I know it's a running joke now, but the first time Simon explained the sum of 1-9 is 45 to make progress on a puzzle I was like, "WOW, that's genius!" Explanations like that or the Phestomefel's ring seem repetitive now, but I bet there are still a lot of people that are new to these types of puzzles that think, "WOW, that's genius!" and I wouldn't want to deprive them of that joy. It's easier for someone 'in the know' to skip forward in the video a bit than for someone 'out of the loop' to have to watch a separate video for an explanation. Simon, please never stop explaining your logic!
Also when Simon says something like “4, 5, or 6 which aren’t a 9” it’s simple logic but it helps explain why certain numbers can be ruled out to new viewer so they don’t go “wait so why can’t that by xyz”
Exactly what I thought, "Oh no, I can't make these the same colour as this as they go in either box, if only I had a way to colour them so they could show properties of both boxes."
I think in his mind multi coloured means both. Where as black is used as one or the other. Which doesn’t really make sense so I’m guessing he just isn’t used to it yet
@@TastyToast1 Agree absolutely! What other use could there be of the multi-colour feature? It is similar to the corner and centre digits. If a cell contains two centre digits, does that mean it must be both digits? I think not ;)
The thumbnails of cracking the cryptic are amongst the dumbest pics on all of RUclips. I believe they are reciprocally correlated to the geniality of the puzzle.
Not sure how reasonable it would be but an interview with a setter of this level would be super interesting. How they come up with an idea and the process they go through etc.
I believe Christoph Seileger (spelling? sorry) did one or maybe more videos on how he constructs puzzles. It may have been on patreon only but it also may be in the back catalogue somewhere. Was probably a year ago I'd say..
Took me 54 minutes to solve. And I think I'm shit at Sudoku. The solving was pretty easy and the route to solve it pretty straight forward. Box 1, Cage equals 22, Arrows sum to 23 minus r3c3, which meant that r1c3, r2c3, r3c3 and r4c3 summed to 23. Box 9, Cage equals 24, Arrows sum to 21 minus r7c7, which meant that r7c9, r7c8, r7c7 and r7c6 summed to 21. Bottom left 9 cell cage, then added up to 45-21-23=1, the point of intersection is 1. The same is true for top right 9 cell cage. That it took Simon over 30 minutes to find that, is beyond me. Then finding that the arrows acting on box 1 and box 9 had to sum to 14, was a bit harder, but doable with similar simple logic. Then it was just basically walking around the grid placing digits in one after another. The actual hardest part is the finding out of the box 1 and 9 acting arrows summing to 14.
@@livedandletdie Well, to be fair, that kind of logic isn't intuitive to everyone, and Simon is continuously trying to explain the logic at every step. Makes sense that he took a bit of time. And I don't blame him at all. Even if it takes a bit of time, it makes for a good video.
If you account for the intro time, the time spent explaining the Mephisto's Ring logic, and all the other explanations he gives, he did it in really impressive time.
I think it's so funny when Simon thinks we're waiting on him not seeing things that he thinks we see, when we're just waiting for him to figure it out so we can finally solve it once it's half filled in.
There is an insanely easier way to get to the first to 1's in the grid. I literally did it in about 5 minutes, though the puzzle as a whole took me hours of fiddling. Here's the simple way: The cells outside the cage in box one sum to 45-22 = 23. The cell outside the cage in box nine sum to 45-24 = 21. Now in box nine you can replace two cells and the arrow with the circle of the arrow to determine the last four cells in row seven = 21. You can do the same in box one to show that the first four cells in column 3 = 23. But where do the first four cells in column three go in the lower left plus cage? They go in c1,c2, and c4,c5 of row seven. That leaves only the central cell of the plus cage unaccounted for in row seven. 45 -23 -21 = 1 = the central cell of the lower left plus cage. Repeat symmetrically to get the central cell of the upper right plus cage. I was absolutely dumbfounded to see Simon go through the most complicated method possible to pick up the same two digits. But then I never used Phistomefel's Theorem, which probably would have helped me through the middle portion of the puzzle.
This is just a BRILLIANT and CLEVER and SIMPLE Deduction (which I also never noticed). It's amazing how many different paths and types of logic can be used on the SAME sudoku puzzle to come up with a solution.
Man, Simon, the break-in was so much easier than you made it xD All you had to do was make the Phistomefel Ring, erase the 2x2s in the top right and bottom left as well as the corresponding cells in the ring which are in the cages (you did this part). Then you erase the arrow circles in the ring and replace them with the arrows and you get sum or purple cells = sum of green cells. But the sum of purple cells in Box1 = 45- 22 since all cells in box 1 are now purple except the top left 2x2 which is green. Similarly sum of purple cells in box 9 = 45-24 = 21 Purples sum to 22+24 = 46, and the purples in box 1 and box 9 sum to 21+23 = 44. So the only remaining purples outside those two boxes are r3c7 and r7c3 which sum to 2 and must both be 1. I just love how you're too smart to do things the simple way mate xD Also, That's 3 in the corner That's 3 in the spotlight, losing its religion
I also spotted that, just after he had set everything as easy as possible for me to spot it... he had talked about replacing arrows with their sums, had colored the Phisto ring, but had just forgotten to remind himself of that little known fact about the 45 sum, which would have solved this conundrum in seconds given boxes 1 and 9...
With such a beautiful and quite difficult break-in to this puzzle, it would be nice to visually see a straight forward demonstration of that logic. I followed your post well, just saying it would be nice to see it too.
You can get the 1s in r3c7/r7c3 without using any hard logic: in row 3, c1-c4 add up to 23, since r3c4 equals r1c3 + r2c3. And c5+6+8+9 in row 3 add up to 21, because they are equal to r6-9c7. 45-23-21=1. Same logic for r7c3
"Purples sum to 22+24 = 46, and the purples in box 1 and box 9 sum to 21+23 = 44." You mean GREENS sum to 22+24, right? As in, the cages in box 1 and box 9 sum to 22+24 and they are green while the rest of their box is purple.
I have found a difference in what my brain sees depending on whether I am watching a solve or actually engaged in solving. It seems like I have an objective omniscience when I watch Simon and can see things that I miss completely when I am focused in on solving the puzzle myself. I can scan the whole puzzle watching Simon, but have tunnel vision when solving. Two different ways of your brain functioning and after a year of watching this channel and doing the sudokus I find my brain is starting to be able to switch tasks. Now throw in teaching as well at the same time and your brain is functioning in a third way, making sure that everyone can be on the same page with you. It is impressive the neurological task switching that is going on inside his melon. When I first found this channel, I thought both Mark and Simon were nuts thinking people were yelling at the screen. A year on I am one of those people and I feel it is a testament to what they have wrought, that I have skilled up enough to be able to yell at the screen "row 4!!!!" The saying "those who can...do. Those who can't...teach." applies here. Rarer are those who can, do and teach. What we are witnessing here is the rarest, someone who can, do and teach all at the same time with great skill when tackling new puzzles with new rule sets and logic. I believe the Sudoku world as a whole is skilling up just to try and stump these two. So it isn't surprising with all those factors going on at the same time that things can get muddled. You have to admit that when you spot something ahead of them you feel pretty good, but I try to remember that I am standing on the shoulders of these two very silly Giants!
I feel like Simon was making this one way more complicated, at least at the start. Let's start with the phistomefel ring. Simon quickly figured out where the top-right and bottom-left corners mapped onto the ring, so let's think about the other cells. The other cells are the cells that appear in the 22 and 24 cage, meaning the other 8 cells of the phistomefel ring add to 46. In box one, we know that the cage adds to 22, and the rest of the cells in the box add to 23 (45-22). Because of the arrows, we now know that the three cells in the corner of the phistomefel ring add to 23 as well. We can do the same in the bottom right, where they add to 21. So we know that all 8 cells add to 46, and we know that those 6 cells add to 44. Therefore, the 2 remaining cells must both be 1's. Simon is brilliant, and solves many puzzles that I could not hope to solve, but in this instance, while watching the video, I knew where those two 1's went while he giving the explanation of how the phistomefel ring works, at about the 14 min mark of the video. It was then quite painful watching him making it so much more complicated, and not figuring out where they were for another almost 20 mins.
Different expert solvers have very different ways of thinking about puzzles which cause them to preform differently on equally difficult puzzles. For example Simon is really good at quickly spotting sudoku techniques such as pairs, triples, wings... which is a strength people who have been to sudoku championships have, but he is not as good at understanding highly theoretical and abstract tactics such as Set Equivalence Theory, which is required for this puzzle. I happen to be the opposite of Simon as I excel in using SET while taking a long time to spot wings, triples, and other simple stuff like that.
@@emmettnelson7260 Exactly. I, for example, am great at noticing exclusionary patterns and performing SET and sum/diff theory, but miss naked singles about half the time
I was gonna post the exact same thing. He had it if he took a few more seconds to map those arrows to the circle clues, then went off on a completely more difficult tangent for 15 minutes. Meanwhile, I did the math in my head while dicking around on my phone. Not that I attempted the puzzle and probably wouldn't have gotten to that point in the first place...
@@emmettnelson7260 The funny thing to for me, is that I learned the idea to look for that trick with the arrows from an earlier video Simon made. I think he just got really caught up in the complicated geometry theories, that he overlooked a much simpler approach.
In your second paragraph, you say "Because of the arrows..." - what do you mean? How do you know that? Not questioning you at all - more just trying to understand your reasoning, as a novice myself!
LOOK, up in the youtubeosphere! It's a nerd; it's a brain; it's Sudoku-Man! Faster than a movie [most of the time]. Generates enough power to illuminate a small village during a cold solve. Able to crack monstrously brutal cryptic puzzles in a single sitting instead of the other way around. I am in awe...
29:00 "There has to be an easier way to see that." Well... here's what I used. First, using Phistomophel's Ring, and subtracting out the cells Simon colored in purple and green (see 16:20 or so), leaves a smaller subset of the ring. Then, using the "trick" Simon alluded to earlier in the video, replace the coloring on the circles with coloring on their respective lines. This results in boxes 1 and 9 being fully colored, along with r3c7 and r7c3. We can then calculate what these cells sum to-- it's 22 plus 24, all doubled thanks to the equivalence Phistomophel's Ring gave us earlier. That gives us 92. But if we subtract out all of the digits from boxes 1 and 9, which sum to 90, we're left with two highlighted digits that sum to 2. Clearly, these must both be 1's.
Once you figure out the two 1s, there's some cool logic you can use with the two longer arrows. Both of them can't have a 1, so one of them has to be a 2-3-4 9 arrow, which eliminates 9s from row 5 in the middle box. Because of that, you can't have 9s in both sets of "double arrows" near box 1 and 9. That would've sped up finding the cage in box 1 as a 1678 fairly significantly.
I was doing progress for a good 15-20 minutes, then I got stuck. I watched Simon going the same route as I did, then he kept going. Thanks for 180-92! That's what I needed to advance. Then I got stuck once more. Simon doing his big realisation at 54:43 solved the rest for me. There are only two 2s in two rows/columns. That helped immensly. I was expecting me to have more trouble than 2 separate points. It went way better than expected.
99:27 is my time. I'm so happy I was actually able to solve this one. There was a much easier break-in for the 1's. In box 1, the cells outside the cage add up to 23. Because of the arrows, the two circles and the corner cell r3c3 on the phisto ring also add up to 23. On the other side, they add up to 21. 23+21 = 44. These are all on the phisto ring. But you must put 46 into the cages. So there is 2 missing on the phisto ring. Hence 1 and 1 in the other two cells in the phisto ring. edit: (You get the stuff below at 1:06:20. I wrote it before I had watched that far) As for the 47 or 69 (r3c3 and r7c7), I haven't yet watched if you figured it out. But if the circles are 89's (or 4 and 6 in the corners of the phisto ring), then you will notice you get a problem with 9's in the center box. You can no longer put 9's in the corners because of the circles. And you can't put 9's on arrows. So the 9's must be on row 5. And you had already figured out that there must be a 9 in one of the circles on row 5. Contradiction!!! So the circles are not 89. They are 68. Said another way, the phisto corners are 7 and 9.
Beautiful puzzle, and I am proud to be able to solve it and that I found a slightly easier break-in then what Simon found at 21:00: Notice that the cells r3c4 and r4c3 have the same sum as the cells on the arrows in box 1, so together with cell r3c3 they add up to 45-22=23. Similarly, cells r7c6 and r6c7 have the same sum as the cells on the arrows in box 9, so together with cell r7c7 they add up to 45-24=21. Thus these cells on the Phistomefel ring add to 23+21=44, while together with cells r3c7 and r7c3 they add to 22+24=46, so cells r3c7 and r7c3 add to 2, so they are both 1. A little later one can also deduce that since r3c3, r3c4, and r4c3 add up to 23 and they are all different (as r3c4 and r4c3 map to different cells in the 22-cage), they also must be 689 just like cells r7c6 and r6c7. Then once one recognizes that 9 must be either in r4c4 or r6c6 in box 5, there can be at most one 9 in the 22- and 24-cages, so there is no 9 in the 22-cage, forcing it to be 1678.
Yeah, I'm not sure why Simon refused to add up the three cells on the corner of the phisto ring in the upper left and also bottom right. It's easy to see they add to 23 and 21 respectively. Together they make 44, but the cages add up to 46. So the other two cells on the phisto ring are 1 and 1 exactly as you said.
Hurricane is a great name for this puzzle because it whips you around thr grid to get a single deduction, and it kept me wondering what the middle of the grid would be
I found the centers of the crosses like this, which was much quicker than what Simon did: r3c3 plus the arrows in box 1 add up to 23 to get the box total to 45. That means that in row 3 and column 3, the cells in the crosses add up to 22 for each cross. r7c7 plus the arrows in box 9 add up to 21 to get the box total to 45. That means that in row 7 and column 7, the cells in the crosses add up to 24 for each cross. Adding the rows and columns of the crosses, double-counting the center cells, gets us 22+24=46 for each cross, which means both the center cells of the crosses must be 1.
Finished in 66 minutes. Been following this channel for about 1 year. Normally I need 2-3x video time to solve most puzzles. This is the 1st time I finished in less than the video time. Took me 2 minutes to find the first 2 “1”s
Simon mistakenly assumed the circles had to be different. The circles pointing with their arrows pointing into box 1 (r3c4 and r4c3) could be the same. But that would mean the circles pointing into box 9 would also have to be the same. But with a concerted effort of mental goodlifism you can work out that they can't be the same and reduce the candidates for the circles.
The deduction at 12:20 is actually not correct. If those two arrow circles are the same, then the cross cages force the number contained therein to appear in _both_ of the corner cages, thus maintaining the equivalence property. What it does mean is that if those circles have the same total, then the two circles on the other side must each sum to a different total, as you would then need to fit 3 copies of that number into two boxes, which would break the rules of sudoku. Now, once it's determined that the black cells are both 1, it's no longer possible for the top-left circles to be the same digit and the bottom-right circles to be two different digits, but it could still be the case that the top-left circles are the same digit and the bottom-right circles are a different same digit. Suppose both top-left circles are 8 and both bottom-right circles are 9, that would put an 8 and a 9 in each corner cage, plus a 1, and then the remaining red and blue cells go into their respective cages, and the grid's validity is maintained.
I personally like to do these really hard sudoku puzzles with you, I go as far as I can then when I get stuck and start playing your solve through; when you make something obvious to me i pause your video and go back to my solving and keep trying with what you made clearer to me. Two heads are better than one, even if we're technically not supposed to collaborate on these sort of things. lol ;p
This puzzle was BRUTAL! Simon did miss a couple of tricks that helped me solve it a bit faster, but then, I never managed to figure out the initial break-in without Simon's help anyways. _ At 38:25, There's a relationship between five of the six red and blue cells that would have helped break the puzzle open. R3C4 and R4C3 add up to either 14 or 17, so they have to be different. R6C7 and R7C6 ALSO add up to either 14 or 17, so they have to be different as well. The 69 in R3C3 and R3C4 both have to appear in the 45 cage in Box 3, but as they can't appear on the wings, they both have to appear down the spine. The 69 in R3C3 and R4C3 both have to appear in the 45 cage in Box 7, but as they can't appear down the spine, they both have to appear on the wings. This means R3C4 = R7C6 AND R4C3 = R6C7! _ Now we ask where does 9 go in Box 5? It can't go on any of the arrows, and it can't go in Row 5 as it already appears in one of the two circles, So in this puzzle, you DO put 9 in one of the two corners of Box 5. We don't have a Scooby Doo which corner the 9 goes into, but it doesn't matter. Because whichever corner it goes into, rules 9 out of the two circles next to it, And because we know from the previous trick that R3C4 = R7C6 AND R4C3 = R6C7, NONE of them can be 9 and they're both 68 pairs!
@52:00 : we can put 9 in two places, either r4c4 or r6c6 If at r4c4 it eliminate 9 from r3c4 & r4c3 and put 9 at r3c3 If at r6c6 it eliminate 9 from r6c7 & r7c6 and again put 9 at r3c3 because blue will have to have 6,8,9 So always 9 at r3c3.
"I'm sure there is an easy way to see it". Well, yes, your explanations clearly went over my head! Instead of the whole grid, just focus on the region 1 cells not in the cage. They sum to 23, and with the arrow trick are equivalent to the digits R1C1:4. Then you use the cross to transfer that sum to region 9.
Ooh, an amazingly fun break-in. I've been doing a lot of killers recently (on the CtC app, among elsewhere!), so I actually placed the 1s quite quickly. I was very happy to see that that wasn't the end of the clever logic, though; just lots of neat deductions throughout. 61:36 for me.
I'm quite proud that I actually managed to get a digit in this one more-or-less on my own! At the beginning, while Simon was puzzling over different geometry possibilities, I spotted that you should be able to do something interesting by subtracting the larger crosses, paused the video immediately, and managed to figure out - after about ten minutes working it out on paper to convince myself the math made sense - those two 1s. After that I got stumped... but I'm glad to have at least gotten that far.
Once Simon deduced that there had to be a 9 in one of the circles in r5, this puts the 9 in box 5 in two positions (r4c4, r6c6). Whichever position the 9 is in will resolve the pairs of circles in the Phistomefel ring to 68 pairs, and place 9 in r3c3 and 7 in r7c7, which allows you to fill the 22 cage in box 1.
Loved the solve Simon. I was surprised that you broke away from the maths at the beginning as that is something you appear to gravitate toward as a solution technique. If you followed through, you would have got the 1s in the black squares much quicker. You and Mark have the best channel on RUclips.
I can’t believe I solved it in 37:14. Just got really lucky to spot the break-in right away I guess. Loved how the 9&6 arrows interacted through box 5 and the two central north-south arrows restricted and defined the four symmetrical arrows pointing towards the corners.
You can easily write down candidates for r3c3, r3c4, r4c3 at the start of the puzzle after marking Phistomefel ring. The killer sudoku clue = 22 and the whole box 1 = 45, thus the rest of the box (expect killer cage) = 23. So, (r3c1+r3c2)+r3c3+(r1c3+r2c3)=23. As soon as r3c1+r3c3=r4c1 and r1c3+r2c3=r3c4, we have 3 squares: r4c1+r3c4+r3c3=23, and as they can't repeat (both r4c1 and r3c4 should appear in killer cage r3c3 see other two) - they are 689. Mostly the same works with Box 9 with sum of 21. Then you see that black squares sum to 24+22-23-21=2, thus both are 1.
Finished in 50 minutes. I confess I may not have finished at all had not my computer, as my cursor passed over the video before clicking on it, not given an image clearly showing Phistomefel's ring. I almost certainly wouldn't have thought of it. And it's only due to videos on CTC that I know what it is. That was an utterly gorgeous puzzle. And to have one starting digit is extremely generous!
An easier break-in, inspired by another comment: Box 1 (I'm told) adds to 45, so the non-caged cells must add to 23. Replace the cells on arrows with their circled cells, and you get that r3c3, r3c4, and r4c3 add to 23. Similarly, r6c7, r7c7, and r7c6 add to 21. Now you have six cells on Phistomefel's Ring that add to 44. The other cells on the Ring are the ones in the 45 cages. However, by geometry, these cells (in rows and cols 3 and 7) must make up the non-caged 2x2s in boxes 3 and 7, save for the two cells on the corner of the ring (r3c7 and r7c3). So, by Phistomefel, we can remove both the non-caged 2x2s in boxes 3 and 7 as well as the caged cells in rows and cols 3 and 7. The cells remaining are r3c3 + r3c4 + r4c3, which by above sum to 23; r6c7 + r7c7 + r7c6, which by above sum to 21; and r3c7 + r7c3. These two must equal the sums of the caged 2x2s in boxes 1 and 9, or 46. Since 23 + 21 = 44, r3c7 and r7c3 must both be 1.
yes this is exactly how i went about it! I am kind of surprised simon didn't go about it this way, as previous videos are the only reason i was able to catch on to this. i think he got too caught up with the arrows?
This slides smoothly into my all time faves. WHAT a puzzle! Such a nice toothsome difficulty all the way. Took me about an hour, but i had done an absolute brainfart with the pencil marks on the arrow in box 7, which didn't inpact anything until the very end, where it suddenly broke everything. And i couldn't for the life of me figure out where the mistake was, so i restarted and obvs did it faster. But i was so close to the end, that i'm counting my original time. For the opening: I used SET, adding rows and collums 3 and 7 (let's make them purple) and subtracting the 9-cell cages and boxes 1 and 9 (let's make them blue). This shows that the 22 and 24 cages equal the 8 scatterend cells in Phistomefel's ring. Then i moved the purple cells on arrow circles, onto their arrows, which the only coloured cells in the grid are two purples in the middle of the two 9-cell cages, and the entirety of boxes 1 and 9. We know the blue cells equal 46 from the killer cages, thus must the purples as well. And blue + purple = 90 (as they fill two boxes) + the two lonely purples. So the lonely purples must both be 1. My explanation in words may not be the clearest, but if you highlight it yourself on the grid, it should be a much easier way to see how the (start of the) break-in works. Or maybe that's just how my brain works, and it's more confusing to others, who knows!
A faster break in: consider that R7C6 is the equivalent of R8C7 + R9C7. Thus, R7C6-9 adds up to the same value as Box 9 minus the 24 cage. That means R7C1-5 has the same value as that cage. The same method applies to each of the arrows in boxes 1 and 9, and gives the value of each of the 9-cell cages except for the center cell as 44. Place 1s in the center of both cages for first digits placed. That then rules 1 out for all 4 arrows, putting 1 in both the 22 and 24 cages. That allows pencil marking 24 as 1689.
@ Simon: Around minute 21 you were almost there. The red, blue and black cells on the ring sum to 22+24=46. Now: the two red cells in the top left of the ring have the same sum as the four arrow cells in box 1. So together with the one blue cell in box 1 they sum up to 45-22=23. Equally, the two blue and the one red cell in the bottom right corner of the ring sum up to 45-24=21. Therefore, the red and blue cells on the ring alltogether sum up to 23+21=44. Ergo, the two black cells must sum up to 46-44=2. So they both must be 1. Ta-da! ;-)
Okay so I've commented a few times about how I got to a deduction before Simon, only to realize my logic was wrong and Simon took the correct path, but for once I'm *certain* I broke into the puzzle faster. So, Phist ring, eliminate the duplicated digits on the crosses and boxes 3/7, you're left with 8 cells in the ring. Simon got us to this point. Then just consider sums. Transfer the circles on the ring to their arrows, which maintains the sum, but now the two halves of the leftover Phist exactly fill boxes 1/9, plus two leftover cells in r3c7 and r7c3 (the black cells Simon marked). We know what the cages in boxes 1/9 sum to (22+24=46), so apply the secret to find what the remaining five cells of this boxes sum to (23+21=44), so the black cells are responsible for the difference (2) and thus must both be 1s! Simon hinted at this as soon as he'd removed the cross overlaps, but then he went in a different direction and had to do a really difficult backwards path to the same result. Thanks for giving me the opportunity to finally feel smart, Simon. ♥️
1:14:14 Simon: "This 1 fixes the 1/6 pair" The 6 which had been sitting in r5c3 for ages: "Am I a joke to you? Am I a bobbing joke to you? I sit hear patiently waiting for you to notice me, and then you give credit to that lousy 1 who just shows up and oh look, instantly noticing the 1/6 pair... I'm right next to it as well! It's not like I'm the other side of the bobbing grid! Seriously, what was even the point of me..." I love following your solves, and I love how sometimes you miss a more obvious fix but go a more complex route around the grid to solve part of the puzzle. But this one really tickled me. And I do post these in jest. While I managed to note the ring and thought it would be useful in the solve, I could barely get past that initial break in. So don't feel bad about not spotting something like this, and stop apologising for taking so long. This really was a doozy of a puzzle and it's still impressive this only took 76 minutes to solve (minus time spent explaining stuff).
WOW. You get the 1’s with that 45 subtract rule out twice. Then I got the center 3 with even odd parity! You can get the central 9 a little sooner, and some other clever central rule outs.
1:16:49 ... but time be darned! I actually had a solid start, figuring out two cells fairly quickly. From then on, I had many moments when I bogged down trying to find the next step (which, in retrospect, was often 'obvious'). That said, I never once guessed ... this is one of my favorite solves. Wonderful puzzle!
Nice solve. However, I struggle to understand, why the 2 arrow-bulbs looking into box 1 and 9 respectively have to be different. Sudoku forces the bulb-digit into the corresponding box cage, and the argument was, that due to the Phistomefel-Ring, there would be only 3 digits for the box-cage. But if you have the same digits on an arrow bulb, you could send one of its values to the box 1 cage, and one to the box 9 cage (maybe the other two arrow bulbs for the other cage also need to be the same then.) And this will not necessarily break the Phistomefel-Ring. Maybe I am missing something, or just have a flaw in thinking this out...
Just because I already posted this illustration elsewhere and had it to hand... imgur.com/a/uv5zB4S Nothing about the arrows, cages or Phistomefel ring gets broken by this arrangement (until you start trying to satisfy the actual cage totals 22 and 24, which Simon wasn't doing at that point).
29:17 yes there is an easier way to find the two 1s: 5cells in box9 outside the cage sum to 21. Due to the arrow c7r6, c7r6-9 also sums to 21. Hence c7r1-5 sums to 24. Same way r3c5-9 adds to 22. For the cross cage, 22+24=46 counts the central cell double which therefore has to be 1.
There is an easier way to go about this puzzle i think : We know thanks to the arrows that the two blue/red corners of the ring add up to 21+23, which is 44, while the blue and red cages add to 46, so the two black squares sum to 2. After that we can't put a 9 in the red cage anymore because then you can't put 9 in box 5 (which Simon figured at 1:06:00). It still took me about 3h to complete this puzzle.
Hey, Simon! I was wondering, since you tend to repeat quite often the explanations for various bits of theory/theorems in your videos, like you did here with Phistomefel's theorem, why don't you just create a new playlist on the channel compiling these short proofs? It would be a great learning tool for less experienced solvers like myself and it would make it easier for you whenever a concept comes up to just link to a different video with the full explanation!
I'm by no means an expert sudoku player... But I think I'm learning a lot by watching your solves. The first thing I noticed was the S.E.T. relationship between the south and west dominos of the upper right cross, where those dominos appeared in the box, and how the 5 was restricting one of the dominos Of course, that's where the puzzle lost me... But I do feel a little proud of myself
I appreciate that you go over Phistomefel's theorem for new viewers, but I wonder if making a quick 5-10-minute video on the subject, linking it in the description, and referring people to it if they want a detailed explanation might be a good idea? Might be good for adsense at least :P
I feel like every time Simon explains something I've heard before, that I'm welcoming a new solver into the fold. New people should always be made to feel welcome in the community. For those less interested in that, the L key will advance the video 10 seconds a pop.
@@Raphael_246 That's overly cynical for you to say. You only need to watch a couple videos to know what the channel is about, they are not "inflating the video length" they are just explaining the process.
@@coffeedude I have watched well over 100 of their videos and I promise you it's all for show. There's a reason Simon's videos are twice the length of Mark's videos - and it's not because Mark is a faster solver.
I've not finished the video yet, but I'd like to highlight some things about the break through. First of all, I sadly think the explanation given by Simon around 12:20 about the sum of arrows entering the same corner square being forced to be different numbers doesn't hold. At this stage of the solve, you could still have them equal. Phistomephel theorem would then just imply that the repeated digit would appear in both the 22 and 24 cages. It took me quite a lot of work afterwards to arrive at the conclusion they had to be different. I think Simon got quite lucky there. I actually didn't use Phistomephel theorem at all, there is much much simpler than this (it took me less than 15 minutes to get the double 1s and I'm not as nearly smart as Simon!) Let focus on row 3 for instance: 4 digits are outside the cross cage : 3 in box 1 and the last one is the sum of an arrow which elements also lie inside box 1. So the sum of these 4 digits is the same as the cells in box 1 outside the 22 cages, which means the 5 digits on the cross sum to 22. By the same logic on column 7, the digits on the cross must sum to 24. Therefore, the cross plus the repeated cell r3c7 sums to 46, thus r3c7 = 1. Similarly, r7c3 is also equal to 1.
At around 55:00, there is also another way of spotting where are the arrows summing to 8. In column 7, where do 8 and 9 go? They can't go in row 1 because of the arrow, they can't go in row 5 because we know there is an 8 in the central box and a 9 in either column 2 or column 8, so they're pushed in rows 2 and 7. Now, can r6c7 be a 9? If so there is a 9 in the cross in row 3, then r3c3 = 6 and r3c4 = 8, then r4c3 = 9, but r6c7 and r4c4 being 9s would push a 9 in row 5 in central box, which is not possible because we've already worked out the 9 in row 5 is in either column 2 or column 8. I think this double interaction of row 5 with the top left and bottom right squares was worth noticing, as it's an absolutely masterful logic!
Very nice proof of the 1s--I've been rewatching this video as it doesn't look like I ever finished it (though I might have just momentarily paused it at that point; it can be hard to tell with RUclips), and Simon's explanation was bugging me... Ah, it's too bad that this somewhat taints the solve.
I got the first digit in r3c7 by considering the two regions you get if you “add” box 1 and 9, then “subtract” r3 and c7, what we could call half a Phistomefel ring (but twice as powerful!). Then you see that the two cages (22 and 24), plus two of the arrow lines, equal the sum of the two associated arrow circles and the plus shaped cage with a double counted center at r3c7. Simply cancel the arrows, and you get that 22 + 24 = r3c7 + 45. r7c3 follows by symmetry.
Getting the first two digits (1 in R3C7 & R7C3) is relatively easy: A. Color the two crosses plus whatever is left from the Phistemofel's ring (two 3-cell shapes). B. Notice that you know the sum of the colored cells (134 = 45 + 45 + 23 + 21). C. Use Phistemofel's Theorem to replace the ring with the four 2x2 regions in the corners of the grid. D. Subtract the cages adding to 22 and 24. E. You are left with box 3 minus one cell (R3C7) and box 7 minus one cell (R7C3). F. The cells in these two regions add up to 134 - 22 - 24 = 88. G. Apply the Knowledge Bomb to boxes 3 & 7 to force R3C7 & R7C3 to be 1.
at 16:00 "These two squares (totals near upper left) have to go in this one (upper left2x2) and these two squares (totals near lower right) have to go in this one (lower right 2x2)" it seems to me this is not a digit set equivalence but merely a requirement that whatever digits appear in the totals must also appear in the same corner 2x2. If, say, the two upper left totals on the ring were equal, indeed, that number must appear in the upper left 2x2, but it of course can only appear once. The other (identical) digit would appear in the lower right 2x2. The two other corners in the ring (eventually coloured black around 18:00) would fill the balance of the two 2x2 corners. I don't see any reason to assume that the two totals near either corner of the ring must be different.
29:15 "I'm absolutely sure there's an easier way of doing that." Easier way I saw: Pause the video at 18:01. Box 1 totals 45. Minus the 22-cage is 23. Because of arrows, we know that the three top-left cells in the ring total 23. Box 9 totals 45. Minus the 24-cage is 21. Because of arrows, we know that the three bottom-right cells in the ring total 21. Therefore the six red and blue cells in the ring total 44. Red + blue + black ring cells must total 46, because they're equivalent to the two corner cages. Therefore the black cells total 2, and are both 1's. edit to add: I only see this after Simon does all the work to set up the ring, cancel the purples/greens, etc.
This is the hardest puzzle I've done on my own, but I have to confess I made one "guess" - well, I was really trying to rule one option out, and it ended up going right to the end. But I was very pleased to find a break in (a bit easier than Simon's, but not as nice as Urvidh Narula's below), and then many more steps, each painstaking to find, and usually leading to the most minor of deductions. I'd worked out the four corners of the Phistomefel Ring, the make-up of the 22 cage, and narrowed down the four key arrow circles (the ones in the Phistomefel Ring) to 6/8 pairs, and had narrowed r2c5 down to 7/8. I thought it might be easy to rule out r2c5 from being 8, as that would force the other 6/8 pairs, but it ended up being correct and the puzzle was surprisingly doable after that.
My first two digits were the same as Simon's, but I got there a lot quicker ... . . . ... Box 1 = 45, so if I label the cells in Box 1 as A-I (left-to-right and top-to-bottom), then C+F+G+H+I=23 (the difference btw. 45 and 22). But C+F=r3c4, so the first four cells in row 3 add up to 23, and so the 5 horizontal cells of the top-right 'plus' cage add up to 22. Similar reasoning using Box 9 gives the 5 vertical cells of the top-right 'plus' cage a sum of 24. Added together, you get a total of 46, and that total repeats the center cell of the 'plus', which therefore must be 1. The exact same reasoning also gives 1 in the center cell of the other 'plus' cage.
What a fantastic puzzle, with lovely symmetry and some cool logic to get going. It's a shame that Simon missed the next step, because it was truly beautiful, and would have got him all emotional, waxing lyrical and bouncing in his seat, which is always a treat to behold. My initial break in to get the 1s was almost the same as Simon's. After that, the key was the 69 in box 1. The way it interacted with the crosses meant that the circled cells leading into each corner box had to be the same - whichever it was (i.e. 6 or 9) not only eliminated that value from the adjacent cells, but forced that value into the crosses in a way which also ruled that value out of the other circles. If it was 6, both pairs of circles would have to be 89, and if 9, both would have to be 68. However, there is a 9 in one of the circles in R5, so 9 is restricted to either R4C4 or R7C7 in box 5, and whichever it is, it rules 9 out of the adjacent circles, making them 68. Therefore both pairs had to be 68, and R3C3 is 9. This meant the arrow cells in both boxes 1 & 9 were 24+35, fixing the cage contents, and giving a 7 in R7C7. The 68s were resolved by C7, where 8 could only go in the circle (because of the arrow cells and the pencil marked 8s in R5 in box 5). Symmetry resolved the other 68 cells. It didn't exactly fall apart after that, but it was very straightforward.
Ok Simon, this video's set theory explanation has shown me why you had people getting confused in the past When you succeed, it's not about bags or scrabble tiles, at least in my mind.. it's when you highlight the sets simultaneously then remove the cells with more than one color. The visual of multi-colored cells makes it very easy to understand how they can be removed while leaving identical sets behind.
The 1's can be gotten much simpler indeed Simon, I saw another explanation using the phistomefel ring, but I found the following the simplest and got the break in quickly with: take e.g. box 1, we know that the digits outside the cage sum to 23, we also know that the digits on the arrow sum to the circle, this means that we know R3 C1-4 sum to 23, that means that C7 R1+2+4+5 sum to 23, just like we could find R3 C1-4 being 23, we can find that C7 R6-9 sums to 21, that means we know that all the digits in C7 except for R3C7 sum to 21+23 for 44 leaving 1, it is symmetrical so same process on the other cage to get that 1.
The 5 squares not in the cage in box 1 must sum to 23. Replace the arrows with their sum boxes, and you have r3c3+r3c4+r4c3=23 without repeating a digit, and must be 689. When Simon said they can't repeat, that'd the first thing I thought of.
Maybe I‘m missing something, but with the nice logic at 44:45 about the 8s in box 5, should Simon not immediately have been able to place 8s in r4c3 and r6c7 at 45:15? Shouldn‘t that have immediately given Simon the 3/5 in rows 3 and 7 (and not much later via the 2s at 56:40)?
This is probably the most advanced sudoku I managed to complete all my own. I owe it all to Simon's past videos really stepping up my logic game. Took me closer to 2 hours since at one point I made a minor misstep in my logic and didn't realize until much later that my entire sudoku was wrong. Once I restarted and corrected that logic I was able to finish.
At 41:56 when the 3/5 pair with an 8 circle was ID'd in box nine, it would have sped things up a lot to see that it couldn't be in row seven because that would push the 3/5 into the pink cells in block two that mirrored the pink cells in block three. The existing 5 in box two prevented this.
I worked hard for an hour and then got stuck. watched the video and saw the 2 logic in column 5. That helped a lot. Then I managed to finish the solve in 1:18:00.
I did it in 34:39. I figured out the center of the crosses pretty quickly it was the third thing i checked (after noting the corner squares was the four digits on the cross outside the box and thinking about how the sum's affect box 1&9). The puzzle seemed strangely easier than I expected it to be, I think I got quite lucky and just happened to look at the right spot multiple times in a sequence and thought about the logic the right way at each of those times. I had expected this to take me 2 hours but was pleasantly surprised. The title did make me think i would have tp spin around the grid and i did in fact have to spin around the grid.
I think there is an easier way für the second step in this puzzle, after getting the 1s in r3c7 and r7c3. As Simon did, you can figure out there has to be a 9 in one of the circles in r5. Therefore, the 9 in the middle box has to be in the top left of the bottom right corner, because it cannot go on an arrow. In case the 9 goes into the top left corner, there is now way to put a 9 into one of the red cells. if on the other hand, the 9 goes into the bottom right corner, the 9 in the blue cells is pushed to r3c3 and again there cannot be a 9 in the reds. So to get to 22 without a using a 9, the reds has to be 1,6,7,8.
Found the 1’s using the Phis-but called segmented the central square into 4 other pairs and subtracted 45 twice (set theory) to get the two Blacks = 2. Then you get the 7891 blue corner. The two cells adding with the arrows was critical. You can get the 9’s in a cross in the center. It didn’t get easier. Used parity in col 5 to get the 3 in the middle then the 17 pairs. Crazy set theory.
Love a good feature length film. Also, not sure if it would have been useful...I certainly didn't fine a use while watching, but I did notice that in the 4x4's at the top/left and bottom/right each, the colored cells totaled to 45. i.e, the red+blue in those 4x4 areas.
I judge whether or not I will invest the time to attempt a puzzle based on the length of the video. 20-30 minutes are my "approachables" which I have a chance of completing. 30-40 are puzzles that I can't do on my own, but as Simon is going I'm able to see a move or two ahead while he's explaining the logic of the current one. I can't do them without his guidance, but I can at least see and understand, so watching the video helps me learn. 40-60 are usually hard enough that I understand the moves Simon makes, but can't really see them on my own ahead of time. But every so often we get one of these. And while I will enjoy it immensely, I will feel the absolute breadth of the gap between what I can do and what can be done. And this excites me.
17:04 Is the logic good here? Both r6c7 and r7c6 must go in the 24 cage in box 9, but do they have to be different? Can't they both be the same digit, and then one of them go into box 9 and one of them into box 1, thus maintaining the 8 digits necessary to fill the two cages?
I know it's a running joke now, but the first time Simon explained the sum of 1-9 is 45 to make progress on a puzzle I was like, "WOW, that's genius!" Explanations like that or the Phestomefel's ring seem repetitive now, but I bet there are still a lot of people that are new to these types of puzzles that think, "WOW, that's genius!" and I wouldn't want to deprive them of that joy. It's easier for someone 'in the know' to skip forward in the video a bit than for someone 'out of the loop' to have to watch a separate video for an explanation. Simon, please never stop explaining your logic!
Also when Simon says something like “4, 5, or 6 which aren’t a 9” it’s simple logic but it helps explain why certain numbers can be ruled out to new viewer so they don’t go “wait so why can’t that by xyz”
I was that guy 2 weeks ago at 3am, and now I’m hooked on this channel. I never even knew it life went further than normal sudoku. Lol
Simon: The software has a cool new multi colour feature.
Also Simon: Let's use black.
Exactly what I thought, "Oh no, I can't make these the same colour as this as they go in either box, if only I had a way to colour them so they could show properties of both boxes."
This hurts me...
I think in his mind multi coloured means both. Where as black is used as one or the other. Which doesn’t really make sense so I’m guessing he just isn’t used to it yet
It seems he doesn't want to use it in the videos until it is available for everyone to use.
@@TastyToast1 Agree absolutely! What other use could there be of the multi-colour feature? It is similar to the corner and centre digits. If a cell contains two centre digits, does that mean it must be both digits? I think not ;)
Simon apologizes for "being slow", yet finishes the puzzle about 7 times faster than the tester. This is soooo Simon.
Can we take a second to appreciate the thumbnail for this video?
Makes me crack up every time I see it 😂
The thumbnails of cracking the cryptic are amongst the dumbest pics on all of RUclips. I believe they are reciprocally correlated to the geniality of the puzzle.
Simon: I think this setter is trying to break my brain.
Simon later: I can't have a 9 with my orange.
Me: I think the setter succeeded.
Not sure how reasonable it would be but an interview with a setter of this level would be super interesting. How they come up with an idea and the process they go through etc.
I believe Christoph Seileger (spelling? sorry) did one or maybe more videos on how he constructs puzzles. It may have been on patreon only but it also may be in the back catalogue somewhere. Was probably a year ago I'd say..
Solves in hour and quarter sudoku, solved by tester in 6 hours.
"Sorry, I'm really slow today"
Typical Simon:)
Took me 54 minutes to solve. And I think I'm shit at Sudoku. The solving was pretty easy and the route to solve it pretty straight forward.
Box 1, Cage equals 22, Arrows sum to 23 minus r3c3, which meant that r1c3, r2c3, r3c3 and r4c3 summed to 23.
Box 9, Cage equals 24, Arrows sum to 21 minus r7c7, which meant that r7c9, r7c8, r7c7 and r7c6 summed to 21.
Bottom left 9 cell cage, then added up to 45-21-23=1, the point of intersection is 1. The same is true for top right 9 cell cage.
That it took Simon over 30 minutes to find that, is beyond me.
Then finding that the arrows acting on box 1 and box 9 had to sum to 14, was a bit harder, but doable with similar simple logic. Then it was just basically walking around the grid placing digits in one after another.
The actual hardest part is the finding out of the box 1 and 9 acting arrows summing to 14.
@@livedandletdie Well, to be fair, that kind of logic isn't intuitive to everyone, and Simon is continuously trying to explain the logic at every step. Makes sense that he took a bit of time. And I don't blame him at all. Even if it takes a bit of time, it makes for a good video.
If you account for the intro time, the time spent explaining the Mephisto's Ring logic, and all the other explanations he gives, he did it in really impressive time.
@@livedandletdie Good job mate .. took me over 2 hours lol
I think it's so funny when Simon thinks we're waiting on him not seeing things that he thinks we see, when we're just waiting for him to figure it out so we can finally solve it once it's half filled in.
There is an insanely easier way to get to the first to 1's in the grid. I literally did it in about 5 minutes, though the puzzle as a whole took me hours of fiddling. Here's the simple way:
The cells outside the cage in box one sum to 45-22 = 23. The cell outside the cage in box nine sum to 45-24 = 21. Now in box nine you can replace two cells and the arrow with the circle of the arrow to determine the last four cells in row seven = 21. You can do the same in box one to show that the first four cells in column 3 = 23. But where do the first four cells in column three go in the lower left plus cage? They go in c1,c2, and c4,c5 of row seven. That leaves only the central cell of the plus cage unaccounted for in row seven. 45 -23 -21 = 1 = the central cell of the lower left plus cage. Repeat symmetrically to get the central cell of the upper right plus cage. I was absolutely dumbfounded to see Simon go through the most complicated method possible to pick up the same two digits.
But then I never used Phistomefel's Theorem, which probably would have helped me through the middle portion of the puzzle.
This is just a BRILLIANT and CLEVER and SIMPLE Deduction (which I also never noticed).
It's amazing how many different paths and types of logic can be used on the SAME sudoku puzzle to come up with a solution.
Man, Simon, the break-in was so much easier than you made it xD
All you had to do was make the Phistomefel Ring, erase the 2x2s in the top right and bottom left as well as the corresponding cells in the ring which are in the cages (you did this part). Then you erase the arrow circles in the ring and replace them with the arrows and you get sum or purple cells = sum of green cells.
But the sum of purple cells in Box1 = 45- 22 since all cells in box 1 are now purple except the top left 2x2 which is green. Similarly sum of purple cells in box 9 = 45-24 = 21
Purples sum to 22+24 = 46, and the purples in box 1 and box 9 sum to 21+23 = 44. So the only remaining purples outside those two boxes are r3c7 and r7c3 which sum to 2 and must both be 1.
I just love how you're too smart to do things the simple way mate xD
Also,
That's 3 in the corner
That's 3 in the spotlight, losing its religion
I also spotted that, just after he had set everything as easy as possible for me to spot it... he had talked about replacing arrows with their sums, had colored the Phisto ring, but had just forgotten to remind himself of that little known fact about the 45 sum, which would have solved this conundrum in seconds given boxes 1 and 9...
With such a beautiful and quite difficult break-in to this puzzle, it would be nice to visually see a straight forward demonstration of that logic.
I followed your post well, just saying it would be nice to see it too.
I had a different break in, still using Phistomefel's Ring - easier than Simon's, but not as easy as yours. Nicely done!
You can get the 1s in r3c7/r7c3 without using any hard logic: in row 3, c1-c4 add up to 23, since r3c4 equals r1c3 + r2c3. And c5+6+8+9 in row 3 add up to 21, because they are equal to r6-9c7. 45-23-21=1. Same logic for r7c3
"Purples sum to 22+24 = 46, and the purples in box 1 and box 9 sum to 21+23 = 44."
You mean GREENS sum to 22+24, right? As in, the cages in box 1 and box 9 sum to 22+24 and they are green while the rest of their box is purple.
I have found a difference in what my brain sees depending on whether I am watching a solve or actually engaged in solving.
It seems like I have an objective omniscience when I watch Simon and can see things that I miss completely when I am focused in
on solving the puzzle myself. I can scan the whole puzzle watching Simon, but have tunnel vision when solving. Two different ways
of your brain functioning and after a year of watching this channel and doing the sudokus I find my brain is starting to be able to switch tasks. Now throw in teaching as well at the same time and your brain is functioning in a third way, making
sure that everyone can be on the same page with you. It is impressive the neurological task switching that is going on inside his
melon. When I first found this channel, I thought both Mark and Simon were nuts thinking people were yelling at the screen. A year on
I am one of those people and I feel it is a testament to what they have wrought, that I have skilled up enough to be able to yell at
the screen "row 4!!!!" The saying "those who can...do. Those who can't...teach." applies here. Rarer are those who can, do and teach. What we are witnessing here is the rarest, someone who can, do and teach all at the same time with great skill when tackling new puzzles with new rule sets and logic. I believe the Sudoku world as a whole is skilling up just to try and stump these two. So it isn't
surprising with all those factors going on at the same time that things can get muddled. You have to admit that when you spot something ahead of them you feel pretty good, but I try to remember that I am standing on the shoulders of these two very silly
Giants!
17:08 : missed opportunity to use the new multi-colored cells!! make them a red/blue split!
was looking for that comment :D
I couldn't figure out why he didn't do exactly that.
I cry when he makes them black instead of half blue/half red
I feel like Simon was making this one way more complicated, at least at the start. Let's start with the phistomefel ring. Simon quickly figured out where the top-right and bottom-left corners mapped onto the ring, so let's think about the other cells. The other cells are the cells that appear in the 22 and 24 cage, meaning the other 8 cells of the phistomefel ring add to 46.
In box one, we know that the cage adds to 22, and the rest of the cells in the box add to 23 (45-22). Because of the arrows, we now know that the three cells in the corner of the phistomefel ring add to 23 as well. We can do the same in the bottom right, where they add to 21.
So we know that all 8 cells add to 46, and we know that those 6 cells add to 44. Therefore, the 2 remaining cells must both be 1's.
Simon is brilliant, and solves many puzzles that I could not hope to solve, but in this instance, while watching the video, I knew where those two 1's went while he giving the explanation of how the phistomefel ring works, at about the 14 min mark of the video. It was then quite painful watching him making it so much more complicated, and not figuring out where they were for another almost 20 mins.
Different expert solvers have very different ways of thinking about puzzles which cause them to preform differently on equally difficult puzzles. For example Simon is really good at quickly spotting sudoku techniques such as pairs, triples, wings... which is a strength people who have been to sudoku championships have, but he is not as good at understanding highly theoretical and abstract tactics such as Set Equivalence Theory, which is required for this puzzle. I happen to be the opposite of Simon as I excel in using SET while taking a long time to spot wings, triples, and other simple stuff like that.
@@emmettnelson7260 Exactly. I, for example, am great at noticing exclusionary patterns and performing SET and sum/diff theory, but miss naked singles about half the time
I was gonna post the exact same thing. He had it if he took a few more seconds to map those arrows to the circle clues, then went off on a completely more difficult tangent for 15 minutes. Meanwhile, I did the math in my head while dicking around on my phone.
Not that I attempted the puzzle and probably wouldn't have gotten to that point in the first place...
@@emmettnelson7260 The funny thing to for me, is that I learned the idea to look for that trick with the arrows from an earlier video Simon made. I think he just got really caught up in the complicated geometry theories, that he overlooked a much simpler approach.
In your second paragraph, you say "Because of the arrows..." - what do you mean? How do you know that? Not questioning you at all - more just trying to understand your reasoning, as a novice myself!
54:41 the moment Simon does that celebration he does when he figures out a portion of the puzzle
This monster took me 5 hours and one full reset, never cease to be impressed by Simon's skill level.
You missed the perfect opportunity to use blue/red multicolour cells!
And he had the opportunity literally 30 seconds after accidentally using the feature. It hurt me, not gonna lie.
LOOK, up in the youtubeosphere! It's a nerd; it's a brain; it's Sudoku-Man! Faster than a movie [most of the time]. Generates enough power to illuminate a small village during a cold solve. Able to crack monstrously brutal cryptic puzzles in a single sitting instead of the other way around.
I am in awe...
29:00 "There has to be an easier way to see that."
Well... here's what I used.
First, using Phistomophel's Ring, and subtracting out the cells Simon colored in purple and green (see 16:20 or so), leaves a smaller subset of the ring.
Then, using the "trick" Simon alluded to earlier in the video, replace the coloring on the circles with coloring on their respective lines. This results in boxes 1 and 9 being fully colored, along with r3c7 and r7c3.
We can then calculate what these cells sum to-- it's 22 plus 24, all doubled thanks to the equivalence Phistomophel's Ring gave us earlier.
That gives us 92. But if we subtract out all of the digits from boxes 1 and 9, which sum to 90, we're left with two highlighted digits that sum to 2. Clearly, these must both be 1's.
Breaking Simon's brain. Yeah, now Simon knows how I feel with EVERY sudoku on this channel.
Serves him right. 🤯
"I do care, but only in a funny way" is a phrase I relate to.
““I know what I am going to do today: I am going to break Simon’s brain.’”
"Do know what we're going to do tonight, Pinky?"
I never know what I'm going to do today.
Once you figure out the two 1s, there's some cool logic you can use with the two longer arrows. Both of them can't have a 1, so one of them has to be a 2-3-4 9 arrow, which eliminates 9s from row 5 in the middle box. Because of that, you can't have 9s in both sets of "double arrows" near box 1 and 9. That would've sped up finding the cage in box 1 as a 1678 fairly significantly.
I was doing progress for a good 15-20 minutes, then I got stuck. I watched Simon going the same route as I did, then he kept going. Thanks for 180-92! That's what I needed to advance. Then I got stuck once more. Simon doing his big realisation at 54:43 solved the rest for me. There are only two 2s in two rows/columns. That helped immensly.
I was expecting me to have more trouble than 2 separate points. It went way better than expected.
99:27 is my time. I'm so happy I was actually able to solve this one. There was a much easier break-in for the 1's. In box 1, the cells outside the cage add up to 23. Because of the arrows, the two circles and the corner cell r3c3 on the phisto ring also add up to 23. On the other side, they add up to 21. 23+21 = 44. These are all on the phisto ring. But you must put 46 into the cages. So there is 2 missing on the phisto ring. Hence 1 and 1 in the other two cells in the phisto ring.
edit: (You get the stuff below at 1:06:20. I wrote it before I had watched that far)
As for the 47 or 69 (r3c3 and r7c7), I haven't yet watched if you figured it out. But if the circles are 89's (or 4 and 6 in the corners of the phisto ring), then you will notice you get a problem with 9's in the center box. You can no longer put 9's in the corners because of the circles. And you can't put 9's on arrows. So the 9's must be on row 5. And you had already figured out that there must be a 9 in one of the circles on row 5. Contradiction!!! So the circles are not 89. They are 68. Said another way, the phisto corners are 7 and 9.
Beautiful puzzle, and I am proud to be able to solve it and that I found a slightly easier break-in then what Simon found at 21:00: Notice that the cells r3c4 and r4c3 have the same sum as the cells on the arrows in box 1, so together with cell r3c3 they add up to 45-22=23. Similarly, cells r7c6 and r6c7 have the same sum as the cells on the arrows in box 9, so together with cell r7c7 they add up to 45-24=21. Thus these cells on the Phistomefel ring add to 23+21=44, while together with cells r3c7 and r7c3 they add to 22+24=46, so cells r3c7 and r7c3 add to 2, so they are both 1. A little later one can also deduce that since r3c3, r3c4, and r4c3 add up to 23 and they are all different (as r3c4 and r4c3 map to different cells in the 22-cage), they also must be 689 just like cells r7c6 and r6c7. Then once one recognizes that 9 must be either in r4c4 or r6c6 in box 5, there can be at most one 9 in the 22- and 24-cages, so there is no 9 in the 22-cage, forcing it to be 1678.
Yeah, I'm not sure why Simon refused to add up the three cells on the corner of the phisto ring in the upper left and also bottom right. It's easy to see they add to 23 and 21 respectively. Together they make 44, but the cages add up to 46. So the other two cells on the phisto ring are 1 and 1 exactly as you said.
Hurricane is a great name for this puzzle because it whips you around thr grid to get a single deduction, and it kept me wondering what the middle of the grid would be
I found the centers of the crosses like this, which was much quicker than what Simon did:
r3c3 plus the arrows in box 1 add up to 23 to get the box total to 45. That means that in row 3 and column 3, the cells in the crosses add up to 22 for each cross.
r7c7 plus the arrows in box 9 add up to 21 to get the box total to 45. That means that in row 7 and column 7, the cells in the crosses add up to 24 for each cross.
Adding the rows and columns of the crosses, double-counting the center cells, gets us 22+24=46 for each cross, which means both the center cells of the crosses must be 1.
Video is 1 minute old. First time catching one this fresh.
Nice and personal version of the intro tune 👍
Finished in 66 minutes. Been following this channel for about 1 year. Normally I need 2-3x video time to solve most puzzles. This is the 1st time I finished in less than the video time. Took me 2 minutes to find the first 2 “1”s
Simon mistakenly assumed the circles had to be different.
The circles pointing with their arrows pointing into box 1 (r3c4 and r4c3) could be the same. But that would mean the circles pointing into box 9 would also have to be the same. But with a concerted effort of mental goodlifism you can work out that they can't be the same and reduce the candidates for the circles.
The deduction at 12:20 is actually not correct. If those two arrow circles are the same, then the cross cages force the number contained therein to appear in _both_ of the corner cages, thus maintaining the equivalence property. What it does mean is that if those circles have the same total, then the two circles on the other side must each sum to a different total, as you would then need to fit 3 copies of that number into two boxes, which would break the rules of sudoku.
Now, once it's determined that the black cells are both 1, it's no longer possible for the top-left circles to be the same digit and the bottom-right circles to be two different digits, but it could still be the case that the top-left circles are the same digit and the bottom-right circles are a different same digit. Suppose both top-left circles are 8 and both bottom-right circles are 9, that would put an 8 and a 9 in each corner cage, plus a 1, and then the remaining red and blue cells go into their respective cages, and the grid's validity is maintained.
Yes, you're right.
Here's an illustration to back it up,
imgur.com/a/uv5zB4S
Repost from an earlier comment
I personally like to do these really hard sudoku puzzles with you, I go as far as I can then when I get stuck and start playing your solve through; when you make something obvious to me i pause your video and go back to my solving and keep trying with what you made clearer to me. Two heads are better than one, even if we're technically not supposed to collaborate on these sort of things. lol ;p
I'm just proud of myself for getting the 1's much faster than Simon. Probably the only thing to celebrate with this puzzle.
This puzzle was BRUTAL!
Simon did miss a couple of tricks that helped me solve it a bit faster, but then, I never managed to figure out the initial break-in without Simon's help anyways.
_
At 38:25,
There's a relationship between five of the six red and blue cells that would have helped break the puzzle open.
R3C4 and R4C3 add up to either 14 or 17, so they have to be different.
R6C7 and R7C6 ALSO add up to either 14 or 17, so they have to be different as well.
The 69 in R3C3 and R3C4 both have to appear in the 45 cage in Box 3, but as they can't appear on the wings, they both have to appear down the spine.
The 69 in R3C3 and R4C3 both have to appear in the 45 cage in Box 7, but as they can't appear down the spine, they both have to appear on the wings.
This means R3C4 = R7C6 AND R4C3 = R6C7!
_
Now we ask where does 9 go in Box 5?
It can't go on any of the arrows, and it can't go in Row 5 as it already appears in one of the two circles,
So in this puzzle, you DO put 9 in one of the two corners of Box 5.
We don't have a Scooby Doo which corner the 9 goes into, but it doesn't matter.
Because whichever corner it goes into, rules 9 out of the two circles next to it,
And because we know from the previous trick that R3C4 = R7C6 AND R4C3 = R6C7,
NONE of them can be 9 and they're both 68 pairs!
@52:00 : we can put 9 in two places, either r4c4 or r6c6
If at r4c4 it eliminate 9 from r3c4 & r4c3 and put 9 at r3c3
If at r6c6 it eliminate 9 from r6c7 & r7c6 and again put 9 at r3c3 because blue will have to have 6,8,9
So always 9 at r3c3.
"I'm sure there is an easy way to see it". Well, yes, your explanations clearly went over my head! Instead of the whole grid, just focus on the region 1 cells not in the cage.
They sum to 23, and with the arrow trick are equivalent to the digits R1C1:4.
Then you use the cross to transfer that sum to region 9.
Ooh, an amazingly fun break-in. I've been doing a lot of killers recently (on the CtC app, among elsewhere!), so I actually placed the 1s quite quickly. I was very happy to see that that wasn't the end of the clever logic, though; just lots of neat deductions throughout. 61:36 for me.
I'm quite proud that I actually managed to get a digit in this one more-or-less on my own! At the beginning, while Simon was puzzling over different geometry possibilities, I spotted that you should be able to do something interesting by subtracting the larger crosses, paused the video immediately, and managed to figure out - after about ten minutes working it out on paper to convince myself the math made sense - those two 1s. After that I got stumped... but I'm glad to have at least gotten that far.
150 minutes for me, glad I didn't give up! Incredible puzzle you got there.
Once Simon deduced that there had to be a 9 in one of the circles in r5, this puts the 9 in box 5 in two positions (r4c4, r6c6). Whichever position the 9 is in will resolve the pairs of circles in the Phistomefel ring to 68 pairs, and place 9 in r3c3 and 7 in r7c7, which allows you to fill the 22 cage in box 1.
Loved the solve Simon. I was surprised that you broke away from the maths at the beginning as that is something you appear to gravitate toward as a solution technique. If you followed through, you would have got the 1s in the black squares much quicker. You and Mark have the best channel on RUclips.
I can’t believe I solved it in 37:14. Just got really lucky to spot the break-in right away I guess. Loved how the 9&6 arrows interacted through box 5 and the two central north-south arrows restricted and defined the four symmetrical arrows pointing towards the corners.
You can easily write down candidates for r3c3, r3c4, r4c3 at the start of the puzzle after marking Phistomefel ring.
The killer sudoku clue = 22 and the whole box 1 = 45, thus the rest of the box (expect killer cage) = 23. So, (r3c1+r3c2)+r3c3+(r1c3+r2c3)=23.
As soon as r3c1+r3c3=r4c1 and r1c3+r2c3=r3c4, we have 3 squares: r4c1+r3c4+r3c3=23, and as they can't repeat (both r4c1 and r3c4 should appear in killer cage r3c3 see other two) - they are 689.
Mostly the same works with Box 9 with sum of 21.
Then you see that black squares sum to 24+22-23-21=2, thus both are 1.
Birds are always singing, I love it
I was so confused why birds were singing at 10 at night in a snowstorm, but just accepted it. Now I feel silly.
Finished in 50 minutes.
I confess I may not have finished at all had not my computer, as my cursor passed over the video before clicking on it, not given an image clearly showing Phistomefel's ring. I almost certainly wouldn't have thought of it. And it's only due to videos on CTC that I know what it is.
That was an utterly gorgeous puzzle. And to have one starting digit is extremely generous!
An easier break-in, inspired by another comment:
Box 1 (I'm told) adds to 45, so the non-caged cells must add to 23. Replace the cells on arrows with their circled cells, and you get that r3c3, r3c4, and r4c3 add to 23.
Similarly, r6c7, r7c7, and r7c6 add to 21. Now you have six cells on Phistomefel's Ring that add to 44.
The other cells on the Ring are the ones in the 45 cages. However, by geometry, these cells (in rows and cols 3 and 7) must make up the non-caged 2x2s in boxes 3 and 7, save for the two cells on the corner of the ring (r3c7 and r7c3). So, by Phistomefel, we can remove both the non-caged 2x2s in boxes 3 and 7 as well as the caged cells in rows and cols 3 and 7.
The cells remaining are r3c3 + r3c4 + r4c3, which by above sum to 23; r6c7 + r7c7 + r7c6, which by above sum to 21; and r3c7 + r7c3. These two must equal the sums of the caged 2x2s in boxes 1 and 9, or 46. Since 23 + 21 = 44, r3c7 and r7c3 must both be 1.
Gorgeous!
yes this is exactly how i went about it! I am kind of surprised simon didn't go about it this way, as previous videos are the only reason i was able to catch on to this. i think he got too caught up with the arrows?
This slides smoothly into my all time faves. WHAT a puzzle! Such a nice toothsome difficulty all the way.
Took me about an hour, but i had done an absolute brainfart with the pencil marks on the arrow in box 7, which didn't inpact anything until the very end, where it suddenly broke everything. And i couldn't for the life of me figure out where the mistake was, so i restarted and obvs did it faster. But i was so close to the end, that i'm counting my original time.
For the opening: I used SET, adding rows and collums 3 and 7 (let's make them purple) and subtracting the 9-cell cages and boxes 1 and 9 (let's make them blue). This shows that the 22 and 24 cages equal the 8 scatterend cells in Phistomefel's ring. Then i moved the purple cells on arrow circles, onto their arrows, which the only coloured cells in the grid are two purples in the middle of the two 9-cell cages, and the entirety of boxes 1 and 9.
We know the blue cells equal 46 from the killer cages, thus must the purples as well. And blue + purple = 90 (as they fill two boxes) + the two lonely purples. So the lonely purples must both be 1.
My explanation in words may not be the clearest, but if you highlight it yourself on the grid, it should be a much easier way to see how the (start of the) break-in works. Or maybe that's just how my brain works, and it's more confusing to others, who knows!
A faster break in: consider that R7C6 is the equivalent of R8C7 + R9C7. Thus, R7C6-9 adds up to the same value as Box 9 minus the 24 cage. That means R7C1-5 has the same value as that cage. The same method applies to each of the arrows in boxes 1 and 9, and gives the value of each of the 9-cell cages except for the center cell as 44. Place 1s in the center of both cages for first digits placed. That then rules 1 out for all 4 arrows, putting 1 in both the 22 and 24 cages. That allows pencil marking 24 as 1689.
Might be the first time I’ve ever got the break-in before Simon! 🤯 just ridiculously difficult today
@ Simon: Around minute 21 you were almost there. The red, blue and black cells on the ring sum to 22+24=46. Now: the two red cells in the top left of the ring have the same sum as the four arrow cells in box 1. So together with the one blue cell in box 1 they sum up to 45-22=23. Equally, the two blue and the one red cell in the bottom right corner of the ring sum up to 45-24=21. Therefore, the red and blue cells on the ring alltogether sum up to 23+21=44. Ergo, the two black cells must sum up to 46-44=2. So they both must be 1. Ta-da! ;-)
That was incredible. 34 minutes of solid ‘barbaric’ logic, and Simon apologises for being slow today 🤯
Okay so I've commented a few times about how I got to a deduction before Simon, only to realize my logic was wrong and Simon took the correct path, but for once I'm *certain* I broke into the puzzle faster.
So, Phist ring, eliminate the duplicated digits on the crosses and boxes 3/7, you're left with 8 cells in the ring. Simon got us to this point.
Then just consider sums. Transfer the circles on the ring to their arrows, which maintains the sum, but now the two halves of the leftover Phist exactly fill boxes 1/9, plus two leftover cells in r3c7 and r7c3 (the black cells Simon marked).
We know what the cages in boxes 1/9 sum to (22+24=46), so apply the secret to find what the remaining five cells of this boxes sum to (23+21=44), so the black cells are responsible for the difference (2) and thus must both be 1s!
Simon hinted at this as soon as he'd removed the cross overlaps, but then he went in a different direction and had to do a really difficult backwards path to the same result.
Thanks for giving me the opportunity to finally feel smart, Simon. ♥️
I find the idea that setters think of puzzles to destroy Simon's brain particularly is hilarious lol.
are you kidding me? what a monster sudoku. really feel for Simon to go through the anguish of solving it. bravo !!
1:14:14
Simon: "This 1 fixes the 1/6 pair"
The 6 which had been sitting in r5c3 for ages: "Am I a joke to you? Am I a bobbing joke to you? I sit hear patiently waiting for you to notice me, and then you give credit to that lousy 1 who just shows up and oh look, instantly noticing the 1/6 pair... I'm right next to it as well! It's not like I'm the other side of the bobbing grid! Seriously, what was even the point of me..."
I love following your solves, and I love how sometimes you miss a more obvious fix but go a more complex route around the grid to solve part of the puzzle. But this one really tickled me. And I do post these in jest. While I managed to note the ring and thought it would be useful in the solve, I could barely get past that initial break in. So don't feel bad about not spotting something like this, and stop apologising for taking so long. This really was a doozy of a puzzle and it's still impressive this only took 76 minutes to solve (minus time spent explaining stuff).
Actually gave this one a go, took around 2 hours. This is my best ever solve
WOW. You get the 1’s with that 45 subtract rule out twice. Then I got the center 3 with even odd parity! You can get the central 9 a little sooner, and some other clever central rule outs.
Simon accidentally coloring squares in two colors, then not doing it when it would make sense two seconds later 😅
Simon's notation just before placing first digit is looking like "Everything is WROGN"
1:16:49 ... but time be darned!
I actually had a solid start, figuring out two cells fairly quickly. From then on, I had many moments when I bogged down trying to find the next step (which, in retrospect, was often 'obvious'). That said, I never once guessed ... this is one of my favorite solves.
Wonderful puzzle!
Nice solve.
However, I struggle to understand, why the 2 arrow-bulbs looking into box 1 and 9 respectively have to be different.
Sudoku forces the bulb-digit into the corresponding box cage, and the argument was, that due to the Phistomefel-Ring, there would be only 3 digits for the box-cage.
But if you have the same digits on an arrow bulb, you could send one of its values to the box 1 cage, and one to the box 9 cage (maybe the other two arrow bulbs for the other cage also need to be the same then.)
And this will not necessarily break the Phistomefel-Ring.
Maybe I am missing something, or just have a flaw in thinking this out...
You're not missing anything. You're right and Simon is wrong.
Just because I already posted this illustration elsewhere and had it to hand...
imgur.com/a/uv5zB4S
Nothing about the arrows, cages or Phistomefel ring gets broken by this arrangement (until you start trying to satisfy the actual cage totals 22 and 24, which Simon wasn't doing at that point).
@@RichSmith77 Thanks for the post
This was brilliant! Exxxxxtra ordinary and absolutely unbelievable puzzle! Congratulations on another great solve - !!
Well done Simon! Well done! Extraordinary solve!
Welcome to another video to make you feel inadequate. Again.
Beautiful puzzle, nori nori, Simon is an absolute genius for being able to solve this.
29:17 yes there is an easier way to find the two 1s: 5cells in box9 outside the cage sum to 21. Due to the arrow c7r6, c7r6-9 also sums to 21. Hence c7r1-5 sums to 24. Same way r3c5-9 adds to 22. For the cross cage, 22+24=46 counts the central cell double which therefore has to be 1.
There is an easier way to go about this puzzle i think :
We know thanks to the arrows that the two blue/red corners of the ring add up to 21+23, which is 44, while the blue and red cages add to 46, so the two black squares sum to 2.
After that we can't put a 9 in the red cage anymore because then you can't put 9 in box 5 (which Simon figured at 1:06:00).
It still took me about 3h to complete this puzzle.
Hey, Simon! I was wondering, since you tend to repeat quite often the explanations for various bits of theory/theorems in your videos, like you did here with Phistomefel's theorem, why don't you just create a new playlist on the channel compiling these short proofs? It would be a great learning tool for less experienced solvers like myself and it would make it easier for you whenever a concept comes up to just link to a different video with the full explanation!
I'm by no means an expert sudoku player... But I think I'm learning a lot by watching your solves.
The first thing I noticed was the S.E.T. relationship between the south and west dominos of the upper right cross, where those dominos appeared in the box, and how the 5 was restricting one of the dominos
Of course, that's where the puzzle lost me... But I do feel a little proud of myself
I appreciate that you go over Phistomefel's theorem for new viewers, but I wonder if making a quick 5-10-minute video on the subject, linking it in the description, and referring people to it if they want a detailed explanation might be a good idea?
Might be good for adsense at least :P
I feel like every time Simon explains something I've heard before, that I'm welcoming a new solver into the fold. New people should always be made to feel welcome in the community. For those less interested in that, the L key will advance the video 10 seconds a pop.
@@CurtisAutery true, but a bookmark would be good
How else are they going to inflate video lengths if they don't repeat things a bunch of times?
@@Raphael_246 That's overly cynical for you to say. You only need to watch a couple videos to know what the channel is about, they are not "inflating the video length" they are just explaining the process.
@@coffeedude I have watched well over 100 of their videos and I promise you it's all for show. There's a reason Simon's videos are twice the length of Mark's videos - and it's not because Mark is a faster solver.
I've not finished the video yet, but I'd like to highlight some things about the break through. First of all, I sadly think the explanation given by Simon around 12:20 about the sum of arrows entering the same corner square being forced to be different numbers doesn't hold. At this stage of the solve, you could still have them equal. Phistomephel theorem would then just imply that the repeated digit would appear in both the 22 and 24 cages. It took me quite a lot of work afterwards to arrive at the conclusion they had to be different. I think Simon got quite lucky there.
I actually didn't use Phistomephel theorem at all, there is much much simpler than this (it took me less than 15 minutes to get the double 1s and I'm not as nearly smart as Simon!)
Let focus on row 3 for instance: 4 digits are outside the cross cage : 3 in box 1 and the last one is the sum of an arrow which elements also lie inside box 1. So the sum of these 4 digits is the same as the cells in box 1 outside the 22 cages, which means the 5 digits on the cross sum to 22.
By the same logic on column 7, the digits on the cross must sum to 24. Therefore, the cross plus the repeated cell r3c7 sums to 46, thus r3c7 = 1.
Similarly, r7c3 is also equal to 1.
At around 55:00, there is also another way of spotting where are the arrows summing to 8. In column 7, where do 8 and 9 go? They can't go in row 1 because of the arrow, they can't go in row 5 because we know there is an 8 in the central box and a 9 in either column 2 or column 8, so they're pushed in rows 2 and 7.
Now, can r6c7 be a 9? If so there is a 9 in the cross in row 3, then r3c3 = 6 and r3c4 = 8, then r4c3 = 9, but r6c7 and r4c4 being 9s would push a 9 in row 5 in central box, which is not possible because we've already worked out the 9 in row 5 is in either column 2 or column 8.
I think this double interaction of row 5 with the top left and bottom right squares was worth noticing, as it's an absolutely masterful logic!
Very nice proof of the 1s--I've been rewatching this video as it doesn't look like I ever finished it (though I might have just momentarily paused it at that point; it can be hard to tell with RUclips), and Simon's explanation was bugging me... Ah, it's too bad that this somewhat taints the solve.
I got the first digit in r3c7 by considering the two regions you get if you “add” box 1 and 9, then “subtract” r3 and c7, what we could call half a Phistomefel ring (but twice as powerful!). Then you see that the two cages (22 and 24), plus two of the arrow lines, equal the sum of the two associated arrow circles and the plus shaped cage with a double counted center at r3c7. Simply cancel the arrows, and you get that 22 + 24 = r3c7 + 45. r7c3 follows by symmetry.
Getting the first two digits (1 in R3C7 & R7C3) is relatively easy:
A. Color the two crosses plus whatever is left from the Phistemofel's ring (two 3-cell shapes).
B. Notice that you know the sum of the colored cells (134 = 45 + 45 + 23 + 21).
C. Use Phistemofel's Theorem to replace the ring with the four 2x2 regions in the corners of the grid.
D. Subtract the cages adding to 22 and 24.
E. You are left with box 3 minus one cell (R3C7) and box 7 minus one cell (R7C3).
F. The cells in these two regions add up to 134 - 22 - 24 = 88.
G. Apply the Knowledge Bomb to boxes 3 & 7 to force R3C7 & R7C3 to be 1.
at 16:00 "These two squares (totals near upper left) have to go in this one (upper left2x2) and these two squares (totals near lower right) have to go in this one (lower right 2x2)" it seems to me this is not a digit set equivalence but merely a requirement that whatever digits appear in the totals must also appear in the same corner 2x2. If, say, the two upper left totals on the ring were equal, indeed, that number must appear in the upper left 2x2, but it of course can only appear once. The other (identical) digit would appear in the lower right 2x2. The two other corners in the ring (eventually coloured black around 18:00) would fill the balance of the two 2x2 corners.
I don't see any reason to assume that the two totals near either corner of the ring must be different.
wow new feature for multicolored cells, that's really cool!
Pity he didn’t use it where it would have been logical to do so.
29:15 "I'm absolutely sure there's an easier way of doing that."
Easier way I saw: Pause the video at 18:01.
Box 1 totals 45. Minus the 22-cage is 23. Because of arrows, we know that the three top-left cells in the ring total 23.
Box 9 totals 45. Minus the 24-cage is 21. Because of arrows, we know that the three bottom-right cells in the ring total 21.
Therefore the six red and blue cells in the ring total 44.
Red + blue + black ring cells must total 46, because they're equivalent to the two corner cages. Therefore the black cells total 2, and are both 1's.
edit to add: I only see this after Simon does all the work to set up the ring, cancel the purples/greens, etc.
This is the hardest puzzle I've done on my own, but I have to confess I made one "guess" - well, I was really trying to rule one option out, and it ended up going right to the end. But I was very pleased to find a break in (a bit easier than Simon's, but not as nice as Urvidh Narula's below), and then many more steps, each painstaking to find, and usually leading to the most minor of deductions. I'd worked out the four corners of the Phistomefel Ring, the make-up of the 22 cage, and narrowed down the four key arrow circles (the ones in the Phistomefel Ring) to 6/8 pairs, and had narrowed r2c5 down to 7/8. I thought it might be easy to rule out r2c5 from being 8, as that would force the other 6/8 pairs, but it ended up being correct and the puzzle was surprisingly doable after that.
My first two digits were the same as Simon's, but I got there a lot quicker ...
.
.
.
... Box 1 = 45, so if I label the cells in Box 1 as A-I (left-to-right and top-to-bottom), then C+F+G+H+I=23 (the difference btw. 45 and 22). But C+F=r3c4, so the first four cells in row 3 add up to 23, and so the 5 horizontal cells of the top-right 'plus' cage add up to 22. Similar reasoning using Box 9 gives the 5 vertical cells of the top-right 'plus' cage a sum of 24. Added together, you get a total of 46, and that total repeats the center cell of the 'plus', which therefore must be 1. The exact same reasoning also gives 1 in the center cell of the other 'plus' cage.
Exactly.
(I provided this illustration in another comment,
imgur.com/a/ksd771v
)
16:50 It looks like a new feature to be released? Multiple colors at the same time in one cell.
*1:52** **_Rules_*
*3:13** **_"Let's get cracking"_*
*31:19** **_First Digit_*
*1:06:25** **_"My orange contains a nine!"_* (#ThingsYouHearOnCTC)
Love when you do this. Any Bobbins in this vid? Lol
@@erikpaige2165 Probably too many to mention LOL
What a fantastic puzzle, with lovely symmetry and some cool logic to get going.
It's a shame that Simon missed the next step, because it was truly beautiful, and would have got him all emotional, waxing lyrical and bouncing in his seat, which is always a treat to behold.
My initial break in to get the 1s was almost the same as Simon's. After that, the key was the 69 in box 1. The way it interacted with the crosses meant that the circled cells leading into each corner box had to be the same - whichever it was (i.e. 6 or 9) not only eliminated that value from the adjacent cells, but forced that value into the crosses in a way which also ruled that value out of the other circles. If it was 6, both pairs of circles would have to be 89, and if 9, both would have to be 68. However, there is a 9 in one of the circles in R5, so 9 is restricted to either R4C4 or R7C7 in box 5, and whichever it is, it rules 9 out of the adjacent circles, making them 68. Therefore both pairs had to be 68, and R3C3 is 9. This meant the arrow cells in both boxes 1 & 9 were 24+35, fixing the cage contents, and giving a 7 in R7C7.
The 68s were resolved by C7, where 8 could only go in the circle (because of the arrow cells and the pencil marked 8s in R5 in box 5). Symmetry resolved the other 68 cells.
It didn't exactly fall apart after that, but it was very straightforward.
Ok Simon, this video's set theory explanation has shown me why you had people getting confused in the past
When you succeed, it's not about bags or scrabble tiles, at least in my mind.. it's when you highlight the sets simultaneously then remove the cells with more than one color. The visual of multi-colored cells makes it very easy to understand how they can be removed while leaving identical sets behind.
The 1's can be gotten much simpler indeed Simon, I saw another explanation using the phistomefel ring, but I found the following the simplest and got the break in quickly with: take e.g. box 1, we know that the digits outside the cage sum to 23, we also know that the digits on the arrow sum to the circle, this means that we know R3 C1-4 sum to 23, that means that C7 R1+2+4+5 sum to 23, just like we could find R3 C1-4 being 23, we can find that C7 R6-9 sums to 21, that means we know that all the digits in C7 except for R3C7 sum to 21+23 for 44 leaving 1, it is symmetrical so same process on the other cage to get that 1.
The 5 squares not in the cage in box 1 must sum to 23. Replace the arrows with their sum boxes, and you have r3c3+r3c4+r4c3=23 without repeating a digit, and must be 689. When Simon said they can't repeat, that'd the first thing I thought of.
Puzzle takes tester seven hours, video is over an hour long “do have a go”... right
I miraculously solved this in 63:26. I think I've never been so proud and satisfied after a solve. 😅 Great puzzle!
Absolutely bonkers. What a puzzle!
Maybe I‘m missing something, but with the nice logic at 44:45 about the 8s in box 5, should Simon not immediately have been able to place 8s in r4c3 and r6c7 at 45:15?
Shouldn‘t that have immediately given Simon the 3/5 in rows 3 and 7 (and not much later via the 2s at 56:40)?
This is probably the most advanced sudoku I managed to complete all my own. I owe it all to Simon's past videos really stepping up my logic game. Took me closer to 2 hours since at one point I made a minor misstep in my logic and didn't realize until much later that my entire sudoku was wrong. Once I restarted and corrected that logic I was able to finish.
At 41:56 when the 3/5 pair with an 8 circle was ID'd in box nine, it would have sped things up a lot to see that it couldn't be in row seven because that would push the 3/5 into the pink cells in block two that mirrored the pink cells in block three. The existing 5 in box two prevented this.
I worked hard for an hour and then got stuck. watched the video and saw the 2 logic in column 5. That helped a lot. Then I managed to finish the solve in 1:18:00.
I did it in 34:39. I figured out the center of the crosses pretty quickly it was the third thing i checked (after noting the corner squares was the four digits on the cross outside the box and thinking about how the sum's affect box 1&9). The puzzle seemed strangely easier than I expected it to be, I think I got quite lucky and just happened to look at the right spot multiple times in a sequence and thought about the logic the right way at each of those times. I had expected this to take me 2 hours but was pleasantly surprised. The title did make me think i would have tp spin around the grid and i did in fact have to spin around the grid.
How does the half and half colouring work?
I think there is an easier way für the second step in this puzzle, after getting the 1s in r3c7 and r7c3.
As Simon did, you can figure out there has to be a 9 in one of the circles in r5. Therefore, the 9 in the middle box has to be in the top left of the bottom right corner, because it cannot go on an arrow. In case the 9 goes into the top left corner, there is now way to put a 9 into one of the red cells.
if on the other hand, the 9 goes into the bottom right corner, the 9 in the blue cells is pushed to r3c3 and again there cannot be a 9 in the reds.
So to get to 22 without a using a 9, the reds has to be 1,6,7,8.
Found the 1’s using the Phis-but called segmented the central square into 4 other pairs and subtracted 45 twice (set theory) to get the two Blacks = 2. Then you get the 7891 blue corner. The two cells adding with the arrows was critical. You can get the 9’s in a cross in the center. It didn’t get easier. Used parity in col 5 to get the 3 in the middle then the 17 pairs. Crazy set theory.
Love a good feature length film.
Also, not sure if it would have been useful...I certainly didn't fine a use while watching, but I did notice that in the 4x4's at the top/left and bottom/right each, the colored cells totaled to 45. i.e, the red+blue in those 4x4 areas.
I judge whether or not I will invest the time to attempt a puzzle based on the length of the video. 20-30 minutes are my "approachables" which I have a chance of completing. 30-40 are puzzles that I can't do on my own, but as Simon is going I'm able to see a move or two ahead while he's explaining the logic of the current one. I can't do them without his guidance, but I can at least see and understand, so watching the video helps me learn. 40-60 are usually hard enough that I understand the moves Simon makes, but can't really see them on my own ahead of time.
But every so often we get one of these. And while I will enjoy it immensely, I will feel the absolute breadth of the gap between what I can do and what can be done. And this excites me.
17:04 Is the logic good here? Both r6c7 and r7c6 must go in the 24 cage in box 9, but do they have to be different? Can't they both be the same digit, and then one of them go into box 9 and one of them into box 1, thus maintaining the 8 digits necessary to fill the two cages?