(aqjhs here:) The way i saw it being logically solved was (ironically) pencilmarking from all tips simultaneously to find a "centroid" of the tree graph. Then the bulb is the closest tip to the centroid. I'm in total awe seeing that Simon quickly picked up the position of the centroid within minutes of solving just from the pressure on row 7 and column 7. Pure genius.
@@catnip420 i wouldn't call a particular way the correct one, but sure, after realizing the nature of r6c6 (which Simon got in a masterful way imho) some pencilmarks might have spared some of the ruckus caused by r4c4.
I was one of the people who watched Mark try to solve and fail yesterday. Honestly, it was very cool to see a puzzle go wrong. I've seen it said before, and I'll say it again. CTC should show some of their failures from time to time. My two cents... :)
looks like this puzzle might require from one to go outside of the box, out of the house, to walk a few blocks, take the bus to a retreat in the mountains, and think there.
Took me two hours almost to the second but I've managed to solve this! It's the first Cracking to Cryptic sudoku I've ever managed to solve, I did it with a LOT of coloring. I've never been prouder of myself. Thank you Simon for the countless hours of fun you've given me over the years.
This man brings joy to so many of us, and I have heard plenty an email stating how Simon and Mark helped people struggling with hard times and mental health. And you Simon Anthony ,dare to call yourself a useless individual! You are not. Thank you for listening to my TED talk
The two parts of the break-in that got me going were the 7 cells on the tree in row 8 and column 8 (thanks to Simon), but also the heavy restriction on R5C5. Because the center cell is not a bulb, it must grow in 3 directions, with the low digits coming from the 4th direction. It's growing in the direction of the diagonals, and we can't put 7 8 or 9 on those diagonals. So since there must be 3 different diagonal digits with a maximum value of 6, the center cell must be 2 or 3. You can then check each bulb to see if it can reach the center while maintaining that 2 or 3, and that eliminates all possible bulbs except R5C3, R6C7, and R7C8. From there R5C3 can be eliminated similarly to how Simon does it, and we're off to the races.
Thanks Simon, for the shout-out to the Asian Sudoku Championship! We are very excited to host the competition, and to see some diverse participation from across the continent! 💖
Oh Simon! at the 90 minutes mark @1:30:30 you finally look at the low digits. I've been waiting for you to do that since @45:00 where you can see that with the 2 in the middle, the 1 and 2 both have to be the outies of the pink and blue regions. which breaks because 1 and 2 are already in column 9 in box 6. So the pink region breaks.
It is not. I solve it in 16 min, and i'm definitely not fast solver. I'm even surprised how Simon could struggle it fo 1.5 hours, must be something overcomplicated logic (didn't saw the video yet).
Simon's amazing. He's solved dozens of puzzles that I haven't even figured out how to start, and of the puzzles I can solve, he usually solves them faster than I do, while slowly and meticulously talking through every step. But he's not perfect. Sometimes he focuses on the wrong thing or misses something entirely. I haven't watched the video yet to see where he went wrong, but I just solved the puzzle in 29 minutes without really getting stuck at any point, so he's definitely gone wrong somewhere. My suggestion when you see a long video is to give it ten minutes and see how you go. Either you make progress and enjoy the puzzle or you get completely stumped and appreciate Simon's solve even more.
I approached this very differently from Simon. Ruling out as many potential bulbs as possible, I quite quickly found that there were only three possible cells. Then I got hopelessly stuck. and had to watch the video until Simon found the insight which I had missed. We live and we learn.
I found the fact that at any junction, only 1 branch will be decreasing to be very helpful. In the central square, it quickly becomes apparent that 3 branches will rise and this is quite constraining as you can't place a 7 or higher in any of the corners - with the next highest digits being 6,5 and 4, you're very quickly down to a max of 3 in the centre.
If the correct bulb were in the "North-west" corner, then, as Simon pointed out, the middle cell of the grid would be forced to be a 2, and therefore R5C3 would have to be a 1. By then you have a 12 pair in R9C7+R9C8. It is then impossible to put 1 and 2 in row 7, since they would have to occupy R7C1 and R7C9, which is now impossible..
I think finding out where 1 and 2 would have to go is a really nice way to disprove the 2 in the centre. If you notice that most of the thermometers have to be greater than 2, you can ask the question of where 1 and 2 go in R7, R8, R9 and C7, C8, C9. This pretty easily and beautifully disproves that R4C4 is lesser than R5C5.
This is a very mysterious comment from you. In all those rows and columns 1 and 2 can go to multiple places. R7 is the most restricted, but even there 1 has has 4 possible places while 2 has more. In C9 and R9 they can go literally anywhere.
I had a similar argument, only asking where 1 and 2 went in row 7 and box 6. 2 in the centre requires the bulb to be r5c3 (only bulb close enough), and this to be a 1. Then every thermo cell in row 7 and in box 6 has to be greater than 2. That only leaves r4c9 and r6c9 in box 6, and r7c1 and r7c9 in row 7. That's three cells in column 9 that can only be 1 or 2.
@@istvanmagi473 Let me explain a different way. If R5C5 is a 2, then the 1 is forced to R5C3. Row 7 and Column 7 would then be forced to have a 1/2 pair at each end. If you try to put 1/2 in Column 8, then box 9 is eliminated because it has 2 1/2 pairs in it already, and row 5 has a 1 and 2 in it, leaving only R1C8 to hold both 1 and 2.
Havent beaten Simon in a while, but this time I beat him by more than an hour. Here is a MUCH simpler path: 1) R4C4 being 1 is impossible, as only way that happens is if R5C3 is also 1 which together with R5C5 (doesnt even matter what digit you put there) creates at least two schroedinger cells (R6C8 and R7C1), since the center cell MUST be greater than all the thermo cells on boxes 6, 7, 8 and 9 and once you follow the sudoku implications through box 6 to 9 to 8&7 you will find the schroedinger cells. So now you can just pencilmark 19 into the two empty squares (assuming you did the 9 logic before lol). 2) Notice that NONE of the corners in box 5 can contain a digit greater than 6, since the extensions (R3C3, R3C7, R7C3 and R7C7) all see 3 digits that are further along the thermo than they are (I will just ignore the low end here, as the low end certainly cannot enter the middle box through 7 lol) 3) That forces 7 and 8 to R5C6 and R6C5, and R5C6 cannot be 8 so you get two digits and can actually fill in those thermos. 4) Now thermo end in R9C5 is under pressure. It cannot be the low end (leads to too high entry digit to middle box), so its now 7 at the highest which pressures R6C6 to be a maximum of 3 (so now you know that is actually 2, since thats the corner you must enter the middle box from). Rest is basically just sudoku, and at the very end you discover the bulb.
This comment is really interesting. Thank you for sharing. My logic was very similar, but did not need your *step 1.* I ruled out the Phistomefelian bulb from *r4c4* easily after discovering that *r5c5 = 3*
Quite pleased I tried to solve it myself first as didn't think I would get it with a 1 h 40 video but I managed to get the answer in a respectable 54 minutes. The break through for me was the realisation that in the centre box the corner cells had to be between 2 and 6 which meant the central cell had to be ... and consequently the 1,7,8,9 were heavily restricted. Now going to watch Simon's solve.
@@RichSmith77 Only after watching Simon's solve did I realise that one could have gone into one corner... How great am I; I must have subconsciously intuitively known that a one there would break, so blocked it from my mind, saving me the trouble of logically disproving it!
@@martingibbs1179 I sometimes have unconscious insights like that too. Ones that only come to light when I watch Simon or Mark's solve afterwards, and find myself forced to ask why I did a step a lot easier than they managed. 😂
I was so happy with my solve until seeing Simon's and realizing that I started off with an incorrect assumption, i.e. that 1 couldn't go in that top corner. I always admire Simon's ability to rarely/never jumping on a conclusion until they are adequately proved.
There is a uniqueness argument that the bulb must be 1 - otherwise deduct 1 from all the thermo cells, which are the only clued cells and that must give a valid solution (you can deduct 1 from the digits 2-9 wherever they appear and replace 9 with 1)
It's easy to prove the bulb must be one - there are no bulbs in box 5, and only two empty cells. There's no cell on the thremo in box 5 that can be a 1 or a 9 (they all either see the next cell on their branch, or will force multiple 1/9s into the next box down their branch). Hence, the empty cells are 1/9. So 2 must be on a thermo line, and the same logic means there's no way to reach a bulb on a branch of only 2s, which means 1 must be on a bulb.
@@eytanz It isn't entirely trivial to rule out 1 from r4c4, with r5c3 being the bulb. Simon struggles with exactly that the first few minutes of the video. Which is to say, it's not immediate to see that 1 in box 5 is in an empty cell. This still makes the bulb 1, so your main point stands.
I’m by no means trying to imply this was easy to see, but after watching you explore the options, I think I found the simplest way of explaining what you found: Once you place 2 in the central square, it sees r5c8. This puts 12 pairs in row 7, column 7, and column 8. This puts 3 cells in box 9 that must be 1 or 2.
The end message roughly translates to: "With a hundred hands and all my strength, and a hundred eyes, keep watch over us." Sounds like a prayer or blessing to some higher power.
"of a hundred hands is our force, by a hundred eyes we're guarded" lyrics from "cento mani e cento occhi" by Banco del Mutuo Soccorso, i read it more as "the power of community"
@@therealAQ The track "Cento mani e cento occhi" is on TH-cam. It's Italian prog-rock from 1972. From the English translation, the lyrics seem to depict an outsider approaching a Stone Age tribe.
@@therealAQ The literal translation is: _"Of a hundred hands is _*_my_*_ force, and by a hundred eyes _*_we are_*_ guarded"_ However, it will have exactly the same meaning if it were as you wrote: _"Of a hundred hands is _*_our_*_ force, and by a hundred eyes _*_we are_*_ guarded"_ or also: _"Of a hundred hands is _*_my_*_ force, and by a hundred eyes _*_I am_*_ guarded"_ Yes, it is about the power of a tribe of cave men, compared to the power of a single one, who is reluctant to join the tribe.
Really nice puzzle. I had a different break-in for it. 1. I checked the maximum digit for the corners in box 5, assuming the values increases from the center. It was 6 for all of them. Since 3 of them will increase it means the center can be a maximum of 3. 2. I checked what happens to the digits 1-3 in row 7, with the bulb in the top half and center being 3. No way to do it. Same with digit 1-3 in column 7 with the buld in left half and center being 3, no way to do it there either. 3. That means the bulb is either in row 5 column 3 (the only way the center can be a 2) or it's in the bottom right branch on r6 c7 or r7 c8. 4. I was sure it would be r5 c3, but that led to lots of the digit 1-3 in box 9 (r7c7, r7c9, r8c9, r9c7, r9c8), and that's simply not possible.
I agree. As usual. It has been already noted in dozens of similar comments in the past. I might be wrong, but his reluctance seems to be due to his strong desire to drastically and stubbornly differentiate his method from Mark's.
Funny how Simon made the exact same mistake I did trying to find the break-in, focussing on eliminating candidates for the true bulb. After I shifted focus to the central box the break-in turned out be a lot simpler. The opening logic is to realize that regardless of where the thermo enters box 5 it has to leave through at least three corners each of which cannot contain a digit higher than 6. This restricts the central digit to be one of [2,3] and you can eliminate 2 as Simon showed in the video. With the central digit known the three exit corners are now a set of [4,5,6] and you can instantly place digits 7 and 8 in the box as well. The way Simon did it this puzzle would easily qualify as a 5* but with the correct logic applied it is really a 3*
Finished quickly via bifurcation, but watched the video and I feel like I've learned a lot about sudoku logic, it was a pleasure and don't worry - you didn't let anyone down. Fun puzzle!
As I wrote in a separate comment, the logic Simon used from 34:33 to 1:17:17 was fascinating, but it was also undeniably, by definition, a *kilometric bifurcation.* It tested an extremely long and complex prong of the "fork," which turned out to be invalid and was hence rejected. Of course, it was not a random bifurcation. It was very well "educated," but it was bifurcation. I needed bifurcation as well, several times, but much shorter and mainly performed in my head.
i took one look at the grid and went "nah there's no WAY that is possible". then proceeded to pause the video at least 10 separate times, try to work through some logic that simon mentioned and i hadnt thought of, only to absolutely CONVINCE myself that i had proven it was indeed impossible. then watched further in just to be proven wrong AGAIN. thoroughly amazed by anyone and everyone able to complete this and hats off to the setter, this is incredible
Loved the fact you did not find out the 'naughty' bulb until the very end of the puzzle. Got there eventually after many false starts. Probably the most difficult puzzle I have ever solved from you guys. Thanks for all you do.
The amount of time it took Simon to deduce that r8c7 could not be green after breaking into the puzzle was ludicrous, and I wouldn't have it any other way ♥
At around 52:00 Simon could just argue that he needs the 1 and 2 in row 7 to be off the thermos, therefore the 2 being in column 9 in Row 7. But also per Sudoku the 2 is placed in Box 6 also in Column 9, this clashes and if you would see this way earlyer on you could disprove the whole central being 2 thing within seconds if you just focus on 1s and 2s by sudoku.
I started quite elegantly. similar to simon, i considered the top left containing the bulb, but focussing on small digits. r5c5 has to be 2 in this case: it cant be 1 since it's too far from any bulb, and it's not 3 or more because that would cause the 6 thermos in box 8 to be at least 5. a central 2 forces a 1 in both r4c4 and r5c3 and also forces at least 3s in all thermos from boxes 6-9. the 1 and 2 in box 6 are forced into column 9. and now the puzzle is broken: r7c1 is the only place in row 7 for both 1 and 2. From there the puzzle solves the same way as simon did it at 1:19:30
i cant believe i crushed simons time. i got it in 65 min. usually i only rarely can beat him on the easier puzzles and its mainly because he has to explain all the logic while solving the puzzle.
The logic i used is a little difficult to explain, but I'll do my best. Basically, i focused on the most obvious thing in a sudoku puzzle. The digits 1-9. Relating this to it being thermo, the question was, will it reach if i keep it to a minimum? This eliminated a few bulbs. Combining it with filling the grid with pencil markings, it gave me a general direction the thermo is going. This put pressure on the low digits in box 5, thus putting the bulb in the lower right. Then, proceed from there.
Also small tip: From a uniqueness perspective, you KNOW that the bulb (wherever it is) MUST contain a 1, just as certain as at least 1 Tip must contain a 9. Because if the values ranged from 2-9 you could just also change all the values to 1 lower (1-8) and change all the 1's in the grid with 9's.
Got it in under 30min. Was very smooth and funily enough, I took Marks approach :D the logic between boxes 3/7/9 was SO beautiful! I was like "Haaaang on, I see what u did there", loved it 😊
Yes, either the question of where 2 goes in r7, or specifically what goes in r7c8 that is both bigger than 5 (for the row) and smaller than 6 (for the column).
Yes. To rule 2 out from r5c5, you only need to look at the effect it has on digits 1 and 2. The only bulb close enough for a 2 in the centre is r5c3, and this would have to be a 1. Then every thermo cell in box 6 and row 7 would have to be greater than 2. This forces 1 or 2 into rows 4, 6 and 7 of column 9.
Solved this in 25 using a different approach. Considering the middle digit, we know 3 of the paths are increasing and all 4 paths have a next max digit of 6. Therefore the middle digit is a 2 or 3. Disproving the 2 case like Simon gets a 3 start. Then because of the pressure on box 8 we know the 2 is one of the bottom paths. We also have a 2456 and a 19 pair in box 5 so the 78 is easily placed to continue
Astounding, Simon. I was immediately and absolutely certain that this would be a viewing experience complete with a large bowl of popcorn. Very interesting puzzle and solve. At one point with 20 minutes or so to go in the video you made a remark that I thought was very interesting, that pencil marking was something that you were going to resort to because the video already was so long. I have often thought that you would find the next step or get a running start toward the solution of some of the complicated puzzles you do if only you would just pencil mark the dang thing - and it seems that you might agree with me. Nevertheless, I love that you don't pencil mark things immediately, because I have learned how to discern where it will be most helpful by watching your solves of these puzzles that are way beyond my abilities for now. Thank you for this video, and for all that you do to make this channel and community so great!
Solved this in 51 minutes flat. Broke it a few times along the way. Every single time was due to forgetting that it's a slow thermo instead of a regular thermo.
The simpler way to rule 2 out from r5c5 is to just look at the effect it has on digits 1 and 2. The only bulb close enough for a 2 in the centre is r5c3, and this would have to be a 1. Then every thermo cell in box 6 and row 7 would have to be greater than 2. This would force 1 or 2 into rows 4, 6 and 7 of column 9.
That seems to be much better. The logic Simon used from 34:33 to 1:17:17 was fascinating, but it was also undeniably, by definition, a *kilometric bifurcation.* It tested an extremely long and complex prong of the "fork," which turned out to be invalid and was hence rejected. Of course, it was not a random bifurcation. It was very well "educated," but it was bifurcation. By the way, I needed bifurcation as well, several times, but much shorter and mainly performed in my head.
The meaning of your comment was not immediately clear to me, but it became clear when I studied it. I believe you meant that, since every thermo in *box 6* and *row 7* would have to be greater than *2,* then: *r4c9 = 1* or *2* *r6c9 = 1* or *2* *r7c9 = 1* or *2* Very elegant and very easy. Thank you for sharing this finding. In my solve, I also needed to rule out *2* from the grid center (my first digit was r5c5 = 3, as I explained in a separate comment), but I cannot remember how I did it, actually... Certainly, I did not need 40 minutes of bifurcation.
This was far easier than the video length would have you believe. I will have to watch the video to find out why Simon struggled with it. I solved it using a question Mr. Anthony often asks: "where can X go in this box/row/column?"
@mattconco No, not far easier for me. It might be hard for someone new to sudoku, but that can hardly be the bar to which we measure a puzzle's difficulty rating - most puzzles presented on the channel would be hard for someone new to sudoku, whether it's 30 minutes or two hours. I think the video length in this case might discourage some solvers, say not beginners, but not yet experts, from attempting it as it suggests a high difficulty rating. I for one hesitated, but decided to have a quick glance. I saw relatively quickly that the center box is very restricted and had a pretty smooth break-in with relatively conventional logic. That is why I was surprised at the length of the video. I do understand that even Simon and Mark have their off days and this was clearly one for Simon.
@@joethornton5321 I would imagine that most people who are novice enough to be put off by a video length would still struggle to complete this quickly. Again, I don't see that fact you were able to spot logic quickly means everyone else will find it easier than Simon too. I think some people don't realise when they get really good at solving sudokus. I find a similar issue on LogicMasters. There is too much use of the 1 star rating because people think "well if I managed to do it then it must be easy", not realising that they are a well above average solver. That's my theory at least. I think what a long video demonstrates is that a puzzle requires some logic that isn't immediately obvious to Simon or Mark and therefore is at least intermediate level. I don't believe so much in 'off days' in this case. Simon found a long piece of logic and decided to persue it. It made for a more interesting video. The only reason I replied was because I found it quite pompous to say "I will have to watch and see why Simon struggled". I don't personally consider failing to notice the fastest route as being struggling.
I finished in 96:57 minutes. This was such an interesting puzzle to figure out. I spent the first 20 minutes doing the puzzle wrong and trying to make each end in a box only have one thermo. It doesn't even make sense trying to describe, so I don't know how I got it so wrong when the rules make it clear. After recovering from that, the center cell became the most interesting. It took me a long time to see the implications of that cell, but focusing on the higher digits helped. The part that broke it open for me was spotting the delightfully simple, but effective way to rule out 2 from the center. A 2 in the center forces a double 3 in r6c6 and r7c7 and a double 1 in r5c3 and r4c4. This really limits the 1's and 2's, especially in column 7. A 1 is forced into both r6c9 and r9c7. This displaces the 2's and force them into r1c7 and r7c9. This breaks as a 2 is forced onto the thermo in box 6. That was delightfully simple, but really hard to spot. From there, all my built up knowledge quickly led to me finishing the puzzle. I really enjoyed this one. It had so much neat geometry. Great Puzzle!
The way I see it (not fully mine solution, I had to watch Simons reasoning to come to the second part of conclusion), the most elegant logic is following: IF we start the thermo from topleft corner, it is at least 2 in R5C5 and thus any values on thermo in R7 and C7 must be 3 or larger, leaving only one way for 12 pair in empty cells. Notice box9 - R7C9 and R9C7 are both 12 (this is until I've got on my own, next paragraph is Simon's thought) Let's continue the thermo logic to R8 and C8. Since they all continue in same boxes from previous row/column, they must contain digits 4-9, gain leaving only 3 cells available in each R8 and C8 for digits 123. (now again my solve, since Simon used different contradiction, dare I say, not so elegant) Look again at box9 - by R8 and C8 logic, we again are forced to have digits 123 in R8C9 and R9C8. Now we have four cells in box9 that can contain only 3 digits - R7C9, R9C7 (both 12) and R8C9, R9C8 (both 123). Since we cannot put only 3 digits in 4 cells, the initial assumption (thermo starts from top-left) is broken.
Am I right in saying that when he has the uniqueness problem at 53:46, he could have disproved it by the bottom yellow domino containing a 2, and therefore not allowing 2 to be placed in box 8 because it couldn’t be on a line and the yellow square was 1
On the note of daily puzzles, have you come across Minute Cryptic Simon? It's a daily cryptic clue, fairly challenging (at least to my beginner efforts), and always has a little explainer video after you're done, I really enjoy them!
Very enjoyable puzzle and video(s). My method (without centroid hunting): the four corners of box five cannot be greater than six, from the thermometer logic in the four corner boxes, and the route from the bulb must enter box five through one of them. The centre cell of box five must be less than all three of the other corners, leaving only two and three as possibilities. There is only the one route for a central two, with a one in the bulb at R5C3 and also in R4C4. Row seven and column seven must now contain seven digits all higher than this central digit on their thermometer cells as the bulb is placed, meaning they must have every digit from three to nine. This would force R7C7 to contain the only possible three in both row and column seven, in turn forcing R6C7 and R7C6 to contain the only possible pair of fours, and then R3C7 and R7C5 to contain the only possible pair of fives. But now apart from the single shared cell R8C8 neither row eight nor column eight can contain a thermometer digit lower than six and there are six distinct thermometer digits needed in each. Therefore the central digit of box five cannot be two and so must be a three. From here the solve is smooth, with forced digits appearing along the thermometers starting with the seven and eight in box five and then box eight forcing the position of the two in box five, until only four bulbs remain with paired options. With all the thermometer logic completed, straightforward sudoku logic on the grid completes the puzzle and identifies the bulb.
What got me passed the bulb being possible in 5,1 without bifurcation was actually Simon's insight that if 2 is placed at 5,5 then the thermo cells in row 7, and columns 7 & 8 are all going to have values greater than 2; therefore where are the possible places for 1 and 2?-- well marking them out there ends up being 3 cells in box 9 needing a 1&2
This was an awesome challenge. I am impressed by *aqjhs's* creativity and enjoied studying the complex geometry of this magnificent ramified thermometer. I probably did not find an elegant solution as I had to rule out every single candidate from the central cell by testing it. I have not seen the video yet, but I am looking forward to learning something more by Simon.
Not much to be learned from Simon. The main steps were similar to mine. However, I learned a lot from many interesting comments posted in this section. Thank you to all the people who shared their logic, including the setter *aqjhs,* and thank you CTC for creating this passionate community.
By the way, the logic Simon used from 34:33 to 1:17:17 was fascinating, but it was also undeniably, by definition, a *kilometric bifurcation.* It tested an extremely long and complex prong of the "fork," which turned out to be invalid and was hence rejected. Of course, it was not a random bifurcation. It was very well "educated," but it was bifurcation. I needed bifurcation as well, several times, but much shorter and mainly performed in my head.
This solve is like how Arsenal used to play football under Arsene Wenger, trying to score the perfect goal and refusing the easier ways to get the ball in the net. Great to watch when it does come off.
46:17 for me - It was hard figuring out where to start. I figured the maximum for the center was 3, and when I tried a 1 in R5C3 (the only starting location that would put a 2 there) and it didn't pan out, I got my first digit.
You can break it in 2 ways in the paradigm you had. 123s in box 9 or you could have asked where the 6789 go in purple. You would have ended up with 3 79s in the row or the column. 🎉 So cool.
There's actually another way to break the puzzle once Simon extended thoughts to row 8 and column after deducing that r8c8 had to be 4. The lowest digits that can be in the yellow cells of box 7 and 8 has to be 6789. Same goes for the red cells in box 3 and 6. This pushes 5 into r8c7 and r7c8 and forces it to repeat in box 9, which breaks the puzzle
90 minutes. I spent like 45 minutes figuring out where the bulb couldn't go, but I ruled it out of the cells in the lower right before fully comprehending the logic, so I incorrectly picked a different cell. Broke it about 15 minutes after that, erased everything, and did it anew. Took me about half an hour once I figured out the correct logic. But, typing this, I realize that I assumed the digit in the bulb was a 1, didn't prove that. So some of my logic may have been wrong in the end.
The moment Simon pointed out r7 and c7, I had a feeling the origin cell would be in the lower right. Of course proving that took much of the same work of Simon in eliminating things.
After 25 mins I got that r6c6 has to be less than 4 and greater than 1 (i.e. 2/3) which forces the starting bulb to be either r5c3 or r7c8 and makes r7c7 the same as r6c6, and the centre forms a 2/3 pair with r6c6 and the other corners can't be greater than 6 - been staring at it quite a while longer and I can't see how to determine which of the two is the starting bulb without brute forcing the rest of it.
I stared at this for long time and finally realized that since you can only hide the one and the nine in the central box that the 8 and 7 have to go somewhere and seemed forced and that wherever the two was would lead to the one and there was only one place that could be. I broke it in the end because I didn't notice that there were multiple spots for the one, but I had the rest of it correct and managed to fix it.
Nice puzzle. 53:08 for me in total. I did screw up on my first attempt, but I think my solve after the reset was sound. My break-in was based on the pressure on the center. Consider the branches that lead into the four corner boxes; three of them must be increasing from the center, but none of those can be more than 6, so the center can be at most 3. If it is 2, there is only one possibility for the 1, and then it is impossible to place the 1 and 2 in boxes 6, 9, and 8. Now consider box 8; the root cannot be in this box, but if it isn’t on this side at all, then all of the cell on the branches must be at least 5. That’s enough to start making more incremental progress.
Once R6 C6 was five or lower, the bulb is either in box 9 or it isn't. If it's not, the puzzle breaks. So the bulb travels up from box 9. There are six squares in box 5 that have to be higher than R6 C6, which means it can only be a 2 or 3 (since it can't be a one). From there the puzzle is much easier to solve.
Just under 45 min for me. I’ll take the vanishingly rare pride of beating Simon, but I do have to say that this is one of those puzzles where “Goodliffing” the possibilities throughout would actually have been the right strategy (and was for me)- it enabled me to see the rows and digits under pressure much quicker, and move to the solution much quicker.
52:20 i think the better argument is that using this configuration you cant place 2 in box 6, since there would be a 1-2 pair in box 8 forcing 2 in box 9 into a domino in row 9
33:06 for me. Easy peasy. I made a mistake, and it took me 10 minute more. It could be done within 25 min. How I solve: 1. You could easily find out what bulbs are not real (not possible to reach the longest ends). You will find out the number in the cell of the box 5 which enters the ends are maximum to 6 (and minimum is 2). 2. Find out the numbers in the X shape of the box 5 are 2/3/4/5/6. The 3 is in the center, and 2 is close to the bulb. 3. The only possible true bulb is in the bottom right (box 6 or box 9). 4. Fill all valid numbers in the slow thermometer. Some cells are squeezed and under pressure. Start solving the sudoku. You will find out the real bulb when you are almost done.
At 50:59, you can prove the puzzle breaks without the uniqueness argument by looking at where 2s are forced after you place the 1s. The row 9 2 is in box 8 and the column 9 2 is in box 6. There is nowhere to place the box 9 2. That's how I saw it. Someone smarter might have noticed earlier that the r7 and c7 combo make a 123 quintuplet between r7c9, r7c7, r8c9, r9c8, and r9c7 when taking into account the restrictions on rows and columns 7 and 8.
23:33. I started with figuring out which prong from the center square had to be decreasing UR and DL were not possible. UL also was because of the two rows going from box 1 to box 4. So bottom right had to be origin. Quick look set cell 6,6 to either 23or4. With 4 being quickly discarded because of the same box 1 and 4 issue. I then did candidates for the cells going up thermometer. That coupled with other sudoku solved a good chunk without knowing which was the bulb (of the 2 possibilities. (After realizing 6,6 had to be 2 which I should have figured out earlier because duh, no where else possible to put a 2) which now thinking of also forces the 3 REGARDLESS of which stem is decreasing. So not a perfect solve path, but very impressed with myself vs video length (which is par for me) My stumbling point was actually not realizing only 1 of the 2 candidates must be bulb and the other must not be. At the almost very end I had a few candidates that I couldn't solve (ie 2 possible answers) once I realized 1 had to be a 1 and the other either a 3 or 5. (I couldn't have 2 1s or a 3 and a 5) figuring out the solution was still difficult but possible.
The extra piece of logic i saw with Simon's method was to focus on 5s in r7 and c7 and 8. Because in box 8, thanks to the forced 4 in r7c6, the 5 could only be in r7 (as it is the lowest digit which can be on a thermo in that box after 4, and all of 4-9 must be on thermos in box 8 as simon showed). This forces r7c8 to be from 6789 as 3 4 and 5 are now accounted for in the row. However, the 3 4 in c7 are forced, and using Simon's logic that in the case where r4c4 is a 1 the 1 and 2 cannot be in a thermo in c7, the lowest a digit can be im the remaining thermo cells in c7 is now 5; where these cells affect thermo cells in c8 therefore, those cells must be 6789. There are 4 such cells, r2c8, r3c8, r4c8 amd r6c8. These 4 must therefore contain all of 6789, and so it breaks because as above r7c8 must also be from 6789, which would require repeating a digit. For all that there may be other methods, I thought Simon's was excellent, andother than doing it this way rather than with the 1s and 2s I actually think it was a genius way to enter the puzzle.
A quick way to prove R5C4 can't be a 1 is to look at where 1s, 2s and 3s go in box 8, column 7 and column 8. You end up with four of them in row 9, which is clearly impossible. Then the numbers must increase from the bottom right. But none of R4C4, R4C6 or R6C4 can be higher than 6, so they form a 456 triple and - that's three in the centre.
I'm quite proud of myself for solving it in 38:25, though I guess I kinda bruteforced it from halfway through the puzzle. I eliminated all the obviously wrong starts and then tried to eliminate the two starts in the bottom right I had left, but it just kept on working and suddenly I was done.
Solved in about 45 minutes. Simon almost gets there around 1:17, but without all the pencil-marking, the idea that row/column 7 has to be 3-9 and row/column 8 has to be 4-9 has implications for 1/2/3s in those rows, putting too many low digits in box 9.
the 1/2 paradigm is broken because on either side of the purple line, it has to be 1 and 2 (since purple is 3 to 9). since box 9 needs 1 and 2 to be at the bottom row, it can't be at the right side of purple.
Google Translate (not perfect, but rough and ready) says that the ending line is: "Of a hundred hands and my strength, And a hundred eyes keep watch over us"
"My strength is that of one hundred hands and a hundred eyes watch over us", the second line is a bit ambiguous, "fanno a noi la guardia" can be "watch over us" or "are watchful of us" as in "we are strong so a lot of people is wary of us"
53:00, probably the quickest relative to the video I've done. Box 8 was the key, basically immediately made it possible to rule out the start being most of the upper circles since you need 4 or lower to appear on the graph within the box.
There was a much simpler way for discarding the 2 in the center cell. If it is a 2 then the only place the bottom cell can occupy is r5c3 and it must be a 1. This forces both 1 and 2 in box 6 to be in col 9 which leaves no choices for r7c9.
My way of breaking the 1/2 issue without uniqueness is that if you mark the 1 2 and 3s in row 7 8 and columns 7 8 you will see box 9 has 4 squares that must be 1 or 2.
![ILLSLICK - KILLSHOT REMIX [FULL VERSION]](http://i.ytimg.com/vi/RIK8RrqIAzE/mqdefault.jpg)
(aqjhs here:) The way i saw it being logically solved was (ironically) pencilmarking from all tips simultaneously to find a "centroid" of the tree graph. Then the bulb is the closest tip to the centroid. I'm in total awe seeing that Simon quickly picked up the position of the centroid within minutes of solving just from the pressure on row 7 and column 7. Pure genius.
Congrats agjhs! You’ve been very busy turning ideas up to 11! Great stuff.
@@kennetsdad thanks!! 😃
The title of the puzzle immediately made me think of hecatoncheires!
Fantastic puzzle. And ironically it turns out Mark's way of approaching thermo puzzles was the Intended and correct way 😀
@@catnip420 i wouldn't call a particular way the correct one, but sure, after realizing the nature of r6c6 (which Simon got in a masterful way imho) some pencilmarks might have spared some of the ruckus caused by r4c4.
I was one of the people who watched Mark try to solve and fail yesterday. Honestly, it was very cool to see a puzzle go wrong. I've seen it said before, and I'll say it again. CTC should show some of their failures from time to time. My two cents... :)
This sudoku messed with Simon's brain so badly that he considered placing 6 before 4 on the thermo and couldn't see the problem with it... 🤣🤣
looks like this puzzle might require from one to go outside of the box, out of the house, to walk a few blocks, take the bus to a retreat in the mountains, and think there.
this is the comment i was looking for 😂
I read the clues, looked at the picture, laughed my head off and then clicked play. This is one I'm happy to enjoy second hand.
Took me two hours almost to the second but I've managed to solve this! It's the first Cracking to Cryptic sudoku I've ever managed to solve, I did it with a LOT of coloring. I've never been prouder of myself. Thank you Simon for the countless hours of fun you've given me over the years.
'My strength is of a hundred hands
And a hundred eyes keep watch over us'
th-cam.com/video/ILSBIJFCIZw/w-d-xo.html ❤️
This man brings joy to so many of us, and I have heard plenty an email stating how Simon and Mark helped people struggling with hard times and mental health. And you Simon Anthony ,dare to call yourself a useless individual! You are not.
Thank you for listening to my TED talk
@@novemberdag1127 100% agree with you.
The two parts of the break-in that got me going were the 7 cells on the tree in row 8 and column 8 (thanks to Simon), but also the heavy restriction on R5C5. Because the center cell is not a bulb, it must grow in 3 directions, with the low digits coming from the 4th direction. It's growing in the direction of the diagonals, and we can't put 7 8 or 9 on those diagonals. So since there must be 3 different diagonal digits with a maximum value of 6, the center cell must be 2 or 3. You can then check each bulb to see if it can reach the center while maintaining that 2 or 3, and that eliminates all possible bulbs except R5C3, R6C7, and R7C8. From there R5C3 can be eliminated similarly to how Simon does it, and we're off to the races.
Thanks Simon, for the shout-out to the Asian Sudoku Championship! We are very excited to host the competition, and to see some diverse participation from across the continent! 💖
Oh Simon! at the 90 minutes mark @1:30:30 you finally look at the low digits. I've been waiting for you to do that since @45:00 where you can see that with the 2 in the middle, the 1 and 2 both have to be the outies of the pink and blue regions. which breaks because 1 and 2 are already in column 9 in box 6. So the pink region breaks.
One of the most satisfying endings I have ever seen... extraordinary!
This puzzle looks like it is going to be hard, I think I shall just watch Simon's solve instead of trying it.
I won’t touch anything over an hour. Even 45mins to an hour is scary
It's actually one of easier puzzles, as long as you are not sleepy like me.
It is not. I solve it in 16 min, and i'm definitely not fast solver.
I'm even surprised how Simon could struggle it fo 1.5 hours, must be something overcomplicated logic (didn't saw the video yet).
It was pretty easy. Simon really over thinks things sometimes. I was done in 35 minutes, and I'm no genius solver.
Simon's amazing. He's solved dozens of puzzles that I haven't even figured out how to start, and of the puzzles I can solve, he usually solves them faster than I do, while slowly and meticulously talking through every step.
But he's not perfect. Sometimes he focuses on the wrong thing or misses something entirely. I haven't watched the video yet to see where he went wrong, but I just solved the puzzle in 29 minutes without really getting stuck at any point, so he's definitely gone wrong somewhere.
My suggestion when you see a long video is to give it ten minutes and see how you go. Either you make progress and enjoy the puzzle or you get completely stumped and appreciate Simon's solve even more.
I approached this very differently from Simon. Ruling out as many potential bulbs as possible, I quite quickly found that there were only three possible cells. Then I got hopelessly stuck. and had to watch the video until Simon found the insight which I had missed. We live and we learn.
I found the fact that at any junction, only 1 branch will be decreasing to be very helpful. In the central square, it quickly becomes apparent that 3 branches will rise and this is quite constraining as you can't place a 7 or higher in any of the corners - with the next highest digits being 6,5 and 4, you're very quickly down to a max of 3 in the centre.
If the correct bulb were in the "North-west" corner, then, as Simon pointed out, the middle cell of the grid would be forced to be a 2, and therefore R5C3 would have to be a 1. By then you have a 12 pair in R9C7+R9C8. It is then impossible to put 1 and 2 in row 7, since they would have to occupy R7C1 and R7C9, which is now impossible..
That’s how I got there!
Oh that finally helped me, the 1/2 pair in r9c7 and r9c8 means both 1 and 2 have to go in r8c6 in box 8.
I'm very proud that I figured this out very quickly
I was wondering why didn't Simon concentrate on where to put 1 and 2 in row 7 and box 6 if r5c5 is 2.
I think finding out where 1 and 2 would have to go is a really nice way to disprove the 2 in the centre. If you notice that most of the thermometers have to be greater than 2, you can ask the question of where 1 and 2 go in R7, R8, R9 and C7, C8, C9. This pretty easily and beautifully disproves that R4C4 is lesser than R5C5.
This is a very mysterious comment from you. In all those rows and columns 1 and 2 can go to multiple places. R7 is the most restricted, but even there 1 has has 4 possible places while 2 has more. In C9 and R9 they can go literally anywhere.
I had a similar argument, only asking where 1 and 2 went in row 7 and box 6.
2 in the centre requires the bulb to be r5c3 (only bulb close enough), and this to be a 1.
Then every thermo cell in row 7 and in box 6 has to be greater than 2. That only leaves r4c9 and r6c9 in box 6, and r7c1 and r7c9 in row 7. That's three cells in column 9 that can only be 1 or 2.
@@istvanmagi473 Let me explain a different way. If R5C5 is a 2, then the 1 is forced to R5C3.
Row 7 and Column 7 would then be forced to have a 1/2 pair at each end. If you try to put 1/2 in Column 8, then box 9 is eliminated because it has 2 1/2 pairs in it already, and row 5 has a 1 and 2 in it, leaving only R1C8 to hold both 1 and 2.
Havent beaten Simon in a while, but this time I beat him by more than an hour.
Here is a MUCH simpler path:
1) R4C4 being 1 is impossible, as only way that happens is if R5C3 is also 1 which together with R5C5 (doesnt even matter what digit you put there) creates at least two schroedinger cells (R6C8 and R7C1), since the center cell MUST be greater than all the thermo cells on boxes 6, 7, 8 and 9 and once you follow the sudoku implications through box 6 to 9 to 8&7 you will find the schroedinger cells. So now you can just pencilmark 19 into the two empty squares (assuming you did the 9 logic before lol).
2) Notice that NONE of the corners in box 5 can contain a digit greater than 6, since the extensions (R3C3, R3C7, R7C3 and R7C7) all see 3 digits that are further along the thermo than they are (I will just ignore the low end here, as the low end certainly cannot enter the middle box through 7 lol)
3) That forces 7 and 8 to R5C6 and R6C5, and R5C6 cannot be 8 so you get two digits and can actually fill in those thermos.
4) Now thermo end in R9C5 is under pressure. It cannot be the low end (leads to too high entry digit to middle box), so its now 7 at the highest which pressures R6C6 to be a maximum of 3 (so now you know that is actually 2, since thats the corner you must enter the middle box from).
Rest is basically just sudoku, and at the very end you discover the bulb.
ditto. I came to watch after solving just assuming it was one of the punk'd video run times!
This comment is really interesting. Thank you for sharing. My logic was very similar, but did not need your *step 1.*
I ruled out the Phistomefelian bulb from *r4c4* easily after discovering that
*r5c5 = 3*
Quite pleased I tried to solve it myself first as didn't think I would get it with a 1 h 40 video but I managed to get the answer in a respectable 54 minutes. The break through for me was the realisation that in the centre box the corner cells had to be between 2 and 6 which meant the central cell had to be ... and consequently the 1,7,8,9 were heavily restricted. Now going to watch Simon's solve.
But one of the corner cells (r4c4) could have been a 1.
@@RichSmith77 Only after watching Simon's solve did I realise that one could have gone into one corner... How great am I; I must have subconsciously intuitively known that a one there would break, so blocked it from my mind, saving me the trouble of logically disproving it!
@@martingibbs1179 I sometimes have unconscious insights like that too. Ones that only come to light when I watch Simon or Mark's solve afterwards, and find myself forced to ask why I did a step a lot easier than they managed. 😂
I was so happy with my solve until seeing Simon's and realizing that I started off with an incorrect assumption, i.e. that 1 couldn't go in that top corner. I always admire Simon's ability to rarely/never jumping on a conclusion until they are adequately proved.
This puzzle looks a lot like a phylogeny tree with the bulb being the last universal common ancestor :) Love it
@@simonpenelle2574 the title comes from a track of the 1972 album "Darwin" from the italian band Banco del Mutuo Soccorso... so not far off
There is a uniqueness argument that the bulb must be 1 - otherwise deduct 1 from all the thermo cells, which are the only clued cells and that must give a valid solution (you can deduct 1 from the digits 2-9 wherever they appear and replace 9 with 1)
@1:15 you need too many 12s in box 9 (two rows and two columns)
btw this is closer to my solving method than your usual video ... trying to find the logic I haven't quite seen
It's easy to prove the bulb must be one - there are no bulbs in box 5, and only two empty cells. There's no cell on the thremo in box 5 that can be a 1 or a 9 (they all either see the next cell on their branch, or will force multiple 1/9s into the next box down their branch). Hence, the empty cells are 1/9. So 2 must be on a thermo line, and the same logic means there's no way to reach a bulb on a branch of only 2s, which means 1 must be on a bulb.
@@eytanz It isn't entirely trivial to rule out 1 from r4c4, with r5c3 being the bulb. Simon struggles with exactly that the first few minutes of the video. Which is to say, it's not immediate to see that 1 in box 5 is in an empty cell. This still makes the bulb 1, so your main point stands.
@@MasterHigure You're right, I should have included that in my logic
I’m by no means trying to imply this was easy to see, but after watching you explore the options, I think I found the simplest way of explaining what you found:
Once you place 2 in the central square, it sees r5c8. This puts 12 pairs in row 7, column 7, and column 8. This puts 3 cells in box 9 that must be 1 or 2.
The end message roughly translates to: "With a hundred hands and all my strength, and a hundred eyes, keep watch over us." Sounds like a prayer or blessing to some higher power.
"of a hundred hands is our force, by a hundred eyes we're guarded" lyrics from "cento mani e cento occhi" by Banco del Mutuo Soccorso, i read it more as "the power of community"
@@therealAQ The track "Cento mani e cento occhi" is on TH-cam. It's Italian prog-rock from 1972. From the English translation, the lyrics seem to depict an outsider approaching a Stone Age tribe.
@@David_K_Booth yes! That album is in my usual rotation
@@therealAQ The literal translation is:
_"Of a hundred hands is _*_my_*_ force, and by a hundred eyes _*_we are_*_ guarded"_
However, it will have exactly the same meaning if it were as you wrote:
_"Of a hundred hands is _*_our_*_ force, and by a hundred eyes _*_we are_*_ guarded"_
or also:
_"Of a hundred hands is _*_my_*_ force, and by a hundred eyes _*_I am_*_ guarded"_
Yes, it is about the power of a tribe of cave men, compared to the power of a single one, who is reluctant to join the tribe.
a bit of artistic liberty on my translation
😅
Really nice puzzle. I had a different break-in for it.
1. I checked the maximum digit for the corners in box 5, assuming the values increases from the center. It was 6 for all of them. Since 3 of them will increase it means the center can be a maximum of 3.
2. I checked what happens to the digits 1-3 in row 7, with the bulb in the top half and center being 3. No way to do it. Same with digit 1-3 in column 7 with the buld in left half and center being 3, no way to do it there either.
3. That means the bulb is either in row 5 column 3 (the only way the center can be a 2) or it's in the bottom right branch on r6 c7 or r7 c8.
4. I was sure it would be r5 c3, but that led to lots of the digit 1-3 in box 9 (r7c7, r7c9, r8c9, r9c7, r9c8), and that's simply not possible.
the resistance against goodlifting really is a disadvantage here.
I agree. As usual. It has been already noted in dozens of similar comments in the past.
I might be wrong, but his reluctance seems to be due to his strong desire to drastically and stubbornly differentiate his method from Mark's.
Funny how Simon made the exact same mistake I did trying to find the break-in, focussing on eliminating candidates for the true bulb. After I shifted focus to the central box the break-in turned out be a lot simpler. The opening logic is to realize that regardless of where the thermo enters box 5 it has to leave through at least three corners each of which cannot contain a digit higher than 6. This restricts the central digit to be one of [2,3] and you can eliminate 2 as Simon showed in the video. With the central digit known the three exit corners are now a set of [4,5,6] and you can instantly place digits 7 and 8 in the box as well. The way Simon did it this puzzle would easily qualify as a 5* but with the correct logic applied it is really a 3*
Finished quickly via bifurcation, but watched the video and I feel like I've learned a lot about sudoku logic, it was a pleasure and don't worry - you didn't let anyone down. Fun puzzle!
As I wrote in a separate comment, the logic Simon used from 34:33 to 1:17:17 was fascinating, but it was also undeniably, by definition, a *kilometric bifurcation.* It tested an extremely long and complex prong of the "fork," which turned out to be invalid and was hence rejected. Of course, it was not a random bifurcation. It was very well "educated," but it was bifurcation.
I needed bifurcation as well, several times, but much shorter and mainly performed in my head.
i took one look at the grid and went "nah there's no WAY that is possible". then proceeded to pause the video at least 10 separate times, try to work through some logic that simon mentioned and i hadnt thought of, only to absolutely CONVINCE myself that i had proven it was indeed impossible. then watched further in just to be proven wrong AGAIN. thoroughly amazed by anyone and everyone able to complete this and hats off to the setter, this is incredible
Loved the fact you did not find out the 'naughty' bulb until the very end of the puzzle. Got there eventually after many false starts. Probably the most difficult puzzle I have ever solved from you guys. Thanks for all you do.
The amount of time it took Simon to deduce that r8c7 could not be green after breaking into the puzzle was ludicrous, and I wouldn't have it any other way ♥
I was shocked at how long it took him to deduce it. Then I was simultaneously shocked that he solved the rest of the puzzle in about 15 minutes.
May I just say how amazing it is just visually? The optical impression is so weird, as if the edges of the puzzle are dissolving in a kind of fog...
At around 52:00 Simon could just argue that he needs the 1 and 2 in row 7 to be off the thermos, therefore the 2 being in column 9 in Row 7. But also per Sudoku the 2 is placed in Box 6 also in Column 9, this clashes and if you would see this way earlyer on you could disprove the whole central being 2 thing within seconds if you just focus on 1s and 2s by sudoku.
Loved to see you solve it after it took me many hours to find all the hidden logic
I started quite elegantly. similar to simon, i considered the top left containing the bulb, but focussing on small digits. r5c5 has to be 2 in this case: it cant be 1 since it's too far from any bulb, and it's not 3 or more because that would cause the 6 thermos in box 8 to be at least 5. a central 2 forces a 1 in both r4c4 and r5c3 and also forces at least 3s in all thermos from boxes 6-9. the 1 and 2 in box 6 are forced into column 9. and now the puzzle is broken: r7c1 is the only place in row 7 for both 1 and 2. From there the puzzle solves the same way as simon did it at 1:19:30
i cant believe i crushed simons time. i got it in 65 min. usually i only rarely can beat him on the easier puzzles and its mainly because he has to explain all the logic while solving the puzzle.
The logic i used is a little difficult to explain, but I'll do my best.
Basically, i focused on the most obvious thing in a sudoku puzzle. The digits 1-9. Relating this to it being thermo, the question was, will it reach if i keep it to a minimum? This eliminated a few bulbs. Combining it with filling the grid with pencil markings, it gave me a general direction the thermo is going. This put pressure on the low digits in box 5, thus putting the bulb in the lower right. Then, proceed from there.
Also small tip: From a uniqueness perspective, you KNOW that the bulb (wherever it is) MUST contain a 1, just as certain as at least 1 Tip must contain a 9.
Because if the values ranged from 2-9 you could just also change all the values to 1 lower (1-8) and change all the 1's in the grid with 9's.
Got it in under 30min. Was very smooth and funily enough, I took Marks approach :D
the logic between boxes 3/7/9 was SO beautiful! I was like "Haaaang on, I see what u did there", loved it 😊
Once you spotted the purple row, if you spotted c7, you'd have pinpointed the quadrant where you need to start the thermo.
52:12 Doesn’t this pattern break without uniqueness when you ask where in that case you put the 2 in row 7?
Yes, either the question of where 2 goes in r7, or specifically what goes in r7c8 that is both bigger than 5 (for the row) and smaller than 6 (for the column).
Yes.
To rule 2 out from r5c5, you only need to look at the effect it has on digits 1 and 2.
The only bulb close enough for a 2 in the centre is r5c3, and this would have to be a 1. Then every thermo cell in box 6 and row 7 would have to be greater than 2. This forces 1 or 2 into rows 4, 6 and 7 of column 9.
@@RichSmith77 or notice that you have to put the 1 in a circled cell from the start...
@@skywiseskchan That doesn't rule out 2 from the centre though, since 1 in r5c3 could be followed by 1 in r4c4 and 2 in r5c5.
The problem I saw was that if you put 2 on row 7 it has to go to r7c9. Which makes it impossible to put 2 in box 6.
Just finished watching - well done Simon that was an amazing feat of endurance on your part. Stop being so hard on yourself - you did it :-)
this tree looks like a biblically accurate angel
biblically accurate thermo, that's hilarious!
Solved this in 25 using a different approach. Considering the middle digit, we know 3 of the paths are increasing and all 4 paths have a next max digit of 6. Therefore the middle digit is a 2 or 3. Disproving the 2 case like Simon gets a 3 start. Then because of the pressure on box 8 we know the 2 is one of the bottom paths. We also have a 2456 and a 19 pair in box 5 so the 78 is easily placed to continue
I used a very similar path. Thank you for sharing.
This is how I solved it too.
Astounding, Simon. I was immediately and absolutely certain that this would be a viewing experience complete with a large bowl of popcorn. Very interesting puzzle and solve. At one point with 20 minutes or so to go in the video you made a remark that I thought was very interesting, that pencil marking was something that you were going to resort to because the video already was so long. I have often thought that you would find the next step or get a running start toward the solution of some of the complicated puzzles you do if only you would just pencil mark the dang thing - and it seems that you might agree with me. Nevertheless, I love that you don't pencil mark things immediately, because I have learned how to discern where it will be most helpful by watching your solves of these puzzles that are way beyond my abilities for now. Thank you for this video, and for all that you do to make this channel and community so great!
Solved this in 51 minutes flat. Broke it a few times along the way. Every single time was due to forgetting that it's a slow thermo instead of a regular thermo.
Wordle in 5 Simon. It was a tough one, but not a streak breaker.
The simpler way to rule 2 out from r5c5 is to just look at the effect it has on digits 1 and 2.
The only bulb close enough for a 2 in the centre is r5c3, and this would have to be a 1. Then every thermo cell in box 6 and row 7 would have to be greater than 2. This would force 1 or 2 into rows 4, 6 and 7 of column 9.
That seems to be much better. The logic Simon used from 34:33 to 1:17:17 was fascinating, but it was also undeniably, by definition, a *kilometric bifurcation.* It tested an extremely long and complex prong of the "fork," which turned out to be invalid and was hence rejected. Of course, it was not a random bifurcation. It was very well "educated," but it was bifurcation.
By the way, I needed bifurcation as well, several times, but much shorter and mainly performed in my head.
The meaning of your comment was not immediately clear to me, but it became clear when I studied it.
I believe you meant that, since every thermo in *box 6* and *row 7* would have to be greater than *2,* then:
*r4c9 = 1* or *2*
*r6c9 = 1* or *2*
*r7c9 = 1* or *2*
Very elegant and very easy. Thank you for sharing this finding.
In my solve, I also needed to rule out *2* from the grid center (my first digit was r5c5 = 3, as I explained in a separate comment), but I cannot remember how I did it, actually... Certainly, I did not need 40 minutes of bifurcation.
3 hours, 18 minutes, 43 seconds, and I got stuck and had to wait on help from Simon's solve.
I was like 'i've got this - imma give it a go' an hour later and i retreat back to the comfort of Simons solving prowess
Great puzzle, I also enjoyed how you kept referring to the only bulb that is actually following the rules as the "naughty bulb"
This was far easier than the video length would have you believe. I will have to watch the video to find out why Simon struggled with it.
I solved it using a question Mr. Anthony often asks: "where can X go in this box/row/column?"
Far easier for you*, I still think video length is a good metric for those newer to sudoku who want to have a go... or not.
@mattconco No, not far easier for me. It might be hard for someone new to sudoku, but that can hardly be the bar to which we measure a puzzle's difficulty rating - most puzzles presented on the channel would be hard for someone new to sudoku, whether it's 30 minutes or two hours. I think the video length in this case might discourage some solvers, say not beginners, but not yet experts, from attempting it as it suggests a high difficulty rating. I for one hesitated, but decided to have a quick glance. I saw relatively quickly that the center box is very restricted and had a pretty smooth break-in with relatively conventional logic. That is why I was surprised at the length of the video.
I do understand that even Simon and Mark have their off days and this was clearly one for Simon.
@@joethornton5321 I would imagine that most people who are novice enough to be put off by a video length would still struggle to complete this quickly. Again, I don't see that fact you were able to spot logic quickly means everyone else will find it easier than Simon too. I think some people don't realise when they get really good at solving sudokus. I find a similar issue on LogicMasters. There is too much use of the 1 star rating because people think "well if I managed to do it then it must be easy", not realising that they are a well above average solver. That's my theory at least.
I think what a long video demonstrates is that a puzzle requires some logic that isn't immediately obvious to Simon or Mark and therefore is at least intermediate level.
I don't believe so much in 'off days' in this case. Simon found a long piece of logic and decided to persue it. It made for a more interesting video.
The only reason I replied was because I found it quite pompous to say "I will have to watch and see why Simon struggled". I don't personally consider failing to notice the fastest route as being struggling.
I finished in 96:57 minutes. This was such an interesting puzzle to figure out. I spent the first 20 minutes doing the puzzle wrong and trying to make each end in a box only have one thermo. It doesn't even make sense trying to describe, so I don't know how I got it so wrong when the rules make it clear. After recovering from that, the center cell became the most interesting. It took me a long time to see the implications of that cell, but focusing on the higher digits helped. The part that broke it open for me was spotting the delightfully simple, but effective way to rule out 2 from the center. A 2 in the center forces a double 3 in r6c6 and r7c7 and a double 1 in r5c3 and r4c4. This really limits the 1's and 2's, especially in column 7. A 1 is forced into both r6c9 and r9c7. This displaces the 2's and force them into r1c7 and r7c9. This breaks as a 2 is forced onto the thermo in box 6. That was delightfully simple, but really hard to spot. From there, all my built up knowledge quickly led to me finishing the puzzle. I really enjoyed this one. It had so much neat geometry. Great Puzzle!
The way I see it (not fully mine solution, I had to watch Simons reasoning to come to the second part of conclusion), the most elegant logic is following:
IF we start the thermo from topleft corner, it is at least 2 in R5C5 and thus any values on thermo in R7 and C7 must be 3 or larger, leaving only one way for 12 pair in empty cells. Notice box9 - R7C9 and R9C7 are both 12
(this is until I've got on my own, next paragraph is Simon's thought)
Let's continue the thermo logic to R8 and C8. Since they all continue in same boxes from previous row/column, they must contain digits 4-9, gain leaving only 3 cells available in each R8 and C8 for digits 123.
(now again my solve, since Simon used different contradiction, dare I say, not so elegant)
Look again at box9 - by R8 and C8 logic, we again are forced to have digits 123 in R8C9 and R9C8. Now we have four cells in box9 that can contain only 3 digits - R7C9, R9C7 (both 12) and R8C9, R9C8 (both 123). Since we cannot put only 3 digits in 4 cells, the initial assumption (thermo starts from top-left) is broken.
Am I right in saying that when he has the uniqueness problem at 53:46, he could have disproved it by the bottom yellow domino containing a 2, and therefore not allowing 2 to be placed in box 8 because it couldn’t be on a line and the yellow square was 1
Yes, I don't understand why Simon didn't see that instantly!
On the note of daily puzzles, have you come across Minute Cryptic Simon? It's a daily cryptic clue, fairly challenging (at least to my beginner efforts), and always has a little explainer video after you're done, I really enjoy them!
Very enjoyable puzzle and video(s). My method (without centroid hunting): the four corners of box five cannot be greater than six, from the thermometer logic in the four corner boxes, and the route from the bulb must enter box five through one of them. The centre cell of box five must be less than all three of the other corners, leaving only two and three as possibilities. There is only the one route for a central two, with a one in the bulb at R5C3 and also in R4C4. Row seven and column seven must now contain seven digits all higher than this central digit on their thermometer cells as the bulb is placed, meaning they must have every digit from three to nine. This would force R7C7 to contain the only possible three in both row and column seven, in turn forcing R6C7 and R7C6 to contain the only possible pair of fours, and then R3C7 and R7C5 to contain the only possible pair of fives. But now apart from the single shared cell R8C8 neither row eight nor column eight can contain a thermometer digit lower than six and there are six distinct thermometer digits needed in each. Therefore the central digit of box five cannot be two and so must be a three. From here the solve is smooth, with forced digits appearing along the thermometers starting with the seven and eight in box five and then box eight forcing the position of the two in box five, until only four bulbs remain with paired options. With all the thermometer logic completed, straightforward sudoku logic on the grid completes the puzzle and identifies the bulb.
What got me passed the bulb being possible in 5,1 without bifurcation was actually Simon's insight that if 2 is placed at 5,5 then the thermo cells in row 7, and columns 7 & 8 are all going to have values greater than 2; therefore where are the possible places for 1 and 2?-- well marking them out there ends up being 3 cells in box 9 needing a 1&2
This was an awesome challenge. I am impressed by *aqjhs's* creativity and enjoied studying the complex geometry of this magnificent ramified thermometer.
I probably did not find an elegant solution as I had to rule out every single candidate from the central cell by testing it.
I have not seen the video yet, but I am looking forward to learning something more by Simon.
Not much to be learned from Simon. The main steps were similar to mine.
However, I learned a lot from many interesting comments posted in this section. Thank you to all the people who shared their logic, including the setter *aqjhs,* and thank you CTC for creating this passionate community.
By the way, the logic Simon used from 34:33 to 1:17:17 was fascinating, but it was also undeniably, by definition, a *kilometric bifurcation.* It tested an extremely long and complex prong of the "fork," which turned out to be invalid and was hence rejected. Of course, it was not a random bifurcation. It was very well "educated," but it was bifurcation.
I needed bifurcation as well, several times, but much shorter and mainly performed in my head.
This solve is like how Arsenal used to play football under Arsene Wenger, trying to score the perfect goal and refusing the easier ways to get the ball in the net. Great to watch when it does come off.
46:17 for me - It was hard figuring out where to start. I figured the maximum for the center was 3, and when I tried a 1 in R5C3 (the only starting location that would put a 2 there) and it didn't pan out, I got my first digit.
You can break it in 2 ways in the paradigm you had.
123s in box 9 or you could have asked where the 6789 go in purple. You would have ended up with 3 79s in the row or the column. 🎉 So cool.
At 32:47 you had i in your hands as you had identified a week digit in column 7 and row 7. This had to be a 2 to make the rest work.
There's actually another way to break the puzzle once Simon extended thoughts to row 8 and column after deducing that r8c8 had to be 4. The lowest digits that can be in the yellow cells of box 7 and 8 has to be 6789. Same goes for the red cells in box 3 and 6. This pushes 5 into r8c7 and r7c8 and forces it to repeat in box 9, which breaks the puzzle
Well done Simon, quite a battle you had to face and you did it with much resilience.
90 minutes. I spent like 45 minutes figuring out where the bulb couldn't go, but I ruled it out of the cells in the lower right before fully comprehending the logic, so I incorrectly picked a different cell. Broke it about 15 minutes after that, erased everything, and did it anew. Took me about half an hour once I figured out the correct logic.
But, typing this, I realize that I assumed the digit in the bulb was a 1, didn't prove that. So some of my logic may have been wrong in the end.
Loved it all the way through!
I absolutely love the new Fog of War puzzles.
Now I want to see Mark solve this! 😅
The moment Simon pointed out r7 and c7, I had a feeling the origin cell would be in the lower right. Of course proving that took much of the same work of Simon in eliminating things.
After 25 mins I got that r6c6 has to be less than 4 and greater than 1 (i.e. 2/3) which forces the starting bulb to be either r5c3 or r7c8 and makes r7c7 the same as r6c6, and the centre forms a 2/3 pair with r6c6 and the other corners can't be greater than 6 - been staring at it quite a while longer and I can't see how to determine which of the two is the starting bulb without brute forcing the rest of it.
First time in probably hundreds of solved sudoku that I managed to solve significantly faster than Simon or Mark!
What a thriller! Simon you are so smart!!
I stared at this for long time and finally realized that since you can only hide the one and the nine in the central box that the 8 and 7 have to go somewhere and seemed forced and that wherever the two was would lead to the one and there was only one place that could be. I broke it in the end because I didn't notice that there were multiple spots for the one, but I had the rest of it correct and managed to fix it.
I'd never imagine I would find a Banco del Mutuo Soccorso reference in a CtC video.
Nice puzzle. 53:08 for me in total. I did screw up on my first attempt, but I think my solve after the reset was sound. My break-in was based on the pressure on the center. Consider the branches that lead into the four corner boxes; three of them must be increasing from the center, but none of those can be more than 6, so the center can be at most 3. If it is 2, there is only one possibility for the 1, and then it is impossible to place the 1 and 2 in boxes 6, 9, and 8. Now consider box 8; the root cannot be in this box, but if it isn’t on this side at all, then all of the cell on the branches must be at least 5. That’s enough to start making more incremental progress.
I do really hope that you'll keep playing Tametsi. I immensely enjoyed watching you play that and would love to see you do the harder levels.
Once R6 C6 was five or lower, the bulb is either in box 9 or it isn't. If it's not, the puzzle breaks. So the bulb travels up from box 9. There are six squares in box 5 that have to be higher than R6 C6, which means it can only be a 2 or 3 (since it can't be a one). From there the puzzle is much easier to solve.
You don't need a uniqueness problem to prove that's wrong. You just need to notice you can't put a 4 in box 5.
Just under 45 min for me. I’ll take the vanishingly rare pride of beating Simon, but I do have to say that this is one of those puzzles where “Goodliffing” the possibilities throughout would actually have been the right strategy (and was for me)- it enabled me to see the rows and digits under pressure much quicker, and move to the solution much quicker.
52:20 i think the better argument is that using this configuration you cant place 2 in box 6, since there would be a 1-2 pair in box 8 forcing 2 in box 9 into a domino in row 9
33:06 for me. Easy peasy. I made a mistake, and it took me 10 minute more. It could be done within 25 min.
How I solve:
1. You could easily find out what bulbs are not real (not possible to reach the longest ends). You will find out the number in the cell of the box 5 which enters the ends are maximum to 6 (and minimum is 2).
2. Find out the numbers in the X shape of the box 5 are 2/3/4/5/6. The 3 is in the center, and 2 is close to the bulb.
3. The only possible true bulb is in the bottom right (box 6 or box 9).
4. Fill all valid numbers in the slow thermometer. Some cells are squeezed and under pressure. Start solving the sudoku. You will find out the real bulb when you are almost done.
24:25 Another way to look at this breaking is once you place the 5 you have 7 more thermo cells in box 5 that need to be higher than 5…
At 50:59, you can prove the puzzle breaks without the uniqueness argument by looking at where 2s are forced after you place the 1s. The row 9 2 is in box 8 and the column 9 2 is in box 6. There is nowhere to place the box 9 2. That's how I saw it. Someone smarter might have noticed earlier that the r7 and c7 combo make a 123 quintuplet between r7c9, r7c7, r8c9, r9c8, and r9c7 when taking into account the restrictions on rows and columns 7 and 8.
23:33. I started with figuring out which prong from the center square had to be decreasing UR and DL were not possible. UL also was because of the two rows going from box 1 to box 4. So bottom right had to be origin. Quick look set cell 6,6 to either 23or4. With 4 being quickly discarded because of the same box 1 and 4 issue. I then did candidates for the cells going up thermometer. That coupled with other sudoku solved a good chunk without knowing which was the bulb (of the 2 possibilities. (After realizing 6,6 had to be 2 which I should have figured out earlier because duh, no where else possible to put a 2) which now thinking of also forces the 3 REGARDLESS of which stem is decreasing. So not a perfect solve path, but very impressed with myself vs video length (which is par for me)
My stumbling point was actually not realizing only 1 of the 2 candidates must be bulb and the other must not be. At the almost very end I had a few candidates that I couldn't solve (ie 2 possible answers) once I realized 1 had to be a 1 and the other either a 3 or 5. (I couldn't have 2 1s or a 3 and a 5) figuring out the solution was still difficult but possible.
20:31 for me. Quite an interesting idea, I did enjoy it!
This is a very rare occasion when I solved it quicker than Simon. I of course pencil Marked without shame, but about an hour.
My 152 day Wordle streak ended today. What an evil word!😢
The extra piece of logic i saw with Simon's method was to focus on 5s in r7 and c7 and 8. Because in box 8, thanks to the forced 4 in r7c6, the 5 could only be in r7 (as it is the lowest digit which can be on a thermo in that box after 4, and all of 4-9 must be on thermos in box 8 as simon showed). This forces r7c8 to be from 6789 as 3 4 and 5 are now accounted for in the row.
However, the 3 4 in c7 are forced, and using Simon's logic that in the case where r4c4 is a 1 the 1 and 2 cannot be in a thermo in c7, the lowest a digit can be im the remaining thermo cells in c7 is now 5; where these cells affect thermo cells in c8 therefore, those cells must be 6789. There are 4 such cells, r2c8, r3c8, r4c8 amd r6c8. These 4 must therefore contain all of 6789, and so it breaks because as above r7c8 must also be from 6789, which would require repeating a digit.
For all that there may be other methods, I thought Simon's was excellent, andother than doing it this way rather than with the 1s and 2s I actually think it was a genius way to enter the puzzle.
Of a hundred hands is my strength. And a hundred eyes guard us (lit.: "do to us the guard")
So please simon, be nice to yourself :)
A quick way to prove R5C4 can't be a 1 is to look at where 1s, 2s and 3s go in box 8, column 7 and column 8. You end up with four of them in row 9, which is clearly impossible.
Then the numbers must increase from the bottom right. But none of R4C4, R4C6 or R6C4 can be higher than 6, so they form a 456 triple and - that's three in the centre.
I'm quite proud of myself for solving it in 38:25, though I guess I kinda bruteforced it from halfway through the puzzle. I eliminated all the obviously wrong starts and then tried to eliminate the two starts in the bottom right I had left, but it just kept on working and suddenly I was done.
Solved in about 45 minutes. Simon almost gets there around 1:17, but without all the pencil-marking, the idea that row/column 7 has to be 3-9 and row/column 8 has to be 4-9 has implications for 1/2/3s in those rows, putting too many low digits in box 9.
the 1/2 paradigm is broken because on either side of the purple line, it has to be 1 and 2 (since purple is 3 to 9). since box 9 needs 1 and 2 to be at the bottom row, it can't be at the right side of purple.
The name of the puzzle is in Italian. The c's are pronounced as the English ch, and the e (word 3) is pronounced as the e in step.
Google Translate (not perfect, but rough and ready) says that the ending line is: "Of a hundred hands and my strength, And a hundred eyes keep watch over us"
"My strength is that of one hundred hands and a hundred eyes watch over us", the second line is a bit ambiguous, "fanno a noi la guardia" can be "watch over us" or "are watchful of us" as in "we are strong so a lot of people is wary of us"
Google translate, translated it to this: of a hundred hands and my strength. And a hundred eyes guard us
53:00, probably the quickest relative to the video I've done. Box 8 was the key, basically immediately made it possible to rule out the start being most of the upper circles since you need 4 or lower to appear on the graph within the box.
There was a much simpler way for discarding the 2 in the center cell. If it is a 2 then the only place the bottom cell can occupy is r5c3 and it must be a 1. This forces both 1 and 2 in box 6 to be in col 9 which leaves no choices for r7c9.
an hour an 19 minutes in Simon somehow proved that he should have read the rules properly and known that the thermo had to start in a bulb!
My way of breaking the 1/2 issue without uniqueness is that if you mark the 1 2 and 3s in row 7 8 and columns 7 8 you will see box 9 has 4 squares that must be 1 or 2.