The Phistomefel Ring Sums To 109

[@TheSonicPerson]

Hello Cracking the Cryptic. PotatoHead21 here. Thanks again for another great feature. When I made the original Antiknight killer, I used the exact same principle to set the cage totals, but this time I did it with the high digits instead of the low digits. That way there are no degrees of freedom and the missing digit is forced to be 1. I had dreamt of making a fog of war sudoku that utilizes this idea and this is what I came up with. Also credit to Riffclown for helping me get the grid fully fogged.

[@trisha2584]

Great puzzle

[@Paolo_De_Leva]

Greatly creative idea. Fascinating geometry. Non-brutal arithmetics. Fairly challenging and properly rewarding from start to end. It can be cracked elegantly with *placeholder digits.* One of my favourite kinds of puzzle‼😏👍 See my separate comment for details about *placeholder digits.*

[@nathanmays7926]

After struggling every step of the way, I have to admit the reveal of the 28 cage in Box 9 made me a bit angry. I think you’re just trying to maintain your statistical reputation. 😂😂

[@TheSonicPerson]

@@nathanmays7926 I added it in because I wanted the solver to know they were solving the puzzle correctly when they discover it. When you add up the cages at the end, you'll see it does in fact sum to 109.

[@minamagdy4126]

I wonder if you had all 4 corner cages initially revealed in an earlier version of the puzzle for the solver to less directly think up your theorem in order to use min-maxing to place in the first digit.

[@rm6176]

This is like the 3rd or 4th time I have seen you explain the Phistomefel ring, and Each time I re-watch it several times and I finally can say "I GET IT!" WOW I actually GET IT! Thank you for taking the time to explain it to me so many times.

[@RL-gl8qg]

Congratulations! All those up thumbs mean that your CTC tribe is proud of you.

[@rm6176]

❤@@RL-gl8qg

[@jorava8768]

I had to hear the explanation several times before getting it, too! :D

[@MarushiaDark316]

This puzzle is actually a perfect way to teach both the Phistomefel Theorem and the Potato Head Theorem. You should give it a place of honor on the channel, maybe with a new playlist for 2024 showcasing the best of the best and / or puzzles that are quintessential at teaching a particular type of puzzle.

[@tamaspolyak5564]

Oh Simon, then he forgets that the red cells are not just for decoration. Still I loved all the logic he knew/found to get the puzzle started.

[@stangerrits6712]

I knew he started creating chaos when giving the 35-cells a red flash rather than just colouring the full cell, as he did for the other cells 😅

[@awilliams1701]

Simon missed a common knights move trick I often use. If you have 3 cells in a row that must contain a specific digit next to another 3 cells that must contain the same digit, that digit must occur in the center. In this case it was R3 C1-3, R4 C1-3 must have 4's in them. This means there is either a 4 in R3 or R4 C2. This makes R7C1 the 4.

[@francoisduez601]

I used it as well 👍

[@stangerrits6712]

I have heard him explaining this strategy in previous anti-knight puzzles, so he sure knows about this trick 😊

[@francoisduez601]

Indeed, but Simon knows too many tricks to remember them all when he needs them 😅

[@theotheo614]

I've been following the channel's videos for over several years now, yet I can't help but feel a little special every time Simon tells us the secret... Another spectacular solve today!

[@stevieinselby]

Fantastic puzzle, thank you PotatoHead21 👍🏻👍🏻 It took me just under an hour, but that was with knowing your theorem about the central cell and not needing to prove it to an audience! I can't help feeling Simon would have found it easier to remember the implications of the red flashed cells if the rest of the grid wasn't still a kaleidoscope of colour, there were definitely some quicker deductions available there, but overall I'm still in awe at how quickly he spots the essential logic (even if not always the basic scanning 😂)

[@Stompound]

I definitely did not have the required knowledge or skills to solve this myself but Simon made it super easy to understand and follow. Excellent work

[@Paolo_De_Leva]

This is one of my favourite kinds of puzzle: 🔹Greatly creative idea. 🔹Fascinating geometry. 🔹Non-brutal arithmetics. 🔹Fairly challenging and properly rewarding from start to end. 🔹It can be cracked elegantly with *placeholder digits,* which immediately reveal the configuration of *8s* and *9s,* relative to each other. After cracking the initial meta-logic and placing *1* in the center cell, you just start placing *8s* and *9s* along the ring and in *box 5* (for instance, you can start with a 9 in r3c4). After studying their configuration, you can properly rotate them and transform them into *8-9* pairs.

[@Paolo_De_Leva]

Of course, before writing a *9* in *r3c4,* I had already figured out that *9s* needed to have a fixed spatial relationship with *8s,* so I knew their correct placement could be eventually obtained from my configuration with a 90°, 180° or 270° rotation about the grid center.

[@facilvenir]

There is no way I could have solved this without help from previous videos. Thanks CTC and PotatoHead21! And what a scary 1-2 "deadly" pattern at the end!

[@ariwizzard]

48:39, this felt amazing to crack! wouldn’t have managed without simon’s wonderful explanation of phistomefel ring

[@marykkuhner6571]

It seemed immediately apparent that the first cell to be revealed must be the center. The puzzle is symmetrical and the 109 clue contains nothing that can break symmetry. The central cell is the only one that does not have symmetrical equivalents elsewhere in the puzzle. However I then got bogged down in maths and did not succeed. Still, just for the first bit it was worth the trip. And listening to you do it is, as always, great.

[@bhavyagarg1264]

31:54 The green digit can't be in r5c7 as it cannot repeat in its cage. You instead found a overcomplicated way to prove the same. Classic Simon!!!

[@nakorbluerider]

Spent the first 62 minutes slowly figuring out the proof that the ring had to have 8 distinct digits on it, and then the rest of the puzzle took another 48 for a 110 time lol. My only advantage was I had already figured out the pattern the 8's and 9's had to form as part of experimenting to figure out what proof PotatoHead21 was trying to get at with the ruleset, so once I finally had the first digit in and saw the cages, it went a bit quicker (for me anyway lol). Really clever construction all the way through. Kept taking advantage of the set theory throughout as well, or at least if it wasn't required, it certainly felt very available. An excellent take on trying to teach an idea like this through constructing a puzzle that requires it.

[@raunakagrawal4041]

30:30 I think simon got a bit lucky here. It seems possible that both r5c7 and r7c5 are not 8s or 9s. And the 8 or the 9 that must exist in the 27 cage lives in box 5 . Although it's easily disprovable as that causes r4c6 which is a 89 to be in r6c5 and then find no home in r456c3 where it must exist being a 8 or a 9

[@MadsOcto7]

Yeah. I noticed that as well. That's annoying

[@LednacekZ]

i think he skipped one step when he explained. he saw that you need 8or9 in the cage, because without it you would need to have 234567 in the cage and that could not be as 1 was already in. i also did this step automatically and had to check hard what you were talking about.

[@MadsOcto7]

@@LednacekZ yeah. It's fine that you need an8 or a 9, but why can't it be in box 5? I know why, but Simon either neglected to entertain the possibility or he neglected to inform us why he discounted it.

[@Posiadam.]

@@MadsOcto7exactly my thoughts! When he skipped this logic I immediately run into the comment section to look if I’m the only crazy to think that Simon missed some logic.

[@pouletbelette]

You're definitely not the only one ​@@Posiadam.

[@looweeg4229]

Simon, I don't usually write comments, but I really enjoy the videos. I also really like the fact that you don't count every subscriber as an expert and take the time to explain the logic behind your reasoning even if for some more experianced viewers it might seems obvious. Keep up the good work. And no you are not boring :)

[@Dysiode]

I've watched dozens of your solves and for some reason I never get tired of hearing the secret! There are so many little perspectives you show (like thinking about the 3 numbers missing from the 27 cage and what they add up to) that make me a better solver and the repetition helps me remember :) Edit: Also this is the first time I've seen this proof of the ring! It's SO simple! I feel like I could tell anyone about it

[@billionai4871]

the best and most impressive mathematical proofs are the ones where, at the end of it you go "gosh I couldn't have thought of that" because they seem so obvious in hindsight, but are so so difficult to figure out when they're in front of you

[@andrewgrant6516]

Fun fact: 109 was also the number of the Croydon bus that was found on the moon. God, I miss the Daily Star sometimes.

[@zadrik1337]

I think Simon has found someone to swap facts with at parties.

[@GuardOfGaia]

Wasn't it still a weekly when that story appeared?

[@michaels4340]

I don't think Simon covered the possibility that there's an 8/9 in r5c6 / r6c5, but it does prove impossible pretty quickly when you place other 8s and 9s (basically you'd end up with two of that digit in col/row 3, depending which one you try).

[@RedBarchetta2019]

Agreed. His proof was not rigorous. He kept saying "IF" this is a 8 or a 9. He didn't try if it wasn't, which quickly disproves anyway. The proof should have continued that because there are 4 of each 8 and 9 in the ring, then in the central box, 8 and 9 can ONLY go in the corners of that box. Then just try the bottom right corner and see that it does not work.

[@JamieDenAdel]

I even backed up and rewatched that part leading into the abrupt 8/9 pair, thinking I might have missed something while slightly distracted, but it was definitely a jumped-to conclusion.

[@aquaticIntrovert]

Yeah he arrived at the conclusion that an 8 or a 9 must be in the 27 cage but then declared that it must therefore be in the wings without investigating whether it could go in the middle box, which was premature. Glad I wasn't the only one who caught that, gotta keep the big man on his toes (as if he really needed our help with that)

[@michaels4340]

​@@JamieDenAdel Oh yeah, I did the same thing :P

[@sasopalma4993]

And I was going through comments solely to find this observation 😊

[@anaayoung9142]

What a great PotatoHead21 puzzle! Thanks for the solve! 🤗

[@arthurcharest9061]

Thanks Simon, I still enjoy the explanations every time!

[@buzzly108]

One of these times I'd like to see you solve a Sudoku without any explanations, then go back and explain the stuff you skipped over. Just for variety's sake, and to see how quickly you can actually solve these difficult puzzles!

[@OmoReturned]

First time that I managed to beat Simons time before watching the video, yay! Keep the videos coming, you guys keep my love for sudoku alive!

[@pedrohernandes3623]

Yes! I came here for my daily watch and just had a fantastic masterclass about Sudoku Theory! Thanks for that!

[@biaberg3448]

I love watching CTC solve fog puzzles! Looking forward to this one.

[@whatsleep17]

I really liked this puzzle. Two cool logic observations. 1) when Simon colored his reds 3s and 5s, you can figure out which one it is without the value of the cage in box 7. By knowing that R9C3 is a 35 you know that it must be red. Then in box 8, the red cell must be R8C6. That cell cannot be a 3 due to the 23 pair in box 5. Thus all the red cells are 5s. 2. Knowing that, you don’t need to use any of the 4 outside cages to solve the puzzle. Which I think is SO cool! Great setting.

[@chacha6993]

46:10 i think this is the first time i saw something before simon - beautiful puzzle and beautiful solve

[@longwaytotipperary]

Enjoyed watching Simon explain the mechanics of the puzzle and then solve it expertly! Also learned a new vocabulary word: otiose! 😁

[@davidrattner9]

Pleasure for us when Simon expands his vernacular!! 😁

[@longwaytotipperary]

@@davidrattner9 and adds to ours!!! 😄

[@RichSmith77]

​Vernacular: "The language spoken by the *ordinary* people in a country or region." Trust me. Ordinary people in the UK are not using the word "otiose". I don't think I've ever heard anyone use it in day to day language other than Simon.

[@davidrattner9]

​@RichSmith77 great response from you. 😁

[@BasWootin]

I dont understand why the 8 or 9 in the 27 cage must be in the phistomophel ring rather than box 5? I followed the logic for why if one of r7c5 or r5c7 is is 8/9 the other must be also. But I dont understand why its impossible for neither cell to be 8/9 and simply have the necessary 8/9 im box 5? (E.g. in r6c5)

[@HunterJE]

If it was it'd be entirely ruling itself off too many positions on the ring to appear four times like it has to. Would have to rewatch to be sure if Simon got lucky or if he saw this and just didn't articulate it clearly. (More specifically - in r6c5 it rules itself off the ring in row 6 by sudoku and in row 5 by knight, which means it can only appear in rs3,4,7 so only three times, with symmetric logic for r5c6; r6c6 it rules itself off r7c4 and r4c7 by knight's, r7c5 r5c7 by cage, and r7c6 r6c7 by sudoku, so can only appear at most once in row 4, once in column 4, and once in r7c7)

[@SirBradiator]

Because of the anti knight rule an 8 or 9 in R5C6 or R6C5 would rule it off the opposite side of the ring. Which would mean not enough 8s or 9s in the ring to add to 109.

[@bhavyagarg1264]

​@@HunterJE Maybe, But it doesn't explain why 89 can't be in r6c6. Remember, Simon didn't prove that both 8 and 9 should be in the cage. He only proved that atleast 1 is in the cage. So having that digit at r6c6 would work.

[@bhavyagarg1264]

Wait, got it. R6c6 can't be 89 as well as it will rule itself out of r456c7 and r7c456

[@RichSmith77]

​@@HunterJEPretty sure Simon didn't see this. He just assumed that one of the wings had to be an 8 or 9, based on his pencilmarks. I found that there was one specific pattern that emerged that allowed the same digit to appear four times in the ring. For that pattern to work for both 8s and 9s, and not repeat a digit in the 27 cage, they had to be in the pattern Simon reached at 38:05. I don't feel this was a completely logical solve by Simon though.

[@Robert_H.]

If, according to PotatoHead's theorem, only one digit may be missing in the Phistomefel ring, then the maximum of these is 16 digits: Σ(2,3,4,5,6,7,8,9 + 7,7,8,8,8,9,9,9) = 44 + 65 = 109 Since this corresponds exactly to the condition, the Phistomefel ring is maximum and contains the listed digits. Since 8 and 9 occur four times, they must not be in the corners. The missing digit, the 1, is therefore placed exactly in the middle of the Sudoku grid. 27:00 START

[@chocolateboy300]

I finished in 64 minutes. I don't think I've heard of the Potatohead Theorem, so this puzzle was completely lost on me. I had figured that the center cell was the key. I didn't realize that every single digit had to go in the Phistomefel Ring but one in a Knight's move restriction. I'm not that good at SET Theory, so I tend to avoid it as much as I can. Otherwise, that was a fun puzzle once I understood it. I did have to watch the video to that point, so Simon got the break-in for me. I enjoyed my time with this, though. Great Puzzle!

[@dannstarrjp]

30:35 I don’t understand why Simon came to the conclusion that at least one of r7c5 or r5c7 has to be 8 or 9. We established the cage has to have one of the 8 or 9, but can’t it be one of the cells in box5 instead? I don’t think we proved it can’t? I can see how we can prove it, but I don’t think it’s obvious unless I’m missing some logic that automatically assumes that if the cage has 8 or 9 it must live outside of box5?

[@grantfraser5430]

I figured there must be something I didn't know about the Phistomefel ring when I started but I had to give up and get the first hint from Simon, who naturally knows all the little secrets about every type of puzzle. I was so flummoxed by then that I was unable to think clearly and had to get a second hint. You win this round PotatoHead. But I will not be fooled again.

[@LorisLaboratory]

Lovely solve, this puzzle deserves a higher rating for sure. Thank you!

[@angec9908]

I always love when Simon gives these long complicated explanations why the 8 in box 3 can’t be on a white dot when there is a 7 and 9 looking at both 😂

[@cable729]

Not one but two great puzzles in a day! This one was fun!

[@markp7262]

29:10 finish. One thing I noticed early on was the pattern of 8-9s around the Phisto ring and into box 5. Both 8 and 9 had to have at least one position in either row or column 5 on the ring, or they would break the center cell. Using one cell as a proof showed the potential pattern, and rotating it around box 5 led to the final position, as the other position would have had two of the same number in the 27 cage. An excellent puzzle, as always!

[@oneeyedman4431]

Thanks for the theory explanation Simon, seen you explain it many times but it makes the solve more explainable to us minions

[@BakingFiend-rl7mo]

Oh snap another fog of war? Simon is bringing us great blessings this new year

[@JohnRandomness105]

I was aware of a certain theorem involving the P-ring and the knight's-move constraint, so it was straightforward arithmetic exercise to determine the middle cell using the P-ring sum. I got stuck for a while. I came up with the simplest combination to fill the 27 cage and the middle block, but what about numerous others? Once I was reminded of the knight's move, I was able to show that anything but the simplest combination violated it. It also gave me another digit, the 7. Overall, it took me a heck of a long time to push through the puzzle. Even at the end, I took a long time finding a knight's-move pair of cells to resolve an apparent deadly endgame. 16:10 It seems as if my calculation was a bit simpler than Simon's. I imagined 6789 in the four corners adding to 120, removed the minimum possible -- three 6s and a 7 adding to 25, and inserting the maximum possible -- 2345 adding to 14. 120 - 25 + 14 = 109. Remove more or insert fewer, and you got a lower total. The ring and the corners had to be this and no other. The digit 1 is omitted and forced into the central cell. On second thought, maybe not simpler: Simon should do this: subtract one each of 567 adding to 18, then insert one each of 234 adding to 9. 118 - 18 + 9 = 109. The result is identical, but has the advantage of beginning with a possible grid. 30:20 Try this out for size: if (say) R5C3 isn't 8 or 9, then you need one of 8 and 9 in row 5, block 5, to place both 8 and 9 in row 5. That digit is not in the central cell, therefore it is in R5C4 or R5C6. That digit now can't go anywhere in column 5 without violating either Sudoku or the knight's move constraint. Therefore, R5C3, R5C7, R3C5 and R7C5 must be 89 cells. The 27-cage: the protruding cells are 89 and the cells in block 5 are the digits 1234. 47:00 For a while, I used C and E for 3 and 5. It got a bit complicated. 52:20 Block 3's cage disambiguated 5 vs. 3 for me, but the other four corner cages -- their contents were determined before the sums appeared.

[@darthrainbows]

This is devilishly tricky to break into without knowing PotatoHead21's theorem. I made practically zero progress on this one (all I could prove was that the ring did not contain 3 sets of digits that repeated 4 times) for a long time before giving in and watching Simon's break in. Once I saw the proof of the theorem, it took about 45 minutes to solve.

[@psiphiorg]

What a fun, fascinating puzzle, PotatoHead21! Lots of interesting logic throughout. My time today was 44:38, solver number 318.

[@LavenderGooms]

I really enjoyed this one. It felt quite magical.

[@Gonzalo_Garcia_]

16:04 for me. Quite an interesting puzzle!

[@floyo]

31:30 Why is that? I don't think that's trivial

[@thesatty]

53:59. Brilliant puzzle! I spent maybe 20 minutes before finishing looking for the last knight's move that I missed. Oh well, I was happy to complete it in under an hour.

[@piarittersporn]

Brilliant puzzle.

[@Paul-cn3ij]

Excellent puzzle. Yet to watch the video but I found it interesting that I had worked out each of the 4 corners contents without the need for the killer totals. Corner marks and poor eyesight meant I didn't notice them until I was down to the last minute of the 52:58 it took me. So I think you could remove those cages if you wanted to make it slightly more difficult.

[@Paul-cn3ij]

Skipped through the video. Shame I forgot the knights move theorem. Took 25 minutes to enter my first digit!

[@DitDede]

same :)

[@rubenssiomusic]

30:34 How do you know that the 8 or the 9 (at least one of which has to be in the 27 cage), should go into the Phistomefel ring part of that cage? Why couldn’t the 8 or the 9, that has to be in the cage, go somewhere in the 5th box part of that cage?

[@rubenssiomusic]

Ok, I’ve tried putting, say 8, in the 3 places in the 5th box where it can go, and using night’s move constraint… it brakes the puzzle, no matter where we put it (b5c6, b5c8, b5c9)… so ok, the 8 or the 9 has to go on the Phistomefel ring part of the 27 cage… but Simon didn’t talk about it. Maybe he thought about it in his head and just didn’t say anything... But if he did not think about it, that means he just got lucky?

[@bernhardrenner5176]

Yes, Simon didn't rule out that possibility. Though it is not hard to prove that any digit which appears 4 times on a Phistomefel ring must appear exactly on two adjacent middle positions of the ring. It might be easier to notice that when the grid is sstill fully covered with fog. The digit has to go into a corner in box 5. The other cases are simple to rule out. As soon as you put it into a corner of box 5 sodoku and knight's move immediately forces the digit into the middle ring cells in the 2 boxes touching the corner cell. Since we have 2 digits appearing 4 times in the ring all ring middle celles becomes a pair of those 2 digits.

[@rubenssiomusic]

@@bernhardrenner5176 Thank you very much for replying! To be honest, I'm not sure if I understand the logic you described. If you have time, I would appreciate if you could describe it with specific examples, maybe by using b5c3 (box 5 cell 3) type of notations? Though, maybe if I read the comment 10 more times I'll get it, lol

[@bernhardrenner5176]

@@rubenssiomusic I'll try. We know that the digits 8 and 9 appears exactly 4 times in the ring each. For a digit that appears 4 times in the ring following is true: (1) It can't appear in a corner cell of the ring (r3c3, r3c7, r7c3, r7c7). (2) It must appear once in one of the 3 cells r3c4, r3c5 and r3c6. (3) It must appear once in one of the 3 cells r4c3, r5c3 and r6c3. (4) It must appear once in one of the 3 cells r4c7, r5c7 and r6c7. (5) It must appear once in one of the 3 cells r7c4, r7c5 and r7c6. Now where does a digit that appears 4 times in the ring go in box 5? (6) It can't go in r5c5, because due to Sudoku and Knight's move neither (2) nor (3) nor (4) nor (5) is possible anymore. (7) It can't go in r4c5 or r6c5, because due to Sudoku and Knight's move there would be only 3 rows of the ring left to put the digit 4 times. (8) It can't go in r5c4 or r5c6, because due to Sudoku and Knight's move there would be only 3 columns of the ring left to put the digit 4 times. (9) Thus it must go into 1 of the 4 corner cells of box 5 (r4c4, r4c6, r6c4, r6c6). (10) If we put it in r4c4 then Sudoku and Knight's move forces the digit into r3c5 and r5c3 to fullfill (2) and (3). (11) If we put it in r4c6 then Sudoku and Knight's move forces the digit into r3c5 and r5c7 to fullfill (2) and (4). (12) If we put it in r6c4 then Sudoku and Knight's move forces the digit into r5c3 and r7c5 to fullfill (3) and (5). (13) If we put it in r6c6 then Sudoku and Knight's move forces the digit into r5c7 and r7c5 to fullfill (4) and (5). We still don't know into which corner cell the digit goes but we do know that whereever it goes it goes into the 2 diagonally adjacent ring middle cells aswell. Since we have 2 digits (8 and 9) which appear 4 times in the ring all 4 ring middle cells (r3c5, r5c3, r5c7, r7c5) must be one of the two digits. Because of the cage we know that r5c7 and r7c5 are different (an 89 pair). From that and considering (10), (11), (12) and (13) we can also deduce that in box 5 r4c6 and r6c4 is an 89 pair and that r6c4 = r7c5 = r5c3 and that r4c6 = r5c7 = r3c5.

[@rubenssiomusic]

@@bernhardrenner5176 First of all, wow, Bernhard! Thank you so much for this outstanding explanation!!! I couldn't ask for more. I went over all the points you made there, and I GOT IT! I now know what you were trying to tell. That's amazing how much deduction can be done. Thanks again, my generous friend! I got it!

[@redstonekid2222]

Very fun puzzle, 64:16 after watching the proofs. I did the 8-9 logic completely differently, looking at where the digits can go in the middle box. The single X was very funny. Also the 3 in the corner.

[@joemucchiello4542]

Question @48:50 "Where is 2 in box 8?" Answer @52:02 Simon needs a little chess knight for his desk.

[@RL-gl8qg]

You aren't wasting my time because I suddenly MUST KNOW the Potato Head Principle. Oh, it's that one. I love that one. Well done, Potato Head.

[@Kinada]

Interesting puzzle. It took me a good while to get the opening setup but once you have it you get a lot of digits and information. Total time was a bit over 90 min.

[@joshuaharper372]

Once Simon placed a red digit, I started shouting at him to notice! And classic Simon, he forgot to use his colors to place the digits...

[@concibar4267]

110 for me. Got stuck a few times and did an oopsie that made me restart but managed it in the end :) very enjoyable

[@henk-ottolimburg7947]

I missed a detailed view about how you could possibly put in 4x9, 4x8, 3x7 etc in such a ring

[@TheFakeMackie]

Solve Spoiler Question: At 28:50 ish with the first cage, how do we initially deduce that there must be an 8 or a 9 in one of the wings of the kite shaped cage? Could they not be flanking each of those cells in box 6 and box 8?

[@RichSmith77]

It's been discussed elsewhere in the comments. Seems that Simon skipped a step.

[@ericpraline1302]

Interesting puzzle, thanks. I think Simon missed his vocation as a teacher - his patience with explaining sudoku for the umpteenth time never seems to wear thin.

[@deszeldra]

Definitely needed that rule for the break-in! But once I knew about rule I managed to solve it myself. 65:19

[@AnnaVahtera]

The 67 and 35 in box six was, again, classic Simon. The red not being able to be 6 OR 7 was of no consequence to Simon, and there was "no other way to do it" :P 3 in box seven: Am I a joke to you?

[@joelstevens5670]

For anyone stuck on the logic of 8s and 9s for box 5, here’s one way to go about it. First of all, note the 4 places where 8s and 9s are located on the Phistomefel ring. These are row 3 columns 4, 5 and 6, column 3 rows 4, 5 and 6 etc. Let’s consider the 8s on the ring in rows 3 and 7. Column 6 can only contain a maximum of one 8, therefore there must be at least one other 8 in columns 4 and 5 which is on the ring. All 4 positions where it could go see row 5 columns 4 and 5 by sudoku and the knights move. By mirrored logic, there must be at least one 8 in columns 5 and 6 which is on the ring and this rules out row 5 column 6. Therefore, 8 cannot go in the middle row of box 5. We can do exactly the same for the 8s on the ring in columns 3 and 7, which rules out 8 from the middle column of box 5. Thus, 8 must be in some corner of box 5 and the same will apply to the 9s. The 27 cage prevents us placing an 8 or 9 in cell 9 of box 5 (the bottom right corner), since it would rule out that digit from the locations on the ring in both row and column 7 using sudoku, killer and knights move logic. The remaining corner positions prevent an 8 or 9 from being in row 3 column 4 and row 4 column 3 (sudoku and knights move) and this allows us to place an 89 pair in both row and column 3. Now look at row 3 column 6. By the same logic it must appear in cell 7 of box 5 (the bottom left) and this in turn stops row 6 column 7 being an 8 or 9. Repeating this step for column 3 row 6 will complete the 89 pairs in the ring.

[@gmjv-1227]

Thanks this helps a lot

[@joelstevens5670]

@@gmjv-1227 You’re welcome, glad you found it useful. :)

[@penningmeestercgkdelft9159]

Solved it in 68:46! I loved the Phistomefel ring maths involved within the break-in :-)

[@adipy8912]

It's been a long time since you proved the Phistomefel ring. So it's not a waste of time doing it again.

[@furbyfubar]

At 45:42 you eliminated the purple corner in the ring saying it no longer could be a 7. But this could be shown just from the start logic you did saying what numbers went in the ring (and thus in the corners of boxes 1, 3, 7, and 9). Because, if any of the corners of the ring was a 7 and we should have exactly three 7's in the ring, this places the two 7's on the ring on the opposite sides. But where do you then place the 7's in boxes 1, 3, 7, and 9? The only answer that doesn't break Sudoku is in the 16 corner squares. But that can't be right as that would be *four* 7's in the corners, and we know that there should be *exactly* three 7's there, because that's how many we should have in the ring.

[@SilverKiMak]

Dear Simon, once again - please clean up unnecessary markup. It stops you seeing obvious things and generally makes the puzzle dirty. Eg @55:42 as soon as you got a 5 in the red-white cell - you have to put 5s all over the place, but you didn't see it as there iOS a ton on old markup.

[@peterpan1886]

31:30 for me, nice logic on the break in. Although I would have taken me a lot longer, if this „Phistomefel“-property of knightsmove sudokus hadn’t been featured on the Chanel before.

[@ingvarsuigin609]

Oh, a new theorem! I love this stuff.

[@TheSonicPerson]

Its not a new theorem. Have you seen "A New Sudoku Discovery Staggers Us"?

[@ingvarsuigin609]

@@TheSonicPerson I'll check it out. Maybe you can also remember some other "bigger Phistomefel's square"explanation, too? I've seen one but don't remember it exactly.

[@N0rdall]

I would love to see a phistomefel ring, knight move constraint, and shroedinger cell (where all digits from 0-9 must follow sudoku rules), where the central box is shroedinger... Just to disprove CTC that you CAN have 2 digits not in the Phistomefel ring under a knight move constraint. Unfortunately, I'm not a setter, unless someone wants to help me build that.

[@nikolakosanovic9931]

Wouldn't be more beautiful proof to highlight row 3 row 7 column 3 column 7 in one color and box 1 box 3 box 7 box 9 in other

[@thibauldnuyten2891]

I found it much more straightforward to use the 2x2 boxes corollary to the Phistomephel ring. If you fill each box with the numbers 6789, you get 120. Which means you have to reduce by 11, while adding in all-but-one numbers. You can add a 2 by reducing one set to 2789. 6->2 = -4 You can add a 3 by reducing another set to 3789. 6->3 = -3 You can add a 4 by reducing a third set to 4789. 6->4 = -2 We've now used 4+3+2=9 out of our 11 allowance to reduce numbers, while doing the smallest possible reductions. We only need to, and can only, add the 5 to to one set of 4 numbers, by reducing a 7 from any set. We've now used our entire allowance by doing the smallest possible reductions, which means this arrangement is forced and the center digit must be a 1.

[@thibauldnuyten2891]

So to rehash, there are 4 sets 2789 3789 4789 6789 Where a seven from any set is actually a 5. ======================= Edit: Well, that didn't quite work out. There was another way to reduce the numbers. Rather than reducing a 7 to a 5, the puzzle got there by reducing the remaining 6 to a 5, an option i discounted too early, and then reduces a 7 from one of the sets to the now necessary 6. Luckily the final conclusion is the same either way.

[@_aullik]

Did i just start a puzzle where it says: Solve counter 1. Well i guess its going to be 2000 by the time i manage to finish this

[@HunterJE]

52:50 Also not 3, since in addition to not being 2 its partner also can't be 4 since there's a 14 pair in the box...

[@jrparker811]

what a fun solve. 88 minutes, while distracted, but fun.

[@filkube8540]

1:00:54 "this must be a 5" sir do your red flashes mean nothing to you? ❤😂

[@MateiGall]

I think the totals for the cages in the corners of the grid are not necessary.

[@DitDede]

That was an interesting puzzle, and I did not make any use of the sums of the four cages in the corners!

[@ralfseewald4897]

Hi. I have one question, if Simon explained this and I just didn’t get it, please post me the timestamp. I totally can follow the logic, that if one digit is missing in the ring it must appear in the center of the grid. But, is it forced that the center digit must not be in the ring? Is it not possible, with the knight‘s move constraint, that we have all digits 1-9 appearing in the ring? Thanks for explaining or pointing me to the explanation given! PS: I love the channel! Thanks for the giving my brain a place to hang out 😂

[@pekrintala]

Did the Phistomefel Ring animation video ever get put on this channel?

[@RedBarchetta2019]

Piling on - my take and explanation: Assumption: We all agree there are four each of 8 and 9 in the Ring. Next: 8 and 9 can ONLY be in the corner cells of box Five. They CANNOT be in the "cross" of box five. Try it and see - you cannot get 4 8's or 9's in the Ring that way. Next: Simon kept saying "IF" green is an 8 or 9. He then assumed it was so and moved on. We all got angry. Next: The B2 Northrop Grumman Bomber shaped cage is what dictates which corners of box Five that 8 and 9 live in. With out that cage, 8 and 9 can go into ANY corner of box Five, including R6C6. Finally: Simon will apologize tomorrow but not admit he skipped this deduction. He will say why it could not work and that he was rigorous, (yes, we know) and will we accept that and move on. Cheers, Simon! And cheers to PotatoHead!

[@johnh2052]

Did Simon explain why the 89 in the 27-cage couldn't be in box 5? It can't, but I don't remember hearing him go over _why_ it can't. Maybe I just missed it?

[@srwapo]

78:01, I spent 30 minutes spinning my wheels before looking at the video to see that 8 of the 9 digits had to be on the ring because of the knight's move constraint. I thought that was the case, but wasn't able to prove it to myself and didn't remember that constraint from previous solves. Still difficult after that.

[@craigroberts7013]

i had the first digit inside of 3 mins and then took over 45 mins to finish the rest of the puzzle. I think I may have missed a trick or two, I already knew all the digits in two of the cages before I ever saw their cage total.

[@kevinmartin7760]

Simon spends a long long time pondering where the 8 and 9 go in the yellow cage. It is much easier to ignore this cage and first show that the 8 and 9 in box 5 must be in the corners of the box (otherwise knight's move and plain sudoku force two identical digits in the same row or column in the ring), then that they must be in opposite corners because of the pattern of the same digits must appear in the ring, then that the 8 and 9 must appear opposite each other in the ring in both row 5 and column 5. This gives an 8/9 pair in the parts of the yellow cage outside box 5, so the cage inside box 5 must be 1/2/3/4 and the rest of box 5 fills in quickly with small choices in each cell.

[@grantfraser5430]

Is it possible to solve this puzzle without the 4 corner killer cages? I had the total known before I even knew there was a killer cage in 3 of the 4 corners.

[@TheSonicPerson]

It is actually, those totals are the only ones that work in this setup

[@rgoyal107]

I don't understand why the 8/9 must appear in R5C6 or R7C5? Why couldn't it just appear in one of the three remaining cells within Box 5?

[@shiftcyclicity]

I KNEW IT WOULD BE POTATOHEAD21!

[@icepyrox]

I'm a little sad that it's been a while since I've heard the term "scrabble tiles"

[@justsignmeup911]

linked puzzle is not displaying correctly. I see shadows of all the variety elements and the fog never clears.

[@ToroHarfang]

"So I've not told anyone the secret in 2024! Well don't worry, I'm going to tell you the secret." 💜

[@awilliams1701]

You didn't rule out 8 or 9 in the 27 cage being only in box 5. I checked and it doesn't work, but you didn't cover it.

[@gmjv-1227]

I was wondering that also. It obviously wasn’t the case because the puzzle was correct but I couldn’t figure out why that wasn’t a consideration at the time

[@awilliams1701]

@@gmjv-1227 I opened sudoku pad and tried to put an 8 or 9 in box 5 in the cage while not having an 8 or 9 in the "outies". No matter what I did, it didn't work.

[@janerobson2297]

Simon you could have done this at least 5 minutes faster if you'd believed yourself regarding the 89 pair. You proved the positions of the 8s and 9s in the PR at least 3 times. Each time from scratch. At least we were all sure that you'd got them right!

[@stumbling]

This is the first puzzle in about a year I just couldn't crack. Over to Simon. :D

[@MariaVlasiou]

10:14 we need a proper theory book with sudoku theorems and practise exercises 42:00 😂😂 Phisto ring?

[@HunterJE]

52:51 Also not 3, since its partner still can't be a 2 and there's a 14 pair in the box...

[@shadeblackwolf1508]

If he'd left the potatohead proof on the grid, it would have resolved many 1s at 43:17

[@Rach881101]

37:12 for me. Interesting puzzle!

[@Kirbyfan87827]

Finished in 44:53 with help from the video.

[@AichoOshiro]

Maybe it would be a good idea to have some sample Puzzle or a screen which contains the most common rules, it's difficult to explain w a fog of war grid 😂

[@CL-rm2iu]

32ish.. why couldn't the green digits exclude both 8 and 9. Why not a 7, 2?

[@thongvu9047]

I have a question. Simon did prove there can only be one digit that's not on the ring, but why can't all 9 digits be on the ring?

[@TheSonicPerson]

If all 9 digits are in the ring, you have a sum of 45 in 9 of its cells meaning the remaining 7 need to sum to 64 which is impossible even if you had seven more 9s.