The Graveyard Of Schrödinger

** TODAY'S PUZZLE **
A Halloween-themed masterpiece today from fjam whose sudoku, The Graveyard Of Schrödinger, is an absolute show-stopper! It is a tricky puzzle but it is standout brilliant from start to finish.
Play the puzzle at the link below:
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Rules:
Normal sudoku rules almost apply. Place the digits 0-9 into cells such that each digit appears once in every row, column and region. To accommodate this, one cell in each row, column and region is a Schrödinger cell, which contains two digits. In graves (cages), digits must sum to the day, month or year of the date on the grave. Digits cannot repeat within graves. Escape the graveyard by carefully plotting an orthogonally connected path between the green and red cells. The path may not cross the highest or lowest digits in a grave.
If you'd like to try Jane Street's Halloween puzzle before Simon's video tomorrow morning then you can do so here:
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** SIMON SOLVES KURT HUGO SCHNEIDER'S NEW SUDOKU **
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▶ Contents Of This Video ◀
0:00 Theme music & introduction
1:18 Try Jane Street's Halloween Puzzle!
2:11 Did someone say ARG? Join The Sudoku Hunters on Discord!
3:03 New DiMono Hunt On Patreon on 1 Nov!
3:49 Happy Birthday
4:27 Rules
6:41 Start of Solve - Let's Get Cracking!
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(Please note to use our real names rather than 'Cracking The Cryptic'.)
▶ 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:
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Android:
play.google.com/store/apps/de...
ALSO on Amazon: Search for “SudokuPad”
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[@fjam]

Well done Simon! This proved a little tougher for you than I anticipated, but I thought you did very well, even if doing a few things in a different order would have made a few steps a little easier. In particular, you made your life a little difficult at the end as you didn't use the 16/06/24 grave in C9. That said, I'm really glad that you found several of my favourite steps, including the sudden parity deductions and the strange way that difficulties summing to 7 in the 07/08/16 grave magically resolves it! One of the things I was happiest about when constructing the puzzle, is that I managed to make every single total possible in isolation, which means, at the start, it is impossible to eliminate any totals without thinking about the path logic. I was desperate to find a final grid which allowed the path to finish the way it does as I agree it is a bit ridiculous. I know other solvers have been convinced they've broken the puzzle when they find they can't draw the path straight from box 8 to box 9. A couple of comments I see have correctly spotted that there was a small logical leap when placing the digit in R6C7, but that doesn't matter much. You'd identified the totals correctly after all and even without using the 16/06/24 grave I think you could have found your way through using the deductions you made in box 9. Hopefully I can come up with a new twist for another graveyard next year. After all, you did give me this idea in the last graveyard feature!

[@davidrattner9]

Absolutely loved this from you!! Beautiful setting!! Thank you for the insight above!!

[@xChikyx]

i can't even comprehend how you could come up with something like this, it was incredible

[@KingOfShadeEmpire]

This puzzle made me say "What? No way!" out loud multiple times. What a brilliant puzzle that plays with expectations.

[@benjaminclegg7109]

I love two things about this video: 1) that half a year ago, you said "and cells cannot be Schrödinger cells - don't get any ideas, setters!". And then a setter took you up on the challenge and 2) instead of exasperation, you ended up enjoying every minute of schrödingerfied cells. :)

[@leojs5673]

i almost panicked until i remembered the uk daylight savings time 😂 glad to have this channel as an indispensable part of my daily routine!! ❤

[@crazypantaloons]

This video looks much longer than it actually is. They use metric time.

[@JRTJunk]

Got lucky at 1:24:00? Simon plonks a 2 in C7R6 even though it it could have been a 0 with the 27 in C8R6.

[@peaoui165]

He always gets lucky on things like that. The conspiracy theorist in me suspects foul play. The realist in me thinks he should just play the lottery

[@RoselynTate]

1:24:09 Simon assumes here that R6C7 must be a 2, which does turn out to be correct, but he skips some lovely logic around proving where the final Schrödinger cells are and how the path must move, which would ultimately end up proving that R6C2 must be a 2. All Simon should have concluded at this point is that R6C8 is definitely a 7, and similarly that R6C9 is a 4, regardless of which cell is Schrödinger. And since the possibility R6C9 being a 0/4 Schrödinger cell doesn’t change the sum for the cage it’s a part of, you can continue from that point into the logic for the cages in box 9, which then rather beautifully forces the completion of the path and the disambiguation of the Schrödinger cells and remaining digits. I honestly don’t blame Simon - by this point in a long solve with lots of complicated steps, it’s so easy to make a little mistake, not to mention the added strain of doing it on camera, and he was clearly a bit tired at this point. And the rest of the solve up to this point was wonderful to watch, and his delight at finding the end of the path mirrored my own. But just in case anyone else was wondering about how to get the 2 logically, I thought I’d share the path I found from this point: So, R6C4 = 4 regardless of whether it’s Schrödinger. Therefore the other two cells in its 16/06/24 cage must add up to either 2, 12, or 20. 20 is impossible with 7,8,9 already used in the box (3456 only add to 18), and 2 is impossible because R7/8C9 would have to be a 0/2 pair, and 2 is already used in the column. Therefore they must be adding to 12, using only 0,1,3,5,6 as options. The only way the math works is to use 1/5/6, requiring that R8C9 is the Schrödinger cell in the box. This makes the 14/11/19 grave 0/2/3/9 summing to 14. (You can also work these cages out in the opposite order as Simon did, but I found it a bit more complicated than starting with the 16/06/24 cage because of the possible totals and the geometry of the cage.) From here, you know the 6 must in the Schrödinger cell in R8C9, with either 1 or 5, which sorts out the 1/3 pair in box 6, and the 0/6 pair in box 3. To disambiguate what goes in the Schrödinger cells in boxes 6&9 with the 7 and 6 respectively, you now have to look back at the path, and how it can now possibly get into R9C9. Since R9C8 is now Schrödinger, and definitely includes a 6 in the 1/4/5/6 cage, it includes the highest digits and can’t be on the path, so the path must go through the 4 in R9C8. It then cannot take the 8 in R9C7, which is the highest in its cage, so it must turn up and get out through the 14 cage. Since it only take at most 2 cells to avoid taking the highest/lowest digits, and and can’t take the 9 in R8C7, it must take R7/8C8, but not R7C7. This disambiguates the order of the 0/2/3 in that cage, as the 0 can’t be on the path and therefore must be in R7C7, proving that R6C7 is a 2. From there everything is resolved, and the path is forced as Simon shows. It’s honestly such a fabulous puzzle, and I thoroughly enjoyed solving it! Congrats to fjam for yet another fabulous puzzle in this series!

[@MrMusAddict]

At 1:24:40, did Simon make a presumption that R6C7 was a 2? Couldn't it have been a 0, and then R6C8 might've been a 2/7 Schrodinger cell?

[@taylorsherry4784]

I think he did, but it was actually deducible at that point. If it was a 0 and R6C8 was a Schrodinger cell, how does the path get into the corner? You can't go through that 0 now so you have to go into the 14/11/19 cage in box 9. You have to go straight through R7C7 and R7C8 because if you turn down into R8C8 you will have to go into the lowest digit of that cage, you can't hide it away with the 9 because it isn't a Schrodinger cell. So you go into the 16/06/24 cage in column 9 and have to go down through 2 of the 3 cells. The only way that would not take the highest or lowest digit in the cage is if both the highest and lowest digit are in R6C9 as a Schrodinger cell, clashing with R6C8 as a second Schrodinger cell in the row and box.

[@dumediat_]

Yes, this seems to have been an unpunished leap. We’ve all been there! :)

[@srwapo]

I came here to ask this, but figured I'd see someone else already did. 😄

[@uigrad]

@@srwapo Same. It seems we all caught it!

[@alienrenders]

Yup, he didn't realize that the 2 could go with the 7 in the schrodinger cell. It's not about if it can be deduced or not. The path is supposed to clear that up. edit: Yeah, he finished the path after he completed the sudoku but you're supposed to use the path to figure out the ending. So because of that mistake, he solved the end backwards from the intended way.

[@slipperybone]

Hello Simon. I have been following your channel religiously for about a month now and it has been one hell of a journey. I had never solved a Sudoku in the 26 years of my life and now I'm solving zipper lines, palindromes, German whispers, and whatnot. Thank you so much for making these videos and helping me find a new hobby. Of all the puzzles I have watched, I had the most fun trying to solve this one. Needed a bit of help from you but I did most of it on my own and it made me feel very proud. @fjam deserves a shoutout for this fantastic puzzle. I don't think I've ever seen anything like it in my life. Absolutely magnificent.

[@Orenotter]

Well this one was really a humdinger. Those gravestones were really a scare bringer. With terror I froze On that path colored rose At the ghost (or the non-ghost) of Schrodinger! (Leaves a virtual cupcake with figurines of Puss in Boots and Dulcinea on top, Fnaire's and my Halloween costumes)

[@Anne_Mahoney]

Props for working Schrödinger into the rhyme scheme!

[@markp7262]

57:13 finish. This puzzle was such fun! I loved how you think the path is going one way, and all of a sudden you are in a completely different area. Masterful!

[@dannstarrjp]

That’s one of the best puzzles this year on the channel, difficult but so much fun!

[@MeriaDuck]

Wow that was great. Loved watching this solve and especially the finish. I was as amazed as Simon was with the final path.

[@inspiringsand123]

Rules: 04:41 Let's Get Cracking: 06:46 What about this video's Top Tier Simarkisms?! The Secret: 5x (11:16, 11:18, 11:29, 16:47, 44:04) Bobbins: 2x (10:50, 10:54) Three In the Corner: 2x (15:05, 17:00) Phistomefel: 1x (58:47) And how about this video's Simarkisms?! Ah: 28x (07:07, 07:42, 07:50, 07:50, 10:37, 16:01, 17:17, 17:17, 18:18, 20:55, 21:49, 23:13, 30:47, 32:34, 35:34, 38:47, 43:28, 48:46, 1:06:03, 1:07:46, 1:07:59, 1:10:49, 1:17:32, 1:19:43, 1:21:14, 1:23:25, 1:30:03, 1:35:43) Hang On: 16x (20:04, 30:47, 31:18, 42:42, 50:17, 55:03, 56:11, 56:11, 1:01:26, 1:09:29, 1:10:16, 1:12:46, 1:14:57, 1:15:28, 1:15:59, 1:30:49) Sorry: 10x (11:04, 25:51, 27:35, 1:06:39, 1:11:31, 1:16:03, 1:19:00, 1:27:17, 1:29:58, 1:35:01) Beautiful: 7x (08:01, 19:05, 19:09, 57:31, 57:52, 1:03:05, 1:20:06) Good Grief: 6x (16:29, 25:09, 54:06, 57:49, 1:02:58, 1:15:03) Clever: 6x (18:38, 18:38, 23:18, 24:46, 27:29, 1:33:00) Goodness: 5x (01:58, 07:14, 27:19, 1:09:01, 1:18:20) Bother: 5x (22:39, 44:22, 55:03, 1:05:05, 1:18:34) Ridiculous: 5x (23:18, 48:21, 1:33:00, 1:33:26, 1:34:13) By Sudoku: 5x (28:08, 46:48, 54:47, 1:05:27, 1:25:27) In Fact: 5x (08:43, 19:22, 1:07:00, 1:24:02) Wow: 5x (30:07, 40:17, 46:52, 48:51, 52:47) Pencil Mark/mark: 5x (35:28, 39:28, 1:09:15, 1:12:49, 1:28:36) What on Earth: 4x (22:54, 41:58, 1:23:45, 1:25:02) Brilliant: 4x (50:17, 50:21, 1:11:52, 1:35:21) Obviously: 4x (23:52, 44:39, 1:11:00, 1:13:57) I've Got It!: 3x (04:25, 1:35:26, 1:35:29) What Does This Mean?: 3x (13:39, 52:53, 1:06:47) Stuck: 2x (1:11:42, 1:12:00) Barbaric: 2x (1:23:38, 1:23:41) Apologies: 1x (10:02) The Answer is: 1x (33:51) In the Spotlight: 1x (17:02) I Have no Clue: 1x (1:31:35) Lovely: 1x (07:57) Incredible: 1x (1:34:39) Extraordinary: 1x (1:34:25) First Digit: 1x (16:54) Gorgeous: 1x (1:17:16) Surely: 1x (46:07) Whoopsie: 1x (18:19) We Can Do Better Than That: 1x (29:51) Phone is Buzzing: 1x (1:10:49) Fabulous: 1x (00:36) That's Huge: 1x (1:05:11) Have a Think: 1x (44:36) Cake!: 1x (04:21) Middly Digit: 1x (28:51) Most popular number(>9), digit and colour this video: Twelve (26 mentions) One (95 mentions) Purple (3 mentions) Antithesis Battles: Low (13) - High (10) Even (16) - Odd (5) Lowest (21) - Highest (17) Row (9) - Column (7) 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!

[@stevesebzda570]

You missed "might'nt it?" ("It might be a 9, might'nt it?") I'm not even sure "might'nt" is in your computer dictionary lol. "Might'nt it," though.. 😂

[@stevesebzda570]

How do ya spell that?? Mightn't? Daresn't (Dare not or Dares not) is spelled something like that-- maybe even dares'nt) in Pennsylvania Dutch. Simon, it cracks me up when you speak like that ("mightn't") lol. Good job ;)) 😂🎃🎃

[@stevesebzda570]

*2ND TIME* @43:21 : "We might be able to get that digit, mightn't we?" Two times. Sand computers probably has no idea that was a strange word or contraction. (Just like "daresn't or "outen" -- "dare not" and "outen the light"

[@stevesebzda570]

"Denise van Outen," is a Cutie. Fantastic on TBFQ a couple times. I like her almost as much as Claudia Winkleman, Although, they're all sweet 😋 😂🎃🎃

[@stevesebzda570]

[For anybody wondering; Denise van Outen and Jack Whitehall on TBFQ (The Big Fat Quiz) one year. They were great. ]

[@istvanmagi473]

At roughly 1:24:30 Simon made a lucky mistake stating that it has to be 2. It could be a 0, and the next cell is a 07. I can't disprove it at that point.

[@chris5619]

You are correct - sort of; you also made a mistake, which I assume was a typo - the next cell would be a 27 (not a 07 as you typed). I noticed that as well. But I will also say that Simon commented at the very beginning that it would be very easy to make a mistake here, and also stated after he finished that he's not even sure if he didn't make a mistake. With that being said, I made two mistakes (that I know of) throughout the solve, but in both cases they were incorrect mistakes, so I didn't get much further and had to backtrack. Simon got lucky with this mistake and it turned out to be correct. He also got to a point where he incorrectly placed a "full" 6 in r8c9, placing a 5 above it, but then realized shortly after that the 5 was pre-mature. It turned out to be correct, but he did think to go back at that point all on his own.

[@BryanLu0]

Simon got lucky, but there is no valid path if it is a 0

[@Vorash00]

Love that the path is how you start and gives so much at the start but by the end you’ve been kind of been forced to abandon it for a while and when you finally get back to path logic your sure you’ve made some kind of error but then it turns out that you haven’t at all. Brilliantly done.

[@alienrenders]

He made a mistake by assuming where the 2 is located in box 6. He got it correct, but only by sheer luck. The path was supposed to clear that up. Since he got the 2, he no longer needed the path and finished the end backwards from the intended way.

[@Vorash00]

I solved the puzzle myself and I didn't make the mistake in box 6 - almost did but caught it. (which is why i did see the mistake whilst watching but didn't feel the need to mention it). my comment still stands as based on my experience as well as Simon's. The routes we took where very different BUT in both instances the path gave so much at the start and then you abandon it midway through to come back to it at the end and I loved that about the puzzle.

[@TheSane42]

1:24:32 . It was too early to assume that the 2 was in r6c7, as it could still have been part of the schrodinger pair at the time. It will get disambiguated based on the path a bit later, knowing that the 0 and 7 would block the path forces them both into the schrodinger cell.

[@rachelrose7027]

Just casually using this British man as soothing background noise

[@G.Aaron.Fisher]

Fantastic puzzle. A little bit high on the work-to-cleverness ratio, but had plenty of cleverness nonetheless. This ruleset clearly gave the constructor a lot of creative leeway while setting, and the effort they put into making each step interesting was very apparent.

[@yadiracamacho499]

I'm also trying to distract myself from the horrible weather. Thanks, Simon, for the company!

[@gagstersps3]

Definitely one of the best puzzles I've solved in a while.

[@ImperialKneazle]

Thoroughly enjoyable to watch. Very well done, and highly entertaining 🙂

[@cgeezy3189]

The daylight savings time in the UK being a week earlier than here in the states had me searching twitter and comments trying to figure out why there wasn’t a video. 😂 thanks for being so consistent for our entertainment and mental health!

[@dinane]

That was a lot of fun. I loved that I could do some of it with the video paused - but I did need some prodding to think differently. Thanks for these videos! They make my brain happy.

[@TheBarnsleylad1995]

this puzzle took me over 5 hours of just having it open trying to take it in. at times i couldn’t see any way to break it down but i made it. it was exhausting!!

[@Swisswavey]

A really enjoyable puzzle, thanks for sharing it. Just over 50 minutes for me, which I'm pretty happy with 😊

[@titusadduxas]

2:11:29 - That was yet another gorgeous puzzle. I did use the checker occasionally along the way just to confirm my logic, which did provide some help, but I’m delighted to have finished it. Loved the snake ending too.

[@jonaclausson6698]

Nothing i could do on my own, but i could follow Simon very well this time and enjoied watching!

[@anaayoung9142]

I tried, and with your help I made it! It was very hard indeed, but fun to watch you doing and quite fun to try it. 🤓🤗

[@Paul-cn3ij]

Well after doing the very enjoyable puzzle yesterday that Simon did and somehow only taking 16 minutes, normal service has resumed! 1:36:55... Brilliant puzzle. Now to watch the video.

[@holliambria9870]

This one was very fun! Difficult, but rewarding. It was so clever how the path led you to the right digits!

[@stangerrits6712]

57:51 Very nice deduction, finding the Schrödinger cell due to parity in the cages!

[@frankjiang1857]

Finished in 81:19. Difficult but satisfying puzzle. The 3 different numbers for the graves threw me a little bit, particularly in the beginning, but if you stick with it and through the schrodinger and no smallest/largest in cage requirements, the logic becomes apparent in eliminating possibilities. I don't think you can try to separate out the path with the schrodinger and killer cages as it seems too interconnected. So, you're going to have to slowly go through both the path and the schrodingers and cages to arrive at the solution.

[@MegaSuperMM]

Dearest Simon, could you please walk me through sour conclusion about R6C7 at 1:24,40? Why couldn’t it have been a 0 and R6C8 be the Schrödinger 2/7? I think it still is correct, as I couldn’t find any feasible path through box 9 directly, but I’d love to see you proof it, just to calm my pedantic nerves (and because I don’t quite trust my own try at it)

[@bethbromley9590]

lovely puzzle!!! Really liked this one, even though I thought I wouldn't do it. 2 hours or so for me.

[@crazypantaloons]

Weird finish for me! After about 150 minutes of slow but steady progress, I was looking at a half complete puzzle and the path just popped out of the screen! The last half of the puzzle took 2 minutes. I believe the thought I had at realization was "you can't reconcile this without a path." Simon proved me wrong, but I got there! LOL.

[@RaniSalminen]

Wow. No halloween movie compares to this. Thank you!

[@ClairvoyantTruth]

1:15:29 Simon struggling over a known 3 and a 6 in the row was painful. Schrödinger cells need some new number placement so it doesn't look like options.

[@tehbertl7926]

Took me just over 2 hours. Solving went very smoothly until I got to box 9; I hadn't appreciated the parity tricks up in the graves in box 3, so the right side of the sudoku ended up very underdeveloped for me. I (unsuccessfully) tried some bifurcation on the box 9 square grave, but let me tell you, when Schrödinger cells are in the mix it becomes messy VERY quickly. Ultimately I deducted that the 16/06/24 grave in column 9 could only be 16, which meant that the two cells above the 7-in-the-corner had to add up to either 12 or 15. It wasn't until after that that I noticed the parity of the graves in box 3 and finishing the puzzle went very quickly after that. A phenomenal puzzle from start to finish; Schrödinger cell puzzles have quickly become my favorite variant.

[@windybeach2184]

Just like Mark needs a chess piece on his desk, I need a neon sign reminding me of PARITY! 57:25

[@thommy4078]

108 minutes for me. I struggled a lot at the beginning and the end. But it was approachable and i really enjoyed it

[@ericpraline1302]

Enjoyed this one though it was quite knotty throughout. At the end I was convinced I'd broken it as the route of the loop seemed so circumlocutory. Took me ages to spot the parity bits but that was very nice.

[@dannstarrjp]

100min for me, what an absolute joy it was! I was terrified in the end as I finished the path and all the sudoku, it was all good but it wasn’t the correct solution.. went through the entire puzzle again checking for mistakes and was able to find two cages being off by 1 in value, fortunately was able to fix it without rewinding back 30min of my solve, phew 😅

[@Oripy164]

At 1:24:30, how does he know that R6C7 is a 2 and not a 0 (with a 27 schrödinger cell in R6C8)? I solved the puzzle differently, and I was surprised by this deduction. By this lucky "guess" I think he skipped the whole path deduction at the end that was very satisfying. Fantastic puzzle!

[@michaelhoffman2011]

Took me a while. 90 mins, but was able to solve it. Pretty hard puzzle, but it just requires a lot of mental gymnastics. Great puzzle

[@MaierFlorian]

Near the end when Simon studies box 9 - was it really true to state that the middle cell can't be Schrödinger? He made that deduction under the assumption of those 5s and 6es (and some more digits) placed...

[@justinharper6909]

What´s the maximum time for a sudoku to still be uploaded here? My sudoku is still unsolved (though verified multiple times, with a guide) but might take around 30-40hrs to solve.

[@Stormorbiter]

I spotted that path about halfway through the puzzle and I was like omg I hope it takes that path :DD

[@miyahollands6136]

zero in the corner. "I need a zero, I'm holding on for a zero until the morning light!" 😊

[@Cid0484]

"We are in Venice, we are going downwards" - definitely something I never expect to hear in a Sudoku solving 😂

[@AlRoderick]

The solution checker in Sven's software must be pretty robust to be able to take a solution like that, I guess it must only check the solid digits and just not check the Schrodinger cells because by process of elimination they must be correct. Well done and happy Halloween Simon.

[@davidkay1612]

I think it checks against a supplied solution if not it says looks good to me

[@RichSmith77]

I have the automatic check on completion turned on, and it told me I had the solution when I entered my final non-Schrödinger cell digit, even though I still had the path and a Schrödinger cell to complete. So it seems it only checks the large digit, non-Schrödinger cells against an entered solution for those positions.

[@RichSmith77]

There was a neat parity argument that could be made in box 2. We could rule 7 out as a possibility from the 16/07/18 cage (minimum sum available was 1+3+4), so that was definitely even. We could also rule 20 out from the 15/11/20 cage (since the two cages in the top row boxes 2 and 3 summed to only 22 combined), so that cage was odd. Since the total for all the cages in box 2 had to be odd (45), the 12/09/27 cage had to be even, and hence 12. Once that was 12, the other two cages had to sum to 33 and therefore had to be 18 and 15.

[@Stormorbiter]

It seems that it so happened that every shrodinger cell contained either the low or high digits, but I don't think that was necessarily a requirement for all cages. I think that Simon made an invalid extrapolation there from the 2-cell cages.

[@HunterJE]

43:15 6+7+9=22 doesn't work here because the path digit would have to be 7, but 6 and 9 are both ruled out of the bottom cell of the grave by sudoku...

[@awilliams1701]

at 1:24:40 how did you put a 2 in R6C7? I see the cage requires a 2, but it could go in either position. If it was a 0 instead that would put a 1 in R1 and a 2 in R7. That seems ok to me.

[@shellmichael9665]

Almost an hour into the solve and parity of all things helps get a deduction. Crazy.

[@RichSmith77]

I managed to solve it in just over two hours. I somehow overlooked that the path disambiguated the 785 triple down c4 in box 5. I don't know why, because I'd been using the path avoiding lowest and highest cage digits logic everywhere else. I completely missed using it on that one cage though. Not sure how I found a path through without it. I may have let slip a mild expletive when I saw Simon enter the 785 with ease. It certainly simplified a few steps that followed.

[@sunriselg]

Interestingly we know that the path doesn't touch itself by uniqueness. Because we only have negative information about the path (other than "go from r1c1 to r9c9). Any points that touch themselves could be replaced by an equally valid shortcut.

[@Now_Or_Nova]

I'm one of the lucky few to have my birthday on one of the graves, my own personal humblebrag

[@chesneymigl4538]

I honestly couldn't even try this one until I watched the video. The instructions are baffling.

[@thiscat4426]

Took me over two and a half hours to solve and I filled a page with notes in the process, but I’ll be honest, I didn’t think I could do it at all. Scanning with the Schrödinger cells got annoyingly hard, but it was really fun anyway

[@David_K_Booth]

Yes, there were moments when I felt completely stuck, and it didn't help that a couple of times I got muddled, thought my labelled cells were doublers and tried to use parity. In a way, it's rewarding to spend over two hours on a puzzle: I just wouldn't have had the patience or confidence two or three years ago. That's something else that Mark and Simon have helped me to develop: thank you again, good sirs.

[@Paul-cn3ij]

I made the mistake of having a Schrödinger cell where I knew 1 number and had 2 possibilities to go with it and when I worked out the correct one, I entered it as a large number rather than leave it pencil marked with the Schrödinger cell. Played havoc with later pencil marking and I did it twice! The first time had me stuck for over 20 minutes...

[@57thorns]

I would sat that at 1:98:53 the just place 7 is also a horizontal gateway. 9 above it is definitely high, and 58 below is also high. And Simon, as usual, has overpopulated the pencil marks.

[@JamesStocks]

Amazing puzzle; but I'm glad I watched it being solved rather than attempted it myself

[@Gonzalo_Garcia_]

48:47 for me. Great puzzle!!

[@steveunderwood3683]

I took me a while to get the last step, as I couldn't believe where it was going and didn't look hard enough. I was just looking for an error in what I had done up to that point.

[@arcanelizard4646]

Should I have started this puzzle just before midnight? Probably not. Was it worth the 177 minutes it took me to solve this behemoth? Probably not. Did I enjoy just about moment of the solve? Absolutely. I’m going to bed now. This puzzle has wrecked my sleep schedule even further, and now I’m going to feel like I should be in one of Schrodinger’s graves when I wake up. Very interesting puzzle and ruleset, and clever setting, especially with the geometry of the path at the beginning of the solve. That break-in was very nice

[@stangerrits6712]

I guess Simon was right in thinking he was a bit slow on figuring out the cages in box 2. After some of the options were ruled out, there was really only one option left. You need to apply a bit of patience to investigate the different options.

[@Igshar]

I really loved this puzzle, but I think you missed the big finish via a couple of accidental shroedinger mistakes in the finale. When you set the 2 in the 02 cell at R6C7, you missed the fact that the 027 triple there was fine, if all 3 digits were in the box, regardless where they were placed. Later, you used some interesting logic to settle box 9 and where the shroedinger cell was. At the point with the 027 triple in the upside-down T, how I ended up solving was reverting to path logic. If you looked at it, you could have reasoned out that the path had to continue up to the 1 - if it turned right, it would either need to take 3 cells of the 4 cage (which fails even if it touches the shroedinger cell, since only 1 digit is untouched) or it would continue through into the triple in C9, which fails because the top cell was 04 and the two cells it had to take *must* contain a 5 - higher than the 4. At that point, you can deduce that the T *does* contain the full 027 triple, since you can't hook left into the 3 or up into the 9, so you have to go right, forcing the 2 deduction and the 07 shroedinger. The path with what you previously deduced would force you up and around and down to the 1-3 pair, which solves since you can't pass through the 1 but need to progress down. Shroedingers are all placed, and the finish is dictated, which then forces the rest of the puzzle to solidify into the final shape, without diving through additional deep logic hoops. Still greatly enjoyed your solve, just figured you would appreciate the beauty of the path coming in clutch to fill the remainder of the puzzle in the home stretch.

[@zacharyhall2012]

Agreed, he jumped the gun just a little bit on that one!

[@matthewread9001]

52:45 if you put the same digit twice in the same cell you can tour 0-9 in the box/row/column. But otherwise a 3 in two different Schrödinger is fine Ig.

[@biaberg3448]

A new video! Happy day 😊😊

[@Zeekfox]

Around 49:00, Simon's looking at square 2. I actually paused and thought about the totals, since all the graves are contained within the square. We need to get to a sum of 45. Naturally, there would be, I believe three ways to do this (20+16+9, 11+7+27, and 15+18+12), but we know we can't have 20 across the top because of the earlier row 1 calculations. And we know we can't have 7 in the smaller grave due to Sudoku blocking smaller digits. So I'm thinking it has to be the 15+18+12 combination. Guess I'll see if that's correct later. :3

[@Zeekfox]

Ooh, another revelation. Around 1:10:55, Simon marks out the 9 as a path blocker, but doesn't immediately deduce anything from it? Doesn't that mean, based off the path and blockades above, that we have to go through the 7 below it? It's slightly non-continuous at the moment since we haven't resolved either the 29 or 35 pairs inbetween, but regardless, we have to get into that 7 by going through the 014 penciled cell to the left, which then cannot be 0?

[@agitatorjr]

Yes, this part was painful to watch because he used the secret along the top and didn't apply it to the second cage. That's how I did it. Although it took me 3.5 hours over two days. 😅 I probably did a lot of things the hard way. I used the secret adding to 225 for 3, 5, 6, 8, and 9 to do some figuring of cage totals. I doubt that was intended, but it was a good check.

[@KeplersDream]

Has anyone ever checked Schröedinger's grave to see if he's okay?

[@donaldsnyder1543]

He was , then he wasn't 😁.

[@David_K_Booth]

Please don't check on Heisenberg's. If we find out just where he is, we'll have no idea where he's going.

[@windybeach2184]

I started in box 1, determining that one of the smaller graves must contain the S-cell for any sum combination to work. Then Simon promptly comes to the same conclusion simply by pointing out how the path must travel 😂

[@robertocurti577]

Before this video I always thought that Schrödinger wasn't for me..... now, I know I shouldn't touch it even with a ten ft pole.

[@jonnybolton4659]

37:51 B4 is much more easily resolved by just asking can the big cage sum to 29? Then the final two cells sum to 12 but all combinations of 12 are impossible. He even asked way back at 31:31 if it was maths, but didn't follow up on that thought!

[@brachypelmasmith]

this was at the top difficulty level for me.

[@57thorns]

55:44 "So this is 18", immediately runs off without cleaning up the pencil marks, possibly placing digits... sure, it was only about 45 seconds, but that is quite a long time when watching.

[@noahvale2627]

This was a rare case where I was ahead of Simon at several points. I'm not used to that.

[@Dragonlord446]

I thought about doing the puzzle, then I saw that it took Simon an hour and a half and I changed my mind

[@Rach881101]

95:37 for me. Magnificent puzzle!

[@flobiish]

The 2 @1:24:30. I'm having trouble understanding other peoples explanations of why that can be determined to be a 2. I've continued on watching the video and I see @1:28:35, once the logic moves the 6 into r8c9, the path is forced to go up r789c8 to avoid that 6 (as it's the largest digit in the cage even if it's an S-cell which Mark isn't considering at this pause point of mine so I'm also not trusting the 5 he just put in) and pushing the 0 or 1 into r7c7 (to be the lowest digit in the cage) whether or not anything had been done to the 2's @1:24:30. And that leaves r6c7 as the only place left in the column for a 2. Of course, where's the other 0 or 1 in that box, I think it has to join the 6 so maybe the 5 is right.

[@Zardoz37]

It's a small pity that Schrödingers actual dates of birth and death do not appear on any grave in the puzzle. But sadly that would be impossible (or at least, inelegant): he was born on 12/08/(18)87 and died on 04/01/(19)61.

[@raysouth1952]

Started watching but the red path on the grey background would not allow my eyes to focus so sadly I had to abandon it. Oh the tyranny of aging eyes!

[@fordprefect9027]

Yes the colour choice of red on gray was unusually unfortunate here. I have only a mild case of red green issue and I almost can't make out that red line.

[@myfyrmadocjones]

58:47 One foot in the graveyard 😂😂😂

[@CaptainSpock1701]

39:37 - Best comment ever! (Love snooker!)

[@dylanbreglio]

Solved it in 100:06, and now I'm frustrated I didn't solve it 6 seconds faster! Great puzzle though!

[@stevesebzda570]

I'm confused there (@1:24:40): "Nine total is needed in there." Yet, Simon says, "I need that to be a 2 (r6c7)." Why can't that go 0 27 (with the "27" Shrödinger totalling "9")? I'm not getting why a 2 right away there. ?? Thanks

[@zacharyhall2012]

I'm seeing this as a common comment; I believe Simon made a mistake here and jumped the gun.

[@robertogonzalezdacosta1836]

How r1c2 or r2c1 are ruled out as Shrodinger cells at the start of the solve?

[@prototypesoup1685]

Lets take one of the cells, say, r1c2. If that one is schrodinger, then you can shade in the rest of Box1, and the path moves downward to r2c1. But since r2c1 is in a domino with 2 non-schroedinger cells, then it WILL be either the highest or lowest. The same logic works for the opposite cell. :) IMO, at this point in the solve, Simon unconsciously creates the rule in his head that "if the path goes into a domino, that domino MUST include a schroedinger cell

[@diederikvandedijk]

The Schrödinger cell has to be at the other end of the domino the path goes through but not on the path because the path can only be on the middle digit.

[@vataksegen2588]

Hi, Because the path will lead through one of the 2-cell cages for sure, so the one with the path will get the schroedinger cell, because it needs at least 3 digits for the high/low constraint. and the high low digits in that cage will be in the schroedinger cell.

[@robertogonzalezdacosta1836]

Ty guys. I completely misunderstood the rules

[@diederikvandedijk]

1:24:38 r6c7 does not have to be 2 because 2 could go with 7 in r6c8

[@diederikvandedijk]

I think the placement should have been figured out by the path.

[@Salman-os7pr]

Can someone tell me how to join the discord server that simon spoke about in the beginning

[@Dragonlord446]

bottom of the description

[@michaelandersen-kk4fc]

wow took him half an hour to see how to get though box 8.. I shouted for so long that the 5-8 pair was always the highest in is grave

[@Probba]

If they used MM/DD/YY, the clues in each grave could have been given in ascending order

[@samanthajames-cooper5529]

Is it just me or was this quite blurry today? The candy for tomorrow we had a taste of was unreadable - maybe i need new specs.....

[@LednacekZ]

1:20:38 for me. i made many small mistakes along the way. not a clean solve at all. but i managed.

[@nikitashaitan9984]

Damn I spent like 10 minutes trying to figure out what’s going on in the 1st box and i’m not going for it…

[@goigadol1420]

1:24:30 why that square cant be a 0? mistake?

[@Ardalambdion]

What does rinky dink mean?

[@yadiracamacho499]

I looked it up because to me it means shoddy (like a rinky-dink carnival), and apparently it also used to mean deceptive, trickery. Maybe that's what he meant?

[@awilliams1701]

I thought that was broken. I thought that the 6 in R4C7 was the high digit missing the 7 in the shrodinger cell. lol So I didn't think that was possible.

[@awilliams1701]

you ruled out 03 in R456C4 incorrectly. You said if it was 03 then it would be 0/1/2 which is correct. But you said that R4C4 would be 1 which is wrong. 1 and 0 are ruled out making it a 2. 2 could go there. update: oh wait I forgot 2 is the largest digit so that is ruled out by the path.

[@tedg1278]

Around 1:24:20 your considering the 02 and 027 cells in box 6, and decide to make the 02 cell a 2 because of what is needed for the cage total. However that cell could equally have been a 0 with the 027 becoming a 27 Schrodinger and you never established any logic to support the decision.

[@timdunkley9173]

Yes, very easy to make a mistake. And another. And another. I backtracked several times until I realised that 4 doesn't lie between 7 and 9!

[@mcpecommander5327]

1:24:31 couldn’t you have a 0 and a shrodingered 27?