The Phistomefel Meta Conundrum

** TODAY'S PUZZLE **
Another chance to bathe in the genius of Phistomefel this evening as Simon takes on his sudoku, Greedy Cages. This is a long video (for which Simon apologises!) but some of the ideas in this puzzle are REALLY interesting. (In particular, drop a comment if you can see a neat way of proving what Simon finds out about one of the cages at the start!) You can try it at the link below:
app.crackingth...
Rules:
Normal sudoku rules apply. Clues in the top left corner of a cell mean that that cell is part of a cage whose digits sum to that clue's value. A cage is an orthogonally connected set of cells in which no digit may repeat. Each cage contains exactly one given clue. Cages do not overlap. If a clue is a question mark it is unclear how large the sum of the respective cage is. The cages are "greedy" ie digits in a cage cannot appear in any orthogonally adjacent cells or cages.
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[@mats1365]

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

[@kathyjohnson2043]

I think it is subconscious rather than accidental.

[@ulkgb]

I think he leaves the low hanging fruit to us so we can feel a little bit smarter

[@LittleBallOfPurr]

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

[@crhodgkin]

Add to that the extra long video length and it feels like we've struck gold. :)

[@hendrikharryg]

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

[@inspiringsand123]

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!

[@sehd3031]

bruh

[@th.nd.r]

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!!

[@radd1865]

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. :)

[@Mega_Tron6891]

Just recently he did a short video cause Mark had a long video, and they didn't want to put out 2 long videos... I felt cheated 🤣🤣🤣

[@emilywilliams3237]

"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.

[@chrismcdonald6195]

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

[@sjm6280]

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

[@andrewgrant6516]

I'd advise both against taking any trips to the Reichenbach Falls.

[@Lucretiel]

Interestingly it’s not technically wrong, every cell that isn’t in a clued cage is in its own 1 cell cage (fulfilling adjacency constraints)

[@NoobixCube]

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!"

[@briancaster2876]

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!

[@Sasha-1313]

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

[@verdiss7487]

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.

[@JohnHazl]

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

[@blackonyx9842]

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.

[@RichSmith77]

You could have just six cells of the 40 cage in box 5 and it wouldn't necessarily break within box 5, XXX XXO XOO (It can be quickly seen that it breaks outside of box 5, in either row 3 or column 3, but that's another matter 🙂)

[@blackonyx9842]

Richard Smith Can’t you only have a maximum of two open adjacent cells within a box, since there are only two available digits that can be adjacent?

[@RichSmith77]

@@blackonyx9842 Two cells adjacent, yes, but you can have a third in the box that is not adjacent to the caged cells, hidden away in a corner, screened from the cage by its two adjacent cells. There are reasons outside the box why that won't work, but it doesn't break within the box.

[@blackonyx9842]

Richard Smith Just realized my mistake and came back to note it, but you’d already explained it. Thanks for catching that.

[@Gonzalo_Garcia_]

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!

[@shalaz1981]

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

[@madeleinearancibia]

Honestly I was really hoping the next step of the break in was going to include the Phistomefel ring. Still a great puzzle though!

[@tomjansen364]

@@madeleinearancibia I actually used it in my solve and I quite quickly got some of the 2’s and 3’s and possible positions. But the brilliant thing is that it didn’t help me all that much in the solve. There just seems to be one path only always in Phistomefels puzzles! Side note; my solve is always together with Simon. I stop the video when I have enough info again and try to carry on. I’m not sure if I could’ve done it without Simon.

[@RichSmith77]

With the blue/orange pointing pairs identified in rows/columns 3 and 7 around the 40 cage, it doesn't need Phistomefel's ring to restrict their placement in the corner boxes. It only needs regular sudoku.

[@elaadt]

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.

[@CaptainSpock1701]

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!"

[@McMxxCiV]

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.

[@elaadt]

Ha! I see what you did there. LOL

[@Jukka70]

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

[@randomusername6]

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

[@bristolrovers27]

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.

[@ahmedbarkallah3130]

he must be gold4

[@nickjayyoung7662]

Phistomefel has actually featured several times on the channel, dig though the older videos to find them.

[@JohnRandomness105]

What do you mean, "a new setter"?

[@DipanGhosh]

Sarcasm people, sarcasm.

[@th.nd.r]

@@DipanGhosh I assume you mean well but I will kindly and gently remind you that Autistic people like myself exist who have a hard time understanding sarcasm, so don’t be surprised or disappointed when we incorrectly interpret a sarcastic moment please :)

[@dwagyn2202]

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)

[@britlion]

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!

[@ian7357]

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.

[@HunterJE]

Yeah, I think the operative part is “the *cages* are ‘greedy,’” which is supposed to indicate that the following rule applies to cages but I definitely read it initially with your interpretation as well.

[@sanabas1]

My initial interpretation was the same, just that orthogonally connected cages can be viewed as one big cage with no common digits, rather than the added restriction that anything within a cage is ruled out of the entire cage border. Clunky rewrite, but maybe 'any digit appearing in a cage (cage A) may not be repeated within the cage, nor appear in any cell/cage which orthogonally borders cage A.

[@korenogamer6075]

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.

[@killamoosdraree730]

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!!

[@RichardDamon]

Phistomefel puzzles are always good.

[@jasond4084]

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

[@cyrille6483]

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

[@maxgreve7856]

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!

[@awilliams1701]

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.

[@LednacekZ]

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.

[@metalmolisher666]

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

[@jimdavis2683]

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.

[@deniz5566]

Please let us know, I'd love to see it.

[@dwilkins499]

Lol. Are you related to Fermat?

[@erikremmelzwaal1225]

I guess we need to wait another 400 years for a proof then...

[@Anne_Mahoney]

@@erikremmelzwaal1225 And it'll be 110 pages long.

[@michielgerretzen9777]

in 16:37 you mean ? :-)

[@RichSmith77]

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.

[@adamheywood113]

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

[@bobh6728]

Only took me 41 minutes. Explanation below. Only 41 minutes to watch the video at 2X speed. So Simon’s apology for taking so long doesn’t apply to me!

[@john-paulwallcraft9362]

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

[@victoriam6569]

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

[@Cloleman]

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.

[@davidmonnerat8598]

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!

[@longwaytotipperary]

Brilliant puzzle - brilliant solve!

[@chrisr6503]

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

[@chitraagarwal8259]

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

[@Pathogenus]

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

[@ConManAU]

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.

[@leojs5673]

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

[@flatfingertuning727]

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.

[@angec9908]

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.

[@LithmusEarth]

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.

[@sanabas1]

@@LithmusEarth Right from the start, the 5s can't touch the 40 cage, which means the r6 5 cages are a single 5 and a horizontal domino, which means the r5 7 cages are a single 7 and a domino within the cage, and the only way to get a 7 domino, a single 5 and the 5 domino you need to have bordering the cage is for the 7 domino to be 16, and the 5 domino bordering the cage to be 23.

[@Moolers]

1:14:37 - "So we can actually..." remove those green lines? Oh never mind, we're doing the opposite instead :D

[@SmartHobbies]

First he had a ring. Now he’s got a cage. What’s next, the Phistomefel arrow?

[@bluji1250]

A ring and a cage... this is starting to sound like a circus, innit?

[@highpath4776]

@@bluji1250 MMA or WWW ?

[@donaldsnyder1543]

Bread & Circuses?

[@themorebeer3072]

A ring is two dimensional. A cage is three dimensional. His next lemma is going to add time, I bet.

[@highpath4776]

@@themorebeer3072 Well could give a completed grid and ask you to complete the clues and ruleset to create that solution.

[@jonathancard4466]

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.

[@dollarsing]

Holy cow!

[@feliomichaels]

@ 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.

[@ohshiditsgriff2793]

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

[@ericpraline1302]

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.

[@isaacthek]

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

[@chrisbattey]

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.

[@DerMarkus1982]

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

[@wazzzuuupkiwi]

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.

[@msolec2000]

1:01:06 Well... 56, actually, but the idea is the important thing. (Let's see who gets the reference)

[@countknersis700]

Imagine you are at a party and the dj plays a better song than before each time. Where will he find himself halfway? If you play snap with traditionalists and newcomers for too long, you will erode the initial core and what you intended. But it is the monetary angle and people involved that changed a simple puzzle into a mindset built on averages.

[@OttzelTV]

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

[@shivmizzet]

"I'm sorry I'm wasting your time" if the video isn't over 1hr I'm out. I'm here to watch you struggle to a solution. :)

[@AussieJohnny]

Just 37 minutes for me....until I gave up and watched the video.

[@Tfin]

Gah! If you really liked the 10 cage, why did you ignore it after filling it? It cleared up r7c2 right off.

[@Daihatski]

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

[@ronjohnson6916]

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

[@yumivt]

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

[@logzmort]

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

[@mickeyjupp1]

I thought cages had to have a clue in their top left corner? Not following the logic of the 7 cage in box 4. The clue is not in the right place aaarrrgghh!

[@msolec2000]

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

[@shteevuk]

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

[@debblez]

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

[@timdunkley9173]

As per a comment below, it was a bit unfair so say "set of cells" if single cell cages could be used.

[@HunterJE]

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…

[@highpath4776]

Sometimes I find explanations are more complicated than the question. !

[@DontMockMySmock]

I spent the first twenty minutes of this solve mad at you for misinterpreting the rules, but then I realized that you made a mistake (?) in writing the rules. Or at the very least, you've left in a semantic ambiguity. You have written "the cages are "greedy" ie digits in a cage cannot appear in any orthogonally adjacent cells or cages." Orthogonally adjacent to what? The antecedent is "digits in a cage," which implies that e.g. a 40 cage with [2,3,5,6,7,8,9] CAN in fact have a 7 adjacent to it, if the 7 in question isn't adjacent to the 7 in the cage. But what you MEANT was "a digit in a cage cannot appear in cells or cages orthogonally adjacent to its CAGE," not to the "DIGIT in the cage." It also makes a lot of sense now why I stared at this puzzle for half an hour accomplishing absolute bupkis.

[@lalalalalalalalal567]

If the rules was "not orthogonally adjacent to the digit" it would pretty much be normal sudoku lol

[@daboffey]

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

[@vladanmajerech4705]

The solution does not have clue on top left corners ... of cages

[@mcdonttube]

Doesn't "clue in the top left cell" mean that the 40 cage cannot go up?

[@tsgsjeremy]

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

[@farmerjohn6192]

Seems Simon you missed the normal cage rule that the cage total is top left. So the 40 cage could not enter Box 1, 2 or 3. The 5 cage in Box 4 Row 6 could never go left as drawn in red shading. So the two digit 5 cage must be the one in Box 6

[@lalalalalalalalal567]

The rules say top left of the cell, not of the cage

[@lexyeevee]

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!

[@Mega_Tron6891]

55:28 r7/c2 also can't be 9, since it's touching the 1/9 domino... has to be 8, which gives you the 7 above it

[@Mega_Tron6891]

1:02:08 he uses the same logic to get the 7 before the 8 🤣🤣🤣🤣

[@davidmonnerat8598]

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...

[@davidmonnerat8598]

Edit: found comments on this, sorry!

[@maverickstclare3756]

Ironic that the first proof puts 4 pairs in the Phistomefel Ring, therefore one pair in each corner and that information was not used later, esp. when they were reduced to 23

[@RichSmith77]

When we get the orange/blue pairs as 2/3, we don't really need anything as advanced as Phistomefel. Just regular sudoku restricts the 2s and 3s to the corner 2x2 squares in the corner boxes. (I feel like Simon used this whenever it was useful, but maybe I missed something.)

[@stevesebzda570]

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.. 😂😎☕[

[@Ratzfaz]

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 ?

[@fresheFresse]

A set can even be empty

[@Ratzfaz]

@@fresheFresse can an empty set be orthoganally connected ? ;-)

[@alienrenders]

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.

[@joelstevens5670]

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). :)

[@RichSmith77]

Only a minor point, but there is a way to only take six cells in box 5 that doesn't break IN box 5, namely XXX XXO XOO It does break either c3 or r3 though, so easily discounted.

[@joelstevens5670]

@@RichSmith77 Good spot, I did wonder if there was a possible 6-cell arrangement in the box. So to correct the proof, once we know that there are at least 2 cage cells in box 5 then at least 3 other cells will be adjacent to the cage unless we take at least 6 cells in box 5. Then (after determining that the 40 cage is 7-cells), we can only take 1 cell or at least 6 cells in the box. The only arrangement for 6 cells is to ignore one of the other corners and the two cells adjacent to it but this breaks the row or column outside of box 5 (based on whichever box the other cell is placed in), so we have to instead take either 1 cell or all 7. So slightly more complex but not too bad. Thanks for the reply.

[@tedg1278]

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.

[@themorebeer3072]

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.

[@Coyotek4]

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.)

[@allgreatfictions]

One of the first videos I've watched where I'm actually shouting at you... I just want you to figure out the shape of the 40 cage. Also, in total fairness, I'm writing this comment before I see you get it, so I have no idea whether I'm even right.

[@stevesebzda570]

I had a mistake [Had to backtrack ]

[@joelstevens5670]

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).

[@Styku]

It took me 4 months but I did it on my own 🤣