I'm looking forward to the puzzle, but even more I was looking forward to Simon's joy about doing another Phistomefel puzzle and seeing him in happiness and anticipation in the first 2 minutes already makes me super happy
39:51 The combination of modularity and region sums (particularly when combined with no repeats) was stunningly powerful. Lines B, C and D force so quickly and give almost a third of the digits for a couple of straightforward deductions. It's devilishly clever how sudoku then leaps in to help build the other lines. Every bit as good as expected.
A thing that Simon could have proved from the start, which would likely have helped in a number of places, is that a modular region sum line MUST have a sum that is 0 mod 3. I think the easiest way to see this is to note that no segment of a modular line can have more than 1 more digit of one modularity than of any other modularity. This means that the only way to have a segment with, say, sum 1 mod 3 is for it to have one more digit that's 1 mod 3 than 2 mod 3. But, then to join two segments that each sum to 1 mod 3, you necessarily get a line that has two more 1s mod 3 than 2s mod 3, which can't be a valid modular line.
Wow, that's a very strong rule! I wish I had seen that. I proved weaker versions of it as I went along (e.g. as soon as one segment was going to have 3 digits on it - it was immediately divisible by 3 and hence the sum as well, etc, which is much faster to think about than the total of all digites along the entire line, or when the A line had one side of 1 digit and the other of 2 digits - it also had to have a sum 0 mod 3), but with the line that had lengths 2 and 4 - I didn't see that straight away.
Most constructors make me feel stupid when doing their hard puzzles. Phistomefel has the uncanny ability of making me feel smart when doing his hard puzzles. Everything feels completely fair to spot, and yet spotting them still feels like an achievement.
15:56 From here it is possible to directly prove orange are 0 mod 3 without enumerating all the options-the two blue + black dominoes will both have the same modularity, which means the orange + orange part must not change the overall mod 3 value of the segment sum, which can only be done with 0mod3 + 0mod3 (1mod3 +1mod3 = 2mod3; 2mod3 + 2mod3 = 1mod3).
Simon touched on something early on and then abandoned it later. He mentioned how the line couldn’t be two cells in each region because of the modular feature prevented it from happening. He missed the general case, that with the modular lines and region sum line combinations, the line in every region is always an even multiple of three. Consider if it’s not. Suppose the first box was a remainder 1, by modular line rules, this is either a single digit with remainder 1 or a two digits with a remainder 1 and a remainder 0 (or 4 digits but that is identical to case 1). For case 1 (single digit) we know there must be at least 2 digits in the next region since they would otherwise need to be the same digit. By modular rules those two will always be a remainder 0 and a remainder 2 which will add to a remainder 2. That doesn’t work as Simon pointed out. But when you add the next digit, it needs to be a remainder 1 which gives you a remainder 0 sum, you can’t get to a remainder 1 unless you could repeat another remain 1 which is against modular line rules. Similar if the first region is a remainder 2 sum. The way these two rules interact is if the sum in each box has a 0 remainder mod 3.
When you are happy, Simon, the whole world smiles. This was a wonderful video and a fascinating puzzle. Thanks so much for this - well worth watching basketball out of only the corner of my eye this evening!
Without a doubt it's puts a smile on all our faces when we see the passion with which Simon solves the one and only Phistomefel!! As usual from you Emily wonderful written!!
Simon missed one key logic about these modular region sum lines that would have made the puzzle massively easier: you can never leave a box with a residue of 1 or 2, as the next box (and region) cannot get to that same residue due to modular series arithmetic. So every region must always be mod 3.
Indeed. Given how Simon is typically quick to go after odd/even patterns even when it's not obvious, I was a bit surprised that he didn't think about doing modulo-3 arithmetic first (as in just putting 1,2,3 (or 0) on lines to figure it out), and only then looking for potential numbers to add up. By doing this, that constraint of "0 modulo-3 in each box" becomes obvious after one or two lines. But instead, he went for number combinations and sums all the way.
@marssang I think he missed it because he's become sooo good at doing these puzzles and commenting them live that he doesn't "need" to find the general implications of the rules anymore. one could almost argue that early CtC Simon would have called the start of the solve " bifurcation" :)
1:03:50 as soon as you got to box 4, and penciled in those numbers, I instantly had a thought of "Well that's just classic Phistomofel right there." What a fantastic setter and a fantastic puzzle.
Lovely puzzle. I was tired when I tried this yesterday, broke it and forgot one of the very important rules. When I tried it this morning it was a joy, thanks.
Wow, an absolutely stunning puzzle with beautiful, satisfying, surprising logic from start to finish. So glad I gave this one a go. One thing I loved about this was that I never really got stuck. I always had somewhere to look and somewhere to whittle down the possibilities. But even though I was never stuck it still took an hour and a half (seems like that's what it took Simon too). Hope we get more Phistomefel puzzles soon!
Wow! I usually struggle to finish the sudoku puzzles on this channel but I found this one surprisingly doable (though it still takes me very long haha). Phenomenal puzzle!
I have been looking forward to Critical Role play testing their new game system for a long time. Video is now up, and I finally have the time to watch. Then I see a 90 minute Simon vs Phistomefel and a new plan emerges 😅
Finished in 108:04. I was stuck for a good at least half an hour in the middle until I realized that I hadn't emphasized the rule that lines cannot have repeating digits. That cleared up a lot of line possibilities which made it relatively trivial to solve once I remembered this rule. Fun puzzle!
26:28 This doesn't work for the same reason a straight line from B to B didn't work - two consecutive two-cell sequences on a mod-3 line will always have different mod-3 values.
Two years ago I could not do any variant puzzles, let alone one this hard. After watching CtC every day for a year and a half, I started attempting every puzzle they put up. To my surprise I was able to do most of them. I thought they were all beyond me. I even finished this one. It took me nearly 3 hours, but I did it. Keep watching, learn why this puzzle solved the way it did, and next time it crops up you will be in a much better place.
The lines are simultaneously region sum lines and modular lines. Because of this the NUMBER of cells in each line has to be a multiple of three. The total on each line has to be a multiple of three; therefore the modulo rule insists that there is an equal number of each modular group in the line. Therefore the number of cells has also to be a multiple of three. This would have helped considerably in solving the paths of the lines as the lines had all to be 3, 6 or nine cells long.
If you do end up making a cookbook with chocolate cakes, please ensure you observe the correct 1:1 ratio of chocolate cake recipes and chocolate icing recipes.
I think the most underrated logic which Simon did at the beginning but then used only if forced to was that if a line has a length of 3 in any box it forces the region sum to be the divisor of three, and, therefore, it requires a one-cell region to be divisible by three and a two-cell region to not contain a digit divisible by three at all.
48:00 If ever I hear Simon saying again (whe he finally opens his eyes and put a seven in R3C1) 'It's outrageous to make me do Sudoku in your sudoku puzzle' I *swear* I'll throw my glass of Châteauneuf-du-Pape at my screen.
I finished in 239 minutes. This was such a tough one. It wasn't tough in the way of logic, but through the sheer amount of possibilities. I could sense the pathway, but proving everything was laborious. There were so many intricate tricks too. The parity between the endpoints for determining length, the parity of the number of regions, and the limitations of the modular rules on some lines. It took me way too long to see why the B line failed by immediately going straight. I think my favorite part was calculating my way through the E line and determining that 15 was the only possible number for each region. That was really fun to do in my head. This one felt like a marathon. None of it was even that tricky, it just required concentration every step. Despite what my time shows, this was fantastic. Great Puzzle!
Dunno, tip for people who struggle with remainder math? I tend to use -1 instead of 2 for remainder 2 modulo 3 It works the same way, it's 1 lower than 0, which is mod 3 and -1 and +1 cancel each other out when you sum them instead of you having to turn a 3 into 0 every time it happens
For what Simon is saying at 58:00, "you can't draw a line that visits 5 different boxes that isn't 9 cells long, that I'm going to claim as true". A more general proof of this would be that in any case where you're trying to do that with a region sum rule where you also can't repeat a digit on the line, only one of those 5 box visits could be 1 digit. All of the rest would have to be at least 2 cells, since another 1 cell visit would require using a repeated digit.
Thoughts early in the watch: two consecutive 2-digit segments along a line cannot have the same sum. In mod3 world (where you only add the remainders), you either have a 1 and 2 and a 3 1, a 2 3 and a 1 2, or a 3 1 and a 2 3. I don't think any of the region sums can be something that isn't mod 3. On both sides you're adding remainders in opposite order, so you can never have them equal.
52:40 The 8 is doing work once more here. If you have a 2:3 cell ratio on a RSL, at least one of the cells on the smaller side has to be from the 258 modularity. But the 8 prevents that
The beautiful thing about Modular RSLs is that the count of cells in adjacent Regions also follows the modularity rules. So you can't have a 2-cell-region followed by another 2-cell-region, nor a 5-cell-region. Which would be a beautiful alternate way to rule out 5 cells at 1:26:30
I was excited by the title for this one and just how pleased Simon would be. I have also confounded my family of maths people by asking why is the trianglulsr number of 10 4? The astonishes faces made me laugh out loud. Love the idea of the recipe book ‘chocolate cake variants around the world’ 😂
"Additionally, digits CANNOT repeat on a line." When I went back and reread the rules, that rule took me from being stuck near the start, to clearing the whole puzzle. It's never a bad idea to recheck the rules if things seem too difficult!
Watching this solve, for me was more entertaining than listening Liverpool cement top 8 last night, Was working at the time just heard the sheer joy in Simons voice at the start of the video and forced myself to wait almost 10hours to watch this
I don't know if it breaks in other ways, but I just found a way to have both ends of the D line as single cell. If you go straight down column 5 as far as you can, then step left one, down one, left one, and then back up to the other D. EDIT: I forgot about the non-repeating requirement. So that's what breaks it.
17:00 so does that mean, each region sum line for green adds up to either 12 or 15, we already know the range is 10-17 (1234 is 10 and 89 is 17) The two extremes, and you just said because it's a region sum line the total of the line is divisible by 2, but each section might be odd adding to a whole, like 11+11 is 22. Also the nature of having 2 pairs of 3-modulus it has to be divisible by 3. the entire sum, the total line is 20-34) So, 21 which isn't 2. etc, so it'll only be the divisible by 6 ones, 24, 30. So.. half of that since it's a region sum, is 12 or 15. not sure how you would mark that. 18:12 it always feels nice when you pause it explain your thought, and then he goes and explains it in less time then it took for you to articulate it in typed words. It's both a, I didn't need to make a comment moment, and a pat on the back you got it just like simon moment.
@52:35, the *grey line* cannot have that shape because of modularity (the segment in *box 3* must add up to a number divisible by *3,* and due to the given *8,* the segment in *box 2* must contain an *orange* cell...). Hence: 🔹The *grey line* must have length three 🔹Both its segments must have *modulo (3) = 0,* i.e they must be both divisible by *3.*
Lemma: A single cell region must be divisible by 3. Proof: The line immediately enters another region (it's a single cell region). The region it enters cannot be 1 cell (by sudoku or line uniqueness or modularity) or 4 cells or greater (T(4)=10>9). If the region is 3 cells, those 3 are from different modular sets, and their sum is divisible by 3 (1+2+3=3 mod 3). If the region is 2 cells, the adjacent 1 cell and 2 cell regions with equal sums are from the 3 different modular sets, and we have three cases: 1=/=2+3 mod 3 2=/=3+1 mod 3 3=1+2 mod 3; therefore, the single cell region is divisible by 3, and the proof is completed. I worked out the casework at the end just about as soon as you started drawing the first line in, and was wondering when you'd do the same. You didn't (why I'm showing it to you), but you did every other bit of reasoning I listed above, so this just sort of cobbled itself together one deduction at a time as I watched. Great puzzle, great video, cheers
You can't go too badly wrong starting with the simple "fifty-fifty-fifty-egg" cake formula. 50g. each of flour, margarine or butter and caster sugar to each egg; plus enough baking powder to raise it (usually 5ml. per 50g. flour, but check what it says on the tub), water to mix (use a half egg shell as a measure; substitute some of the water with glycerin to avoid the cake drying out, although this might not be a problem anyway since you probably will eat it before it gets the chance) and enough cocoa powder to bring it to the desired shade of brown. Divide between baking tins and cook at no. 4 (electric 180 degrees) until a knife inserted into the cake comes out clean (no liquid mixture on it). If you start from this solid foundation, even a failed experiment should not be completely inedible. A one-egg mix with chocolate chips is enough to make three muffins which will fit perfectly into one drawer of an air fryer! Don't run it too hot, or you will scorch them on top; 150 or 160 should be plenty. Beware, baking can be addictive .....
The principle Simon is describing at 15:00 is also known as the Pigeonhole Principle. It seems intuitively obvious, but is hard to explain concisely why it must be true. And, there are some fun Pigeonhole Principles, for example: given that the number of Londoners is greater than the max number of hairs on a person's head, there must be at least one pair of people in London with precisely the same number of hairs. Probably several :)
Lovely puzzle. I think that Simon made it a bit harder than necessary because of the order of lines he choose. A lot of sudoku could have been done, resolving the upper left before tackling E-line. Which does make the deductions that he was able to make rather amazing.
This was quite a challenger! But I made my way through it logically, using some external tools to enumerate possibilities along routes (specifically, the Sukano Helper tool). My time was 53:33, solver number 140.
I can’t believe I solved this, usually 90min+ puzzles I just skip right away, but this one being phistomephel (whose puzzles I’ve only solved a few of) I decided to give it a shot. 85min, so similar time to Simon, very happy with that 😊 Btw it took me a while to realize how the modularity affects the length of the lines and their region sums, but once I got it it went pretty smooth.
45:50 Okay, this is the moment, when you eliminate 4 from r3c1, see that blue is MODULO 1, thus r5c2 is 14, thus r4c2 is the 2 and r5c2 is the 4! As you said, that the c5 domino is the same: r4c5 = 4, r5c5 = 2.
Just finished watching the solve and 🎉🎉 the A clue in box 2-3 kept curveballing me as everytime i had an idea.. the puzzle itself proved me wrong 😂😂.. genius stuff. 10/10
1:23:48 how I resolved the question what r9c8 was, I did the following: suppose it is a 9, then the sum would be 16, or 1 mod 3. However, r9c7 would need to be 2 mod 3. But then one can prove by modular arithmatic that the sum in box 8 would either be 2 mod 3 or 0 mod 3. We conclude that r9c8 cannot be 9 and must be 8.
No 😢😢 I came to see u solve sudoku and at the beginning of rules at 3:41 you show me a very delicious looking cake😭😭 I want it now so badly😭😭 now got to rush to nearest bakery to get it, I want such chocolate cake
I solved it in 30m exactly, but I spent some time beforehand looking at how modular lines work. I'd never seen them before. Short story is that all 3 cell sum sections must have remainder of 0. The only sums that can be non zero remainder are 2 cell sums attached to a 4 cell sum. But in this puzzle, ALL sums ended up being zero remainder. Anyhow, this creates a HUGE restriction on lines and the top section basically solves itself. I actually spent more time on the bottom section. BTW, the purple line could not have an 9 on the circle in box 9 because of what I just mentioned. 97 would leave a remainder of 1. So the sum in box 8 would also need a remainder of 1 and that's only possible with a line of length 4. Three of the cells will have a remainder of 0 by definition and the last one will have a remainder of 1. You cannot do non-zero lengths smaller than that when attached to a length of 2 in the next box since the same remainder group would be too close. Anyways, you can't do a length of 4. The line length must be odd. So r9c8 must be an 8.
You can draw line A really early. It can't go through 8 because of modularity. 8 mod 3 is 2. That would put 0 and 1 in the first segment, and 0,1,2 in the second segment. The first segment would have a modular value of 1, and the second of 0, so they couldn't be the same number.
How do I draw these solid lines and type stuff on the grid (like x in between cells) in the sudokpad that opens when I click on the link in the description?
57 minutes for me but I checked my work by looking at the video a few times and saw mistakes because of it. Was an interesting puzzle. Wonder how much faster I'd have solved it if I realized each region sum has to be divisible by 3 at the start rather than 50 ish minutes in. I generally think of modulo as +0 = 3,6,9; +1 = 1,4,7; -1 = 2,5,8. A full set of 3 is 0+1-1. If you have 2 cells without a 3,6,9 it it's 1-1 = 0. With this in mind, you can see it's impossible for a region sum with a +1 or -1 charge to be adjacent to a line with the same charge. If it is +1, the cycle in the next box's next two cells will be -1 and 0 which will net to -1. Even when the +1 comes around it can only take it from -1 to 0 and it'll go back to -1 before it comes back around. It can never reach +1 no matter how many boxes are used. That means all region sum lines MUST be +0 charge. Given that there are no repeats that means at most one region sum can include a single cell. From there it's actually pretty easy and quick to solve. Wouldn't be surprised if someone picked up on that at the start solving this in just 15 minutes or so.
I got to about where you were 51:00 in (with less digits but the lines were the same and I had the pencil marks in) but I moved into E and made a bad deduction that ruined everything 😢
73:04, but I looked at the video about five times to see what I had to look at next. Seeing how box 4 affected 1 was what was keeping my stuck at the start. I just didn't see at all how to do the final purple line.
I spent more than 3 hours on this puzzle because I put the 15 pair of line E to the wrong place. I don't know why I didn't think for a while and just put them to column 9 immediately. Of course I got a paradox later. Then I tried some other routes on line E and still can't solve it. I thought maybe there's something wrong in other lines. Finally I have to go through all my logic again. What a disaster!
26:25 counting parity: 0,1,2,0,1,2 1 2 0 not possible now shifing this sequence by 2 spots we can get the conclusion that starting at 2 and 1 wields the same 3 different modulus in all sums so not a single one sum can be the same as any other sum in that line if it goes down like that
When I saw that title, I stopped immediately what I was doing and put watching this video on top of my queue. 🎉
Same.
Same!
Same 😅
Me too.
I'm looking forward to the puzzle, but even more I was looking forward to Simon's joy about doing another Phistomefel puzzle and seeing him in happiness and anticipation in the first 2 minutes already makes me super happy
Agreed
Same!
@@emilywilliams3237ditto
@@longwaytotipperaryditto ditto ❤🫠
@@davidrattner9 ❤️💕
As a math PhD, watching this is bordering on procrastination but I’ll count it as my ‘reading’ for the day 😂
I am far from a math PhD and my brain was butter after “rules” 😂
i have a munching cat phd and got 40:02
We're in the same boat... Good luck with your research!
Petition to call the collaborative dessert recipe book Cracking the Biscuit!!
Baking the Biscuit?
yeah but: Cracking the Chocolate
39:51
The combination of modularity and region sums (particularly when combined with no repeats) was stunningly powerful. Lines B, C and D force so quickly and give almost a third of the digits for a couple of straightforward deductions. It's devilishly clever how sudoku then leaps in to help build the other lines. Every bit as good as expected.
For he's a Phistomefellow, which nobody can deny!
Rules: 03:47
Let's Get Cracking: 09:25
What about this video's Top Tier Simarkisms?!
Phistomefel: 9x (00:23, 00:29, 01:03, 01:17, 01:25, 03:42, 05:53, 51:20, 1:30:19)
The Secret: 5x (59:24, 59:38, 59:43, 1:03:37, 1:20:32)
Bobbins: 3x (21:18, 47:37, 1:09:41)
And how about this video's Simarkisms?!
Ah: 16x (09:48, 21:28, 34:39, 38:01, 42:03, 45:12, 50:03, 1:03:46, 1:03:59, 1:08:46, 1:08:46, 1:10:26, 1:10:26, 1:11:17, 1:20:17, 1:28:26)
Sorry: 13x (03:00, 30:26, 34:48, 34:48, 34:48, 38:39, 42:05, 43:21, 47:37, 51:13, 1:09:44, 1:10:11, 1:22:14)
By Sudoku: 9x (44:38, 46:29, 48:34, 1:10:47, 1:13:30, 1:17:46, 1:21:42, 1:28:20, 1:28:36)
Hang On: 7x (25:16, 35:29, 47:11, 54:26, 1:00:11, 1:04:03, 1:23:34)
Pencil Mark/mark: 7x (19:03, 24:16, 45:33, 1:04:06, 1:11:21, 1:13:37, 1:13:37)
Obviously: 6x (02:44, 08:52, 42:54, 56:44, 1:18:13, 1:29:36)
In Fact: 5x (06:08, 32:59, 35:02, 1:15:12, 1:24:32)
Clever: 4x (17:58, 1:29:18, 1:29:21, 1:29:21)
Brilliant: 4x (03:04, 1:26:11, 1:26:13, 1:29:14)
Triangular Number: 4x (13:40, 15:16, 52:13, 53:34)
Good Grief: 3x (1:15:30, 1:30:10, 1:30:10)
Goodness: 3x (23:39, 1:21:58, 1:28:13)
Wow: 3x (18:14, 23:39, 23:39)
What Does This Mean?: 3x (04:40, 32:32, 44:49)
Naughty: 2x (43:01, 43:05)
Phone is Buzzing: 2x (10:05, 37:53)
Nature: 2x (48:11, 50:19)
Cake!: 2x (03:18, 03:25)
Useless: 1x (1:03:59)
Bother: 1x (1:12:43)
The Answer is: 1x (1:29:18)
Recalcitrant: 1x (43:01)
Stuck: 1x (47:41)
Horrible Feeling: 1x (1:19:06)
Lovely: 1x (1:03:46)
Beautiful: 1x (43:46)
Extraordinary: 1x (49:39)
Astonishing: 1x (23:57)
Gorgeous: 1x (1:05:18)
Shouting: 1x (02:22)
Of All Things: 1x (48:02)
Checkerboard: 1x (1:01:03)
Thingy Thing: 1x (44:17)
Weird: 1x (56:17)
Most popular number(>9), digit and colour this video:
Fifteen (28 mentions)
Three (133 mentions)
Orange, Blue (7 mentions)
Antithesis Battles:
Low (5) - High (0)
Even (11) - Odd (2)
Column (10) - Row (8)
FAQ:
Q1: You missed something!
A1: That could very well be the case! Human speech can be hard to understand for computers like me! Point out the ones that I missed and maybe I'll learn!
Q2: Can you do this for another channel?
A2: I've been thinking about that and wrote some code to make that possible. Let me know which channel you think would be a good fit!
Is "longest pause" possible to identify reliably? It would be an interesting one.
What about "the secret"
This is the first computer/bot feature i really like. Thank you very much and please go on.
@@dg_dos the secret is in the Top Tier Simarkisms
"Same modularity" was repeated at least a 1000 times.
A thing that Simon could have proved from the start, which would likely have helped in a number of places, is that a modular region sum line MUST have a sum that is 0 mod 3. I think the easiest way to see this is to note that no segment of a modular line can have more than 1 more digit of one modularity than of any other modularity. This means that the only way to have a segment with, say, sum 1 mod 3 is for it to have one more digit that's 1 mod 3 than 2 mod 3. But, then to join two segments that each sum to 1 mod 3, you necessarily get a line that has two more 1s mod 3 than 2s mod 3, which can't be a valid modular line.
Well, no segment that has fewer than four digits can have more than one modularity on it.
Wow, that's a very strong rule! I wish I had seen that. I proved weaker versions of it as I went along (e.g. as soon as one segment was going to have 3 digits on it - it was immediately divisible by 3 and hence the sum as well, etc, which is much faster to think about than the total of all digites along the entire line, or when the A line had one side of 1 digit and the other of 2 digits - it also had to have a sum 0 mod 3), but with the line that had lengths 2 and 4 - I didn't see that straight away.
Yes please I need the CtC Chocolate Cake Book!! Thank you Phistomefel! Thank you Simon!
Most constructors make me feel stupid when doing their hard puzzles. Phistomefel has the uncanny ability of making me feel smart when doing his hard puzzles. Everything feels completely fair to spot, and yet spotting them still feels like an achievement.
Great to have the one and only Phistomefel return!! The love you have Simon in solving his puzzles is so genuine!!
15:56 From here it is possible to directly prove orange are 0 mod 3 without enumerating all the options-the two blue + black dominoes will both have the same modularity, which means the orange + orange part must not change the overall mod 3 value of the segment sum, which can only be done with 0mod3 + 0mod3 (1mod3 +1mod3 = 2mod3; 2mod3 + 2mod3 = 1mod3).
A new Phistomefel AND it's a region sum lines puzzle - can't wait to watch this solve!
Saving this for tonight! Very excited
Simon touched on something early on and then abandoned it later. He mentioned how the line couldn’t be two cells in each region because of the modular feature prevented it from happening. He missed the general case, that with the modular lines and region sum line combinations, the line in every region is always an even multiple of three. Consider if it’s not. Suppose the first box was a remainder 1, by modular line rules, this is either a single digit with remainder 1 or a two digits with a remainder 1 and a remainder 0 (or 4 digits but that is identical to case 1). For case 1 (single digit) we know there must be at least 2 digits in the next region since they would otherwise need to be the same digit. By modular rules those two will always be a remainder 0 and a remainder 2 which will add to a remainder 2. That doesn’t work as Simon pointed out. But when you add the next digit, it needs to be a remainder 1 which gives you a remainder 0 sum, you can’t get to a remainder 1 unless you could repeat another remain 1 which is against modular line rules. Similar if the first region is a remainder 2 sum. The way these two rules interact is if the sum in each box has a 0 remainder mod 3.
When you are happy, Simon, the whole world smiles. This was a wonderful video and a fascinating puzzle. Thanks so much for this - well worth watching basketball out of only the corner of my eye this evening!
@emilywilliams “When … Simon is happy, the whole world smiles!” So well said! And so true! 😃
Without a doubt it's puts a smile on all our faces when we see the passion with which Simon solves the one and only Phistomefel!! As usual from you Emily wonderful written!!
@longwaytotipperary without a doubt it is one of life's great pleasures. 😁
@@davidrattner9 absolutely! 😃
The legend returns
wow - my birthday shouteded out yesterday and then I appear in the snipped LMD comments today! I love to appear on CTC two days in a row 😅
My favorite part is where he said "if you divide 3 by 3 you get 0." I love these videos and I love little tiny things like that.
Yeah, he meant a remainder of 0
We love Phistomefel.
AND my bedtime viewing is sorted!!
Simon missed one key logic about these modular region sum lines that would have made the puzzle massively easier: you can never leave a box with a residue of 1 or 2, as the next box (and region) cannot get to that same residue due to modular series arithmetic.
So every region must always be mod 3.
Indeed. Given how Simon is typically quick to go after odd/even patterns even when it's not obvious, I was a bit surprised that he didn't think about doing modulo-3 arithmetic first (as in just putting 1,2,3 (or 0) on lines to figure it out), and only then looking for potential numbers to add up. By doing this, that constraint of "0 modulo-3 in each box" becomes obvious after one or two lines. But instead, he went for number combinations and sums all the way.
@marssang I think he missed it because he's become sooo good at doing these puzzles and commenting them live that he doesn't "need" to find the general implications of the rules anymore.
one could almost argue that early CtC Simon would have called the start of the solve " bifurcation" :)
1:03:50 as soon as you got to box 4, and penciled in those numbers, I instantly had a thought of "Well that's just classic Phistomofel right there."
What a fantastic setter and a fantastic puzzle.
Lovely puzzle. I was tired when I tried this yesterday, broke it and forgot one of the very important rules. When I tried it this morning it was a joy, thanks.
That’s my evenings entertainment sorted 😅
Wow, an absolutely stunning puzzle with beautiful, satisfying, surprising logic from start to finish. So glad I gave this one a go. One thing I loved about this was that I never really got stuck. I always had somewhere to look and somewhere to whittle down the possibilities. But even though I was never stuck it still took an hour and a half (seems like that's what it took Simon too). Hope we get more Phistomefel puzzles soon!
Wow! I usually struggle to finish the sudoku puzzles on this channel but I found this one surprisingly doable (though it still takes me very long haha).
Phenomenal puzzle!
I have been looking forward to Critical Role play testing their new game system for a long time. Video is now up, and I finally have the time to watch.
Then I see a 90 minute Simon vs Phistomefel and a new plan emerges 😅
Both are well worth the watch, hope you're having a good evening!
I couldn't decide, I've been swapping back and forth every few minutes :D
@@PhilBoswellThank you, I was! But I had to go to bed halfway through after all, because kids 😂
@@GuatemalanJediI will hopefully finish both today, as I have the day off! (And so nice to know there is overlap between the two viewerships! 😊)
I’m here for both of these, I’m glad to know I’m not the only critical role watcher inhabiting the ctc chats.
Finished in 108:04. I was stuck for a good at least half an hour in the middle until I realized that I hadn't emphasized the rule that lines cannot have repeating digits. That cleared up a lot of line possibilities which made it relatively trivial to solve once I remembered this rule.
Fun puzzle!
26:28 This doesn't work for the same reason a straight line from B to B didn't work - two consecutive two-cell sequences on a mod-3 line will always have different mod-3 values.
Great puzzle, as expected! Thank you for the solve! 😀
I'm always so surprised by how well Simon can solve these puzzles. I tried but could barely start! Humbling and inspiring.
Two years ago I could not do any variant puzzles, let alone one this hard. After watching CtC every day for a year and a half, I started attempting every puzzle they put up. To my surprise I was able to do most of them. I thought they were all beyond me. I even finished this one. It took me nearly 3 hours, but I did it.
Keep watching, learn why this puzzle solved the way it did, and next time it crops up you will be in a much better place.
Phistomefel and 1.5 hours long? Gonna be a great vid
The lines are simultaneously region sum lines and modular lines. Because of this the NUMBER of cells in each line has to be a multiple of three. The total on each line has to be a multiple of three; therefore the modulo rule insists that there is an equal number of each modular group in the line. Therefore the number of cells has also to be a multiple of three. This would have helped considerably in solving the paths of the lines as the lines had all to be 3, 6 or nine cells long.
Faulty logic. Apologies
If you do end up making a cookbook with chocolate cakes, please ensure you observe the correct 1:1 ratio of chocolate cake recipes and chocolate icing recipes.
I dream of the day that I'll be able to solve a Phistomefel puzzle on my own. Thanks for taking us along the solve, Simon!
@30 mins - the line can't be 7 because it would give 4 consecutive digits from two modular sets (no 369) - I found that easier to see.
I think the most underrated logic which Simon did at the beginning but then used only if forced to was that if a line has a length of 3 in any box it forces the region sum to be the divisor of three, and, therefore, it requires a one-cell region to be divisible by three and a two-cell region to not contain a digit divisible by three at all.
48:00 If ever I hear Simon saying again (whe he finally opens his eyes and put a seven in R3C1) 'It's outrageous to make me do Sudoku in your sudoku puzzle' I *swear* I'll throw my glass of Châteauneuf-du-Pape at my screen.
I finished in 239 minutes. This was such a tough one. It wasn't tough in the way of logic, but through the sheer amount of possibilities. I could sense the pathway, but proving everything was laborious. There were so many intricate tricks too. The parity between the endpoints for determining length, the parity of the number of regions, and the limitations of the modular rules on some lines. It took me way too long to see why the B line failed by immediately going straight. I think my favorite part was calculating my way through the E line and determining that 15 was the only possible number for each region. That was really fun to do in my head. This one felt like a marathon. None of it was even that tricky, it just required concentration every step. Despite what my time shows, this was fantastic. Great Puzzle!
I haven't watched the video fully yet, but I will immediately offer to share recipes for the Cracking the Cryptic Chocolate Cake Compendium
Dunno, tip for people who struggle with remainder math?
I tend to use -1 instead of 2 for remainder 2 modulo 3
It works the same way, it's 1 lower than 0, which is mod 3 and -1 and +1 cancel each other out when you sum them instead of you having to turn a 3 into 0 every time it happens
For what Simon is saying at 58:00, "you can't draw a line that visits 5 different boxes that isn't 9 cells long, that I'm going to claim as true". A more general proof of this would be that in any case where you're trying to do that with a region sum rule where you also can't repeat a digit on the line, only one of those 5 box visits could be 1 digit. All of the rest would have to be at least 2 cells, since another 1 cell visit would require using a repeated digit.
Yaaay, I managed to solve one of Phistomefel's puzzles! :))
148:43 for me.
A wonderful puzzle! Modular RSLs offer so many different solve paths
Thoughts early in the watch: two consecutive 2-digit segments along a line cannot have the same sum. In mod3 world (where you only add the remainders), you either have a 1 and 2 and a 3 1, a 2 3 and a 1 2, or a 3 1 and a 2 3.
I don't think any of the region sums can be something that isn't mod 3. On both sides you're adding remainders in opposite order, so you can never have them equal.
Having finished watching it now, this would really simplify r9c8.
52:40 The 8 is doing work once more here. If you have a 2:3 cell ratio on a RSL, at least one of the cells on the smaller side has to be from the 258 modularity. But the 8 prevents that
The beautiful thing about Modular RSLs is that the count of cells in adjacent Regions also follows the modularity rules. So you can't have a 2-cell-region followed by another 2-cell-region, nor a 5-cell-region. Which would be a beautiful alternate way to rule out 5 cells at 1:26:30
Easy way to think about it is to treat every cell as 0, 1, or 2
I was excited by the title for this one and just how pleased Simon would be. I have also confounded my family of maths people by asking why is the trianglulsr number of 10 4? The astonishes faces made me laugh out loud. Love the idea of the recipe book ‘chocolate cake variants around the world’ 😂
24:25 Simon's phrasings truly crack me up 😂😂😂😂
I think Simon should be banned from using colours until he proves he can use them responsibly...
I was just looking to see if I had any unwatched phist videos last night and I had none left. Truly a wonderful day!
"Additionally, digits CANNOT repeat on a line."
When I went back and reread the rules, that rule took me from being stuck near the start, to clearing the whole puzzle. It's never a bad idea to recheck the rules if things seem too difficult!
Watching this solve, for me was more entertaining than listening Liverpool cement top 8 last night, Was working at the time just heard the sheer joy in Simons voice at the start of the video and forced myself to wait almost 10hours to watch this
Insane setting as usual.. I almost forgot how exceptional Phistomefel puzzles are..
I go to sleep every night watching CtC. Tonight there will be no sleep for I must watch😊 this to the end
I don't know if it breaks in other ways, but I just found a way to have both ends of the D line as single cell. If you go straight down column 5 as far as you can, then step left one, down one, left one, and then back up to the other D.
EDIT: I forgot about the non-repeating requirement. So that's what breaks it.
This made a pretty terrible friday turn into quite a good one. Thank you!
17:00 so does that mean, each region sum line for green adds up to either 12 or 15, we already know the range is 10-17 (1234 is 10 and 89 is 17) The two extremes, and you just said because it's a region sum line the total of the line is divisible by 2, but each section might be odd adding to a whole, like 11+11 is 22. Also the nature of having 2 pairs of 3-modulus it has to be divisible by 3. the entire sum, the total line is 20-34) So, 21 which isn't 2. etc, so it'll only be the divisible by 6 ones, 24, 30. So.. half of that since it's a region sum, is 12 or 15. not sure how you would mark that.
18:12 it always feels nice when you pause it explain your thought, and then he goes and explains it in less time then it took for you to articulate it in typed words. It's both a, I didn't need to make a comment moment, and a pat on the back you got it just like simon moment.
@52:35, the *grey line* cannot have that shape because of modularity (the segment in *box 3* must add up to a number divisible by *3,* and due to the given *8,* the segment in *box 2* must contain an *orange* cell...). Hence:
🔹The *grey line* must have length three
🔹Both its segments must have *modulo (3) = 0,* i.e they must be both divisible by *3.*
Lemma: A single cell region must be divisible by 3.
Proof:
The line immediately enters another region (it's a single cell region). The region it enters cannot be 1 cell (by sudoku or line uniqueness or modularity) or 4 cells or greater (T(4)=10>9).
If the region is 3 cells, those 3 are from different modular sets, and their sum is divisible by 3 (1+2+3=3 mod 3). If the region is 2 cells, the adjacent 1 cell and 2 cell regions with equal sums are from the 3 different modular sets, and we have three cases:
1=/=2+3 mod 3
2=/=3+1 mod 3
3=1+2 mod 3; therefore, the single cell region is divisible by 3, and the proof is completed.
I worked out the casework at the end just about as soon as you started drawing the first line in, and was wondering when you'd do the same. You didn't (why I'm showing it to you), but you did every other bit of reasoning I listed above, so this just sort of cobbled itself together one deduction at a time as I watched. Great puzzle, great video, cheers
You can't go too badly wrong starting with the simple "fifty-fifty-fifty-egg" cake formula. 50g. each of flour, margarine or butter and caster sugar to each egg; plus enough baking powder to raise it (usually 5ml. per 50g. flour, but check what it says on the tub), water to mix (use a half egg shell as a measure; substitute some of the water with glycerin to avoid the cake drying out, although this might not be a problem anyway since you probably will eat it before it gets the chance) and enough cocoa powder to bring it to the desired shade of brown. Divide between baking tins and cook at no. 4 (electric 180 degrees) until a knife inserted into the cake comes out clean (no liquid mixture on it). If you start from this solid foundation, even a failed experiment should not be completely inedible.
A one-egg mix with chocolate chips is enough to make three muffins which will fit perfectly into one drawer of an air fryer! Don't run it too hot, or you will scorch them on top; 150 or 160 should be plenty.
Beware, baking can be addictive .....
The principle Simon is describing at 15:00 is also known as the Pigeonhole Principle. It seems intuitively obvious, but is hard to explain concisely why it must be true. And, there are some fun Pigeonhole Principles, for example: given that the number of Londoners is greater than the max number of hairs on a person's head, there must be at least one pair of people in London with precisely the same number of hairs. Probably several :)
I kicked my legs in the air like a teenage girl when I saw the title! Well, I AM a teenage girl but still
Lovely puzzle. I think that Simon made it a bit harder than necessary because of the order of lines he choose. A lot of sudoku could have been done, resolving the upper left before tackling E-line. Which does make the deductions that he was able to make rather amazing.
This was quite a challenger! But I made my way through it logically, using some external tools to enumerate possibilities along routes (specifically, the Sukano Helper tool). My time was 53:33, solver number 140.
1:53:03 - Yay - I finished a Phistomofel puzzle. Some brilliant logic there, though I found it not easy to spot.
It's funny that at the very beginning he proved that you can't have two consecutive 2-lines, and then promptly forgot about it.
Brilliant puzzle.
I can’t believe I solved this, usually 90min+ puzzles I just skip right away, but this one being phistomephel (whose puzzles I’ve only solved a few of) I decided to give it a shot. 85min, so similar time to Simon, very happy with that 😊
Btw it took me a while to realize how the modularity affects the length of the lines and their region sums, but once I got it it went pretty smooth.
45:50 Okay, this is the moment, when you eliminate 4 from r3c1, see that blue is MODULO 1, thus r5c2 is 14, thus r4c2 is the 2 and r5c2 is the 4!
As you said, that the c5 domino is the same: r4c5 = 4, r5c5 = 2.
It took me almost 3 hours, but it was worth it ^^ What a nice puzzle :)
The lines felt like Islands of Insight's Logic Grids about character separation
Just finished watching the solve and 🎉🎉 the A clue in box 2-3 kept curveballing me as everytime i had an idea.. the puzzle itself proved me wrong 😂😂.. genius stuff. 10/10
waiting to contribute to chocolate cakes cookbook, have just reserved shelf space in the cookbooks cabinet!
Reminded me of those flow free phone app puzzles.
1:23:48 how I resolved the question what r9c8 was, I did the following: suppose it is a 9, then the sum would be 16, or 1 mod 3. However, r9c7 would need to be 2 mod 3. But then one can prove by modular arithmatic that the sum in box 8 would either be 2 mod 3 or 0 mod 3. We conclude that r9c8 cannot be 9 and must be 8.
No 😢😢 I came to see u solve sudoku and at the beginning of rules at 3:41 you show me a very delicious looking cake😭😭
I want it now so badly😭😭 now got to rush to nearest bakery to get it, I want such chocolate cake
I solved it in 30m exactly, but I spent some time beforehand looking at how modular lines work. I'd never seen them before.
Short story is that all 3 cell sum sections must have remainder of 0. The only sums that can be non zero remainder are 2 cell sums attached to a 4 cell sum. But in this puzzle, ALL sums ended up being zero remainder. Anyhow, this creates a HUGE restriction on lines and the top section basically solves itself. I actually spent more time on the bottom section.
BTW, the purple line could not have an 9 on the circle in box 9 because of what I just mentioned. 97 would leave a remainder of 1. So the sum in box 8 would also need a remainder of 1 and that's only possible with a line of length 4. Three of the cells will have a remainder of 0 by definition and the last one will have a remainder of 1. You cannot do non-zero lengths smaller than that when attached to a length of 2 in the next box since the same remainder group would be too close. Anyways, you can't do a length of 4. The line length must be odd. So r9c8 must be an 8.
idk about others but I barely hear the maverick, your mic is just that good, youre fine.
Tonight is going to be a deep sleep.
Thanks Simon for the long video.
CtC is an essential part of my mental health and insomnia routines 😊
You can draw line A really early. It can't go through 8 because of modularity. 8 mod 3 is 2. That would put 0 and 1 in the first segment, and 0,1,2 in the second segment. The first segment would have a modular value of 1, and the second of 0, so they couldn't be the same number.
How do I draw these solid lines and type stuff on the grid (like x in between cells) in the sudokpad that opens when I click on the link in the description?
Click on the cog icon and check "enable pen tool". You then select the additional icon on the right of the screen.
Simon: Do have a go. Me: no thanks I'd rather watch you battle phistomofel.
how do i draw lines and symbols?, i cant do it?
I did it in 168:49. 🥴 That was quite some puzzle and the first Phistomefel puzzle I have ever solved. Just happy I got through it...
57 minutes for me but I checked my work by looking at the video a few times and saw mistakes because of it. Was an interesting puzzle. Wonder how much faster I'd have solved it if I realized each region sum has to be divisible by 3 at the start rather than 50 ish minutes in.
I generally think of modulo as +0 = 3,6,9; +1 = 1,4,7; -1 = 2,5,8. A full set of 3 is 0+1-1. If you have 2 cells without a 3,6,9 it it's 1-1 = 0. With this in mind, you can see it's impossible for a region sum with a +1 or -1 charge to be adjacent to a line with the same charge. If it is +1, the cycle in the next box's next two cells will be -1 and 0 which will net to -1. Even when the +1 comes around it can only take it from -1 to 0 and it'll go back to -1 before it comes back around. It can never reach +1 no matter how many boxes are used. That means all region sum lines MUST be +0 charge.
Given that there are no repeats that means at most one region sum can include a single cell. From there it's actually pretty easy and quick to solve. Wouldn't be surprised if someone picked up on that at the start solving this in just 15 minutes or so.
I'm happy I predicted the A, even if it was a complete guess without evidence.
"When you divide 3 by 3 you get zero" - Simon, 2024.
36:43 for me. So happy to finally see another Phistomefel puzzle. Loved it!
I had to use a spreadsheet for this one 😂
I got to about where you were 51:00 in (with less digits but the lines were the same and I had the pencil marks in) but I moved into E and made a bad deduction that ruined everything 😢
73:04, but I looked at the video about five times to see what I had to look at next. Seeing how box 4 affected 1 was what was keeping my stuck at the start. I just didn't see at all how to do the final purple line.
Here’s a vote for the CtC’s book of chocolate cakes. I offer to be a tester!! 😋
I will be a taster with you and we can corner the market. 😁
@@davidrattner9 ❤️that idea!!
RSL + ML is much more restricted than Simon is giving it credit
E.g. the A line can't go down and through the 8, because 8 is the wrong modularity for such a line
Fun and challenging. Finished in 66:57 which felt surprisingly fast.
I spent more than 3 hours on this puzzle because I put the 15 pair of line E to the wrong place. I don't know why I didn't think for a while and just put them to column 9 immediately. Of course I got a paradox later.
Then I tried some other routes on line E and still can't solve it. I thought maybe there's something wrong in other lines. Finally I have to go through all my logic again. What a disaster!
26:25 counting parity:
0,1,2,0,1,2
1 2 0 not possible
now shifing this sequence by 2 spots we can get the conclusion that starting at 2 and 1 wields the same 3 different modulus in all sums
so not a single one sum can be the same as any other sum in that line if it goes down like that
Dear Simon, plz clean up used markup. Like the X's once the Lines are full. Thnks.
Very fun puzzle. Took me just over an hour.
I always think when i solve it quicker than the ctc guys I have made an assumption that just worked 😅