Let's Get Cracking: 04:49 What about this video's Top Tier Simarkisms?! Three In the Corner: 3x (42:26, 42:28, 42:34) Bobbins: 2x (21:20, 43:44) The Secret: 2x (02:24, 03:58) Knowledge Bomb: 1x (43:13) Cooking with Gas: 1x (22:37) And how about this video's Simarkisms?! Sorry: 11x (11:30, 12:05, 21:20, 26:26, 31:51, 34:41, 35:24, 35:32, 35:32, 39:37, 54:58) Ah: 10x (00:18, 00:35, 08:12, 16:00, 21:20, 30:58, 42:56, 49:34, 55:50, 58:53) By Sudoku: 8x (19:39, 28:33, 40:28, 48:34, 51:07, 55:12, 58:33, 58:57) In Fact: 6x (42:28, 55:20, 55:31, 57:21, 57:23, 58:13) Hang On: 5x (15:39, 19:29, 34:23, 35:29, 55:46) Pencil Mark/mark: 5x (08:31, 47:19, 49:47, 52:52, 55:50) Surely: 4x (36:06, 39:16, 43:21, 48:04) Obviously: 4x (06:57, 07:00, 14:22, 19:54) Clever: 2x (51:53, 1:00:59) Brilliant: 2x (01:10, 02:33) Shouting: 2x (03:18, 03:28) The Answer is: 1x (13:11) Missing Something: 1x (09:18) Bingo: 1x (29:39) Lovely: 1x (03:31) Beautiful: 1x (16:11) Fascinating: 1x (1:01:43) Gorgeous: 1x (57:41) Come on Simon: 1x (52:17) Proof: 1x (56:47) Full stop: 1x (18:56) Progress: 1x (40:54) Wow: 1x (1:00:01) Almost Interesting: 1x (29:02) That's Huge: 1x (29:16) Cake!: 1x (04:19) Most popular digit this video: One (145 mentions) Antithesis Battles: High (10) - Low (7) Even (13) - Odd (2) Higher (4) - Lower (2) Column (4) - Row (2) 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!
Some piece of logic I used extensively but didn't see used in the puzzle nor in the comments: When you go from one digit to another (let's take the 6 in r6c4 as the example), you can either jump to a next digit by two or by one. However, two adjacent digits on a line that are 1 apart form a WALL. if you were to put a 5 next to a 6 then your next digit can be either higher or lower than your wall (e.g. 4 or 7), but you can't get to the other side of the wall anymore. You need a 5 or a 6 to go back from 4 to 7. So what you need to do is keep a gap for the return journey. With this logic you could instantly see that the 7 at r1c1 couldn't have a 6 nor an 8 next to it and thus needed a 5 and 9 to it's sides with the 9 obviously followed by 8,6,4 (which I would call the return journey).
The tricky thing here is to figure out when having a wall is okay (because you can just put higher and lower digits on both sides, or have space outside the line), and when it's not okay (e.g. because you have a loop, or all your digits need to be on the line and there is not enough space).
I kept wondering how in the world this puzzle is supposed to be solved without bifurcation, since even Simon had to resort to it over and over. This makes sense
Simon asks if there is easier way to solve box 3 at 29:57. I believe this is easier: you know (r3c8,r3c9) is either some combination of (5,7) or (6,8). Any combination of (6,8) doesn't work because you need to put 4 in either r2c8 or r2c9. (7,5) doesn't work because you need to put 6 in r2c7, so the only option is (5,7). The rest follows directly
Yeah because the long string needed either 5-7 or 6-8. Box 3 had one of those pairs, couldn't be 68 cause that would leave them both connected to a 7 due to 456-89 taken by then. Easy work when you put 57 pair to know they need to go with 3 and 8, the 8 connecting to a 6 rules left side out so 786 in right side.
great puzzle as always. i started this differently, using a trick that was pretty useful in a few places. if you put 2 consecutive digits together, they make a wall that can't be crossed until you leave the box. this means that every digit on one side of the wall has to be all higher digits than the wall and the other side has to be all lower digits than the wall. you can use the wall trick right at the start. if you put a 6 or 8 next to the 7 in box 1, then you make an either 67 or 78 wall. this means that you have to put all high digits on one of the 2 branches (either going right or down from the 7). since you have to fill one of the branches with only high numbers plus the wall digit, this is impossible (you have 3 digits, 689, but 4 cells to fill). this gives us a 59 pair next to the 7 which lets us easily fill the first box like simon did. another example: you can apply this "wall" trick at 56:48, you can say that r6c5 can't be 7 because you only have 1 branch and so couldn't put either any high digits or any low digits in the center box.
23:43 for me which shocked me in hindsight, my times are usually comparable to the video length or worse. I skimmed the video to see if I had noticed techniques that were more efficient than Simon's. The two things I noticed were: (1) heavier focus on 1s and 9s when first looking a box, it was surprisingly easy to whittle those down (2) noticing that if you hit a 9 or 1 part way along a path, they have to "get out" so to reach them you had to go even-in odd-out or vice versa (helpful for example in box 1 -- you could see that it needed 7986 one direction or the other, and the given 6 immediately told you which way).
Did it in 35 minutes. Found it not that hard. Key for me was lot of pencilmarking. And staying aware that every canceled pencilmark can have an unexpected effect on the other possible digits. Quite like Black cropcidots work. And Simon isn`t the best in sudoko as we all know :)
First time I've attempted a puzzle featured on here: grand total time of 53:18! I'm a little surprised I managed to do it (and faster. Maybe it must be because I was also listening to the video while I was working it out? Or it could just be "beginner's luck" lol)!
One day I will say: "First time I ever solved a CtC puzzle faster than the length of the video", but today is not that day. In an old puzzle I managed to solve it in 10 seconds more than Simon. For now, I'm still taking too long, but learning and enjoying every step. Anyway, this is another excellent puzzle and solution. My compliments to the puzzle maker as well as Simon.
Today was the first time I solved it faster by 10 minutes. Of course I didn't have to explain everything and I just rushed it. Nevertheless a milestone for me 🎉
I find much easier to reason on the 9's in box one. This way we deduce almost immediately that there can't be an 8 neighbouring the 7. And then we find that R2C1 can't be 6, so these neighbours are 5 and 9, and the two branches are 753 and 7986, which can be placed using the 6 below.
If you allow yourself a pencil and paper writing the numbers 1-9 out in staggered columns of odd & even and drawing lines between numbers that are within 2 makes a neat graph, covering any numbers that see the line shows many restriction and tricks. Total Time 90min Everything Simon struggled with was easy for me while everything that was easy for Simon was so difficult. Such as placing the 9 in box 9 took me 30 min and 2 restarts to see if I made an error. It took Simon like 2 min flat... While from the very start covering the 7 on my graph shows instantly that the only way to put 4 digits either side is 59 38 16, the 6 sees the given 6 and box 1 is filled, it took me about 3-4 min. I finished box 3 in about 10-15 but box 9 and 7 took me an hour. I was perfectly out of sync with Simon on this puzzle. I relied too much on trying to use the graph to solve the puzzle, it helps but is not the "cheat sheet" I though I found.
I usually don’t try the puzzles that are longer than an hour, but I liked the simple rule set, and it went excellently. 29:43 for me, and that’s on mobile (I tend to have slower times on here than with a real mouse and keyboard)
No mate, me too, I noticed 9 must be next to 7 or 8, if you put 9 in middle in box 1, by the time you reach 7, you can’t go 4321, so 9 can’t be in middle, and if you go anticlockwise, 7986 will clash with the 6 in box 4, so 798 must be on the first 3 cells of row1/box1, and 6 must be the next on the spiral
@@WilliamSpieth I thought the same. I liked when Simon drew the line on the screen showing how a 1 would have to be flanked by a 2 and a 3, with a 9 being flanked by 7 and 8, but then he seemed to dismiss the idea as being helpful as it would only work for straight lines, where digits can 'see' each other. But like you, I noticed that all of box 1 was on the same line, so the principle could be applied much more easily.
It's interesting to see just how utterly different our approaches were to this puzzle! In almost *every* box and chokepoint, you found a different way to get to the solution than I did.
Thanks for the birthday shout-out. Can't believe it's taken this long for me to get round to watching it. Work really gets in the way. And for those wondering, I'm part of a choir. And we just so happened to be performing with the Welsh Association of Choirs (even though we're an English choir. It's a big honour) on the day before my 18tg birthday. The compere mentioned this and audience members started singing, so all the choir members joined in. Approximately 5500 audience and 800 choir members. And it's all filmed and released on DVD.
I managed to solve this in 34:19. The logic was absolutely sublime (as is typical of Aad's puzzles), and I really enjoyed the way it "recycled" logic from other whisper variations.
I think Simon wasn’t resorting to normal Sudoku early enough for this puzzle. If he did he may have completed this puzzle much earlier, and I think at some point he even realized that but wasn’t able to resist doing the adjacent cell calculations instead of going back to sudoku.
i really appreciated the stages of the challenge, you get the first box done, and then it opens up a little, you solve the next box, so on so forth until the youre looking all over to make sure
I watched on a Saturday morn A spiral-shaped tempest was roar'n. Simon was first to solve it. Took one hour to resolve it. And a brand new solution was born.
31:26, I hit a wall and peaked at the video to see that I messed up my pencil marks in Box 6, but was doing the logic correctly (I put in 68 in r456c8, then used the give 6 in Box 4 to (incorrectly) delete the 8 from r4c8). I found the puzzle fun, but a lot of my work was to limit digits 1 and 9 to a few cells, then just try each possibility to see which didn't work.
Any easier, less bifurcat-y way to resolve box 5, once we get to just past 51:00 is: Once we know a 5 is in the top corners of box 5, the top middle cell has to be from 34 (as 67 is in the row). With the highest r4c5 being a 4, the 8 in row 8 has to be in c8. That puts 1 in r6c8 and removes a 1 from r6c1. We can then rule out a 4 from r6c5 because that would required a 2 in r6c6 (as 356 are all used in the row), and that would make row 6 columns 1, 5, 6, and 9 all come from 249, which of course breaks. Once you remove the 4 from r6c5, the 78 pair in row 6 puts a 9 in r6c6, then the box/puzzle resolves quickly.
I got the even/odd split concept early but in a much less elegant way. I kept thinking about the effect of changing between jumps of 1 and jumps of 2 multiple times in a full non-repeating sequence of 1-9. Because of how I got to it, I didn't consider that you could reach a point where the sequence could break from that pattern. That mistake led me to needing to rewind a nearly filled grid. Once I realized that the line leaving the box created that breakpoint on the end still in the box as long as the digit on the tip of the spiral was either 1 or 9, I was able to solve the puzzle in just under an hour. I'm surprised I got it at all, but I'm especially proud of myself for fully completing boxes 1 and 3, plus the line running between them in 22 minutes (I think if I hadn't made the mistake, I would have had it complete in 40 minutes total)
Solved in 22:09. I’m not really sure what it was about this puzzle that clicked with me, but once I figured out the mechanics of the ruleset from box 1, everything just seemed to fall into place. Excellent puzzle!
34:58 for me! Surprising that I was this much quicker than Simon on this one! After watching, looks like I got the 9 in box 1 in just a couple of minutes, much quicker break in for me
Had to watch to 31:53 to solve. What an amazing puzzle. Getting that 5 into the lower-right corner of box4 felt absolutely magical. Absolutely recommend :D
The softest of whisper lines creates a beautiful puzzle - but one far beyond my ability. Well, I could have solved it probably with an aide memoire on paper (rather than one I had to erase from time to time, the way you erased yours, Simon), and a lot of pencil marks. But I had much more fun watching you solve it! I love it that you are persistent and thorough in your exploration of possibilities. Thanks, Simon. Well worth my returning to this video 10 days late because I did not have time to watch it on the day you released it.
I also don't usually attempt puzzles with videos longer than an hour, and when I do, they usually require several hours and frequently a peak at the video for a hint. Not so this time! I got started quickly and solved box 1 in just a few minutes and box 3 not much later - it really is all about the 1s and 9s and realizing that once you have 3-1-2 then that determines the next digits as well because 2 can now only go to 4, which means 3 can only go to 5, etc. Anyway after that I was slowed down a bit until I figured out how to crack the rest, so after a slower middle bit of the solve the rest fell together in about 10 minutes. 45 minutes total and I never felt stumped or stuck. Nifty ruleset!
26:37 for me. Nice puzzle. Usually with new rules, you have to look at what you can do with them. A loop with 8 digits MUST have alternating digits. So that opened up the puzzle. Lots of alternating digits with this one.
Simon seemed to always gravitate towards the longer path for deduction. :) E.g. box 5 would've been far easier to start looking at where 9 would go, and to look at it with the next digits following it. The r4c2 6 and r4c7 7 and r3c6 8 really limit the options for 9.
49:44 for me! Super fun and interesting concept. I focused on the pattern of 1 a lot to help me place the 1 correctly in boxes 5, 7, and 9 and that really benefited me personally.
Although Simon found it his own way, I loved the way 456 played within rules in box 3. (One of them was taken in r3, forcing two into r1. Bob’s yer uncle!)
28:06 for me. I always admire how Simon tries to find a good logic path to solve puzzles, but I have to say for this puzzle, pencil marks are really my friends, because it very much helps me to either restrict the place where 1 and 9 can go, or restrict possible digits in a cell, and form some useful pairs or triples. I don't even think of how the line works if I put 1 or 9 on middle of a line.
This took me 78:07, so very slow, but for me to have solved this at all would have been impossible a few months ago. Wonderful channel, thank you both.
40:27. Box 1 was very straightforward. Then box 3 realizing that either 1 or 9 couldn't be on the loop. Had to be 9 so finishing that box was approachable. Box 9 next all but 1234s were easy. Leading to the 9 in box 7. Got held up not realizing r7c6 could be a 6 if 67c7 was a 4, which led to dead end on box 7. But soon realized and back to the races. Very fun puzzle, with a good level of difficulty that didn't want me to give up. And love the ruleset! 9.5/10
39:30 "I can't see it, why brain do you let me down in these crucial moments" - an excellent example of feeling incapable for not being able to progress when you have not yet been provided the tools to progress; an allegory for many parts of life.
I got off to a good start, filled in all the lines in the upper 3 boxes. Got quite a bit of pencil-marks in box 4 and 7. Then I turned to box 9, and got a conflict between boxes 9 and 7 (after ~ 80 minutes). I'm pausing for now, maybe I'll retry later, or just watch Simon do it. Second try (after a big break, a restart and being a bit more careful), I got it in 81:38, solve counter 5491.
11:20 8 next to 7 on either side breaks the puzzle. (9 must follow it, then what follows 9?) 27:00 If you place 68 in the highlighted cells, where do you place 4? Therefore, the highlighted cells must be... 32:40 I almost finished the puzzle, when the puzzle revealed its breakage. I'd broken it somewhere filling out the lines. 34:00 I hope that Simon notices the 78 pair soon. 36:30 He noticed it. Now, where does the 8 go in block 9, if not next to the 9. 52:00 If R6C5 is 7 or 8, it has to be part of the 798 triple. If the cell is 4, it will approach the 213 triple. (Then you miss the pencilmarked 5s.)
This was interesting to watch. I failed this puzzle, though I got off to strong start. I actually figured out a lot of stuff about 5's early on, but then ran into a brick wall. And on a watch, my error was obvious: I completely neglected Box 3, because I assumed it would be important late in the solve. I was trying to unravel Boxes 1, 5, & 9, then move to Boxes 2, 4, 6, & 8, with the assumption that 3 & 7 would be the last part of the sequence. Blinded myself to some important clues. Thanks for the video, Simon!
Simon very seldom used the negative constraints - eg in box 7 you can tell 1 is not on that line so it must be in lower right pretty much right away. Considering that makes 1 and 9's a lot easier to do. With that I actually beat Simon for the first time ever.
In the beginning of the solve you asked how to get one of the extreme digits in the middle, I think asking how to get both into the middle of the line would have shown you more about how this puzzle must work. There's a few comments here about how you build a wall when you place 2 consecutive digits next to each other. I think trying to get them both in the middle of a line helps highlight this problem.
I try the puzzle cause it's a simple ruleset, and it's interesting to watch someone solving it, cause yes there is a Meta, 1 and 9 on a line have there neighbour force, so where does 9 go in box 1 ? not between 2 cells wich are not 7/8, so only 3 place to look after 1) on the tip of the line : if we try to put a 9 here, the sequence is forced, 9->8->6->5->7 and then the puzzle is broken, cause 7 have no number to chain with 2) on the exit of the box : it's the same sequence, from the over side, so the puzzle is broken 3 ) next to the seven, and then you can start to solve Another way to find this idea was : could the number next to the 7 be two higher or two lower number ? in this situation a pair of 5/6 or 8/9, and the answer is no, cause with a 8/9 9 have no linked number next to him, and 5/6 will make the sequence 7->6->8->9 otherwise there will be no 9 in the box, but this sequence is broken cause we miss the number who exit the box PS after watching : if Simon didn't forget the sequence of 7->9->8 the puzzle will be easier
50:06 The 5 weirdly gives a 34 pair in row 4 -- the cell in between the 5 pencil marks will always be adjacent with a 5, so must be from 3-4-6-7, and cannot be 6 or 7 by sudoku; all this does at this point is eliminate a 4 pencil mark from r4c9 but it's something?
THat's what I was seeing at that point too, but I'm cursed with not being able to then project 3 or for steps ahead.. which is why I'm useless at chess.
56:26 for me. spoiler/hint: 1s and 9s are the most restricted digits, espcially if they are in the middle of a line. 1 must be flanked by a 23 pair, and 9 must be flanked by a 78 pair.
@ 27:30 - you can rule out 68 from R3 in box 3, because there'd be a 4 in R2. It must therefore be 57, with 38 above, and because the next two digits on the loop are 16, and the 6 can't go in R2, you know the direction of the loop. @32:25 - pencil-marking 5678 in a column that has a 78 pair. Simon, sudoku rules always apply, not just when you're forced to think about them. @ 36:23 - Stop thinking about imaginary lines and do some basic sudoku. You have a 78 in C7, you can't have either 7 or 8 in C9, so there's a 78 in C8. A 78 can't go next to a 34, so it goes next to the 56. Now, 5 can't go next to 7 and 8, so that's a 6, making R5C7=5. Now R4C8 can't be 8, so it's 7 and R8C7=8, R9C8=7, R9C9=5, R8C9=34, and R7C8/9 are a 12 pair. Sudoku is your friend, stop abusing it. Now, in C3, where does 9 go? It can only be in R7 or R9. If it went in R9, it would have to be next to 8, but the 8 can't be next to 7 or 6, so R7C3=9. On the outer line, you can't have 7 or 5, which must go on the inner line, and there's a 6 in R7/8C1 which must join 4 and/or 8. 1 can now only go in R9C3, with 23 in R9C2. It doesn't take much to finish off the box. You now have 567 to join a 9 to a 4. The 7 can only go next to the 9, the 5 and 6 can't be resolved. @ 39:43 "why brain do you let me down at these crucial moments" - why Simon, do you keep imagining that everything can be resolved before you move on? The only way your brain is letting you down is that it's not telling you that you need more information. There are two possible resolutions, and there isn't enough information to resolve it yet. @ 45:37 "7 and 5 are on this line" - so where does 1 go in the box? Now 2, 4, and 6 can't go on the line with the 57, so it's either 578 or 357. @ 45:57 "8 is in one of those two squares" - so why not make R7C4=7, and remove your 7 pencil-mark from R4C2? This is exactly what I keep saying - follow up on your deductions. @ 58:14 "in fact this is going to be a 12 pair" - Really? What about the 12 in R8C4? There must be a 3 in there, which means you can remove 3 from C5, giving you a 124 triple in the column. If you'd done some sudoku, you'd have finished R9 and all of box 2. I normally love Aad's puzzles, but this one not so much. It was typically clever, but I felt it was more like a jigsaw, where the pieces can only go one way, and one just had to work it out. I thought it was interesting how the line sequence only applied until one of the ends ran out of the box or hit the end of the line. For instance, in box 5, the 9 had to be sandwiched by 78, and they had to be sandwiched by 56, but then because the line ran out of the box, the remaining digits were less constrained.
56:40 Solving R6C5 here is actually quite easy to me. It can't be 4 because then it has to be 421357 in the spiral, but 67 is already in R4, so you can't get to 89. It can't be 7 because 8 is already in C6, and then you can't do 56 because they're already in the row. So it's 7.
The easier way to solve box 5 at the end is to just ask yourself what you're gonna put next to 5. There's a 6 in the box so it has to be flanked by a 7 and a 3or 4. There is only one place you can put 7 next to 5.
I don't think this is fully true; you've missed off the possibility of the 5 being flanked by 3 and 4. It turns out you're correct, but your logic doesn't work
Right around 51:14, you see that r4c5 needs to be within 2 of 5 since there will be a 5 next to it. What’s available are 3,4 so you get a 3,4 pair in row 4, and things just sort of fall into place
This is the first one I’ve every solved, I would do a few boxes and then press play to check my work based on Simon 😅 but I did it! Didn’t make any mistakes either
Some bad pencil marks and some failure to notice some important pairs. He normally spots the implications of pairs from nowhere. Didn't today. Happens.
30:29 for me. That was quick, but unfortunately I found the comment about the 9 in box 1 that helped my starting path. Still, it was boxes 3 and 9 that were more useful for me, anyway.
I got the top left box in two minutes, it's easy once you realise that there has to be a 9-8-6 coming off of the 7, and you can't do that by going down off the 7 as the 6 clashes via sudoku. It was the rest of it I had trouble with 😄 Still, one of only two times I've beaten Simon in total time
The one bit of logic I have noticed you not using, even though you skirt around it was the idea of the 123 & 789 sets. It is the idea of in a box or along a straight line you have to be very careful of consecutive digits. For example if you have to duck up to grab the 9 but also need to get back down, how you would end up achieving that.... Yeah I made a dogs breakfast of explaining that idea. In my solve I focused a good bit on not cutting myself off and found the logic went quite smoothly with the logical check points that Aard had build into this beautiful puzzle for us.
I feel like Simon took the long way on getting started. I think the shorter answer is to say above and below 7: can the digits next to the 7 be both above or both below? No, because there aren't 4 digits above 7, so whichever it is the line will need to switch to the other side of 7 at some point which will requireboth digits next to 7. The requirement of the digits next to 7 is key and comes from when switching to the other side of 7 you must pass 7 and use the digit immediately on the other side of it. This requirement also leads to the conclusion that neither digit next to the 7 can be an 8 or 6 because the 8 side isn't ready to go down yet, the 9 is still needed. So which side is the 9? Well if R2C1 is the 9 then we have a problem with the 6 resolving that question and providing the necessary tools to continue.
I say this because so often I'm astounded by how clean Simon's proofs are and so this is a rare case. Indeed, the rest of the puzzle his proofs continued to be very clean even if sometimes it might have taken a second for him to see the proof.
The easier way to resolve box 5 is to ask if R4C4 can actually be 5. It would be flanked by 3 and 4, so R4C6 would have to be from 12, but then R5C6 would be broken
39:33 "Why Brain do you let me down?" I'm curious why in these situations where you have clear one or the other options, you never consider picking one or the other and run with it until it breaks? If it does break then you can be certain the other option is correct.
Let's Get Cracking: 04:49
What about this video's Top Tier Simarkisms?!
Three In the Corner: 3x (42:26, 42:28, 42:34)
Bobbins: 2x (21:20, 43:44)
The Secret: 2x (02:24, 03:58)
Knowledge Bomb: 1x (43:13)
Cooking with Gas: 1x (22:37)
And how about this video's Simarkisms?!
Sorry: 11x (11:30, 12:05, 21:20, 26:26, 31:51, 34:41, 35:24, 35:32, 35:32, 39:37, 54:58)
Ah: 10x (00:18, 00:35, 08:12, 16:00, 21:20, 30:58, 42:56, 49:34, 55:50, 58:53)
By Sudoku: 8x (19:39, 28:33, 40:28, 48:34, 51:07, 55:12, 58:33, 58:57)
In Fact: 6x (42:28, 55:20, 55:31, 57:21, 57:23, 58:13)
Hang On: 5x (15:39, 19:29, 34:23, 35:29, 55:46)
Pencil Mark/mark: 5x (08:31, 47:19, 49:47, 52:52, 55:50)
Surely: 4x (36:06, 39:16, 43:21, 48:04)
Obviously: 4x (06:57, 07:00, 14:22, 19:54)
Clever: 2x (51:53, 1:00:59)
Brilliant: 2x (01:10, 02:33)
Shouting: 2x (03:18, 03:28)
The Answer is: 1x (13:11)
Missing Something: 1x (09:18)
Bingo: 1x (29:39)
Lovely: 1x (03:31)
Beautiful: 1x (16:11)
Fascinating: 1x (1:01:43)
Gorgeous: 1x (57:41)
Come on Simon: 1x (52:17)
Proof: 1x (56:47)
Full stop: 1x (18:56)
Progress: 1x (40:54)
Wow: 1x (1:00:01)
Almost Interesting: 1x (29:02)
That's Huge: 1x (29:16)
Cake!: 1x (04:19)
Most popular digit this video:
One (145 mentions)
Antithesis Battles:
High (10) - Low (7)
Even (13) - Odd (2)
Higher (4) - Lower (2)
Column (4) - Row (2)
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!
Wow this is cool
Some piece of logic I used extensively but didn't see used in the puzzle nor in the comments: When you go from one digit to another (let's take the 6 in r6c4 as the example), you can either jump to a next digit by two or by one. However, two adjacent digits on a line that are 1 apart form a WALL. if you were to put a 5 next to a 6 then your next digit can be either higher or lower than your wall (e.g. 4 or 7), but you can't get to the other side of the wall anymore. You need a 5 or a 6 to go back from 4 to 7. So what you need to do is keep a gap for the return journey. With this logic you could instantly see that the 7 at r1c1 couldn't have a 6 nor an 8 next to it and thus needed a 5 and 9 to it's sides with the 9 obviously followed by 8,6,4 (which I would call the return journey).
The tricky thing here is to figure out when having a wall is okay (because you can just put higher and lower digits on both sides, or have space outside the line), and when it's not okay (e.g. because you have a loop, or all your digits need to be on the line and there is not enough space).
I kept wondering how in the world this puzzle is supposed to be solved without bifurcation, since even Simon had to resort to it over and over. This makes sense
This was a piece of logic I used extensively as well, especially with the spirals whilst figuring out possible combinations
Simon asks if there is easier way to solve box 3 at 29:57. I believe this is easier: you know (r3c8,r3c9) is either some combination of (5,7) or (6,8). Any combination of (6,8) doesn't work because you need to put 4 in either r2c8 or r2c9. (7,5) doesn't work because you need to put 6 in r2c7, so the only option is (5,7). The rest follows directly
Yep, saw the issue with 6 being in Row 3 in box 3 early so it was painful watching him take the long way around
I saw that quickly and began my usual yelling at the screen.
I just tried each of the three options for 4 (R1C789) and two of them break fairly quickly
Yeah because the long string needed either 5-7 or 6-8. Box 3 had one of those pairs, couldn't be 68 cause that would leave them both connected to a 7 due to 456-89 taken by then. Easy work when you put 57 pair to know they need to go with 3 and 8, the 8 connecting to a 6 rules left side out so 786 in right side.
Yeah thats how i spotted it as well
great puzzle as always. i started this differently, using a trick that was pretty useful in a few places. if you put 2 consecutive digits together, they make a wall that can't be crossed until you leave the box. this means that every digit on one side of the wall has to be all higher digits than the wall and the other side has to be all lower digits than the wall.
you can use the wall trick right at the start. if you put a 6 or 8 next to the 7 in box 1, then you make an either 67 or 78 wall. this means that you have to put all high digits on one of the 2 branches (either going right or down from the 7). since you have to fill one of the branches with only high numbers plus the wall digit, this is impossible (you have 3 digits, 689, but 4 cells to fill). this gives us a 59 pair next to the 7 which lets us easily fill the first box like simon did.
another example: you can apply this "wall" trick at 56:48, you can say that r6c5 can't be 7 because you only have 1 branch and so couldn't put either any high digits or any low digits in the center box.
23:43 for me which shocked me in hindsight, my times are usually comparable to the video length or worse. I skimmed the video to see if I had noticed techniques that were more efficient than Simon's. The two things I noticed were: (1) heavier focus on 1s and 9s when first looking a box, it was surprisingly easy to whittle those down (2) noticing that if you hit a 9 or 1 part way along a path, they have to "get out" so to reach them you had to go even-in odd-out or vice versa (helpful for example in box 1 -- you could see that it needed 7986 one direction or the other, and the given 6 immediately told you which way).
I noticed this too: It is all about the nines! Sometimes the ones. It got so much done.
I felt the same, but my looking at the box got me the solution without really "teaching" me the logic that you seem to have learned!
Did it in 35 minutes. Found it not that hard.
Key for me was lot of pencilmarking. And staying aware that every canceled pencilmark can have an unexpected effect on the other possible digits.
Quite like Black cropcidots work.
And Simon isn`t the best in sudoko as we all know :)
First time I've attempted a puzzle featured on here: grand total time of 53:18! I'm a little surprised I managed to do it (and faster. Maybe it must be because I was also listening to the video while I was working it out? Or it could just be "beginner's luck" lol)!
One day I will say: "First time I ever solved a CtC puzzle faster than the length of the video", but today is not that day. In an old puzzle I managed to solve it in 10 seconds more than Simon. For now, I'm still taking too long, but learning and enjoying every step.
Anyway, this is another excellent puzzle and solution. My compliments to the puzzle maker as well as Simon.
Today was the first time I solved it faster by 10 minutes. Of course I didn't have to explain everything and I just rushed it. Nevertheless a milestone for me 🎉
This puzzle is very aad 😀
i guess you could call it aad-ly satisfying.
Perfect comment
Phenomenal puzzle
Excellent solve
This video should have been titled "Breaking Aad."
I find much easier to reason on the 9's in box one. This way we deduce almost immediately that there can't be an 8 neighbouring the 7. And then we find that R2C1 can't be 6, so these neighbours are 5 and 9, and the two branches are 753 and 7986, which can be placed using the 6 below.
Yea, can't put 7 next to a 6 or 8 in that box from the start; isolates the 9 out of the sequence
Took me just about 2 hours, but it's another I was able to solve on my own. Loving this channel!
My first time ever solving a CtC puzzle faster than the video length! Of course I don’t have to do birthday greetings or a running explanation.
If you allow yourself a pencil and paper writing the numbers 1-9 out in staggered columns of odd & even and drawing lines between numbers that are within 2 makes a neat graph, covering any numbers that see the line shows many restriction and tricks.
Total Time 90min
Everything Simon struggled with was easy for me while everything that was easy for Simon was so difficult.
Such as placing the 9 in box 9 took me 30 min and 2 restarts to see if I made an error. It took Simon like 2 min flat...
While from the very start covering the 7 on my graph shows instantly that the only way to put 4 digits either side is 59 38 16, the 6 sees the given 6 and box 1 is filled, it took me about 3-4 min.
I finished box 3 in about 10-15 but box 9 and 7 took me an hour.
I was perfectly out of sync with Simon on this puzzle.
I relied too much on trying to use the graph to solve the puzzle, it helps but is not the "cheat sheet" I though I found.
I usually don’t try the puzzles that are longer than an hour, but I liked the simple rule set, and it went excellently. 29:43 for me, and that’s on mobile (I tend to have slower times on here than with a real mouse and keyboard)
Is it just me, or would it be easier to start by asking where 9 can go in box 1?
I feel like Simon went thru a very difficult method of getting started in Box 1. 9 seemed easier to me.
No mate, me too, I noticed 9 must be next to 7 or 8, if you put 9 in middle in box 1, by the time you reach 7, you can’t go 4321, so 9 can’t be in middle, and if you go anticlockwise, 7986 will clash with the 6 in box 4, so 798 must be on the first 3 cells of row1/box1, and 6 must be the next on the spiral
@@WilliamSpieth I thought the same. I liked when Simon drew the line on the screen showing how a 1 would have to be flanked by a 2 and a 3, with a 9 being flanked by 7 and 8, but then he seemed to dismiss the idea as being helpful as it would only work for straight lines, where digits can 'see' each other. But like you, I noticed that all of box 1 was on the same line, so the principle could be applied much more easily.
It's interesting to see just how utterly different our approaches were to this puzzle! In almost *every* box and chokepoint, you found a different way to get to the solution than I did.
I was blown away by this video! ...The Scorpions are still performing?!! 😮
Thanks for the birthday shout-out. Can't believe it's taken this long for me to get round to watching it. Work really gets in the way.
And for those wondering, I'm part of a choir. And we just so happened to be performing with the Welsh Association of Choirs (even though we're an English choir. It's a big honour) on the day before my 18tg birthday. The compere mentioned this and audience members started singing, so all the choir members joined in. Approximately 5500 audience and 800 choir members.
And it's all filmed and released on DVD.
I managed to solve this in 34:19. The logic was absolutely sublime (as is typical of Aad's puzzles), and I really enjoyed the way it "recycled" logic from other whisper variations.
This seems to be the odd puzzle that is not that hard but sort of stumped Simon (obviously he got there).
I think Simon wasn’t resorting to normal Sudoku early enough for this puzzle. If he did he may have completed this puzzle much earlier, and I think at some point he even realized that but wasn’t able to resist doing the adjacent cell calculations instead of going back to sudoku.
i really appreciated the stages of the challenge, you get the first box done, and then it opens up a little, you solve the next box, so on so forth until the youre looking all over to make sure
I watched on a Saturday morn
A spiral-shaped tempest was roar'n.
Simon was first to solve it.
Took one hour to resolve it.
And a brand new solution was born.
This is yet another brilliant puzzle - the idea of it is just so good
31:26, I hit a wall and peaked at the video to see that I messed up my pencil marks in Box 6, but was doing the logic correctly (I put in 68 in r456c8, then used the give 6 in Box 4 to (incorrectly) delete the 8 from r4c8).
I found the puzzle fun, but a lot of my work was to limit digits 1 and 9 to a few cells, then just try each possibility to see which didn't work.
Very nice, it just flows when it gets started
Any easier, less bifurcat-y way to resolve box 5, once we get to just past 51:00 is: Once we know a 5 is in the top corners of box 5, the top middle cell has to be from 34 (as 67 is in the row). With the highest r4c5 being a 4, the 8 in row 8 has to be in c8. That puts 1 in r6c8 and removes a 1 from r6c1. We can then rule out a 4 from r6c5 because that would required a 2 in r6c6 (as 356 are all used in the row), and that would make row 6 columns 1, 5, 6, and 9 all come from 249, which of course breaks. Once you remove the 4 from r6c5, the 78 pair in row 6 puts a 9 in r6c6, then the box/puzzle resolves quickly.
I got the even/odd split concept early but in a much less elegant way. I kept thinking about the effect of changing between jumps of 1 and jumps of 2 multiple times in a full non-repeating sequence of 1-9. Because of how I got to it, I didn't consider that you could reach a point where the sequence could break from that pattern. That mistake led me to needing to rewind a nearly filled grid. Once I realized that the line leaving the box created that breakpoint on the end still in the box as long as the digit on the tip of the spiral was either 1 or 9, I was able to solve the puzzle in just under an hour.
I'm surprised I got it at all, but I'm especially proud of myself for fully completing boxes 1 and 3, plus the line running between them in 22 minutes (I think if I hadn't made the mistake, I would have had it complete in 40 minutes total)
Solved in 22:09. I’m not really sure what it was about this puzzle that clicked with me, but once I figured out the mechanics of the ruleset from box 1, everything just seemed to fall into place. Excellent puzzle!
34:58 for me! Surprising that I was this much quicker than Simon on this one!
After watching, looks like I got the 9 in box 1 in just a couple of minutes, much quicker break in for me
It's a welcomed sight to see a sudoku that doesn't have a small novel as the ruleset
Had to watch to 31:53 to solve. What an amazing puzzle. Getting that 5 into the lower-right corner of box4 felt absolutely magical. Absolutely recommend :D
The softest of whisper lines creates a beautiful puzzle - but one far beyond my ability. Well, I could have solved it probably with an aide memoire on paper (rather than one I had to erase from time to time, the way you erased yours, Simon), and a lot of pencil marks. But I had much more fun watching you solve it! I love it that you are persistent and thorough in your exploration of possibilities. Thanks, Simon. Well worth my returning to this video 10 days late because I did not have time to watch it on the day you released it.
I was at the scorpions concert in Hannover on 19th, the Thundermother support act (2:11) was awesome. Definitely worth giving a try!!
Just watched the recording of them playing at Wacken in 2022. Really great band! Hoping to see more of them in the future in Germany!
Not that it detracts from the puzzle's excellence - but didn't we have a "Spirals" previously?
19:45 ... I've been struggling on the Simon videos as of late, so I surprised myself on this one
Nice puzzle!
I also don't usually attempt puzzles with videos longer than an hour, and when I do, they usually require several hours and frequently a peak at the video for a hint. Not so this time! I got started quickly and solved box 1 in just a few minutes and box 3 not much later - it really is all about the 1s and 9s and realizing that once you have 3-1-2 then that determines the next digits as well because 2 can now only go to 4, which means 3 can only go to 5, etc. Anyway after that I was slowed down a bit until I figured out how to crack the rest, so after a slower middle bit of the solve the rest fell together in about 10 minutes. 45 minutes total and I never felt stumped or stuck. Nifty ruleset!
26:37 for me. Nice puzzle. Usually with new rules, you have to look at what you can do with them. A loop with 8 digits MUST have alternating digits. So that opened up the puzzle. Lots of alternating digits with this one.
36:54 For me, this went better than I expected albeit with few errors around, surprisingly forcing ruleset if you know what to look for
Simon seemed to always gravitate towards the longer path for deduction. :)
E.g. box 5 would've been far easier to start looking at where 9 would go, and to look at it with the next digits following it. The r4c2 6 and r4c7 7 and r3c6 8 really limit the options for 9.
279! Though I couldn't even start this myself. Genuinely impressed so far (apart from mixing up 4 and 7 :)
I enjoyed this one, it fit well with my musical brain. Numbers can either move stepwise, in an arpeggio, or in a thirds scale exercise (1324).
I hope you get another with this rule set and Mark has a go at it to see other approaches and ways to use the lines.
49:44 for me! Super fun and interesting concept. I focused on the pattern of 1 a lot to help me place the 1 correctly in boxes 5, 7, and 9 and that really benefited me personally.
Although Simon found it his own way, I loved the way 456 played within rules in box 3. (One of them was taken in r3, forcing two into r1. Bob’s yer uncle!)
Very fun. I imagine it like running along stepping stones that sink when you step off of them.
Very nice puzzle. Way easier than most 1 hour videos.
As usual love watching and hearing you solve it.
I wish the software would let you type in the large numbers in different colors, or at least a light grey.
28:06 for me. I always admire how Simon tries to find a good logic path to solve puzzles, but I have to say for this puzzle, pencil marks are really my friends, because it very much helps me to either restrict the place where 1 and 9 can go, or restrict possible digits in a cell, and form some useful pairs or triples. I don't even think of how the line works if I put 1 or 9 on middle of a line.
Finished in 26:44 by following along with the video.
00:53:31 for me. Enjoyed solving this. Bedankt Aad voor weer een geweldige puzzel!
This took me 78:07, so very slow, but for me to have solved this at all would have been impossible a few months ago. Wonderful channel, thank you both.
40:27. Box 1 was very straightforward. Then box 3 realizing that either 1 or 9 couldn't be on the loop. Had to be 9 so finishing that box was approachable. Box 9 next all but 1234s were easy. Leading to the 9 in box 7. Got held up not realizing r7c6 could be a 6 if 67c7 was a 4, which led to dead end on box 7. But soon realized and back to the races.
Very fun puzzle, with a good level of difficulty that didn't want me to give up. And love the ruleset! 9.5/10
I love the first digit moment that would be a great time stamp
39:30 "I can't see it, why brain do you let me down in these crucial moments" - an excellent example of feeling incapable for not being able to progress when you have not yet been provided the tools to progress; an allegory for many parts of life.
I got off to a good start, filled in all the lines in the upper 3 boxes. Got quite a bit of pencil-marks in box 4 and 7. Then I turned to box 9, and got a conflict between boxes 9 and 7 (after ~ 80 minutes). I'm pausing for now, maybe I'll retry later, or just watch Simon do it.
Second try (after a big break, a restart and being a bit more careful), I got it in 81:38, solve counter 5491.
11:20 8 next to 7 on either side breaks the puzzle. (9 must follow it, then what follows 9?)
27:00 If you place 68 in the highlighted cells, where do you place 4? Therefore, the highlighted cells must be...
32:40 I almost finished the puzzle, when the puzzle revealed its breakage. I'd broken it somewhere filling out the lines.
34:00 I hope that Simon notices the 78 pair soon.
36:30 He noticed it. Now, where does the 8 go in block 9, if not next to the 9.
52:00 If R6C5 is 7 or 8, it has to be part of the 798 triple. If the cell is 4, it will approach the 213 triple. (Then you miss the pencilmarked 5s.)
This was interesting to watch. I failed this puzzle, though I got off to strong start. I actually figured out a lot of stuff about 5's early on, but then ran into a brick wall.
And on a watch, my error was obvious:
I completely neglected Box 3, because I assumed it would be important late in the solve.
I was trying to unravel Boxes 1, 5, & 9, then move to Boxes 2, 4, 6, & 8, with the assumption that 3 & 7 would be the last part of the sequence.
Blinded myself to some important clues.
Thanks for the video, Simon!
Simon very seldom used the negative constraints - eg in box 7 you can tell 1 is not on that line so it must be in lower right pretty much right away. Considering that makes 1 and 9's a lot easier to do. With that I actually beat Simon for the first time ever.
What negative constraints? When did Simon state them? I understood it that that the rule doesn't constrain cells that are *not* adjacent on a line.
Silly as this sounds but it feels like an Aad vd Wetering puzzle. Not very difficult but certainly takes a good amount of time 45-50 mins in my case.
In the beginning of the solve you asked how to get one of the extreme digits in the middle, I think asking how to get both into the middle of the line would have shown you more about how this puzzle must work. There's a few comments here about how you build a wall when you place 2 consecutive digits next to each other. I think trying to get them both in the middle of a line helps highlight this problem.
I try the puzzle cause it's a simple ruleset, and it's interesting to watch someone solving it, cause yes there is a Meta, 1 and 9 on a line have there neighbour force, so where does 9 go in box 1 ? not between 2 cells wich are not 7/8, so only 3 place to look after
1) on the tip of the line : if we try to put a 9 here, the sequence is forced, 9->8->6->5->7 and then the puzzle is broken, cause 7 have no number to chain with
2) on the exit of the box : it's the same sequence, from the over side, so the puzzle is broken
3 ) next to the seven, and then you can start to solve
Another way to find this idea was : could the number next to the 7 be two higher or two lower number ? in this situation a pair of 5/6 or 8/9, and the answer is no, cause with a 8/9 9 have no linked number next to him, and 5/6 will make the sequence 7->6->8->9 otherwise there will be no 9 in the box, but this sequence is broken cause we miss the number who exit the box
PS after watching : if Simon didn't forget the sequence of 7->9->8 the puzzle will be easier
50:06 The 5 weirdly gives a 34 pair in row 4 -- the cell in between the 5 pencil marks will always be adjacent with a 5, so must be from 3-4-6-7, and cannot be 6 or 7 by sudoku; all this does at this point is eliminate a 4 pencil mark from r4c9 but it's something?
THat's what I was seeing at that point too, but I'm cursed with not being able to then project 3 or for steps ahead.. which is why I'm useless at chess.
38:50 for me. Very proud because this is the first time I beat Mark or Simon on a difficult puzzle!
It took me a long time, but I solved it 100% by myself. I am so pleased!
It actually took me only 2 min to figure out the first 3x3 when it took Simon around 15. I feel very proud of myself!
51:40 for me. Interesting puzzle!
Kurouzu-cho Whisper Lines
Are you sure they are called whisper lines? Whisper usually means difference *at least* X. In this case it is difference *at most* 2.
... or this might be a joke about Uzumaki: en.wikipedia.org/wiki/Uzumaki
56:26 for me.
spoiler/hint: 1s and 9s are the most restricted digits, espcially if they are in the middle of a line. 1 must be flanked by a 23 pair, and 9 must be flanked by a 78 pair.
20 minutes on the dot! That was a great, fun puzzle, simple ruleset indeed!
Ps. Personally, I think a simpler ruleset would just be "Normal Sudoku Rules apply."
This was a very hard puzzle to put the first digit (almost 20 minutes) but after I put some numbers in box 1 it all felt quite doable .
@ 27:30 - you can rule out 68 from R3 in box 3, because there'd be a 4 in R2. It must therefore be 57, with 38 above, and because the next two digits on the loop are 16, and the 6 can't go in R2, you know the direction of the loop.
@32:25 - pencil-marking 5678 in a column that has a 78 pair. Simon, sudoku rules always apply, not just when you're forced to think about them.
@ 36:23 - Stop thinking about imaginary lines and do some basic sudoku. You have a 78 in C7, you can't have either 7 or 8 in C9, so there's a 78 in C8. A 78 can't go next to a 34, so it goes next to the 56. Now, 5 can't go next to 7 and 8, so that's a 6, making R5C7=5. Now R4C8 can't be 8, so it's 7 and R8C7=8, R9C8=7, R9C9=5, R8C9=34, and R7C8/9 are a 12 pair. Sudoku is your friend, stop abusing it. Now, in C3, where does 9 go? It can only be in R7 or R9. If it went in R9, it would have to be next to 8, but the 8 can't be next to 7 or 6, so R7C3=9. On the outer line, you can't have 7 or 5, which must go on the inner line, and there's a 6 in R7/8C1 which must join 4 and/or 8. 1 can now only go in R9C3, with 23 in R9C2. It doesn't take much to finish off the box. You now have 567 to join a 9 to a 4. The 7 can only go next to the 9, the 5 and 6 can't be resolved.
@ 39:43 "why brain do you let me down at these crucial moments" - why Simon, do you keep imagining that everything can be resolved before you move on? The only way your brain is letting you down is that it's not telling you that you need more information. There are two possible resolutions, and there isn't enough information to resolve it yet.
@ 45:37 "7 and 5 are on this line" - so where does 1 go in the box? Now 2, 4, and 6 can't go on the line with the 57, so it's either 578 or 357.
@ 45:57 "8 is in one of those two squares" - so why not make R7C4=7, and remove your 7 pencil-mark from R4C2? This is exactly what I keep saying - follow up on your deductions.
@ 58:14 "in fact this is going to be a 12 pair" - Really? What about the 12 in R8C4? There must be a 3 in there, which means you can remove 3 from C5, giving you a 124 triple in the column. If you'd done some sudoku, you'd have finished R9 and all of box 2.
I normally love Aad's puzzles, but this one not so much. It was typically clever, but I felt it was more like a jigsaw, where the pieces can only go one way, and one just had to work it out.
I thought it was interesting how the line sequence only applied until one of the ends ran out of the box or hit the end of the line. For instance, in box 5, the 9 had to be sandwiched by 78, and they had to be sandwiched by 56, but then because the line ran out of the box, the remaining digits were less constrained.
It is always funny to see, that Simon inevitably has to take the same way of solving the puzzle, but gets there way faster 😄👏🏻👏🏻👏🏻
I managed to fill box 1 and 3 (and the rest of row 3) in under 5 minutes. But then I got very stuck.
56:40 Solving R6C5 here is actually quite easy to me. It can't be 4 because then it has to be 421357 in the spiral, but 67 is already in R4, so you can't get to 89. It can't be 7 because 8 is already in C6, and then you can't do 56 because they're already in the row. So it's 7.
"It was solvable, but it required a lot of thinking!"
The creator: "Good. :) "
The easier way to solve box 5 at the end is to just ask yourself what you're gonna put next to 5. There's a 6 in the box so it has to be flanked by a 7 and a 3or 4. There is only one place you can put 7 next to 5.
I don't think this is fully true; you've missed off the possibility of the 5 being flanked by 3 and 4. It turns out you're correct, but your logic doesn't work
Right around 51:14, you see that r4c5 needs to be within 2 of 5 since there will be a 5 next to it. What’s available are 3,4 so you get a 3,4 pair in row 4, and things just sort of fall into place
56 mn? I can’t believe I got through an Aad puzzle, let alone within an hour. Loved it.
19:16 for me. Very interesting idea!!
Yes I also solved this while thinking there must be a cleverer way of doing this.
33:52 Great puzzle !
25:13 for me, very cool puzzle. Thank you for sharing :)
Nice puzzle. I was a bit slow, 46 mins - i should have left it til later really but i was too impatient to have some fun!
This is the first one I’ve every solved, I would do a few boxes and then press play to check my work based on Simon 😅 but I did it! Didn’t make any mistakes either
01:11:47 for me! proud of this one and wish id recorded since i managed to articulate all my logic aloud
Some bad pencil marks and some failure to notice some important pairs. He normally spots the implications of pairs from nowhere. Didn't today. Happens.
30:29 for me. That was quick, but unfortunately I found the comment about the 9 in box 1 that helped my starting path. Still, it was boxes 3 and 9 that were more useful for me, anyway.
Just wondering, when will you get back to live streamin' and continuing the puzzle games you haven't finished yet?
His refusal to do sudoku in sudoku puzzles is mind boggling
I got the top left box in two minutes, it's easy once you realise that there has to be a 9-8-6 coming off of the 7, and you can't do that by going down off the 7 as the 6 clashes via sudoku. It was the rest of it I had trouble with 😄
Still, one of only two times I've beaten Simon in total time
a down solve for simon is a career solve for most… still my GOAT!!!
The one bit of logic I have noticed you not using, even though you skirt around it was the idea of the 123 & 789 sets. It is the idea of in a box or along a straight line you have to be very careful of consecutive digits. For example if you have to duck up to grab the 9 but also need to get back down, how you would end up achieving that....
Yeah I made a dogs breakfast of explaining that idea.
In my solve I focused a good bit on not cutting myself off and found the logic went quite smoothly with the logical check points that Aard had build into this beautiful puzzle for us.
40:38 That would be a V pentomino, thank you very much 😊
Took me 01:25:22 but it's my first time solving one of those without watching the video!!
I feel like Simon took the long way on getting started. I think the shorter answer is to say above and below 7: can the digits next to the 7 be both above or both below? No, because there aren't 4 digits above 7, so whichever it is the line will need to switch to the other side of 7 at some point which will requireboth digits next to 7. The requirement of the digits next to 7 is key and comes from when switching to the other side of 7 you must pass 7 and use the digit immediately on the other side of it. This requirement also leads to the conclusion that neither digit next to the 7 can be an 8 or 6 because the 8 side isn't ready to go down yet, the 9 is still needed. So which side is the 9? Well if R2C1 is the 9 then we have a problem with the 6 resolving that question and providing the necessary tools to continue.
I say this because so often I'm astounded by how clean Simon's proofs are and so this is a rare case. Indeed, the rest of the puzzle his proofs continued to be very clean even if sometimes it might have taken a second for him to see the proof.
First time I solved something faster than Simon, this is a big moment for me.
The easier way to resolve box 5 is to ask if R4C4 can actually be 5. It would be flanked by 3 and 4, so R4C6 would have to be from 12, but then R5C6 would be broken
"I can read the ruleset in two seconds!" - *reads the rules in eight seconds*
Took me 32 mins to solve the puzzle, I'm proud of this one.
Not that difficult to solve once you get the hang of how digits interact
39:33 "Why Brain do you let me down?" I'm curious why in these situations where you have clear one or the other options, you never consider picking one or the other and run with it until it breaks? If it does break then you can be certain the other option is correct.
22:45 Gas hobs, maybe, but definitely not GAS!