Is It Possible To Solve This Very Quickly?

** TODAY'S PUZZLE **
Simon's respect for the_cogito only grows with every puzzle of theirs he tries. Tonight's sudoku, Onion, contains some intricate and brilliant logic BUT Simon has a feeling it might be possible to express what he finds in a succinct way. Let him know if you manage it! If you enjoy these longer videos, Simon's solve of same author's Roller Coaster Nurikabe is available on Patreon now (and is 2.5hours long!).
Play the puzzle at the following link:
sudokupad.app/i9wx9vdy41
Rules:
Normal sudoku rules apply. A digit in a circled cell indicates the total number of times that digit appears in circled cells in the grid. Lines connecting circled cells are “Between Lines”. Digits on a “Between Line” must have values between those of the digits on that line's circled endpoints.
You can watch us in the latest @numberphile video here:
• A Sudoku Secret to Blo...
If you'd like to see Simon's solve of TotallyNormalCat's Broken Secrets on the main channel, drop a comment and let us know.
** ROLLER COASTER NURIKABE by the_cogito **
Simon's 2.5 hour solve of Roller Coaster Nurikabe by the_cogito is now available on Patreon. Check the puzzle out at the link below:
logic-masters.de/Raetselporta...
Also there's our January Patreon competition: All The Fun Of The Fair featuring sudokus themed around fairground rides!
Join us on Patreon (for as little as $2):
/ crackingthecryptic
** KICKSTARTER - DELIVERED 24 December 2023 **
Our Kickstarter has finished and we've created the finished novella/puzzle collection for you. It was delivered on Christmas Eve so check your inboxes!
www.kickstarter.com/projects/...
** ORDER OUR NEW BOOK AT THIS LINK **
If you would like to get hold of a physical copy of our new book, Cracking The Cryptic's Greatest Hits - Volume 2, there is still a chance. Visit the link below:
bean-games.pledgemanager.com/...
▶ SUDOKU PAD - Use Our Software For Your Puzzles ◀
You can input classic sudoku puzzles into our software and help support Sven, the programmer responsible for the wonderful user interface we all use to play these puzzles everyday. The app also comes with 12 handmade puzzles from us:
iOS:
apps.apple.com/us/app/svens-s...
Steam: store.steampowered.com/app/17...
Android:
play.google.com/store/apps/de...
ALSO on Amazon: Search for “SudokuPad”
▶ Contents Of This Video ◀
0:00 Theme music & Puzzle intro
1:37 Simon vs Roller Coaster Nurikabe
2:27 Crossword Masterclass
2:55 Hexcells stream coming soon
3:05 Happy Birthdays
4:33 Ludvig
5:25 Rules
7:00 Start of Solve: Let's Get Cracking
▶ Contact Us ◀
Twitter: @Cracking The Cryptic
email: [e-mail]
Our PO Box address:
Simon Anthony & Mark Goodliffe
Box 102
56 Gloucester Road
London
SW7 4UB
(Please note to use our real names rather than 'Cracking The Cryptic'.)
**************************************************************
▶ CTC FAN DISCORD SERVER◀
/ discord
▶ OUR BACK CATALOGUE - ALL CATEGORISED WITH LINKS!◀
tinyurl.com/CTCCatalogue
▶ NEW CRACKING THE CRYPTIC MERCHANDISE◀
teespring.com/en-GB/stores/cr...
▶SEND US PUZZLES TO SOLVE/CONTACT US◀
[e-mail]
▶FOLLOW US◀
Twitter: @crypticcracking
Instagram (for how to solve daily clues from The Times): crackingthe...
We also post the Wordle In A Minute videos on TikTok.

[@the_cogito5170]

the_cogito here. Thanks so much for another wonderful feature, I love this puzzle. The puzzle title came after I'd finished setting it, I felt like the entire break-in had so much to unpack, it felt like peeling layers off an onion :P Swordfish jokes aside, lovely solve by Simon, and much quicker than his humility could possibly let him take credit for. Beautifully worked out. Without spoiling too badly (hopefully), after seeing the trick with counting low digits, one can realize that using any pair actually adds at least 4 digits. In general the counting ideas Simon found were completely spot on, and I felt fortunate to have stumbled onto this puzzle :D Thank you again

[@emilywilliams3237]

It is so much fun to watch Simon solve something that is difficult like this, but clearly gives him so much pleasure. Thanks for creating this puzzle!

[@HCR2_Actuary]

I thought it was called onion because you can see the word “ONION” in the circles. O = all circles in columns 1-5 excluding the diagonal ones N = same as as O but add diagonal and remove rows 1 and 9 I = column 5 Then the last O&N are the same as the first two but in other side.

[@aonpl]

Great puzzle mate

[@bobbie0976]

Awesome puzzle!!!

[@BozoTheBear]

It's a pleasure to have experienced this puzzle - thanks for creating it (the_cogito) and showcasing it (Simon)! The break-in was too tough for me to discover, but beautiful to appreciate. I then enjoyed the solve afterwards, but still found it a little tough. It seems to be to be a case of a puzzle that exists in nature, and is what it is, so I don't think you could have forced it to be any easier.

[@debrabowen4276]

I always love hearing the secret, and I especially always love hearing that I am one of Simon’s favorite people!

[@MatthewBouyack]

I love that the first 40 minutes of this is just Simon relaying to us the inner workings of his brain, with almost nothing going on in the grid, and somehow it's riveting!

[@mbarbh]

My heart broke when Simon removed the 2 in R5C5 originally. Great solve overall though, it took me four times longer to solve this myself. Keep up the great original solves!

[@Melissa_Hz]

I just reached this point in the video and had to pause to come and check comments. Hopefully it’s somehow ok or he catches it quickly! 🥺

[@Hussfromthearea]

Same here, I was doubting myself when thinking about it. Fingers crosses he rewinds!

[@thisnthat3530]

Yeah, just got to that point and am thinking: Only _one_ of the twos must be "efficient"...

[@windybeach2184]

I had to rewind a few times at 47:10 to check if ruling the two out here was plausible.

[@359Aides]

The tricky thing is that the composition of the yellow digits seems to be 3x9, 1**, 1**, 2**, 2*, 3**, 4* where the number of * represents the number of colors covered by it. Because of that it is valid to remove 1 and 3 since there are only ** variants. With 2 however there is a ** and a * variant which could go into the center cell (or any of the middle row yellow cells that is). Actually since the same logic applies to any of the 1 color covering cells (R5C1, R5C5, R5C9) repeating Simons mistake would cause 3 cells to form a 4, 9 tuple which would immediately show that can't be correct.

[@jabber_likes_games]

I understood that 9 was the missing digit in a different way. If you look at the circles not in the swordfish there are 6 of them. The numbers 6, 7, 8 and 9 can only fit 5 of themselves into the swordfish if they are in the circles. This means that three of them have to fit the remainder of their digits into the 6 circles not in the swordfish. If 9 was one of those three digits (using two of them leaves more than a value of 9 out of the circles) then the other two digits cannot fit their spares into the remaining 2 circles after nine takes up 4 of them.

[@marpocky]

I saw it yet another way. {2,3,4} is ruled out easily by noticing it leaves too many 1s in circles. Thus we know exactly one of 5,6,7,8,9 is missing from the circles, meaning the other four can only fill up 20 of the 30 circles in the colored rows (swordfish). The remaining 10 circles can only come from 1,2,3,4 so we need all of the small digits to be present in circles and the only possible way this works is to have 9 be the unique absent digit.

[@Sviter]

exactly the way i went around it, glaad to see it here ! :)

[@torinpena288]

I figured out 9 was the missing digit the same way, but I *also* went further and figured out the exact contents of the 6-circle rows and columns (with rows as a whole and columns as a whole), plus the value of the swordfish digits in the middle row. From there I was able to determine the swordfish digits in the outer rows and make a few handy deductions that way.

[@easternbrown]

This is the way I saw that nine was the uncircled digit too. It immediately gives that the six other circles are 678, as well as placing the first 9. It felt likely given the way the bottom right box is configured, but that seems to be the easiest way to prove it.

[@maavinkayi]

How I got there as well

[@HunterJE]

It's not significant since it ends up being a dead end either way, but Simon seems to have gotten twisted up in the swordfish discussion early on, he keeps saying the swordfish digit(s) must be in blue but pretty sure the actual swordfish logic is saying they must *not* be in blue...

[@CrackingTheCryptic]

Yes you're right. I go a bit mad - although in my defence I think it is a little complicated. The existence of the swordfish rules 9 out of blue... and, yes, you're also right that this doesn't help :)

[@57thorns]

@@CrackingTheCryptic I think it helped to _not_ rule out 9 or 18, as there would be space for those digits outside the swordfish. But with between lines removing 19 from the lines, it could have.

[@shawncarter7188]

Thank you so much for this channel. There's nothing but pleasantness and fun in every video.

[@Orenotter]

Now that one was simply a slayer. It wore out this sudoku player. We were at it all night. And yes, "Onion" is right. For I stank at it layer by layer.

[@chrissolnordal9421]

I had to watch the first 25 minutes of Simon, to sort out the five rows of six circles in my head. I then went away and solved it. It took me countless hours, but I solved it!

[@warrenbrodsky7409]

What's so crazy about this one is that I think I could have probably stared at this for hours and not seen the yellow square swordfish thingy, but as soon as Simon pointed it out, the 1,2,3,4,9 logic became immediately clear to me and then I was waiting for Simon to catch up to me. But I wouldn't even have been at that step, if it weren't for Simon first. Really fascinating the way different brains see different things

[@douglasrogers3918]

Agreed - wanted a scratchpad that showed which digits of 1,2,3 & 4 had to be able to cancel a row and column and which just a column - then in a spreadsheet like way you can quickly prove the need for only removing 9 and also show that row 5 yellow contains a 2,4 and 9

[@evelynda5235]

Same!!!! I was suprised how long it took him when in most videos he is far far far ahead of me.

[@Vedvart1]

The way I determined the contents of the circles is by thinking from the get-go about how to put 6s, 7s, 8s, and 9s in the circles. If you excluded none of 6, 7, 8, and 9 from the circles, then you can can fit 5 of each into green, which leaves one 6, two 7s, three 8s, and four 9s left to put in the non-green circles. However, there are only 6 non-green circles, and you can only exclude *one* of 6, 7, 8, or 9 from the circles, so you have to exclude 9 - leaving the non-green circles to contain one 6, two 7s, and three 8s.

[@IlliniRocket]

Yes, this is the cleanest way to deduce that 9 was missing number. I didn't get it while I was solving this (and as a result this took me over 2 hrs), but I did upon reflection afterwards.

[@Gonzalo_Garcia_]

50:44 for me. Wow, what a break-in!! It took me around 30 minutes to get to understand it, and even then it wasn't easy to finish. Awesome puzzle!

[@KittSpiken]

In case the video length wasn't an indicator, this comment let me know the puzzle's above my paygrade. 😆

[@inspiringsand123]

Rules: 05:33 Let's Get Cracking: 07:04 What about this video's Top Tier Simarkisms?! The Secret: 7x (07:21, 07:55, 08:09, 08:47, 08:52, 08:55, 09:02) Bobbins: 4x (37:31, 38:08, 46:03, 58:52) Three In the Corner: 2x (47:43, 58:01) Knowledge Bomb: 1x (12:23) Maverick: 1x (1:00:26) And how about this video's Simarkisms?! Sorry: 12x (11:11, 13:54, 18:15, 18:34, 30:34, 36:24, 36:46, 43:26, 50:17, 52:59, 1:06:27, 1:14:46) Ah: 8x (19:49, 39:33, 45:54, 58:52, 1:00:07, 1:07:07, 1:07:07, 1:12:54) By Sudoku: 7x (53:01, 54:21, 1:06:46, 1:09:34, 1:09:59, 1:10:55, 1:15:46) What Does This Mean?: 6x (05:52, 22:23, 38:30, 45:54, 53:19, 1:03:04) Cake!: 6x (04:03, 04:20, 04:26, 04:40, 04:56, 05:01) Brilliant: 4x (1:20:10, 1:20:13, 1:20:15, 1:20:18) Pencil Mark/mark: 4x (59:23, 1:11:18, 1:13:06, 1:13:22) Beautiful: 3x (52:03, 1:12:41, 1:20:29) Break the Puzzle: 3x (54:02, 56:34) Hypothecate: 3x (38:59, 39:02, 40:36) Shouting: 3x (03:24, 05:04, 05:22) Hang On: 3x (18:15, 34:49, 38:47) In Fact: 3x (39:33, 42:07, 1:05:43) The Answer is: 2x (14:50, 49:31) Lovely: 2x (57:25, 1:05:16) Gorgeous: 2x (1:08:44, 1:08:52) Obviously: 2x (06:32, 16:29) Full stop: 2x (31:29) Whoopsie: 2x (32:11, 1:09:00) Fabulous: 2x (52:41, 1:18:31) Good Grief: 1x (52:03) What on Earth: 1x (54:26) Bother: 1x (1:12:37) Naked Single: 1x (1:17:23) Nonsense: 1x (52:06) In the Spotlight: 1x (58:03) Fascinating: 1x (1:20:15) Incredible: 1x (00:51) Astonishing: 1x (01:45) Take a Bow: 1x (1:19:48) Irritating: 1x (10:08) Surely: 1x (1:06:50) I've Got It!: 1x (1:02:25) Propitious: 1x (16:50) Phone is Buzzing: 1x (20:05) Wow: 1x (1:18:18) Weird: 1x (11:05) Most popular number(>9), digit and colour this video: Eighteen, Thirty Six (7 mentions) One (167 mentions) Yellow (85 mentions) Antithesis Battles: Low (8) - High (2) Even (3) - Odd (0) Column (54) - Row (46) 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!

[@aceman0000099]

Holy crap, I don't think I've ever seen autism of this calibre before

[@chocolateboy300]

I finished in 144 minutes. This is one of the most brilliant puzzles I have ever done. It took me about 113 minutes to spot the break-in. I realized the digits not in circles in columns 1,5, 9 and rows 1, 9 were limited due to the circles clues. I figured that there must be at least one 4, one 3, two 2s, and two 1s. That adds up to 6 cells being used from the 9 cells not in circles. This makes two and three digits adding up to 9 impossible. The missing digit must be 9 itself. I was so happy when I spotted that. It made me feel like Simon when he spots something gorgeous. I loved this one. Great Puzzle!

[@franzvanpoppin9418]

"Well done, puzzle!"😂

[@reeseparker2423]

Every time Simon tells us that we are his favorite people, it makes me smile so wide haha

[@dwebb2805]

Dear Simon, First of all, thank you as always for the video, I quite enjoyed this one (as I usually do). Second, while I believe your presumption about it being possible to understand things quickly, and in turn have a much quicker solve of the puzzle, I firmly believe from both an educational and entertainment standpoint, the way you did the puzzle and the resulting video were actually amazing (near-perfect I might say). Because you consider every option possible in the yellow sword fish and their far reaching implications, you allow the viewer to understand along with you, and possibly come up the the logic themselves as well. You may not like it when you feel like viewers can “see” things before you, but when it happens in this way it really feels rewarding, at least to me. And finally, if I were to try to describe e the break-in succinctly, it would be something like this: Part 1: any digit not represented in the yellow cells is circled at least 5 times, which therefore states that every digit under 5 is present. Part 2: all un-circled digit(s) are present in 3 yellow cells (each). Given those 2 axioms, once you do your logic about the minimum required number of 1s, 2s, 3s, and 4s, you get to the end point.

[@emilywilliams3237]

Well said. I love the rigorous logic, too!

[@MCarrington01]

Simon:”You can’t put a 1 or 9 on a between line.” Setters:”Hold my beer.”

[@LuanMerlin]

Challenge idea: Fog of War Sudoku, with Between Lines that also act as Zipper Lines, with negative numbers and Schrödinger Cells

[@MaierFlorian]

Put the digits 0 to 10 into each row, column and 3x3 box. Place two Schrödinger Cells into each row, column and box. :'D

[@MCarrington01]

It seems all the great and latest innovations have come from CTC challenges

[@byrearejohnson7698]

@@MaierFlorian Place 25 Schrödinger Cells into each cell

[@Mephistahpheles]

Heh. No doubt, one of these brilliant setters will (or maybe already have) come out with a suitable rule set for it.

[@piarittersporn]

After two weeks I tried again today and finally succeeded. Nice puzzle.

[@Mephistokles333]

Donauwelle - on of my favourite cakes (back home we called it Snow White cake) ^.^ its made with a filling of vanillapudding and sweet cherrys and a chocolate icing on top - very, very tasty ^.^(and not hard to make - you can even do it without baking the cake base - if you have the oportunity, try it ^.^) I´m not that good at analyzing - so I´ve never could have solved this puzzle without your input - thank you, Simon

[@RoderickEtheria]

Edit: Nothing here.

[@riziko2]

And row 5 was proven all yellow cells to be 249

[@jake_a_g]

He fixes later

[@iceberg54321]

Yah, I saw that as well, but he got unlucky and 2 was in the middle, it didn't have to be wrong, there just had to be a one two anywhere in row 5 at that point, so he missed on a 67% chance to be right.

[@emilywilliams3237]

Thanks for editing! I appreciate it when commenters who find mistakes that Simon (or Mark) go back and fix also "fix" their comments!

[@RoderickEtheria]

@@emilywilliams3237 I should probably fix this comment more than I already have. No point leaving a comment when it turns out the solvers never made a mistake since they went back and fixed it. In fact, it was my mistake for questioning their logical path.

[@duder1328]

Dear Simon, The speech you were referring to: "One day we will have to stand before the God of history and we will talk in terms of things we've done.Yes,we will be able to say we built gargantuan bridges to span the seas... It seems I can hear the God of history saying: 'That was not enough!’ But I was hungry, and ye fed me not. I was naked and ye clothed me not. I was devoid of a sanitary house to live in, and ye provided no shelter for me, and consequently, you cannot enter the kingdom of greatness." Martin Luther King Cheers!

[@anaayoung9142]

Well done puzzle 😅

[@zorandrndarevic1444]

Simple rules, great idea, beautiful look, elegant solution path. Love this! Bravo!

[@puritan7473]

An excellent solve of a fascinating puzzle, thank you Simon.

[@bilbf7881]

Great work as always. Enjoyed the puzzle very much.

[@HunterJE]

A generalized way of looking at the math preventing two off-circle digits once you know you need at least two 1s, two 2s, one 3 and one 4 in the yellow swordfish without needing to exhaust all the options - as established any digit that's fully off the circles will need to appear three times in yellow. Any pair of digits summing to 9 will have one digit four or less and one five or higher. So the swordfish will have three copies of the high digit and three of the low digit. But that leaves the other three low digits, which at best will use 1+1+2=4 cells. 3 low off circle digits + 3 high off circle digits + at least 4 other low digits = at least 10, which means that can't fit in the 9 cells of the swordfish.

[@CrackingTheCryptic]

Yes, that's a nice way of putting it.

[@tinus7036]

The first time in years of watching this channel that I get the same deduction in 5 minutes as Simon does in half an hour, haha. I feel on top of the world! Yeah, I simply counted how many 9's you could fit in the puzzle if there were supposed to be 9 9's. And there could only be 8. Of course, I got stuck immediately after that, but I take the small w!

[@Morgnanana]

1 minute into solve: Oh, it's a swordfish 10 minutes into solve: oh, it's a movie

[@emilywilliams3237]

As always, Simon, you are brilliant and totally captivating as you solve this kind of complex puzzle. It would be all too easy to just assume that of course the digit 9 must be the answer - it was very interesting to watch you work though all of the possibilities and prove what the circles must contain, and how to achieve it. Thanks as always.

[@Paolo_De_Leva]

Great title. It aroused my curiosity, without spoiling my solve.

[@ouwebrood497]

Here is my attempt to explain the 'Swordfish'-thing in a more straightforward way. To be honest: I didn't discover it by myself, at least not 'completely' so to say. We have 5 lines with 6 circles where the gaps form the Swordfish pattern. For the sake of argument, lets suppose we put all numbers in the circles. There is no problem with numbers 5 to 9, because they can each be placed in a circle on one of those lines. When it comes to 4: we have to 'free' at least one line, which can be obtained by placing 1 4 in a SF-gap. When it comes to 3: we have to 'free' at least two lines, which can be obtained by placing 1 3 in a SF-gap on the crossing of 2 lines When it comes to 2: we have to 'free' at least three lines, which can be obtained by placing 2 2s in SF-gaps, at least one on a crossing When it comes to 1: we have to 'free' at least four lines, which can be obtained by plaing 2 1s in SF-gaps on the crossings of two lines So, no matter what we will find out eventually, in the SF-gaps at least we have 6 of them filled with digits (2x1, 2x2, 1x3, 1x4). Which leaves us with 3 free places. So lets go back to the fact that we actually have 36 circles instead of 45. We can fill the 3 remaining gaps with one of the digists 5 to 9. In case of a nine, we are out of the woods. In case of any other of those 4 digits, we need to add 1 or 2 more digits from the numbers 1 to 4. This is not possible. It is even less so possible to add more of those digits 1-4 to make a sum of 3 adding to 9. Then we have our last option: 3 digits adding to 9, all from the digits 1-4 (actually this is only 2-3-4). We need to fill all 9 SF-gaps with those numbers 3 for each digit of the sum. But we need at least one more for the fourth digit as we found out earlier (every digit 1-4 has to be at least once in a SF-gap, no matter what). So our conclusion is that 9 has to be in the SF-gaps and is the only digit not appearing in the circles. As a bonus, we can also conclude that 2 of the 9s have to be on a crossing of lines, because one has to be on row 1 and one on row 9. Which leaves us with 4 SF-gaps on a crossing and 2 SF-gaps not on a crossing. This forces digit 4 and one digit 2 on the gaps without crossing, since the other digits need to be on a crossing to 'free' enough lines.

[@rickysmith5867]

The way I thought about it is that there are four digits from 5-9 that appear in all 5 green rows. That uses up 20 of the 30 green circles. This means we need to fill in the remaining 10 green circles with small digits so we cannot be missing any of them from the circles as a whole.

[@jdkemsley7628]

ah, yep, starting thinking from high digits makes the process much quicker than Simon's route. I only realized that once he pointed out the 678s in the remaining 6 circles.

[@karlmortenlunna2417]

Really hard, at least the break-in. Finished in 29:13. Loved the puzzle!

[@Zombiemask]

I would honestly make puzzles with 3's in the corner so he'd sing the little jig he does. XD Love it!!

[@angec9908]

I love Simon’s knowledge bombs

[@neil2796]

Did Simon completely change track on his description of swordfish, or did I miss something. He says 9 can't be in blue in row 5, then he says 9 has to be in blue in row 5?

[@CrackingTheCryptic]

I seem to go a bit mad :) The existence of the swordfish rules 9 out of blue but (I don't think) this helps anyway alas.

[@neil2796]

@@CrackingTheCryptic Thank you for the response.

[@cz8189]

Loved this puzzle, noticed that a lot of circles knocked out other boxes/rows/columns as possibilities I saw fairly quickly that it was impossible to put 9 9s in circles, then I started looking where 8 had to go. I may be daft, but I could find only one 8 pattern that would work with 8 8s in circles that didn't break the between lines or the normal sudoku rules then 6s and 7s.... it was only after that that I figured out the 3 nines in x wings without circles on rows and columns 1,5 and 9, and the remaining digits needed to fill the gaps and was able to see the 45 pairs on those between lines and sudoku the rest out, admittedly some trial and error working out the 8s so not as pure an approach as yours - but I did get a solution that my sudoku pad said was right in a time that pleased me. Thank you for your clear exposition - I learn something new with every one of your videos that I watch - and find them easier to understand and work through than those of other sudoku masters. Btw also loved the 159 miracle sudoku a couple of days ago.

[@MaierFlorian]

What Simon missed with yellow (even after he saw the easy fact that you need to put 1234 into it and even after he deduced how many of them each): You need 6 cells for the low digits, giving you a spare of 3 cells within a swordfish. Every non-1-digit constillation will cost you 6 digits for that, even if you take one of 1234, as none of them are used 3+ times, you will always exceed your limit.

[@FrancisFjordCupola]

Needing two 1's, two 2's, one 3 and one 4 makes a minimum of 6 out of 9 occupied yellow cells. In order to leave out a duo, both duo numbers (from 18, 27, 36 or 45) need three occurrences. Since none of the low numbers appeared at least three times, you'd need more iterations of them than the minimum. Thus the low numbers would need more than 6 cells total. But just one more makes 7 plus the 3 from the high number makes 10 cells. You only have 9. Thus they don't fit.

[@Cthulhus_Mum]

There’s an old adage: “A good conversationalist leaves you feeling that they are the most interesting person in the world. A *great* conversationalist leaves you feeling that YOU are the most interesting person in the world.” I bring this up because I feel that you are a *great* Sudoku youtuber - you do amazingly, but always just slow enough that we can keep up and feel that we’re extremely smart and could have done that ourselves lol. Thank you. And I’m sure you’re wonderful to talk to at parties, too 😊😁

[@SourabhDas95]

What an incredible puzzle! An absolute masterpiece! I'm confident I'm going to remember this one vividly! Simon often says that a good constructor draws your attention in a way that each step looks like the natural place to look next, and it leads you on this narrative... this journey of discovery... and it IS incredibly satisfying to solve a beautiful puzzle, but what makes puzzles really memorable, I think, is when you feel like the solve path closely followed the ideation and construction path, and you can see the story of the ideas that formed in layers behind the scenes, as you solve it. Now I don't know if this follows the actual story behind the construction of the puzzle, but the narrative I have in my head, is one of a constructor playing around with a lattice of circles taking up rows and columns 159 other than the intersection points. Usually, this brings up interesting set possibilities, so it seems like a natural place to start. Now, this would immediately use up 36 circles, and lead to 1 having to appear in 3 spots on the swordfish (even if there were a circled 1 somewhere), 2 requiring at least 2 spots, 3 requiring 2 spots, 4 requiring 1 spot, and 5 requiring 1 spot, which uses up all 9 spots. Then, since there's no space in the swordfish for the digits 6-9, 3 more circles would need to be added for the 9's that don't appear in rows and columns 159, 2 more for 8's, and 1 more for 7, bringing up the total to 42 circles. Also, since none of the digits other than 1, appear in 3 spots in the swordfish, and there's no room to add any, all the other digits would HAVE to be in circles, leading to either 44 or 45 circles in the final puzzle, which would trivialize the break-in, and circumvent any logic that arises from the lattice layout. But removing the circles from one of the rows or columns, leads to there still being 9 spots open, but more wiggle room for what the final number of circles would be, and only the required two 1's, two 2's, one 3, and one 4 to appear in those spots. At this point with 30 circles already placed, the digits that don't already appear in circles can add to a theoretical maximum of 15, but with a little more observation, the actual limit comes down to 9. With 6 of the 9 spots already taken up by 112234, the digits that don't appear in circles either have to be: - a single digit from 5-9 (which uses up the 3 remaining spots on the 9 yellow cells), - a single digit between 1-4 (which uses up at least 1 more spot in the 9 yellow cells, leaving no space for a swordfish to be possible for the digits 5-9) - or a combination of at most two of the digits between 1-4 (which could be a maximum of 6 using 2 and 4 as uncircled digits, leading to the 9 cells being 112223444) Of those, a single 9 is the most interesting, as it doesn't automatically eliminate any digit from from not being included in the circles count from the starting count of 36, and really allows a showcase of the lattice logic. From there, the challenge becomes figuring out how to take the idea further, to either start placing digits, or getting pairs or triplets, that start affecting other cells. With this idea, and the 112234999 group of digits that appear on the yellow cells, using between lines feels like an almost natural evolution of the puzzle. Of course, with the benefit of hindsight, and knowing how the story ends, that's easy to say, and it's absolutely an inspired idea. But given the disposition of the nine yellow digits JUST from the circles logic, it feels natural to start asking the question, "how can i restrict 1's and 9's?" and arrive at between lines. it feels natural to see the result that arises from JUST the circles logic (and with the benefit of knowing how the story ends) feels like almost the most natural conclusion to arrive at. From there, the one 6, two 7's and three 8's that must appear in circles not yet placed, along with the between line constraint, becomes a tool to then differentiate the rest of the grid to an unique solution.

[@ruckmanikrishnan4221]

What probably helps articulate the logic better and sonething Sinon didnt concentrate on (infact leading to his 49 error in the middle cell) was that... The 2 1s and 1 3 in yellow needed to be at junctions, while the 4 in yellow needed ro be in R5 (no junction)... Similarly 1-2 needs to be in the junction and 1-2 in R5... And there are exactly 6 junction and 3 non junction cells in yellow. Appreciating that would have allowdd for a reduced and more accurate pencil mark and easier articulation

[@Telixion]

I haven't started watching yet but I just wanted to let you know that I literally laughed out loud after I read the title "Is it possible to solve this very quickly?" and then looked at the video length duration. haha I will take that as a no :D

[@msgeryjo]

I can't believe that I will stop whatever I'm doing now, including working under a deadline, to watch Simon's daily sudoku. Never been happier. I dread the future day he might miss a posting. Re Simon's comment that he might have to do a second video if he gets stuck -- if Simon gets stuck long enough that it makes the video unpostable, I assume Mark can give it a go, but if Mark also gets stuck (or vice versa) does than mean it never shows up on CTC? Does Simon still try to solve the puzzle even though the delay in solving made it ineligible for posting? (I'm assuming yes, just because it would bother him not to). Do Simon and Mark ever give up on a puzzle totally? I think a list of puzzles that were tried by both and didn't make it onto CTC because of a solve delay, or puzzles that were tried by both and never successfully solved at all (if there are any of those, probably aren't), would be an irresistible challenge for some folks. Would be a good way to turn folks on to those especially tricky puzzles. Also, for the ones that did eventually get solved, I wouldn't mind seeing the solve path even though it wasn't a "live" solve, as there might be some interesting solving points in those that would be worth knowing. Seems a shame to never know what those super-diabolical puzzles were or who the setters were, or how the puzzles could be solved. (Maybe just on the Patreon platform, maybe just a link to a video by the setter re the solve path if it wasn't made by S or M.)

[@RichSmith77]

Yes. I've often wondered if there are puzzle masterpieces that didn't get their moment in the spotlight because both Mark and Simon couldn't solve them the first time, while recording.

[@ericpraline1302]

Very slick puzzle. I can't answer Simon's question, but I can confirm that it is possible to solve this very slowly. I made one deduction about the uncoloured circles of which I'm quite proud.

[@frankjiang1857]

Finished in 79:43. I figured out the first part pretty quickly where the digits in the circles had to be certain numbers. But then I got stuck for around 30 minutes with all the between lines stuff until I realized that two of the circles eliminated possibilities for the between lines stuff and everything flowed from there. Very fun puzzle, but it was tough to spot that 2nd break-in!

[@starrmayhem]

about the break in, the way i see it is if there are 2 digits that are missing in circles then there are only 3 vacant cells in swordfish, if "1" is in the circle then it needs 2 vacant cells, "2" also needs 2 vacant cells, "3" needs 1, "4" also needs 1; so even if 1 of the low digit doesn't need to be in a circle, 134 or 234 still need 4 vacant cells, thus disproving the circle is missing 2 digits or 3; i would not have gotten this without Simon, tygs

[@ServantOfSatania]

40 minutes of thinking before even earning the first digits, what a puzzle

[@Rubrickety]

Amazing puzzle, which unfortunately I failed to break into. I found the swordfish, and saw some implications for 1s in it, but was unable to visualize the dispensations of numbers as a whole.

[@briancouchman5658]

For the breakin, you can have a circled 9 in r1/9 and c1/5/9 so that’s 5, you need four in the six remaining circles… the same is true for 7 but only leaving two more in the spare circles, thus if 9 is circled 8 and 6 can’t be in circles… but 8+6 is more than 9… those six extra circles can satisfy 6 (one) 7 (two) and 8 (three) but not 9, therefor 9 isn’t circled… you were running the logic backwards with low number in the three columns and two rows. (Comment left at 44 minutes into the video)

[@anaphysik]

41:50 It's nice to know that 11 is technically-speaking "very many more" than 9. I forget which video this is from, but it reminds me of Simon declaring a potential sum to be "an ASTRONOMICALLY large number (in the context of sudoku)" (I don't think it was counted out, but the sum in question would've been 15).

[@NotWhiteRice]

This is how i did the first part Since there are 36 circles in the puzzle and 81 squares in a sudoku grid, there are 45 squares without a circle. The sum of each number that exists in a circle is 36, there are 45 numbers outside of circles, and at most 3 unique digits don't exist within circles due to row 1 requiring 6 unique digits within circles. Suppose that 9 is in the circles, Let 9+b+c+d+e+f+g+h+i=45: Case 1 (2 missing digits): Let 9+d+e+f+g+h+i=36, b+c=9, and b+c+(9-d)+(9-e)+(9-f)+(9-g)+(9-h)+(9-i)=45 Simplifying the third equation gives us 63-d-e-f-g-h-i=45 -> d+e+f+g+h+i = 18 Plug in the new equation in the first equation: 9+18=36 -> 27=36 Therefore, either only 9 or 3 digits are outside of the circles. Case 2 (3 missing digits): Let 9+e+f+g+h+i=36, b+c+d=9, and b+c+d+(9-e)+(9-f)+(9-g)+(9-h)+(9-i)=45 Simplifying the third equation gives us 54-e-f-g-h-i=45 -> e+f+g+h+i = 9 Plug in the new equation in the first equation: 9+9=36 -> 18=36 Therefore, only 9 is outside of the circles.

[@bobbie0976]

I love these hard puzzles that don’t require set theory (my weak spot). And so satisfying to know that, after figuring out the break in, there will be a 3 in a corner!!

[@LuanMerlin]

Cannot start solving the puzzle because my brain says it won't focus on anything else than wondering if there's a way to get a Donauwelle (the German name for Danube Wave Cake) during the night.

[@longwaytotipperary]

I am now craving chocolate!

[@levb258]

90:37, got the break-in logic on my own but got a bit stuck on the fiddly bits figuring out which cells were important

[@cheesybeard2980]

is there a bot that can keep track of "time to digitage" for us? 😂

[@jRoy7]

I've played all of the Hexcells series of games, lots of fun!

[@RazzlePhoxx]

It's a bit refreshing to see how tied up in knots Simon is on this because I can't understand it even WITH his help

[@corbness]

If you consider the yellow digits that cater for 2 sudoku units (i.e. rows 1 and 9) to count as being “twice” in yellow, I think it’s a little easier. We can say that there are then exactly four 1s, three 2s, two 3s, and one 4 as an absolute minimum. That’s “10” of the yellow digits already gone, but we also need to include one high digit or else we can’t add up to nine (you can quickly rule out 3 digits summing to 9 because you need 4 digits in yellow at a minimum). That digit must appear in yellow in rows 1+5+9, which is 5 times in yellow (because rows 1+9 count twice), so we have already accounted for our “15” yellow cells, meaning that the number of 1-4 in yellow MUST be minimized, so each must appear in a circle at least once, leaving only space for 9 outside of the circles. qed To take things a little further, you can also note that 4 can only appear in yellow in row 5 since it counts exactly once, so we can immediately rule 4 out of yellow in row 1+9, likewise, 1 must appear in row 1+9 in yellow to count for 4, and since two appears “3” times it must appear in row 5 to get an odd count, so we can also say from that pretty quickly that the yellows in row 5 must be 249, and rows 1+9 each contain 1, 9, then one of 2 or 3

[@fredgoodyer4907]

I like this, it gives me “let’s systematically overcount and then adjust at the end” vibes which is a common way to solve so-called counting problems 😊

[@SailSmBi]

@51:00 I feel like this should have been symmetric logic. When he ruled out the 1 in gray there could have been a 1 in r5c1 and r1c9 and it would have worked. Did he get lucky here?

[@urkbar8062]

I paused the video to look at the comments to see if anyone else spotted that, can't find an explanation on why it's fine one way and not the other?

[@SailSmBi]

@@urkbar8062 it was one of those times I got stuck in my own solve and wanted to see how they got through it and did not get a good answer.

[@gumbarius]

First digit within almost an hour into the puzzle is what I consider very quickly Sudoku solving (for my standards with easier puzzles)

[@aonpl]

Wow, you foud that mistake quick. In that kind of cases it usually takes me a few years of my expected life lenght to find it..

[@andremouss2536]

How is it possible that this horrible 1 in the upper right corner left thoughtlessly for soooo long didn't hurt Simon's eyes ?

[@_-_-Sipita-_-_]

45:47 for me. was stuck for long before finding the break in at almost 20 minutes.

[@gianlumndrll5999]

I think that the way of seeing that 9 is the uncircled, following the logic you used in the initial solving, is by aknowledging that 1, 2 , 3 and 4 need 6 Yellow squares out of 9 in the swordfish. Then you have only 3 Yellow swordfish sqares remaining so you must put a single digit in every r/c of the swordfish which sums to nine...which is 9 :D

[@jonh6585]

Glad he chnaged track as i spent nearly 10min trying to work out why if middle blue (c5) had to only be 4 or 9 then wouldnt same logic apply to green (c1) and red(c9) but then you have three cells in r5 with a "4/9 pair"

[@michaelawilliams]

I did get there faster. The very first thing you must do in a circles puzzle is COUNT THE CIRCLES. If it is a triangular number, then you instantly know the composition of the circles. If not, then you have a clue to their composition and can begin to narrow it down to potential combinations. Once you work out the numbers that are NOT in the circles, what is left must add up to the number of circles present. [SPOILER ALERT] Simon went down the swordfish rabbit hole which worked out fine for him but for me, it was the fact that there are five groups of six circles with six left-over circles that made me ask myself what digits needed to go into those left-over circles. 6, 7, 8, and 9 all have too many digits to fit into just the five groups, with 6 having one extra, 7 having two extra, 8 having three extra, and 9 having four extra. But 1+2+3+4 is greater than the six leftover circles. Since we also know that there are 36 circles altogether, the culprit or culprits must add up to nine. There is only one combo of 6, 7, 8, and 9 that's adds up to 9. So 9 is our odd number out of all the circles and the remaining leftover circles contain exactly one six, two sevens, and three eights. Now let's reverse the thinking. Since 1, 2, 3, 4, are ALL exclusively in the five groups of linear circles, how do you fit them in? Only one can contain a 1, only two can contain a 2, etc. This is where we have now gotten back to Simon's swordfish and the way he got to the conclusion that it is two 1's, two 2's, one 3, and one 4 in the yellow sections (with the rest being our 9's) was amazing. I got there through brute force instead by first seeing the only place where a 1 could go which forces the 2 out of the opposite corner and the rest was suduko, without ever having looked at the swordfish, which apparently was a faster approach.

[@RichSmith77]

"If it's a triangular number, then you instantly know the composition of the circles." Doesn't sound right. 36 is a triangular number, the eighth. You didn't instantly know 9 is the digit missing from the circles though.

[@michaelawilliams]

@@RichSmith77 Oh! You're right! I was so excited about presenting my alternative way to get to a solution that I made a major misstatement about triangular numbers. My bad.

[@SenselessUsername]

Simon very much starts from the circle count, seeing 9 are missing. He questions whether it's all 9s, two digits ("e.g. 1 and 8") counting up to nine, or even three digits (he lists the possibilities, and at 16:00 he tries to rule that out).

[@SenselessUsername]

I avoided Simon's mistake around 55:00 (wrong central digit) by excluding more earlier on, and not excluding the 2 by keeping two thoughts in mind: (1) The six 'swordfish lattice' lines are mostly interchangable -- the three columns (1,5,9) are, and rows 1 & 9 are; but not row 5: Putting a digit on the junctures of row 5 saves you from only one circle where putting them on the other six junctures saves you from two, and (2) Simon worked out the nine junctures are three 9s, one 4 and 3, and two 2s and 1s. Putting these two ideas together, the two 1s must exclude four circles so they must be on rows 1 & 9; the single 3 similarly must exclude two so must be on either row 1 or 9... But the 4 has to only exclude one circle so MAY be on row 5, and the two 2s must exclude three circles so one goes on row 1 or 9 and the other MAY go to row 5; and each row has a 9. Now fitting these 9 in 9 spaces, the "MAY"s are "must"s, and on row 5 we have (2,4,9) allowing us to remove 4 from rows 1 and 9 too.

[@rainhunter5546]

The way I got to 9 being the digit not in the circles was by double counting the yellow squares that see both a row and a column, which gives me fifteen takeouts. Of those fifteen takeouts I need at least four 1s, three 2s, two 3s and one 4, which leaves only five takeouts, exactly enough to remove one more digit.

[@vlad1209palovic]

46:55 - The yellow cells should include: - three 9's - two 1s (both "crossed") - two 2s (at least: one "crossed", one "single") - one 3 ("crossed") - one 4 (at least: "single"). So removing option 2 from "single" r5c5 is unjustified because one 2 still could be on "single yellow".

[@vlad1209palovic]

OK... 2 is in cell r5c5, so that was found 10 minutes later. 🙂

[@crazypantaloons]

Is this the fast explanation? There are 9 spots open in 5 sudoku groups (3 columns 1/5/9, 2 rows 1/9). 6 spots can count towards both a row and column. 15 is the max representation of uncircled. Each number needs 5 representation. The numbers 1 through 4 will take up 5 x 4 - (1+2+3+4) = 10 of 15 uncircled positions, if those numbers show up in the circles, and 5 each if they don't. The remaining positions can either go towards "9 not in circles" or "2 numbers that add up to 5 not in circles." Only the nines work. Values for uncircled in those 5 groups known: 5x 9, 4x 1, 3x 2, 2x 3, 1x 4. I've just finished filling those 9 in, working on the rest of the grid. Not sure if another surprise awaits me...

[@karsaanita]

This one took me FOREVER

[@Rach881101]

66:29 for me. Brilliant puzzle!

[@MichaelGreen831]

Title: "Is It Possible To Solve This Very Quickly?" Video length: 1hr 20min Answer: No.

[@garfnob4832]

47:20 sorry, the middle can be 2. you only need to remove 3 not 4 like with 1s. so, 2 in middle and a corner would work. inappropriately removing 2 was used at 53:05 never mind he found it and eliminated 4 a valid way :)

[@timdunkley9173]

Ten minutes shy of 14 hours for me. The longest I've ever spent on a sudoku which I have solved without guessing.

[@BakingFiend-rl7mo]

According to this video length it seems like yes you can solve it quickly

[@MarkBennet10001]

At 1h:10 ff is the deduction that caused me most problems just casually noticed - I did the logic on high digits before the logic on low digits and I think that clarified the solve, but I am in awe of this step, which was far from obvious to me.

[@MarkBennet10001]

BTW the answer to the question in the title is "yes, but if you do you have either been lucky or the headhunters should be after you as a genius"

[@RazzlePhoxx]

I made the mistake of just assuming that the 9 had to be missing from the circles so when I got stuck and watched the video I was extremely worried when Simon pointed out it could be any number of digits that add to 9

[@Xitsleachim]

I saw pretty much imidiatly that you need 3 of the digets 6,7,8 and 9 and you can only 5 of each in r1,r9,c1,c5,c9 so the six extra circles have to fill up the rest being one 6 two 7s and three 8s. Leaving 9 as the diget outside the circles. This took me Like 3.5 minutes but the rest of the puzzle took 3.5 hours lol

[@francoisduez601]

The question I ask myself is more like "Is it possible to solve this puzzle at all?" Or even to find a digit for that matter... 😵‍💫

[@Zwangsdemokrat]

Donauwelle ("Danube wave") is one of the most delicious treats you can eat in this world.

[@anibotic]

Using basic circle rules, I came to the conclusion that I'll need 2 1s, 2 2s, 1 3 & 1 4 in the non-circled cells on Rows 1 & 9 & Columns 1, 5 & 9. So, the last 3 cells need be occupied by only another digit which turned out to be 9, all other combos failed. After that, it's a simple circle sudoku.

[@woif00]

There's a logical error around 47:00 You rule out the 1 and 2 from the middle on the basis that they have to take care of two colors, but that's not true for 2. It's true for 1 and 3 but there can (and will) absolutely be one 2 that takes care of only one color and one 2 that takes care of two colors. The same logic that applies to the middle digit here also applies to the middle digit in columns one and nine

[@woif00]

Nevermind, he clears it up later in the video

[@Mn0ty]

The clearest way of seeing it: you need at least six low digits in yellow (1,1,2,2,3,4) therefore you need three of the same digit to make three sums of nine. Well those sums cannot be pair because you don't have three of the same digit in your low digits and if you were to put one more, there wouldn't be room for your third big digit.

[@GaimeGuy]

I'm so stuck but I feel like I'm so close to breaking into this... the digits 1-9 add up to 45. I see that there's 36 circles, so the digits that don't appear in circles must add up to 9... which means the high digits (5-9) have, at most, one digit that doesn't occupy any circles. Either every box, row, and column contains a circle with a 9, or a non-circle with a 9. My eyes are drawn to box 9... there is only one cell that isn't an endpoint or on a between line. If 9 is not a circle digit, then it must appear in R7C7. If 9 is not in R7C7, then it must appear only in circles. 1 and 9 can't be on between lines.... Argh. Maybe first I should consider if it's possible for the non-circle digits to be a 3 tuple (6 2 1, 5 3 1, 4 3 2).... or consider the number of rows/columns/boxes with 6 circles...

[@MattYDdraig]

49:04 Gruellingly clever.

[@ericsjoberg8167]

One of the hardest ones I’ve managed to complete without help 😅

[@chrisk9620]

Row 1 and 9 are high value: Remove 2 colors Row 5 is low value: Only removes 1 color Digits left out of the circles required 2 high, and 1 low For those in circles you also needed: 1 in 2 high 2 in 1 high, and 1 low 3 in 1 high 4 in 1 low So 1-4 needed 4 high, and 2 low This only left 2 high and 1 low

[@chrisk9620]

From here the minor mistake was taking 2 out of the center square thinking it needed 2 high.

[@BakuSudoku]

1:17:55 well done puzzle hehehe

[@jacob9692]

i didnt want to spend hours on this puzzle i just guessed all the Right far side 6 Digets in the Circles than the rest of the puzzle was approachable and Fun took 30 min

[@AngelWedge]

This kne took me a long time staring at the grid before putting anything in. First pencilmark at 5:01, first digit (7) at 5:59.

[@Escviitash]

The deduction that 2 couldn't be in R5C5 was really bad. Not only was it wrong even at this point but he failed to realise that whatever logic applies to this cell applies to all yellow cells in R5. His logic would leave a 4-9 triple in these cells. Wait, a 4--9 triple?!?, maybe double (nah, there are three cells), so maybe drible or, yes of course, a 4-9 trouble. None of them can be 1 or 3 as both 1s and the 3 in yellow must each see two colors each, and these three cells only see one color each, so they are a 2-4-9 triple with nothing else preventing any of the digits going in any of the cells.

[@pickledcucumber8693]

I was so close but I must have made a mistake because I can’t get boxes 2 and 3 to work 😅

[@noxumbra173]

At the 28 minute mark, where you put the 3 does not work. That space need to be either 1 or 8. How row 9 columns 9 would work for 3.

[@Mephistahpheles]

Pleased with myself that I got through the swordfish logic quickly. I broke it down as "1 must have 2 junctions, 2 must have 1 junction & 1 non-junction, 3 must have 1 junction. 4 must have 1 non-junction." minimum....forcing the 9 as the only option for the remaining 3 junctions. Filling them in was easy thinking about it that way. After that....returned to my usual (very slow) pace and made an repairable error along the way somewhere. Oops? Heh.

[@tiesz109]

If you can solve it slowly, than you can solve it quickly!

[@mariahbaetzing972]

If you worked out that 1,2,3 and 4 have to appear in yellow, you know that there is only one digit missing in the circles. Otherwise, 6 yellow cells would be occupied with the missing digits, so you would need 10 yello cells.

[@RichSmith77]

Not if one of your "1234" four digits is also one of your six missing digits from circles in the sum to 9 (which it would have to be if 9 is made up of multiple digits). You need the fact that there's at least four digits from 1234 even after you take one of them out of consideration, because it's used in your six digits missing from circles.

[@sirgeremiah]

The swordfish explanation left me confused. He went from saying a 9 could never go in those cells in row 5, to saying a 9 would have to go in those cells in row 5. I rewatched it twice, and my brain just couldn't follow. Anyone able to help?