The way he ignores his own logic at 13 minutes, just to spend another 10 minutes coming to the same conclusion with a much more convoluted, (or as he describes it: "beautiful" way) is astonishing.
I saw this comment before watching the video.. I forgot about it and when it got to that part I was like what is he doing? it's 19.. and then remembered this comments again 😂 Now I need to see what weird mind gymnastics does to get to that conclusion 😂
@@ceevio_art I don’t know how many times I’ve almost shouted at the screen at Simon (and Mark too for that matter) to use his earlier logic again, but it’s a lot.
13:50 By the very logic Simon had just gone over orange can be limited to 19, since all 8 2-digit renban options must appear, and orange can only appear on the little diagonal one, and digit "in the middle" appears on two different 2-renban options, once with each of its neighbors
Always so awesome to see one of my puzzles appear on the channel, very happy to see it, thanks Simon! This one was particularly a blast for me to watch for two reasons: 1) Simon found a path that was very different than I intended to get through the break-in, and it is so elegant and natural looking it looks intentional. You could even say it is telegraphed by the cross. 2) I was streaming when this video came out, so I got to watch this video together with all of my stream viewers, and that was awesome, seeing live chat react to Simon's solve! I'll be sure to share a video of this live watch party soon.
@@goliathcleric Note that r5c5 only appears on one two cell renban, were as the digits 2-8 each need to appear on two. So r5c5 is 19, and that digit must appear at r1c4. It also shows up on the renban in box 8 by sudoku. Meanwhile, ask where the other 19 is in row 5. Well, its on a renban. So it can't be on a two cell renban in column 5. That places the other 19 on a length 3 renban in column 5. Now do sudoku on the other 19, and you place it it r5c6, disambiguate highs and lows, and from there you pretty much converge with Simon's path. On a logical level, both amount to counting 19s on renbans and using that to force the places of 19s, but Simon does it in a more abstract way by noting that (completely by coincidence) all of these lines are concentrated in this cross. I love Simon's break in, but if that was the intended path I don't think the testers would have called it approachable!
@@zetamathdoespuzzles I stared at this one for 20 minutes or so thinking there was no way, then I saw your intended 1-9 breaking and I thought "....you bastards, that's beautiful" Well done.
Even though the way Simon parses it is not exactly what I intended, he homed straight in on the important thing: 19s are weird in this puzzle. His intuition is unbelievable.
@@balkthor I was just making a pun. Let's get cracking. It's hard for any expansion team to be good right from the start, except Vegas. I think the GMs learned what not to give up after the Vegas expansion.
@@tamarockstar45 I know. I'm big on self-deprecation when it comes to my fandom. It's what's allowed me to be a Mariner fan for all these years. I tell people all the time, it's far easier to be a fan of a terrible team then a good one. That's what made the Seahawks tough the last ten years. That, and having to endure Russell Wilson's super creep factor *shudder*. Nice thing about hockey, unlike other sports you still have a great time going to the game even if your team is getting demolished. There's always fights, fancy shots, great saves, and let's face it: Hockey fans are the greatest sports fans inthe world.
The same for me, about 58 minutes, I'm normally never that close. But with a totally different path, I started with the 2-cell renbands, establishing the content of the vertical and horizontal (starting with odd or even), and then continuing from there with the 456-triple in box 5, and then similar deductions as Simon, but a lot slower... :) And without Simon's sudoku class I take almost every evening, I would _never_ have been even close to finishing. Fun fact, my first language is Swedish, however, I find myself thinking a lot in English (in Simon's accent) when I try these puzzles...
Something that might have been useful to note at a few points: Once the 1 is placed in the center, you know the horizontal part must contain the "odd-high" options 23, 45, 67, and 89 (since they all see the 1 and that's the only way to break the remaining digits in to four 2-renbans), and therefore the vertical/diagonal must contain all the "even-high" ones.
Long way, after he posts it soon, watch zeta watch Simon solve his puzzle in live chat, or can watch it now from his stream yesterday. It begins at 3:34..and watch it with chat comments.
@@davidrattner9 Thanks, David - I'll do that! Now that you've told me about zeta's channel, I sometimes tune in to see what he's up to. Not for 8 hours at a time, though.
I would have proven that the center cell was a 1 or 9 a lot quicker by just saying that every digit between 2 and 8 inclusive must appear twice within the eight two cell lines. As the center cell can only appear once and both 1 and 9 can only appear once, the center cell is either a 1 or 9.
Loved this puzzle and video. I actually "got" the 1-9 logic, as or even before Simon was beginning to explore it, which seems to me like a huge advance in my ability to observe and draw reasonable conclusions. Thanks so much for this! (I credit 100% of my growing understanding of the more complex logic in Sudoku puzzles to watching the videos on this channel, and doing the daily GAS puzzles on the discord channel. If you have not checked those out, you really should. Instructive and fun!)
The "viewer yelling at the screen" rate in this one was probably super high! Props to Zetamath for twisting our brain into a fine little knot everytime they come up with a new puzzle.
After rather quickly figuring out R1C3&4 were a 19 pair, I put the web app over to test mode and used the alphabet to note one of the options as A and one of them as I. Solved the puzzle in this mode until an H was placed in box 6 and since H is equivalent to 2 or 8 and there is already a 2 in the box. Kinda fell apart from there. Very enjoyable puzzle! And thanks to Sven for this option in the app!
I must say, I always felt releasing a new app for every puzzle type was a little cash grabby. But after seeing the last few puzzles you've solved and watching the video about setting you released a while ago, I sorta realized that it's not cash grabby. It's to pay the people who made these incredible puzzles. I only realized after getting absorbed into the classic sudoku app. Without a doubt the best puzzles I have ever solved in a sudoku app. I assume the quality is the same for the rest. Keep up the great work.
I'm very proud of myself that this took me under an hour to finish on my own. Renban is one of the constraints that I see most clearly, so that helps. This was an amazing puzzle! Thank you, Zetamath, for the renban exercise, and thank you, Simon, for the wonderful videos! I solved Philip's puzzle, but I don't even understand how I did it, so I did not venture to submit a solution. Also, watching the video now, and the nearly Stone Temple Pilots reference made me pause, unintentional though it probably was.
If all cells are pointing at the center cell, and they all have duplicate numbers except 1 and 9, doesn't that limit the center cell to 1 or 9? instead of all odds? I think so.
@@zetamathdoespuzzles Just wanted to say a big thank you for this puzzle, I found it incredibly satisfying to solve. It starts out seeming impossible but then there's a pretty clear path once you start thinking about how it has to look.
Just after saying 1 and 9 are the only digits that appear once in the length two renban lines, he pencilmarks the orange cells with all odd digits, 3, 5 and 7 included
@@joosepmattiseriste8455 Classic Simon. I always give a go at the puzzles, but frankly i never could do, most of the time cant even start any of them unless they are an actual GAS puzzle, but i still somehow figured out in the first 2 minutes the 19 logic, which took him 20 minutes with some much harder logic.
I am not usually one to sit at the screen, but I did twice this time. First, the 1-9 in the middle, then the 3-4-5-6 because of the 2s in c2 and c3 looking at the whole of the horseshoe. Great work!
finished in 42:48 without referring to the video!! very satisfied solve, thank you for honing my deductible skills (looking back, few months ago I needed to refer to the video numerous times and I took exceptionally long to solve too)
40:53 for me. That was an incredibly satisfying puzzle, I started out looking at the board wondering "how could this ever be possible?" and then slowly a few things began to drop into place and then a few more.
What an incredible puzzle! Geez how utterly elegant! Loved your solve of it, hats off to zetamath and Simon once again. Such a cool idea for a ruleset, and the weird quasi-SET thing is AWESOME.
47 minutes for me. I really enjoyed that puzzle a lot. I sat scratching my head for a while at the outset but once I worked out how to start everything flowed quite beautifully. Excellent stuff, thanks for sharing it
I saw the 19 in the center before Simon, but I did not see how to utilize it until he explained all the set. So I was quite stuck and started the video after 10 or so minutes of banging my head. So with a little help I got a 67:16 time. And considering I had to backtrack with the box6 logic, I'm pretty pleased with it all. So as Simon said at the start, it's not the easiest break in but quite approachable once you get past that part.
Simon sure did this the hard way. Just need the 2-cell renbans to get 19s and once you know the middle cell, you know which set of renbans run horizontally and vertically.
15:05 Although I've never seen this used in a puzzle, we actually do know about the sum of the digits on a 4-cell renban. It's always an even number that's not a multiple of 4. The options are 10, 14, 18, 22, 26, 30.
The thing I have used about length 4 renbans is they always have a digit that is each of 0, 1, 2, and 3 mod 4. But if you look at the digits 1-9, there are actually three numbers that are 1 mod 4 and 2 that are each other congruence class. That imbalance often lets you pin down a lot of cells that have to be 159 from the start.
I got right to the end but realised I had unresolvable 34 pairs in columns 8 and 9 of rows 1 and 9...then spotted that I had two renbans with 3456. Fortunately, I only had to backtrack a little to fix my error. Another brilliant puzzle from Zetamath.
Wow! I heard approachable and decided to try this puzzle. After a couple of errors in my logic, I was able to complete it. Had I seen your break-in I would have steered clear. Kudos for seeing that, but it was far simpler for me.
Very clever puzzle, ζ-math 👍🏻 It took me 53 minutes, which I thought was really slow and that Simon would blitz through in half the time, so it's really gratifying to see that he struggled as well! I was particularly impressed with the way that he seemed to have worked out that the orange pair in r1c4 and r5c5 had to be 1 or 9, and then completely ignored that logic and pencilled them in as 13579 🤔 _Edit:_ Also, can confirm that if you follow through with the deduction that the middle cell has to be 1 or 9 to avoid knadgering up the renban dominoes, and remember that renbans must be unique, you can pencil the 1s and 9s into the blue cells without even thinking about set theory and the maltese cross.
I got it in 28:50, super proud of this one!! I very quickly realized that all possible 2- and 3-renbans must occur, and it was a very fun solve from there. This is probably the only time I'll ever beat Simon, so I'll cherish it for long time ;)
18:42 This is the first time I feel like I understood the puzzle faster. I instantly looked at the arrangement of 2-cell renbans and thought "Theres 4 were the lower digit is even and 4 where the lower digit is odd" Then noticed that there were seven 3-cell renbans which must have a lowest digit of 1-7 once each. Fun!
In terms of the break-in and the logic that you did at around 21:00, couldn’t you postulate that the center cell had to be a 19 care because they each only appeared at one time on the eight 2-digit lines and so you needed it one in the middle to guarantee that row five had all of the digits? (Since 3, 5, and 7 were already twice accounted for)
It felt like being on a trip in wonderland solving this puzzle. I mean really. The beginning was absolutely stunning and I was expecting the rest to be just a regular sudoku already satisfied with the beautiful beginning. But NO, the rest was not a downfall at all. Just hard enough to be fullfilling but not so hard to be painful. Perfect match for my current state of mind.
@8:55 Correct me if I'm mistaken but there is 1 detail to be gleaned immediately here. There are 7 different 3-cell renban lines. But a key feature of the "3x" rule of 3-cell renbans is that a digit must HAVE a x-1 and x+1. So within the digits 1-9, only the digits 2-7 are valid Xs. So EVERY unique version of a 3-cell renban must appear in this puzzle exactly 1 time, no more and no less. This may get appreciated later and used to narrow some logic down, I just happened to notice it out the gate. :)
you could also say, that since every 3 cell renban is equal to a multiple of 3, the highest 3 cell renban (9,8,7) is equal to 24, 24/3 = 8 different options, now you can't have 3 in a 3 cell renban so that option is ruled out, leaving you with 7 different options
*2-8 are valid Xs in your example. I took my X to be the lowest of the three digits, so X, X+1 an X+2. Then X could be any digit from 1 to 7, so 7 unique combinations. Boils down to the same thing, just slightly easier to see there are precisely 7 of them.
Got it in 57:11. Not a bad time for me considering the video length! I liked coming to the realization that every possible two-cell and three-cell renban was represented in the puzzle and I could figure out which was which that way.
16:55 -- How long will it take him to figure out that the two oranges must be 1 or 9? 21:00 -- Found it. The method was rather roundabout. It really was a lot easier to discover that the oranges were 19. Think 12-34-56-78-9 and 1-23-45-67-89. One's vertical and one's horizontal. 43:50 -- Where is 2 in cell seven? (Can't be on the line.) 49:00 -- Congratulations!
well, I tried this and I think I actually succeeded without needing Simon's help (yay!) and just skimming the comments I think I need to watch it now, because my break-in was definitely working out the 1-9 pair... in row 1.
It seems that the comments may have already addressed this but your in to the solve was far more complex than necessary as you could think think immediately that all the 2 cell Renbans will use all digits twice except for 1 and 9 as they cant pair lower or higher respectively
You could have gotten the middle as 1 or 9 straight from the start if you realised there was only 8 possible 2 cell renban options and that 7 of them could see the central cell meaning anything other than an extreme digit would use up 2 of the options. E.g. if it was 4 then that take out 34 and 45 leaving only 6 options Edit: oops.... Just seen everyone else has said the same thing but I hadn't loaded comments yet 😂 they all explained it better than myself so go read them comments!
Solved in 37:48 - this was really neat! I haven't watched Simon's solve yet so I don't know how he did it, but figuring out the center square was a 1-9 was a great a-ha moment, and there were plenty more as I worked through the puzzle.
46:25 solve, the first time I've ever beaten the video solve time! Like most of the comments I found the 19 much more simply and of course not needing to speak the logic aloud as I use it helps, but it's still a huge milestone for me.
34:35 finish. The way I looked at the start was: each digit appears on a 2-cell renban twice except 1 and 9 (once each). Since the center cell sees all but one of the two-cell renban cells, it must be a 1-9. Since it has to go onto a line once, it must be in r1c4. Now that the center cell is an extreme digit, all of 1-2/3-4/5-6/7-8 must be horizontal or vertical, with 2-3/4-5/6-7/8-9 being the other. Therefore, r1c5 (seeing all of the vertical two-cell renbans) must be an extreme digit, forming a 1-9 pair in the box/row. I then identified r1c5 as matching with r5c6, the only non-three-cell renban in column 6, and r7c4 as the only place for it in box 8. This, of course, differentiates the 1-9 pairs, due to r5c7. Next, you don't actually need set to establish the 1-9s in the cross. You've placed them in columns 4-5-6 (all placed except the 1-2-3 renban in box 8). In box 6, neither the 1 nor the 9 can go on any of the renbans (already have 1-2-3 and 7-8-9 placed elsewhere, and they are already in row 5 elsewhere), so they must be in r4c8/9. This finally pencils them in box 4 in row 6. That having been said, there's nothing wrong with a good set in a sudoku. More maths!
Wow.. I really solved this one differently to Simon again... I figured out that 19 must in the middle due to it being the only number that appears once in double pairs. and then I was able to figure out the triple at the top had to be the other 19, and that that basically started to solve the puzzle for me Very fun puzzle. great work =)
In the 20 minutes I took attempting this, the one useful fact I realised is that the middle digit needs to be a 1 or 9, because there's only one 1-2 and one 8-9 set in all possible 2-renban pairs. And r1c4 needs to be the exact same number. All that to say, I don't think I'll be solving this one 😅
43:53 before watching the video!! I never could have done that without the tutoring I have received watching these videos. Now to see if I found the same path as Simon. 😁
Something about the break-in I'm not sure other people have pointed out yet: whichever of 19 is not in r5c5 must go in r1c5 because it has to appear twice in r5 and c5, since it cannot be in r5c5. It will appear once on a 2 cell renban and then it appears in the only cell which is not a 2 cell renban
That was a very long and complicated way of figuring out that break in. The way I did it is to note that there has to be at least one 89 or 12 pair in row 5 and whichever of 1 or 9 was in that pair has to be in r1c5 because it can no longer be on a 2 cell renban. After that it becomes obvious that no other renban in row 5 can contain the other of 19 so it must be r5c5. And r5c5 must be on a 2 cell renban somewhere, therefore must go on r1c4.
I've been enjoying the app updates. I did encounter one in killer though where I got a lot done but then I got stuck so I used the smart hint. It was talking about outies can't contain a 5 therefore this specific digit must be a 4. So I put in a 4 not understanding it because I'm like the outies HAVE TO contain a 5. And it marked it wrong!!!! That specific digit was in fact a 5. The smart hint gave me an invalid digit. I had no choice but to bifricate (I refused to put in a 5 when it was a 4/5 digit but the 4 that the smart hint gave me was marked wrong). Fortunately I didn't have to bifricate much. After I managed to get a digit it filled in rather quickly. I don't remember the number but it had a massive ring on the outer perimeter and they all had 1 possible fill or quickly resolved to 1 possible fill.
Somehow, despite finding the 2-cell renban logic much faster than Simon, I still end up much slower ;-) (60 minutes). The 2-cell renbans were a work of beauty in their own right. I realised quickly that they had to be every combination, that 19 could only appear once and therefore had to be place in r5c5 and r1c5 (and as a by-product r1c4). Also the vertical and horizontal 2-cell renbans had to respectively either all have an odd low digit or an even low digit, which eliminated options quickly. Then I failed to apply effectively the same logic to the 3-cell renbans, where Simon shone.
As I expected, many others on this channel have already pointed out that Simon did not catch the quickest break in to the puzzle even though he makes the point that all of the digits one to nine appear in the 2-digit renbans twice except 1 and 9. He said it more than once and completely missed the significance of that in reference to the center square. Naturally, being Simon, he found a substantially more obscure route to the same result.
I've been a subscriber for over two years. That's not even close to the most unnecessary overcomplicated thing Simon has done. It's amazing how he manages to solve the most complicated break ins but manages to take the scenic route on some of the more simple deductions, and yet I still love it every time!
Have to agree. You can work out the 1-9 pair in r1c4/5 far more easily and that either 1 or 9 has to go in r5/6c6 which has to be the 9. Admittedly it then takes longer to also know r7c4 is a 1 or 9. The puzzle is actually approachable once you spot the 1-9 in the top row and realise how much that gives you in column 6. No need for anything as complicated as Simon does.
@@adrianhead6272 I'm still impressed how he and Mark just manage to attack all these puzzles and solve them logically even if the method is sometimes over the top.
A way I like to explain the 3 cell line is: "If you take one from the big and put it in the small, you get the same number 3 times, because they are consecutive"
some cool logic id like to throw out is that in box 1/4, the 8 sees the entire renban. since we have to have 2 evens on the renban, it’s either 24, or 46 so we know there’s a 4 on the renban.
That's a cool way to see that! I usually think of it as, I need four of the digits 123 4 567 and if I omit 4 then I can't get to four of them, but your way is a bit easier to think about in my head!
@@zetamathdoespuzzles I usually imagine putting the lowest set of digits on the line (1234) and the highest set of digits (7654) and seeing if anything exists in the overlap (4), but I guess it comes down to the same thing. It's just how I approach it. Very enjoyable puzzle, Zetamath.
@@zetamathdoespuzzles damn. acknowledgement from the puzzle maker himself. what an honor. i love your way of seeing it too! i just tend to better see parity interactions especially on a puzzle so heavily renban’d
I found the analysis of the last two 4-cell renbans frustrating. The 3-cell renbans had been worked out by looking at the set of various compositions, and the same could be done for the 4-cell cases. Simon noted that the last two 4-cell renbans had to be selected from 1-6, so they both included 3s and 4s. He could also have noted he'd already used the 1234 case, so the last two renbans had to be 2345 and 3456, and the one in box 7 couldn't include a 2. So, both compositions were known.
I think the renban lines might too dark/opaque. Simon seemed to miss a number of 4s that had the renban going through them while scanning? The same might have been true with the 2s in the upper left, since the purple is so similar in value to the blue. Perhaps Sven can add some transparency to the renbans? It won't need to be very dark at all to still be visible. Perhaps something with a similar value to the thermo grey? (though that could be bad in terms of color blindness, where pink and grey are sometimes difficult to distinguish when the value is similar.)
Hmm, 25 mins here, and usually I'm much slower than Simon. I wonder if I was helped by lack of tricks to apply, so had to see that all the 2- and 3- length renban were in use, and hence the sequence 19 [ R5C5=R1C4= 1 or 9, and then C5 gives R1C5 is the other of 1,9 which in turn must also be in R5C6 (considering R6)].
I've spent ten minutes or so trying to get *any* logic to even get started beyond the simplest deductions ever (for example: there is not another 8 in this box) and I am intrigued to see how it actually plays out.
looking over this i thought how many sets are used. all 2 and 3 sets are used! thats gonna make it easier since you know the consecutive numbers must be on the lines
![[UNCUT] The Loyal Pin ปิ่นภักดิ์ EP.15 (2/4)](http://i.ytimg.com/vi/qbHbshrA1f0/mqdefault.jpg)
Came to the comments knowing that EVERYONE would be noting the same 19 thing. Love it when Simon finds a deeply complex route to the simple!
Its funny how he states the orange cell appears only on one renban, then didn't mark it 19.
He even mentioned it, which is the most offending part grrr
The way he ignores his own logic at 13 minutes, just to spend another 10 minutes coming to the same conclusion with a much more convoluted, (or as he describes it: "beautiful" way) is astonishing.
I saw this comment before watching the video.. I forgot about it and when it got to that part I was like what is he doing? it's 19.. and then remembered this comments again 😂
Now I need to see what weird mind gymnastics does to get to that conclusion 😂
@@ceevio_art
I don’t know how many times I’ve almost shouted at the screen at Simon (and Mark too for that matter) to use his earlier logic again, but it’s a lot.
13:50 By the very logic Simon had just gone over orange can be limited to 19, since all 8 2-digit renban options must appear, and orange can only appear on the little diagonal one, and digit "in the middle" appears on two different 2-renban options, once with each of its neighbors
And then proceeds to figure out that very same fact in a way more complicated way.
@@WormHunter Pure Simon
Simon being Simon, and it’s hardly the last time in the video 😁
@@ragnkja Right? I couldn't stop shouting "SIMON USE YOUR OWN PENCIL MARKS" for the whole tail end of the video
Exactly :-D
Always so awesome to see one of my puzzles appear on the channel, very happy to see it, thanks Simon!
This one was particularly a blast for me to watch for two reasons:
1) Simon found a path that was very different than I intended to get through the break-in, and it is so elegant and natural looking it looks intentional. You could even say it is telegraphed by the cross.
2) I was streaming when this video came out, so I got to watch this video together with all of my stream viewers, and that was awesome, seeing live chat react to Simon's solve! I'll be sure to share a video of this live watch party soon.
Indeed..was awesome and such a treat to be part of.
I loved this, thanks ZM!! It was very natural, and I think Simon managed to over-complicate it a bit.
What was the intended path? That break-in was so elegant I would have assumed that was the intended path.
@@goliathcleric Note that r5c5 only appears on one two cell renban, were as the digits 2-8 each need to appear on two. So r5c5 is 19, and that digit must appear at r1c4. It also shows up on the renban in box 8 by sudoku. Meanwhile, ask where the other 19 is in row 5. Well, its on a renban. So it can't be on a two cell renban in column 5. That places the other 19 on a length 3 renban in column 5. Now do sudoku on the other 19, and you place it it r5c6, disambiguate highs and lows, and from there you pretty much converge with Simon's path.
On a logical level, both amount to counting 19s on renbans and using that to force the places of 19s, but Simon does it in a more abstract way by noting that (completely by coincidence) all of these lines are concentrated in this cross. I love Simon's break in, but if that was the intended path I don't think the testers would have called it approachable!
@@zetamathdoespuzzles I stared at this one for 20 minutes or so thinking there was no way, then I saw your intended 1-9 breaking and I thought "....you bastards, that's beautiful"
Well done.
One of the most amazing things about Simon (among many) is how quickly he's able to zero in on the break-in of these puzzles. Absolutely stunning.
Even though the way Simon parses it is not exactly what I intended, he homed straight in on the important thing: 19s are weird in this puzzle. His intuition is unbelievable.
Let's get Kraken.
@@tamarockstar45 At least we're stockpiling draft picks!
@@balkthor I was just making a pun. Let's get cracking. It's hard for any expansion team to be good right from the start, except Vegas. I think the GMs learned what not to give up after the Vegas expansion.
@@tamarockstar45 I know. I'm big on self-deprecation when it comes to my fandom. It's what's allowed me to be a Mariner fan for all these years. I tell people all the time, it's far easier to be a fan of a terrible team then a good one. That's what made the Seahawks tough the last ten years. That, and having to endure Russell Wilson's super creep factor *shudder*.
Nice thing about hockey, unlike other sports you still have a great time going to the game even if your team is getting demolished. There's always fights, fancy shots, great saves, and let's face it: Hockey fans are the greatest sports fans inthe world.
Simon the 1/9 in the middle was clear from just the 2-renbans because any other digit would have to appear on the horizontal too
Yeah, even I figured that out. lol. But it's typical for Simon to figure something out in a more complex way than anyone else could figure out.
47:16 vs Simon’s 40:14. That’s the absolute closest I’ve ever come to Simon’s solve time. Proud of myself, and consider myself well-trained by Simon.
The same for me, about 58 minutes, I'm normally never that close. But with a totally different path, I started with the 2-cell renbands, establishing the content of the vertical and horizontal (starting with odd or even), and then continuing from there with the 456-triple in box 5, and then similar deductions as Simon, but a lot slower... :)
And without Simon's sudoku class I take almost every evening, I would _never_ have been even close to finishing. Fun fact, my first language is Swedish, however, I find myself thinking a lot in English (in Simon's accent) when I try these puzzles...
these videos have become part of my evening winddown routine and i love it
Something that might have been useful to note at a few points: Once the 1 is placed in the center, you know the horizontal part must contain the "odd-high" options 23, 45, 67, and 89 (since they all see the 1 and that's the only way to break the remaining digits in to four 2-renbans), and therefore the vertical/diagonal must contain all the "even-high" ones.
That’s the logic I used
@@zacharyhall2012 Same here. I was waiting for Simon to deduce this.
And since the 9 was on a 2 cell renban line in row 5, it forces the 9 into r1c5 (or earlier it was a 19 pair.
@@cluso9985 And he only deduced it to place the last pair of numbers. That's was the first thing I see and my key to solve the entire puzzle.
I ❤ how Simon cheerfully explains the intricacies of solving the puzzle as he solves it.
Long way, after he posts it soon, watch zeta watch Simon solve his puzzle in live chat, or can watch it now from his stream yesterday.
It begins at 3:34..and watch it with chat comments.
@@davidrattner9 Thanks, David - I'll do that! Now that you've told me about zeta's channel, I sometimes tune in to see what he's up to. Not for 8 hours at a time, though.
@@longwaytotipperary cool..hope you enjoy it
@@davidrattner9 I'm sure I will! I enjoy listening to setters explain their craft. Even when (mostly) I still can't solve their puzzles unaided. 😉
I would have proven that the center cell was a 1 or 9 a lot quicker by just saying that every digit between 2 and 8 inclusive must appear twice within the eight two cell lines. As the center cell can only appear once and both 1 and 9 can only appear once, the center cell is either a 1 or 9.
Loved this puzzle and video. I actually "got" the 1-9 logic, as or even before Simon was beginning to explore it, which seems to me like a huge advance in my ability to observe and draw reasonable conclusions. Thanks so much for this! (I credit 100% of my growing understanding of the more complex logic in Sudoku puzzles to watching the videos on this channel, and doing the daily GAS puzzles on the discord channel. If you have not checked those out, you really should. Instructive and fun!)
Beautiful puzzle Zetamath. Another joyful solve. Thank you Simon.
The "viewer yelling at the screen" rate in this one was probably super high! Props to Zetamath for twisting our brain into a fine little knot everytime they come up with a new puzzle.
After rather quickly figuring out R1C3&4 were a 19 pair, I put the web app over to test mode and used the alphabet to note one of the options as A and one of them as I. Solved the puzzle in this mode until an H was placed in box 6 and since H is equivalent to 2 or 8 and there is already a 2 in the box. Kinda fell apart from there. Very enjoyable puzzle! And thanks to Sven for this option in the app!
I must say, I always felt releasing a new app for every puzzle type was a little cash grabby. But after seeing the last few puzzles you've solved and watching the video about setting you released a while ago, I sorta realized that it's not cash grabby. It's to pay the people who made these incredible puzzles. I only realized after getting absorbed into the classic sudoku app. Without a doubt the best puzzles I have ever solved in a sudoku app. I assume the quality is the same for the rest. Keep up the great work.
I'm very proud of myself that this took me under an hour to finish on my own. Renban is one of the constraints that I see most clearly, so that helps. This was an amazing puzzle! Thank you, Zetamath, for the renban exercise, and thank you, Simon, for the wonderful videos! I solved Philip's puzzle, but I don't even understand how I did it, so I did not venture to submit a solution.
Also, watching the video now, and the nearly Stone Temple Pilots reference made me pause, unintentional though it probably was.
If all cells are pointing at the center cell, and they all have duplicate numbers except 1 and 9, doesn't that limit the center cell to 1 or 9? instead of all odds? I think so.
This was my intended break-in.
i was sitting here watching thinking the same thing. i think even simon said it but then didn’t correlate it with that cell
Of course it does. Hence Simon's missing it.
@@zetamathdoespuzzles Just wanted to say a big thank you for this puzzle, I found it incredibly satisfying to solve. It starts out seeming impossible but then there's a pretty clear path once you start thinking about how it has to look.
Yes, you're absolutely right. Simon surely just wanted to leave something for us to discover 😉
Just after saying 1 and 9 are the only digits that appear once in the length two renban lines, he pencilmarks the orange cells with all odd digits, 3, 5 and 7 included
exactly. It frustrated me so much
@@joosepmattiseriste8455 Classic Simon. I always give a go at the puzzles, but frankly i never could do, most of the time cant even start any of them unless they are an actual GAS puzzle, but i still somehow figured out in the first 2 minutes the 19 logic, which took him 20 minutes with some much harder logic.
I am not usually one to sit at the screen, but I did twice this time. First, the 1-9 in the middle, then the 3-4-5-6 because of the 2s in c2 and c3 looking at the whole of the horseshoe. Great work!
finished in 42:48 without referring to the video!! very satisfied solve, thank you for honing my deductible skills (looking back, few months ago I needed to refer to the video numerous times and I took exceptionally long to solve too)
40:53 for me. That was an incredibly satisfying puzzle, I started out looking at the board wondering "how could this ever be possible?" and then slowly a few things began to drop into place and then a few more.
What an incredible puzzle! Geez how utterly elegant! Loved your solve of it, hats off to zetamath and Simon once again. Such a cool idea for a ruleset, and the weird quasi-SET thing is AWESOME.
Excellent puzzle watching zetamath react live was a great treat. 👋👋👋👋👋
I normally never have to yell at Simon for missing things but today I felt like I did the whole puzzle! Always fun to watch though
47 minutes for me. I really enjoyed that puzzle a lot. I sat scratching my head for a while at the outset but once I worked out how to start everything flowed quite beautifully.
Excellent stuff, thanks for sharing it
Love the puzzle and I solved it on my own for once. I even found the quicker 19 break-in :)
Thanks ZM and Simon.
I saw the 19 in the center before Simon, but I did not see how to utilize it until he explained all the set. So I was quite stuck and started the video after 10 or so minutes of banging my head. So with a little help I got a 67:16 time. And considering I had to backtrack with the box6 logic, I'm pretty pleased with it all. So as Simon said at the start, it's not the easiest break in but quite approachable once you get past that part.
Simon sure did this the hard way. Just need the 2-cell renbans to get 19s and once you know the middle cell, you know which set of renbans run horizontally and vertically.
15:05 Although I've never seen this used in a puzzle, we actually do know about the sum of the digits on a 4-cell renban. It's always an even number that's not a multiple of 4. The options are 10, 14, 18, 22, 26, 30.
The thing I have used about length 4 renbans is they always have a digit that is each of 0, 1, 2, and 3 mod 4. But if you look at the digits 1-9, there are actually three numbers that are 1 mod 4 and 2 that are each other congruence class. That imbalance often lets you pin down a lot of cells that have to be 159 from the start.
41:00 Vertical scanning strikes again, yes that 4-renban sees 7s both its rows, it ALSO sees 2s in both its columns...
I got right to the end but realised I had unresolvable 34 pairs in columns 8 and 9 of rows 1 and 9...then spotted that I had two renbans with 3456.
Fortunately, I only had to backtrack a little to fix my error. Another brilliant puzzle from Zetamath.
Took me just under an hour, fairly happy with the break-in logic even though it was slower than I'd hoped.
Wow! I heard approachable and decided to try this puzzle. After a couple of errors in my logic, I was able to complete it. Had I seen your break-in I would have steered clear. Kudos for seeing that, but it was far simpler for me.
Very clever puzzle, ζ-math 👍🏻
It took me 53 minutes, which I thought was really slow and that Simon would blitz through in half the time, so it's really gratifying to see that he struggled as well! I was particularly impressed with the way that he seemed to have worked out that the orange pair in r1c4 and r5c5 had to be 1 or 9, and then completely ignored that logic and pencilled them in as 13579 🤔
_Edit:_ Also, can confirm that if you follow through with the deduction that the middle cell has to be 1 or 9 to avoid knadgering up the renban dominoes, and remember that renbans must be unique, you can pencil the 1s and 9s into the blue cells without even thinking about set theory and the maltese cross.
I got it in 28:50, super proud of this one!!
I very quickly realized that all possible 2- and 3-renbans must occur, and it was a very fun solve from there.
This is probably the only time I'll ever beat Simon, so I'll cherish it for long time ;)
27:20 for me - what an interesting break in with the pairs intersecting on R5C5, except for the diagonal one, which tells you what it is (almost).
18:42 This is the first time I feel like I understood the puzzle faster. I instantly looked at the arrangement of 2-cell renbans and thought "Theres 4 were the lower digit is even and 4 where the lower digit is odd" Then noticed that there were seven 3-cell renbans which must have a lowest digit of 1-7 once each. Fun!
Solved it with much help from the video.
This puzzle is very satisfying to solved! Absolutely brilliant!
In terms of the break-in and the logic that you did at around 21:00, couldn’t you postulate that the center cell had to be a 19 care because they each only appeared at one time on the eight 2-digit lines and so you needed it one in the middle to guarantee that row five had all of the digits? (Since 3, 5, and 7 were already twice accounted for)
19 pair*
And it makes it a much simpler break-in.
19 pair in centre cell also r1c4 and r1c5,
That’s exactly how I broke in. :-D
Of course! The method he found was fascinating, though.
12:56 for me. Beautiful puzzle, I always enjoy puzzles using renban lines and this one, of course, was no exception.
It felt like being on a trip in wonderland solving this puzzle. I mean really. The beginning was absolutely stunning and I was expecting the rest to be just a regular sudoku already satisfied with the beautiful beginning. But NO, the rest was not a downfall at all. Just hard enough to be fullfilling but not so hard to be painful. Perfect match for my current state of mind.
Even though I also started with the same logic most of you decribe here to lock the 19 in R5C5, I absolutely loved Simon's break in. Just genius!
@8:55 Correct me if I'm mistaken but there is 1 detail to be gleaned immediately here. There are 7 different 3-cell renban lines. But a key feature of the "3x" rule of 3-cell renbans is that a digit must HAVE a x-1 and x+1. So within the digits 1-9, only the digits 2-7 are valid Xs. So EVERY unique version of a 3-cell renban must appear in this puzzle exactly 1 time, no more and no less.
This may get appreciated later and used to narrow some logic down, I just happened to notice it out the gate. :)
you could also say, that since every 3 cell renban is equal to a multiple of 3, the highest 3 cell renban (9,8,7) is equal to 24, 24/3 = 8 different options, now you can't have 3 in a 3 cell renban so that option is ruled out, leaving you with 7 different options
*2-8 are valid Xs in your example.
I took my X to be the lowest of the three digits, so X, X+1 an X+2. Then X could be any digit from 1 to 7, so 7 unique combinations. Boils down to the same thing, just slightly easier to see there are precisely 7 of them.
27:49 ... surprisingly approachable, yet a lot of fun to solve!
It was a pleasure solving this puzzle!
Got it in 57:11. Not a bad time for me considering the video length! I liked coming to the realization that every possible two-cell and three-cell renban was represented in the puzzle and I could figure out which was which that way.
17:45
Very surprisingly approachable. At first blush this looked impenetrable but it generously lays bare its secrets in ingenious fashion
Rules: 04:00
Let's Get Cracking: 06:17
Simon's time: 41m15s
Puzzle Solved: 47:32
What about this video's Top Tier Simarkisms?!
The Secret: 2x (03:26, 48:45)
And how about this video's Simarkisms?!
Beautiful: 13x (12:21, 20:07, 20:09, 22:33, 22:35, 28:39, 28:42, 29:54, 39:01, 39:03, 46:17, 48:03, 48:20)
Ah: 11x (11:14, 16:06, 16:08, 18:48, 21:35, 23:23, 24:57, 26:43, 28:09, 33:28, 44:03)
Sorry: 9x (08:53, 16:08, 17:29, 18:48, 19:40, 21:20, 22:58, 41:07, 47:36)
By Sudoku: 9x (25:56, 30:54, 35:38, 37:25, 38:05, 38:42, 41:07, 47:00)
Hang On: 8x (06:56, 13:55, 19:45, 19:47, 25:54, 41:03, 44:03, 44:05)
Obviously: 4x (04:33, 08:06, 16:38, 27:00)
Stuck: 3x (25:51, 33:28, 40:27)
Pencil Mark/mark: 3x (37:21, 37:46, 45:11)
Useless: 2x (10:45, 34:06)
Goodness: 2x (10:49, 47:15)
Clever: 2x (17:58, 38:32)
Break the Puzzle: 2x (14:16, 30:23)
Approachable: 2x (00:42, 22:28)
Magnificent: 2x (47:45, 47:48)
I've Got It!: 2x (17:54)
In Fact: 2x (01:04, 14:05)
Nature: 2x (09:46, 18:17)
Good Grief: 1x (07:25)
Bother: 1x (40:27)
The Answer is: 1x (43:29)
Naughty: 1x (22:28)
Going Mad: 1x (21:39)
Surely: 1x (39:40)
Stunning: 1x (01:34)
QED: 1x (22:18)
Grind to a Halt: 1x (40:33)
Wow: 1x (07:27)
That's Huge: 1x (24:17)
Most popular number(>9), digit and colour this video:
Ten (3 mentions)
One (119 mentions)
Blue (12 mentions)
Antithesis Battles:
Even (12) - Odd (10)
Black (7) - White (0)
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!
2:11 - Big slice of cake
@@th.nd.r Added, thanks! I'm curious how often that one will pop up
@@inspiringsand123 it seems to pop up every time someone has a birthday, so I’ll guess probably a couple times a week?
This should be the top comment!
16:55 -- How long will it take him to figure out that the two oranges must be 1 or 9?
21:00 -- Found it. The method was rather roundabout. It really was a lot easier to discover that the oranges were 19. Think 12-34-56-78-9 and 1-23-45-67-89. One's vertical and one's horizontal.
43:50 -- Where is 2 in cell seven? (Can't be on the line.)
49:00 -- Congratulations!
well, I tried this and I think I actually succeeded without needing Simon's help (yay!) and just skimming the comments I think I need to watch it now, because my break-in was definitely working out the 1-9 pair... in row 1.
It seems that the comments may have already addressed this but your in to the solve was far more complex than necessary as you could think think immediately that all the 2 cell Renbans will use all digits twice except for 1 and 9 as they cant pair lower or higher respectively
You could have gotten the middle as 1 or 9 straight from the start if you realised there was only 8 possible 2 cell renban options and that 7 of them could see the central cell meaning anything other than an extreme digit would use up 2 of the options. E.g. if it was 4 then that take out 34 and 45 leaving only 6 options
Edit: oops.... Just seen everyone else has said the same thing but I hadn't loaded comments yet 😂 they all explained it better than myself so go read them comments!
Solved in 37:48 - this was really neat! I haven't watched Simon's solve yet so I don't know how he did it, but figuring out the center square was a 1-9 was a great a-ha moment, and there were plenty more as I worked through the puzzle.
46:25 solve, the first time I've ever beaten the video solve time! Like most of the comments I found the 19 much more simply and of course not needing to speak the logic aloud as I use it helps, but it's still a huge milestone for me.
Amazing puzzle. I felt good about understand those two squares in the beginning had to be the same, but then I was stumped!
34:35 finish. The way I looked at the start was: each digit appears on a 2-cell renban twice except 1 and 9 (once each). Since the center cell sees all but one of the two-cell renban cells, it must be a 1-9. Since it has to go onto a line once, it must be in r1c4. Now that the center cell is an extreme digit, all of 1-2/3-4/5-6/7-8 must be horizontal or vertical, with 2-3/4-5/6-7/8-9 being the other. Therefore, r1c5 (seeing all of the vertical two-cell renbans) must be an extreme digit, forming a 1-9 pair in the box/row. I then identified r1c5 as matching with r5c6, the only non-three-cell renban in column 6, and r7c4 as the only place for it in box 8. This, of course, differentiates the 1-9 pairs, due to r5c7.
Next, you don't actually need set to establish the 1-9s in the cross. You've placed them in columns 4-5-6 (all placed except the 1-2-3 renban in box 8). In box 6, neither the 1 nor the 9 can go on any of the renbans (already have 1-2-3 and 7-8-9 placed elsewhere, and they are already in row 5 elsewhere), so they must be in r4c8/9. This finally pencils them in box 4 in row 6.
That having been said, there's nothing wrong with a good set in a sudoku. More maths!
Simon: "We don't know..."
Me: "Oh but you do!"
After a bunch of crazy tough puzzles glad to have one where i was shouting through the video :)
Wow.. I really solved this one differently to Simon again...
I figured out that 19 must in the middle due to it being the only number that appears once in double pairs.
and then I was able to figure out the triple at the top had to be the other 19, and that that basically started to solve the puzzle for me
Very fun puzzle. great work =)
24:51 for me. I thought that was a very interesting rule set - and quite approachable.
In the 20 minutes I took attempting this, the one useful fact I realised is that the middle digit needs to be a 1 or 9, because there's only one 1-2 and one 8-9 set in all possible 2-renban pairs. And r1c4 needs to be the exact same number.
All that to say, I don't think I'll be solving this one 😅
37 minutes! I’m very happy with that! This was a very clever ruleset. :-D
47:32 you can see the happiness drain so fast from the shock of the check box being not the normal one
43:53 before watching the video!! I never could have done that without the tutoring I have received watching these videos. Now to see if I found the same path as Simon. 😁
THAT WORKED OUT COOL
Great job
Something about the break-in I'm not sure other people have pointed out yet: whichever of 19 is not in r5c5 must go in r1c5 because it has to appear twice in r5 and c5, since it cannot be in r5c5. It will appear once on a 2 cell renban and then it appears in the only cell which is not a 2 cell renban
This was absolutely my intended step 2!
That was a very long and complicated way of figuring out that break in. The way I did it is to note that there has to be at least one 89 or 12 pair in row 5 and whichever of 1 or 9 was in that pair has to be in r1c5 because it can no longer be on a 2 cell renban. After that it becomes obvious that no other renban in row 5 can contain the other of 19 so it must be r5c5. And r5c5 must be on a 2 cell renban somewhere, therefore must go on r1c4.
Yes that was definitely a Simonism (finding the hardest way to get the solution).
This looks doable... See you in two hours then I given up!
15 minutes! I got my first digits!
@@Jonas.Nilsson You're doing well then!
@@zetamathdoespuzzles nope, I fell asleep after 30 min!
And now I just scan the video... I was stuck at the point that Simon were on at 24.48
So many favorite moments on this channel....... 18:48 is my current favorite :-)
Brilliant break-in even though I hate it when I get de-blue-ified.
Great deduction by Simon leading @21m30s to the 1,9 pair in box 5 and 7. My solution path was more like the "intended" one (see below)
Got it in 35:50. I really enjoyed solving this puzzle.
Excellent puzzle, to me this lies at the intersection between challenging and doable, and it's all about logic.
Cheers Simon, your'e impressive, the genius is the solver.
I've been enjoying the app updates. I did encounter one in killer though where I got a lot done but then I got stuck so I used the smart hint. It was talking about outies can't contain a 5 therefore this specific digit must be a 4. So I put in a 4 not understanding it because I'm like the outies HAVE TO contain a 5. And it marked it wrong!!!! That specific digit was in fact a 5. The smart hint gave me an invalid digit. I had no choice but to bifricate (I refused to put in a 5 when it was a 4/5 digit but the 4 that the smart hint gave me was marked wrong). Fortunately I didn't have to bifricate much. After I managed to get a digit it filled in rather quickly. I don't remember the number but it had a massive ring on the outer perimeter and they all had 1 possible fill or quickly resolved to 1 possible fill.
Simon successfully found the most complicated way of breaking into this Sudoku. :-D
Somehow, despite finding the 2-cell renban logic much faster than Simon, I still end up much slower ;-) (60 minutes).
The 2-cell renbans were a work of beauty in their own right. I realised quickly that they had to be every combination, that 19 could only appear once and therefore had to be place in r5c5 and r1c5 (and as a by-product r1c4).
Also the vertical and horizontal 2-cell renbans had to respectively either all have an odd low digit or an even low digit, which eliminated options quickly. Then I failed to apply effectively the same logic to the 3-cell renbans, where Simon shone.
because the middle digit sees 7 out of the 8 two-liners it has to be 1 or 9, since all other digits appear twice.
As I expected, many others on this channel have already pointed out that Simon did not catch the quickest break in to the puzzle even though he makes the point that all of the digits one to nine appear in the 2-digit renbans twice except 1 and 9. He said it more than once and completely missed the significance of that in reference to the center square. Naturally, being Simon, he found a substantially more obscure route to the same result.
Very cool puzzle. 39:55
The 1 and 9 clue in the middle can also be gotten by realizing that digit must only be in the renban once so it must be an extreme digit.
I got this done without hints! The 1/9 logic noted in other comments was my path.
That break-in has to be one of the most unnecessarily overcomplicated things I’ve ever watched Simon do on the channel.
I've been a subscriber for over two years. That's not even close to the most unnecessary overcomplicated thing Simon has done. It's amazing how he manages to solve the most complicated break ins but manages to take the scenic route on some of the more simple deductions, and yet I still love it every time!
Have to agree. You can work out the 1-9 pair in r1c4/5 far more easily and that either 1 or 9 has to go in r5/6c6 which has to be the 9. Admittedly it then takes longer to also know r7c4 is a 1 or 9. The puzzle is actually approachable once you spot the 1-9 in the top row and realise how much that gives you in column 6. No need for anything as complicated as Simon does.
Agreed. Completed this in 14m31s. Sometimes he's very blind to the seemingly obvious.
@@adrianhead6272 I'm still impressed how he and Mark just manage to attack all these puzzles and solve them logically even if the method is sometimes over the top.
Simon's brain:
The simple way? Nah 👎
The most complex way? Hell yeah! 😎👍
The middle digit only appear in the purple lines once. It can only be 1 or 9.
A way I like to explain the 3 cell line is: "If you take one from the big and put it in the small, you get the same number 3 times, because they are consecutive"
Simon does an incredible job explaining things
Couldn't put a single digit in this bad boy.
Time to watch while trying to sleep
I definitely COULD spend a few hours trying to break in to puzzles like these, but I’d rather watch Simon connect the circuits in my head instead lol
Nice puzzle. Solve would have been quicker if I'd observed that the puzzle had each of the 7 possible 3-long renbans.
some cool logic id like to throw out is that in box 1/4, the 8 sees the entire renban. since we have to have 2 evens on the renban, it’s either 24, or 46 so we know there’s a 4 on the renban.
That's a cool way to see that! I usually think of it as, I need four of the digits 123 4 567 and if I omit 4 then I can't get to four of them, but your way is a bit easier to think about in my head!
@@zetamathdoespuzzles I usually imagine putting the lowest set of digits on the line (1234) and the highest set of digits (7654) and seeing if anything exists in the overlap (4), but I guess it comes down to the same thing. It's just how I approach it.
Very enjoyable puzzle, Zetamath.
@@zetamathdoespuzzles damn. acknowledgement from the puzzle maker himself. what an honor. i love your way of seeing it too! i just tend to better see parity interactions especially on a puzzle so heavily renban’d
Sven, you are amazing on colored cells can you do a white number by holding ctl or shift, to signify a number that cannot go in that cell.
I found the analysis of the last two 4-cell renbans frustrating. The 3-cell renbans had been worked out by looking at the set of various compositions, and the same could be done for the 4-cell cases. Simon noted that the last two 4-cell renbans had to be selected from 1-6, so they both included 3s and 4s. He could also have noted he'd already used the 1234 case, so the last two renbans had to be 2345 and 3456, and the one in box 7 couldn't include a 2. So, both compositions were known.
15:33
"Let's highlight all of them again.."
"In a, particularly, GARISH YELLOW.."
LMAO
[You cracked me up, Simon. ]
;))😂☕😎
19:50 for me. using colors did this kinda easy,
At 13:28 you had already declared the exact reason orange was 1 or 9 and I was really surprised that you didn't pencil mark it right then.
I think the renban lines might too dark/opaque. Simon seemed to miss a number of 4s that had the renban going through them while scanning? The same might have been true with the 2s in the upper left, since the purple is so similar in value to the blue.
Perhaps Sven can add some transparency to the renbans? It won't need to be very dark at all to still be visible. Perhaps something with a similar value to the thermo grey? (though that could be bad in terms of color blindness, where pink and grey are sometimes difficult to distinguish when the value is similar.)
Adding a white shadow or outline to the numbers would also work for this. I was having trouble while watching too.
Hmm, 25 mins here, and usually I'm much slower than Simon. I wonder if I was helped by lack of tricks to apply, so had to see that all the 2- and 3- length renban were in use, and hence the sequence 19 [ R5C5=R1C4= 1 or 9, and then C5 gives R1C5 is the other of 1,9 which in turn must also be in R5C6 (considering R6)].
Am I correct that as soon as those 2s were placed in b7c1 the 4-cell renban had to contain a 6?
You are.
I've spent ten minutes or so trying to get *any* logic to even get started beyond the simplest deductions ever (for example: there is not another 8 in this box) and I am intrigued to see how it actually plays out.
looking over this i thought how many sets are used. all 2 and 3 sets are used! thats gonna make it easier since you know the consecutive numbers must be on the lines
29:21 for me! much easier than expected