The Imposter Digit

[@DyingFlutchman]

Flutchman here: woohoo, just in time to ruin bedtime :) But seriously, thrilled to see this featured. Will watch asap!

[@unrealed]

Flutchman: Idk if you came up with the impostor mechanic, but I could see that becoming a new norm, like thermometers and chess moves are

[@lauraiss1027]

@Anna ♪ Please don't spam. This link has nothing to do with sudoku or anything relevant or interesting. People, don't click this.

[@giladzxc17]

Sir, this puzzle was a pleasure to solve. One of the best I have ever solved. Keep it up!

[@santaisreal]

@@lauraiss1027 telling them not to do it isnt gonna do anything, its a bot. best thing to do is just repot it and move on

[@DyingFlutchman]

After watching: Wow, what a very good solve. Your ability to focus on exactly the right spots for progress is almost uncanny ;) Indeed, I really liked the way 4 is forced into the central 9-cage as well, even through it hardly helps in solving the puzzle... Phrase of the day: "naughtiness indicator". That made me laugh 😂 And thanks for letting me take part in this continuing CTC puzzle discovery journey.

[@dingleford1611]

No way could I have done this puzzle, I just don't have Simon's logical acumen. But I still get great satisfaction from shouting at Simon every time he misses an obvious write-in.

[@Shif80]

I think this sums up 99% of people watching these videos!

[@masheroz]

Never am I so confident until Simon has done half the grid.

[@MargarineMeadow]

When he placed the “4” in box 6 which finally revealed the true and imposter cages I was internally screaming but Simon was so focused on other logic that he didn’t apply it immediately. By the time he finally got back to that section he had so much else worked out that it didn’t matter. So fun to watch that realization be mundane for him after it was such a focus

[@dingleford1611]

@@MargarineMeadow Very controlled to internalise your scream. Mr D in the next room is much amused by my soundtrack to Simon's solving.

[@uigrad]

If Simon's videos are over an hour, I pretty much know I can't do it alone. Those over 50 minutes are usually fails for me, but somehow, this weird puzzle is one that I did all by myself. Every time I made a deduction, I felt like I had squeaked out one last victory, but surely this was the point I was going to get stuck. Yet, there was always just enough to get me to the next step. It was extremely satisfying!

[@radstarmuffin]

So I decided to actually give this one a try before watching the video, and. Oh boy. After *16.5 hours*, I've finally done it! Hahaha. I can't wait to finally get to watch the video now and see Simon kick my ass completing it in under an hour 😂. Very proud of my accomplishment regardless, and it just proves even further to me how brilliant Simon is and how impressive these puzzles are!! Love you guys' videos (and puzzles), thanks for the (many) hours of entertainment!!

[@inspiringsand123]

Rules: 03:27 Let's Get Cracking: 04:59 What about this video's Top Tier Simarkisms?! The Secret: 3x (05:11, 05:13, 05:22) Bobbins: 2x (19:18, 54:12) Maverick: 2x (07:38, 07:43) And how about this video's Simarkisms?! Ah: 17x (06:09, 06:32, 06:36, 14:29, 19:18, 19:37, 20:01, 21:09, 21:23, 30:40, 41:57, 48:46, 52:18, 53:00, 53:00, 53:10, 54:24) By Sudoku: 16x (12:58, 16:04, 20:49, 28:43, 38:26, 41:18, 42:10, 43:04, 44:15, 45:36, 48:41, 49:44, 49:48, 50:23, 53:24, 53:33) Beautiful: 10x (00:33, 01:51, 01:51, 11:45, 17:17, 25:03, 30:19, 37:43, 37:43, 38:40) Good Grief: 6x (15:02, 17:14, 30:16, 40:48, 42:49, 46:23) Clever: 5x (17:19, 29:47, 38:12, 38:42, 40:52) Hang On: 5x (07:51, 12:36, 24:02, 34:28, 45:23) Obviously: 5x (00:53, 13:05, 26:50, 31:19, 32:01) Naughty: 4x (12:00, 12:07, 12:20, 12:39) Lovely: 4x (15:05, 22:24, 22:29, 22:33) Pencil Mark/mark: 4x (28:59, 43:37, 43:41, 45:05) In Fact: 3x (14:52, 20:14, 41:12) Sorry: 2x (09:54, 32:27) Stuck: 2x (02:54, 33:38) Brilliant: 2x (03:15, 55:00) Gorgeous: 2x (48:06, 48:06) Of All Things: 2x (49:28, 51:45) Magnificent: 2x (55:56, 55:59) Wow: 2x (36:20, 56:15) Aplomb: 1x (50:47) What on Earth: 1x (30:52) Bother: 1x (51:48) The Answer is: 1x (31:44) Ridiculous: 1x (46:11) Deadly Pattern: 1x (54:39) Approachable: 1x (02:45) Alacrity: 1x (53:48) Progress: 1x (55:38) Cake!: 1x (03:18) Most popular number(>9), digit and colour this video: Forty Five (16 mentions) One (144 mentions) White (12 mentions) Antithesis Battles: Odd (5) - Even (3) Highest (3) - Lowest (0) White (12) - Black (11) Column (12) - Row (10) FAQ: Q1: You missed something! A1: That could very well be the case! Human speech can be hard to understand for computers like me! Point out the ones that I missed and maybe I'll learn! Q2: Can you do this for another channel? A2: I've been thinking about that and wrote some code to make that possible. Let me know which channel you think would be a good fit!

[@Luwi1996]

I saw a three in the corner, but didn't hear it mentioned, so I came down here to check and it seems like indeed it wasn't mentioned, which is a bit of a shame xD

[@aristotelismetaxas7802]

Me in the begining of the puzzle: I am stupid. Me when I see things Simon isn't noticing: I am GENIUS.

[@th.nd.r]

Fifth incredible puzzle in a row! Dying Flutchman, take an absolute bow! The way it goes around in a circle, the geometry, the logic around one distinct digit missing from each cage, the novel ruleset, the fact that the numbers almost go up IN ORDER with the 1, 2, 3, and 5 in the opening followed by the 4 near the bottom right and the 6 and 7 in the middle, just beautiful. Simon, absolutely stellar solve, well done!

[@yagamilight08]

At this point I sometimes think if Simon purposefully leaves the most important clue which he was looking for 2 mnts ago n we all keep shouting at the 4 in row 4 col 9 looking at the real cage🤣🤣 P.S.: jk Simon it's a pleasure to see u solve these n it's a delightful experience that I mentioned 😄

[@kylenorris2836]

Finally up to date with CTC, these are the most calming videos ever. I watch them when I can’t stop my mind from spinning after a day of work🤔 keep keeping me sane, thank you

[@09h13]

i love to try these hard sudokus to see how much i have improved my logic since i began watching the channel! months ago, i'm sure i would never be able to spot a single thing, but now i figured that 4 was the imposter fairly quickly! still wasnt able to solve the puzzle after that tho lmao i gotta a long way to go still but i sure am enjoying the ride

[@solfeinberg437]

I can definitely relate. I'm starting to get the occasional hard puzzle and just getting minimal information from Simon. Like he just starts asking a question and I can unravel a bit more.

[@michaelawilliams]

The leap of logic that there is exactly one cage without an imposter was impossible for me to see even after listening to Simon's explanation repeatedly. I'm thinking it goes like this: every square is in a cage; there are 10 cages; one of them (the only 9-digit cage) has ALL of the digits including the imposter. Therefore, there are 9 cages left and, by necessity of Suduko, they must all be missing a DIFFERENT digit and, as a consequence, one of them is missing the imposter digit. It is the realization that they must all be missing a different digit that continues to be too big a stretch for my limited mind even after figuring that out. Why can't two of the cages be missing the same digit? Answer: because then that missing digit would have to appear twice in one of the other cages. Wow. That's a lot to take in.

[@ProdigalGeek]

Took me a fair bit of time to figure this out as well, so you're not alone on this Michael!

[@JosueSantiagoG]

Thank you for this! I paused around 12:00 to look for what turned out to be your comment. I still don't really understand, but the simple fact that I wasn't alone helps me continue watching his stream.

[@muferguson]

To borrow the wording of another commentor, it is the 9-cell cage that 'robs' one digit from each other cage, because it must be composed of the digits 1-9 once each. If the 9-cell cage was to rob two of the same digit, it would no longer be a cage. Hope this helps!

[@Hussfromthearea]

@@muferguson Omg, legit this morning (Aus time) I was reading through the comments trying to understand how this works, how he thought that each cell is missing one specific number each, until I read this....... I know you took reference from another comment, but thankyou so much!!!

[@Hussfromthearea]

@@muferguson on top of those now i understand the wording where "One of the digits from 1 to 9 is an imposter". Example is that let's say the imposter is a 7, in the 9 cell, there is 100% a 7. Therefore in 8 of the 9 for the other cages, there is a 7 imposter. *face slap*

[@justarandomdood]

I'm not looking at the thumbnail but if there's no lines in the shape of an amongus crewmate, then I dunno, I'll find this title even funnier than if there was xD

[@tylersoldo]

Plot twist: you are Dying Flutchman being an imposter, thus already knowing the imposter digit 😉😂

[@alanschutz7981]

Rule 1: The sum of the digits in a cage ADD UP to the total in the top left corner. Rule 2: The sum of the digits in a cage DO NOT ADD UP to the total in the top left corner.

[@crystalgehrt8861]

Definitely a mind bender, but what else can we expect from this setter? I'm very proud to have solved it, and I enjoyed watching Simon's delight, as always.

[@alasyon]

So many beautiful logical conclusions! And wonderfully explained! Thank you Paul and Simon!

[@andrewnelson2525]

Very happy to solve this in just under the video time. I saw right off the bat that r1c4=r3c3, r3c7=r4c9, r6c1=r7c3, r7c7=r9c6; all of which is provable from the cage shapes, and doesn't need the laborious process Simon used. I also saw quickly that nine eight cell cages and one nine cell cage meant every cage missing a digit was missing a different one which makes the progression from missing a 1 to missing a 2 to missing a 3 to missing a 5 much smoother. It's not often I see pure logic quicker than Simon! Usually I'm the one stumbling in a round about matter, then when I watch Simon I'm like well sure, it's easier when you notice that -- how'd he see that so fast?

[@typeoneauthor]

I have to give you a huge compliment. The way you think and the way your brain works is absolutely incredible, and I could sit here for hours listening to you solve puzzles. You're incredibly talented at this, and I love learning from you. My brain doesn't work the same way, and having learning diasbilities means I may never be bale to solve these sorts of puzzles, but I will never stop being fascinated by watching others complete this, nor will I ever stop finding enjoyment. I love this channel way more than I thought i would, and I don't regret clicking on it fir a second!

[@HunterJE]

24:37 Feels like an unnecessarily roundabout approach here, you could get straight to "10 cages total, 9 1s, no 1 in this cage so every other cage has a 1" without need for the "we can ignore this cage with all digits so there are 8 1s in the remaining 9 cages" step

[@huhusmremre]

Simon is a slow-and-steady, very careful and methodical, out-loud thinker. Which means sometimes he states the obvious, but it's what allows him to take on these extreme puzzles in the first place.

[@the29monkeys]

Another way to get there from where Simon is at the 24min mark is to ask where 1s go in boxes 5 and 6. Then see what happens if r3c4 is a 1 or not. Turns out the 47 cage has a 1 in it.

[@ytvids7347]

Such a beautiful puzzle, amazing construction. There's no way I could have solved this myself, but I love following Simon's logic and filling in the puzzles in a second window and pretending I'm clever too. :)

[@lunardancer6047]

I finally finished this puzzle after nearly two weeks of opening it up, looking at it fot 10 minutes, having no idea how to start and then closing it again. I finally managed to break in today and once I figured out the secret of the cages, I was able to finish it. But that break-in really was brutal, at least for me but very satisfying.

[@Ulija100]

Oh, love it - so good I did it twice to see how it worked... first time was half following Simon, half pausing to scuttle ahead. Wonderful puzzles when you can look back afterwards and see the ideas with clarity. Some observations afterwards: 1. Really neat that without the cage totals or imposter rule, the cage size distribution alone is enough to fully determine what the sets of digits in the cages must be (one of size 9, all nine options for size 8). 2. Easier to think of cage total as usual sum of digits inside, plus an adjustment from the imposter. So for example. if the imposter is 3, the possible adjustment must be between -2 and +6. Like Simon, I was expecting the imposter to be an extreme digit (1 or 9) forced by an extreme cage total, but here's an interesting thing: what is the highest imposter total for a cage of size (for any imposter)? You want it to be 52: largest possible true 8-cage (44) plus possible largest imposter shift (+8 from a 1 contributing a 9 to sum), but this is not achievable as it requires both the missing digit to be 1 and also the imposter to be 1 and in the cage. So largest 8-cage total is 51, but this can be achieved with EITHER the 1 or 2 as the imposter (and the other missing from the cage)! Similar thing with low total: 28 not possible but 29 is with 8 or 9 as the imposter. So, not possible to signal the imposter from a single extreme cage total. This surprised me! (Resolved here with the true sum without the imposter must be a given cage total, but now I wonder if there are other ways to make a puzzle work, would be interested to know if others can think of any.) Thank you DF and Simon!

[@Dcramer710]

Thank you for the happy birthday wish, Simon!!! Kim and I are watching at the beach! Such a nice surprise to hear this while watching :-)

[@pelahnar4]

I'm surprised Simon even considered that the imposter digit could be a 9. How can you change the value of 9 to make the cage total greater than it already could be using the regular digits from 1 to 9?

[@aBel-f6i]

in this case yes but if values were less than min sum of 8 numbers (which is 36) then it could have been 9

[@pelahnar4]

@@aBel-f6i Right. At the point in the puzzle where he mentioned he expected it to be 1 or 9, he'd already gone through how the cages that were greater than 45 needed an imposter to get that high. He clearly hadn't considered what an imposter 9 would mean and, since he concluded the imposter needed to be 4 or 5 by other logic, he never needed to.

[@aquaticIntrovert]

He also said putting a cage total on the 9 cell cage would be useless because it's a 9 cell cage, but it could have given info about what number the imposter digit could possibly be based on how different that total is from 45.

[@twentysEvan]

I found the comment odd too. Every number has the same potential range of “wrongness,” just in different directions. If you consider positive and negative adjustments as different, then 5 is the most “unpredictable” number. So I think (?) he just meant his instinct (poetic maybe) was 1 or 9 before having done the logic… ruleset alone. 1 and 9 do often have the tightest constraints in cage puzzles, so letting them take a different value would “seem” to have a lot of power to complicate a solve.

[@ruokothyan7658]

the thumbnails this week are really creative, i like that. Good work you two!

[@jamessweatt985]

I'd love to watch them each do a quick vid on creating the thumbnail.

[@lemlem35]

I'll be honest I forgot about this channel for a bit, just quirks of life and youtube being the way it is but man this was an absolute joy to watch

[@barebos]

I love to watch these video’s, but they go way above my head. Always a joy to watch Simon solve these difficult sudokus. From time to time I learn a thing or two, so thanks for that. Unfortunately I have to say that this sudoku isn’t solved. By sudoku rules it is correct and also the dots are correct, but if you add up the numbers in the cage you won’t get the total of that cage. None of the cages reach the total. For example, in the cage bottom left, it has to be a total of 49 with 8 numbers. Simon has a 1 and 8 outside the cage. To get to 49, you have to use a 7, 8 or 9 for the imposter in box 4, so the 2 remaining numbers outside the cage in box 7 must be low digits 1, 2, 3 or 4. Sorry Simon, but you have to redo this one :)

[@glee0rwj0nt]

None of the cages are less than 8 digits, so it's impossible that any of them are missing 2 digits since repeating digits within a cage is not allowed

[@RichSmith77]

"Simon has a 1 and 8 outside the cage". 8 in r6c1 (box 4) is inside the cage. (I mean, I guess technically there's an 8 outside the cage too, but then there are 73 digits outside the cage, eight of which are 8's. But what's important is that there's an 8 inside the cage). 🙂 In the cage, 4 is the imposter digit. It has an actual value of 9. In the cage, we therefore have values 9+8+7+6+5+9+3+2=49, so the cage total is satisfied. Not sure if I've missed your point, or you didn't understand the rule about an imposter digit? The solution is good, and satisfies the stated rules.

[@barebos]

@Sarah@@RichSmith77 thanks for explaining. The rules of the sudoku were puzzling to me. But your explanation makes it clear. Thanks!

[@MattYDdraig]

25:54 Fun concept and some beautiful geometric wrangling.❤

[@rogerrienstra1174]

At 17:45, you can prove there is a 1 in the top right 47 cage. First, there is a 1 in r4/5 c3 (in the 41 cage). That means that the 1 in the center box has to be in the nine cell cage. Then, the 47 cage and the 40 cage are the only two cages available to contain the needed two 1s for boxes 3 and 6.

[@wibblywobblyidiotvision]

Jeebus on a pogo stick, that was hard. Took me 2 hours, although I did spend 20 minutes at the end staring at the grid and not seeing the white dot between r3c3 and r4c3 that disambiguated it all. Well played, Flutchman. Well played.

[@MitchHamer98]

This is gorgeous, absolutely baffling, but gorgeous

[@JSully169]

Lol Dying Fluchman! That had me laughing for a good 5 minutes

[@timothydoyle3380]

I'm happy to hear that I'm one of Simon's favourite people. He told me the secret about 1-9 adding to 45.

[@roryparle]

Amusing moment at about 46:40. You immediately see that if r4c3 is a 6 then the cage total is correct, but then decide to add up all the cells again to work out the 4 would have to count as a 6 to make it work.

[@solfeinberg437]

You can see that R7C7 = R9C6 right away by considering where the first cell goes in Box 8. Not in the cage. Not in the row. That leaves only the second cell. This is similar in each corner.

[@Bokkie100k]

Please add a second camera so we can have a view of Simon's hands rattling the keyboard.

[@JohnDBlue]

It's always fascinating to realize how different people think so differently. I could never solve this puzzle, but I found it fairly obvious that the imposter digit can't be very high or it would not be able to provide the necessary extra value for the cages above 45 - meanwhile Simon, who is much smarter than me by any measure, said early in the video he thought it would be 1 or 9 (before any logical deductions were made) Unless of course my thinking is completely wrong, which I'm fully aware is a possibility... But I am fairly sure 9 being able to count for any sudoku digit wouldn't help at all.

[@Gonzalo_Garcia_]

35:37 for me. Awesome puzzle!! I’ve never seen any kind of logic like this before, I loved it. And I finish just in time to go and watch the stream!!

[@leojs5673]

su(s)doku

[@johnrupesh4535]

I absolutely love this new idea.

[@Jodawo]

I think what makes this puzzle difficult is figuring out that one of them doesn't have the imposter digit. It seems very logical because of Simon's explanation but without the explanation finding the logic seems daunting. Once you get that piece of logic you get to narrow down which one is the imposter digit. It's brilliant how it was set up because even in the rules it doesn't say that there will be a cage without an imposter gidget. It's the pure logic that steers you into that fact. It's almost like a riddle was put inside of a Sudoku puzzle and yet you were not told there is a riddle. I didn't realize how the setting of this puzzle was as genius as it was until I saw Simon explain, what now seems so obvious.

[@asktheraccoon]

When you go fetch some logic from behind the gates of hell at 39:00 and it does nothing for you. I feel you Simon…

[@justinvance9221]

Lol

[@parkerbond9400]

My first shout out in a Simon video!

[@jstarks21]

Very nice, I enjoyed this one!

[@adam70501]

I’m struggling to see a gap in my thinking: - I came to same conclusion as Simon at the beginning, 4 being the impostor and behaving like 9 so impostor factor is adding 5 to the number - I also took other way and counted the sum of 9 marked cages which is 392 and subtracted 8*45 (there is only 9 digits from unmarked cage but they are equal to 45+impostor bonus) so all the rest cages should add up to 360+8*impostor bonus. 392-360=32 and 32/8=4, so impostor bonus should add 4 to the digit Those 2 ways doesn’t have same solution so one is wrong. I just can’t see where PS. I haven’t yet seen full video (just first 13mins to check which way Simon took) as I am intending to complete the puzzle on my own so if the answer is later then I’m sorry

[@adam70501]

Ok, i read the rules - the value of impostor can be different, it is just the same number being the impostor but with different values. Sneaky

[@MariaVlasiou]

Dear God, alacrity! As a foreigner who never heard the word, I read it as "alarcity" decades ago, which made sense as an alarm gives me a feeling of briskness, and it stuck that way...

[@Alsadius]

80:09 for me, and I got a bit of help from Simon getting there. Tough, but interesting.

[@ChrisRemo]

Do the rules require that the imposter digit never count as its stated value? Some of the solving logic in the video seems to rely on that assumption, but if it’s part of the rules it isn’t clear to me. The rules state the imposter can be ANY value from 1-9, it doesn’t say any value from 1-9 other than its stated value, or something like that. I suppose it’s sort of implicit in the word “imposter” but as far as I can see it isn’t explicitly stated.

[@10bd1]

It can still be a 4. However none of the logic in the video relies on the imposter digit not being itself.

[@huhusmremre]

No but it's never relevant

[@RichSmith77]

For some of the cages it could be deduced that they had to contain an imposter digit with a different value to itself. I don't believe it was ever just assumed the imposter digit would have a different value.

[@littlebreadlive6232]

Ooh, another long one! Time to stay up for an extra hour!

[@Tof0986]

I found this puzzle somewhat different from the others... it was puzzling 🙂

[@bluerizlagirl]

The way to look at it was, the 9-cell cage is robbing one digit out of each of the nine 8-cell cages which comprise the other 72 digits of the puzzle. So they must each be missing different digits; and one is missing the impostor digit, so must have a correct total. You probably could do some stuff "out of the grid" with the cage totals and what each combination would add up to, but for the rule, and how to match the cage totals by altering only one digit .....

[@isavedtheuniverse]

Ahhh, of course. I rewatched the part where Simon concluded that one cage did not have an imposter like 5 times and could not grasp what logic he used to do so. But stating it this was was perfectly clear to me. Thank you.

[@RussellPaez]

As Simon puts the 3 in the corner, I wait for him to sing "That's 3 in the corner," but it never came. No confetti coming out I guess means he doesn't notice anymore.

[@Fredrick046]

When both his arms go in the air, you know its a great puzzle

[@Theolynn]

This was brilliant!

[@gregind01]

To work out that there needed to be a 4 in untotalled cage in box 5 I saw that there needed to be two 4s in 47 and 40 cages in boxes 3 and 6 ruling 4 out of cells r5c7 and r6c7 and used the same logic in boxes 8 and 9. To work out the 12 pair on the white dot in box 5 I saw that 1 was ruled out of row 4 by cage logic (1 in r5c3) and row logic by the same 1. BTW Simon it IS possible to pencil mark the lack of a digit (if you were to pencil mark 8 digits 😉🤣).

[@SpeedyBozar]

In my eyes the imposter digit is a true hero saving so many cages and saving the puzzle. Imagine how much the hero must have sacrificed to keep the lie alive in order to keep it all together.

[@TheFreeBro]

Today I didn’t get my ~$4 coffee and instead bought the domino app. Great deal for hours of puzzles

[@jaydyer3999]

115:15 for me. I'm glad, and just a little surprised, I managed to solve this one. The logic felt so alien, in the best possible way.

[@teafanatic8452]

Can't go one video with out that Maverick flying by

[@santaisreal]

38:38 why couldnt a 4 go on the whit dots in box 5? the 4 can pose as any number, right?

[@RichSmith77]

It only poses as another value when adding up for cage totals. The white and black dot relationships had to hold for the entered digits.

[@Furuchooo]

4:23 i was 100% sure he was going to say "sus" XD

[@MarushiaDark316]

Thanks to Simon, I've been conditioned like Pavlov's Dogs to associate threes in the corner with losing their religion.

[@NabsterHax]

Surprised myself by solving this one, even if it took me two hours. It took me about an hour of messing about to figure out the imposter was 4 and get the first digit, but it was so satisfying when I did.

[@NabsterHax]

I also go the one in the 47 cage a different way. A 1 has to go in the bottom of box 5 because sudoku and a 41 cage eliminates row 5 and 6. Then in box 6, 1 goes in the top. Either it's in the 40 cage, which forces a 1 in the 47 cage in box 3, OR it's r4c9 which is already in the 47 cage. You can use a similar trick for 1s and 2s using the 9 cell cage and the bottom cage 40 cage to get the 45 cage.

[@10bd1]

78:47 that was a really tough puzzle! Had a lot of fun with solving it though!

[@ruokothyan7658]

i tested the new app. The auto erase of pencilmarks is a nice and handy feature

[@chitraagarwal8259]

Wow! I slept so much better at night coz it took me just 10 mins to get to the epiphany that Simon had at 26:00! Took me much longer to do the math though

[@welcomeblack]

Love the pidgeonhole principle at work here

[@isnimshchikov]

I had a different logic (but based on the same idea) about 4 in central block. If a left 41 cage contains 4 then middle 41 cage doesn't contain 4. But if a left 41 cage doesn't cantain a 4 then middle 41 cage has 4 in one of it's two left cells (in left middle 3x3 block)

[@technoneko8108]

With the logic you had, it being four was a lucky mess up, since it could have been 3, 2, or 1 , you deducted 49 cell couldnt be more then 5 as the digit, but it could have been impostor 3 as an 8 with a normal 2 or a 9 with a normal 1, and 2 or 1 could both have been impostors as a 7

[@technoneko8108]

Oh wait i see now, because you deduced that one of the 8 cages with a 41 or 40 must be imposterless, that alones deduced your options to 4 or 5 and thats where the 49 taking the 5 out makes sense, dang thats embarrassing for me

[@gobsvensen]

Ok, I completely missed the logic of each 8-cell cage missing a different digit. However, the 1 in r5c3 excludes a 1 from r4c4-6 and places one in r4c7-9. if it's in 7or8, it forces a 1 into the 47 cage, or it is in c9 and that is the 1 in the 47 cage, so you can get through the puzzle without that revelation

[@bradfooks4485]

I'm not sure if anyone else picked this up but since he found the 4 as the imposter and he replaces it with 9 in each cage it was obvious that the missing total was the difference between 50 and the cage total

[@RichSmith77]

Isn't this back to front. The only reason the imposter digit had to be 9 was because of the lowest digit that could be missing from the cages.

[@kadlerio]

The way you solved the imposter digit at 10:58 seems more complicated to me. Here's what made more sense to me: The highest cage adds up to 49 and is an 8 cell cage. The largest 8 cell cage could be is one without a 1, thus 45-1=44 is the largest it could be normally. This gives a deficit of 5, since 49-44=5. The largest value you could replace to make up this deficit is 4, since 9-5=4. This means that the imposter must be 4 or less. Since you already concluded that the value must be 4 or 5, the imposter must then be 4.

[@kadlerio]

Or maybe it just made more sense to me because I'd already worked out the difference before I saw you determine it must be 4 or 5.

[@rubenreyes2258]

Ei i can't access the link in the description to try the sudoku, someone who had the same problem??

[@lewsouth1539]

[31:15] “... I can't put 4 on a white dot [in b5]....” The central 41 cage-the one containing r4c3-can't have a 4 in any cell other than r4c3, else it would force a 4 into the left 41 cage as well. So the 4 in b5 must go in r6c6. [39:28-39:55] *OR* one could just ask, where does 1 go in b5? [46:52] “... I wonder what total [sic] the 4 has to have if this is a 4.” You mean, what value it has to have (or what digit it has to impersonate), if it's substituting for a *SIX?* Hmm, lemme think.... [46:57] “I'm just going to check that just in case it feels interesting.” This sounds like a potentially dangerous attitude. [47:12] “What about that black dot? That's not 3,6.... It's either 2,4 or 4,8... which means that that [r7c9] is definitely not a 3.” Never mind that it's not a 3 *by sudoku,* as they say.

[@fulltimeslackerii8229]

i’m very confused by this rule set. the rules say “one of the digits from 1 to 9 is an impostor”. yet by the end of the puzzle, we have more than 1 impostor digit that changes based on the cage it’s in. the rules also say “whenever that digit appears in a cage” yet the entire puzzle is caged. so why specify this? why did he need to specify 1-9 as the impostor? that’s all we can put in this puzzle. my understanding is: “in each cage there is a possibility of an impostor digit which can have any normal sudoku digit value of 1-9 inclusive”? if this is the case, at around 8:11, simon says there are only 9 impostor digits for 10 cages. why can’t there be 10 impostor digits? is it because the center cage has no sum? PLEASE HELP ME

[@sirjmo]

Sudoku rules all digits appear 9 times. Like the impostor digit.

[@steve470]

Of the nine digits 1, 2, 3, 4, 5, 6, 7, 8, and 9, eight of them (the same eight throughout the puzzle) behave as themselves when adding up cage totals. The other digit (the same digit throughout the puzzle) is an impostor. The puzzle contains nine instances of the same imposter digit. (For this puzzle, Simon identifies this digit at the 12:43 mark.) The impostor digit behaves as itself when you're looking at the sudoku, but can act like any digit from 1 to 9 when adding up the total of the cage it appears in. For example, if you figure out that 6 is the impostor digit in the puzzle, and you have a cage containing 12346789, then the 1, 2, 3, 4, 7, 8, and 9 contribute 34 toward the cage sum, and the 6 contributes anywhere from 1 to 9, so a 12346789 cage would have to have a total of somewhere between 35 and 44 inclusive. Because 6 is the impostor in this (hypothetical example) puzzle, a 23456789 cage would then add to somewhere between 39 and 47 inclusive, and a 12345789 cage would add to exactly 39 (because it's missing the imposter digit).

[@fulltimeslackerii8229]

@@steve470 thank you!

[@awilliams1701]

"it will take a couple more weeks for steam" awwwwwwww lol

[@stevieinselby]

At 7:00, you're not quite right on how valuable a clue for a 9-cell cage _could_ be, eg if the cage total was 37 then that would be a slam-dunk giveaway that the impostor digit was 9, or likewise if it was 53 then the imposter could only possibly be 1.

[@muratyagc4082]

At 53 ish minute you could figure it out that purple is 8 before all that sudoku.Because 8 is missing from 40 cage on the right side and it has to be in all the other cages soo purple is 8 :D

[@tangrybird04]

So far I’ve watched up until the “Let’s get cracking” but I’m going to predict the imposter digit is a 4…let’s see if I’m right 👀 Edit: Ok I promise y’all I did not comment this after finding out 😂 I had a 1/9 chance of guessing correctly

[@paulmdt1]

No spoilers in comments please!

[@Bruno-xi3jm]

yo that's sus, I think ur that 4

[@LittleBallOfPurr]

Now I regret reading the comments.

[@5t757]

Anyone reading the comments before watching the video but still intending to watch the video without "spoilers"... that's on you.

[@andremouss2536]

I find it *very* odd that Simon didn't use the *secret* to deduce instantly from the 49 cage that the impostor was a 4. Come back, genuine Simon, we miss you ,-)

[@bennywoo891]

i dont understand how could he know the first 1 must be outside the bottom left box, cant it be 9+8+7+6+5+4+3 = 42 and 1 being the imposter digit equal to 7 which add up to 49? or may be i miss something in the rules :p

[@RichSmith77]

From Simon's earlier reasoning (9:10 - 10:05) he already knows the imposter digit is either 4 or 5. He goes on to show the 49 cage cannot work if the imposter digit is 5, so it must be 4. It can only work with 4 if the missing digit is 1, and 4 counts as 9.

[@jensschmidt]

I would never in my life have gotten to the 42:00 mark, but from then on, I take great joy in the usual abysmal scanning ;-)

[@siyuanhuo7301]

This should be called "Shapeshifter Digit"

[@nickellis999]

Pure poetry.

[@fuxpremier]

This sudoku felt really good but I wouldn't say it was that hard because even if it had quite deep logic, the path to the solution was quite natural. This is exactly the kind of puzzle I enjoy. Kudos to the brilliant setter!

[@danielmarkelon7807]

what are the odds of there being another Dan from Washington DC that watches this channel? lol

[@fisher00769]

Once you prove the 49 cage doesn't have a 1 in it, you know that every other cage HAS to have a 1 in them, because the 9 cage has a 1 in it, and you need to put 8 1s into 9 cages without repeating a 1 in any of the cages. Basically, the logic you did with the imposter digit at the beginning extends to all digits 1-9 regardless of whether they are imposter or not.

[@luizzeroxis]

The impostor digit in this puzzle sure does look sus

[@matejlaskomulej]

3:23 rules

[@mse326]

It only took me about 20 minutes once I realized the unnumbered cage was 9 cells and not 8. Should I have counted? Yup. Should I have realized basic math says 10 8 cell cages doesn't equal 81 cells in a sudoku? Yup. But I am a dummy and was for 40+ minutes

[@brenthoskisson6979]

Simon's logic appears to be based on the "fact" that 4 cannot actually count as 4 in any cage, I think. The rules do not explicitly state that.

[@RichSmith77]

I don't remember seeing Simon assume 4 cannot count as 4 anywhere that it isn't deduced. There are some cages that can only work with the imposter digit being a different value - those 8 cell cages with a cage total > 44. So clearly in these cages, the imposter digit (4) cannot have a value of 4. Do you have an example?

[@babyroseofvongola]

Among us

[@susannahdrazin220]

I managed to solve it but only after going wrong five times!

[@JosueSantiagoG]

Am I the only one who, the first time viewing, listened reeeeeal carefully when he shared "the Secret"? 😂

[@Illeea]

did you voice danger mouse? you're amazing at saying 'good grief'.

[@jjflamedash]

If you wanna skip ahead to the rules 3:25 :)

[@victorfinberg8595]

among us ... not a game i would ever play, but hilarious to watch

[@thecompanioncube4211]

Don't say it... Don't say it... There is a Sus digit

[@sunriselg]

A couple more WEEKS for the app!?!?

[@MrSchesam]

I still dont get why 4 had to be the imposter which counts as 9 instead of 3 which counts as 8. Sounds similar to me

[@dukdog]

By the lower left cage alone that would seem to be an option, but 3 is off the table. Here's why. You have 10 cages, and each digit can only appear up to 9 times in the whole puzzle. So every number will not be present in one of the 10 cages, and thus the imposter will not be present in one of the cages. This will be the only cage we know is telling the truth. After eliminating all the boxes that must contain the imposter and is lying (the corner cages and the 3rd one down in the middle since it contains all 9) we are left with 4 cages, one of which has a correct sum. All 4 cages are comprised of 8 digits and add up to either 41 or 40, which means the one with a correct total and missing the imposter is one of these cages, and the value it would be missing is either 4 or 5. It doesn't matter that 3, 2, and 1 are other possible liars in the bottom left, because they aren't an option when this logic is considered, leaving 4 the only number that meets both of these conditions.

[@Sam_on_YouTube]

I started this puzzle by figuring out the imposter had to he 1, 2, 3, or 4. Then you step in and find it is 4 or 5. Suddenly, I know the digit.

[@ngdjohn2563]

That damn red 4