The fact that there is an interesting puzzle everyday consistently is insane to me! How are there so many interesting puzzles and how are you guys finding them all🤭
Interesting puzzles help create interesting puzzles. They inspire old and new setters alike so more and more puzzles are being made. CtC gets many of these puzzles submitted directly from the setters are recommended by other puzzle enthusiasts. There are literally thousands of amazing puzzles that this channel doesn't have the time to showcase because of this. It's a really fun hobby and a blast to make puzzles
Logic masters Germany has heaps of puzzles and people will email in to suggest the ones they find interesting/beautiful/challenging. There are certain constructors you know are absolutely going to provide amazing puzzles. They also have their discord channel and such where people discuss their favourite puzzles. Famously recently there was one from reddit submitted too, so there’s quite a few places to find puzzles it’s just wilting then down to brilliant ones but that’s also not too difficult to do either. Plus ctc do have testers who will ensure that good puzzles are passed along - although some videos start with Simon saying the testers couldn’t crack it and want him to give it a go (probably for their sanity, to see the break in)
That was nuts!!! Took me over 4 hours of timed thinking plus some extra time over 4 days to figure that out. But I persevered and finally conquered the hardest puzzle I've ever done without Simon's help!! I know that doesn't mean much to the pros on this site but I'm damn proud of myself. Still gotta give credit to all the setters and solvers that I've learned from over the last 2-3 years.
Thanks for the birthday shoutout! Jill here. In order the names of the birds are Kiwi, Mocha, and Lemon. I suppose Lemon was smiley because she was a baby when I took the picture. And yes, the chantilly cake was delicious, or passable in your terms 😅
Amazing solve and amazing puzzle❤. At 59:08 the 4 line could be from bottom to top 1 2 / 3 / 3, this gets disproved later by sudoku but I think Simon missed this possibility so there is a bit flawed logic because he didnt mark 3 as a candidate in R3C1
This is exactly the type of puzzle that Simon's brain is simply way more adept at dealing with than mine. I get lost in the possibilities and don't ever manage to make any meaningful deductions.
I believe a similar type of scenario happens at 59:10, where it was concluded a 6 must be in r3c1 if there is a 1 in r6c3. However, there could actually be another 3 in r3c1 if the line total was 3 rather than 6.
I finished in 165 minutes. There was no way I was able to do this in my head. I had to write down on a piece of paper all the possibilities and reference it. It worked out quite well and I was able to reduce it down to 15, 21, and 35. I knew 35 had to belong, because a 9 had to be used. Ruling out 21 was clever, due to the six cell region having to go in the center flanked by two three cell regions consisting of 678. This left no 78's on any of the flank positions for the 35 side. After that, I slowly made my way to the finish, partially messing up due to forgetting that 5 is a possibility for the line in box 4. Great Puzzle!
That's pretty much how I got along on this puzzle also. Except the possibility of the 4 cell lines being split into three segments did not leap out at me immediately. Imagine my consternation on Day 4 of this solve.
@@LovethectcHe overlooks 9-45-9 as an option at around 33:00, when pencil-marking the options along the four-cells line in box 8. There isn't a 45 pair in r8c2 at this stage.
😮Yes, overlooked but it's easily not possible since it would require a 9 in box 2/column 6 and a 4 in R8C5 leading to 456 in column 4/ rows 6-8 but R8C4 must be a 2 or 3. I'm sure he thought it was blindingly obvious and just neglected to mention it. Or maybe he just got lucky.
At 48:55 he talks about that possibility on the other line and then says "but that's different than this middle line" but doesn't explain why. So yeah it does seem like it's an oversight
He doesn't address this possiblility, but it can be ruled out. By the 5 x-wing, it would have to be r8c5=4 and r7c5=5. This forces a 2/3 pair in r8c4 and r8c6. The 4 in r8c5 means that r8c2/r8c3 cannot be a 4/5 pair, and instead are a 7/8 pair. This is a full segment, and so the squares r8c4 and r7c4 are also part of the same segment. r8c4 is 2/3 meaning it must be a part of the segment 1,2,3,4,5. However r7c4 must be one of 6,7,8 since they are the only remaining values to be placed in box 8, which is not possible if r7c4 is in the 1,2,3,4,5 segment.
Okay, regarding Simon missing the 9-4-5-9 option at 33:00 - a way of removing that option is the following. We first get a 19 pair at the ends of the column. That puts Row 8's 2 and 3 in Box 8, along with a 4 or a 6. The other 4 or 6 accompanies the 5,7, and 8 in R8 C2378. This removes 2 and 3 from R7C5, which in turn removes the option of 1 from R9C4, placing 9 there (our first digit). This now means that the R8 4 or 6 that's not in Box 8 is either a 4 in Box 7 or a 6 in Box 9. (Note that at this point, R8C5 is either 2, 3, or 4.) If the 4 is in Box 8, that removes 4 from R8C5. if the 6 is in Box 9, this must go with 5 in Box 9 and 4 in R8C6, also removing the 4 from R8C5. This removes the missed 9-4-5-9 option, and we may now proceed with the puzzle without despairing.
32:29 Fascinatingly clever, and the ending where it kept looking like it would crack and didn't, and didn't, and didn't and finally did was marvellous.
Yes but that still leaves a 2 and 3 in the middle 2 boxes of the line which is the deduction that mattered at the moment, idk if leaving the posibility of a 1 at r6c3 and a 3 in r3c1 would habe made an impact later
Intriguing puzzle. Having watched Simon's solve, glad to see I got most of the logic, if not always quiet as elegantly! Like him was tripped up by the line in box 8, and box 4. Unlike Simon, nobody will deem it worthy of comment, I'm sure.
As a non-English speaker, I just love learning new words by watching these videos. Today's word was "kibosh". Past words include "recalcitrant" and "alacrity". Thank you, Simon, for expanding my vocabulary. :)
To those who have been struggling with the 9-45-9 possibility on the middle line, I found the solution: -You can't put a 5 in R8C5, so it must be a 4, with a 5 in R7C5. -This forces a 78 on the left hand line "wing" (R8C2/3) -R8C4/6 must be a 23 pair to complete the row. -We already have a 1 in the box. -The 2/3 in R8C4 must be part of the 12345 sequence, but there is no digit left to put in R7C4, because all of the digits have been used in box 8 already.
Okay, regarding Simon missing the 9-4-5-9 option at 33:16 - a way of removing that option is the following. We first get a 19 pair at the ends of the column. That puts Row 8's 2 and 3 in Box 8, along with a 4 or a 6. The other 4 or 6 accompanies the 5,7, and 8 in R8 C2378. This removes 2 and 3 from R7C5, which in turn removes the option of 1 from R9C4, placing 9 there (our first digit). This now means that the R8 4 or 6 that's not in Box 8 is either a 4 in Box 7 or a 6 in Box 9. (Note that at this point, R8C5 is either 2, 3, or 4.) If the 4 is in Box 8, that removes 4 from R8C5. if the 6 is in Box 9, this must go with 5 in Box 9 and 4 in R8C6, also removing the 4 from R8C5. This removes the missed 9-4-5-9 option, and we may now proceed with the puzzle without despairing.
I missed a little bit here: 6 in Box 9 puts 4 on the left line in Row 1 with 5, thus making Row 8 Box 7 a 78 pair and Box 9 the 6 goes with a 5. Now if we put a 4 in Row 8 Column 5 it forces a 5 above it, making for 9-4-5-9 (as we had limited that cell - R7C5 to 4 or 5), we break Row 8 Column 4, as this needs to be part of a 456 or a 12345 (due to the 78 in Box 7 seeing it). it can't be 456, as it is seen by all of these. it can be on of the 23s from 12345, but the cell above it then sees 12345. This means that when there is 6 in Box 9 on that line, there still can't be a 9-4-5-9 run on the line. (which we previously removed in the 4 in Box 8 version). I hope that there's a more fluid way to prove that, because it felt a little clumsy to me, but I hope that does the trick for you :)
59:48 for me. Wow what a puzzle. I got very stuck on a wrong assumption and it wasn't until I took a break and came back that I realized what I was doing wrong. Fantastic puzzle, but it almost broke me.
My favorite part of solving this puzzle especially towards the end is I felt like I had to keep going in circles around the puzzle to resolve a clue back in the side I started with and just round and round and round it unwound itself.
Simon got lucky with the line in box 8. It could have a 9 in both ends, a 4 in R8C5 and a 5 in R7C5 which would create a 5678 quadruple in row 8 and a 4578 quadruple in row 2. It would quickly break the puzzle as it would force two 6s into box 2 and two 1s into box 8. But he didn't even consider that possibility, and therefore never made the deduction that it would break the puzzle.
N-Sum Line: Divide the line into at least two non-overlapping segments. The digits in each segment sum to the same total ‘N’. I’ve seen this used several times before… even on this channel. Although this is the first that combined it with segmented renbans.
Wow! What a puzzle! That one was way above my pay grade. It's such a relief when Simon stops doing the really hard stuff and does that arcane thing called "Sudoku". A little know art form, known only to a few cogniscienti. (I probably have spelt that incorrectly, because I am not one).
Oh, Simon... Such a genius in the opening stages, I was looking in awe how he was navigating the breakthrough, I myself gave up on that. And then second part, when I was literally screaming at the screen, but Simon refused to see any of the single step or even zero-step deductions...
I did not need to use Simon's brilliant initial deduction (see video @16:32). I used an *aide memoire* to list the possible totals for 2-cell, 3-cell, ..., 7-cell segments (8-cell and 9-cell segments are incompatible with the grey lines). Then I coloured the *common totals:* *15, 21, 30* and *35* from which I easily ruled out *21* and *30* due to incompatibility with the long *grey lines.* *Common totals* are needed because it is impossible to divide a long line into segments of the same size.
Simon's way to show (@24:01) that the long line containing a *9* was divided into segments with *total = 35* was sound, but I think it was not clear enough to be understood by everybody. The same is true for the long line with *total = 15.* I think he should have used an *aide memoire* to make it crystal clear, although he personally does not need it because of his powerful working memory.
2:02:41 - Phew! That was HARD! I worked out the two big lines pretty quickly but it took me ages to prove which way round they went! It didn’t get much easier even when I’d got that! Excellent puzzle though.
how bizarre and wonderful. GPT helped me write a python script to get all the possible combinations. Here they are: 3: 12/3 5: 23/5 6: 123/6 7: 34/7 9: 234/45/9 15: 12345/456/78 18: 3456/567 21: 123456/678 30: 45678/6789 35: 2345678/56789
Wish I could have given this puzzle extra hearts. Such lovely satisfying interplay of lines, and lots of subtle knots after the initial break in. Needed to refer to the solve to give me a hint or two and would have missed a couple of line options without it. Such is the way with new rule sets!
Nothing so far, mmm, there could be two solutions then. I guess Simon needs to revisit this puzzle and make sure and to do it without that assumption and see what happens.
No reason other than he missed it. I have to hold my hand up and admit I also missed it. At least this one doesn't skip too much logic. The three-cell line on the right, between boxes 3 and 6, still has to be split into two digits summing to one digit. The one digit has to be odd. The whole line sees 9's, so it sums to 3, 5 or 7. In all cases, it requires a 3 on the line (3=1+2, 5=2+3, 7=3+4). The 3 in r6c7 forces that 3 into r3c9. That removes 3 as an option from r3c1. (A harder oversight to get past is that he also overlooked 9-45-9 as a possibility at around 33:10. More work needs to be done to rule out this as an option. It's been discussed in other comments).
Sorry, missed what you meant, it could be a three line with a 3 in r5c2. The resolution would come a bit furhter ahead, once the three cell line on the right hand side of the grid is also considered and a 3 is forced into r3c9. So there are not two solutions. But it was an oversight at this stage of the solve.
The beauty of this puzzle is i started a 9x10 empty grid ...the 8 midrows centers on the long 12cell lines....then plopped the other lins in after discovering what ur suppose ti....just glad.i didnt verrtically flip If the oddcircle was evensquare(an 8) there are way to many solutions
55:40 "How do we do this then?" Simon wonders as there is a 12-pair just right above a cell labelled "19". Totally missed that one when he found out the 12-pair... almost 2 minutes earlier. Then finally uncovering what should be where through sudoku in a totally different box. Never change, Simon, never change! :P
Thanks to watching so many of your awesome sudoku videos I'm actually starting to remember some of the triangular numbers (up to 6 now) Because of this I actually figured out the possible segment totals for the 12 line with the one before Simon while watching him ponder it - which is a rare occurrence, that something maths related in these clicks for me so fast! It's all thanks to your wonderful teachings 😊
1:23:25 for me. I made a mistake. Didn't account for an option and had to unwind half the puzzle to see what I'd done wrong. Nonetheless an amazing puzzle. Fun solve
I used a different approach in solving the two long lines, I just brute forced the lines to be 2(78)-2(78)-3(456)-5(12345) long and 5(56789)-7(2345678) long through a lot of mental calculations. Then I noticed where the 78's should go 33:00 9-45-9 could be possible, but can be ruled out by following If it is 9-45-9, the colored squares in row 8 would be 78 56 (cannot be 45 78). colored squared in row 1 would be 45 78. 9 in column 6 must be in box 2, which means it is 56789 up there, which means box 2 column 6 would be 569 triple. The top segment on the left can be 456 or 45123, but 456 is ruled out by 569 triple, so it would be 45123. This forces 78 pair in column 4 to be in row45, and now the 78 pair in column 6 has only 1 square to go, contradiction.
Hi Simon, great puzzle and solve. At 1:02 you eliminated the option of 1 in box R6C3. I think this is a mistake because you missed an option of the segments being 1 & 2 - 3 - 3. This would have probably been removed as an option by other logic as the puzzle progresses, but I think it is a mistake in the solve.
This was very interesting - kind of like region sum lines, but with region boundaries that one must find oneself. I really liked watching you solve this, Simon.
For anyone who didn't quite get it, the logic right after getting the 4 in r7c5 involves the fact that 5 is eliminated from r8c5 by the blue-orange "x-wing". That said, I still don't understand why 4 was forced into r7c5 in the first place, rather than a 5 with a 4 below it. If anyone will reference the 6 markings at that point, they are completely bogus, and the "x-wing" works on neither 4 nor 5. EDIT: While Simon doesn't explicitly disprove it, placing 4 in row 8 in box 7 forces 5 with it per prior logic. This is what ultimately eliminates 5 from r8c5.
@@minamagdy4126 Yes, but that is revisionist history on Simon's part, proving that you were right back then when you had not right to claim it at the time.
The sequence 9, 4, 5, 9 in box 8 (R7C4, R7C5, R8C4, R8C5) breaks Sudoku rules because the number 9 appears twice in the same 3x3 box. This repetition violates the rule that each number from 1 to 9 must appear exactly once in each row (R7, R8), column (C4, C5), and box (B8). Placing a 9 in both R7C4 and R8C5 disrupts the unique placement of 9 in rows 7 (R7) and 8 (R8), as well as columns 4 (C4) and 5 (C5). For example, R7C4 and R8C4 would both have a 9, which conflicts with columns C4 and C5.
@cypher686 what are you talking about? The rltwo cells that would have 9's in them would be r8c4 and r6c5, which share neither row, column, nor box. EDIT: all good by me
After reading the ruleset and the name of the constructor, I can predict this is *yet another* cosmic class construction. 😏👍 I am looking forward to solve it and then watching Simon's solve.
Sorry... answered already.... Original post: The dark green line from the pink 9 at minute 30 could be 9-5-4 too instead of 2-3-4 only. Later it would be excluded because of missing space for 2-3-6 in row 8.
At 1:01:34 Simon removes a 1 form r6c3, because he excluded 6 from r3c1, but couldn´t r3c1 still be another 3? (It´s late maybe I am missing something...)
It could. He (and I) missed it. I think the next step, getting 3 in r3c9, can still be done, but you need to allow for that line to be one of 3=1+2, 5=2+3 or 7=3+4. They all put 3 in r3c9 though, which rules out the missed 12,3,3 in box 4.
Maybe there was beautiful logic in this one, but it was very hard to see. It was a lot of guessing and checking all the options and very easy to miss one for me.
At 1:12:33, the grid's been reduced to a Deadly Pattern, and Simon reduces the 45 to a 5 in r3c1, turning the line into a 5-sum line with 5,3+2,5 line. However, what's stopping that line from being a 7-sum line with 4+3,2+5 ? I can't figure out what I'm missing there.
I made a mistake on the box 4 4-cell line, which led to me actually getting a 3 in the corner (and then breaking the puzzle). The only thing worse than not getting a 3 in the corner is thinking you've got one and then losing it.
Incredible puzzle, Nahelion! Really interesting ideas. I think Simon maybe prematurely gets his four on the bottom crooked line, because he doesn't seem to consider the possibly it could be a 5 instead, i.e 9 at the bottom, then 4,5 in the middle and 9 at the top?
When it says divide into non-overlapping segments, it's ruling out things like dividing a sequence 12412 into 124 and 412, where the two segments both sum to 7 but overlap on the 4 in the middle. Each digit on the line has to belong to one and only one segment. There might be a case for saying it's a bit redundant, as I don't believe you've truly *divided* the line into segments if the segments overlap. But it's spelling it out for clarity. 5235 is divided into the segments (5)(23)(5). As they don't share any instances of the digits on the line, they're said to be non-overlapping.
Wonderful solve Simon! Friendly tip, try and cross-reference your pencil marks with the digits you’ve correctly deduced, you may not get stuck as much and the solve will be more natural.
So, I figured out 1. A renban line of length n will have a sum divisible by n if n is odd, and n mod n/2 if n is even, and 2. you can't have segments of the same length in a row of the same length, so 3. The only way to have segments with the same length is if you hide the overlap in the wings, so you can have up to 3 length 2s and 2 length 3s, but no more, so 4. A segment of length 4 can only have even sums, and none of the other lengths can except for 8, and the largest length 4 sums to 30 and the smallest length 8 sums to 36, so there isn't a length 4 (or 8) segment. So I pretty quickly got to the small number of partitions of 12 that can follow these rules. And then spent 40 minutes failing to figure out an arrangement, because I didn't read the instructions that said the consecutive digits can be in any order. Lol. Oh, well. I'll finish either the puzzle or the video tomorrow.
Except that's not true. Odd-length renbans will have alternating odd-even sums as you increase the numbers. Even-length renbans will either always have odd or even sums, depending on whether they are divisible by 4.
Another ingenious puzzle. It's amazing how many great puzzles this channel manages to find. It's a measure how good they are that one puts up with Simon's mucky solves and fatuous patter.
Amazing puzzle It took me 53 minutes to solve and it was wild and i had the time of my life Thank you for featuring this and also for the setter that built it 😊
The break-in was considerably easier if only you used sudoku. After working out the composition of the two lines, you can ask where 1 and 9 go in R8, and they have to be at the ends. You then have 2 and 3 in R8 in box 8. This rules out the 123=6 option, forcing R9C4=9. The arms of the long lines at the bottom are either 456 on the left and 78 on the right, or 78 on the left and 56 on the right. On the short line, you can rule out 9,45,9 (which you didn't even consider), because that would put a 2 and 3 on the corners of the lines, and R7C4 would be 678 next to 23, but the 23 goes on a 1-5 line. The 9 line must therefore be 23/4/23. Now, in R7 4 must be on the left arm with 5, and it can't be part of the 1-5 segment because you can't have three 23s in a box, so 6 is in R8C4, and 78 is on the lower-right arm. You can then put 78 and 56 on the left and right upper arms. You can't put the second 78 at the top of the left line, so it must go on top of the 6, and the rest of the line is 1-5 . The 23 in R8C6 means the 2-8 segment is at the bottom of the C6 line, and the top needs 789. Putting in the options for the short lines in R6/7, you get a 1237 quad in R7 and a 34 pair in R6. Most of the rest is just sudoku. You tried to make the 45 on the bottom left arm be part of the 12345 segment, happily proclaiming you can put a 23 pair on the line just moments after getting your first digit by noting that both 2 and 3 were in R8, and already having one 23 in R8C5. It can't be the 1-5 segment and therefore must be the 456 segment, putting 78 on the bottom-right arm and 56 on the top-right (which you could have got after putting 78 on the opposite arm). You might only be able to think about one constraint at once, but having come up with what you think is plausible (i.e. a 23 pair in R7/8C4 and 1 in R6C4), stop and ask yourself whether it breaks any of the other rules, like only having two 23s in a box. I'm 40 minutes in, you've just broken in (the hard part), but it's been 5 minutes since you got the 4 and it's already going downhill. After seeing there's still another 33 minutes, when it should only take you more like 10-15 (from getting the 4), I can't bear to watch the rest of the video. There's no way I'm sitting through that; my blood pressure can't take it. The central three columns virtually fill themselves in, and writing candidates for the short lines leads to doubles/triples/quads, which resolves the rest with alacrity. It really is simple from this point on. Bring back 2020 Simon, who could still do sudoku. This Stepford Simon who is programmed to think only about 3s in the corner, Maverick, flies and poltergeists isn't a patch on the Real Simon who taught me a lot about the way to attack harder puzzles.
This is funny, Supposing Tn is the triangular number with n digits you have that T9 - Tn = Tm (the leftover of a triangular 9 - another triangular sequence being also a triangular total) or Tn + Tm = T9 That's the equation of a circle in -1/2, -1/2 This has a bunch of negative integer solutions. I wonder if you can make a new puzzle with subtracting totals with a similar renban region sum
35:35 Wĥy did R8C46 have to include a 6? It could have been in the wings on one side or the other as pencil marked. I know that there was another logical error immediately afterwards but this one stumped me.
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The fact that there is an interesting puzzle everyday consistently is insane to me! How are there so many interesting puzzles and how are you guys finding them all🤭
And how are there so many cosmic-class constructors?
Interesting puzzles help create interesting puzzles. They inspire old and new setters alike so more and more puzzles are being made. CtC gets many of these puzzles submitted directly from the setters are recommended by other puzzle enthusiasts. There are literally thousands of amazing puzzles that this channel doesn't have the time to showcase because of this. It's a really fun hobby and a blast to make puzzles
Logic masters Germany has heaps of puzzles and people will email in to suggest the ones they find interesting/beautiful/challenging. There are certain constructors you know are absolutely going to provide amazing puzzles. They also have their discord channel and such where people discuss their favourite puzzles. Famously recently there was one from reddit submitted too, so there’s quite a few places to find puzzles it’s just wilting then down to brilliant ones but that’s also not too difficult to do either.
Plus ctc do have testers who will ensure that good puzzles are passed along - although some videos start with Simon saying the testers couldn’t crack it and want him to give it a go (probably for their sanity, to see the break in)
They're finding but most important can solve them 😊
There are a lotttttt of talented setters working today
I now think in Simon's voice when I'm solving these Sodoku 😂
Nahileon wins the award for Most-Aptly-Named Puzzle.
"where did I get a 6 pencil mark from?"
I had that very question. LMAO
It was right and deducable though: there's a floating 578-triple in row 8 and only 3 squares left for them in box 8 (that are not in row 8).
@@Illithien He didn't explain it. And he didn't even see the logic at that point. So it was a "right answer for a wrong reason" type issue.
That was nuts!!! Took me over 4 hours of timed thinking plus some extra time over 4 days to figure that out. But I persevered and finally conquered the hardest puzzle I've ever done without Simon's help!!
I know that doesn't mean much to the pros on this site but I'm damn proud of myself. Still gotta give credit to all the setters and solvers that I've learned from over the last 2-3 years.
Thanks for the birthday shoutout! Jill here. In order the names of the birds are Kiwi, Mocha, and Lemon. I suppose Lemon was smiley because she was a baby when I took the picture. And yes, the chantilly cake was delicious, or passable in your terms 😅
Lovely happy birds!!
@@longwaytotipperaryand happy birds ❤ you. 😁
@@davidrattner9 🦜🦜❤️
Amazing solve and amazing puzzle❤.
At 59:08 the 4 line could be from bottom to top 1 2 / 3 / 3, this gets disproved later by sudoku but I think Simon missed this possibility so there is a bit flawed logic because he didnt mark 3 as a candidate in R3C1
YES! I was looking at that option as well ... I was going to leave a comment asking if I was missing something.
This is exactly the type of puzzle that Simon's brain is simply way more adept at dealing with than mine. I get lost in the possibilities and don't ever manage to make any meaningful deductions.
At 33:00, is there a trivial reason why the 4 cell line in box 8 couldnt have been 9, 4 5, 9? The 5 is immediately ruled out of r8c5 but the 4 isnt.
The 2 and 3 in row 8 need to be in box 8, so r7c5 is a 4 or 5. This still makes r9c4 a 9 which makes r8c2 and r8c3 a 45 pair.
No, Simon should absolutely have considered 9-45-9. He would eventually have discarded it, but it wasn’t trivial.
The answer you are looking for is, there is no reason Simon had to get rid of the possibility, he makes this kind of mistake all the time.
@@wrightn9 How does r9c4 being 9 make r8c2 and r8c3 a 45 pair? They can be a 78 pair too
I believe a similar type of scenario happens at 59:10, where it was concluded a 6 must be in r3c1 if there is a 1 in r6c3. However, there could actually be another 3 in r3c1 if the line total was 3 rather than 6.
I finished in 165 minutes. There was no way I was able to do this in my head. I had to write down on a piece of paper all the possibilities and reference it. It worked out quite well and I was able to reduce it down to 15, 21, and 35. I knew 35 had to belong, because a 9 had to be used. Ruling out 21 was clever, due to the six cell region having to go in the center flanked by two three cell regions consisting of 678. This left no 78's on any of the flank positions for the 35 side. After that, I slowly made my way to the finish, partially messing up due to forgetting that 5 is a possibility for the line in box 4. Great Puzzle!
That's pretty much how I got along on this puzzle also. Except the possibility of the 4 cell lines being split into three segments did not leap out at me immediately.
Imagine my consternation on Day 4 of this solve.
The way Simon overlooked that the 4 cell 9 line could have been 9 4 5 9 was terrifying lol
He didn’t overlook it….he mentioned it couldn’t be a 9 45 9 line because of the 45 pair in R8C2
@@LovethectcHe overlooks 9-45-9 as an option at around 33:00, when pencil-marking the options along the four-cells line in box 8. There isn't a 45 pair in r8c2 at this stage.
came here to say the same thing... properly distressed by that missed option
😮Yes, overlooked but it's easily not possible since it would require a 9 in box 2/column 6 and a 4 in R8C5 leading to 456 in column 4/ rows 6-8 but R8C4 must be a 2 or 3. I'm sure he thought it was blindingly obvious and just neglected to mention it.
Or maybe he just got lucky.
@@grantfraser5430 "box 3/column 6" ?
In the upper line in box 4, there was a missing possibility : 3,3,2,1. Otherwise the new rule and the solve are awesome as usual
33:08 Am I mistaken or could it have been 9 - 4,5 - 9? (in general)
I was thinking the same thing I’m sure it might cause issues on the big lines but I would want to explore.
At 48:55 he talks about that possibility on the other line and then says "but that's different than this middle line" but doesn't explain why.
So yeah it does seem like it's an oversight
Yeah, when he places the first digit, I don't see why it couldn't have been a 5 instead
He doesn't address this possiblility, but it can be ruled out. By the 5 x-wing, it would have to be r8c5=4 and r7c5=5. This forces a 2/3 pair in r8c4 and r8c6. The 4 in r8c5 means that r8c2/r8c3 cannot be a 4/5 pair, and instead are a 7/8 pair. This is a full segment, and so the squares r8c4 and r7c4 are also part of the same segment. r8c4 is 2/3 meaning it must be a part of the segment 1,2,3,4,5. However r7c4 must be one of 6,7,8 since they are the only remaining values to be placed in box 8, which is not possible if r7c4 is in the 1,2,3,4,5 segment.
Okay, regarding Simon missing the 9-4-5-9 option at 33:00 - a way of removing that option is the following. We first get a 19 pair at the ends of the column. That puts Row 8's 2 and 3 in Box 8, along with a 4 or a 6. The other 4 or 6 accompanies the 5,7, and 8 in R8 C2378. This removes 2 and 3 from R7C5, which in turn removes the option of 1 from R9C4, placing 9 there (our first digit). This now means that the R8 4 or 6 that's not in Box 8 is either a 4 in Box 7 or a 6 in Box 9. (Note that at this point, R8C5 is either 2, 3, or 4.) If the 4 is in Box 8, that removes 4 from R8C5. if the 6 is in Box 9, this must go with 5 in Box 9 and 4 in R8C6, also removing the 4 from R8C5. This removes the missed 9-4-5-9 option, and we may now proceed with the puzzle without despairing.
If only the segments always ended at region boundaries so you could call them Rengion Sum Lines.
22:24 "thats Mark calling me ill have to"
oh dang he has to leave and come back
"make it go away"
😮 lmao
😂
32:29
Fascinatingly clever, and the ending where it kept looking like it would crack and didn't, and didn't, and didn't and finally did was marvellous.
The deduction on 1:01:30 is wrong. The line could have been a [1,2][3][3] without breaking the rules.
I saw that too, he just got lucky I guess
No it can’t … there is a 3 in R6C7 that prevents that
yeah, at that moment he could have deducted row 3 col 9 to be 3 and removed this possibility
@@Lovethectc I was moving from R6C3 - 1. To R3C1 - 3. And then 2 and 3 in R4-5C2
Yes but that still leaves a 2 and 3 in the middle 2 boxes of the line which is the deduction that mattered at the moment, idk if leaving the posibility of a 1 at r6c3 and a 3 in r3c1 would habe made an impact later
Rules: 04:55
Let's Get Cracking: 07:15
Simon's time: 1h5m28s
Puzzle Solved: 1:12:43
What about this video's Top Tier Simarkisms?!
Three In the Corner: 2x (1:07:32, 1:10:00)
The Secret: 2x (15:39, 15:46)
Bobbins: 1x (18:36)
Phistomefel: 1x (00:30)
And how about this video's Simarkisms?!
Hang On: 18x (10:15, 10:46, 13:25, 13:25, 15:04, 26:31, 29:41, 35:39, 35:39, 39:17, 39:17, 41:55, 55:57, 56:16, 56:54, 56:58)
Triangular Number: 14x (10:59, 13:44, 13:49, 14:05, 14:53, 14:55, 15:06, 15:10, 16:56, 18:03, 18:13, 24:59, 33:18, 59:01)
Ah: 11x (11:41, 11:41, 18:26, 22:06, 29:41, 40:38, 40:38, 41:16, 1:06:45, 1:06:45, 1:07:55)
Sorry: 9x (00:40, 15:42, 22:16, 23:44, 32:18, 56:31, 58:12, 58:24, 1:02:05)
Weird: 7x (18:36, 31:51, 36:21, 41:16, 41:16, 42:42, 50:07)
Obviously: 6x (04:25, 17:42, 21:44, 28:01, 31:56, 56:31)
Cake!: 6x (02:41, 02:52, 03:16, 04:06, 04:13, 04:15)
Bizarre: 5x (00:47, 04:51, 36:21, 36:24, 1:13:18)
Surely: 4x (47:12, 51:27, 56:39, 57:51)
Nonsense: 3x (16:26, 39:29, 58:28)
Beautiful: 3x (27:07, 27:09, 28:29)
Brilliant: 3x (02:47, 28:33, 28:35)
Masterpiece: 3x (00:58, 01:26, 1:12:45)
Goodness: 2x (59:37, 1:05:06)
The Answer is: 2x (31:10, 35:57)
By Sudoku: 2x (42:16, 1:05:44)
Pencil Mark/mark: 2x (1:07:27, 1:08:53)
Good Grief: 1x (1:12:36)
What on Earth: 1x (57:55)
Bother: 1x (59:28)
Naked Single: 1x (43:25)
Missing Something: 1x (50:28)
Naughty: 1x (41:10)
Lovely: 1x (16:41)
Ridiculous: 1x (1:00:22)
Going Mad: 1x (39:20)
Gorgeous: 1x (36:18)
Take a Bow: 1x (1:13:28)
Discombobulating: 1x (1:07:30)
Think Harder: 1x (1:08:49)
In Fact: 1x (11:07)
Whoopsie: 1x (07:59)
Wow: 1x (26:37)
Fabulous: 1x (04:37)
What Does This Mean?: 1x (25:03)
That's Huge: 1x (35:47)
Have a Think: 1x (22:50)
Most popular number(>9), digit and colour this video:
Fifteen (14 mentions)
One (141 mentions)
Orange (4 mentions)
Antithesis Battles:
Odd (11) - Even (10)
Row (15) - Column (11)
FAQ:
Q1: You missed something!
A1: That could very well be the case! Human speech can be hard to understand for computers like me! Point out the ones that I missed and maybe I'll learn!
Q2: Can you do this for another channel?
A2: I've been thinking about that and wrote some code to make that possible. Let me know which channel you think would be a good fit!
The robot told me to help it, so here you go. You missed a "Mark" at 22:23
Intriguing puzzle. Having watched Simon's solve, glad to see I got most of the logic, if not always quiet as elegantly! Like him was tripped up by the line in box 8, and box 4. Unlike Simon, nobody will deem it worthy of comment, I'm sure.
As a non-English speaker, I just love learning new words by watching these videos. Today's word was "kibosh". Past words include "recalcitrant" and "alacrity". Thank you, Simon, for expanding my vocabulary. :)
To those who have been struggling with the 9-45-9 possibility on the middle line, I found the solution:
-You can't put a 5 in R8C5, so it must be a 4, with a 5 in R7C5.
-This forces a 78 on the left hand line "wing" (R8C2/3)
-R8C4/6 must be a 23 pair to complete the row.
-We already have a 1 in the box.
-The 2/3 in R8C4 must be part of the 12345 sequence, but there is no digit left to put in R7C4, because all of the digits have been used in box 8 already.
Okay, regarding Simon missing the 9-4-5-9 option at 33:16 - a way of removing that option is the following. We first get a 19 pair at the ends of the column. That puts Row 8's 2 and 3 in Box 8, along with a 4 or a 6. The other 4 or 6 accompanies the 5,7, and 8 in R8 C2378. This removes 2 and 3 from R7C5, which in turn removes the option of 1 from R9C4, placing 9 there (our first digit). This now means that the R8 4 or 6 that's not in Box 8 is either a 4 in Box 7 or a 6 in Box 9. (Note that at this point, R8C5 is either 2, 3, or 4.) If the 4 is in Box 8, that removes 4 from R8C5. if the 6 is in Box 9, this must go with 5 in Box 9 and 4 in R8C6, also removing the 4 from R8C5. This removes the missed 9-4-5-9 option, and we may now proceed with the puzzle without despairing.
I missed a little bit here: 6 in Box 9 puts 4 on the left line in Row 1 with 5, thus making Row 8 Box 7 a 78 pair and Box 9 the 6 goes with a 5. Now if we put a 4 in Row 8 Column 5 it forces a 5 above it, making for 9-4-5-9 (as we had limited that cell - R7C5 to 4 or 5), we break Row 8 Column 4, as this needs to be part of a 456 or a 12345 (due to the 78 in Box 7 seeing it). it can't be 456, as it is seen by all of these. it can be on of the 23s from 12345, but the cell above it then sees 12345. This means that when there is 6 in Box 9 on that line, there still can't be a 9-4-5-9 run on the line. (which we previously removed in the 4 in Box 8 version). I hope that there's a more fluid way to prove that, because it felt a little clumsy to me, but I hope that does the trick for you :)
Amazed I managed this (albeit with a couple of careless missteps) -can't usually do the videos over 1hr
59:48 for me. Wow what a puzzle. I got very stuck on a wrong assumption and it wasn't until I took a break and came back that I realized what I was doing wrong. Fantastic puzzle, but it almost broke me.
58:19 I think what Simon was looking for here is that the sum of three consecutive digits _must be divisible by 3_, which rules out the 8 sum.
My favorite part of solving this puzzle especially towards the end is I felt like I had to keep going in circles around the puzzle to resolve a clue back in the side I started with and just round and round and round it unwound itself.
It’s a really good puzzle that loops the logic right to the end like this.
Simon got lucky with the line in box 8. It could have a 9 in both ends, a 4 in R8C5 and a 5 in R7C5 which would create a 5678 quadruple in row 8 and a 4578 quadruple in row 2. It would quickly break the puzzle as it would force two 6s into box 2 and two 1s into box 8. But he didn't even consider that possibility, and therefore never made the deduction that it would break the puzzle.
Great to see this one featured - interesting puzzle
N-Sum Line: Divide the line into at least two non-overlapping segments. The digits in each segment sum to the same total ‘N’.
I’ve seen this used several times before… even on this channel. Although this is the first that combined it with segmented renbans.
An amazing puzzle! 5 bags of popcorn out of 5!
though I must say: I don't know why you even bother with puzzles that don't have a 3 in the corner... 🙃
41:20 for me. that 4 line at the left was very bs, but the break in was satisfying
I really like your channel! Makes me chill and have been helping a lot with my English and anxiety at the same time 😂 Hello from Brasil ❤
Wow! What a puzzle! That one was way above my pay grade. It's such a relief when Simon stops doing the really hard stuff and does that arcane thing called "Sudoku". A little know art form, known only to a few cogniscienti. (I probably have spelt that incorrectly, because I am not one).
Thanks Simon
this was fantastic! and indeed bizarre
Unrelated to the puzzle, but I love cockatiels. They're so precious!
Oh, Simon... Such a genius in the opening stages, I was looking in awe how he was navigating the breakthrough, I myself gave up on that. And then second part, when I was literally screaming at the screen, but Simon refused to see any of the single step or even zero-step deductions...
I did not need to use Simon's brilliant initial deduction (see video @16:32). I used an *aide memoire* to list the possible totals for 2-cell, 3-cell, ..., 7-cell segments (8-cell and 9-cell segments are incompatible with the grey lines). Then I coloured the *common totals:*
*15, 21, 30* and *35*
from which I easily ruled out *21* and *30* due to incompatibility with the long *grey lines.*
*Common totals* are needed because it is impossible to divide a long line into segments of the same size.
Simon's way to show (@24:01) that the long line containing a *9* was divided into segments with *total = 35* was sound, but I think it was not clear enough to be understood by everybody. The same is true for the long line with *total = 15.*
I think he should have used an *aide memoire* to make it crystal clear, although he personally does not need it because of his powerful working memory.
2:02:41 - Phew! That was HARD! I worked out the two big lines pretty quickly but it took me ages to prove which way round they went! It didn’t get much easier even when I’d got that! Excellent puzzle though.
how bizarre and wonderful. GPT helped me write a python script to get all the possible combinations. Here they are:
3: 12/3
5: 23/5
6: 123/6
7: 34/7
9: 234/45/9
15: 12345/456/78
18: 3456/567
21: 123456/678
30: 45678/6789
35: 2345678/56789
Wish I could have given this puzzle extra hearts. Such lovely satisfying interplay of lines, and lots of subtle knots after the initial break in. Needed to refer to the solve to give me a hint or two and would have missed a couple of line options without it. Such is the way with new rule sets!
59:36 Is there a reason the 4 cell line segment could not have been a 1 2 pair, and a 3 and a 3?
Nothing so far, mmm, there could be two solutions then. I guess Simon needs to revisit this puzzle and make sure and to do it without that assumption and see what happens.
Just sudoku on row 6
No reason other than he missed it. I have to hold my hand up and admit I also missed it.
At least this one doesn't skip too much logic. The three-cell line on the right, between boxes 3 and 6, still has to be split into two digits summing to one digit. The one digit has to be odd. The whole line sees 9's, so it sums to 3, 5 or 7. In all cases, it requires a 3 on the line (3=1+2, 5=2+3, 7=3+4). The 3 in r6c7 forces that 3 into r3c9. That removes 3 as an option from r3c1.
(A harder oversight to get past is that he also overlooked 9-45-9 as a possibility at around 33:10. More work needs to be done to rule out this as an option. It's been discussed in other comments).
Sorry, missed what you meant, it could be a three line with a 3 in r5c2.
The resolution would come a bit furhter ahead, once the three cell line on the right hand side of the grid is also considered and a 3 is forced into r3c9. So there are not two solutions. But it was an oversight at this stage of the solve.
It ends up breaking the remaining line, which either needs to have a 3 or be a 4,5,9.
Says something about the commutity that I tuned in 4 minutes after the vid dropped and it already has 28 likes!
Absolutely brilliant and adventurous puzzle.
The beauty of this puzzle is i started a 9x10 empty grid ...the 8 midrows centers on the long 12cell lines....then plopped the other lins in after discovering what ur suppose ti....just glad.i didnt verrtically flip
If the oddcircle was evensquare(an 8) there are way to many solutions
55:40 "How do we do this then?" Simon wonders as there is a 12-pair just right above a cell labelled "19". Totally missed that one when he found out the 12-pair... almost 2 minutes earlier. Then finally uncovering what should be where through sudoku in a totally different box. Never change, Simon, never change! :P
Thanks to watching so many of your awesome sudoku videos I'm actually starting to remember some of the triangular numbers (up to 6 now)
Because of this I actually figured out the possible segment totals for the 12 line with the one before Simon while watching him ponder it - which is a rare occurrence, that something maths related in these clicks for me so fast! It's all thanks to your wonderful teachings 😊
1:23:25 for me. I made a mistake. Didn't account for an option and had to unwind half the puzzle to see what I'd done wrong. Nonetheless an amazing puzzle. Fun solve
I used a different approach in solving the two long lines, I just brute forced the lines to be 2(78)-2(78)-3(456)-5(12345) long and 5(56789)-7(2345678) long through a lot of mental calculations. Then I noticed where the 78's should go
33:00 9-45-9 could be possible, but can be ruled out by following
If it is 9-45-9, the colored squares in row 8 would be 78 56 (cannot be 45 78). colored squared in row 1 would be 45 78. 9 in column 6 must be in box 2, which means it is 56789 up there, which means box 2 column 6 would be 569 triple. The top segment on the left can be 456 or 45123, but 456 is ruled out by 569 triple, so it would be 45123. This forces 78 pair in column 4 to be in row45, and now the 78 pair in column 6 has only 1 square to go, contradiction.
Horrific pencil marking, but would Mark pencil mark it? Ha.
Hi Simon, great puzzle and solve. At 1:02 you eliminated the option of 1 in box R6C3. I think this is a mistake because you missed an option of the segments being 1 & 2 - 3 - 3. This would have probably been removed as an option by other logic as the puzzle progresses, but I think it is a mistake in the solve.
This was very interesting - kind of like region sum lines, but with region boundaries that one must find oneself. I really liked watching you solve this, Simon.
1:05:06 for me - wow, that was tricky, I feel so good for figuring it out.
35:23 why do these squares have to include 6?
also, couldn't the 9 line be 9, 45, 9?
oh i should just keep watching
For anyone who didn't quite get it, the logic right after getting the 4 in r7c5 involves the fact that 5 is eliminated from r8c5 by the blue-orange "x-wing".
That said, I still don't understand why 4 was forced into r7c5 in the first place, rather than a 5 with a 4 below it. If anyone will reference the 6 markings at that point, they are completely bogus, and the "x-wing" works on neither 4 nor 5.
EDIT: While Simon doesn't explicitly disprove it, placing 4 in row 8 in box 7 forces 5 with it per prior logic. This is what ultimately eliminates 5 from r8c5.
@@minamagdy4126 Yes, but that is revisionist history on Simon's part, proving that you were right back then when you had not right to claim it at the time.
The sequence 9, 4, 5, 9 in box 8 (R7C4, R7C5, R8C4, R8C5) breaks Sudoku rules because the number 9 appears twice in the same 3x3 box. This repetition violates the rule that each number from 1 to 9 must appear exactly once in each row (R7, R8), column (C4, C5), and box (B8). Placing a 9 in both R7C4 and R8C5 disrupts the unique placement of 9 in rows 7 (R7) and 8 (R8), as well as columns 4 (C4) and 5 (C5). For example, R7C4 and R8C4 would both have a 9, which conflicts with columns C4 and C5.
@cypher686 what are you talking about? The rltwo cells that would have 9's in them would be r8c4 and r6c5, which share neither row, column, nor box.
EDIT: all good by me
Fabulous puzzle from the creative mind of Nahelion!! Every day we get to see something special unfold before our very eyes!!
After reading the ruleset and the name of the constructor, I can predict this is *yet another* cosmic class construction. 😏👍
I am looking forward to solve it and then watching Simon's solve.
I was right, of course
This is just a delightful puzzle
Sorry... answered already....
Original post: The dark green line from the pink 9 at minute 30 could be 9-5-4 too instead of 2-3-4 only. Later it would be excluded because of missing space for 2-3-6 in row 8.
At 1:01:34 Simon removes a 1 form r6c3, because he excluded 6 from r3c1, but couldn´t r3c1 still be another 3? (It´s late maybe I am missing something...)
It could. He (and I) missed it.
I think the next step, getting 3 in r3c9, can still be done, but you need to allow for that line to be one of 3=1+2, 5=2+3 or 7=3+4. They all put 3 in r3c9 though, which rules out the missed 12,3,3 in box 4.
Yeah the 2 3s along with the 9s in boxes 5 and 9 break the remaining line
Well shoot, time to start this one over. Just noticed the words "in any order."
Maybe there was beautiful logic in this one, but it was very hard to see. It was a lot of guessing and checking all the options and very easy to miss one for me.
How can you possibly do a break in like that in your head and not be able to scan a row or column or box?
At 1:12:33, the grid's been reduced to a Deadly Pattern, and Simon reduces the 45 to a 5 in r3c1, turning the line into a 5-sum line with 5,3+2,5 line. However, what's stopping that line from being a 7-sum line with 4+3,2+5 ? I can't figure out what I'm missing there.
The segments need to contain consecutive digits, so 2 and 5 wouldn't work.
33:47 Why is the green line not a 4/5 in box 8 with another 9 in box 5?
I made a mistake on the box 4 4-cell line, which led to me actually getting a 3 in the corner (and then breaking the puzzle). The only thing worse than not getting a 3 in the corner is thinking you've got one and then losing it.
Can we do 2000 more before summer solstice?
Incredible puzzle, Nahelion! Really interesting ideas. I think Simon maybe prematurely gets his four on the bottom crooked line, because he doesn't seem to consider the possibly it could be a 5 instead, i.e 9 at the bottom, then 4,5 in the middle and 9 at the top?
I guess I don't understand the non overlapping rule, isn't the 2 and 3 in column 2 overlapping with the 5 on each end??
When it says divide into non-overlapping segments, it's ruling out things like dividing a sequence 12412 into 124 and 412, where the two segments both sum to 7 but overlap on the 4 in the middle. Each digit on the line has to belong to one and only one segment.
There might be a case for saying it's a bit redundant, as I don't believe you've truly *divided* the line into segments if the segments overlap. But it's spelling it out for clarity.
5235 is divided into the segments (5)(23)(5). As they don't share any instances of the digits on the line, they're said to be non-overlapping.
Incredible solve!
8:30 It would, in fact, be a Goodliffe of Phistomefellian proportions!
This rules, at least, are understandable !
Wonderful solve Simon! Friendly tip, try and cross-reference your pencil marks with the digits you’ve correctly deduced, you may not get stuck as much and the solve will be more natural.
Whoa!!!!!!!!!
You didn't actually use that gray dot clue in the solve, did you?
You seem really happy today
It’s so sad when Simon /wants/ to do sudoku 😢
@59:05, can't it be a 1,2 pair with a 3 in r4c2 and another 3 in r3c1?
Simon missed it, but it breaks the remaining line, (see if you can figure out why 😉)
So, I figured out 1. A renban line of length n will have a sum divisible by n if n is odd, and n mod n/2 if n is even, and 2. you can't have segments of the same length in a row of the same length, so 3. The only way to have segments with the same length is if you hide the overlap in the wings, so you can have up to 3 length 2s and 2 length 3s, but no more, so 4. A segment of length 4 can only have even sums, and none of the other lengths can except for 8, and the largest length 4 sums to 30 and the smallest length 8 sums to 36, so there isn't a length 4 (or 8) segment. So I pretty quickly got to the small number of partitions of 12 that can follow these rules. And then spent 40 minutes failing to figure out an arrangement, because I didn't read the instructions that said the consecutive digits can be in any order. Lol. Oh, well. I'll finish either the puzzle or the video tomorrow.
Except that's not true. Odd-length renbans will have alternating odd-even sums as you increase the numbers. Even-length renbans will either always have odd or even sums, depending on whether they are divisible by 4.
this video is great but the thumbnail on marks video was horrifying i hate snakes so much
Another ingenious puzzle. It's amazing how many great puzzles this channel manages to find. It's a measure how good they are that one puts up with Simon's mucky solves and fatuous patter.
52:43 am I missing something or could there also have been 3|5|8? With the 5 in r6c2. Is it better to be lucky sometimes?
5 and 3 are not consecutive.
I was wondering about the 1, could it possibly have been 1233? I can't see why it wouldn't, anything I missed?
Sorry, the line above I mean.
Amazing puzzle
It took me 53 minutes to solve and it was wild and i had the time of my life
Thank you for featuring this and also for the setter that built it 😊
Welp, it doesn't often happen anymore, but this one hurt my brain.
The break-in was considerably easier if only you used sudoku. After working out the composition of the two lines, you can ask where 1 and 9 go in R8, and they have to be at the ends. You then have 2 and 3 in R8 in box 8. This rules out the 123=6 option, forcing R9C4=9. The arms of the long lines at the bottom are either 456 on the left and 78 on the right, or 78 on the left and 56 on the right.
On the short line, you can rule out 9,45,9 (which you didn't even consider), because that would put a 2 and 3 on the corners of the lines, and R7C4 would be 678 next to 23, but the 23 goes on a 1-5 line. The 9 line must therefore be 23/4/23. Now, in R7 4 must be on the left arm with 5, and it can't be part of the 1-5 segment because you can't have three 23s in a box, so 6 is in R8C4, and 78 is on the lower-right arm. You can then put 78 and 56 on the left and right upper arms. You can't put the second 78 at the top of the left line, so it must go on top of the 6, and the rest of the line is 1-5 . The 23 in R8C6 means the 2-8 segment is at the bottom of the C6 line, and the top needs 789.
Putting in the options for the short lines in R6/7, you get a 1237 quad in R7 and a 34 pair in R6. Most of the rest is just sudoku.
You tried to make the 45 on the bottom left arm be part of the 12345 segment, happily proclaiming you can put a 23 pair on the line just moments after getting your first digit by noting that both 2 and 3 were in R8, and already having one 23 in R8C5. It can't be the 1-5 segment and therefore must be the 456 segment, putting 78 on the bottom-right arm and 56 on the top-right (which you could have got after putting 78 on the opposite arm). You might only be able to think about one constraint at once, but having come up with what you think is plausible (i.e. a 23 pair in R7/8C4 and 1 in R6C4), stop and ask yourself whether it breaks any of the other rules, like only having two 23s in a box.
I'm 40 minutes in, you've just broken in (the hard part), but it's been 5 minutes since you got the 4 and it's already going downhill. After seeing there's still another 33 minutes, when it should only take you more like 10-15 (from getting the 4), I can't bear to watch the rest of the video. There's no way I'm sitting through that; my blood pressure can't take it. The central three columns virtually fill themselves in, and writing candidates for the short lines leads to doubles/triples/quads, which resolves the rest with alacrity. It really is simple from this point on. Bring back 2020 Simon, who could still do sudoku. This Stepford Simon who is programmed to think only about 3s in the corner, Maverick, flies and poltergeists isn't a patch on the Real Simon who taught me a lot about the way to attack harder puzzles.
This is funny,
Supposing Tn is the triangular number with n digits
you have that T9 - Tn = Tm (the leftover of a triangular 9 - another triangular sequence being also a triangular total) or Tn + Tm = T9
That's the equation of a circle in -1/2, -1/2
This has a bunch of negative integer solutions.
I wonder if you can make a new puzzle with subtracting totals with a similar renban region sum
68:15 for me
I gave up watching after Simon overlooked the 45,9 just before placing first digit. Not up to his usual standard
I'm going to skip this one. Looks too difficult.
35:35 Wĥy did R8C46 have to include a 6? It could have been in the wings on one side or the other as pencil marked. I know that there was another logical error immediately afterwards but this one stumped me.
Keep it simple, Simon . A 1/2 pair in box 9 deduce the 2/9 cell just below to a 9 😉🙃🫡