No Given Digits in this one!

Normal sudoku rules apply; wherever neighbouring digits are consecutive, a bar is shown; the two green cells sum to more than 10.

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Play the puzzle in the video by clicking the link under the video (above). We are building a website which will allow you to enter your own sudoku puzzles into the software and this is coming soon!

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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

[trombonedavid]

“I hate myself sometimes”
Don’t be too hard on yourself! Everybody makes mistakes. You’re an incredible solver and your viewers are always impressed by your speed in these puzzles. :) At least I know I always am.

[MisterNohbdy]

Comments a couple months ago: "At this rate they'll be solving puzzles with no given digits soon!"
 
Comments now: "Oh, another blank puzzle?"

[tone]

It's amazing to me how often parity-coloring pops up in variants, even when they nominally have nothing to do with parity.

Another way to work out the central box: Once you had everything down to two options, r3c5 sees two cells with 6/8, one of which it's consecutive with and the other it's not. That means they must be different, which means one of them must be a 6, and since they both see r5c4, that must be a 4, unwinding the box.

[victormanjarinsala]

Amazing. I couldn't even understand where to start. The way Mark approached this left me stunned. OF COURSE it was the fuken parities! And then, the speed is overwhelming. Again, amazing.

[jrparker]

It is amazing that you can take 2 very different paths to arrive at the solution. Mark had a completely different set of cells empty toward the end than I did.

[Afterthoughtbtw]

I may be wrong, but I think you got lucky - row 6 column 4 should have been labelled 79 rather than 17, which would have invalidated your breaking in logic.

Yup: for people wondering about missing logic to proceed in the puzzle then look at R4C3.

Out of the possible values of 468, you can quickly deduce it can't be 4. If it's a 4 then R4C4 must be a 1 which means that R6C4 must be a 7 and thus R5C4 must also be a 4. However, R5C4 and R4C3 each border R5C3, and only 1 of them is allowed to be consecutive to it. Hence a contradiction, and R4C3 cannot be a 4.
So, R4C3 must be a 6 or 8. Now take a look what happens if it is a 6. The exact same problem! R4C6 this time is an 8, forcing a 6 into R4C5. So again, you are left with R5C4 and R4C3 being the same digit, but only one being allowed to be consecutive to R5C3.
Therefore R4C3 is an 8.

[flsal]

5:05 Explanation: if it is 3, 5 or 7, they would be two consecutive numbers next to the cell, you would see two consecutive markers. Or there is only one. That’s why it’s only 1 or 9, those are the only two digits that has only one consecutive number.

[AJ-dwlw]

"No Given Digits in this one!" Finally a clickbait I can get behind lol

[victora.colonna]

Lately I've been loading the puzzle, staring at it for 5 minutes with no ideas, watching Mark solve for 5 minutes, going, "Ok, I got the idea now, " and then solving the puzzle in 30-40 minutes. This one was fun, I kept on with the "odd-even" coloring and that let me get a lot of clues quick.

[gordonglenn]

I really wanted to like this one, but... Having to do the "what if" with the 4s in the middle box so early on just felt like a defeat for logic. I know that there is a time and a place for "following the trail, " but it usually feels like a last-ditch technique, and is unsatisfying.

[lauramcastro]

I did such a turnabout to fill in the centerbox, but I am so proud I managed by myself!! And I used a lot of logical thinking I learned watching other Mark solves with parity puzzles!

I noticed by accident that C7, R4 and R6 couldn't be evens; no matter how the central box looked like, all evens got cancelled out of those. So I did some parity-playing around the central box, counted numbers of odd and even cells on lines, backtracked a couple times, but I did it!!!

[yunghei]

there's a mistake at 7:44, but after combing through the comments, seems like others have already spotted it. r6c4 is 79 instead of 17, but luck prevailed. so... yay?

[OlafDoschke]

10:30 Needing a 7 in the left lower cell is a wrong conclusion from wrong pencil-marks, a 9 there would solve that.
It would have helped to pencil-mark the two different solutions of the center box in different notations, as putting both in the center pencil mark makes it quite impossible to follow which of them belong to which alternative,
I find the second central solution with a 1 in the center not needing a 1 or 7 in the lower-left corner, but a 9:
385
612
947
So your 17 pencil mark is wrong, that has to be 79. The cell above the 3 fails to have an even digit solution, that's the weak point. We know that's necessary from the parity logic of box 1. No even digit works in the cell above the 3. 8 is ruled out by the 9 or 7 in the green cell, 6 by the column below, and both 2 and 4 by the 3 itself.
So the even/odd parity was still an important role, not only in the center. You're a bit lucky picking the right alternative for the wrong reason, but the wrong alternative would have turned out impossible very fast.

[nadeemhajiiqbal]

The central box can be resolved at 9:30, when we discover both cells marked 68 have to be different.
The two 68s are BOTH seen by R5C4, and that rules out 6 from that cell.

[manudude]

The logic at 10:12 works, I resolved it a different way though. r3c4, r4c3, r4c5 and r5c4 are clearly at least 2 different even numbers, but r4c5 can't be the same as r3c4 and r5c4 so it is 3 different evens (i.e 4, 6, 8) therefore r4c4 must be a 1 as it is adjacent to all those previously mentioned squares.

[jvcmarc]

10:55 wrong logic, but luckily correct result, the (17) you used to determine that the r3c5 had to be 5 was incorrectly marked, it should be (79) which wouldn't let you deduce the same
however, having a nine in r3c5 would collapse the center box and made so the r3c4 would have no possible values
that's how I solved that part, and I believed that was the intended solution

[Blindpig]

“Business life goes on, sorry...”. I hate it when work gets in the way of Sudoku...

[BigAsciiHappyStar]

I think the most elegant way to begin is: stick a 1 in the middle of the grid. If it's wrong the only possible error could be the two green digits sum to less than 10. But then we replace every digit X with 10-X and the world is again a happy place :)

Another fantastic solve as always :)

[Martykun]

I tried this myself. I am quite happy that I finished it without bifurcating, but it certainly was a pyrrhic victory.
Took me about 7 hours, not counting breaks.

[Coyotek]

Took a 25-minute period that ended in a break, followed by a restart … but once I got rolling, I got there without much further issue. Nice puzzle.