A New Sudoku Masterpiece

Phistomefel is back with (yet) another stunning creation. You can play it here:
cracking-the-c...
Rules:
Normal sudoku rules apply. A black dot between two cells means that one of the digits in the cells is exactly double the other. A white dot between two cells means that the two digits in the cells must be consecutive. A V between two cells means that the two digits in those cells add up to five. An X between two cells means that the two digits in those cells add up to ten. Not all possible black and white dots, X and V are indicated.
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Our Patreon page has lots of extra content! Our solve of Cosmology by Bobuardo Da Vinci is just out and Mark's solve of the Times Monthly Special Cryptic Crossword! Coming soon (1 October) is a new puzzle hunt (which includes sudoku) by Scott Strosahl.
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Hi! We're Simon Anthony and Mark Goodliffe, two of the UK's most enthusiastic puzzle solvers. We have both represented the UK at the World Sudoku Championships and the World Puzzle Championships. We're also "cryptic crossword" aficionados. Mark is the twelve-time winner of The Times championship and Simon is the former record holder for most consecutive correct solutions to The Listener crossword. We hope we can help your puzzle solving while also introducing you to some of the world's best puzzles.
Thank you for watching!
Simon and Mark

[@anomalousresult]

A masterclass in British English: "not very many black dots" = 2 black dots "a few white dots" = 16 white dots

[@Zombie-lx3sh]

Always understating everything, aren't they?

[@andreww4473]

A fine example of British understatement.

[@Varksterable]

"A fair few" = anything over 140.

[@HonkeyKongLive]

Me, after finishing up a medium puzzle in the newspaper: "Well let's see what my fellow sudoku experts are doing."

[@ghosttwo2]

Showing off with impossible puzzles, like usual.

[@azpcox]

Getting trolled by Phistomephel into using 5’s to start this puzzle must be humbling. Another brilliant set and brilliant solve.

[@leftysheppey]

I thought it was genius too. Definitely trolling...

[@boydegg]

"Phistomefel brings you crashing down to Earth again!" - it always sounds like he's talking about his arch-nemesis, the supervillain of the puzzle world.

[@josiahtasich6073]

One he can never catch either BTW; while he might always prevent the villain’s plan from succeeding, he inevitably gets away.

[@benconlon875]

I remember when they used to do sudokus on this channel that had more than 1 given digit... those were the days

[@Iteria]

What are you talking about? There are TWO whole givens in this puzzle. A luxury really.

[@SengokuTheGouda]

They still do a fair share of classic sudoku, but there are only so many ways to make a classic puzzle interesting or entertaining to watch, or even to solve, hence why all these variants were created in the first place. When the logic gets interesting, that makes for a better video.

[@benconlon875]

@@SengokuTheGouda yh i know i was only joking haha

[@ignion2163]

Next week: Schroedinger`s Sudoku by Phistomefel

[@andrewnelson2525]

That was my immediate thought. I want a Schroedinger Sudoku!

[@conexant51]

Careful what you wish for!.. ;D

[@prahas777]

Marvelous idea. Might be the first game made for a quantum computer! :-)

[@conexant51]

@@prahas777 Our brains are quantum computers according to some highly regarded physicists.

[@charlottestang9462]

I feel like every time Simon mentions schroedinger cells he's just trying to tempt setters

[@Caljkusic1]

That fist ‘em fella seems like a briliant sudoku constructor!

[@aaronmorris1513]

Far and wide one of the best.

[@HonkeyKongLive]

It's an accurate description of how I feel after trying one of their puzzles.

[@mratkovich]

He really stretches the boundaries of puzzle making

[@aaronmorris1513]

Quite possibly the GOAT, see?

[@rancid83]

Simon: "We have a 2-3 pair here" *ignores the 4 right next to them, that would fully solve the V in row 3, for another two minutes* Me: AAAHHHH SIMON PLEASE

[@petemagnuson7357]

The logic there is that corner marks indicate restrictions within the 3x3 box, in the notation style they like to use. Mixing the box-based notation with rows/column restrictions runs the risk of getting ambiguous and confusing later.

[@rancid83]

That has nothing to do with what happens at 12:40 though.

[@petemagnuson7357]

@@rancid83 oops, I think I clicked the wrong comment's reply section. I thought I was responding to someone complaining about the use of coloring on the right side instead of normal pencil marks.

[@oakleyg2870]

This channel has helped me get so much better at logic puzzles. I used to just stare blankly at Phistomefel puzzles. This was one of the only ones I've ever solved of his and even though it took 1 hour and 23 minutes it feels like progress!

[@Rubrickety]

Something I've been wondering for a long time, Simon: If you lean back in your chair, do you smack your head on the angled wall?

[@matthewc.7301]

I just wanted to comment on the solve for the 2-3-4-5-6 chain at the lower left corner. (shown at around 8:45 in the video) It would have been so much easier if you had pencil-marked the Vs for the 1st box(top left box) because it would show that those four boxes use 1-2-3-4. Since you have 5 placed already and three of those numbers would be in the 1st box, that would leave 6-7 and 8-9 as the only possible consecutive numbers to place at the top left. So once you pencil in 6-7-8-9, you know that those four boxes sum to 30. Combined with the X-marks, V-marks and the 5 already on the board, you would then see that 70 out of 90 for the first two columns have been used, leaving only 20. Since these five numbers in the lower left are consecutive, you just take 20 and divide it by 5 to find the central number in the string of numbers, which is shown to be 4. And, as you pointed out, 5 had to be placed in the 2nd column, which would then place all five numbers without the need to guess.

[@crossgibson6986]

Simon needs a gaming keyboard with colorful lights

[@Ishant4President]

Only if it still makes that satisfying clicky sound when he taps it tho

[@msr0685]

simon needs to learn to use numpad and implement right ctrl

[@awilliams1701]

@@msr0685 yeah I don't get that. How do you need light to use a numberpad. I can use a numberpad in my sleep. lol

[@WhoStoleMyAlias]

21 minutes in I found myself at a complete deadlock. As the clock came to a full hour I sort of gave up and started watching the video to find that Simon had a stall at the exact same point where I had stranded. Don't feel ashamed having followed the master's lead out of this position, that amazing odd-even evaluation combined with the possible positioning of the 5 would have never crossed my mind. Until now of course, which is why I frequent this channel.

[@steef6074]

Exactly the same

[@ragnkja]

Easy way to break into the last two columns after you’ve placed the first few digits in box 3: The last two columns have to contain exactly eight even digits. Two are used in the 4-5-6 chain, and the rest are needed in consecutive dominos, so the X domino and r1c8 all have to be odd.

[@RichSmith77]

I like this suggestion. One of those ideas that seems obvious once you hear it, but not something I was able to spot. It does indeed make columns 8 and 9 a lot easier to solve. Thank-you.

[@fuxpremier]

I was pretty sure Simon's logic was not what was meant to be done. Your suggestion is more than probably what Phistomefel had in mind.

[@eclectichoosier5474]

In columns 8 and 9, you can use parity to get the order of the 12 pair in row 1. Notice that there are an even number of consecutive pairs, an X, and a consecutive triple that fill all but one cell of the two columns. The remaining cell has to be the same parity as the consecutive triple. Therefore, it must be 1, (since you have figured out the triple) which places a 2 in col 7. That sorts out the 2 in row 9.

[@yichen6313]

Finally got it after struggling for 1:46:17 - and then came back and see Simon zip through all the logic in a much better way. Brilliant!

[@richardfarrer5616]

21:05 "As usual, I apologise... Me at the same position: OK, I'm stuck. Better watch the video.

[@andrewnelson2525]

Simon always apologizes for getting stuck, but that's literally the best part of the video. The slow build up of tension followed by the joyous release. I was quite happy to solve this, but I got there by a lot of complicated hi-low parity logic (coloring 1-4 green, 6-9 blue, & 5 gray) that slowly led me to what Simon spotted straight away. I completely missed that 28 in the x was impossible in row 8 and got there the looong way around. I also didn't do sums in the lower left corner. I spotted (once the 5 was placed in box 1) that none of 1234 could be on the white dots in box 1 without breaking the box with at least five low cells, putting 6789 in all those cells. Meaning the run in box 7 couldn't have three high digits in a row without breaking column 1 with five high cells, so the straight had to be descending from 5. So there were two paths to the break-in. Uncharacteristically generous of Phistomefel. That mid-puzzle block though was brutal.

[@eclectichoosier5474]

You know what else has a slow buildup of tension followed by a joyous release? . . . . . . . Morgenstimmung, by Edvard Grieg kzbin.info/www/bejne/g4u1ZaaEi7J9m68

[@rebmcr]

I broke in a different way: • All the X clues in box 4 must be different • r7c2 and r6c4 must then be replicated (in some order) in r4c3+r5c3 • c1 in box 7 cannot contain (123) nor (789) because it would break too many X clues above and to the side - this places the 5 • (46) is now locked into box 7, surrounding the 5 • Our replicated cells now prove which X clue is the (46) - r6c3+r6c4 • (19) is forced into c3 in box 7 • (23) is forced into r7c1+r7c2+r8c1 (because of the replicated cells acting on box 7) - and again the X clues are broken if any of it ends up in r7c2 • Once we have fixed the box 7 sequence as (23456) the rest flows from there, starting with the high-low split in box 1

[@89roddy]

Solving this was a great deal of fun! The way these different sudoku types interact is gorgeous! It is a brilliantly set puzzle! To solve it I used more calculating than just pure logic. For example here my beginnings: Set A as the smaller digit of {r1c1,r1c2}, B as the smaller digit of {r2c1,r2c2}, and R as the smallest digit of the run of five cells from r7c1 to r8c2. Columns 1+2: A+A+1+B+B+1+5+10+10+5+10+R+R+1+R+2+R+3+R+4=90 2*(A+B)+5*R=38 => R must be even R={2,4} => A+B={14,9} r3c3 is 5. Set S as the sum of r1c3+r2c3. Box 1: A+A+1+B+B+1+S+5=45 2*(A+B)+S=33 A+B={14,9} => S={5,15} The maximum of r2c3 is 4. S=5, A+B=14, R=2 => The run is 2+3+4+5+6 and begins at r7c1. The minimum of A and B is 6 . => A+B=6+8 After some sudoku: Because of a 9 in row 2, B=6 and A=8.

[@hendrik9644]

The two rightmost columns at 16:23 can be solved in an easier way by using the consecutive numbers clue. Two columns must contain 10 odd and 8 even numbers. Since we already have the 4, 5 and 6, we only need 9 more odd and 6 more even digits. The white dot cells must contain 6 even and 6 odd digits. Thus the remaining three digits must be odd, i.e. the x clue must contain only odd digits and the 1-2 cell at the top must be a 1.

[@eclectichoosier5474]

Even easier: All but the top 4 cells of the columns are filled with an X (even) and an even number of consecutive pairs, which must have an even total. Therefore, the consecutive triple and the remaining cell must be the same parity. Since he knew that the parity was odd, the remaining cell must be a 1. (That's how I found it, anyway.)

[@RoderickEtheria]

@@eclectichoosier5474 But given the fact that there are 6 adjacent-2 Kropkis, and the adjacent-3 Kropki has an excess of evens, this fact not only solves r1c8 but also limits the X value on c8.

[@joeyw.7131]

I’ve never been afraid of how much a man can shake a stick at, but the way Simon said it at 0:37 makes me concerned for my safety in a way I can’t explain.

[@WheatGrinding]

I was excited to break into the puzzle, and got as far as about the 18 minute mark on Simon's solve...but the fact that there's like 5 different revelations needed to decipher the right side broke my puny brain. Phistomefel wins again

[@clevermorepuzzles4557]

Every time I see that devil in the thumbnail, I know I'm in for a hell of a ride.

[@sherylkoenigsberg5421]

Loved this one. Also love how appreciative Simon is of the quality of this puzzle.

[@PriffyViole]

I decided to try this puzzle along with the video, pausing every time the guy took a step, trying to get as far as I could before I unpaused again. At 27:29, I managed to solve the entire rest of the puzzle on my own.

[@splode7414]

Two restarts, three sets of mistakes and some hair-tearing later, I finished a Phistomefel puzzle for the first time. Thanks to this channel, because before I started watching I wouldn't even have known where to think about the possibility of starting. Superb!

[@GodShiru]

I totally bypassed the "35" logic at the beginning, as I knew that, in box 1, once three of the four V digits and the 5 were used, to have two pairs of consecutive numbers, you had to have 6789, and that insured you could use no more than two of the 6789 digits in column 1 for box 7 as two were already used in box 1, thus, it was 12345, or 23456, and I went from there.

[@grahamrskelly6042]

me too. :)

[@RoderickEtheria]

I was shouting for a long time, "Count the even digits on the right side." If you have 4 and 6, in the 3 digit adjacent Kropki, how many even digits are in columns 8 and 9 due to adjacent Kropki patterns? In column 8 and 9, you must have exactly two 2s, 4s, 6s, and 8s, totalling 8 even digits. How many of these have been accounted for by the Kropki patterns?

[@Qazqi]

I like that a logic a lot more than what I did with noticing in a roundabout way that all of the options for the X in column 8 (i.imgur.com/6At3iBx.png ) force a 5 in r8c8. Simon's was a bit better, but this feels like what was actually intended.

[@SnapDragon128]

Ahhh. Very non-obvious, but definitely the prettiest way to break through the logical speedbump in the middle. Nice!

[@jga5821]

This comments needs more likes. Simple and elegant logic that cracks the puzzle wide open.

[@RoderickEtheria]

@@SnapDragon128 To me it was a very obvious thing to do, given the number of white dots on columns 8 and 9 was so massive compared to the rest of the puzzle. I narrowed down the possibilities of the left side first, but and then all the consecutive Kropki patterns were on the right side except for the 2 connected to Box 8. All of these Kropkis were also quite visibly not connected to other Kropki patterns, except for the 3 digit adjacent one, which I had solved before approaching columns 8 and 9.

[@ChocolateJesii]

It's time for Phistomefel to make a Schrodinger cell sudoku.

[@cappyjing8150]

A trick similar to the one Simon used early on in boxes 4 and 7 can be used in columns 8 and 9, around 21 minutes in. Those columns sum to 90; holding aside the 4, 5, 6 and the X domino leaves us with boxes summing to 65. Twelve of those thirteen boxes make up six white circle dominoes, which sum to 6 x (even + odd) = even. Since the one remaining tile is either a 1 or 2, our choices are limited :).

[@57thorns]

My opening (after the initial 5) was to look at the top left and see that those had to be 6789 because three of 1234 are in the three V cells and the 5 I just had plonked in. This did reduce the possibilities for the 5 long snake (might even have solved it, one hour in before I started to look at the video). The 45 sum Simon does at 9:07 helped me resolve an X box. I think, unless I made more than one mistake along the way. Really fun to find so many clever tricks along the way, but as always, eventuelly, I get stuck.

[@57thorns]

The 45 sum (for X:s and a 5) has two outside crlls. These have to add to the same as the two cells inside the box, and if these to add up to ten, both the X dominoes that goes outside the square have to be the same pair, which is impossible.

[@57thorns]

And after seeing Simon do his full mental arithmetics, I still find my breakin better. For once!

[@RichSmith77]

@@57thorns I did similar for my break-in. Got the 6789 quad in box 1 as you did. Then looked at numbers of high cells v low cells in columns 1 and 2 for the various options of the 5 consecutive digits in box 7. Only found one option that worked. Avoided all the mental arithmetic that Simon used.

[@randomstrategy7679]

I used an alternative break in, using "high digits" (6,7,8,9) and "low digits (1,2,3,4) 1) Place the 5 in box 1 as in the video 2) There are 3 cells with V clues in box 1 which therefore contain three of the four low digits. 3) A low digit is not adjacent to a high digit so the digits in the four cells with white dots have to all be high digits (ie: they form a 6,7,8,9 quadruple) 4) The three digits in row 1 of box 7 have to either all be high digits or all be low digits. Since we have already placed two high digits in row 1 and we can't have five high digits in a column, they must be low digits. 5) We have now placed 4 low digits in column 1 and the 5, so the final two cells of column 1 (c1r4 and c1r5) have to be high digits 6) c1r4 and c1r5 are high digits in X clues, so c2r4 and c2r5 are low digits 7) the X clue in c2r6 - c2r7 contains exactly one high digit and one low digit 8) We have now placed four low digits and three high digits in column 2, so the two remainig slots (r2c8 and r2c9) contain 5 and a high digit 9) the high digit in the 6 in r2c8 while the 5 is in r2c9

[@nortonmalcontent8778]

The easier way that I found to orient the five consecutive numbers in box 7 was to realize that with one 5 sum and half of another, 3 numbers from 1-4 are already used in the box, so the only way for there to be to separate sets of consecutive digits in the upper left of the box was with 6, 7, 8, and 9. Then you can sum all of the givens in the first two columns (the 6, 7, 8, 9; the 5, the ten-sums and the five-sums) to get 70. Since two columns need to sum to 90, that leaves 20 left for the five consecutive numbers, which can only be 2, 3, 4, 5, and 6; and the 5 being forced into the second column orients the direction of the series.

[@charliegrant5061]

New app, coming soon: "Schrödinger Sudoku"

[@clarematthews4051]

I solved a Phistomefel! I didn't expect that miracle to happen. 32 minutes.

[@archdiangelo7930]

So, from what I've seen from these videos, it seems like a Phistomefel puzzle is like a highway with speed bumps. Brief moments of exhilarating speed and progress dotted with several major slowdowns that you have to take carefully or you'll crash and burn (read: screw up the puzzle)

[@paulcook2961]

I found a different, possibly faster opening: Consider the parity of columns 1&2 together. Note that white-dot dominos have an odd sum. Add everything up and you see that the box7 pentomino must start and end with even numbers. So it's 23456 or 87654. Now take the sum of columns 1&2: you find that the two box1 white dominos must be 2378 or 6789. Then notice 2378 breaks the two V dominos. So it must be 6789 and box7 must be 23456. A similar trick can be applied at 14:14 looking at the parity of columns 8&9 together to get R1C8 = 1.

[@paulcook2961]

I was super pleased to finish this in 35 min (I'm hardly ever anywhere near Simon's time). I feel my logic path was neater and more efficient yet I'm still a few minutes slower than him (even allowing him time to explain everything). Coming back to watch the video, I see the main reason for this is that what he achieves in his last 4 minutes (i.e. all the tidying up and applying "normal" sudoku techniques) takes me 10+ minutes.

[@fbrume]

Horrible time (45:48) because I messed up what I thought was a really bright bit of logic: After I got 456 in the upper right, I figured out the parity of R1C8. (15+6 Kripkes+ an X clue = an odd number, and the sum of the two rightmost columns are even.) (My first pass I mistakenly thought it was even, instead of odd.) Having the 12 pair sorted out helped a lot.

[@eclectichoosier5474]

Have to admit: I did the exact same thing.

[@stevenape377]

I worked out the left two columns differently than Simon. I used low/high colouring instead of arithmetic. It's a beautiful puzzle and a beautiful solve!

[@dcay5424]

I really do enjoy watching both of you solving these hard puzzles. I try solving these for myself sometimes and most of the time i manage but with some struggle, especially with Phistomefels puzzles. I wanted to add something i spotted which might have been a somewhat easier way to solve for the end. When you had the 1-2 pair in box three, it could have been solved by using some geometry using column 8 and 9 since all the cells in those two columns contain either an X or a white dot. The only one not connecting to a dot or an X is R1C8. By adding the parity of all the white cells (plus adding the X=10), it could be deduced that R1C8 had to be an odd digit, therefore it had to be 1.

[@piarittersporn]

The 5s made some things clear very quickly, but then I had a long period of recurring errors. After I finally discovered this, things got better again. Overall it took me triple Simon's time at least, but I really enjoyed it.

[@anibotic]

42:45....I'm very happy to be able to solve a phistomefel puzzle all by myself without looking the video before. The start of the puzzle is very great but after the 45 set in column 8 n 9, it really comes at a stand still...I was only able to finish it assuming 8 in box 8 which led to put 1 n 3 both in same cell in column 9 which disambiguated the 45 set n the puzzle opened up completely...

[@futurefox128]

There is actually a clever way finding the 46 pair in the r6-"X" much earlier (it doesn't help progression much though). We can do it from the position @15:12 in the video. In box 4: We are still missing 4&6 which can only go in c3. We have a 5 and two Xs already in the box... so we can think about it as needing 2 more ten-combinations to make the box sum to 45. 4&6 is one of those combos but r6c2 also needs a complementary partner in the grid. However, if it's next to it in the second X clue (r6c3) that X clue would break as r6c2 & r6c4 would be identical. So we are left with 46 which gives as the 46 pair.

[@Coyotek4]

First attempt: spent roughly 35 minutes on the puzzle, ended up breaking, gave up. Second attempt: armed with previous knowledge, I clocked in a time of 33:09 ... but that included some bifurcation passed the halfway mark, and I never deduced the 'uniqueness' of the solution. Tough puzzle!

[@PerStarback]

Yet another way to think about this after placing r3c3=5: Box 1 has 3 of the 4 low digits in the V'd cells, so does only have 1 more. That can't go in any of the o'd cells because then it would need to be with another low digit. So the four o'd cells are 6789. Now count the sum of c1+c2. It is 30+5+10+10+5+10 + the 5-cell ladder in box 7, so the ladder has sum 20, i.e. 23456.

[@RCPlanes59]

I'm only partway into the video but I think you can solve this without either black dot. At 13:55 you are left with 1246 in row 1 and 2469 in box 2, but the 4 can't be in row 3 because of the 14 pair. So 4 must go in r1c5. Further, 1 and 2 can't be consecutive to any of 4, 5, or 6, which places the 6 in r1c9 which gets you to the state at 14:16. I can see why he'd leave some black dots in, but just an interesting piece of logic!

[@neilramsey1368]

Another great puzzle from a genius setter and great solver

[@rdbasha5184]

I got to where simon is at 12:30 in a different (simpler?) way. On the top-left box we have a 5, and three 1-2-3-4 cells, and two pairs of white-dot cells. It you look at it, youll find that 1,2,3,4 cannot appear in the white-dot cells (no one to pair them with), so we get a 1-2-3-4 quadruplet. thus, the upper-left cells are 6-7-8-9, which forces the configuration in the bottom-left box, as we cannot use more than two numbers from 6-7-8-9

[@Da-mD]

13:20 the moment I realized that the white dot in r12c2 does nothing, I knew Phistomefel wasn't focused when he constructed that pazzle

[@danielrhymer1762]

37:44 for me. Preferred not having the anti-constraint because so often I forget about it anyway 😂

[@alienrenders]

1:25:41 while taking a lunch break. Really liked this one. I solved it on my own using only logical deductions which I'm happy about. edit: Very strange.... I had a completely different break-in. I colored even odd pairs. If you notice the two rows and columns around the grid, there are LOTS of white dots and X's. Any single white dot means there's one even and one odd number. Pairs on X's are the same parity. V's are different parities just like white dots. The black dots are more ambiguous, but must have at least one odd. In the two right column, you can see there are 6 white dots that form pairs and a triple with 2 white dots. So we know there are at least 7 even numbers in those. Since X pairs must be the same parity, it cannot be even because that would give 9 even numbers and there are only 8 even numbers in two columns. So the X pair is odd. You can do the same thing on the left side and eventually determine the 23456 sequence. You can at least deduce that the ends of the sequence are even right away. 45678 is easy to eliminate. edit2: Yeah, after watching the whole video, ALL the logic Simon does on the rightmost two columns can be explained visually if he colored in the parities (even/odd cells). He wouldn't even need to explain anything. The colouring would do all the talking. There'd be no need for "if 3 is here, you can't use 2 or 4 because the 5 needs the 4 and the 2 needs to be paired with something else...". The colouring just explains it much more clearly.

[@BigAsciiHappyStar]

Don't bother colouring in. Just enter repeated 1's and 2's to represent odd/even digits. It violates the laws of Sudoku but you can always delete them later.

[@daijutsi]

when you keep saying 5 is useless and then a puzzle comes where 5 is the key to the solution, not just once, but several times.

[@nityantagarwal2532]

Where you got stuck at about 24 minutes in the video, you could've combined the last two columns of the grid. We know that there are 4 evens in any row/column, so there will be 8 evens in 2 columns and each consecutive pair has to have exactly 1 even and 1 odd. so by counting the number of evens, we can easily get that there will be no more evens in the 2 columns which would give us a 1 at r1c8, and also the 37 x pair.

[@thesolarfutureenthusiast1102]

I was right with you up to 20 mins, even a little ahead and feeling good about how well I was doing at a Phistomefel. Then totally stuck, had to come see how it was done. I did have a different break in to the puzzle. Place the 5 in box 1 just as you did. Then you have 3 cells in box 1 containing 1234 and one containing a 5 so only one digit left under 6. Therefore the consecutive numbers are 67 & 89. Although just 6789 in the cells. You now must have a 6 or 7 in column 1 in box 1 so the consecutive numbers in box 7 must increase 1-5 or 2-6. I think I ruled the 1-5 out by using the X's as you did.

[@uy-ge3dm]

21:40 Consider the parity of the sum of columns 8 and 9. Every white dot pair is odd and the X is even, and we can add in the 6+5+4=15 at the top. We find that the sum of all the squares other than r1c8 is odd, so r1c8 must be odd to make the columns sum to the even total of 90.

[@TheGerkuman]

Once again, we have been Phist-ed

[@SienAppelsien]

I needed a little push to be able to do this one because i never think of the maths business, but once i got properly started i could figure out the rest

[@thumper8684]

That went remarkably smoothly for a Phistomefel puzzle! Glad I caught the odd-even parity check half way through. 33:38

[@thumper8684]

Maybe Simon spotted it eventually but I can't hold back. The sum of digits in columns 8 an 9 must be even. There are 7 2-cell consecutive pairs, a 10-cage and a 4-5-6 sequence. These must sum to an odd number. That tell you r1c8 is odd making it a 1. That puts a 2 in row 9. Phistomofel uses this trick precisely once in the puzzle.

[@alonamaloh]

The last two columns contain lots of white dots, mostly in separate dominos. Each of those dominos contains an even number and an odd number. Once you have the 4-5-6 sequence in the top right corner, you can see that all other even numbers are contained in dominos with white dots, which means that r1c8, r6c8 and r7c8 are all odd. That could have simplified the logic somewhat.

[@Raven-Creations]

Another Phistomefel corker. Did you notice there was a 5 right in the centre? I'm inclined to agree that this was a dig at Simon. I got off to a quick start using arithmetic on the first two columns to get the 23456 and 6789, but like Simon, I got a bit bogged down in the mid-game. The interplay of the consecutive pairs was exquisite. The end game was a bit of a cruise, which is unusual for Phistomefel. I must say that we've been treated to some absolutely stunning puzzles over the last few weeks. I'm in complete awe of the setters. I'm sure I'm not alone in wanting to see more videos where the setters explain their approach, not necessarily a video of them setting a puzzle - an interview would be great.

[@smahoney1212]

It would be interesting to see Mark do a Phistomefel puzzle.

[@matthewender3473]

Interestingly different break-in than I used on box 7 -- I missed the application of sums, but harnessed the two white dots in box 1 instead to get the start in the chain in box 7. I got about as far as Simon did by the 20m mark in the video, but bogged down and needed to brute force a step to proceed. At least that step took him a while to logic out, so I don't need to feel *too* bad about not figuring it.

[@alibabapirce9782]

small tip. the part from 31:47 could be done much sooner. if you look at box 4 you have 2x and than 2x on edge. when you got 4 and 6 in row 1 and 2 you can see that 4 or 6 has to go in r3c6 because otherwise you cant make 4x10

[@sarahr.1076]

Phistomefel didn't want to be mean about the 5s for sure. Look, he even wanted to help Simon, because there is no negative constraint. Phistomefel knew, that Simon likes to forget them :'D

[@LimitedRegicide]

"You don't get Schrodinger cells in Sudoku. At least not in this Sudoku." So, countdown to when we get a video involving that has already started, right?

[@samus88]

LOL, he rules out 2 and 3 from Box 1 because the column already contains 2 and 3, but doesn't rule out 4 even tho it's also in the same column. I love how sometime we just don't notice the most obvious things.

[@JalebJay]

21:30 you can use parity of white dots always being odd (ignoring the given of 456 already known) and the only cell not part of an X or dot is R1C8. Count the modulus and you get that it has to be 1

[@FirstLast-gw5mg]

There is an easier way to get the initial run of digits in box 7. Coloring the high/low (5-9 and 1-4) digits will help. Because there is already a 5 in box 1, the white dots must contain *either* two consecutive high digits (larger than 5) or two consecutive low digits (lower than 5). But there are V clues affecting 3 of the cells in box 1, which means that 3 cells must be in the range 1-4, so making either of the white dots contain two more low digits would break the puzzle because you can't have five 1234 cells in the box. So they must be high; one of the white dots must be 6-7 and the other is 8-9. Either way, we now know that the first column contains two digits in the range 6-9 and also the given 5 in box 4, so three digits in the range 5-9. We can place, at most, two more high digits in that column. We know that the sequence of white dots in box 7 must contain a 5 in column 2, so either the cells in column 1 of that box are all high (6, 7, 8 or 7, 8, 9) or they are all low (1, 2, 3 or 2, 3, 4). But we can only put two high digits in the column. So all 3 of them must be low digits. But there's already 1 low digit in the column, on the V clue in box 1, so we have a 1234 quadruple, which means that both of the X clues in box 4 now must have high digits in column 1. If they have high digits in column 1, the digit on the other side of the X must be low. There's also a low digit in column 2 from the V clue in box 1, and there's an X clue in the column which must contain one low and one high digit. That gives us a total of 4 low digits in the column, giving us a 1234 quadruple in that column as well. That forces both of the cells in column 2 of the white dot sequence to be high digits, and they must connect to 3 low digits in column 1. So the sequence is 1, 2, 3, 4, 5, 6, with the 1, 2, 3 in column 1 and the 5, 6 in column 2.

[@FirstLast-gw5mg]

Unfortunately, to get that far, I needed Simon to jog my solving skills by deducing the position of the 5 in column 2 which fixed a 5 in box 1. And I was quite pleased to find the 4578 quadruple in row 8 which left a naked single 2 in the middle before continuing the video to see how Simon found that, but promptly found myself stuck again, so back to the video I go now...

[@solaciasolis669]

Because of the Vs and the 5 the cells with white dots in box 1 cannot contain low digits. So they add up to 30. Using that with the X clues in column 1 and 2 made it easier to find the values of the cell with white dots in box 7. I think I didn't use the black dots at all

[@jetzeschaafsma1211]

There's an elegance to the negative constraint. This puzzle does not have it.

[@alkabur1079]

29:12 to find r1c8 *just think only for c8 and c9* all digits add up to 90 which is even digit 2 consecutive digits add up to a odd digit. if you add all white dot pairs below row 2 and the X pair which is even (10) you get even number. you also got 4-5-6. add them too and you got an odd number in total. we got last 1 cell to add (r1c8) to our odd total to get even total. so it must be odd. so it is 1

[@Da6moose6]

Come for the logic. Stay for the bobbins.

[@michaelhoffman2443]

I did have to watch the video for a hint... using the dominos on the right to group... i was doing so well until then... great puzzle as always Phistomefel

[@cheakmate401]

We should make a broken puzzle. All the testers should say its a bit hard but solvable and then send it to phistomefel.... REVENGE...

[@CCHdennis1]

It can solve the consecutive in box 7 without using 35 clues. In box 1, there are 3 cells in V clues. which means they are 1234, and 5 is given. therefore, the 2 consecutive dots could only be 6789, because it will be repeated if it includes 1234. Then we know that C1R789 can't be 6789 because it will have 5 cells could only be 6789 then, and it is impossible. then C1R789 must be 1234. Then as C1R45 is 6789, and they lies on X clues, therefore, the cell nearby are 1234. Then if we look at C2, C2R234 must be 1234, and the last 1234 possibility is lying on the X clues. Therefore C2R9 can't be 4 as it will break. Therefore, the consecutive clues in box 7 must be 2,3,4,5,6. The tricks in C8C9 is AWESOME.

[@shadout]

18:13 gives Phistomefel crazy new ideas for the next puzzle

[@TheGreatPurpleFerret]

The only thing I really spotted well before Simon was that after the majority of the left two columns are done, the X in r3c34 can't be 28 37 because of c3, and it can't be 19 because it ends up excluding a 1 from box 7. Not sure why that jumped out at me.

[@TheGreatPurpleFerret]

This would be starting around 15:10 and it doesn't really help much lol

[@DipanGhosh]

More cleverness than you can shake a stick at... I am just laughing imagining shaking a stick at cleverness, while looking like a complete idiot.

[@vince9565]

And at the start if you shade in all the squares with no constraints .... thats the cup you win for solving it ....

[@bertbergers9171]

Hour and a half and three times needing to see a break from Simon. I am tired so i will excuse myself :P

[@carljohanr]

Upper left white dots had to be 6789 a lot earlier to avoid some of the roundabout logic.

[@alarictaqiarentza9451]

That logic on 26:11 was hard. Manage to solve it until that point but then have no idea what to do next. But that was a very beautiful puzzle and very brilliant solved!

[@Majoris239]

"You don't get Schrödinger's cells in sudoku...at least not this sudoku"...meanwhile, Phistomefel is rushing to grab his pen to write that down

[@victorolosaurus]

so... apparently there was no negative constraint? damn

[@shantanunene4389]

Very nice puzzle. Took 55 mins. Had to use some clever parity tricks to break in

[@wolffyhowl5622]

I have no clue what's happening but I love watching it anyways

[@mikepictor]

27:00 boy I needed that help, everything just unfurls after that

[@alvaropallete]

2:34:12 I started with the low- high parity, and then in the middle of the puzzle, I had to stare a lot, not being able to put more digits (for like an hour) till I saw the constraints of 1 and 9 with white dots (1 forces 2; 9 forces 8). Surely a long way to go.

[@DaShikuXI]

Right from the start you can conclude that r7c2 and r6c4 have to be the same digits as r4c3 and r5c3. In the final solution they ended up being the numbers 8 and 4. Not sure how useful this would be for solving the puzzle. It's just something I noticed. To explain: box 4 has four 'X' clues given, a total of 8 numbers with 2 outside of the box. The six numbers that are in the box + the 5 thats already given mean 7/9 numbers are accounted for. The remaining two cells cannot also be 5, therefore they must be the same as two of the digits in the X clues. The only way this is possible without repeating a number is if they are the same as the two cells outside box 4. Therefore r7c2 and r6c4 = r4c3 and r5c3

[@AfonsodelCB]

33:09 how did he write a 6 there "by sudoku"? after looking at it a while I could only see the 6 be forced inplace because of the 3 by sudoku and the 2 by sudoku after that. Is that what went through his mind? didn't sound like it at all...

[@laurv8370]

12:40 That four went like "Am I a joke to you???" :D

[@jdkemsley7628]

The lefthand X dominos all must be unique from each other, which would have gotten you a lot further early on

[@Kelarys]

I just want to say thank you for including the classic devil again, I know it's been like that for a bit but I still remember when you stopped including it. Maybe that one was just an accident, or maybe you saw my comment and changed it back. Regardless, I'm happy it's there.

[@gustavostabe2490]

Watching these videos makes me think rocket science has got to be easier then this...

[@jamesrempel8522]

At about the 24:00 mark I did a bit of logic that I think is interesting. Consider the parity of C8C9. We know that any consecutive domino is odd, and there are 6 such dominoes, resulting in an even total. The X domino is even, leaving only the 2x2 block in the NE corner. The 3-cell 4-5-6 chain is odd, meaning that R1C8 must be odd to make an even total for C8C9 (i.e. 90).

[@spubbler]

More cleverness than you can shake a stick at -- how much cleverness is that? Well it largely depends on your stick-shaking abilities

[@johnlorenz377]

Needed Simon's help. Just not disciplined enough. Was working through the 1 & 2 logic in box 8 and R8 and figured out the one and placed the 2 but because I started with the 1 jumped to the incorrect conclusion that the domino in box 8 was 2,1. Spent way too much time proving this was wrong, Way too many ways to break a puzzle. Once I watched the video and was reminded that 3 was an option was able to complete the solve.

[@jacksonpineau6387]

18:14 that sounds like a challenge for Phistomephel!

[@davidebellomo358]

Took me 30 min to put the first digit... and after that another 15 to go further! but this is just a great puzzle!