Irregular Frame Sudoku

[@2001herne]

"Sorry if I'm not solving it as efficiently as you are" - I have no hope in hell of solving these on my own, my mind just doesn't work that way. But seeing how your mind works is really interesting, which is why I watch these videos.

[@ragnkja]

I used four colours for clarity: blue for horizontal sum, yellow for vertical sum, green for both, and red for neither.

[@Vearru]

Nillie I also used 4 colours except I went with blue for vertical, red for horizontal, purple for both, and green for neither

[@msscoventry]

Yellow and blue. Then green for both.

[@gizel4376]

i used blue for vertical, yellow for horizontal, green for both.

[@mbakenemdusink9757]

@@gizel4376 Anybody interested in the colors I prefered...? :-)

[@12tone]

This was a really fun one, but I think your sandwich sudoku experience may have actually hurt you here: You were focused for a long time on whether squares could be in or out, but the crux of the problem was in the fact that squares in the corner boxes could be both simultaneously. All the corners were relatively vulnerable to attack by comparing the values of the two sets of outies, and once you have those the board opens up pretty quickly. I was definitely worried when I saw the video was over half an hour long, but I think the fact that I'm not as good as Simon may have actually worked to my advantage here 'cause I wasn't as diligent in chasing down the initial red herring.

[@samslack4885]

I started on the bottom right. Summing vertically, 6+10+11=27, leaving 18 for the remainder cells not in the vertical frames - 18 requires 3 cell sums, so must be the entire top row. Top row is either 9+x or 9. Obviously can't be 9, so the top left of the bottom right is a 9. Seems like you can do a similar trick with the bottom left, but have to be a bit careful. 8+6+17=31, so the remaining squares not in the horizontal frames are 14. This *could* be the entire third column, or just the top 2 cells of the third column. If its the entire third column, obviously the top right square is 4. If its just the top 2 cells of the third column, theres a few options.. but run through the options (5-9-1, 6-8-2, 8-6-4) and it's easy to find a contradiction with the assumed three square horizontal 17. Fun puzzle. Liked the variety of starts.

[@Tehom1]

Nobody's pointed this entry point out yet: The lower left square. 8+6+17=31, so you need 14 more and they have to all come from column 3. But column 3's vertical subtotal is 10, so it can't be a 3-cell column, and the remaining cell r7c3 must be 4.

[@Zerotan]

Ditto the bottom right square, r7c7 must be 9.

[@paulwatson746]

Try lower right square, total of horizontals is 30 leaving 15, but lh column is only 6 so tl cell in lr square must be 9 and the two cells below sum to 6, all 3 horizontal sums are 2 cell, Top row must therefore sum to 18, but the vertical sums are 27, leaving 18, so the vertical sums are also 2 cell. applying standard Sudoku logic gets you down to pairs in each cell. Now go to top right (a bit harder) and work round the corner squares. the pairs start to cancel out as you go round.

[@renedekker9806]

Actually... the remaining cell r7c3 does not need to be 4. There is no guarantee that the 14 covers all three of the column 3 cells. Column 3 could be 8, 6, 4, for example. The same logic does work in box 9, though. The vertical clues 6, 10, and 11 add up to 27. So there is 18 missing, which is larger than 9+8, so it must cover all three cells of row 7. Therefor the 9 clue on row 7 puts a 9 into r7c7.

[@Tehom1]

@@renedekker9806 No, you did not understand. The 14 sum does not cover all three of the column 3 cells, by the reasoning I gave. It covers two of them.

[@renedekker9806]

@@Tehom1 I still don't understand how you can get to the conclusion that r7c3 must be 4. It can only be 4 if the 14 sum covers all three cells of column 3 (which it does in the final solution, but you cannot draw that conclusion at that time).

[@Pyromonkey83]

This took me about an hour with just some trotting around the grid focusing on row logic, and it was solvable. I then watched the video expecting you to do a similar take, but instead found that you went about the solve in a COMPLETELY different manner. I really, really enjoyed this solve from you as it taught me a new way to look at box logic in these types of puzzles, well done! Thanks as always for showing me a new way to look at a sudoku grid.

[@zwaffel]

you not solving the 2,4 really grinds my gears

[@Wecoc1]

31:16 He pointing the only 5 in the box and not solving it rubs my ribs

[@CrackingTheCryptic]

Lol. I've just watched that bit back. Sorry. I know it must be frustrating but I'm trying to do a lot of things at the same time so seeing what's staring me in the face is clearly asking too much ;)

[@glennmelven3414]

@@CrackingTheCryptic I am waiting for you to do a guest solve where you and/or Mark comment on all of the obvious numbers the guest misses. Or would this video be to painful for you to make?

[@zwaffel]

@@Wecoc1 yea that also got me

[@zwaffel]

@@CrackingTheCryptic it's still fun to watch and it's really not uncommon to not see things when you're focused

[@RangeWilson]

Under an hour despite one screw-up, I'm happy with that. The 23 constraint on the right caught my eye right away, then the shapes in the bottom right. Getting the whole right side was pretty easy, and the logic flowed slowly but steadily from there.

[@Darkness3827]

In the top center 3x3 box when you got to the 589 triple you were talking about how only one cell could not be in the frame, and that one cell must have been an 8. You also said that either way whichever one of collumn 4 or 5 had a 3 cell frame would have to have a 5 in row 3. You missed that this meant there was no way a 9 could have been in either of those squares and therefore must have been in collumn 6 of row 3 (and therefore collumn 4 & 5 of row 3 was a 58 pair). I think if you had found that it would have saved some trouble with that 3x3 box but either way really nice job solving it.

[@TheQuicksilver115]

Wow, that was great to watch!! I love watching you think! If I could be so bold as to give you a bit of constructive criticism that I have been noticing: with these variants you seem to tunnel vision pretty hard on using the non-typical restrictions to solve it (rather than looking for normal sudoku rules), this time in particular you missed a lot of easy givens by focusing too hard on using the sums.

[@londonbobby]

In the top left the first two column totals (11 + 13) adds to the same as the first two row totals (7 + 17) which tells you that the MUST represent a 2x2 square. You can do the same trick in the bottom right. Really helps to know that early on.

[@dsch772]

Yes, it was some pretty awesome logic.

[@blackauron85]

This is false. Consider the solution: ------ 2 4 1 3 9 5 6 8 7 ------ This satisfies all the constraints but the sum is not confined in the 2x2 square

[@britboy3456]

52 min solve for me. I'm loving these difficult variants, great to get stuck into - more please!

[@Tehom1]

This was a very fun puzzle. Not as much count-grinding as I had feared. Keeping track of the known 2-or-3's was tricky, but there was a lot of logic going on with them. Towards the end, it had more violations of normal-sudoku-uniqueness than any variant puzzle I've seen before.

[@ericpraline1302]

"Really complicated" - writing two years after this video, this comment of Simon's seems rather quaint. Enjoyed the puzzle. With the benefit of experience, didn't find it overly difficult.

[@phlogiston6292]

In terms of notations I started immediately using yellow for horizontal, blue for vertical and green for overlaps between the two to keep things clearer... still haven't finished though because far too tired for some of this logic. Gonna dig in again tomorrow

[@ForeverDayGreen]

I love how different you can approache this sudoku and still end at the same solution. In my solve for example, I didn't find the 2 square 13 (top right) via the row until close to the end, but started out in the bottom left giving me a 4 rather early

[@Tehom1]

Perhaps an easier way to get into the right middle box: You've already used 23. 45 minus 23 gives 22 for the remaining cells. We still have to place a 6 and an 8 in that box, so the sum of the squares outside of that area must be 8. But we've already used the 8 to make 23 so we can't make it in one cell. So the area outside must be two cells. So both 6 and 8 are two cells long.

[@littleschnitzel8226]

Such a marvelous puzzle full of unusual logic, thank you very much for sharing it with us! I really enjoyed solving this one.

[@AMagicalTortoise]

I did this before watching, and was surprised to see us doing it two completely different ways. I rotated from top right and clockwise until I got to bottom left. Then I went back and went counter clockwise. Before the last 3 sections, I was able to just solve using basic process of elimination.

[@michaelhinkson6038]

I got lucky and found a great way to break into the puzzle, which opened everything up much quicker. If you look at column 6, you'll see that at the top, it's looking for 19, and at the bottom, it's looking for 17. That leaves 9 as a remainder, and as you can't make 9 with 4 cells, I know that the 17 at the bottom has to be a 3-cell sum. Since I know that the 18 at the bottom in column 5 has to be 3 cells as well, and the 5 in column 4 has to be 2 cells, that leaves just one remaining cell in that sector, which is a 5 (45 - (17 + 18 + 5)). Figuring that bit out opened up some great in-roads to the rest of the puzzle.

[@snigbylife]

@23:35 you apologize for not being efficient. LOL. I can’t even get to the first number. You logical abilities fascinates and humbled me.

[@R.a.t.t.y]

I really appreciate puzzles where all the numbers are given (in this case all edge numbers) rather than just some forcing you to solve it in the way the designer wants

[@inf0phreak]

Over 28 minutes in: Finally fills in the 9 in the bottom right that was technically possible to fill in right from the start due to the 6, 10 and 11 clues on the bottom right. Not that it did me much good in trying to solve it though. Great job, Simon.

[@yoavlev8962]

I SUCCEEDED!!!! God damn I didn't think I could. Took me some 43 minutes, but I stuck through!! Thank you so so so much, I never thought I could do such puzzles, and it's all thanks to you!

[@threoples]

I got stuck on the center left cell. I got the digits wrong and continued on until I was at an impass, where I could not place a 9 in the center left box... I had to give up and watch your solution. I feel like I learned something from it, even though I don't play these sorts of puzzles. Thanks for the video! Great solve.

[@tiemen596]

I did this in 50 minutes, which I think is the closest I have ever gotten to your time :). I think I was "lucky" to look at the right places a couple of times. WHen using the 45 sums, I tend to look for adjacent clues that are either very small or very large, as they tend to give the most information early. The bottom right 3 clues, 6 10 11, add up to 27, meaning the remaining total is 18, which is at least 3 squares and the 6 10 11 were all 2-square clues. That was a powerful starting point for me.

[@maxfriis]

Loved this solve after solving it my self. Never seen a hard puzzle with that much liberty in the logical path to solving it. Great video.

[@docneuro7477]

Great puzzle. One trick I notice was very useful is to focus in the corner. Top left and both lower corners you can use the sum of horizontal and vertical numbers as they overlap to solve the corners. I hope this trick helps others. For example: Top left corner. 17+7=13+11 so all numbers are in a 2x2 square. Use same logic in both lower corners.

[@LoveYaAngelino]

As i am not as clever as you are, obviously, i had a different starting move. As i found out that not all sums were 10 or larger, i noted that i could mark ALOT of spots where you can't put a 9 in it. In that way that i found out that a 9 could only be in 3 different spots in the bottom-right block. After some puzzling what matches i could make, i could fill in almost the entire bottom-right block and continue the entire bottom row. Going on to the left collumn and filling in the rest of the puzzle. Very great puzzle! Thanks for sharing this one! :)

[@TheNewAccount2008]

Really nice puzzle. Took me 45 minutes, but I was actually able to get through it. Thanks a lot to you and your channel here, without all your explaining I would have never been able to do that.

[@livedandletdie]

I've done some testing and there are a few ways to solve it, depending on where you begin to solve, the upper right corner is really good to start with, but the middle right edge works too although it's a bit more constrained.

[@Flati36]

I have watched about 10 videos on your channel now. This is the first Sudoku puzzle I have been able to finish on my own before watching your video. (with 2 hints from the video)

[@Flati36]

10 videos in 3 days, I might add.

[@Bennatte]

This took me so many tries! I eventually did manage it without consulting the video. It probably took 3 or 4 hours total, final attempt was 48:50. What a challenge!

[@corsothewolf8911]

I solved this in 50 minutes without watching you for clues. I made some hopeful guesses on the corners first based on what math made the most sense. but then I had the most amazing triples/doubles to solve down the middle columns. Awesome puzzle

[@SherlockSage]

26:19! This was great fun to disentangle. I rather wish I'd come up some of the color notation some of the others in this comment section had (one for horizontal sum, one for vertical sum, one for both, and one for neither), as I just went with green for a 3-cell sum and blue for a 2-cell sum and only colored one side in in rather random directions. In the corners where it was all 2-cell sums, I just colored the four cells that were in all the 2-cell sums blue as a compromise. Notation troubles aside, what a wonderful puzzle and thank you for sharing it with us!

[@Lord_Volkner]

Another great puzzle. Enjoyed solving this one, thanks.

[@matejkovacic9310]

What an interesting puzzle! I had such trouble getting started so I had to look at the first couple of minutes of the video just to see how to get started, but after that it all went so nicely. In the end it took me 24 minutes to solve, with around 8 being spent on just figuring out how to start. Thank you very much for this puzzle Simon!

[@yash08100]

Hello! I tried solving this on my own and got the solution but what was interesting was that i reached a stage where there are multiple possible solutions to the puzzle. I've taken a screenshot but i do not know how to share it. But it's a very interesting scenario in a very interesting puzzle.

[@nadeemhajiiqbal9214]

I feel like this is more of a Mark puzzle as you get the solution much quicker if you focus mostly on counting the outies. The bottom right box is the easiest once you compare the outies, and another thing Simon missed was that C6 clues added up to 36 meaning the outies had to add up to 9 so they can't be 4 squares, hence the 17 is a 3 square sum and the 5 has to be in C4R7.

[@benwhitaker9888]

Not gonna lie you were definitely slower than normal but then again I'm not the one solving. I think you made it more difficult in some places than it was. Good vid

[@timeomnivore]

Really enjoyed working through this puzzle. Took me 1:11:41, though I had to watch the first 3-4 minutes of the video to see how to get started.

[@72DerekB]

Wow, that was great. I've never seen this variant before and it took me well over an hour to work out the trick - I actually placed the 5 in the bottom middle box by looking at all the permutations of 17 & 18 and noticing I needed somewhere to put the 5. I was working on the bottom left box when the trick dawned on me and after a quick test in a couple of other boxes looked promising I reset the puzzle and solved it in just over 23 minutes. So it was either a 2 hour solve, or a 23 minute solve, depending on how boastful I'm feeling :p

[@jamierogers294]

Great (but difficult) puzzle. The 4 in the lower left box was my first number. The 16 and 15 could never be 2 digits as that would leave at least a 7 and 6 in a 2-digit 6 'cage' leaving the 4 to exist above the 10. As you saw, that was very useful for unravelling the top left of the grid later on.

[@andremouss2536]

I think the easiest method to start that puzzle would be to sum the three vertical or horizontal numbers outside a corner cell to determine how much is missing from 45 in the three cells on the furthest row or column in that box. For example for the upper right cell the three horizontal numbers add up to 42, so there is 3 missing which should be in the lowest row, in the form of a lone 3 or of one 1 and one 2. Since there is a 6 at the right, there must be a 3 in that row. If the 3 is on one of the two first cells, then the row is (12)(12)3 or (12)3(12). If the 3 is afar, the row is 3(12)(12). Thus the row is a triple 123. You can also repeat this approach with the side cells, though that will collect much less information.

[@Radinden]

Really pleasing puzzle. 48 minutes, via a couple of detours.

[@XoIoRouge]

I actually started off by blacking each square the couldn't have a 9 in it. The bottom right 3x3 square (box nine) would only have three possible positions for a 9: it's top left, middle center, or middle right. From there, it seemed odd to simply do "If this is a 9..." check for every square, but it always broke itself relatively quickly. The summation restrictions become really relevant when all the other number options are gone. The 9 quickly found itself in the top left tile of box nine, which helped me force a 9 into Box Seven's 17 summation (bottom-left 3x3 square). I completed both of those boxes first. I was also able to use this same restriction to check for a 9 in Box Three (top-right 3x3 square) due to how the 10s interacted. 9 Couldn't go in the left most column of that box, nor the last row of that box -- and it couldn't go in the top right corner because that'd break both 10s. I had the 9 from Box 6 solved (middle-right 3x3 square) because of Box Nine, thus the 9 must be within the 19 summation in Box Three -- I just couldn't place which row it was in until I shoved it in the middle, and ran by the possible options. As I mentioned, the chains quickly break in those corner boxes.

[@kenanderson3954]

You're often so good at seeing where the most restrictions are, but on this one I think you missed out on the bottom right 3x3 after the middle-right and top-right ones got fairly sorted. 6, 10 and 11 means you need an 18 outside , which also fixes them as all being 2 square vertical clues. I think the puzzle might unwind a bit better from that point but I was certainly not nearly as efficient as you were in the solve and it was great watching this puzzle get attacked from a different angle and seeing it continue to put up resistance.

[@StrategicGamesEtc]

The bottom right 3x3 was the first one I completed (Though I got the 85342 part of the upper right first).

[@Tehom1]

I found that a different coloring scheme made sense. For the 4 side boxes, a single color will do, but for the 4 corner boxes, I labelled the ones that were used both horizontally and vertically with one color and the ones that were used only horizontally or only vertically in another color. So far I haven't needed a third color but we'll see.

[@ano_nym]

It interesting to see these older videos referred to as being long at 35 min, now that we have recently got multiple 1-2h ones.

[@pedrodias7839]

using the same color for vertical and horizontal sums was a bad idea

[@ragnkja]

Pedro Marte I figured that out pretty quickly, and ended up using yellow and blue, with green meaning both.

[@forestroads]

@@ragnkja I used red and yellow, with orange meaning both :)

[@lucaslorenz2887]

@@forestroads You guys are smart.. I ended up using yellow for columns, green for rows, and then I got confused at multiple occasions. :D

[@biscuitz7422]

2:45:25 was my final time, though that included significant periods of time away from the screen, so I would guess it was more like two hours. This was a very enjoyable puzzle all the same with some very lovely bits of logic thrown in (column 3 and the bottom middle box in particular had some nice moments).

[@methethpropbut8519]

Very interesting how different our approaches were - I very quickly realised column 6 had 19+17 = 36, i.e. 9 left over, and 9 can't fill 4 squares, so the 17 at the bottom must be sum of 3, fixing the 5 (45-5-18-17) in row 6 column 4 - which helped put some of the bottom row in the right place.

[@brianrichter8255]

This was hard but very rewarding to solve! But I made two errors; in both cases it was a combination of 2-cell and 3-cell sums that I forgot to consider within a 3x3 block. Also, it hadn’t occurred to me that you could use the possible total of the center box rows/columns to deduce whether the edges were 2-cell or 3-cell sums. I was able to solve it without using any really obscure summing knowledge beyond 6/8/9 is 23 and 1/2/4 is 7, no need to write down all the possible ways of getting a sum, it was straightforward. This is a great sudoku variation!

[@lunafaekaida1772]

38:37! Really clever puzzle. took the first 10 minutes to really get a foothold into how to go about solving, and i used a lot of pen and paper to note possible combinations and whittle them down instead of doing it in my head, but this was a lot of fun

[@BlueCyann]

If I had a dollar for every restart I'd have enough for a decent coffee. I absolutely love this kind of puzzle, though. Breaks the brain, but in a good way. Will say that I my (final) start of this was completely different, in the lower right corner. Sum of 6, 10 and 11 is only 27, leaving 18 "out" of the sum of the lower right box as given by the numbers underneath. You can't get 18 with just two numbers, so r7c7, r7c8 and r7c9 together are exactly 18. The sum given on the right for r7 is only 9. So, r7c7 is a 9. You can further move on to realize that not only are the top three cells of that box all "out" of the lower sums, the left-most three cells of that box are also "out" of the right-hand sums. From there, most of the right-hand side of the puzzle becomes accessible.

[@sunriselg]

The 5 you found at 30:52 could be resolved from the very beginning - because the vertical 17 had to be 3-cell -otherwise the middle 4 spaces would have to add up to 9.

[@mememalistic3257]

Beside the 3x3 area with the 23, there’s another solution for the bottom left and bottom right 3x3 area that I realized at the very beginning. On the bottom left, notice the 6 from the outside, it helps to identify that 15 & 16 will definitely have at least 3 cells, since the maximum of the 6’s single cell can only be 5, it’ll force 16 & 15 to have 3 cells to sum it up. Which means 10 can only have 2 cells, because 16+15+10=41, there has to be 1 cell left for the 4. I can then identify the rests bases on the 4. Move to the bottom right 3x3, 6+10+11=27, 45-27=18, which means we will need at least 3 cells left to add up to 18, therefor the top row has to add up to 18. With 9 from the outside, the left cell of the top row has to be 9. Again, base on the 9 (and bas on the bottom left 3x3) I can solve the rests in that 3x3. Then, thanks to those 2 bottoms I managed to solve the middle bottom 3x3. Then I move to the top, just like you did. The bottoms really helped a lot in the process. I thought you could see it as well but guess I was wrong.

[@ZERO126PS2]

The 8s and 6s in the same box forces them to be a two cells sum. Once you realize that the puzzle goes pretty easy.

[@ratchet1freak]

The bottom right box could be tackled a lot earlier, 6,10 and 11 make 27 meaning you need 18 for the outies, but 2 are already summing to 9 meaning the last square had to be a 9. A bit of further logic can restrict the rest to pairs

[@Vearru]

This took me almost an hour and a half to solve at a leisurely pace. It was fun though.

[@TheBlackZodiacGhost]

I see a 30+ minute video, I instanly pres like :D

[@bristolrovers27]

Kasper Nielsen same

[@JohnJillky]

I see a 30+ minut video, I instantly give up on trying to solve the puzzle myself lmao

[@stuartmcconnachie]

6:50 all the squares in the top right nine, except the bottom one in column 7, MUST be in the sum, since it is not possible to have the second cell in in a row or column in a sum and not also the one that borders the edge (sums of first two or three NEIGHBOURING cells).

[@CrimFerret]

I hope eventually the new Sudoku app eventually gets ported to Android. I generally don't do sudoku at my computer other than trying some of the puzzles on this channel. I suppose Steam is a necessary evil to have any decent chance of distribution though I hate having it as a nanny for my gaming.

[@jibbiddy]

In the lower left section you know that 16 and 15 both have to be three cells because if they are two, they break the 6 sum which runs horizontally (there isn't a way to make 15 and 16 with two cells with numbers less than 6) Then using the box sum of 45 you understand that 10 is two squares, and the remaining blank square is 4. I think that 4 was the first number I plugged into the grid.

[@dylandugan76]

A very interesting approach. I completed almost everything in the third, sixth, seventh, eighth, and ninth blocks before I could make any progress on the top left. I'm not sure if I made it harder than it needed to be or Simon did.

[@LithmusEarth]

3 minutes in- that's what I did when I started it, I highlighted 2 horizontal and 4 vertical is what I got, 18,19,23,5 were the digits. I missed the horizontal 18 up there.

[@TheLuczyw]

Since the very beginning you could've noticed in the column 6,that 17+19=36,so other cells sum up to 9,so 17 is a three number cell. This would have helped a lot later on.

[@StrategicGamesEtc]

Well, I took 1 hour, 43 minutes, and 41 seconds to solve it, and I had to get a few hints from the video in the process, so you're way faster than me, at the very least. :D

[@njhcomposer]

It took me 1:35 to solve it, with a few hints as well. What a cool but mind-bending puzzle :)

[@YayapLives]

For stuff like the obvious 2/4 I was wondering if you would ever consider doing a co-op puzzle where someone else is watching and talking to you over Skype/discord(or really any screen sharing service) and you work together? It would be interesting to see if it helps or is distracting.

[@craftyraf]

The only cell I found was the 9 in row 7. Simon needed 28+ minutes for that. I feel proud :-) My logic: 6+10+11=27 of the 45. So 18 is missing. You need at least 3 cells to get 18, so 6, 10 and 11 are all 2-cell clues, adding up to 27. Now add up the 9 (above the 13), and you get 27+9=36. Therefor, ths missing digit in the upper left corner of the bottom right box is a 9!

[@DavidGlendinning]

It was a good thing I was an only child growing up -- I used to take so long finding the treats my parents set out for Easter Sunday that if I'd had brothers and/or sisters I would have gone without every year. :( So I can understand not seeing something that's about to slap you if you get any closer to it. :D For me at least it's a good thing that some of these solutions take extra time... that time gives my brain a chance to follow along, learn, and process clues in different ways... that is a big reason why I'm subscribed.

[@McMxxCiV]

I know how you often say "sorry if you're screaming at the screen right now", and though I rarely am, seeing you figure out that logic about the 6 and the 8 both having to be two squares by 4:30 and then not filling anything in the bottom left 3x3 box until 20 minutes later was indeed painful to watch.

[@edensaquaponics1941]

I started at the bottom and used the 6 + 8 logic in the bottom left box as well as the middle right. However I managed to mess up the middle left box, which I only noticed when I got down to the top center and middle center boxes remaining at my 45 minute mark. I tried backtracking but gave up at 50 minutes.

[@thomasdalton1508]

Very nice puzzle, but it took me 1hr 20mins to solve. I had difficulty switching between thinking about horizontal sums and thinking about vertical sums - probably could have saved a lot of time by improving that.

[@Zerotan]

I found a path from the start that went unexplored - the bottom right box has 6-10-11 along the bottom which sum to 27, which is 18 less than 45 - three digits are required to sum to 18, so the 6-10-11 must all be two-cell. The 9 on the right must also be two-cell which forces the top left cell of the bottom right box to be a 9.

[@Qazqi]

Actually managed to finish this in around half an hour without using negative sums like you were doing with your "outie" logic (although it definitely would have given me some numbers earlier). One nice bit of logic is the 6 clue in box 7: The 6 in box 9 uses 2/4, so this one can't be 2/4, and the 5 in box 8 must be 1/2/3/4, so making this 6 a 1/2/3 would cause 5 cells with 1/2/3/4 (not possible). Therefore, it must be 1/5. That immediately makes the 8 clue right above it a 2/6 and gives an opening into that part of the puzzle.

[@57thorns]

E.g at 30:56, you need to make 5 by combining 3 5 and 7, which of course means it has to be a 5. At the same time on row 3, you know there is a single 8 that is out, and the number to chose from are 5,8,9. Since the 19 is three long, it can not contain the 8, The start was hard enough that I gave up, you found some really nice items there. However, as you progressed you missed a lot of simple reductions, and did not follow through on all of them which lead to complicated deductions for number that were given.

[@RogueLich]

24:01 after a lot of false starts, probably a couple of hours proper time, but I kept being unsure of everything I did previously. Glad it's over!

[@pizzapm]

Trial and error on the corner boxes that are more restricted led to a relatively easy solution, I didn't need had sound logic on this one so i have a bit of mixed feelings for this puzzle, it took me just over half an hour I solved the 4 corner boxes first (starting with the bottom right box) then the 4 edge boxes then the center last. if i had started with this stratagy right off the go it would have taken 15 minutes

[@olegpovkh5581]

Simon - "sorry if I'm not solving it as efficiently, as you are". *me finishing this puzzle in 1:03:26 - yeah, it's okay.

[@prabalsingh3022]

For C6 sum of frame = 19 + 17 = 36 sum of non frame = 45 - 36 = 9, which can only be 3 cell long and all these cells are in Box5. Now in Box8 there is only one cell(R7C4) which is not part of frame and that has to be 5

[@howardpayne4128]

Bottom right 3 columns add up to 27 leaving 18 so they are all 2 high

[@glennmelven3414]

I was kind of disappointed. That is about the only logic I was able to spot, and Simon didn't even look at it.

[@stephenbeck7222]

And the 9 sum on row 7 means you can immediately pen a 9 on r7c7. That’s an immediate digit with 30 seconds of logic. Surprised he missed it, since he normally focuses on the variant constraints so well.

[@coolpapabell22]

Also could have gotten the 9 easily after the 24 pair in c7. The right frame numbers in r789 add to 30, so the c7 boxes in the lower right cage add to at least 15. Thus the 9 must be there.

[@RoderickEtheria]

This puzzle kept catching me up, making me think that two pairs consisting of the same two numbers along the same two lines shouldn't be able to happen due to uniqueness and proving me wrong multiple times.

[@colinfun]

I am shocked that he fails to realize that you can do logic like a 19 opposite a 17 means both have to be 3 long because you can't have 4 squares totalling only 9!

[@Jokseej]

The first thing I noticed was that 19 and 17 column6 have to be 3 spaces each. Having that, it's obvious that 18 must be 3 squares and 5 - 2 squares. Leaving the last square to be a 5.

[@VinniePaul91]

It's basically the same as saying 5+6+7+8+9 < 17+19

[@CasualGraph]

10:32 look at how that effects the 9 at the left. If the 9 is in three cells, then it contains 6, 7, 8, or 9. The only one it could contain is 6, which would make the other two digits 1 and 2, which contradicts the 2, 4 in the 6. So, the 9 then has to be in two cells. 18:24 Yes, it could be 1, 5. So, whether it's two cells or three, the it must always contain a 1 in one of those two cells. This means the 8 right above it must be in two cells.

[@TheNewAccount2008]

And just the question as I am watching along: I made an assumption that you are not: In the left lower block the row total is 31, which means that colum 3 in that box needs to add up to 14. (You said that in minute 18 of the video.) Now... The column clue for that column is 10, which in my opinion means that the 10 clue is as a 2-cell clue and R7C3 has to be a 4. Am I missing something here?

[@skasperl]

At around 17:50. c6 the 17 can't be a 2 cell 17 (I think). If so, you would have made the sum of c6 in block 5 smaller than 9 (45-19-17-c6r7). But the lowest value for the c6 in block 5 is 9 (5/1/3). If you try to lower it to 5/1/2 you would have a repeated 1 in c6r7. Am I correct?

[@christophseeliger5551]

Really nice puzzle! ( But obviously I love math variants) I wonder how this is such an unknown author. That name seems to be from a german speaking country, and if one has fun constructing such unusual variants, it is very likely to get in touch with the national or international puzzle community at some point. However, I have never heard from Florian before.

[@vincet68]

I think you got snared in the sums during the last section, Simon, as just moving backwards through your pencil marks would have unwound things sooner. That said, AMAZING job!

[@antonyduhamel1166]

12:58 - From the moment he filled in that 2 & 4, I was waiting for him to fill in the **other** 2 & 4 right below it. And he did...SEVENTEEN MINUTES LATER!!!! I always hear him say "you must be screaming at the screen right now" and finally, **FINALLY** I understand why he says that! Seven. Teen. Minutes. My throat hurts. (And if you don't believe me, here's the time stamp to prove it - 28:50.)

[@rabidsamfan]

Argh! 40 minutes in and I have messed up somehow. Time to watch the vid.

[@tbriecheese]

59:38 for me, I definitely need to work on these kinds of puzzles, still waiting for more thermo and sandwich sudokus

[@richardfarrer5616]

Two things I spotted early were the vertical sums in the bottom right box left 18, so they must all be two digit sums; and that the sums in column 6 only allowed for 3 missing digits, hence the 17 was a 3-digit sum, and there was a 5 in r7c4. Shame I missed other inferences so I still took over 40 minutes.

[@robinshen5888]

Very interesting, not only did I go about it a completely different way, I also got a different (but correct) answer! My slight variation in my answer was that the 73 / 37 pairs in rows 6-7, columns 5-6 was reversed (making it 37 / 73). Also I took 1hr 14mins to complete the puzzle, hmmmm.

[@codingbloke]

So Simon is human then. ;) Really enjoyed the video and puzzle although it seemed labor he was still 3 times quicker than my solve. However I was able to solve with very limited complexity (I just don't see all the possible sums so quickly like Simon, I have work them out). My solve followed Simons on block 3 and 6 but I was then able to get block 9 to set of double pairs which let me solve block 7 similarly and identify the 8 and 6 sum horizontals as only needing 2 cells. That was key because it opened full solve on block 1, after that, so much was restricted that basic Sudoku and simple arithmetic solved it. I have a suggestion on using colour. I was highlight horizontal sums in Cyan and vertical in Magenta, where they overlap I used red, this looks quite natural (the same colours are used by colour printers to produce red). This allowed me to keep track of both the horizontal and vertical "innies" at the same time.

[@Ian_Hay]

If I’m any kind of a representative sample size, I think most of your audience is watching most of the time to come back and figure out where they went wrong AFTER COMPLETING 7 BOXES AND FINDING THAT ONE SQUARE THAT NEEDED A 6/8 AND BOTH 6/8 ARE BLOCKED DAMMI... sorry. Lost myself there. Love your channel.

[@RandomBurfness]

24:56 for me! Good solve, good puzzle!

[@TalathRhunen]

I didn't solve this myselft but I noticed that you could have solved the bottom right box a lot earlier than you did by noticing that the 6+10+11 sum to 27 which requires 18 more, so three more cells so all three have to be 2-clues.

[@angusl1694]

18:04 Is it confirmed that there is a 4 in the upper right corner of the box? Because a 14 sum is forced into the right and there is a 10 clue at the bottom.

[@markp7262]

I was solving it as I played the video... I was way ahead of him for the first time EVER, then I second guessed myself when I got to the 2-6 pairs on the left section horizontal 8s... thought I had messed something up, forgetting about the vertical numbers, and deleted half of my solution... just to redo the whole thing 5 minutes later... BAH.