgetting about such a weirdly specific clue as the greedy cages, and he just immediately manages to work it into the way he processes the grid.

Always a pleasure to watch you crack Phistomefel's gems. Could never imagine doing it on my own. Thank you for sharing these !

Simpler way to look at the break-in: if more than one cell in the central box is in the cage, then all 7 must be in the box, or else it breaks within the bounds of the box. However, this same logic holds true for the other boxes as well, and since one of the cells in the cage has to be in the central box, you can’t get to a count of 7 in any other box. (And there’s no way to create an orthogonal connected region of 7 cells with each cell in a different box.) So, you have to build the entire cage within the central box.

No mention of wedding cake? I am shocked and dismayed, Simon!

Solved it! Took me 137 minutes but happy enough to get to a solution. One thing Simon overlooked was that r7c2 couldn't be a 9 since it was adjacent to a 1-9 cage. That would have fixed it as an 8 much earlier and simplified the options in box 1 a little. However, Simon realising the 25 and 18 cages couldn't form a corner, leaving a cell touching both cages and seeing a 2, was something I missed and allowed him to sort out that top left area far more elegantly than I managed. I thought the rules were perfectly well expressed too, despite seeing comments complaining about them.

The shape of the 40 cage really restricted the corners from the Phistomefel Ring Theory. 🤯 just brilliant.

Since English is not my first language, I'm blaming accent for the fact that every time Simon says "By the power of Sudoku", I'm hearing "By the power of Grayskull!"

That was beautiful. The effects rippling from bottom-right to top-left at the end. *Chef’s kiss*. I took a moment while watching the solve to make sure that the 2s and 3s in the Phistomefel ring had counterparts in the corners.

Yet another genius puzzle, and a fascinating constraint. At first sight it looks so threadbare that I could do nothing except stare, uncomprehending. Gradually some light dawned, the mists lifted, and the brain cranked up a gear and I staggered to the end, taking a mere half an hour to resolve the 25/18 cages. Also I left one of Simon's rinky-dinks as I cannot say that I rigorously proved that there was only one way to configure the 40 cage.

Was watching this and for some reason I thought it was an older video and I was like "oh yeah the pen tool is sort of a new thing isn't it" and then Simon starts using it and I had a "wait, why weren't you using it the whole time?!" Awesome video, the Phistomefel videos are certainly some of my favorites. Thank you CTC~! (By the way, for all those who use CTC for these guys, I can't help thinking of Cinnamon Toast Crunch cereal every time I read that)

58:15 "Let's resort to Sudoku if we absolutely have to." What a gem! 😄

Phistomefel puzzles are always good.

Brilliant puzzle - brilliant solve!

An 80 minute Phistomefel solve. The great man is going easy.

Very excited i figured out the structure of the box 5 cage before Simon!

"mystery digit" sounds like a prompt someone gonna soon use as part of ruleset for their sudoku

Que maravilloso canal. Es bueno que lo haya encontrado Te quiero Simón 🥰 Saludos desde Mexico 👍

Started this about 20 mins ago and it's absolutely blowing my mind

I chipped one of my teeth recently and I dread going to the dentist to get it fixed, so every time Simon mentions a two-three pair, I am severely triggered.

67 minutes. Another great puzzle. Strange as after finally working out the middle box (at least 10 minutes of screen staring) and feeling like I had a handle on the logic, I forgot it each time and started thinking about the cages rather than what couldn't go where.

struggled with it for over an hour before realizing that single-cell cages were allowed, which was a little odd since the rules define a cage as an "orthogonally connected set of cells." Put it aside after realizing that there was no way to make the 40 cage work without putting a 5 into one of the 5 cages, which sort of invalidated everything I thought I'd figured out to that point. Disappointing, but still love the channel!

11:50 Listening to Simon talk about constructing shapes with a maximum boundary length makes me want to take my books and do mathematics.

That time when Simon went beast mode. Incredible solve!

I somehow solved this puzzle and I loved it! My favorite part also was the ? Placing 3 in box 9, that is so neat. The break-in is kind of mean, you have to do some trial and error to find the shape of the 40 cage. I couldn't come up with a short elegant proof for the cage but convinced myself that everything else was impossible. Any quick proof found by someone else? Another brilliancy from the devil. Thanks Simon for the great solve!

What a brilliant puzzle again! I wonder how Phistomefel discovers all these

I wish I knew a way to prove the 40 cage orientation in the puzzle. For me it was simply intuitive that it had to be a particular shape and position in order to follow all of the rules. I imagine any proof would require a bunch of "If/Then" type arguments. For a bit of fun, try using closed captions and watch the various ways Phistomefel is spelled by the software. Thanks!!

just imagining the Sudoku Devil himself watching this video and cackling away while Simon struggles with these cages

I think the 40-cell cage situation is easier to prove if you look at it from the point of view of placing border cells instead of placing the cage... Suppose you have a cage with only two bordering cells in some box. Either every other cell in the box is also in the cage - i.e. all 7 cells of the cage are in the box - or the two border cells divide the box into two different sections. The only way to do that is to have the two border cells touch diagonally and cordon off a single cell from the other six. So if you have two bordering cells in a box, the cage occupies either 1, 6, or 7 cells of that box. If it's 6, the cage cells within the box occupy a triangular stair-step shape, with two three-cell-long borders in the two adjacent boxes, and the seventh cell of the cage can not reduce either of them to two cells, let alone both of them. (However, this situation would work if the box in question were in the corner!) If it's 1, then an adjacent box has to have the remaining 6 (or fewer) cells in the cage and we have the same problem there. (You cannot construct the seven-cell cage from cells from seven different boxes.) So if the cage has to include a cell from the center box, it has to include all of them - and you have to make sure the two border cells within the box touch all four of the box's borders with the other boxes to avoid putting three border cells into any of those boxes either, which requires the two border cells within the box go in opposite corners.

Never in a million years would I have figured out that the 40 cage was useful at the start

“Anyway, do have a go…” *Me (Looks at length of puzzle) 🫣oh no no no no 🤦‍♂️ (Settles in)

i'm not usually great at building polyominoes (or at phistomefel puzzles generally), so i'm rather pleased that i solved this in nearly the same time, and the only hint i needed was that the 40 couldn't possibly span 8 cells :) i was very curious what that ? cage could possibly be useful for, too, and that was a clever surprise. lovely puzzle, thank you both for setting and sharing it!

I know most people would consider this cheating, But I am not as good at math as most of you, but I enjoy doing these. Do you know of a hidden sudoku calculator where you can put in a total (like 33) and the number of cells in a cage, and it tell you all the possible combinations? With some effort I could probably learn to make one in Visual Basic. Have a box for the total. A box for Number of Cells. And checkboxes for excluded digits. Hit go and it would list you all the combinations. Currently I use a notepad and write them down then cross them out as I disqualify them.

196 minute marathon for me, and I barely had to, use Simon for it, only till 16:22 in the video to get me started thinking about the 40 cage after phaffing about for an hour. but eventually got there! now it's nearly 5am and I need to go lie down.

Sounds like we’re in for a delightfully devilish video 🤓

Since there are max 2 cells left out from the it can only have 2 cell boundaries in any row, column or region

@ 36:45 the obvious step would be to note that if the border 2 isn't part of the 7 (and it can't be because then the 7/40 would overlap digits), it would immediately also break the 7 as the 3-cell 7 cage necessarily contains a 2 and would be bordering a 2, against the rules. Equally, it can't be a 3, because then you'd have to fetch the 2 of the 7 cage from out of the box, which could only be in row 3 where it's impossible. In fact, this puzzle is filled with dazzling logic, some of which I assume is very incidental, due to the monstrous restriction, but the 40 cage is a beautiful break in.

Is this a new setter ? He seems to have a good grasp of the basics, he'll probably develop into a decent setter. I'll be watching this one out, over an hour for Simon is 3 hours over two days for me at best.

I found the phrasing of the rules very confusing. I saw the problem with the 40 cage and didn't find the shape Simon did, and the revisited the rules and it doesn't say what we (correctly, I guess) assumed it did. I read it as "digits *that are* in a cage cannot be in cells or cages orthogonally adjacent *to the cell the digit is in*." If you have a cage covering r1c1,r2c1-3, and r3c1-2, then r2c1 would be a digit in a cage not orthogonal to any digits outside the cage or any other cages, so it poses no restrictions and could be in, say, r4c2. It should read, "digits in a cage cannot appear in any cells orthogonally adjacent to the cage, or in a cage orthogonally adjacent to its cage," or "orthogonally connected cages cannot share digits, and digits in a cage cannot appear in cells orthogonally adjacent to the cage" or "the set of digits in a cage cannot appear in cells or cages orthogonally adjacent to the cage." Maybe I'm being a sore loser; I obviously thought the rules said what Simon did at first, and I wouldn't have gone this way if I'd thought of the 40 cage shape that worked, but once I revisited the rules and thought things were less restricted than they were, I definitely had to return to the video. I just think the rules aren't clear what the cells are restricted because it doesn't say to what they are orthogonally adjacent. Thank you for these wonderful videos. I enjoy them very much.

is there any chance you could add a min/max calculator to arrow sudoku? Those digit pills are really hard for me since I can't do the math in my head. I have to use my phone to do the match while I use my laptop to solve it.

Goodie, another movie to watch. I don’t think I will be trying this one today.

Fabulous puzzle. 90:10 for me. Proving the layout of the cage in box 5 is tricky. I can give you a sketch that centers around the concept of a cage's "surface area" within a box, i.e. the set of cells in the box which aren't in the cage but are adjacent to it: 1. The cage starting in box 5 has to have at least seven cells, which means it can’t have a surface area of more than two cells in any given box: the digits in its surface area in any one box must be distinct from each other and from the cage, and there can only be at most two such digits. 2. A cage with at least two cells in a box has to have a surface area of at least three cells in that box, unless it takes up seven or more cells in the box so that there aren't three cells left in it to be surface area. 3. A cage with five or more cells has to have two cells in at least one box: at four cells, you can have adjacent corners in four boxes, but at five cells you need to take a second cell in one of those boxes. 4. So the cage starting in box 5 has to have seven cells in at least one box. 5. Now, a greedy cage with cells in box A can only contain cells in another box B if there are cells in A which are part of neither the cage nor its surface area, because the digits in the cage in box B can't be in either the cage or its surface area in box A. 6. That is only possible with at most 6 cells in a box, because if a cell is not part of the cage or its surface area, then the cells adjacent to it must also not be part of the cage, and that has to be at least three cells in the box. 7. The cage starting in box 5 has to have at least seven cells in a box, so it cannot cross boxes and must be wholly contained in box 5. 8. The cage starting in box 5 also can’t include all three cells on any given edge of the box, or else it will have a surface area of 3 in the box across that edge, so it has to exclude a cell on all four edges. 9. Since the cage has to have at least seven cells, it can only exclude at most two cells. Therefore, it has to exclude exactly two cells, and they have to be in opposite corners. 10. Since we know that r4c4 is in the cage, the excluded cells have to be r4c6 and r6c4.

I'm going to attempt a proof of the location of the 40 cage, let's see how far I get before I mess something up. I think Simon basically had it, but was struggling to prove the details rigorously enough for his own tastes. Notice that we can't have more than 2 non-cage cells touching the cage in the same row, column *or* box of the sudoku. Also, at least one box the cage crosses must include at least 2 cage cells, because otherwise the biggest cage you can make is a 2x2 square. If we look at one of the boxes that the cage goes into, we know that the two excluded digits must divide the box so that the cage is isolated from any other digits, but it's pretty easy to see that with only 2 digits your options are (a) to have the cage fill the rest of the box, or (b) to fence one cell in a corner away from the remaining 6 digits. But that means that we're down to two possibilities: either the entire cage is contained in a single box, or one box contains 6 cage digits and the remaining digit is in an adjacent box. But if you look at the arrangement of 6 digits, it forms a glider shape with two adjacent edges of 3 cells each, so one of those edges must touch the border of the puzzle and the other must either also touch the border or be broken up by the extra cell. Since we must include at least one cell of box 5 in the cage, that would involve putting the glider against one of the middle borders (e.g. the top border in box 2), but then the other long side would not be facing box 5 and so that arrangement is impossible. (If we needed to place a 7-cell cage elsewhere in the puzzle, however, this might be feasible.) Hence, we're left with only the possibility that the entire cage must fit within a single box of the puzzle, i.e. box 5. The two excluded cells have to be in opposite diagonals, since we need to ensure that each side of box 5 contains no more than 2 cage cells, and then since we have the 40 clue in r4c4 there's only one orientation for us to do that, giving the final position of the 40 cage.

I feel like the rules were not presented very well, I interpreted the last line as meaning that greedy digits cannot appear in cells or cages orthogonally connected to that cell, rather than those connected to the entire cage.

could not solve it, because i thought the clue has to be in a top left square, like any other killer cage. When simon solved this problem for me in the video. I finished it in 56:44.

Despite taking close to an hour, I thought I was making good progress ... but I erred with the 33-cage (ruling out r8c6 as a possible cell for that cage, forgetting that the cell above didn't have to be what I thought it would be). Eventually I saw the error, but not before at least beginning to watch Simon's solve. Though I didn't take anything tangible from that, it counts as a 'miss' for me. Tough puzzle! (Too tough for me, mostly.)

I was able to figure out the shape and orientation of the 40 cage fairly quickly (plus the checkerboard surrounding it), but couldn't be certain that was the only configuration possible. No chance I could have solved the entire puzzle. Utterly diabolical.

That 8 in Block7 - the 91 next to it solved it 30 minutes earlier :)

I completely this puzzle in 15:23 using an incredibly elegant break-in that depended only on the "?" clue. Unfortunately the size of this comment field is too small to contain the proof, I'll try to come back and fill in the details later.

25:28 minutes in, well done, good entry. I'm pretty sure the 5s can touch the 40 cage, but the 7s cannot, because of the greedy rule. The makeup of the 5s has to be perfect though BUT! wait can you even the 5s because there would be double in the row... So either the 5 cages are the opposites (1/4) if the orange/blue is (2/3) or... they are lone 5s, they also can't touch, oh man! Just pondering out loud, you are almost certain to stumble onto this in the next couple of minutes. 34:01 minutes in, I'm also pretty sure since the 7 on the right is touching the 5 and it can't escape, it can't use any of the digits in the 5, so if that's a 2 cell 5 on the right it's a 7, if it's a singular 5 on the right then it's a 1-6 and the orange blue are a 2-3. it's very constrained whereas the 7 on the left can always escape if it wants, since it's not directly touching the 5. which since there is ANOTHER 7 touching all of these fives, we are about to run into some more problems. 41:48 Yay the right side as I guessed based on a feeling was going to be the 2 cell, the left was just touching too many things. 46:43, look above the seven, it cannot have a 2, 3, or 5. It sees a 1 in its box, so... 7. Only 7. The other one can't be a 34, it can't be a 16, it cant be 124 it can't be a 7, it must be a 25. with the 5 first and the 2 underneath it. which resolves the 10 cage as a 91 since it would see the 2. 51:27 well he missed the 7 that is a 7, but he'll get there. Got everything else like the champ he is, god that break in was great. So this yellow cage has to go left, and on the bottom probably, it could take the middle left only for one cell as long as orange is a 2. since it can't have a 3 touching it, since it uses a 3. Which means for sure it takes the bottom left cell just does it go up or continue along the bottom, hmm. So far he's seeing the same stuff as me. Makes ME feel good that I'm spotting all the right stuff. 1:16:12 Column 2, the 489 next to red can't contain an 8 so it's 49, the 89 next to purple can't contain a 9 so it's 8. so the grey cell is 49 as well but since there is a 69 in the green cage above it, it's a 4, and the cell next to red is 9.

Feel like there was a quicker route to ruling out any but the required shape of the 40 cage - specifically, the edge of the grid was never going to make it any easier in the other boxes because the arrangements for 2-6 cells examined in box 5 all left three orthogonal connections *in that box,* which wouldn’t be helped by just getting rid of adjacent cells outside the box…

As Phistomefel sudoku ideas go, this one was both incredibly hard to get your head around and incredibly clever once the solve got going. The proof about the cage in the middle box is actually relatively simple. First of all, a 40 cage must be 7 or 8 cells (since there is a digit sum of exactly 5 missing). If it is 8 cells, then because of the position of the 40 clue in the middle box, at least one of the cells adjacent in the same box must be part of the cage (or else both would be the missing 5, which would repeat in the box). The key is to realise that having found 2 cage cells in box 5, these will be adjacent to at least 3 other cells in the box unless we take at least 7 cells in box 5. Since we know that only 1 cell can be adjacent (a 40 cage in 8 cells has only a 5 missing), we are forced to take all 8 cells in the same box and the puzzle breaks as now the 2 cage cells which we previously found in box 5 are adjacent to 2 cells outside of the box which share a row or column, repeating the 5 again. Therefore the 40 cage is 7 cells large and by similar logic to before, we can only ever take 1 cell or all 7 in a box (taking 2 cells will force the other 5 cells to be in the same box in order to avoid 3 cells being adjacent to the cage, as we know there are only 2 digits missing). Taking only 1 cell clearly will not work as the cage could only reach a maximum of size 4, so all 7 cells must be from box 5. Now the only thing left to do is to determine the shape of the 40 cage within box 5. Not taking the bottom right corner cell will force the top row or left column of the box to be part of the cage, breaking row 3/column 3 respectively (2 digits in 3 cells), so that cell must be in the cage. The other 2 corner cells now cannot be cage (if either of them is in, the 2 cells adjacent to the corner must both be out but that breaks the box on the other side using the row/column logic from before) and thus the shape of the 40 cage is forced. Very pleased to get through this one, it needed a bit of help from Simon at the start (it turns out I was trying to jump ahead several steps, which came in handy later on but wasn’t quite so useful for trying to start a Phistomefel puzzle solve!) to point me to the 40 cage but from there I managed to find my way without breaking the puzzle for a change (though I did have a nervous moment where I thought I might have done). My favourite part of the solve has to be box 9, I loved the way that the 7 cage and the 33 were so powerful when considered as a combined 40 cage. I enjoyed finding the initial proof as well [there is in fact a fun alternative route for proving the 40 cage cannot be 8 cells. Since you cannot hit the 7 clue (47>45), the 8 cage cannot extend left so it is forced to extend all the way right but this hits the 7 clue in box 6). Thanks as always to Phistomefel for another intriguing sudoku challenge and Simon for an incredibly useful solve (which meant I didn’t miss out on such an amazing puzzle). :)

Would love to see an interesting chess move puzzle on Saturday for my birthday.

Well, I think I did it. Just over an hour. But this is another one of those puzzles where you have to know the implications of the ruleset. So most of the time is spent just trying to figure out what the starting restriction is. Still, interesting puzzle. edit: Almost done watching the video. Very enjoyable solve by Simon. Not sure why, but this one was different than many other solves. More relaxed I think. Even though it was a difficult puzzle, it flowed more naturally on video.

I’m only 12 minutes in but it seems like the 7/8 cell restriction for the 40 indicates the cells on the borders are NOT part of a cage unless they’re the ones that contain a clue. My guess is the Left 5 is a cage of 1 cell

Gee I'd love to take a crack but I'm wrapped up in this blanket and I can't reach the mouse ;);)

Brutal, but very fun. Took me almost an hour and a half.

Phistomefel makes haiku. Simon is perplexed. It's raining on Mt. Fuji.

Let’s get cracking!

That poor 25 cage is still awaiting for its enclosure to be finished...

First time I've seen an uncaged variant on CTC :)

I’m so confused. I angled the 25 cage (1-3-5-7-9) down to disambiguate the 2-3 in box 2 and solved the puzzle correctly according to Sven. But this put the 3 in box 8 touching the 33 cage. So the correct solution is incorrect? And this “correct” solution is different to Simon’s.

At 54:35, I thought the odd wording in the rule about the question mark cage would be justified (as opposed to simply saying "the value of the question mark cage is must be deduced". It would IMHO have been more elegant, giving the wording of that rule, if the size of the question mark cage had genuinely been ambiguous, or perhaps (to be truly evil) if there were two ways of filling in the numbers in the grid that would be consistent with all of the rules except "the value of the question mark cage is unclear", but the correct solution would be the one which left the size (and thus total) of the question mark cage ambiguous.

This one broke my brain. So many options caused paralysis-by-analysis and I missed LOTS of simple digits.

Thanks!

I was so upset that Simon wasn't using the pen tool throughout the solve. Glad he at least broke it out near the end. Is there a puzzle hall of fame that Phistomefel can be inducted into?

Holy cow!

How fit the 4 single cell cages in the rules ? Rule: ''a cage is an orthogonally connected set of cells'' means ''othogaonally connected'' and ''cells'' not at least 2 ?

yesterday I watched Bremster solve Genetics by Angelo, interesting set of rules and beautiful puzzle, it would be great if you did a sudoku with a similar set of rules on CTC

The blue and oranges are 4x on the Phistomefel ring, which forces them into the corners. Not really necessary for the solve, but it speeds up some deductions later. I did not see that mentioned in other comments, so i share...

Didn’t solve myself. But while watching I realized the blues and yellows were on the phistomefel ring, which restricted the remaining blues and yellows to the corners. I think this helps solve the early 7s not being 3 cages long as it must at least have one digit from either possible blue yellow pair, but it can’t reach the corner. I wonder if he intended that or was just a consequence of the 40 cell shape. I also think this fact was useful in other portions but not sure.

Wow alright I guess I’m joining the discord. I’m going to figure this sudoku stuff out.

Having just watched the middle of the solve video, I am mildly confused by Simon’s path of logic. Firstly, there appear to be several instances where digits are used in an example but only checked in one order and not the other, which could be significant although in this case it isn’t hard to disprove. Second, you can immediately establish the nature of the 5 cages by looking at the 10 clue, rather than using it after. Simon seemed to get so distracted by the 5 and 7 clues that he ignored it for over 10 minutes! It is worth stating that Simon is able to solve most sudokus because of his flexible approach in ways that many cannot, it can just be tough to watch when he appears to have done most of the hard work and gets caught up in the minor details. The logic in box 6 is not needed once the position of the single 5 cage is determined, which could have saved a fair amount of time! But any solve of a Phistomefel puzzle, especially live, is extremely impressive so all is definitely forgiven (especially since I wouldn’t have known where to start if not for Simon mentioning the restriction on an 8 cell 40 cage).

i keep hearing simons stomach rumble. eat, simon!!

@ 49:00 I had to go back and watch this part 3 times, to understand the logic about the 33 cage (as to how it must be a certain number of cells).

I'm yet to try the puzzle, the rules didn't quite make sense to me, so I watched the start of the video to get a better understanding of them, watched a couple of minutes of musings about that first cage, then went and had breakfast while having a think about it, and came up with something. I think a neat (and rigorous enough to convince me to use it, though might need a bit of work for a proper proof) way of proving it is... Just think about a cage entirely within a box, connections only within the box, so it is uncomplicated by things like puzzle borders. I'm writing it out in a quite long-winded way, I think it's shorter/more intuitive if you can visualise it. If there is only one cell that is part of the cage, and it is in a corner, then it has 2 connected cells within the box. That's allowed.(Obviously a single cell not in the corner has 3 or 5 connections, and isn't, but that's not needed for the proof) As soon as you increase the number of cells in the box to 2, you *must* have at least 3 cells connected in the box. Because any 2 cell domino has a perimeter of 6 cells, and a maximum of 3 cells at the edge of the box, so a minimum of 3 cells looking into the box. 3 cells in the box, same problem. Perimeter is 8, maximum of 5 at the edges/concave corners, so minimum of 3 looking into the box. Anything bigger, easy to visualise the same problem. It will *always* see at least 3 cells in the box, or it will see the entire remainder of the box. Therefore, as it is only allowed to see a maximum of 2 cells in any box, it can only have either 1 (to only see 2) or 7+ (to only have 2 or less left to see) cells within any box. The cage is 7 cells, and using 1 cell in 7 different cages is obviously impossible. So it must have 7 cells in one box, which leaves zero cells remaining outside of the box. Identical logic for the possibility of an 8 cell cage if you let it see 2 cells, it must use 1 or 8 within each box, therefore it must be 8 in a single box. (and as it could only see 1 different digit maximum, 1 cell in a cage isn't allowed anyway, even if you could have a disjoint cage that was 1 cell in 8 different boxes. But a 7 cell disjoint cage across 7 boxes would be possible. Or an 11x11 construction like yesterdays would allow an 8 cell cage in a single 3x3 box. Just in case anyone wants to set something even more insane...) Now, because either 7 or 8 can't see 3 different numbers outside the cage, and we've proved the cage is entirely within a box, the cage can't occupy an entire edge of the box (unless it is the puzzle border), because then it would see 3 different cells in the neighbouring box. That means the cage can't be 8 cells. So it is 7 cells, and as the 2 cells not in the cage have to touch all box edges that aren't puzzle edges, those 2 cells must be in opposite corners of the box (this isn't just true for box 5, btw, it would also hold for boxes 2, 4, 6, 8), and the location of the given clue tells us which 2 corners of the box are the non-cage cells. Looking forward to doing the puzzle now.

Around 1:08:00 when you start contemplating the 18 cage I screamed. The logic you did much earlier should have proved that the red 13 cage can't touch the 18 cage (18+25=43, only a single 2 or 1 & 1 could be left out of the 18 & 25 cage junction). There was nothing to think about, the 18 cage must expand to the right.

Don’t apologize for a long Phistomefel video. He’s diabolical.

Phistomefel and Simon [NO, not sitting in a tree -- but MAYBE.. lol] and an "Hour 22" video.? [And, "Have a go."]? NO, LOL. Sittin' and watching this one though.. [And watching THEE..] TE-HE, TE-HE.. 😂😎☕[

Don’t all cages have the clue in the top left corner?

A phistomefel puzzle where the cage borders are already in the grid? Pah! Can't be too hard then ... :p

So the rule about the clues having to be located in the top left corner is just a formality now? For literally half the puzzle I was sure that the leftmost 7 cage had to go downwards to abide by the rule

Simon, I think that yellow [6789] cage has to creep along the bottom [away from those 3s in box8] That would have pleasurable [it was for me] creeping one at a time. It gives you a 4 in r9c3 and a 25-25 pair in row9 I think. A little Sudoku helps though you're right [there couldn't be an extra 14 in column nine of box9 , that fixed the 1s and 4s ] I think that yellow 6789 cage had to creep along the bottom though. [No idea how you didn't see that -- and I hope I'm not wrong lol ]. [[@52:11 or abouts, btw. COOL ONE ]]