A Sandwich Sudoku That Took Us 2 Hours!!

You can try Christoph Seeliger's brilliant and brutal puzzle here:
The puzzle appeared on Logic Masters Germany here:
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▶ABOUT US◀
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 eleven-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

[Kaiser]

4 weeks ago, I wasn't interested in sudoku. I was randomly suggested one of your videos, and now I'm sitting here in awe admiring this incredible puzzle for 50 minutes.

Thank you for kindling my interest in sudoku.

[SgtElevn]

Im gonna try this, see you guys in 2021

[vikingslayer]

You know this puzzle is gangster when it takes 50 minutes just to EXPLAIN the logic. I commend anyone who got this on their own. Hats off to you!

[yahya]

Oh my. A 50 minute video...that’s awesome and it also implies that I can’t do the puzzle, haha
Edit: wow, just wow. Amazing solve and beautiful logic. Worth every minute

[Deathranger]

Cracking the Cryptic on December 6th: “This is the hardest sudoku variant we’ve ever seen!”

Cracking the cryptic a month later: “But wait, there’s more!”



That being said, this was a great video. Enjoyed it a lot, and I appreciate the amount of time that went in from both the setter and the solver. Occasional ones like this would be wonderful.

[Rangar]

There's a word for this type of puzzle: humbling. This has some incredible construction.

[ItsSansom]

11:42 Sounds a lot like the sound I made when I first saw this puzzle

[kinnaruto]

If it takes you THAT long to solve.... I'm not gonna try. Love watching you do it tho!

[DanielFoerstner]

maybe have an interview with Christoph and have him explain his method of being a setter.

[mightymaniac]

Great explanation. I understand it even though I would never even consider approaching a puzzle of such difficulty

[timdunkley]

This was brilliant, Christoph. A tad over four hours for me. We love these brutal puzzles. Simon, go back in your Tardis and do it live.

[Socialdogma]

Simon’s joy about deducing the H clue at around 18:00 is like when I find an old $20 bill in my pants pocket that I didn’t know was there.

[hypnotized]

I wish if there was a page in ur web software that we can browse all the puzzles that were shown in the channel

[tomcheuk]

I know this puzzle is an year old on the channel. But I found an alternate logic after 27:18 when Simon got the 1's locked in box 5. Here is how.

If H=9, then the sandwich clue in row 8 forces C1R8 to be a 1. Now, look at where a 9 can go in R9. 9 cannot go on C4R9, C6R9 because there is a H=9 in the box. 9 cannot go on C7R9, C8R9 because there is no way to place a 1 that will satisfy the sandwich clue in row 9. Since the sandwich clue is at least 20, if 9 is on either C7R9, C8R9, then 1 has to be either on C2R9 or C4R9 but 1 can't be on C2R9 because of the 1 in the box and C4R9 can't be 1 because of the 1 in box 5. So, 9 can only goes on C2R9.

On the other hand if H is not a 9, then G has to be a 9. In either cases, the 9 in box 7 have to be locked in C2. Next, the 9 in box 1 is locked in C3 (either on C3R2 or C3R3). Since the sandwich clue on C3 is a 0, it forces C3R3 to be either 1 or a 9. So, there is a 19 pair on R3 (C3R3 and C9R3). Hence, C2R3 can't be a 1.

Moreover, because of the sandwich clue in C2, C2R7 can't be a 1 (too close to 9) and C2R1 can't be a 1 as well (too far away to 9) . And so the 1 in box 4 has to be locked in C2R4 and C2R6. That will give the same outcome as Simon does on 30:50.

[JohnSmith-kcov]

A quick tip to help add consecutive digits
Take the middle one, and multiply by the number of digits
4+5+6+7+8 = 6+6+6+6+6 = 6×5 = 30

[reaperskill]

Took me just over two hours. The logic is... intricate. Had to think at least 7 steps ahead at several points.

[the_cogito]

So fairly early into the puzzle I think it's possible to deduce that H must be 9. If the 1 and 9 are in the same box in the far-right column, it prevents the FI column from being multiple digits. Therefore the 1 and 9 in the far-right column must be in different boxes, and the only way that's possible is if the middle box, far-right column is 2+3+4=9, giving the minimum value for H. Not sure if that speeds the solve up that much, but just something I noticed.

[Phantom]

There is a 2nd way of eliminating F = 3, that I ended up using because I didn't see the much easier FJ clue. So here is an alternative and it is beautiful in its complexity, and I was in awe until I watched the video. Column 8 FI = 35 puts 19 in r1c8 and r9c8 as well as 5 in r8c8. FE in row 1 is either 31 or 32 and puts a 19 in r1c1. Column 1 C, puts a 19 in either r3c1, r4c1, or r5c1. We can eliminate r5c1, since that would make C = 9, and place a 9 in r2c8, but we already have a 19 pair in that column. Now we can disambiguate A and E. If E = 1 and A = 2, then r5 = 21. If you go left of E in r5, that puts the only possible 19 in r5c1 but we just eliminated that possibility, and if you go right, you can't put a 19 in the row since F = 3 and max there is 18, therefore E cannot = 1 and A cannot = 2. Therefore E = 2, and A = 1. A = 1 makes r1c1 = 9, r1c8 = 1, r9c8 = 9, eliminates 19 from r3c1 since we already have a 19 in that box, so only possiblity left is 1 in r4c1. Now we can use the F? column 2 clue to ask if A = 1 where does 9 go in the column? F? >= 30 puts 9 in either r8c2 or r9c2. r9c2 is impossible since we already have a 9 in the row at r9c8. Therefore G = 9 at r8c2. Now use the F? clue on row 8. Where can we put a 1 on row 8 if G = 9? Only at r8c9 since I = 5, so r8c8 is not a possibility. Now where can we put a 9 in c9? Well H <= 8, and > 3, since G = 9, therefore only 2 locations for a 9 is r5c9 and r6c9, and we can prove that both are impossible therefore resolving that F != 3. for r5 we know AE = 12, so where do we put the 1 in that row? Well is c6 possible? No since biggest you can make c7/c8 = 11 (8 + 3). c5 is not possible since E = 2, so how about c4? Nope since even 4 squares at its smallest would be 2 + 3 + 4 + 5 = 14, therefore a 1 is impossible to put in r5 if 9 is at r5c9, and therefore there can't be a 9 in r5c9. Now for r6c9 we now have to look at c3 J clue = 0. Where does the 1 and 9 go in column 3? Well we have 1s at r2c2 and r4c1, therefore the 1 has to be in the bottom box. We have r1c1 = 9 and r8c2 = 9 therefore 9 must go in the middle box in column 3. How can we make J = 0? Well only possibility is if r6c3 = 9 and r7c3 = 1. But now we have 2 9s in r6 at r6c3 and r6c9, so this can't be true, and there is now nowhere to put the 9 in c9. Therefore proving that F != 3. Therefore if you wanted to make an even harder version of this puzzle and force this logic you could easily do so by changing the c7 FJ clue to a (* + F) clue and put * in r5c1.

[DavidNayer]

Lovely to watch. Thank you for including this puzzle. I value the novelty.

[lakarto]

Brutal, brutal puzzle... Took me multiple sessions (probably several hours combined) to solve, and once again, I'm just pleased that I managed to solve it without watching the video first.


I used a slightly different pathway in the initial 1/9 filling part; after figuring out the initial 0/1/2/3, it is possible to deduce, using very complex sandwich chains, that the 9 out of the A-I has to be either B or H (i.e. in column 5). From there, it requires slightly less complicated chaining to figure out B cannot be 9, and thus H has to be 9, r1c5 has to be 1, B has to be 7 or 8... and so on. I think the logic I used from thereon and throughout the sandwich part was almost identical to the video solve.
But I have to admit, the video solve was much more elegant and impressive... the logic used to rule out a 1/9 from r9c9 was simply stunning. (17:05 in the video)