I'm so excited to see Simon solve my puzzle! and an hour long, too! Now to watch the video... And if the title is anything to go by, he likes it too! A dream come true, Thank you Simon!
This variant was really fun. I love how it required a bunch of tricks to solve, but at the same time, it wasn't insanely hard. Thank you so much for sharing it!
Brilliant work, Lizzy. Congratulations on putting this together. And I know I'm not supposed to say this, but it's always a pleasant surprise for me to be reminded that women (and you too Clover) can design a puzzle as clever and original as anything a man can do.
You can really tell how disjoint sets trip up experienced solvers because it's a dimension that they're not used to using. For those of us with less experience, it's no more or less unusual than the regular types of scanning.
Simon, you got that 7-8 pair in box 1 in a rather difficult way. There's an easier way to get that. Because of the disjoint sets you actually know the value of the bottom left to top right cel diagonal in box 1. Those same 3 cells in every other box have a little killer clue looking at them. The value of all the bottom left to top right cells in all boxes combined is 3× 45 because of the disjoint sets rule. You have a 17 clue, an 8 clue and a 41 clue for those cells and you know the bottom left to top right diagonal in the grid is 45. Therefor the value of the diagonal in box 1 = 3×45 - 45 - 41 - 17 - 8 = 24, making it a 7-8-9 triple you could have pencilmarked before you filled in anything in the sudoku. Then as soon as you find the 9 in box 1 you know the other 2 cells are that 7-8 pair.
The best reasoning for the 17 diagonal is this: 17 diagonal, 8 diagonal and the main diagonal consist only of three different squares in their respective boxes and thus can contain at most three of each digit. Since main diagonal contains exactly one of each digit, we are left with at most two of each digit on the 17 and 8 diagonal together. They consist of 9 digits thus the minimum is 2*1+2*2+2*3+2*4+5=25 which is precisely what their sum is. So there have to be two ones, twos, threes and fours as well as one five.
Or you do the maths on the diagonals that consists of 3 disjoint sets. So 17+45+41+8+(diagonal in box 1) = 3*45 giving us 24 for the diagonal in box 1 - I admit I have only seen this after seeing the solve.
I missed the interaction between the diagonals but I was still able to get the 9 worked out and a good break into the puzzle. Loved it, brilliant logic with how the disjoint sets and diagonals work together!
@@lizzyvv447 The end is so pleasing with the 7/8 pairs and the only at first glance deadly 2/5 pair. Beautiful how the the 22 diagonal resolves the 2/5 and the 7s and 8s fall at last. Thank you very much!
54:00 Simon notices the forbidden positions of the 6's in any box except 3 and 7, and instantly forgets that, which could have done a great deal of job in box 1 with the diagonal. This would reveal the position of all 6's.
That was definitely a brilliant puzzle and a very beautiful ending. There was also a lot of logic we could learn from and I was very happy to see the video was over an hour. Simon, I think you did it brilliantly and I do agree it looks like a very difficult puzzle but still a very elegant solve.
I'm SO glad I didn't give up on this one! The solve was so satisfying! Took me 115 mins, but still, worth every minute. My solve was pretty much exactly the same as Simon's up until 1:01:20, Kudos to the constructor for creating such a linear path. Once we've done all we can with the 5,6,7,8,and 9s, All the remaining cells contain only the numbers 1,2,3 and 4 (apart from an X-wing of 5s) At this point, I changed my 78 coloring to light grey and black, and used 4 colors to mark all the remaining cells, This helped disambiguate the remaining numbers (the new multi-color function in particular is a very handy tool).
One look at the video's time, and my heart sank. Nonetheless, I gave this a shot ... and darned if I didn't solve it successfully, with no guesswork, in under an hour (53:03 to be exact). I *loved* this puzzle!
@Simon - you could have found the 789 triple in box 1 right at the start. Not the sum of all the / diagonals has to be 3 * 45 the ( 3 sets of 1-9) . You were given the sums of all the other diagonals, leaving box 1 to be 24.
There is an easier way of finding the 7,8-pair in box 1. The diagonals marked with 8,41,17 add to a total of 66. Adding the main diagonal gets us to 66+45=111. Because of the relative position constraint we know that the diagonals that we have counted, along with the diagonal in box 1 contain exactly 3 sets of the digits from 1-9. Therefore the sum of the diagonal in box 1 is 45*3-111=24. The only way to make 24 with three separate digits is 7,8,9 :)
Anyone else screaming while pulling there hair out around 44:06when Simon puts 2 possible purple cells in box 5 and not paying attention to the purple cell in box 1. Which would give you the purple cell in box 2 then box 3 and then box 6. lol, but it's still fun watching.
Also slightly frustrated because by 55:00 I had worked out that since the middle and bottom-right squares had sixes in boxes 3 and 7, the main diagonal starting in r1c1 needed to have a six in the upper right of either box 1 or 4. Since 6 in row 1 was assigned in box 2, that means the only place in that diagonal a 6 could go was in r4c4.
That feeling when CTC has to think about writing their 3rd book before the first one is released. The puzzles that have been featured in the last couple of months have been sublime.
What possessed me to try a puzzle that took Simon over an hour?! My time was 252:17! I think Lizzy's puzzle was fun and challenging, but not impossible (even for someone who is barely competent like myself), so I'm really thankful. The combination of disjointed sets and the main diagonal constraints manipulated the way the numbers saw each other in a really cool way. It created a lot of interesting and varied types of logic, so I got to use a lot of tricks (not just the same trick over and over again) throughout the solve which made the puzzle really fun. I found the 78 pair in box 1 pretty fast because I just added up all the positive-sloping middle diagonals (8+41+45+17+9=120) and subtracted it from 3 sets of 1-9, which I've learned is 3*45 (135) from watching this channel for a difference of 15. Since there is already a 9 in box 1, the other two cells had to be a 78 pair. But then I started working on the 41 clue and did my math wrong, so it took me forever to figure out lol. After the 3-hour mark, I was worried I'd get to the end and find I had made a mistake, but thankfully everything worked out and when I watched the video I could totally relate to Simon's excitement and satisfaction when he colored in all the 78 pairs!
With the puzzles getting increasingly harder and the video lengths getting increasingly longer, I've found it a lot more pleasing to go back into the archives where they tackled some easier puzzles. I just watched a puzzle from 3 years ago where Simon intros an apology for the hard puzzle and the video was only 41 minutes long. My how times have changed.
The Disjoint rule on the Green/Purple section at around 45:00 eliminated every extraneous option but wasn't spotted until 47:15. It's such a tricky rule.
Yeah, saw it too; eventhough I‘m not that good at sudoku (compared to Simon, at least xD) I can spot patterns and symmetry rather easily. Because of that I know it‘s unfair, but I still couldn‘t stop screaming atz my screen 😅 But I still love this video, especially the complexity of the puzzle :D
@@Schambes yep I'm still in minute 45 and have been screaming for a few but I'd have never gotten this logic so I dint really have room to be too critical. Actually I like to come check the comments to make sure I'm not screaming at something that's not there. Also gives me a chance to let my blood pressure go down.
i just watched the video to see how long it would take you to realize that the diagonal eats up one of each numbers for the 17,41,8 clues. It blew my mind when i was solving.
Wow, that was insane! I got the initial break in at the bottom left without any trouble, but the rest of it just left me broken ... I could have stared at it for _weeks_ without thinking about how the leading diagonal used up digits from the disjoint sets ...
all of the big triangles with the 1234 digits near the end of the solve made me very happy, after seeing some incredible stuff earlier in the puzzle it still had more left to give!
One of my favourite puzzles so far. 2hr solve for me and I'm very happy with that. It's not just difficult because of the intricate logic - you also have to use very different logic at different points which makes it hard to spot.
The much more simple logic to figure out 39:40 that on the diagonal of the first box we have 7-8 pair: Look at 41, 8, 17 diagonal and diagonal of the first box. it consist of 3 sets of digits 1-9(because this is a right-top corner, center, and left-bottom corner of each box. But we know sums of all diagonal except one in the first box: main diagonal consists of 1-9 set, so we have 3*45 = 135 - 45 - 41 - 8 - 17 = 24. So the diagonal in the first box sums up to 24 and it is only one way to do it by using 7, 8, 9.
I was somehow able to solve this one all on my own. Yes it took me coming back to it over the course of several days, but I was able to do it. This channel has taught me so much.
There's an easier way to get the 78 pair in box 1: sum the diagonals 17+45+41+8=111. Add the same diagonal positions in box 1 gives three lots of 45, which is 135. 135-111=24, which must be 7,8,9. (That might have made it easier to get the composition of the 41 diagonal as well: two sets of digits 1-9, minus those diagonal cells in boxes 1,2,4, and 9.)
Simon *almost* had the right set of cells highlighted at 24:37 to apply the set theory logic (highlighted diagonals plus main diagonal plus R1C3,C2C2,R3C1 = 3 sets of 1-9). That plus the colouring would then have given the location of purple in box 8.
This close up view of Simon's face in this video certainly does illustrate the mad delight he revels in - when solving a puzzle that would reduce me to a puddle of tears and gnashing of teeth.
Much quicker way to find the 78 pair in box 1: the 8 + 41 + 17 diagonals, plus the main diagonal from top right to bottom left, account for 8 out of 9 from the cell-3, cell-5, and cell-7 disjoint sets (all but box 1). And they add to 111, so the cells in box 1 that finish those sets must be a 789 triple to make up the remaining 24.
At 40 minutes, a much easier way of establishing the 78 pair in box 1 is to maths on the diagonals. 8+41+45+17 = 111. Due to the disjoint sets constraint, these diagonals must contain 3 sets of the digits 1-9. Therefore the blank diagonal in box 1 must equal 135-111 (24). 9 is already placed, so gives us the 78 pair. It’s actually available from the beginning.
Happy to have completed this while watching a couple of movies. I also colored 23 pairs and did almost as much sudoku on them as on the 78 pairs (but couldn't take them to the end like the 78s). I got the 78 pair in box one more simply with maths. The diagonals are a total of 3 disjoint sets totaling 3 x 45. Subtract the main diagonal of 45 minus the 17 minus the 41 minus the 8 minus the 9 in the box leaves 15, hence a 78 pair.
I'm amazed Simon didn't find the 78 by maths with the secret. 17 on boxes 2 and 4, 45 on boxes 3, 5, and 7, 41 on boxes 6 and 8, and 8 on box 9 makes 111. Together with box 1 they should add to 135, therefore, box 1 needs 24, and the only way is 789.
I thought I would be able to do this puzzle - I couldn't. I watched the video and followed along. I had to rewatch parts of the video several times to understand some of the logic. A very difficult puzzle.
That was tough, but I was always able to make progress, which kept frustration at bay, and some of the logic was very enjoyable. My solve was broadly similar to Simon's, although I seem to be able to see the disjoint sets a bit easier. I thought the logic with the disjoint sets, particularly around the box diagonals, was especially clever in this one, allowing us to eliminate candidates from some cells even though the number hadn't been placed yet.
28:00 (or earlier) the 8-diagonal in box 9, 41-diagonal in box 8-6, the "secret" 45-diagonal in box 7-5-3, the 17-diagonal in box 2-4, and the corresponding unknown diagonal in box 1, all contain 3 sets of 1-to-9. the total is 3×45=135 the unknown diagonal in box 1 is 24, and must be from 7-8-9 24 and 41-diagonal can't have a 3, and the 45 have one. there must be 2 3s in 17+8-diagonals, but not in box2: there must be a 3 on the diagonals in box 4 and box 9 the 8-diagonal in box 9 must be from 1-3-4
Blimey that fried my brain! It’s not often I spot stuff that you don’t, but you could have got to diagonals in box 1 being 789 a lot quicker by subtracting the given diagonals in that direction summing to 111, which is 24 less than 135 (3 lots of 1-9 from disjoint sets from positions 3-5-7)
Idea for a new feature: The ability to make pencil marks in a different colour (red?), to indicate when a particular digit *cannot* go into a square. And maybe even "self-rubbing" (normal blue) pencil marks, which disappear when the matching "big" digit is placed anywhere in the same row, column or box as the pencil mark. (But perhaps that is too close to the computer solving the puzzles all by itself .....)
I might be better at spotting the disjoint sets and screaming at my screen about them, but the logic you use with them is still far beyond what I can do!
Great job!! As usual I cursed a few times whenever i found some stuff way earlier and even when you where close you missed it, but jeez, you found everything and i found maybe half :) Insane setting btw!! Great puzzle.
39:30 the 789 diagonale in box 1, u can see from the beginning. all diagonales of all 9 boxes, are 3 sets of all numbers, and big sudoku hint, all numbers 1 to 9 = 45 times 3 = 135, and main diagonale 45 + the 3 hints 41,8,17 summs to 111 and left 24 for the top left diagonale
39:30 I had a simpler way of discovering the 78 pair here. I noticed that the bottom-left/center/top-right cells in all the boxes except one fell on either a marked diagonal or a main diagonal. This allowed me to sum up those diagonals. Since the set of all the bottom-left cells in a complete set of the digits 1 to 9, and the same is true of the set of all the center cells and the set of all the top-right cells, I knew I was working with 3 complete sets of the digits 1 to 9, minus the three of those digits that appear in box 1. Using the Secret, we know the sum of 3 complete sets of the digits 1 to 9 is (45x3) = 135, from which we can subtract 17, 41, 8, and (because the Secret applies to the main diagonal) 45, which leaves us with 24. The only way to make 24 in 3 cells that are all in the same box is with 7, 8, and 9. We already have 9 in r3c1, so r2c2 and r1c3 must be a 78 pair. QED.
57:12 for me. It took me quite awhile (35+ minutes) to get to coloring possible 78 pairs, but things started moving quicker once i did that. It enabled me to fill in all the 6's. Fun puzzle as usual for CTC
Imagine how much less joy there would be in the world if Simon Anthony only did crossword puzzles? What a horrific tragedy that would be. Watching Simon solve this particular puzzle counteracts all the suffering in human history.
Using the disjoint sets on the diagonals to find the 134 triple in box nine was really clever, and it set everything going. It was funny how getting into that state kept Simon's mind sort of looping in that disjoint thought process. Great solve! On of the best puzzles here on CtC, Lizzy01!
EDIT: This comment is incorrect... Disregard. 22:33 I am going to teach you something about disjoint sets... Not that it's going to help at this point of the solve but it's still worth mentioning. Consider 7's and 8's... Where can they go in cell #7 in each box? Your 78 pairs in column 1 and row 9 eliminates a lot of 78's from cell #7 in boxes 1, 4, 7, 8 and 9. So in the final solution, there MUST be a 7 and an 8 in the cells R3C4, R3C7, R6C4 and R6C7. That is a virtual X-wing. So you cannot have any more 7's and 8's in those affected rows or columns. So i suggest you keep an eye out for any similar patterns. Because they might matter.
Interestingly from the very start I think you can conclude that the 7-8-9 diagonal in box one is 7-8-9 since all of the other diagonals that make up the whole disjoint set for those boxes are all clued.
17:13 ... and you have a 9 in both doubles, so they form an "X-wing equivalent". That kills the middle top 9 pencil mark in box 3, forcing the 9 there to be in the top left corner. That forces the 9 into the middle of box 6 (column rule), which forces the 9 into the bottom left corner of box 1 via the disjoint madness, which forces the 9 in box 4 to be in the middle down (again, column) and thus the other 9 in that "42 diagonal" (in box 8) to be the left cell of the center line of that box (via again, the disjoint madness). Lots of nines pretty hidden in plain sight from an expert solver :D
20:00 Which gives me idea what would happen if I removed the diagonals constrains and forced the correct solutions with a few more given digits instead. That would force you to have to see the "warped X-wing madness" to solve the 9-s.
Um, i tried to play the puzzle ... and I realized that I am wrong in this comment. The "X-wing equivalent" is shifted a row down in its box, so that line of reasoning would fail miserably (it does not hit the top 9 there). But using an "X-wing equivalent in the disjoint madness" to disambiguate such a set of "forced 9 pairs" (instead of an easy to see diagonal) would make for really neat puzzle. I may try to construct such a puzzle but it will take a lot of time (as my brain alone is a very poor tool, I need an assistance of a computer for this).
The point was to try and ignore the diagonal line that you used to crack the nines and see where I get. Then I (after making a clear grid with only the nines pencil-marked) realized the shift of the "X wing madness" pattern.
And it actually uses the "X-wing madness" to disambiguate the stuff, it just uses a diagonal to drag two of these monsters into killing that one of the nines in the center box. Neat.
The 78 pair at 39:33 was easier to get. Take all of the diagonals going up/right. There is the 17, 8, 41, long 45, and the unnumbered in box 1. 8+17+41+45 = 111. The sum of the all must be 3 lots of 45, so 135, meaning box 1 is 24. Must be a 7 8 9
45:00 For some time now purple is excluded in box 5 from top-right because of box 1, and therefore placed in bottom-left. Simon warned that he has trouble scanning the disjoint set. He was right 😎. This is merely an observation and no criticism of Simon, heaven forfend. I am not worthy to set his operational parameters.
Good use of the 'Pigeonhole Principle', to restrict the number of ones in the 17 diagonal, and again to determine the 8 diagonal hasn't got a 2 in it. What is the Pigeonhole Principle, you ask? Simple. If you have more letters than pigeonholes, you cannot place the letters in the pigeonholes such that each pigeonhole has at most one letter.
There was some logic on the 7-8 pairs that I finally got. If a 7-8 pair goes in the middle of a box exactly one other 7-8 pair will go in the middle of another box. In fact this has to happen for all 9 positions. So somewhere in the grid two 7-8 pairs will go in the top left corner of a box and so on.
Whenever they get a disjoint set puzzle they should just focus on that before going off on other tangents. It seems like the second they go off on the normal path they completely lose track of the ruleset and never get back on track. The rule is added because it is crucial and should be scanned first whenever you get a digit
Cool Thumbnail (is that a representation of the Antithikarian Device, or an exploded version of a Watch/Clock.?). Very cool, either way. :) 😅 [[PS: an "Exploded 'View' " I think I meant to say. I'll explain the Antithakarian Device (or however it's spelled) if it's needed, lol. [It was discovered on a shipping route (possibly from Greece ) from the old days [[in a Shipwreck]] ( so probably from a Greek Mathematician/Inventor who made it for someone) deep under the ocean (or somewhere on a Bay/Ocean floor) to show the different seasons , other astronomical events, the phases of the moon of course, and probably the phases/placements of the sun and planets in the sky, - it was a Calender/Clock in a way though - that several people/Engineers have made working models/duplications of it from it ] That's what that is/was sorta' though, lol. [Look it up lol] I would say an Exploded View of a Clock/Watch Calendar though (like a Pocket Watch for a Watch example [[improbable though]] ). Very cool Thumbnail :) 😅 ]]
@@Jeff121456 , I was just looking, if I turn my device sideways, and enlarge the screen, I lose that front face. It looks like Roman numerals though (or 12 designations) so maybe a clockface (without the hands). At first (after I looked this 2nd time) it looks like there might be a second designation on the very inside (spaced out for a 5 or 7 day workweek). I think it's just Roman numerals of different lengths though (so a clockface). [[But I can't be sure]] Lol, Cheers. :) 😅 [[PS: I'll have to take a screenshot and enlarge it (maybe I'll be able to see) lol ]]
@@stevesebzda570 I am looking at it on a 50" TV screen and it most definitely is outer ring= day of month, middle ring= month, inner ring= day of week.
@@Jeff121456 Cool. How do they divide 7 days (or 5 days if earlier) into that many designations though.? They double up on a couple.? Thanks. Cool. :) 😅 [[PS: Ah, I just realized, 14 designations then maybe. Cool]]
52:35 I kind of not notice the plane before mentioned. I just believed it was here. Because I live just 2-3 km from the local airport, and it is not uncommon to hear a plane like that, but not like it is annoyingly common either.
![[LIVE] : ONE ลุมพินี 86 | คู่เอก "คมเพชร vs ชาติพยัคฆ์"](http://i.ytimg.com/vi/m3q_dRYDuU8/mqdefault.jpg)
I'm so excited to see Simon solve my puzzle! and an hour long, too! Now to watch the video...
And if the title is anything to go by, he likes it too! A dream come true, Thank you Simon!
it took an hour to solve but how long did it take to make it?
Thank you very much!
This variant was really fun. I love how it required a bunch of tricks to solve, but at the same time, it wasn't insanely hard. Thank you so much for sharing it!
Brilliant work, Lizzy. Congratulations on putting this together. And I know I'm not supposed to say this, but it's always a pleasant surprise for me to be reminded that women (and you too Clover) can design a puzzle as clever and original as anything a man can do.
Your puzzle was beautiful, Lizzy! Thank you :))
You can really tell how disjoint sets trip up experienced solvers because it's a dimension that they're not used to using. For those of us with less experience, it's no more or less unusual than the regular types of scanning.
Simon, you got that 7-8 pair in box 1 in a rather difficult way. There's an easier way to get that.
Because of the disjoint sets you actually know the value of the bottom left to top right cel diagonal in box 1.
Those same 3 cells in every other box have a little killer clue looking at them.
The value of all the bottom left to top right cells in all boxes combined is 3× 45 because of the disjoint sets rule. You have a 17 clue, an 8 clue and a 41 clue for those cells and you know the bottom left to top right diagonal in the grid is 45. Therefor the value of the diagonal in box 1 = 3×45 - 45 - 41 - 17 - 8 = 24, making it a 7-8-9 triple you could have pencilmarked before you filled in anything in the sudoku.
Then as soon as you find the 9 in box 1 you know the other 2 cells are that 7-8 pair.
Aye, did the same thing, summing all the forward diagonals leaves 24 for in box 1.
Next year's April Fools should be the inverse of this year's: an easy classic with a three-hour video.
Yeah, you can solve an easy sudoku using a long chain of very hard & weird techniques if you ignore the obvious steps
Okay, but imagine how interesting it would be to design an easy classic sudoku where it is possible to follow an extremely difficult path!
No
Next year april fool
Would be he does a genuine video
1hr + video
Just before he is about the crack the puzzle
The video stops
That would be one hell of a commitment tho I can't imagine anyone doing that
So, just record me doing a solve?
"And as an extra challenge, I'm going to ignore the digit '4'."
"and so far, it's been clever, but not impossible" famous last words
Looked at the video length when he said this. nods head in agreement with Cameron
Run times like this are always beautifully terrifying
My favorite thing is the excitement in Simon and Mark's voices when they are making a break through.
Mine too.
The best reasoning for the 17 diagonal is this: 17 diagonal, 8 diagonal and the main diagonal consist only of three different squares in their respective boxes and thus can contain at most three of each digit. Since main diagonal contains exactly one of each digit, we are left with at most two of each digit on the 17 and 8 diagonal together. They consist of 9 digits thus the minimum is 2*1+2*2+2*3+2*4+5=25 which is precisely what their sum is. So there have to be two ones, twos, threes and fours as well as one five.
Or you do the maths on the diagonals that consists of 3 disjoint sets. So 17+45+41+8+(diagonal in box 1) = 3*45 giving us 24 for the diagonal in box 1 - I admit I have only seen this after seeing the solve.
@hydra147147 That is the intended reasoning!
I haven't watched the video yet, only done the puzzle and I did exactly this.
I missed the interaction between the diagonals but I was still able to get the 9 worked out and a good break into the puzzle. Loved it, brilliant logic with how the disjoint sets and diagonals work together!
@@lizzyvv447 The end is so pleasing with the 7/8 pairs and the only at first glance deadly 2/5 pair. Beautiful how the the 22 diagonal resolves the 2/5 and the 7s and 8s fall at last. Thank you very much!
54:00 Simon notices the forbidden positions of the 6's in any box except 3 and 7, and instantly forgets that, which could have done a great deal of job in box 1 with the diagonal. This would reveal the position of all 6's.
Making some popcorn to enjoy this one 😎 much love from Brazil Simon, happy Easter!
That was definitely a brilliant puzzle and a very beautiful ending. There was also a lot of logic we could learn from and I was very happy to see the video was over an hour. Simon, I think you did it brilliantly and I do agree it looks like a very difficult puzzle but still a very elegant solve.
Thank you!
I'm SO glad I didn't give up on this one!
The solve was so satisfying!
Took me 115 mins, but still, worth every minute.
My solve was pretty much exactly the same as Simon's up until 1:01:20,
Kudos to the constructor for creating such a linear path.
Once we've done all we can with the 5,6,7,8,and 9s,
All the remaining cells contain only the numbers 1,2,3 and 4 (apart from an X-wing of 5s)
At this point, I changed my 78 coloring to light grey and black, and used 4 colors to mark all the remaining cells,
This helped disambiguate the remaining numbers (the new multi-color function in particular is a very handy tool).
One look at the video's time, and my heart sank. Nonetheless, I gave this a shot ... and darned if I didn't solve it successfully, with no guesswork, in under an hour (53:03 to be exact).
I *loved* this puzzle!
Hi! Yes! Another hour relaxing after dinner...love it! Thanks Simon!
Simon, you DO need a tool to automatically highlight all the cells in the same positions!
@Simon - you could have found the 789 triple in box 1 right at the start. Not the sum of all the / diagonals has to be 3 * 45 the ( 3 sets of 1-9) . You were given the sums of all the other diagonals, leaving box 1 to be 24.
Thank you for saving me from having to write that.
There is an easier way of finding the 7,8-pair in box 1. The diagonals marked with 8,41,17 add to a total of 66. Adding the main diagonal gets us to 66+45=111. Because of the relative position constraint we know that the diagonals that we have counted, along with the diagonal in box 1 contain exactly 3 sets of the digits from 1-9. Therefore the sum of the diagonal in box 1 is 45*3-111=24. The only way to make 24 with three separate digits is 7,8,9 :)
Anyone else screaming while pulling there hair out around 44:06when Simon puts 2 possible purple cells in box 5 and not paying attention to the purple cell in box 1. Which would give you the purple cell in box 2 then box 3 and then box 6. lol, but it's still fun watching.
Yes I could not help but scream at the television
Also slightly frustrated because by 55:00 I had worked out that since the middle and bottom-right squares had sixes in boxes 3 and 7, the main diagonal starting in r1c1 needed to have a six in the upper right of either box 1 or 4. Since 6 in row 1 was assigned in box 2, that means the only place in that diagonal a 6 could go was in r4c4.
He pre-warned us that he won’t see disjoint sets and that this will drive us (as the puzzle title says) to disjoint madness
Main diagonal combined with disjoined sets was incredible setting!
That feeling when CTC has to think about writing their 3rd book before the first one is released. The puzzles that have been featured in the last couple of months have been sublime.
Book two will be a brick, won't it?
Simon: What do I do now?
Me: I know the answer. Watch Simon solve the puzzle.
Not only stayed tough throughout, but several beautiful logical steps buillt deep into the solve. Very nice.
Top class puzzle, what a brilliant construction, congratulations to the setter!
I appreciate that you take on this challenging rulesets even when solving for an audience! Thanks for your hard work
I'm impressed that I got the 42 and 15 break in with the 7 8 9 s before Simon had even finished his introduction.
What possessed me to try a puzzle that took Simon over an hour?! My time was 252:17!
I think Lizzy's puzzle was fun and challenging, but not impossible (even for someone who is barely competent like myself), so I'm really thankful. The combination of disjointed sets and the main diagonal constraints manipulated the way the numbers saw each other in a really cool way. It created a lot of interesting and varied types of logic, so I got to use a lot of tricks (not just the same trick over and over again) throughout the solve which made the puzzle really fun.
I found the 78 pair in box 1 pretty fast because I just added up all the positive-sloping middle diagonals (8+41+45+17+9=120) and subtracted it from 3 sets of 1-9, which I've learned is 3*45 (135) from watching this channel for a difference of 15. Since there is already a 9 in box 1, the other two cells had to be a 78 pair. But then I started working on the 41 clue and did my math wrong, so it took me forever to figure out lol. After the 3-hour mark, I was worried I'd get to the end and find I had made a mistake, but thankfully everything worked out and when I watched the video I could totally relate to Simon's excitement and satisfaction when he colored in all the 78 pairs!
With the puzzles getting increasingly harder and the video lengths getting increasingly longer, I've found it a lot more pleasing to go back into the archives where they tackled some easier puzzles. I just watched a puzzle from 3 years ago where Simon intros an apology for the hard puzzle and the video was only 41 minutes long. My how times have changed.
A very impressive solve and very impressive puzzle. Amazing work by Lizzy 😊
The Disjoint rule on the Green/Purple section at around 45:00 eliminated every extraneous option but wasn't spotted until 47:15. It's such a tricky rule.
Yeah, saw it too; eventhough I‘m not that good at sudoku (compared to Simon, at least xD) I can spot patterns and symmetry rather easily. Because of that I know it‘s unfair, but I still couldn‘t stop screaming atz my screen 😅 But I still love this video, especially the complexity of the puzzle :D
Disjoint is def his weakness. Which is odd for how he can logically work thru these difficult puzzles. Like saying Ali has a weakness.
Would you say this frustrating phenomenon could be called Disjoint Madness?
@@Schambes yep I'm still in minute 45 and have been screaming for a few but I'd have never gotten this logic so I dint really have room to be too critical. Actually I like to come check the comments to make sure I'm not screaming at something that's not there. Also gives me a chance to let my blood pressure go down.
The relief at 47:30 when he finally spotted the disjoint set constraint on r4c6! I was screaming "disjoint set" at my monitor for a few minutes there!
The second time I tried it took me slightly less than two hours, not looking at the video and only logic! Thank you very much!
i just watched the video to see how long it would take you to realize that the diagonal eats up one of each numbers for the 17,41,8 clues. It blew my mind when i was solving.
Was so painful to see him mark all of the diagonals and not noticing that almost all of them are marking the same sets
Wow, that was insane! I got the initial break in at the bottom left without any trouble, but the rest of it just left me broken ... I could have stared at it for _weeks_ without thinking about how the leading diagonal used up digits from the disjoint sets ...
"So far this is not too difficult". *over an hour to go*
Famous last words :D
But he did consciously avoid "cooking with gas".
all of the big triangles with the 1234 digits near the end of the solve made me very happy, after seeing some incredible stuff earlier in the puzzle it still had more left to give!
One of my favourite puzzles so far. 2hr solve for me and I'm very happy with that. It's not just difficult because of the intricate logic - you also have to use very different logic at different points which makes it hard to spot.
The logic with the 9s was exquisite.
I love it when Simon genuinely celebrates ! (not that I've seen him fake it afaik)
The much more simple logic to figure out 39:40 that on the diagonal of the first box we have 7-8 pair:
Look at 41, 8, 17 diagonal and diagonal of the first box. it consist of 3 sets of digits 1-9(because this is a right-top corner, center, and left-bottom corner of each box.
But we know sums of all diagonal except one in the first box: main diagonal consists of 1-9 set, so we have 3*45 = 135 - 45 - 41 - 8 - 17 = 24. So the diagonal in the first box sums up to 24 and it is only one way to do it by using 7, 8, 9.
I was somehow able to solve this one all on my own. Yes it took me coming back to it over the course of several days, but I was able to do it. This channel has taught me so much.
Same here. I started it when it first came out and finished 3 days later lol
I really enjoyed the disjointed rule. That's a very interesting mechanic
All th diagonals sum to 135 and the central diagonal sums to 45, so you can find the diagonal in box one by sums.
This is how i did it too.
Yeah, I was constantly screaming at Simon: come oooon, just add up all the diagonals!!!! Pleaseee!!!
ABSOLUTELY UNBELIEVABLE! You are the most amazing logician.
There's an easier way to get the 78 pair in box 1: sum the diagonals 17+45+41+8=111. Add the same diagonal positions in box 1 gives three lots of 45, which is 135. 135-111=24, which must be 7,8,9.
(That might have made it easier to get the composition of the 41 diagonal as well: two sets of digits 1-9, minus those diagonal cells in boxes 1,2,4, and 9.)
Simon *almost* had the right set of cells highlighted at 24:37 to apply the set theory logic (highlighted diagonals plus main diagonal plus R1C3,C2C2,R3C1 = 3 sets of 1-9). That plus the colouring would then have given the location of purple in box 8.
This close up view of Simon's face in this video certainly does illustrate the mad delight he revels in - when solving a puzzle that would reduce me to a puddle of tears and gnashing of teeth.
Much quicker way to find the 78 pair in box 1: the 8 + 41 + 17 diagonals, plus the main diagonal from top right to bottom left, account for 8 out of 9 from the cell-3, cell-5, and cell-7 disjoint sets (all but box 1). And they add to 111, so the cells in box 1 that finish those sets must be a 789 triple to make up the remaining 24.
Really glad to see this one featured. I played this one awhile ago and it was one of my favorites, and was sad that it didn’t get rated
At 40 minutes, a much easier way of establishing the 78 pair in box 1 is to maths on the diagonals. 8+41+45+17 = 111. Due to the disjoint sets constraint, these diagonals must contain 3 sets of the digits 1-9. Therefore the blank diagonal in box 1 must equal 135-111 (24). 9 is already placed, so gives us the 78 pair. It’s actually available from the beginning.
It never gets old to see Simon celebrating at the end of every puzzle ❤️😂😂. Loved it!
Happy to have completed this while watching a couple of movies. I also colored 23 pairs and did almost as much sudoku on them as on the 78 pairs (but couldn't take them to the end like the 78s). I got the 78 pair in box one more simply with maths. The diagonals are a total of 3 disjoint sets totaling 3 x 45. Subtract the main diagonal of 45 minus the 17 minus the 41 minus the 8 minus the 9 in the box leaves 15, hence a 78 pair.
I did it! 1:12:02 . I wanted to give up so many times.
What an amazing puzzle! Simon clearly had a blast as well, so satisfying
6-green pair is my new favorite color.
I'm amazed Simon didn't find the 78 by maths with the secret. 17 on boxes 2 and 4, 45 on boxes 3, 5, and 7, 41 on boxes 6 and 8, and 8 on box 9 makes 111. Together with box 1 they should add to 135, therefore, box 1 needs 24, and the only way is 789.
Yeah, he ended up with the same result but definitely did it the hard way.
I thought I would be able to do this puzzle - I couldn't. I watched the video and followed along. I had to rewatch parts of the video several times to understand some of the logic. A very difficult puzzle.
Rules 1:47
Let’s get cracking 4:00
That was tough, but I was always able to make progress, which kept frustration at bay, and some of the logic was very enjoyable. My solve was broadly similar to Simon's, although I seem to be able to see the disjoint sets a bit easier.
I thought the logic with the disjoint sets, particularly around the box diagonals, was especially clever in this one, allowing us to eliminate candidates from some cells even though the number hadn't been placed yet.
28:00 (or earlier)
the 8-diagonal in box 9, 41-diagonal in box 8-6, the "secret" 45-diagonal in box 7-5-3, the 17-diagonal in box 2-4, and the corresponding unknown diagonal in box 1, all contain 3 sets of 1-to-9.
the total is 3×45=135
the unknown diagonal in box 1 is 24, and must be from 7-8-9
24 and 41-diagonal can't have a 3, and the 45 have one.
there must be 2 3s in 17+8-diagonals, but not in box2: there must be a 3 on the diagonals in box 4 and box 9
the 8-diagonal in box 9 must be from 1-3-4
but then I got stuck...
@39:40 "Thats a 78 pair". Unbelievable! Sometimes I feel like my IQ is just 78 when I watch Simon solve these puzzles.
Blimey that fried my brain! It’s not often I spot stuff that you don’t, but you could have got to diagonals in box 1 being 789 a lot quicker by subtracting the given diagonals in that direction summing to 111, which is 24 less than 135 (3 lots of 1-9 from disjoint sets from positions 3-5-7)
Idea for a new feature: The ability to make pencil marks in a different colour (red?), to indicate when a particular digit *cannot* go into a square.
And maybe even "self-rubbing" (normal blue) pencil marks, which disappear when the matching "big" digit is placed anywhere in the same row, column or box as the pencil mark. (But perhaps that is too close to the computer solving the puzzles all by itself .....)
I might be better at spotting the disjoint sets and screaming at my screen about them, but the logic you use with them is still far beyond what I can do!
Great job!! As usual I cursed a few times whenever i found some stuff way earlier and even when you where close you missed it, but jeez, you found everything and i found maybe half :)
Insane setting btw!! Great puzzle.
39:30 the 789 diagonale in box 1, u can see from the beginning. all diagonales of all 9 boxes, are 3 sets of all numbers, and big sudoku hint, all numbers 1 to 9 = 45 times 3 = 135, and main diagonale 45 + the 3 hints 41,8,17 summs to 111 and left 24 for the top left diagonale
Chuffed to have finished this one, very clever puzzle, thanks.
The desire to shout "DISJOINTS!" was really strong at about 45:00
A typical Simon moment:
"You can't put 1s, 2s, 3s, and 4s there." (35:39)
38 seconds later: "So if we were to put 3 here...) (36:17)
39:30 I had a simpler way of discovering the 78 pair here. I noticed that the bottom-left/center/top-right cells in all the boxes except one fell on either a marked diagonal or a main diagonal. This allowed me to sum up those diagonals. Since the set of all the bottom-left cells in a complete set of the digits 1 to 9, and the same is true of the set of all the center cells and the set of all the top-right cells, I knew I was working with 3 complete sets of the digits 1 to 9, minus the three of those digits that appear in box 1. Using the Secret, we know the sum of 3 complete sets of the digits 1 to 9 is (45x3) = 135, from which we can subtract 17, 41, 8, and (because the Secret applies to the main diagonal) 45, which leaves us with 24. The only way to make 24 in 3 cells that are all in the same box is with 7, 8, and 9. We already have 9 in r3c1, so r2c2 and r1c3 must be a 78 pair. QED.
57:12 for me. It took me quite awhile (35+ minutes) to get to coloring possible 78 pairs, but things started moving quicker once i did that. It enabled me to fill in all the 6's. Fun puzzle as usual for CTC
So glad to finish a puzzle faster than Simon!
1:04:54 finish. Wow, what a beast. Finding the 7-8 pairs was fairly easy, and then the REAL work began.
Imagine how much less joy there would be in the world if Simon Anthony only did crossword puzzles? What a horrific tragedy that would be. Watching Simon solve this particular puzzle counteracts all the suffering in human history.
"This is a 6-green pair". WAT? :)
Using the disjoint sets on the diagonals to find the 134 triple in box nine was really clever, and it set everything going. It was funny how getting into that state kept Simon's mind sort of looping in that disjoint thought process. Great solve! On of the best puzzles here on CtC, Lizzy01!
EDIT: This comment is incorrect... Disregard.
22:33 I am going to teach you something about disjoint sets... Not that it's going to help at this point of the solve but it's still worth mentioning.
Consider 7's and 8's... Where can they go in cell #7 in each box?
Your 78 pairs in column 1 and row 9 eliminates a lot of 78's from cell #7 in boxes 1, 4, 7, 8 and 9.
So in the final solution, there MUST be a 7 and an 8 in the cells R3C4, R3C7, R6C4 and R6C7.
That is a virtual X-wing. So you cannot have any more 7's and 8's in those affected rows or columns.
So i suggest you keep an eye out for any similar patterns. Because they might matter.
Interestingly from the very start I think you can conclude that the 7-8-9 diagonal in box one is 7-8-9 since all of the other diagonals that make up the whole disjoint set for those boxes are all clued.
4:29 --> the secret!
Gorgeous puzzle. I've got to start trying (and failing at these).
He was 100 percent right when he said I would spot things he wouldn’t, I was yelling “ the purple is right there!” For a second😂
17:13 ... and you have a 9 in both doubles, so they form an "X-wing equivalent". That kills the middle top 9 pencil mark in box 3, forcing the 9 there to be in the top left corner. That forces the 9 into the middle of box 6 (column rule), which forces the 9 into the bottom left corner of box 1 via the disjoint madness, which forces the 9 in box 4 to be in the middle down (again, column) and thus the other 9 in that "42 diagonal" (in box 8) to be the left cell of the center line of that box (via again, the disjoint madness). Lots of nines pretty hidden in plain sight from an expert solver :D
20:00 Which gives me idea what would happen if I removed the diagonals constrains and forced the correct solutions with a few more given digits instead. That would force you to have to see the "warped X-wing madness" to solve the 9-s.
Um, i tried to play the puzzle ... and I realized that I am wrong in this comment. The "X-wing equivalent" is shifted a row down in its box, so that line of reasoning would fail miserably (it does not hit the top 9 there). But using an "X-wing equivalent in the disjoint madness" to disambiguate such a set of "forced 9 pairs" (instead of an easy to see diagonal) would make for really neat puzzle. I may try to construct such a puzzle but it will take a lot of time (as my brain alone is a very poor tool, I need an assistance of a computer for this).
The point was to try and ignore the diagonal line that you used to crack the nines and see where I get. Then I (after making a clear grid with only the nines pencil-marked) realized the shift of the "X wing madness" pattern.
And it actually uses the "X-wing madness" to disambiguate the stuff, it just uses a diagonal to drag two of these monsters into killing that one of the nines in the center box. Neat.
It was painful to watch , but still a joy to see Simon's reaction on finishing it
The 78 pair at 39:33 was easier to get. Take all of the diagonals going up/right. There is the 17, 8, 41, long 45, and the unnumbered in box 1. 8+17+41+45 = 111. The sum of the all must be 3 lots of 45, so 135, meaning box 1 is 24. Must be a 7 8 9
I like when ctc honours me with their colouring
Fitting for the best turtle of the team!
45:00 For some time now purple is excluded in box 5 from top-right because of box 1, and therefore placed in bottom-left. Simon warned that he has trouble scanning the disjoint set. He was right 😎.
This is merely an observation and no criticism of Simon, heaven forfend. I am not worthy to set his operational parameters.
One of the best puzzles I've ever solved.
Superb solve of a supremely tricky puzzle.
Very nice puzzle and logic!
Oh Bobbins! I thought for a bit there you would give up. Well done.
Yoo, first time this early. Eager to be amazed by Simon
Very nice puzzle. 72:46 for me; very pleased with that time.
Good use of the 'Pigeonhole Principle', to restrict the number of ones in the 17 diagonal, and again to determine the 8 diagonal hasn't got a 2 in it.
What is the Pigeonhole Principle, you ask? Simple. If you have more letters than pigeonholes, you cannot place the letters in the pigeonholes such that each pigeonhole has at most one letter.
1:06:07 Very beautiful puzzle. The disjoint sets were hard, but not unpleasant and so strangely 'harmonious' with the rest of the constraints.
I'm amazed how far Simon got before realizing that the digits in R1C3 R2C2 had to be 7 & 8 due to the diagonal totals.
Now I see that at 40:14 he doesn't even do the adding ! 17+8+45+41 = 111 and 135 - 111 = 24
I am bowled over.
Very impressive, Simon. You teach me to try variants that I really do not care for. I like disjointed, but I am not crazy for killers or sandwiches.
You did say at the beginning you weren't very good with disjoint rules. True true.
There was some logic on the 7-8 pairs that I finally got. If a 7-8 pair goes in the middle of a box exactly one other 7-8 pair will go in the middle of another box. In fact this has to happen for all 9 positions. So somewhere in the grid two 7-8 pairs will go in the top left corner of a box and so on.
Whenever they get a disjoint set puzzle they should just focus on that before going off on other tangents. It seems like the second they go off on the normal path they completely lose track of the ruleset and never get back on track. The rule is added because it is crucial and should be scanned first whenever you get a digit
Cool Thumbnail (is that a representation of the Antithikarian Device, or an exploded version of a Watch/Clock.?).
Very cool, either way. :) 😅
[[PS: an "Exploded 'View' " I think I meant to say.
I'll explain the Antithakarian Device (or however it's spelled) if it's needed, lol.
[It was discovered on a shipping route (possibly from Greece ) from the old days [[in a Shipwreck]] ( so probably from a Greek Mathematician/Inventor who made it for someone) deep under the ocean (or somewhere on a Bay/Ocean floor) to show the different seasons , other astronomical events, the phases of the moon of course, and probably the phases/placements of the sun and planets in the sky, - it was a Calender/Clock in a way though - that several people/Engineers have made working models/duplications of it from it ]
That's what that is/was sorta' though, lol.
[Look it up lol]
I would say an Exploded View of a Clock/Watch Calendar though (like a Pocket Watch for a Watch example [[improbable though]] ).
Very cool Thumbnail :) 😅 ]]
A close look at he face shows day month and day of the week rings. So it is a calendar. Though not as complex as the Antikythra device.
@@Jeff121456
Lol, I knew i was spellin' that wrong.
Thanks:) 😅
@@Jeff121456 , I was just looking, if I turn my device sideways, and enlarge the screen, I lose that front face.
It looks like Roman numerals though (or 12 designations) so maybe a clockface (without the hands).
At first (after I looked this 2nd time) it looks like there might be a second designation on the very inside (spaced out for a 5 or 7 day workweek).
I think it's just Roman numerals of different lengths though (so a clockface).
[[But I can't be sure]]
Lol, Cheers. :) 😅
[[PS: I'll have to take a screenshot and enlarge it (maybe I'll be able to see) lol ]]
@@stevesebzda570 I am looking at it on a 50" TV screen and it most definitely is outer ring= day of month, middle ring= month, inner ring= day of week.
@@Jeff121456
Cool.
How do they divide 7 days (or 5 days if earlier) into that many designations though.?
They double up on a couple.?
Thanks. Cool. :) 😅
[[PS: Ah, I just realized, 14 designations then maybe. Cool]]
The 1 logic at 27 minutes is amazing!
52:35 I kind of not notice the plane before mentioned. I just believed it was here. Because I live just 2-3 km from the local airport, and it is not uncommon to hear a plane like that, but not like it is annoyingly common either.