I always like Simon solving symmetrical puzzles because he always finds two different ways to solve the puzzle instead of just doing the same thing twice.
I found it strange how long it took him to place the 13 pair, when he had aldready placed the symmetrical 14 pair and the logic applies the same way in both cases.
@@HappyLooter Exactly. Its ubelievable. He talks about symetry, then he finds a beautiful logic, and then he is not able to spot the exact same symmetrical logic in other box...
Thank you so much for mentioning Zelda's birthday! She pointed at the screen when you talked about her (although that might have been by coincidence). I find the tradition of congratulating viewers to their birthdays so charming.
There's an elegant break-in at the start that uses some SET. Highlight the two L-shaped cages. Those are two sets of 1-9, or 90 total. Remove the four circles in those cages and add the 3-cell lines in boxes 6 and 8 to compensate. Still 90. That leaves only one cell each in box 6 and 8 not highlighted, and the maximum they can be are 4 and 5 since the are in the quad, and the four highlighted cells in box 5 and 9 must equal the total of the un-highlighted cells in 6 and 8. The lowest the highlighted cells can be are 1,3 and 1,4, which equals the 4,5. That gives you 1,3 in column 4 of one quad, 1,4 in column 9 of another quad, and the positions of 4/5 vs 2/3 in the 2345 quad. The rest of Simon's logic about where the digits in the circles go flows smoothly from there.
I did something similar: highlight box 6, swap lines for the end circles and compare to L cage going through the box, proving: [8,9] + [9,9] = [6,7] similarly: [4,4] + [5,4] = [6,5] then likewise, you needed the values of 1,4,5 and 1,3,4
And surprisingly, I went to this SET break in almost instantly, by instinct. I am a bit surprised, knowing how much Simon loves using SET, he didn't do it this way. It was quite obvious to me, for some reason i dont know. maybe just lucky.
every puzzle is the hardest, the best puzzle of the year, the most brilliant puzzle ever made, etc etc etc don't read too much into youtube's click-bait meta. hyperbolic video names are a big aspect of big success on this platform. it's also in this case a direct quote from the LMD comments that are shown barely a minute into the video, so the clickbait was basically handed to them on a plate
@@nb2vcxz The puzzle was finished. The aim was to fully shade the grid in a certain way and to draw a path through the grid. Maybe read the rules of the puzzle?
I’ve seen a couple explanations of set theory, and I’ll add that I think it makes this puzzle much easier (though I’m normally very bad at identifying useful sets!). The 2 9-cell cages are one set, boxes 6 and 8 are the other. Eliminate all overlapping cells, and all lines/circles that have equal sums.
I have never seen such an intricate and at the same time beautiful *lattice* of geometric restrictions by clueless cages, and I totally understand the reason why *cam* wrote on LMG _"I am extremely confident I followed the intended solving path"._ This is a *world class masterpiece.* Every step, from start to end, was challenging, but never brutal and always divinely beautiful. Quads were magnificently used as disambiguators. How the hell can a human being be smart enough to conceive such a *divine logic creature?* For instance, how can just a bunch of neurons 💪🧠 conceive the multiple stratospherically clever and totally unexpected interactions between elementary restrictions that force circle in the UR corner of *box 3* = circle in the UR corner of *box 5* = just *4* or *5?* And that's only the first step in my solve, followed by dozens of equally mind-blowing others‼ I would love to see a video by Math Pesto about the developement of this apparently superhuman achievement. Thank you *Math Pesto.* Thank you *Simon.* 😏👍 You both deserve a standing ovation 👏👏👏👏👏👏👏👏👏👏👏👏👏👏👏👏👏👏
I am also fascinated by the beautiful *mirror-symmetric* arrangement of most of the clues (all of them except for the L-shaped cages and the line in box 5). The *quads* aligned along the axis of symmetry look like an *arrow* notched to the bowstring of a fancy professional arch. This is a *logic masterpiece* and a *beautiful picture* at the same time 😍👍
If you use SETS to compare the 9 digit cages with the BOX6 and BOX8 (all sum to 45) you can determine R6C7 is 5 and R7C6 is 4. You also determine the missing digit is 1 in both the quads 378 and 458,
34:36 for me. I'd vote for 3 stars difficulty, personally. Really fun puzzle. Simon would have made his life easier if he'd 1) done the same logic on the left L-nonomino that he'd just finished on the right one (the same logic that got him the 1 and 4 around the 458 circle and the 5 around the 2345 circle could immediately be used on the other side to get the 1 and 3 around the 378 circle and the 4 around the 2345), and 2) left his roping colours in the top three rows, which would have resolved the 10-pair in box 3 much earlier, by noticing that the 1 couldn't go in r3c7 because of the quad clue in box 1.
Yep - I always love watching his solves but I was willing him on to spot the 1/3 and 1/4 issue and that the logic was identical to the other cage, which I would never have spotted in a million years. Especially funny because 2 mins in he is excited about the symmetry of the puzzle :-)
Knowing where the 1-3 go in column 4 is pretty easy using this logic, and would have saved him probably 15 minutes on his solve. Like you said, it is the exact same logic that he found with the other 9-cell cage, so I don't see how he didn't notice it applied!
That's very typical for his solves. He almost always refuses to believe that the setter used the same trick more than once and goes to hunt for something else in the puzzle after solving one part of it. It does make his solves quite frustrating to watch because he is also quite sloppy with many of basic sudoku mechanics and doesn't follow a rigorous organized plan in his solves.
I finished in 94 minutes. That break-in was so beautiful. I surprised myself on how I was actually able to see it. The geometry of all at all made it so satisfying. Utilizing the surrounding circles clues in box 9 and 5 to disambiguate the non-caged cells in box 6 and 8 were incredible. I did that part quite fast only to get stuck near the end, but realized that roping became very powerful and easily finished the puzzle. This goes down as one of my favorites. I actually didn't find the difficulty to be that hard. It was hard, but nowhere near the hardest I have ever done. I loved this one. Great Puzzle!
17:43 for me. Fairly approachable once you spot the break-in. I would encourage people to give it a go despite the title of the video, 'cause it's a very enjoyable one to solve. Great puzzle!
27:37 for me. Haven't watched Simon's solve yet but I thought it solved beautifully. The clues flowed together perfectly and I never felt lost at any point.
At 40:00, 1 and 3 in box 5 are row 4, because they have to be in the cage, but they are outside the cage in boxes 8 and 9. Combine this with the deduction at 41:15 and you can place 1 and 3 and 7 and 8 in box 5
I couldn't understand why he wasn't applying the exact same logic he'd just applied on the other side, with the 1,4 pairs on the line in box 6 having to go in the 9 cell cage in r8-9c9. It's the same logic, just rotated! 😂
65:05 This was brutally, brilliantly, difficult. The instantly obvious roping made it look like it would be a bit simpler, but instead it was quite the grind of logical steps to chip away and this beauty's secrets until the roping appropriately tied up the final loose ends. A masterpiece.
33:08 and I echo some of the other comments, this never felt anywhere near the level of difficulty that I was lead to expect from the description. Some tricky scanning with the overlapping ropes, and I suspect if you miss the break-in you could probably go in circles for ages without getting much. Very interested to watch Simon now and see where he might have gotten held up. (quality puzzle tho! Don't get me wrong)
This video is proof once again that Simon is a master solver. Much like top mountain climbers pick the challenging routes to the peak that no one else dares try he consistently picks the harder route to a solve that us mere mortals are left in awe of. I found the box 6 and box 8 set break in that others have mentioned.
So I think I found a slightly easier method with set theory. The 9 cell cage and box 6 are both sets of all 9 digits. Take away the 5 shared in box 6. Then you can get rid of the double arrow. Now you are only left with 1 cell in box 6 which has to be 2, 3, 4, or 5 and the two bottom cells which have to now add up to 2, 3, 4, or 5. Can't add up to 2, and you have two use at least 1 of the quad of 4, 5, and 8 so the only thing you can do is 5 which adds up with 1 and 4. You can do the same with the other 9 cell cage and box 8.
I love watching Simon's solves. :) There was a simple deduction which should have come when he was placing orange and red in box 8 (20:32): the same rules applied to the 2 cells above orange in the cage.They had to be in that 2x2. When he used that principle to solve the gray/green situation (30:21) and didn't apply the same thing to orange/red when he switched back (34:28), I broke a little lol I had to wait until 52 minutes in. Love the video!
Oh, this puzzle was fun. I didm't find it that difficult, but it was very much a different style of difficult. Very intricate and it can absolutely wall you if you don't see some stuff.
there is a very easy break in. Take the L cage in boxes 5&8 as one group and subtract the box8. Then end subtract the double sided arrow and you end up with R4c4+R5c4=R7c6. Do similar for the other L cage and you get the 2345 positions, and marginal positions of the other circled digits, too.
I'm an avid user of the Vivaldi browser for its unmatched tab handling and memory saving features, and it has just increased my enjoyment of the CTC channel as well. Its page tiling feature makes it a cinch to display the video and the app for solving the puzzle side-by-side on a simple laptop screen, without having to manually adjust window sizes or anything. And when done with the puzzle I just turn the tiling off again. Easy as pie!
At 37:37, isn't there the exact same problem with R7C6 being 3 as was with the other side where R6C7 couldn't be anything other but 5? With R7C6 set to 3, R9C7 would be forced to be a 3 and then there has to be both 1 and 2 on the arrow in box 8 and they would be forced to be a pair in R4R5C4 which is impossible.
There was a much easier break in ...... do not read unless you want spoilers > > > > > > > > > > Using set on each of the two 9 cell cages against box 8 and box 6 respectively allows you to cancel 5 cells in the cages and cancel the two end cells against the three line cells leaving 1 cell equal in box 6 and 8 vs two cells in the cages. This forces 5 into the NE and 4 into the SW corners of the 2345 quad.
Do not worry about spoilers. This video is about solving a puzzle, so the discussion section is supposed to discuss the solve, and suggest alternative ways of solving. This is why I read this section after solving. People who do not want to be spoiled should just ignore the comments.
52:00 Finally the logic which has been done identically for the other 9-cage like 20 minutes ago. That would have made some of the logic in-between easier.
99:31 for me. Basically two internconnected puzzles (boxes 5,6,8,9 and boxes 1,2,3,4,7). The former solved very quickly but it took me quite a while to resolve the latter. Great puzzle.
53:03 finish. This puzzle was determined to try and break me. First, I kept forgetting that they weren't between lines. Then my computer froze multiple times and ran off about five minutes of my clock (so technically I finished in 45-50 minutes). I finally settled in and pushed through, however, and enjoyed that ending very much. I only colored the ropes that were outside of the 1-6 quad, to save colors and not make it harder to spot the individual ropes. Fun fun fun!
Always so interesting, Simon - thank you for this video! I thought about SET myself, (not while solving - I am not going to try this one) because of the layout of the grid. But you followed a great path, I think, and I enjoyed watching it.
It took me multiple sessions and over an hour's total time, but I was able to chip away at this one before finally completing this diabolical creation. Very challenging puzzle!
Took me a surprisingly-brisk 25 minutes or so. One interesting trick I came across was that regardless of whether the line in box three was 1&9 or 3&7, it resolves the 7&9 in box nine in the same order. In turn, having those digits early made some of the other logic along the bottom rows faster.
35:50 "If you can see what's going on..." shows Simon's genius. He can make such complicated deductions, but he's always looking out for them as well. Now in box nine there's that 58-pair. That one can be resolved easily. Imagine r8c8 getting the five. Then the red and grey squares would need to add up to five. They can't. There's that 14-pair in the box and r7c7 is either a two or a three. So then red and grey have to add up to eight instead. Three options for that: 17 (eliminated by the 14-pair), 35 (eliminated by the 5 that ends up in r9c8) and thus they must be a 26 pair. Since the red/orange square in r7c6 cannot be either a two or a six, it cannot be red and must become orange. Orange cannot be 3 because of the 378-circle in the center box.
11:15 The (16) circle can be useful exactly because you know there's roping. It means the row/column that sits outside the (16) clue can't have a 1 or a 6 in it in box 1, which then propagates to boxes 2/3 and 4/7. That is, neither 1 or 6 can appear in row 3 of box 3 and column 3 of box 7. You also know, however useful that is, that at minimum one of the cages in boxes 1-3 or 1-7 will have a 1 or a 6 in it. No matter how you twist and turn it, BOTH cages will never have 1 and 6 in them at the same time. Only one of the cages can have both 1 and 6 in it, the other is then forced to have at least one of the digits. Of course it's entirely possible for one cage to have one of 1 or 6 in it and the other cage having neither. I don't see how this would get the puzzle started but it's worthy of note.
Simon, you missed something fundamental that would have cut a half hour off your solve time. You discovered that R6C4 and R9C7 both must fit into the four squares in Box 8 that are not in the cage, but you failed to realize that R4C4 and R5C4 must also go into those same four squares. In other words, you could have known all four occupants. By so doing, you can immediately detemine which numbers they are due to the circles.
I think that Simon has tunel vision, this must be the reason he can't see some numbers while doing sudoku. He only focus on some box and forget the others. 😂 But I didn't get it why he didn't do the same logic in boxes 5 and 7 (the logic he used to find 14 pair in column 9, could be used to find 13 pair in column 4) 🤔... but well, he really likes to go to the hardest path!
67:04 for me. Enjoyed the puzzle. My solving times are often similar to the length of the video, so this puzzle felt like a normal CTC level of difficulty. Definitely not "the hardest puzzle I've ever seen".
You came really close to cracking this open around the 27 minute mark, where you were considering whether orange or red was in R7C6. You rightly pointed out that if it were red, it would have to be 2 or 3, and also concluded that it couldn't be 2 in that case. What you failed to spot, is that if it were a red 3, that would put both 1 and 2 into R4/5C4, leaving no room for the 378 that must be on the quad. Therefore orange is in R7C6 and is 45. This also means that the counterpart of orange on the line in box 8 must be 13, or 23, so the 3 on the quad is in C4, along with 1 or 2. This pushes the 78 on the quad into C5, and puts a 123 and a 78 on the line in box 5, which means the circles sum to at least 8, and R4C6 cannot be 2. You could now repeat the same logic on the similar construction in box 6. The R8/9C9 domino needs at least a 4 in it, which means that the counterpart to R6C7 on the bent line in box 6 must include 4 and R6C7 must be 5 with 14 on the line and in that domino at the bottom of C9. Now there's 58 in C8 on the quad in box 9, but 5 can't be on the one-cell line, so it's 8. Grey can't be 5, because red can't be 3, so green is 5, orange is 4, and R9C7/R7C9 are a 26 pair. Because orange is 4, the domino on the quad is 13. You now have enough to resolve all of the 2345 quad. The line in box 5 is 18, with 37 above it, which must go in R5/6C9, putting 68 in blue and 9 in purple. After this, it all becomes much easier, especially if you use the roping to place digits in R1-3 and C1-3. You could probably have cut 15 minutes off your time if you'd spotted what was available at 27 minutes, but it certainly wasn't an obvious thing to spot unless you can built a complex picture mentally and see where the contradictions lie. Fortunately, for my work I need to deal with chains of consequences so I've had plenty of practice building mental pictures. This was an incredibly intricate construction, which does require you to be able to keep lots of things in mind at once to solve efficiently. I'm glad you didn't forget the roping, because that was crucial to the end game. I find with roping that you only really need to colour one strand. The other two are then easy to visualise. When you've got two ropes crossing like here, it helps keep the grid relatively clean.
Still love you guys! Sorry for almost never watching, I feel like a faithless spouse - but I have a lot going on in my life. I am working on a tough maths problem that may well become a rather interesting paper. The spirit of patiently chipping away at a problem and rejoicing in occasional Aha-moments was reinforced by my love affair with CtC in the dark moments of lockdown. I haven't forgotten the warm friendship I had from a few very dear co-obsessives. When you see this comment, you will know who you are...
Funny how after talking about all the symmetry at the start Simon proceeds to utterly discard the logic of symmetry while resolving the squiggly lines and the quadruple clues. The 3 in r7c6 could be immediately eliminated with the exact same logic applied to r6c7 - and having the 4 in r7c6 meant that r45c4 was a 13 pair (same as before), and then 1378 could all be placed.
This one clicked for me pretty quickly, another day perhaps I would have ended up tearing my hair out. Very enjoyable, thanks. Unusually my path was quite different from Simon's, I used a more SET-like approach which I assumed was the intended path, but perhaps Simon's was.
38m08s. I feel like I had an early break-in with a cleaner version of the logic Simon used in 26-31 minutes. The four digits of box 6 that are not in the cage are the same as the four digits of that cage that are not in box 6, and so in particular they have the same sum. But the two circles add up to the three digits on the line, and so the remaining digit in box 6, R6C7, must equal the sum of the remaining two digits in the cage, R8C9 and R9C9. But the largest R6C7 can be is 5, and the smallest R8C9 + R9C9 can be is 1+4=5, so each is exactly 5. And then a similar argument puts 4 in R7C6, and you're off to the races
This was hard. Nice geometric effects here, and for many it took me quite a while to find them, even though I knew the theory from watching earlier videos. At some point, after I had pencilmarked the whole grid, I noticed that the circled 1 in box 1 has an impact on box 7, and then it fell apart. 102:53 for me (that's slightly under the total video length), solve counter 551.
Simon: when you find symmetry in a puzzle, and then you find logic, try to apply it to the symmetrical part of the puzzle. Also Simon: doesn't do that in this puzzle.
Solved this one in a little over an hour so definitely far from the most difficult on the channel. Was a little bit hesitant to start given the title but broke in fairly quickly.
How did you eliminate the possible for a 2 or 3 in box 6 cell 4/5/6? You said that the 5 in box 5 had to be made using 4 and 1 in box six. But couldn’t you have made the five with 2 and 3?
Whichever two digits digits add up to that 5, the only place they can go in the L-shaped cage is in the two bottom cells. Those two cells are on a 458 quadruple clue. If you were to use 2 and 3, you'd have to put five different digits (23458) in the bottom-right 2x2.
For those of you who are confused (as I was), the puzzle is NOT symmetrical apart from the kink. The nine cages are oriented differently. This is why thy break in works, and why the logic in the top right does not apply in the bottom left (and vice versa).
1:10:19 today. i had a very complicated start, i missed the obvious break in. then i struggled a lot in the middle and the ending with the triples was easy.
the timer ticked over 240 minutes today at work while working on it. i found the breakthrough myself which was nice, but i couldn't finish it after that
“That’s so annoying it’s not done anything!” - it would have if you’d done it 20 minutes earlier before getting the other stuff the hard way 🙈🙈🙈 (When you got the first 5, it was instantly “same logic puts 4 on the other side with 1 and 3 above, which fills out the 1378 because two have to add to 9” - which would have made the 869 in that row a lot easier! Impressed that you got it anyway, was almost convinced you’d disambiguate the four digits without noticing the repeated logic at all 🤣🤣🤣)
I don't understand Simon's mind. He tends to ignore obvious things, like how the digit at r3c7 affects r8c7 while exploring things that are beyond a normal sudoku solver.
I don't think it is as hard as other puzzles on this channel. It's a good geometrical puzzle. The break in is simpler to see if approached by comparing the totals of the nine cell cages and the totals of boxes 6 and 8. And I didn't get too stuck once I'd seen this.
I’m determined to solve this, I have 23 digits after 90 minutes of work, which I’m shocked at because I don’t normally do sudokus and the one is difficult
How can you successfully eliminate 3 as option from one side of a puzzle, with a high degree of symmetry, and not look to see if the same logic can be used to eliminate 3 on the otherside, straight after? A 3 in r7c6 forces 1+2 on the arrow in box 8, which could then only go in the domino r4-5c4 in the 9-cell cage running through box 8. This is prevented by the quad in box 5. It's the same logic you just used in box 6!
To be fair, the title is a quote, and he tells us where the quote is taken from in the intro. (Is it click-baity? Maybe, but it doesn't bother me like it seems to for some people. I'm already hooked. I'm watching, whatever the title is. 🙂)
Not a hardest puzzle.. We can start with colouring Boxes 5, 6, 8 &9 with total sum as 4x45 and then with different colour of long cages 2x45 and boxes 5 & 9. With that you will find value of cages in box 5 & 9 as 1&3 and 1&4. Rest is simple
Simon has a habit of forgetting logic he just figured out and not applying it to another area of the puzzle where the logic fits perfectly. The logic he figures out to place 1 and 4 in box 9 could have been used to place digits in box 5.
I always like Simon solving symmetrical puzzles because he always finds two different ways to solve the puzzle instead of just doing the same thing twice.
I found it strange how long it took him to place the 13 pair, when he had aldready placed the symmetrical 14 pair and the logic applies the same way in both cases.
@@HappyLooter Exactly. Its ubelievable. He talks about symetry, then he finds a beautiful logic, and then he is not able to spot the exact same symmetrical logic in other box...
Thank you so much for mentioning Zelda's birthday! She pointed at the screen when you talked about her (although that might have been by coincidence). I find the tradition of congratulating viewers to their birthdays so charming.
There's an elegant break-in at the start that uses some SET. Highlight the two L-shaped cages. Those are two sets of 1-9, or 90 total. Remove the four circles in those cages and add the 3-cell lines in boxes 6 and 8 to compensate. Still 90. That leaves only one cell each in box 6 and 8 not highlighted, and the maximum they can be are 4 and 5 since the are in the quad, and the four highlighted cells in box 5 and 9 must equal the total of the un-highlighted cells in 6 and 8. The lowest the highlighted cells can be are 1,3 and 1,4, which equals the 4,5.
That gives you 1,3 in column 4 of one quad, 1,4 in column 9 of another quad, and the positions of 4/5 vs 2/3 in the 2345 quad. The rest of Simon's logic about where the digits in the circles go flows smoothly from there.
This was my break in as well. one small typo: when you said boxes 6 and 9, you meant 6 and 8.
I did something similar:
highlight box 6, swap lines for the end circles and compare to L cage going through the box, proving:
[8,9] + [9,9] = [6,7]
similarly:
[4,4] + [5,4] = [6,5]
then likewise, you needed the values of 1,4,5 and 1,3,4
@@mstmar thank you, corrected.
And surprisingly, I went to this SET break in almost instantly, by instinct. I am a bit surprised, knowing how much Simon loves using SET, he didn't do it this way. It was quite obvious to me, for some reason i dont know. maybe just lucky.
Clever!
Title, the hardest puzzle I've ever seen. Simon "do have a go"
Love the conference you have in us!
Wait, so yesterday's puzzle wasn't the hardest? We've been had!
Didn't watch it, but maybe Simon had a slow day yesterday and fast one today.
every puzzle is the hardest, the best puzzle of the year, the most brilliant puzzle ever made, etc etc etc
don't read too much into youtube's click-bait meta. hyperbolic video names are a big aspect of big success on this platform.
it's also in this case a direct quote from the LMD comments that are shown barely a minute into the video, so the clickbait was basically handed to them on a plate
Yeah, the hardest puzzle ever will come tomorrow.
@@nb2vcxz The puzzle was finished. The aim was to fully shade the grid in a certain way and to draw a path through the grid. Maybe read the rules of the puzzle?
It's like Gregg Wallace saying "cooking doesn't get tougher than this"
I’ve seen a couple explanations of set theory, and I’ll add that I think it makes this puzzle much easier (though I’m normally very bad at identifying useful sets!). The 2 9-cell cages are one set, boxes 6 and 8 are the other. Eliminate all overlapping cells, and all lines/circles that have equal sums.
I have never seen such an intricate and at the same time beautiful *lattice* of geometric restrictions by clueless cages, and I totally understand the reason why *cam* wrote on LMG _"I am extremely confident I followed the intended solving path"._
This is a *world class masterpiece.* Every step, from start to end, was challenging, but never brutal and always divinely beautiful. Quads were magnificently used as disambiguators.
How the hell can a human being be smart enough to conceive such a *divine logic creature?* For instance, how can just a bunch of neurons 💪🧠 conceive the multiple stratospherically clever and totally unexpected interactions between elementary restrictions that force
circle in the UR corner of *box 3* =
circle in the UR corner of *box 5* =
just *4* or *5?*
And that's only the first step in my solve, followed by dozens of equally mind-blowing others‼
I would love to see a video by Math Pesto about the developement of this apparently superhuman achievement.
Thank you *Math Pesto.* Thank you *Simon.* 😏👍
You both deserve a standing ovation
👏👏👏👏👏👏👏👏👏👏👏👏👏👏👏👏👏👏
I am also fascinated by the beautiful *mirror-symmetric* arrangement of most of the clues (all of them except for the L-shaped cages and the line in box 5).
The *quads* aligned along the axis of symmetry look like an *arrow* notched to the bowstring of a fancy professional arch.
This is a *logic masterpiece* and a *beautiful picture* at the same time 😍👍
If you use SETS to compare the 9 digit cages with the BOX6 and BOX8 (all sum to 45) you can determine R6C7 is 5 and R7C6 is 4. You also determine the missing digit is 1 in both the quads 378 and 458,
@@sampathkumar-ej7xl Yes, we discussed this in a separate thread started by *bengrabow3019.*
Gouge Away is the magnificent final track on the altogether magnificent album "Doolittle" from Pixies, released 1989.
And its subject is the story of Samson and Delilah.
Thanks for sharing your knowledge!
I've actually had this song stuck in my head lately.
And magnificently delivered on the 2004 Toronto gig, showing how the track builds in intensity.
Always loved playing it live, too.
I love the Pixies, one of my favorite bands as a little kid.
34:36 for me. I'd vote for 3 stars difficulty, personally. Really fun puzzle.
Simon would have made his life easier if he'd
1) done the same logic on the left L-nonomino that he'd just finished on the right one (the same logic that got him the 1 and 4 around the 458 circle and the 5 around the 2345 circle could immediately be used on the other side to get the 1 and 3 around the 378 circle and the 4 around the 2345), and
2) left his roping colours in the top three rows, which would have resolved the 10-pair in box 3 much earlier, by noticing that the 1 couldn't go in r3c7 because of the quad clue in box 1.
Indeed, Simon missing the obvious symmetry between the two cages resolving the 14 and 13 was really frustrating!!
Yep - I always love watching his solves but I was willing him on to spot the 1/3 and 1/4 issue and that the logic was identical to the other cage, which I would never have spotted in a million years. Especially funny because 2 mins in he is excited about the symmetry of the puzzle :-)
Knowing where the 1-3 go in column 4 is pretty easy using this logic, and would have saved him probably 15 minutes on his solve.
Like you said, it is the exact same logic that he found with the other 9-cell cage, so I don't see how he didn't notice it applied!
That's very typical for his solves. He almost always refuses to believe that the setter used the same trick more than once and goes to hunt for something else in the puzzle after solving one part of it. It does make his solves quite frustrating to watch because he is also quite sloppy with many of basic sudoku mechanics and doesn't follow a rigorous organized plan in his solves.
Brilliant puzzle. The way it revealed what the line in box 7 had to be is one of my favourite sudoku moments ever! ❤
I finished in 94 minutes. That break-in was so beautiful. I surprised myself on how I was actually able to see it. The geometry of all at all made it so satisfying. Utilizing the surrounding circles clues in box 9 and 5 to disambiguate the non-caged cells in box 6 and 8 were incredible. I did that part quite fast only to get stuck near the end, but realized that roping became very powerful and easily finished the puzzle. This goes down as one of my favorites. I actually didn't find the difficulty to be that hard. It was hard, but nowhere near the hardest I have ever done. I loved this one. Great Puzzle!
17:43 for me. Fairly approachable once you spot the break-in. I would encourage people to give it a go despite the title of the video, 'cause it's a very enjoyable one to solve. Great puzzle!
27:37 for me. Haven't watched Simon's solve yet but I thought it solved beautifully. The clues flowed together perfectly and I never felt lost at any point.
At 40:00, 1 and 3 in box 5 are row 4, because they have to be in the cage, but they are outside the cage in boxes 8 and 9. Combine this with the deduction at 41:15 and you can place 1 and 3 and 7 and 8 in box 5
Okay, he got there at 52:15 😀
I couldn't understand why he wasn't applying the exact same logic he'd just applied on the other side, with the 1,4 pairs on the line in box 6 having to go in the 9 cell cage in r8-9c9. It's the same logic, just rotated! 😂
65:05
This was brutally, brilliantly, difficult. The instantly obvious roping made it look like it would be a bit simpler, but instead it was quite the grind of logical steps to chip away and this beauty's secrets until the roping appropriately tied up the final loose ends. A masterpiece.
33:08 and I echo some of the other comments, this never felt anywhere near the level of difficulty that I was lead to expect from the description. Some tricky scanning with the overlapping ropes, and I suspect if you miss the break-in you could probably go in circles for ages without getting much. Very interested to watch Simon now and see where he might have gotten held up. (quality puzzle tho! Don't get me wrong)
This video is proof once again that Simon is a master solver. Much like top mountain climbers pick the challenging routes to the peak that no one else dares try he consistently picks the harder route to a solve that us mere mortals are left in awe of.
I found the box 6 and box 8 set break in that others have mentioned.
So I think I found a slightly easier method with set theory. The 9 cell cage and box 6 are both sets of all 9 digits. Take away the 5 shared in box 6. Then you can get rid of the double arrow. Now you are only left with 1 cell in box 6 which has to be 2, 3, 4, or 5 and the two bottom cells which have to now add up to 2, 3, 4, or 5. Can't add up to 2, and you have two use at least 1 of the quad of 4, 5, and 8 so the only thing you can do is 5 which adds up with 1 and 4. You can do the same with the other 9 cell cage and box 8.
Did it the same way
"more appropriate to being muttered by a chibbering chimp" 😂😂😂 love it
I love watching Simon's solves. :)
There was a simple deduction which should have come when he was placing orange and red in box 8 (20:32): the same rules applied to the 2 cells above orange in the cage.They had to be in that 2x2. When he used that principle to solve the gray/green situation (30:21) and didn't apply the same thing to orange/red when he switched back (34:28), I broke a little lol I had to wait until 52 minutes in.
Love the video!
You know it's a tough puzzle when Simon stops talking (I still wish I could find the one with 15 straight minutes of silence)
Oh, this puzzle was fun. I didm't find it that difficult, but it was very much a different style of difficult. Very intricate and it can absolutely wall you if you don't see some stuff.
there is a very easy break in.
Take the L cage in boxes 5&8 as one group and subtract the box8. Then end subtract the double sided arrow and you end up with R4c4+R5c4=R7c6. Do similar for the other L cage and you get the 2345 positions, and marginal positions of the other circled digits, too.
What a beautiful puzzle. I needed Simon's break-in but after that I'm quite proud that I managed to do the rest.
Simon: "Diagonal symmetry"
Also Simon: Not using the logic he used to place the 1/4 in box 9 to do the same thing with the 1/3 in box 5.
I'm an avid user of the Vivaldi browser for its unmatched tab handling and memory saving features, and it has just increased my enjoyment of the CTC channel as well. Its page tiling feature makes it a cinch to display the video and the app for solving the puzzle side-by-side on a simple laptop screen, without having to manually adjust window sizes or anything. And when done with the puzzle I just turn the tiling off again. Easy as pie!
At 37:37, isn't there the exact same problem with R7C6 being 3 as was with the other side where R6C7 couldn't be anything other but 5? With R7C6 set to 3, R9C7 would be forced to be a 3 and then there has to be both 1 and 2 on the arrow in box 8 and they would be forced to be a pair in R4R5C4 which is impossible.
There was a much easier break in ...... do not read unless you want spoilers
>
>
>
>
>
>
>
>
>
>
Using set on each of the two 9 cell cages against box 8 and box 6 respectively allows you to cancel 5 cells in the cages and cancel the two end cells against the three line cells leaving 1 cell equal in box 6 and 8 vs two cells in the cages. This forces 5 into the NE and 4 into the SW corners of the 2345 quad.
Do not worry about spoilers. This video is about solving a puzzle, so the discussion section is supposed to discuss the solve, and suggest alternative ways of solving. This is why I read this section after solving. People who do not want to be spoiled should just ignore the comments.
52:00 Finally the logic which has been done identically for the other 9-cage like 20 minutes ago. That would have made some of the logic in-between easier.
99:31 for me. Basically two internconnected puzzles (boxes 5,6,8,9 and boxes 1,2,3,4,7). The former solved very quickly but it took me quite a while to resolve the latter. Great puzzle.
53:03 finish. This puzzle was determined to try and break me. First, I kept forgetting that they weren't between lines. Then my computer froze multiple times and ran off about five minutes of my clock (so technically I finished in 45-50 minutes). I finally settled in and pushed through, however, and enjoyed that ending very much. I only colored the ropes that were outside of the 1-6 quad, to save colors and not make it harder to spot the individual ropes. Fun fun fun!
I didn't watch yesterdays video (haven't had time!) but if this one is harder, does that mean there were 2 hours of birthday messages in yesterdays?
LOL
Always so interesting, Simon - thank you for this video! I thought about SET myself, (not while solving - I am not going to try this one) because of the layout of the grid. But you followed a great path, I think, and I enjoyed watching it.
Did you end up using SET or solving a different way?
Hi Emily and David 👋🏻. I haven’t watched this yet - it’s been a long day and I’m tired. Just looking through comments and stopped by to say hi.
@@longwaytotipperary Hi, Longway, good to see you!!
@@davidrattner9 I did not try it at all. But perhaps, someday, when I have a spare four or five hours, I will give it a try! 🤣
@@emilywilliams3237 😁
It took me multiple sessions and over an hour's total time, but I was able to chip away at this one before finally completing this diabolical creation.
Very challenging puzzle!
Took me a surprisingly-brisk 25 minutes or so. One interesting trick I came across was that regardless of whether the line in box three was 1&9 or 3&7, it resolves the 7&9 in box nine in the same order. In turn, having those digits early made some of the other logic along the bottom rows faster.
35:50 "If you can see what's going on..." shows Simon's genius. He can make such complicated deductions, but he's always looking out for them as well. Now in box nine there's that 58-pair. That one can be resolved easily. Imagine r8c8 getting the five. Then the red and grey squares would need to add up to five. They can't. There's that 14-pair in the box and r7c7 is either a two or a three. So then red and grey have to add up to eight instead. Three options for that: 17 (eliminated by the 14-pair), 35 (eliminated by the 5 that ends up in r9c8) and thus they must be a 26 pair. Since the red/orange square in r7c6 cannot be either a two or a six, it cannot be red and must become orange. Orange cannot be 3 because of the 378-circle in the center box.
You made the logic with the L cages much more complicated than I did. I just considered the cage, the box, sets, sums and the possible digits.
Another day, another fabulous puzzle with another fabulous solve.!!
Brilliant puzzle. Quite difficult at the beginning but not the hardest at all. Anway I could enjoy it without loosing my mind.
11:15 The (16) circle can be useful exactly because you know there's roping.
It means the row/column that sits outside the (16) clue can't have a 1 or a 6 in it in box 1, which then propagates to boxes 2/3 and 4/7.
That is, neither 1 or 6 can appear in row 3 of box 3 and column 3 of box 7.
You also know, however useful that is, that at minimum one of the cages in boxes 1-3 or 1-7 will have a 1 or a 6 in it.
No matter how you twist and turn it, BOTH cages will never have 1 and 6 in them at the same time. Only one of the cages can have both 1 and 6 in it, the other is then forced to have at least one of the digits.
Of course it's entirely possible for one cage to have one of 1 or 6 in it and the other cage having neither.
I don't see how this would get the puzzle started but it's worthy of note.
This has to be one of the greatest sudoku puzzles ever made.
43:39, today is my birthday and it was a treat better than chocolate cake to spot the SET break-in immediately! 😊
I love that Simon put the roping up again and then ignored it when one part was completed.
If he had picked the other roping first, the 1-6 limitation was beautiful.
At 1:09:35
"Now 6 has moved into the green rope..."
Yes...
(Uses the *light* green rope)
...No, the other green!!!
Simon, you missed something fundamental that would have cut a half hour off your solve time. You discovered that R6C4 and R9C7 both must fit into the four squares in Box 8 that are not in the cage, but you failed to realize that R4C4 and R5C4 must also go into those same four squares. In other words, you could have known all four occupants. By so doing, you can immediately detemine which numbers they are due to the circles.
I think that Simon has tunel vision, this must be the reason he can't see some numbers while doing sudoku. He only focus on some box and forget the others. 😂
But I didn't get it why he didn't do the same logic in boxes 5 and 7 (the logic he used to find 14 pair in column 9, could be used to find 13 pair in column 4) 🤔... but well, he really likes to go to the hardest path!
Simon’s logic included:
A: This is symmetric around the negative diagonal.
B: I make some progress in box 3
C: Therefore, I ignore box 7
???
It's hilarious actually 🤣 It looks as he absolutely forgets his own logic, that he made before (
67:04 for me. Enjoyed the puzzle. My solving times are often similar to the length of the video, so this puzzle felt like a normal CTC level of difficulty. Definitely not "the hardest puzzle I've ever seen".
You came really close to cracking this open around the 27 minute mark, where you were considering whether orange or red was in R7C6. You rightly pointed out that if it were red, it would have to be 2 or 3, and also concluded that it couldn't be 2 in that case. What you failed to spot, is that if it were a red 3, that would put both 1 and 2 into R4/5C4, leaving no room for the 378 that must be on the quad. Therefore orange is in R7C6 and is 45. This also means that the counterpart of orange on the line in box 8 must be 13, or 23, so the 3 on the quad is in C4, along with 1 or 2. This pushes the 78 on the quad into C5, and puts a 123 and a 78 on the line in box 5, which means the circles sum to at least 8, and R4C6 cannot be 2.
You could now repeat the same logic on the similar construction in box 6. The R8/9C9 domino needs at least a 4 in it, which means that the counterpart to R6C7 on the bent line in box 6 must include 4 and R6C7 must be 5 with 14 on the line and in that domino at the bottom of C9. Now there's 58 in C8 on the quad in box 9, but 5 can't be on the one-cell line, so it's 8. Grey can't be 5, because red can't be 3, so green is 5, orange is 4, and R9C7/R7C9 are a 26 pair.
Because orange is 4, the domino on the quad is 13. You now have enough to resolve all of the 2345 quad. The line in box 5 is 18, with 37 above it, which must go in R5/6C9, putting 68 in blue and 9 in purple. After this, it all becomes much easier, especially if you use the roping to place digits in R1-3 and C1-3.
You could probably have cut 15 minutes off your time if you'd spotted what was available at 27 minutes, but it certainly wasn't an obvious thing to spot unless you can built a complex picture mentally and see where the contradictions lie. Fortunately, for my work I need to deal with chains of consequences so I've had plenty of practice building mental pictures.
This was an incredibly intricate construction, which does require you to be able to keep lots of things in mind at once to solve efficiently. I'm glad you didn't forget the roping, because that was crucial to the end game. I find with roping that you only really need to colour one strand. The other two are then easy to visualise. When you've got two ropes crossing like here, it helps keep the grid relatively clean.
Still love you guys! Sorry for almost never watching, I feel like a faithless spouse - but I have a lot going on in my life. I am working on a tough maths problem that may well become a rather interesting paper. The spirit of patiently chipping away at a problem and rejoicing in occasional Aha-moments was reinforced by my love affair with CtC in the dark moments of lockdown. I haven't forgotten the warm friendship I had from a few very dear co-obsessives. When you see this comment, you will know who you are...
Funny how after talking about all the symmetry at the start Simon proceeds to utterly discard the logic of symmetry while resolving the squiggly lines and the quadruple clues. The 3 in r7c6 could be immediately eliminated with the exact same logic applied to r6c7 - and having the 4 in r7c6 meant that r45c4 was a 13 pair (same as before), and then 1378 could all be placed.
For me solution went smoothly after the large box, with horizontal roping. It absolutely magically resolves options in box 3, almost out of nowhere.
This one clicked for me pretty quickly, another day perhaps I would have ended up tearing my hair out. Very enjoyable, thanks. Unusually my path was quite different from Simon's, I used a more SET-like approach which I assumed was the intended path, but perhaps Simon's was.
38m08s. I feel like I had an early break-in with a cleaner version of the logic Simon used in 26-31 minutes. The four digits of box 6 that are not in the cage are the same as the four digits of that cage that are not in box 6, and so in particular they have the same sum. But the two circles add up to the three digits on the line, and so the remaining digit in box 6, R6C7, must equal the sum of the remaining two digits in the cage, R8C9 and R9C9. But the largest R6C7 can be is 5, and the smallest R8C9 + R9C9 can be is 1+4=5, so each is exactly 5. And then a similar argument puts 4 in R7C6, and you're off to the races
Solved in 60:04! I have good teachers :). Not the hardest puzzle, but a really nice one!
This was hard. Nice geometric effects here, and for many it took me quite a while to find them, even though I knew the theory from watching earlier videos.
At some point, after I had pencilmarked the whole grid, I noticed that the circled 1 in box 1 has an impact on box 7, and then it fell apart.
102:53 for me (that's slightly under the total video length), solve counter 551.
Actually, 102 minutes is 1 hour and 42 minutes, so a lot more than the video length. Not sure what I thought here.
I just feel the title doesn’t quite match yesterdays video…
Simon forgets to use the 9-cell Lregion in boxes 5 8 and 9 to confine the 1-3 pair in box 5
Early today. Looking forward to dozing off to this (not because Simon is boring, but because his voice is soothing)
Simon: when you find symmetry in a puzzle, and then you find logic, try to apply it to the symmetrical part of the puzzle. Also Simon: doesn't do that in this puzzle.
RIGHT??? I was shouting "WHY ARE YOU NOT PUTTING THE 1 AND 3 IN BOX 5!?" and then he gets them a totally different way omg
Solved this one in a little over an hour so definitely far from the most difficult on the channel. Was a little bit hesitant to start given the title but broke in fairly quickly.
Fun puzzle! 41:08
It really helped me to keep the roping colored even though it doesn't help until later.
How did you eliminate the possible for a 2 or 3 in box 6 cell 4/5/6? You said that the 5 in box 5 had to be made using 4 and 1 in box six. But couldn’t you have made the five with 2 and 3?
Whichever two digits digits add up to that 5, the only place they can go in the L-shaped cage is in the two bottom cells. Those two cells are on a 458 quadruple clue. If you were to use 2 and 3, you'd have to put five different digits (23458) in the bottom-right 2x2.
Sudoku blindness was so real in this one... never ceases to amaze me lol
For those of you who are confused (as I was), the puzzle is NOT symmetrical apart from the kink. The nine cages are oriented differently. This is why thy break in works, and why the logic in the top right does not apply in the bottom left (and vice versa).
I found the solve pretty smooth. 33:36.
Simon would have saved himself a lot of trouble by comparing each 9 cell L shape with the box containing the 5 shared digits.
For being "the hardest puzzle ever" I somehow managed to do this without help! Used the SET break-in others have mentioned here.
1:10:19 today. i had a very complicated start, i missed the obvious break in. then i struggled a lot in the middle and the ending with the triples was easy.
37:09 for me, Phistomefel's are much tougher at times
22:16 for me. Very nice puzzle!
the timer ticked over 240 minutes today at work while working on it. i found the breakthrough myself which was nice, but i couldn't finish it after that
“That’s so annoying it’s not done anything!” - it would have if you’d done it 20 minutes earlier before getting the other stuff the hard way 🙈🙈🙈
(When you got the first 5, it was instantly “same logic puts 4 on the other side with 1 and 3 above, which fills out the 1378 because two have to add to 9” - which would have made the 869 in that row a lot easier! Impressed that you got it anyway, was almost convinced you’d disambiguate the four digits without noticing the repeated logic at all 🤣🤣🤣)
Yup, Kitchener Ontario is correct Simon
What a fun puzzle!
I don't understand Simon's mind. He tends to ignore obvious things, like how the digit at r3c7 affects r8c7 while exploring things that are beyond a normal sudoku solver.
I don't think it is as hard as other puzzles on this channel. It's a good geometrical puzzle. The break in is simpler to see if approached by comparing the totals of the nine cell cages and the totals of boxes 6 and 8. And I didn't get too stuck once I'd seen this.
Was there a link to the snake egg video?
You could use the 7/9 pair in box 9 , to solve the 3/7 in box 3
Fun puzzle. My time is 62:23 and I think this is the first time I have solved faster than Simon!
where‘s the link to the old snake egg puzzle?
64:15 for me. Nice puzzle!
42 minutes whilst watching news feeds on my 2nd screen. I'm confused about the title 🤔
I’m determined to solve this, I have 23 digits after 90 minutes of work, which I’m shocked at because I don’t normally do sudokus and the one is difficult
Much easier compared with yesterday's one
How can you successfully eliminate 3 as option from one side of a puzzle, with a high degree of symmetry, and not look to see if the same logic can be used to eliminate 3 on the otherside, straight after?
A 3 in r7c6 forces 1+2 on the arrow in box 8, which could then only go in the domino r4-5c4 in the 9-cell cage running through box 8. This is prevented by the quad in box 5.
It's the same logic you just used in box 6!
Wow, I thought it was pretty easy, it took me 36 min
Was that the end of the birthday catch up? I sent one during the vacation 😢
49:33 double nice
Like how 2 cells one belong on the line 1 needs to break into smaller sigits
one of the more frustrating videos to watch as a viewer I think lol just flip the logic across the diagonal Simon!!
46:45 nice
Yaaay solved in 110:25
Boooo ads in the pause menu? Ads on the check card? When did we start doing that?
I don't know if the title is ironic. I found it surprisingly simple: 23 minutes 🙂
To be fair, the title is a quote, and he tells us where the quote is taken from in the intro.
(Is it click-baity? Maybe, but it doesn't bother me like it seems to for some people. I'm already hooked. I'm watching, whatever the title is. 🙂)
I actually found this puzzle the easiest in a while. I usually get so stuck in all the easier ones. 😅
Definitely not the hardest I've seen, but not trivial either. 68:29 gg
32:19 with no help
Not a hardest puzzle..
We can start with colouring Boxes 5, 6, 8 &9 with total sum as 4x45 and then with different colour of long cages 2x45 and boxes 5 & 9. With that you will find value of cages in box 5 & 9 as 1&3 and 1&4. Rest is simple
41:00 for me. definitely not the hardest puzzle ive ever solved
Simon has a habit of forgetting logic he just figured out and not applying it to another area of the puzzle where the logic fits perfectly. The logic he figures out to place 1 and 4 in box 9 could have been used to place digits in box 5.
61:48 for me
51:39 for me