I think this channel and for that matter the rest of the sudoku community have to advance the difficult setting to 6 stars, especially if all these videos continue to be full length movies.
I spent a lot of this video screaming “what is the only possible composition of that three length renban in box 3/6 if you can’t put 3 or 4 on it?!” Great puzzle
That's a nice deduction! I don't blame him for not spotting it though, that's pretty specific Literally just got to that part of the video when I replied here...
yay! new hard puzzle. Can't wait to solve this one with you. Sometimes I find a digit before you and tell you, many times I deduce some elements and tell, but they turn out to be wrong, yet you still manage to solve. Sometimes I fall asleep in the middle but nonetheless I'm always delighted to wake up seeing you finishing the puzzle. I think we're a great team. thank you for doing the puzzles.. JE
48 mins in and ive seen something quite beneficial.. Look at the 3 cell renban and ask if it can work without a 6 on it?... The answer is no as it will force a 3 but the 3s are circled 🎉
@chipsounder4633 actually you can deduce it is a 5, 6, 7 triplet because it is well establiched the only uncircled 4 is in row 6. And that confirma the 7 in the "powerful square"
It's even more restricted. Right from the moment he found that 7 is not circled: There's no uncircled 3 and only one 4 which is needed in R6 and one 8 which is already used in R5. So the renban must be 567.
Made more obvious when Simon highlighted the cells, but does anyone else think the uncircled digits look a bit like Metroid? Almost certainly unimportant and/or coincidental, but my childish brain was amused... 😂
I never could have even started this puzzle without Simon's help, but with the assistance of his brilliance when I got stuck, it was a really fun solve.
I'm just always amazed when Simon misses something I think is crucial to the solve and comes around the long way to a solution anyway. Even with a miss like yesterday's, you're still a brilliant solver, Simon! I was quite proud of catching that min=11=max bit yesterday, but I think I only got that because of how much I've learned watching this channel.
The interaction between renbans and circles in this puzzle is amazing! At this stage, 1:08:54 The 6s can be disambiguated by considering the implications of placing 6 at r8c6. This makes both renban cells green, and the other cell can’t be 7, so must be 5. But if the circled 5 is green from this position, then green 5s are eliminated from boxes 3, 4, 5, 6, and 8. Leaving at most 4 possible green 5s. Therefore the 6 in row 8 is in column 5.
Took me almost 2.5h with a couple of mistakes but I did it. I focused much more on boxes than rows and I can see it how it was much less efficient than Simon's approach. Terrific puzzle and a gorgeous solve!
Another way to eliminate red sixes pretty early on was to count the number of possible red circles (19) and there would have to be 20. As it had to have 1,2,3,6,8. Just something I did while I waited for Simon to see the renban had to be 5.6.7.
Just absolutely incredible setting from you Jeff!! How do you even come up with this idea. Even more astonishing solving from you Simon! Never have to apologize for not seeing something right away and taking time to work things out. What a treat everyday to see such brilliant puzzles be conjured up and then see them cracked!!
I had second thoughts and decided to give it a go - what an amazing puzzle with some clever logic. Now looking forward to Simon explaining to me things that I missed. Though the logic seems tight and the final resolution in my solve was quite extraordinary.
54 mins - the one thing I saw more quickly (and there are two things Simon has that definitely beat me) - 3 and the three-cell Renban. I think I got that because I had notes and wasn't doing it all in my head.
Simon, i just wanted to say that your videos are very interesting as well as brilliant, and you'd be more than welcome in any party of mine. Whoever keeps telling you you're no fun at parties can come see me on the mat if they wanna argue
Trying to do this last night when I was tired was probably a mistake. I made some progress though, and finished it this morning 🎉. Almost four hours - I'm not quick, but I can be persistent 😂. My path was almost identical to Simon's up to the point he coloured those two cells yellow in boxes 8 and 9, and that gave him the 5 in in box 8. I wasn't very efficient towards the end.
My trouble with this sort of ruleset is that I read "colour all circles" as exactly that, but with 45 circles in 3 colours and 15 being not a sudoku number, I can't see how to proceed. Because I read the last rule as "only one number can appear in circles of colour x", rather than "it's fine to have two 2s, three 3s, four 4s and six 6s, because that fulfils the colouring and number rules". tl;dr I read rules in too exclusionary a fashion as if that's the only property a thing can have.
No the last rule means that if you have a number x in a circle it has to appear x times in the same coloured circle. So for Example a 5 in a green circle means that you have to have 5 5s in green circles
You said that you can't see everything that other people can (which is true), but right after you said that you feel bad for letting down those people who wrote the logic to you. Therefore it would make sense to feel bad for everything you do because no matter what you do people will have other thougths. Think about it, if you had done the thing that got mentioned many times, someone would tell you another logic, and you would feel bad for not spotting them. The world is this way that not everyone can be happy about everything.
Looking at row 8 will immediately show that there can only be 2 un-circled numbers, pointing out at 7 being the second one (after 9 was found). Later edit: The comments are correct, and I was wrong. In my mind, all the other digits HAD to be in those circles, when (at that point) they absolutely didn't have to be.
the maximum count is based on duplicate circled digits wherever possible. so you might have 2 circled 2s in one color but none in the others, giving you 4 of the missing circled digits (comparing with the maximum). so row 8 alone doesn't disprove that possibility. you still have to show that 7 is circled to force the maximum number of circled digits for everything smaller.
@@CassandraComar You need to know the absolute maximum (3*1+3*2+3*3+2*4+5+6+7+8+9) = 61 and there are 45 circles, so the missing digits sum to 16. Row 8 shows that this 16 has to be the sum of only one or two different numbers (but could be something like a+a+a or a+b+b). With a=9 found you can't write the missing 7 as b+b, so the second number is 7.
That doesn't sound valid to me, you could have 1-8 circled and have a different combination of duplicates. All it shows is that there are not more than 2 un-circled numbers.
Meanwhile im struggling with both of today's NYT sudokus... I usually solve med in 8 minutes and hard in under 20. Usually. Im at like 45 minutes on each with multiple leaves and come backs.
1:22:58 Check your pencil marks. Box 9 has 4's in row 9. There is no 4 in R9C3. This makes R7C3 a 4. Can R7C3 be green? No? Cool, then it's blue or red. Conclusion, 4's do not appear in green. R9C7 is not a 4. R9C8 is a red 4. EDIT: Ok, that conclusion was a silly mistake by me and i fluked it. Of course if R7C3 was blue, it wouldn't make green impossible to be 4. Since row 6 is the only row in the grid that has an uncircled 4, one of box 4 or 6 will have an uncircled 4. All other boxes will have circled 4's. Meaning box 2 will have a 4, that is circled and NOT green. R1C4 or R2C4. You get an X-Wing on 4's so one of R3C7 or R3C9 will be a 4. Here's a magic trick... Box 5 will need a circled 4 in it. R4C5 is that circled 4. This because of where 4 is in box 2. Now R4C1 is not a 4. So R6C1 becomes the uncircled 4. R5C9 becomes a 4. So R3C7 becomes a 4.
11:55 It's pretty obvious that counting the number of circles is going to be pointless at this stage. In your rules explanation, you didn't put forth the idea that if a 7 appears in a red circle, there are AT LEAST 7 red circles. But if a 4 also appears in a red circle, there will be 11 red circles. 7 with sevens in them and 4 with fours in them. Then green could have a 4 in four of its circles and a 5 in five other green circles and a one in one green circle. And so there would be 12 green circles etc etc etc... What you want to start with is the orthogonal rule. R8C3 is green R9C2 is green Both see blue and red circles. Here i am wishing there was a way to just color a circle so you don't end up with just square colors everywhere. Anyways, R8C7 and R0C8 are red circles with the same logic. R7C3 and R8C4 are not green. R7C7 and R8C6 are not red. If those renban lines are both blue, then R8C5 is not very restricted, could be any of two colors. Otherwise, R8C5 is going to be forced to be whatever color the two lines are not. Only if both lines are blue is R8C5 not restricted and the only case where the lines can be the same color is if they are blue. In R2C1 and R3C1, neither of those circles can be red. This will form a reg, green, blue pair. I.E, one cell is green and one cell is blue. Similar logic applies for R2C9 and R3C9, they will be red and blue. And yes, there are 45 circles in the grid. But again, that doesn't mean they are all different digits. 3 can appear in red, green AND blue. And thus relieve circles from having 9's in them.
What's harmful about this comment? TH-cam eats my comments all the time. This platform... I am so done with it! 32:50 Symmetry Simon, there's symmetry in the puzzle. Please make R2C1 and R3C1 a green/blue pair. And yes, i want to question whether 7 is in circles. If not, 7 would be in R1 C1, C5 or C9. Along with a 9. And for R8, there would be a [79] pair in the unciricled cells C1 and C9. It would displace either 7's or 9's from the rest of C1 and C9. Anyways, if 7 went in green. You'd know that neither R4C9 or R6C2 would be green 7's. And so there would have to be a 7 in every other box on a green circle. But in box 5, there are no circles outside of row 4 and that sees a [79] pair. So green does not have a 7 in them! If a red circle has a 7 in it, then there's already 15 red circles. You have 9 red circles so far. You'd need 6 more. And again, R4 eliminates the possibility for a red circled 7 in that row. Box 4 and box 5 does not have circled 7's in it at all. So all other boxes would need a circled 7 that is red. Let's focus on rows 7, 8 and 9. If R9C3 was a red 7, then R7C9 would be a red 7. And R8C4 would HAVE to be a red 7. Now R7C3 is a red cell that is consecutive with 7 and isn't an 8, so it would be a red 6. And if R9C3 was a red 7, then box 9 can't have a red 7 in it and make it impossible for red to have a 7 in it. Either way, if a 7 were to appear in red. You'd automatically end up with a 6 in red. And in box 2, you can only have 2 red circles. One would be a 7 and the other is 8. In box 6, R5C9 would be a 7 and there would be no 6. in row 4, there would be a 6 in C1, C4 or C6. But one of boxes 4 and 5 would not have a red 6 in them. And row 8 is restricted because you now have no place to put a 6 in a red circled cell. You've run out of places for 6's. Red does NOT have a 7 in it. Blue has a similar situation where, box 4 cannot have a 7 in blue. Box 5 could have a 7 in blue, but then box 6 cannot have a 7 in blue. And vise versa. So all other boxes/rows would need a blue 7 in them. Well, row 9 cannot have a blue 7 in it so blue does not have a 7 in any of its circles. 7's do not appear in circles. All other digits, aside from 7 and 9, do appear in circles the maximum number of times allowable. To add to 45. 1+1+1+2+2+2+3+3+3+4+4+5+6+8 = 45. Addition: 36:28 That's a brilliant way of simplifying why blue can't have a 7. I'll give you that! And let me rephrase why green has no 7's in it. There are no circles that can be green and have 7's in them in boxes 4, 5 and 6. So you'd get at most 6 circled 7's in green. The complex logic for why red can't have a 7 still is... complex.
1:43:07 Look at R2C6. If it's a 3, it's not a green 3 because you already have 3 green threes in the grid. If it's a 5, it's not green because 5's are blue. In fact, as far as greens go: 1x1 2x2 3x3 6x6 4, 5, and 8 are not green. All your greens are finished! R1C7 is blue, R1C8 is red. R2C4 is blue or red. In column 4 you have a [13] pair. R2C4 cannot be a red 1, since you already have your red 1 elsewhere. And by elimination, you will have a blue 1 in column 4. So neither R1C7 or R1C8 can be 1's. This places a 1 in R1C9. (And i now see that R9C7 looks at R6C7 which would resolve that the same) But this makes R1C7 a blue 3. So all other 3's are red. And thus, R2C4 is a red 3.
1:09:05 If R8C6 was a green 6, R7C7 would be a green 5. Quick and dirty... Boxes 4, 5, 6 can't have a green 5. Box 8 could not have a green 5 because of adjacency rule and R7 already having a 5. Boxes 1, 2, 3, 7 and 9 would have green 5's in them. Box 9 already has its 5. Where do you find a green 5 in box 3? Nowhere, that's where. It is therefore impossible for R8C6 to be a green 6. Or even green for that matter. R8C5 is the green 6. Could 5's be in green now? No, for the same reason that 5's cannot be green in boxes 3, 4, 5, 6 and 8. Box 3 still can't have a green 5 because R9C7 would have to be a green 5 making no cell in box 3 able to have a green 5. Box 8 can't have a green 5 because R8C5 is a green 6 and is adjacent to all other circles in the box. Box 4 has no empty green circles and box 5 sees a 5 in the row where a green could go and box 6 already has a 5 that isn't circled. EDIT: 1:15:50 I mean, look at box 9. If 5 can't go in green and 5 will be in column 7 ox box 9. Then R7C7 is a 5. And it is blue. This gives you a 4 at R8C6.
The amount of concentration and sheer brain power that goes into solving these type of puzzles just boggles my brain.
I think this channel and for that matter the rest of the sudoku community have to advance the difficult setting to 6 stars, especially if all these videos continue to be full length movies.
I spent a lot of this video screaming “what is the only possible composition of that three length renban in box 3/6 if you can’t put 3 or 4 on it?!”
Great puzzle
That's a nice deduction! I don't blame him for not spotting it though, that's pretty specific
Literally just got to that part of the video when I replied here...
Good on you, but that's really hard to spot.
Simon, watching your mind work is a delight. I don't write comments very often, but I enjoy watching ctc and wanted to let you know. ;)
Rules: 01:35
Let's Get Cracking: 11:42
Simon's time: 1h35m37s
Puzzle Solved: 1:47:19
What about this video's Top Tier Simarkisms?!
Bobbins: 4x (22:06, 29:29, 32:09, 45:38)
The Secret: 3x (06:55, 12:58, 13:08)
Three In the Corner: 1x (1:34:11)
Maverick: 1x (30:34)
And how about this video's Simarkisms?!
Sorry: 18x (06:42, 15:34, 15:42, 22:10, 30:32, 32:56, 42:09, 45:38, 49:09, 52:51, 1:03:05, 1:29:01, 1:29:01, 1:29:38, 1:29:55, 1:34:26, 1:36:56, 1:37:00)
Ah: 17x (10:36, 15:42, 30:46, 30:46, 32:19, 32:19, 34:01, 42:57, 1:06:33, 1:27:58, 1:30:57, 1:34:44, 1:35:45, 1:36:56, 1:37:16, 1:38:59, 1:42:34)
Brilliant: 10x (00:39, 02:47, 27:51, 28:50, 28:55, 1:07:09, 1:07:12, 1:40:54, 1:48:23, 1:48:23)
Pencil Mark/mark: 9x (42:05, 44:09, 1:10:08, 1:11:55, 1:13:16, 1:14:17, 1:14:19, 1:14:21, 1:31:27)
Hang On: 7x (12:33, 23:45, 29:56, 32:56, 37:03, 1:00:46, 1:21:37)
In Fact: 7x (09:04, 1:15:10, 1:19:02, 1:19:55, 1:23:00, 1:29:34)
Wow: 7x (23:52, 45:21, 49:03, 1:09:22, 1:15:37, 1:47:21, 1:47:21)
By Sudoku: 6x (38:47, 1:16:01, 1:26:15, 1:29:55, 1:32:02, 1:39:02)
Obviously: 6x (08:59, 16:31, 17:04, 20:39, 1:27:00, 1:42:12)
Nonsense: 5x (46:26, 46:39, 1:10:22, 1:39:33, 1:44:15)
Beautiful: 5x (04:30, 23:52, 59:37, 1:06:37, 1:31:02)
Bother: 4x (29:42, 1:22:10, 1:33:00, 1:41:19)
The Answer is: 4x (24:33, 27:32, 38:33, 39:34)
Weird: 4x (14:26, 50:41, 1:05:02, 1:20:18)
Good Grief: 3x (46:59, 58:31, 1:31:02)
What on Earth: 3x (15:34, 47:11, 1:09:11)
Goodness: 3x (20:01, 59:32, 1:14:07)
Apologies: 3x (04:04, 08:51, 09:34)
Wake Up: 3x (07:23, 07:23, 08:30)
Have a Think: 3x (47:25, 50:03, 1:19:06)
Cake!: 3x (06:58, 08:59, 09:15)
Clever: 2x (06:30, 1:46:16)
Stuck: 2x (01:52, 01:55)
Lovely: 2x (07:44, 08:42)
Fascinating: 2x (00:51, 1:07:14)
Incredible: 2x (04:07, 04:21)
Phone is Buzzing: 2x (34:01, 49:38)
Nature: 2x (27:48, 27:53)
Useless: 1x (1:05:33)
Naughty: 1x (1:35:39)
Horrible Feeling: 1x (1:24:20)
Elegant: 1x (06:14)
Ridiculous: 1x (1:46:20)
Come on Simon: 1x (1:09:29)
Shouting: 1x (08:38)
Flurry of Activity: 1x (1:09:44)
Surely: 1x (1:14:07)
Stunning: 1x (1:47:17)
Unbelievable: 1x (1:35:06)
We Can Do Better Than That: 1x (1:29:26)
I Digress: 1x (1:05:33)
That's Huge: 1x (1:13:39)
Symmetry: 1x (20:05)
Most popular number(>9), digit and colour this video:
Forty Five (12 mentions)
Nine (94 mentions)
Red (128 mentions)
Antithesis Battles:
High (2) - Low (0)
Even (7) - Odd (0)
Row (73) - Column (23)
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!
Simon has a coloring puzzle - today is his day, for sure!
The amount of times Simon said that doing a little bit of basic sudoku wont do anything is amazing. And he was wrong every time.
I wish we could be there to gently remind him.
"Three is like the new nine" 😅 His children will grow accustomed to hearing him say weird things while working in his youtuber office
Except nobody puts nine in the corner.
After 84 minutes, I finally did it! Now I would have my potato chips ready to watch Simon solve this puzzle.
Out with "Orange is the New Black". In with "3 is the new 9"
yay! new hard puzzle.
Can't wait to solve this one with you.
Sometimes I find a digit before you and tell you, many times I deduce some elements and tell, but they turn out to be wrong, yet you still manage to solve. Sometimes I fall asleep in the middle but nonetheless I'm always delighted to wake up seeing you finishing the puzzle.
I think we're a great team.
thank you for doing the puzzles..
JE
At nearly 2 hours, of course only 7 people have solved. I enjoy watching the solve and learning the logic!
Grabbing the popcorn.
Should I watch a movie? Nah, too long.
Should I watch Simon be amazed at another sudoku for 90 min? COunt me in
48 mins in and ive seen something quite beneficial..
Look at the 3 cell renban and ask if it can work without a 6 on it?... The answer is no as it will force a 3 but the 3s are circled 🎉
@chipsounder4633 actually you can deduce it is a 5, 6, 7 triplet because it is well establiched the only uncircled 4 is in row 6. And that confirma the 7 in the "powerful square"
It's even more restricted. Right from the moment he found that 7 is not circled: There's no uncircled 3 and only one 4 which is needed in R6 and one 8 which is already used in R5. So the renban must be 567.
@@ivogarcia7762 awesome
Made more obvious when Simon highlighted the cells, but does anyone else think the uncircled digits look a bit like Metroid? Almost certainly unimportant and/or coincidental, but my childish brain was amused... 😂
A. N n qa. Ai ini iII. J QAak q9’ 💓💞 o Moonoooooo😅ko😅😅😅 ooookonoookookoo
A. N n qa. Ai ini iII. J QAak q9’ 💓💞 o Moonoooooo😅ko😅😅😅 ooookonoookookooo oooo😅o😅😅😅o😅😅😅 😅ooo😅😅okooó oooooooo ok
That was a fun one to watch. I got nowhere so it was neat to see the progress. Slow feels an unfair way to describe the progress -- this was hard.
Watching Simon solve this puzzle is like watching an archeologist dismantle an historical site using tweezers and a magnifying glass. Exquisite!
I never could have even started this puzzle without Simon's help, but with the assistance of his brilliance when I got stuck, it was a really fun solve.
I'm just always amazed when Simon misses something I think is crucial to the solve and comes around the long way to a solution anyway. Even with a miss like yesterday's, you're still a brilliant solver, Simon! I was quite proud of catching that min=11=max bit yesterday, but I think I only got that because of how much I've learned watching this channel.
The interaction between renbans and circles in this puzzle is amazing!
At this stage, 1:08:54 The 6s can be disambiguated by considering the implications of placing 6 at r8c6.
This makes both renban cells green, and the other cell can’t be 7, so must be 5.
But if the circled 5 is green from this position, then green 5s are eliminated from boxes 3, 4, 5, 6, and 8.
Leaving at most 4 possible green 5s.
Therefore the 6 in row 8 is in column 5.
Took me almost 2.5h with a couple of mistakes but I did it. I focused much more on boxes than rows and I can see it how it was much less efficient than Simon's approach.
Terrific puzzle and a gorgeous solve!
Another way to eliminate red sixes pretty early on was to count the number of possible red circles (19) and there would have to be 20. As it had to have 1,2,3,6,8. Just something I did while I waited for Simon to see the renban had to be 5.6.7.
Brilliant long video, long videos are my favourite. The longer the better. Amazing puzzle.
Just absolutely incredible setting from you Jeff!! How do you even come up with this idea. Even more astonishing solving from you Simon! Never have to apologize for not seeing something right away and taking time to work things out. What a treat everyday to see such brilliant puzzles be conjured up and then see them cracked!!
I had second thoughts and decided to give it a go - what an amazing puzzle with some clever logic. Now looking forward to Simon explaining to me things that I missed. Though the logic seems tight and the final resolution in my solve was quite extraordinary.
Congratulations!
54 mins - the one thing I saw more quickly (and there are two things Simon has that definitely beat me) - 3 and the three-cell Renban. I think I got that because I had notes and wasn't doing it all in my head.
I unwound the end in a slightly different order, but whichever way you do it, the fact that the logic lasts till the end ia a feat by itself.
I was ons of the green people from the last puzzle. Love love love your solves even if you do itdifferently. Dont ever change!
Happy birthday Alexis! Particularly because we got to hear Simon play guitar in celebration!! 🎸
Interesting an application of quantum chromodynamics - nice!
Simon, i just wanted to say that your videos are very interesting as well as brilliant, and you'd be more than welcome in any party of mine. Whoever keeps telling you you're no fun at parties can come see me on the mat if they wanna argue
Several caffeine breaks and 160 minutes later, I finally solved this one! I had some ridiculous markings for this one.
What a great challange. After ages I finally could solve it.
"Oh, so that is the only candid eight in the puzzle..."
Trying to do this last night when I was tired was probably a mistake. I made some progress though, and finished it this morning 🎉.
Almost four hours - I'm not quick, but I can be persistent 😂.
My path was almost identical to Simon's up to the point he coloured those two cells yellow in boxes 8 and 9, and that gave him the 5 in in box 8. I wasn't very efficient towards the end.
Waiting for Simon to notice the renban in box 2-6 can’t have 3 and must have 4-5-6-7 which would be the last uncircled 6 and no 4.
If I am not mistaken, the rule about circles on same line is not required to solve sudoku, but is needed only to color all the 4s circles?
Yep. This is a puzzle where sudoku is only one of the rules.
Simon's stubborness around 1:40h of getting all the digits without doing any Sudoku whatsoever is truly impressive
Simon, just use letters for your 1’s and 2’s 🤦🏻♂️
But then what is the colour palette for? 😂
I'm just kidding, we know Simon loves a good colour
My trouble with this sort of ruleset is that I read "colour all circles" as exactly that, but with 45 circles in 3 colours and 15 being not a sudoku number, I can't see how to proceed. Because I read the last rule as "only one number can appear in circles of colour x", rather than "it's fine to have two 2s, three 3s, four 4s and six 6s, because that fulfils the colouring and number rules".
tl;dr I read rules in too exclusionary a fashion as if that's the only property a thing can have.
No the last rule means that if you have a number x in a circle it has to appear x times in the same coloured circle. So for Example a 5 in a green circle means that you have to have 5 5s in green circles
The simpler way to resolve the 9s is when you highlighted where 9s could go there was a skyscraper in rows 2 and 5
solved in 56 minutes and 45 seconds. That was really really hard.
LETSGOOO ANOTHER ONEEE DAY:8
You said that you can't see everything that other people can (which is true), but right after you said that you feel bad for letting down those people who wrote the logic to you. Therefore it would make sense to feel bad for everything you do because no matter what you do people will have other thougths.
Think about it, if you had done the thing that got mentioned many times, someone would tell you another logic, and you would feel bad for not spotting them.
The world is this way that not everyone can be happy about everything.
Please dont use black as backgroundxclour 😊
Looking at row 8 will immediately show that there can only be 2 un-circled numbers, pointing out at 7 being the second one (after 9 was found).
Later edit: The comments are correct, and I was wrong. In my mind, all the other digits HAD to be in those circles, when (at that point) they absolutely didn't have to be.
the maximum count is based on duplicate circled digits wherever possible. so you might have 2 circled 2s in one color but none in the others, giving you 4 of the missing circled digits (comparing with the maximum). so row 8 alone doesn't disprove that possibility. you still have to show that 7 is circled to force the maximum number of circled digits for everything smaller.
You could skip 9 and 5, and have two colors of 2s instead of three colors of 2s. And probably some other variants as well.
@@CassandraComar You need to know the absolute maximum (3*1+3*2+3*3+2*4+5+6+7+8+9) = 61 and there are 45 circles, so the missing digits sum to 16. Row 8 shows that this 16 has to be the sum of only one or two different numbers (but could be something like a+a+a or a+b+b). With a=9 found you can't write the missing 7 as b+b, so the second number is 7.
That doesn't sound valid to me, you could have 1-8 circled and have a different combination of duplicates. All it shows is that there are not more than 2 un-circled numbers.
It just seemed that the other 3-cell purple line couldn't have a 123 so it has a 6.
I lived next to Pope Air Force base for 2.5 years so aircraft don't both me.
"My eyes are circles"
last half of hour was exyremely hard to see
Meanwhile im struggling with both of today's NYT sudokus... I usually solve med in 8 minutes and hard in under 20. Usually. Im at like 45 minutes on each with multiple leaves and come backs.
Coming back to say I reset both and got em in my usual times 🤦♀️
3 was the trick, first r7 c5
Horrible hot mess of colour but an interesting brain bender of a puzzle for sure
Bother.
Never been this early before :)
Yey 1st
51:12 for me. What a tough one today. Great puzzle!
58:38 - This is what genuine discovery looks like after much struggle...! It's the ultimate catharsis.
57:35 it's most definitely a Narwhal on 9's 😂.
7:58 A little warm up music before the main event. Primed and ready!
52:00 minute mark, 6 has to be on line. Because 3 can not be on the line.
1:36:40 just unbearable
1:22:58 Check your pencil marks.
Box 9 has 4's in row 9.
There is no 4 in R9C3. This makes R7C3 a 4.
Can R7C3 be green? No? Cool, then it's blue or red.
Conclusion, 4's do not appear in green. R9C7 is not a 4. R9C8 is a red 4.
EDIT: Ok, that conclusion was a silly mistake by me and i fluked it. Of course if R7C3 was blue, it wouldn't make green impossible to be 4.
Since row 6 is the only row in the grid that has an uncircled 4, one of box 4 or 6 will have an uncircled 4.
All other boxes will have circled 4's. Meaning box 2 will have a 4, that is circled and NOT green. R1C4 or R2C4.
You get an X-Wing on 4's so one of R3C7 or R3C9 will be a 4.
Here's a magic trick... Box 5 will need a circled 4 in it. R4C5 is that circled 4. This because of where 4 is in box 2.
Now R4C1 is not a 4. So R6C1 becomes the uncircled 4. R5C9 becomes a 4. So R3C7 becomes a 4.
40:48 So what we have to do now is... Well I don't know what we have to do :))))))))
11:55 It's pretty obvious that counting the number of circles is going to be pointless at this stage.
In your rules explanation, you didn't put forth the idea that if a 7 appears in a red circle, there are AT LEAST 7 red circles. But if a 4 also appears in a red circle, there will be 11 red circles. 7 with sevens in them and 4 with fours in them.
Then green could have a 4 in four of its circles and a 5 in five other green circles and a one in one green circle. And so there would be 12 green circles etc etc etc...
What you want to start with is the orthogonal rule.
R8C3 is green
R9C2 is green
Both see blue and red circles.
Here i am wishing there was a way to just color a circle so you don't end up with just square colors everywhere.
Anyways, R8C7 and R0C8 are red circles with the same logic.
R7C3 and R8C4 are not green. R7C7 and R8C6 are not red.
If those renban lines are both blue, then R8C5 is not very restricted, could be any of two colors. Otherwise, R8C5 is going to be forced to be whatever color the two lines are not.
Only if both lines are blue is R8C5 not restricted and the only case where the lines can be the same color is if they are blue.
In R2C1 and R3C1, neither of those circles can be red. This will form a reg, green, blue pair. I.E, one cell is green and one cell is blue.
Similar logic applies for R2C9 and R3C9, they will be red and blue.
And yes, there are 45 circles in the grid. But again, that doesn't mean they are all different digits.
3 can appear in red, green AND blue. And thus relieve circles from having 9's in them.
What's harmful about this comment? TH-cam eats my comments all the time.
This platform... I am so done with it!
32:50 Symmetry Simon, there's symmetry in the puzzle.
Please make R2C1 and R3C1 a green/blue pair.
And yes, i want to question whether 7 is in circles. If not, 7 would be in R1 C1, C5 or C9. Along with a 9.
And for R8, there would be a [79] pair in the unciricled cells C1 and C9.
It would displace either 7's or 9's from the rest of C1 and C9.
Anyways, if 7 went in green. You'd know that neither R4C9 or R6C2 would be green 7's.
And so there would have to be a 7 in every other box on a green circle.
But in box 5, there are no circles outside of row 4 and that sees a [79] pair. So green does not have a 7 in them!
If a red circle has a 7 in it, then there's already 15 red circles. You have 9 red circles so far. You'd need 6 more.
And again, R4 eliminates the possibility for a red circled 7 in that row. Box 4 and box 5 does not have circled 7's in it at all.
So all other boxes would need a circled 7 that is red.
Let's focus on rows 7, 8 and 9.
If R9C3 was a red 7, then R7C9 would be a red 7. And R8C4 would HAVE to be a red 7. Now R7C3 is a red cell that is consecutive with 7 and isn't an 8, so it would be a red 6.
And if R9C3 was a red 7, then box 9 can't have a red 7 in it and make it impossible for red to have a 7 in it.
Either way, if a 7 were to appear in red. You'd automatically end up with a 6 in red.
And in box 2, you can only have 2 red circles. One would be a 7 and the other is 8.
In box 6, R5C9 would be a 7 and there would be no 6.
in row 4, there would be a 6 in C1, C4 or C6. But one of boxes 4 and 5 would not have a red 6 in them.
And row 8 is restricted because you now have no place to put a 6 in a red circled cell. You've run out of places for 6's.
Red does NOT have a 7 in it.
Blue has a similar situation where, box 4 cannot have a 7 in blue. Box 5 could have a 7 in blue, but then box 6 cannot have a 7 in blue. And vise versa.
So all other boxes/rows would need a blue 7 in them. Well, row 9 cannot have a blue 7 in it so blue does not have a 7 in any of its circles.
7's do not appear in circles.
All other digits, aside from 7 and 9, do appear in circles the maximum number of times allowable. To add to 45.
1+1+1+2+2+2+3+3+3+4+4+5+6+8 = 45.
Addition:
36:28 That's a brilliant way of simplifying why blue can't have a 7. I'll give you that!
And let me rephrase why green has no 7's in it.
There are no circles that can be green and have 7's in them in boxes 4, 5 and 6. So you'd get at most 6 circled 7's in green.
The complex logic for why red can't have a 7 still is... complex.
1:43:07 Look at R2C6.
If it's a 3, it's not a green 3 because you already have 3 green threes in the grid.
If it's a 5, it's not green because 5's are blue.
In fact, as far as greens go:
1x1
2x2
3x3
6x6
4, 5, and 8 are not green.
All your greens are finished!
R1C7 is blue, R1C8 is red.
R2C4 is blue or red.
In column 4 you have a [13] pair. R2C4 cannot be a red 1, since you already have your red 1 elsewhere.
And by elimination, you will have a blue 1 in column 4. So neither R1C7 or R1C8 can be 1's. This places a 1 in R1C9. (And i now see that R9C7 looks at R6C7 which would resolve that the same)
But this makes R1C7 a blue 3. So all other 3's are red.
And thus, R2C4 is a red 3.
1:09:05 If R8C6 was a green 6, R7C7 would be a green 5.
Quick and dirty...
Boxes 4, 5, 6 can't have a green 5.
Box 8 could not have a green 5 because of adjacency rule and R7 already having a 5.
Boxes 1, 2, 3, 7 and 9 would have green 5's in them. Box 9 already has its 5.
Where do you find a green 5 in box 3? Nowhere, that's where.
It is therefore impossible for R8C6 to be a green 6. Or even green for that matter.
R8C5 is the green 6.
Could 5's be in green now? No, for the same reason that 5's cannot be green in boxes 3, 4, 5, 6 and 8.
Box 3 still can't have a green 5 because R9C7 would have to be a green 5 making no cell in box 3 able to have a green 5.
Box 8 can't have a green 5 because R8C5 is a green 6 and is adjacent to all other circles in the box.
Box 4 has no empty green circles and box 5 sees a 5 in the row where a green could go and box 6 already has a 5 that isn't circled.
EDIT: 1:15:50 I mean, look at box 9. If 5 can't go in green and 5 will be in column 7 ox box 9. Then R7C7 is a 5. And it is blue.
This gives you a 4 at R8C6.