That was fantastic. My fave vid for a long time. Simon's vacillation between the best of times and the worst of times was wonderful - and hilarious at times. He was clearly deep in the zone and absolutely loving it. It epitomises to me that wonderful dance a designer demands of a player. That's the best hour I've spent recently. Such a pleasure to be transported into a hard core battle between humans where the only element is the joy shared in discovery. Cracking!
i have been wathcing CtC off and on for a couple of years. and i have to say that this is the first one for which i didn't have a clue as to what was going on.
He starts with a fictional axiom of checkerboard logic and then keeps it consistent and it magically works!!!! I saw a different proof. That is, once box 1 is checkerboarded than you get green orange green in the 4th row. Then you have two orange and a green left and then once r4c2 is green, than the other two are orange in that box. Then it checkerboards down to the bottom box. Then his logic is consistent for the rightmost 3 boxes. But why does the checkerboard logic work in the first place? I have no clue. Undisclosed rules, I suppose. Anyway I got caught up in the beginning, so I hope this helped!
@@jdog2531 It's not undisclosed -- they're being defined by colouring them. You say, whatever three digits are in these boxes, let's colour them orange, and the others green. Then you notice how that colouring spreads.
@@lloydwalker6965 Thanks a bunch. I was just raised to be really careful with manipulating digits so that values aren’t change, Through some jiggery pokery as Simon would say! So as he manipulated the values later on in the solve, it got a little hard to follow. However he normally does a very good job and I enjoy his videos and I really liked his eureka moment when he found the breakin to the right 3 boxes, using the knights move restriction to limit the 2 into a single location in box 2.
I guess my problem with it was why can he get away with assigning elements to a probability distribution ( I used and defined 6 colors on my solve) because these elements shift in the grid but i guess it doesn’t matter if you keep the set right and treat the set as 1 element.
It is so interesting to read the comments for this one. (I usually don't read comments until after I have left mine, but I was so mystified by this one that I was very curious!) I must say that I felt the way I used to feel way back at the beginning of my watching this channel, when most of the puzzles and constraints were somewhat like foreign language and I just had to hope that I would be able to store away some facts and strategies to use in the future. I'm glad it got edited and posted, Simon, because despite that I was at sea the whole time, you were clearly enjoying yourself, and that is always fun to see.
15:20 I think you can continue the proof without using numbers by looking at the cell Simon has highlighted (r4c5). It sees all three green cells in column 4, so it has to be orange. Similarily, r3c5 has to be green. The rest of the pattern should then just unfold.
The impatience of Simon coming into play here. He tried two options in c5 unsuccessfully and did not want to try a third one, which would have been an easy and obvious 3-seconds-solve.
I had a dream last night where I met Simon and we had a very pleasant conversation. He had inexplicably replaced one of my coworkers. I think Mark was there too. Is this a sign that I'm watching too much Cracking the Cryptic? Surely not.
Another straightforward way to prove the checkerboard pattern when you got stuck after column 4 was to look at R3C5 and R4C5. Both those cells see all the opposite colored cells in row 4. That gives you all of box 4 and away you go.
Once you've got the first four columns of the checkerboard, there is a less painful way to claim victory on proving your checkerboard lemma. The first four columns can be checkerboard-colored according to box 1. By symmetry, the last four columns can be checkerboard-colored according to box 6. The middle two columns are colored both ways, so the colorings are the same.
21:51 Simon looking at the knights move constraint in column 2, but disregarding column 1, which would've proved the 3 in the corner together with that 1st 5. 47:22 "what you're meant to do is revert back to them a bit later to see only one of them can be a 3"... sure simon ;) I love how Simons brain is able to do such incredible and elegant solves of puzzles I'd get stuck on forever, and yet be so blind in other cases that are obvious to me xD
I can't decide if it's just Index rules in general, or if it's just the unfortunate mish-mash of markings, but this puzzle *broke* Simon's brain. "If this orange cell is orange, then it can't be orange"
Even after he finally understood that logic, he stopped thinking about it immediately, and did not consider what other parts of the puzzle the same logic could be applied to. (It applied to row 5 in the reverse direction and would disambiguate the 3s in box 6)
Simon usually worries about choosing colours and still sometimes ends up picking wrong ones. He has spent ages training us to accept orange for odd because both words begin with the letter "o", but today uses orange for even digits. I will still watch, obviously!
@ He is not using orange for odd digits in this solve. He is using the colours only to investigate the positioning of certain digits, very early in the solve, before anything has been found on even and odd digits. Further on in the solve the two colours doesn’t mean odd/even whatsoever. So my comment ‘he is not’ referred to your statement ‘… uses orange for even digits’.
One of the weirdest complaints I've seen under these videos. He has "trained" you to "accept" orange for odds? But you will still watch even though he used "wrong" colours? Look, Simon uses orange-blue scheme for odd-even puzzles for the sake of consistency, in other words because he wants to. This is not an odd-even puzzle so he doesn't. There is no rule or law or sacred oath to never put an even digit in green or whatever.
Another way to prove the checker boarding for the right side is to use a second set of two colours to checkerboard the right side, using the same original logic, then use the fourth column to map the left colours onto the right colours.
I don't think that's a valid proof. It's only proving that the alignment of the two checkerboard patterns is fixed IF there are two checkerboard sets. This is based on the *assumption* that the second set IS ALSO a checkerboard (following the same distribution of digits). So to complete this proof you also need to disprove any other alignment of coloring or digits in the second set.
@@BrinaaSM no, I think the "assumption that the second set IS ALSO a checkerboard" is not needed, as the checkerboard coloring is arbitrary just like the left side. We can use the same logic with different colors (say red and blue) and use column 4 to conclude that red = orange and blue = green.
@@williamnathanael412 It's not actually arbitrary coloring because it's a fact that this state of coloring exists in a single box in any 6x6 puzzle. We just don't know the distribution of its digits in this puzzle. Let's call this distribution ABC/DEF. Simon then follows logic that proves that the adjacent box is also a checkerboard following the same distribution ABC/DEF. Proof complete: Any arbitrary yet definitely existing checkerboard pattern ABC/DEF in a 6x6 Sudoku box leads to the entire grid being the same checkerboard pattern distribution ABC/DEF. No assumptions here. We can state this as: A = There exists a checkerboard pattern in a box. B = the entire grid is this checkerboard with the same distribution of digits. If A is true, then B is true. A is always true. Meanwhile OP suggested you start with another INDEPENDENT set which is also a checkerboard. This leads to this problem: We know there definitely exists a checkerboard pattern ABC/DEF. What we DON'T KNOW is that second INDEPENDENT set follows the SAME distribution. This is an assumption! (In reverse we can also say that the distribution is always the same but then the checkerboard is the assumption.) We can state this as: A = There exists a checkerboard pattern in a box. B = There exists another checkerboard pattern in a different box. C = The second checkerboard follows the same distribution of digits as the first. (Assumption!) D = the entire grid is this checkerboard with the same distribution of digits. If A is true AND B is true AND C is true, then D is true. A is always true. B is always true. C is NOT always true, from our knowledge, so we cannot make a general statement about D, until we've proven that - If A is true AND B is true AND C is NOT true, then D is still true. OR - C is always true. Both options need proof first. Example where this does not work. Digit distribution: 123 465 456 213 With the coloring: ABA XYX BAB YXY by coloring we see that A = Y and B = X However the distribution of digits in these sets are: A = 153 B = 426 X = 415 Y = 263 Both boxes are in a checkerboard pattern, that's all OP asked for, but the digit distribution is different between those.
I finished in 241:43 minutes. This was a brutal one. The only thing I was able to get was the limitation of 3's being next to a circle that wasn't next to a corner. I tried for a long time to play around with those to see any limitation at all. I got nothing. After 3 hours, I finally gave up to watch the video. I was unaware of the checkerboard theory for this ruleset. Utilizing that made my 3's appear much more restrictive, down to a set of two possibilities. The limitation of the 5's on the cells next to circles wasn't something I was expecting and disambiguated my 3's instantly. The rest was just as difficult, mainly due to the brain fog that developed through the puzzle. The final nail in the coffin was when r1c3 couldn't be a 1. That disambiguated everything. It sucks that I had to rely on the video for this solve. I was so close with my understanding of the 3's and their limitations. All I was missing was the checkerboard theory and using the same principle for 5's to finish. I wonder if I knew about that theory, would I have been able to do it on my own. This must be how beginners feel when struggling with a puzzle only to find out about the Phistomefel Ring. It was a very cool demonstration of a puzzle with some sick ideas. I wish I could have giving it a better solve. Oh well. Great Puzzle!
1:47:48 - Crumbs! I’d remembered the checkerboard from a Mark video a while back and started by marking box 3 with letters. After about 50 minutes I had an epiphany and realised that marking columns two and five instead gave me 50% more knowledge. I still didn’t find it that easy because my scanning tends to be all over the place but I got there in the end. Excellent puzzle.
Very nice puzzle. I didn't do the checkerboard, but got going by colouring box 6 with six different colours and trying to propogate them based on the knight's move. This got me 1246 in r6c5. I then deleted the colours and did the same thing in box 2 to confirm that the same logic worked for the 1246 in r1c5. So then I kept deleting the colours and starting again to target different cells. I was never really stuck but kept making small deductions until I got a 5, and then gradually the other digits fell out. Maybe fixing the checkerboard colouring, for all its elegance, might not actually be the easiest solve path?
Nice puzzle, but without the help from Simon with the checkerboard pattern (or rather, parts of its proof which did give even a bit more information, up to 19:32) I wouldn't have gotten anywhere. Total time 119:20, solve counter 1118.
Solved in 47:04. Took me half an hour to remember that checkerboard restriction on anti-knight 6x6 sudoku (that it exists but not how to prove it), but by then I'd already proved that 3s next to a numbers room clue are very restricted so the rest was pretty quick. I think I learnt the checkerboard thing from this channel and would have been stuck forever without it.
right after you get the checkerboard, it feels to me like the next natural step is the special relationship between columns 2 and 5. r1c2 and r2c5 see the same two orange cells in box 1 and are therefore the same. this kind of logic lets you get to work on a more informative 6-color.
20:26 for me. I vaguely remembered that trick with the checkerboard pattern being a thing, don't think I would have ever solved it if not for that. Very tough puzzle.
This one was brutal. Was able to get all the fives but got stuck at the point of determining threes. I did finish it at that point but without proof of a couple steps including the threes.
I started by coloring for each digit. I got to the part where all my 5s were done and I had 3 or 4 digits penciled in for each but couldn't go further. This one was too hard for me.
I still don't understand how come I was able to solve that. 1/2h to paint the grid and miraculously found a digit. And another 1/2h to finish it. Incredible puzzle
16:27 Did I miss the part where he explains why we're checkerboarding the grid and why/how we're putting 123456 into box 1 like that? Maybe he gets into it later in the video, but man, at this point, I'm frustratingly lost.
@@katiekawaii The checkerboard idea came from a memory of something he was told about 6x6 sudoku logic that he thought could be a useful place to start. As for the numbers that went into box 1, they are not final, just placeholders to test the checkerboard logic so he can use it to apply later when he does start finding the actual numbers.
Oh, already watched half of the video, but poor horsey in the circle haven't got half of its moves still... I mean, circle in R0C3 sees not only R2C2 and R2C4, but also R1C1 and R1C5 that makes it either orange or 1. And it was going for circles since the beginning of the video - even in the example at 7:33.
The only thing I managed to deduce before deciding to throw in the towel on this one is that 3 is actually very tricky to place in the numbered room cell next to a circle. If you place a 3 next to (not in) any circle other than the corners, whatever digit is in the circle will be eliminated from one of the boxes. For example, a 3 in r3c1 will place its circled digit in r3c3, and whatever digit that is becomes fully eliminated from box 1 by a combination of normal sudoku (eliminates r2c3 and r1c3), knight's move on the digit in the grid (eliminates r2c1 and r1c2), and knight's move on the circle (eliminates r1c1 and r2c2). Note for clarity that it is not the 3 getting eliminated from that box, but rather whatever digit was circled. You have to check that in a few different positions to ensure it works the same everywhere, but if you test you'll see it applies to the columns too (a 3 in r1c5 will eliminate its circled digit from box 2, for example). The only place a 3 can go next to a circle turns out to be in a corner. That's the only progress I ever made, alas, but it does feel like it could have been very helpful had I deduced basically anything else lol.
Maybe because I'm watching on mobile but by flipping the puzzle 90 degrees. It's again the same pattern and 5 isn't possible for the top clue index anymore either
the easy way to solve that checkering without numbers is to ask how to solve the middle box for either color. both see all 3 of their color when placed adjacent. From there the tops and bottoms can be solved and the last row is pretty easy if you start from the center as well. It's just not intuitive to start in the middle for most people. I'm awful at these puzzles but I'm at least okay at these kinda proofs.
I think he was only referring to that particular numbered rooms clue being considered against the established checker boarding of the column. ie. The green cells are an even number of steps away from the edge of the grid.
Strange solve for me. I started after Simon revealed the chevkerboarding, and from there proceded to a completely different path, starting to feel the grid from the bottom left.
I've changed my pallettes so I have dark/medium/light versions of 9 different colours, so I can use light green/dark green for pairs I'm colouring, rather than flashes, with medium reserved for "don't know which yet", which is better overload wise for me than flashes or letters.
As a novice in sudoku - what does the checkerboard have to do with evens and odds? In the beginning simon used that to propagate color into the circles but neither green nor orange is solely odd or even?
Seems to be to give the colors a fixed property (assign a number and a color to a a space) that can propagate through the grid. There's no inherent odd or even, it's just to use logic on actual numbers.
Simon was using numbers as placeholders to prove the checkerboard pattern, where they were actually arbitrary digits (it's more clear to use letters for that). So there's no relation to even/odd.
There is a relation to parity in this puzzle because it involves indexing, and a given color is always going to appear either in the 1st, 3rd, and 5th, or the 2nd, 4th, and 6th positions in any given row or column.
In a sudoku variant with only set theoretic rules (numbers cant repeat in certain given sets - as opposed to arithmetic based rules like german whispers and killer cages) the numbers themselves are completely arbitrary. You can substitute any number for any other and it would not make any difference to the puzzle. In a sense, numbers in sudoku are always just placeholders for "these things are the same" and "these things are different".
The checkerboard made it easier, I suppose, to check for the anti-knight constraint and to propagate a digit found to the entire grid. The odd/even numbers refer to the position of the digit in an odd or even column/row. That's the best I can formulate it.
To be fair, the fact that column clues were at least as restricted as row clues seemed obvious to me after a moment's thought, but I had no idea about that initial checkerboard pattern 😅 (or should I say chessboard pattern?)
You don't need to prove the entire checkerboarding from first principles. If you start doing any sort of coloring and basic knight's move eliminations, it just falls out.
I managed to solve it in 39 minutes without checkerboarding the grid. I ended up assigning unique colours to the six cells in column 2 and propagating these around the grid by knight's move, so that I had column 5 uniquely coloured, and every other cell had dual options. (I suppose the checkerboarding pattern was hidden in there, but I didn't spot it). Finding the 5, then the 3 in column 1 was the break through. Those propagated around the grid and allowed me to fully resolve my colouring.
Difficult and nice. Marking cells in box 3 with small middle letters ABCDEF, with checkboard in mind, put all cells in strict restrictions. latter, using pencilmarks for possible numbers give me a chance to follow logic.
No you didnt prove the checker boarding. The top cell is proved because the 2s and 5s have 2 rows to eliminate the centre, but thebmiddlenrow does not. Im sure a little more steps would prove it.
This is so frustrating to watch. You broke into this puzzle so effortlessly compared to my attempt, using the knowledge of the checkerboard. And then you just continued to ignore half of the Knight's moves for what seemed like the next hour and a half.
Interesting puzzle. However, Simon once again managed to make life hard for himself by not following up on the simple implications of his deductions using the rules. Once he proved the checkerboard, the horizontal 5 and the double-1 whenever the first cell to a clue is a 1, though the puzzle would still not be _solved,_ it would have been relatively easy to propagate knight‘s move implications into the perimeter clues. And, strangely, sometimes he _did_ follow up on this, but only in places it didn‘t really help. Instead, in true Simon fashion, he abandoned the thought and moved to the opposite side of the grid to find some wild logical bounces to progress. The knight‘s move rule affecting the Numbered Rooms clues outside the grid is an extraordinarily powerful rule _if_ you are willing to actually apply it. *Don‘t get me wrong,* I love CtC and Simon is still 100 times the solver I can dream to become, and I do see that solving live, explaining your thoughts coherently in real time and still being entertaining is exceptionally hard - and also, Simon might be a bit biased against logic that appears obvious due to the amount of deviously hard puzzles he solved in the past, so maybe that‘s why he is quite prone to ignore that. But still, it amuses me that while I am usually quite pleased with my solve when I manage to stay within 2.5x Simon‘s time, I tend to match (and at one point even _beat)_ his time more closely the „easier“ the puzzle is, simply because of my utter ineptitude to spot viciously hidden logic and Simon‘s peculiar habit of doing the polar opposite. One such instance is the question whether the Numbered Rooms clue in r4 could be orange - if it _was_ it would automatically be the same digit as (and I use the usual nomenclature only for the „inner“ 6x6) r4c1 because Knights Moves eliminate both r2c1 and 6c1, therefore, both would be one. At this point, he already deduced (though he failed to properly mark it) that r4c1 must be B and, because of c2 (and the relationship between the NR clue in c2 with B in r5c2) could not be 1, thus immediately removing the option of the NR clue in r4 to be orange. I know I use a lot of Simon‘s deductions to make the point (and some of those were acquired in equally convoluted ways instead of using the knight‘s move rule properly). But it was one of the most obvious ones that made me yell my best McGonnagall impression at my screen.
Simon, I'm sorry, but you're not doing something that's sooooo obvious for someone who's watching, not only is it 2 left/right and 1 up/down, but it's also 1 left/right and 2 up/down for knights move. "Sigh"
That was fantastic. My fave vid for a long time. Simon's vacillation between the best of times and the worst of times was wonderful - and hilarious at times. He was clearly deep in the zone and absolutely loving it. It epitomises to me that wonderful dance a designer demands of a player. That's the best hour I've spent recently. Such a pleasure to be transported into a hard core battle between humans where the only element is the joy shared in discovery. Cracking!
i have been wathcing CtC off and on for a couple of years. and i have to say that this is the first one for which i didn't have a clue as to what was going on.
With you! Marvellous, wasn't it?!
He starts with a fictional axiom of checkerboard logic and then keeps it consistent and it magically works!!!! I saw a different proof. That is, once box 1 is checkerboarded than you get green orange green in the 4th row. Then you have two orange and a green left and then once r4c2 is green, than the other two are orange in that box. Then it checkerboards down to the bottom box. Then his logic is consistent for the rightmost 3 boxes. But why does the checkerboard logic work in the first place? I have no clue. Undisclosed rules, I suppose. Anyway I got caught up in the beginning, so I hope this helped!
@@jdog2531 It's not undisclosed -- they're being defined by colouring them. You say, whatever three digits are in these boxes, let's colour them orange, and the others green. Then you notice how that colouring spreads.
@@lloydwalker6965 Thanks a bunch. I was just raised to be really careful with manipulating digits so that values aren’t change, Through some jiggery pokery as Simon would say! So as he manipulated the values later on in the solve, it got a little hard to follow. However he normally does a very good job and I enjoy his videos and I really liked his eureka moment when he found the breakin to the right 3 boxes, using the knights move restriction to limit the 2 into a single location in box 2.
I guess my problem with it was why can he get away with assigning elements to a probability distribution ( I used and defined 6 colors on my solve) because these elements shift in the grid but i guess it doesn’t matter if you keep the set right and treat the set as 1 element.
It is so interesting to read the comments for this one. (I usually don't read comments until after I have left mine, but I was so mystified by this one that I was very curious!) I must say that I felt the way I used to feel way back at the beginning of my watching this channel, when most of the puzzles and constraints were somewhat like foreign language and I just had to hope that I would be able to store away some facts and strategies to use in the future. I'm glad it got edited and posted, Simon, because despite that I was at sea the whole time, you were clearly enjoying yourself, and that is always fun to see.
15:20 I think you can continue the proof without using numbers by looking at the cell Simon has highlighted (r4c5). It sees all three green cells in column 4, so it has to be orange. Similarily, r3c5 has to be green. The rest of the pattern should then just unfold.
The impatience of Simon coming into play here. He tried two options in c5 unsuccessfully and did not want to try a third one, which would have been an easy and obvious 3-seconds-solve.
Rules: 05:18
Let's Get Cracking: 08:05
Simon's time: 55m14s
Puzzle Solved: 1:03:19
What about this video's Top Tier Simarkisms?!
Three In the Corner: 2x (46:08, 47:40)
Bobbins: 1x (51:12)
And how about this video's Simarkisms?!
Sorry: 21x (03:17, 05:41, 12:48, 15:21, 15:21, 16:52, 19:33, 20:10, 23:00, 23:00, 23:29, 24:11, 30:44, 37:43, 38:16, 39:37, 43:13, 44:21, 53:07, 1:01:12, 1:01:12)
Hang On: 13x (19:33, 20:05, 20:05, 20:07, 20:52, 21:08, 21:08, 21:08, 22:32, 24:44, 29:30, 31:48)
Obviously: 9x (06:11, 10:04, 10:33, 11:06, 12:45, 13:36, 27:11, 27:14, 45:51)
Ah: 7x (13:31, 34:42, 41:25, 41:25, 41:25, 43:26, 45:12)
In Fact: 5x (02:54, 11:22, 11:22, 15:02, 47:14)
Clever: 4x (44:55, 1:02:01, 1:02:06, 1:02:58)
By Sudoku: 4x (11:46, 24:28, 38:43, 1:01:57)
Weird: 4x (20:52, 22:37, 47:11, 47:16)
Useless: 3x (11:39, 30:48, 39:40)
Goodness: 3x (18:49, 36:57, 44:29)
The Answer is: 3x (17:09, 17:25, 40:01)
Brilliant: 3x (00:59, 04:37, 04:40)
Symmetry: 3x (13:07, 30:22, 30:31)
Good Grief: 2x (36:17, 1:01:57)
Lovely: 2x (49:15, 49:19)
Beautiful: 2x (53:35, 1:00:38)
Ridiculous: 2x (1:02:06, 1:03:19)
Going Mad: 2x (32:30)
Proof: 2x (11:42, 14:19)
Wow: 2x (17:02, 57:21)
Lunacy: 2x (1:03:03, 1:03:03)
Pencil Mark/mark: 2x (34:02, 43:15)
Cake!: 2x (03:39, 04:42)
What on Earth: 1x (39:53)
In the Spotlight: 1x (47:44)
I Have no Clue: 1x (10:47)
Horrible Feeling: 1x (34:00)
Incredible: 1x (01:00)
Whoopsie: 1x (50:52)
Jaw-dropping: 1x (1:03:16)
What Does This Mean?: 1x (19:24)
That's Huge: 1x (1:00:30)
Nature: 1x (27:20)
Most popular number(>9), digit and colour this video:
Forty Six (3 mentions)
One (94 mentions)
Orange (87 mentions)
Antithesis Battles:
Even (15) - Odd (3)
Outside (5) - Inside (0)
Row (11) - Column (8)
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!
I had a dream last night where I met Simon and we had a very pleasant conversation. He had inexplicably replaced one of my coworkers. I think Mark was there too. Is this a sign that I'm watching too much Cracking the Cryptic? Surely not.
I don't know why, but I found Simons struggle to get a grasp on anything in the first half hour very entertaining haha
A 6x6 sudoku with more than an hour of video time - this'll be fun!
...as you predicted!
I know why it didn't trigger. Place a 3 on an outside corner of the grid, outside the puzzle. That's 3 in the corner.
Quickly tested this: You are right.
astonishing, mesmerising and most enjoyable. Thank you Simon.
Another straightforward way to prove the checkerboard pattern when you got stuck after column 4 was to look at R3C5 and R4C5. Both those cells see all the opposite colored cells in row 4. That gives you all of box 4 and away you go.
All cells in *column 4. Great observation! This helps
Once you've got the first four columns of the checkerboard, there is a less painful way to claim victory on proving your checkerboard lemma.
The first four columns can be checkerboard-colored according to box 1. By symmetry, the last four columns can be checkerboard-colored according to box 6. The middle two columns are colored both ways, so the colorings are the same.
Genius! If there's a gallery of every perfect puzzle this belongs there in my opinion.
(Cheers from 99%Sneaky)
Thank you for posting this, even though you struggled. It gives us numpties, who love your channel, a glimmer of hope. But only a glimmer….🙄❤
47:40 "That's not fair" - The corner of the puzzle, as seen by the program, is one cell left south-west of where you put the 3.
21:51 Simon looking at the knights move constraint in column 2, but disregarding column 1, which would've proved the 3 in the corner together with that 1st 5.
47:22 "what you're meant to do is revert back to them a bit later to see only one of them can be a 3"... sure simon ;)
I love how Simons brain is able to do such incredible and elegant solves of puzzles I'd get stuck on forever, and yet be so blind in other cases that are obvious to me xD
I can't decide if it's just Index rules in general, or if it's just the unfortunate mish-mash of markings, but this puzzle *broke* Simon's brain.
"If this orange cell is orange, then it can't be orange"
Even after he finally understood that logic, he stopped thinking about it immediately, and did not consider what other parts of the puzzle the same logic could be applied to. (It applied to row 5 in the reverse direction and would disambiguate the 3s in box 6)
Simon said do-do as part of the rules and made me laugh outloud… I’m a child
Simon usually worries about choosing colours and still sometimes ends up picking wrong ones. He has spent ages training us to accept orange for odd because both words begin with the letter "o", but today uses orange for even digits. I will still watch, obviously!
Well, at 38.30, he enters an odd digit in an orange cell. I have completely lost the plot (not an unknown occurrence).
He is not.
@@stangerrits6712I admire people who can be as succinct as you, a skill I struggle with, but on this occasion, I don't understand. He is not what?
@ He is not using orange for odd digits in this solve. He is using the colours only to investigate the positioning of certain digits, very early in the solve, before anything has been found on even and odd digits. Further on in the solve the two colours doesn’t mean odd/even whatsoever. So my comment ‘he is not’ referred to your statement ‘… uses orange for even digits’.
One of the weirdest complaints I've seen under these videos. He has "trained" you to "accept" orange for odds? But you will still watch even though he used "wrong" colours?
Look, Simon uses orange-blue scheme for odd-even puzzles for the sake of consistency, in other words because he wants to. This is not an odd-even puzzle so he doesn't. There is no rule or law or sacred oath to never put an even digit in green or whatever.
34:25 Simon, the other one sees 2 in its box as well 😂
Another way to prove the checker boarding for the right side is to use a second set of two colours to checkerboard the right side, using the same original logic, then use the fourth column to map the left colours onto the right colours.
I don't think that's a valid proof. It's only proving that the alignment of the two checkerboard patterns is fixed IF there are two checkerboard sets. This is based on the *assumption* that the second set IS ALSO a checkerboard (following the same distribution of digits). So to complete this proof you also need to disprove any other alignment of coloring or digits in the second set.
The fourth column contains all digits 1-6, so it does determine the sets on both sides. @@BrinaaSM
@@manoftheforest7505
You're missing the point of my comment.
@@BrinaaSM no, I think the "assumption that the second set IS ALSO a checkerboard" is not needed, as the checkerboard coloring is arbitrary just like the left side. We can use the same logic with different colors (say red and blue) and use column 4 to conclude that red = orange and blue = green.
@@williamnathanael412
It's not actually arbitrary coloring because it's a fact that this state of coloring exists in a single box in any 6x6 puzzle. We just don't know the distribution of its digits in this puzzle. Let's call this distribution ABC/DEF.
Simon then follows logic that proves that the adjacent box is also a checkerboard following the same distribution ABC/DEF.
Proof complete: Any arbitrary yet definitely existing checkerboard pattern ABC/DEF in a 6x6 Sudoku box leads to the entire grid being the same checkerboard pattern distribution ABC/DEF.
No assumptions here.
We can state this as:
A = There exists a checkerboard pattern in a box.
B = the entire grid is this checkerboard with the same distribution of digits.
If A is true, then B is true.
A is always true.
Meanwhile OP suggested you start with another INDEPENDENT set which is also a checkerboard. This leads to this problem:
We know there definitely exists a checkerboard pattern ABC/DEF. What we DON'T KNOW is that second INDEPENDENT set follows the SAME distribution. This is an assumption! (In reverse we can also say that the distribution is always the same but then the checkerboard is the assumption.)
We can state this as:
A = There exists a checkerboard pattern in a box.
B = There exists another checkerboard pattern in a different box.
C = The second checkerboard follows the same distribution of digits as the first. (Assumption!)
D = the entire grid is this checkerboard with the same distribution of digits.
If A is true AND B is true AND C is true, then D is true.
A is always true.
B is always true.
C is NOT always true, from our knowledge, so we cannot make a general statement about D, until we've proven that
- If A is true AND B is true AND C is NOT true, then D is still true.
OR
- C is always true.
Both options need proof first.
Example where this does not work.
Digit distribution:
123 465
456 213
With the coloring:
ABA XYX
BAB YXY
by coloring we see that A = Y and B = X
However the distribution of digits in these sets are:
A = 153
B = 426
X = 415
Y = 263
Both boxes are in a checkerboard pattern, that's all OP asked for, but the digit distribution is different between those.
I finished in 241:43 minutes. This was a brutal one. The only thing I was able to get was the limitation of 3's being next to a circle that wasn't next to a corner. I tried for a long time to play around with those to see any limitation at all. I got nothing. After 3 hours, I finally gave up to watch the video. I was unaware of the checkerboard theory for this ruleset. Utilizing that made my 3's appear much more restrictive, down to a set of two possibilities. The limitation of the 5's on the cells next to circles wasn't something I was expecting and disambiguated my 3's instantly. The rest was just as difficult, mainly due to the brain fog that developed through the puzzle. The final nail in the coffin was when r1c3 couldn't be a 1. That disambiguated everything. It sucks that I had to rely on the video for this solve. I was so close with my understanding of the 3's and their limitations. All I was missing was the checkerboard theory and using the same principle for 5's to finish. I wonder if I knew about that theory, would I have been able to do it on my own. This must be how beginners feel when struggling with a puzzle only to find out about the Phistomefel Ring. It was a very cool demonstration of a puzzle with some sick ideas. I wish I could have giving it a better solve. Oh well. Great Puzzle!
1:47:48 - Crumbs! I’d remembered the checkerboard from a Mark video a while back and started by marking box 3 with letters. After about 50 minutes I had an epiphany and realised that marking columns two and five instead gave me 50% more knowledge. I still didn’t find it that easy because my scanning tends to be all over the place but I got there in the end. Excellent puzzle.
Very nice puzzle. I didn't do the checkerboard, but got going by colouring box 6 with six different colours and trying to propogate them based on the knight's move. This got me 1246 in r6c5. I then deleted the colours and did the same thing in box 2 to confirm that the same logic worked for the 1246 in r1c5. So then I kept deleting the colours and starting again to target different cells. I was never really stuck but kept making small deductions until I got a 5, and then gradually the other digits fell out. Maybe fixing the checkerboard colouring, for all its elegance, might not actually be the easiest solve path?
Nice puzzle, but without the help from Simon with the checkerboard pattern (or rather, parts of its proof which did give even a bit more information, up to 19:32) I wouldn't have gotten anywhere.
Total time 119:20, solve counter 1118.
I've grown scared of 6x6 sudokus over time. You would think they would be easier but most of them are evil.
Myxo is such a genius!!! Brilliant and loved seeing Simon tackle this!!
Using the knight's move rule between rings did not occur to me for a very, very long time. :-) Nice puzzle!
Solved in 47:04. Took me half an hour to remember that checkerboard restriction on anti-knight 6x6 sudoku (that it exists but not how to prove it), but by then I'd already proved that 3s next to a numbers room clue are very restricted so the rest was pretty quick.
I think I learnt the checkerboard thing from this channel and would have been stuck forever without it.
Always fun to see a minimal sudoku board
46:59 for me - I loved this puzzle, interesting logic, and then I made the final realization that circles also see other circles from knight's moves.
A myxo puzzle, brilliant.
right after you get the checkerboard, it feels to me like the next natural step is the special relationship between columns 2 and 5. r1c2 and r2c5 see the same two orange cells in box 1 and are therefore the same. this kind of logic lets you get to work on a more informative 6-color.
Really nice puzzle - caught it in the end. Those knights moves are used so precisely.
20:26 for me. I vaguely remembered that trick with the checkerboard pattern being a thing, don't think I would have ever solved it if not for that. Very tough puzzle.
47:35 it's not in the corner, no song allowed! (with the circles in play the grid is not the full extent of the puzzle)
14:47 another way to carry on from here is to rotate the colouring 180deg, and the middle 2 columns overlap
I Love 6x6s! And I love setting them
This one was brutal. Was able to get all the fives but got stuck at the point of determining threes. I did finish it at that point but without proof of a couple steps including the threes.
I started by coloring for each digit. I got to the part where all my 5s were done and I had 3 or 4 digits penciled in for each but couldn't go further. This one was too hard for me.
13:17. All you need to do is propagate cells 3 and 4 in box 4, against column 4, then cells 5 and 6 - finish the rows.
19:14 the Chess* Board emerges.
I still don't understand how come I was able to solve that. 1/2h to paint the grid and miraculously found a digit. And another 1/2h to finish it.
Incredible puzzle
16:27 Did I miss the part where he explains why we're checkerboarding the grid and why/how we're putting 123456 into box 1 like that? Maybe he gets into it later in the video, but man, at this point, I'm frustratingly lost.
@@katiekawaii The checkerboard idea came from a memory of something he was told about 6x6 sudoku logic that he thought could be a useful place to start.
As for the numbers that went into box 1, they are not final, just placeholders to test the checkerboard logic so he can use it to apply later when he does start finding the actual numbers.
@@hellsfoxes Thank you. :) I love how kind and helpful this community is, especially when it comes to responding to questions.
Oh, already watched half of the video, but poor horsey in the circle haven't got half of its moves still...
I mean, circle in R0C3 sees not only R2C2 and R2C4, but also R1C1 and R1C5 that makes it either orange or 1.
And it was going for circles since the beginning of the video - even in the example at 7:33.
There needs to be a Sorry counter whilst he's entertaining us so well
21 sorry's according to the brilliant bot created by @inspiringsands123. I love that bot.
Can you click the cog icon and put the confetti on ?
The only thing I managed to deduce before deciding to throw in the towel on this one is that 3 is actually very tricky to place in the numbered room cell next to a circle. If you place a 3 next to (not in) any circle other than the corners, whatever digit is in the circle will be eliminated from one of the boxes. For example, a 3 in r3c1 will place its circled digit in r3c3, and whatever digit that is becomes fully eliminated from box 1 by a combination of normal sudoku (eliminates r2c3 and r1c3), knight's move on the digit in the grid (eliminates r2c1 and r1c2), and knight's move on the circle (eliminates r1c1 and r2c2). Note for clarity that it is not the 3 getting eliminated from that box, but rather whatever digit was circled.
You have to check that in a few different positions to ensure it works the same everywhere, but if you test you'll see it applies to the columns too (a 3 in r1c5 will eliminate its circled digit from box 2, for example). The only place a 3 can go next to a circle turns out to be in a corner.
That's the only progress I ever made, alas, but it does feel like it could have been very helpful had I deduced basically anything else lol.
Maybe because I'm watching on mobile but by flipping the puzzle 90 degrees. It's again the same pattern and 5 isn't possible for the top clue index anymore either
the easy way to solve that checkering without numbers is to ask how to solve the middle box for either color. both see all 3 of their color when placed adjacent. From there the tops and bottoms can be solved and the last row is pretty easy if you start from the center as well. It's just not intuitive to start in the middle for most people. I'm awful at these puzzles but I'm at least okay at these kinda proofs.
47:39 That's 3 in R7 C2, That's 3 in the 8x8 grid, losing it's confetti...
I'm confused at 35:50 where he proved that green was odd and orange was even.
I think he was only referring to that particular numbered rooms clue being considered against the established checker boarding of the column. ie. The green cells are an even number of steps away from the edge of the grid.
Strange solve for me. I started after Simon revealed the chevkerboarding, and from there proceded to a completely different path, starting to feel the grid from the bottom left.
Let's get cracking
I just watched Simon do nothing for half an hour lol…. I was wondering why it was an hour long😂
In terms of information overload, I much prefer pencil marked letters over "flashed" colors, even if it does produce pretty patterns at the end.
I've changed my pallettes so I have dark/medium/light versions of 9 different colours, so I can use light green/dark green for pairs I'm colouring, rather than flashes, with medium reserved for "don't know which yet", which is better overload wise for me than flashes or letters.
As a novice in sudoku - what does the checkerboard have to do with evens and odds? In the beginning simon used that to propagate color into the circles but neither green nor orange is solely odd or even?
Seems to be to give the colors a fixed property (assign a number and a color to a a space) that can propagate through the grid. There's no inherent odd or even, it's just to use logic on actual numbers.
Simon was using numbers as placeholders to prove the checkerboard pattern, where they were actually arbitrary digits (it's more clear to use letters for that). So there's no relation to even/odd.
There is a relation to parity in this puzzle because it involves indexing, and a given color is always going to appear either in the 1st, 3rd, and 5th, or the 2nd, 4th, and 6th positions in any given row or column.
In a sudoku variant with only set theoretic rules (numbers cant repeat in certain given sets - as opposed to arithmetic based rules like german whispers and killer cages) the numbers themselves are completely arbitrary. You can substitute any number for any other and it would not make any difference to the puzzle. In a sense, numbers in sudoku are always just placeholders for "these things are the same" and "these things are different".
The checkerboard made it easier, I suppose, to check for the anti-knight constraint and to propagate a digit found to the entire grid. The odd/even numbers refer to the position of the digit in an odd or even column/row. That's the best I can formulate it.
Not me screaming "the 3s!" for half this video. Nuh-uh. Nope. Not at all.
Beautiful puzzle! But not easy - took me 58:48. 😄
Stares at 6x6 grid. Just can't a start. Settles in to watch.
The start of the puzzle solve didn’t make sense to me
About half an hour into the video... Hopefully Simon notices the other knight's move into the box before too long!
33:56 Somehow he's noticed that but not realized there's only one place for green-purple in the box
To be fair, the fact that column clues were at least as restricted as row clues seemed obvious to me after a moment's thought, but I had no idea about that initial checkerboard pattern 😅 (or should I say chessboard pattern?)
this puzzle is clearly out of my league, but I've a question for who've completed it. Is the cherckerboarding really necessary to solve the puzzle?
You don't need to prove the entire checkerboarding from first principles. If you start doing any sort of coloring and basic knight's move eliminations, it just falls out.
I managed to solve it in 39 minutes without checkerboarding the grid.
I ended up assigning unique colours to the six cells in column 2 and propagating these around the grid by knight's move, so that I had column 5 uniquely coloured, and every other cell had dual options. (I suppose the checkerboarding pattern was hidden in there, but I didn't spot it).
Finding the 5, then the 3 in column 1 was the break through. Those propagated around the grid and allowed me to fully resolve my colouring.
I solved it by fillling in the letters a,B,C,d,E,F. It worked, but I needed bifurcations.
5 in the top corner was the first digit, I found on that way,too.3 was the second.
Difficult and nice. Marking cells in box 3 with small middle letters ABCDEF, with checkboard in mind, put all cells in strict restrictions. latter, using pencilmarks for possible numbers give me a chance to follow logic.
This looks so confusing, i wouldnt even be able to solve it
At this point I just assume all comments are AI generated unless proven otherwise
No you didnt prove the checker boarding. The top cell is proved because the 2s and 5s have 2 rows to eliminate the centre, but thebmiddlenrow does not. Im sure a little more steps would prove it.
This is so frustrating to watch. You broke into this puzzle so effortlessly compared to my attempt, using the knowledge of the checkerboard. And then you just continued to ignore half of the Knight's moves for what seemed like the next hour and a half.
THEY CAN'T BE 3 EITHER, come on Simon :(
The color pallet u chose to color the grid was ugly!
Shouldn't the chess board be expanded to 8×8
Actually this is an 8x8 grid. That is the reason Simon didn't get his 3 in the corner confetti in box 5. That cell is actually R7 C2 of the 8x8 grid.
A++
2 hours.
Interesting puzzle. However, Simon once again managed to make life hard for himself by not following up on the simple implications of his deductions using the rules. Once he proved the checkerboard, the horizontal 5 and the double-1 whenever the first cell to a clue is a 1, though the puzzle would still not be _solved,_ it would have been relatively easy to propagate knight‘s move implications into the perimeter clues. And, strangely, sometimes he _did_ follow up on this, but only in places it didn‘t really help. Instead, in true Simon fashion, he abandoned the thought and moved to the opposite side of the grid to find some wild logical bounces to progress. The knight‘s move rule affecting the Numbered Rooms clues outside the grid is an extraordinarily powerful rule _if_ you are willing to actually apply it.
*Don‘t get me wrong,* I love CtC and Simon is still 100 times the solver I can dream to become, and I do see that solving live, explaining your thoughts coherently in real time and still being entertaining is exceptionally hard - and also, Simon might be a bit biased against logic that appears obvious due to the amount of deviously hard puzzles he solved in the past, so maybe that‘s why he is quite prone to ignore that. But still, it amuses me that while I am usually quite pleased with my solve when I manage to stay within 2.5x Simon‘s time, I tend to match (and at one point even _beat)_ his time more closely the „easier“ the puzzle is, simply because of my utter ineptitude to spot viciously hidden logic and Simon‘s peculiar habit of doing the polar opposite.
One such instance is the question whether the Numbered Rooms clue in r4 could be orange - if it _was_ it would automatically be the same digit as (and I use the usual nomenclature only for the „inner“ 6x6) r4c1 because Knights Moves eliminate both r2c1 and 6c1, therefore, both would be one. At this point, he already deduced (though he failed to properly mark it) that r4c1 must be B and, because of c2 (and the relationship between the NR clue in c2 with B in r5c2) could not be 1, thus immediately removing the option of the NR clue in r4 to be orange. I know I use a lot of Simon‘s deductions to make the point (and some of those were acquired in equally convoluted ways instead of using the knight‘s move rule properly). But it was one of the most obvious ones that made me yell my best McGonnagall impression at my screen.
Simple implications to you, witchcraft to me 😂
An hour for a 6x6???? I'll just be watching then
Simon, I'm sorry, but you're not doing something that's sooooo obvious for someone who's watching, not only is it 2 left/right and 1 up/down, but it's also 1 left/right and 2 up/down for knights move. "Sigh"
29th comment!
First!