World-Class Sudoku Setter Creates A Monster

[@markaustin3858]

"We won't do a better puzzle this month!" Well you've only got tomorrow to go lol Fantastic video and solve once again!!!

[@Novas12390]

when I saw the thumbnail I thought: oooow a sandwich sudoku, this is easy. but when I saw the clues on each other end against to each other, I was like "nevermind, I'll just watch it."

[@schnupferr]

I usually decide by the time he takes, so if it's over 30 min I don't even bother trying

[@mantaspauza526]

@@schnupferr I don't even trying his puzzles at all :D. I simply don't understand.

[@uhmarcel7417]

@@schnupferr me too

[@thesolarfutureenthusiast1102]

You marvel at how the puzzles are set in the same way many of us marvel at how you solve them. I played with this puzzle for30 mins, made the 456 determination to R1C1 and then incorrectly ruled them all out.

[@timsullivan4566]

"Sorry it's taken me a while - 36 minutes." REALLY? Apologizing for taking 36 minutes on THIS?! I doubt this self-isolation could last long enough for me to solve this one myself.

[@JaharNarishma]

I tried it, found an inconsistency. Looked at the video to see were I went wrong. I wrote a 3 instead of a 2, I calculated 9+6+3=21, 3+6+9=16 and I swapped positions on the 78 pair in the lower right corner. It drained my brain too much for me to not make mistakes. 4 mistakes was 3 more than I believed I made.

[@timsullivan4566]

@@JaharNarishma Sorry to be the bearer of bad news, bt if you THOUGHT you'd only made 1 mistake but ACTUALLY made 4, that in itself is ANOTHER mistake for a revised total of 5 mistakes! But if it's any consolation to you, I'd rather have made 10 mistakes instead of simply standing with my nose pressed against a brick wall. ;-)

[@abob821]

​@@timsullivan4566 I can't solve many puzzles on this channel, but I have solved this puzzle. I wrote down every combination in advance that can be in a row / column. e.g. 21 | 22 4,8,9 | 5,1,3,6,7 6,7,8 | 5,1,3,4,9 5,1,3,4,8 | 6,7,9 It took a long time, but after that it was no longer extremely difficult.

[@IvoHristov1]

@@abob821 Mark, is that you?

[@abob821]

@@IvoHristov1 Nope, who is Mark?

[@samtravis4822]

Sensational. So glad I get to live vicariously through Simon as I’d never have a chance with a puzzle like this on my own

[@ilanknafo]

What to you mean " i'm not as brilliant" my jaw just droped for half an hour !! Amazing work!!!

[@stevewebber707]

While I also experienced a jaw drop, considering how demonstrably smart this guy is about these puzzles, I will quite comfortably defer to his judgement on who is more brilliant in this case. On the one hand it's a humbling thought, and on the other it's wonderful that such minds exist.

[@christophseeliger5551]

You can use the 22/23 pair in the lower right for an immediate break-in. They can’t both be three cells, so one is an Xsum. And because of the 1s and 2s in C9 and R9 this Xsum clue has to be a 5.

[@CrackingTheCryptic]

Ah yes, nice spot!

[@RoderickEtheria]

or look at the x-wings formed by the rows and columns that are missing 1 and 2 respectively to add up to 45. And how only one of the 5s could work for both x-wings in the bottom box, and similarly look at the x-wings on 6s that are on both columns and rows afterward, and how they also lock a 6 in the top left corner.

[@BlueCyann]

If I logicked it right, you don't even need the 1s and 2s penciled in. I believe that the 23 clue there can only be an X-sum. If the 22 is a frame clue, it can't be a frame clue, and if the 22 is an X-sum, it also can't be a frame clue. Reason being, if the 22 is an X-sum, its first cell can only be 4 or 5, and neither 4 nor 5 is in a 3-cell frame adding to 23. So the 23 is an X-sum. First digit 4 is ruled out since it would place a 1 in C9R5, and 6 ruled out because it doesn't leave space for the 1. So it's a 5, and the 1 follows.

[@hudaboombeach2189]

Took me almost 3 hours of solving but I couldn’t be happier I managed to pull this one off! I found this channel just two months ago and ever since I’ve been completely hooked. Thank you for this amazing content Simon and Mark.

[@mahatmagandhi5692]

3 hours, that’s a long time for a puzzle! You must have been pretty determined to solve it!

[@Coyotek4]

19:34 This might very well be my proudest Sudoku-solving moment! :D

[@glennmelven3414]

Sam Cappleman-Lynes is in a league with Phistomefel which, sad to say, I am not in. I do enjoy watching Simon solve there puzzles though.

[@Daiephir]

Plot twist, Sam IS Phistomefel. I've never seen both in the same room...

[@mindassassin]

*Makes pencil mark* Well you made it further than me.

[@lakarto19]

Absolutely amazing puzzle... took me 46:58, and although it was fairly slow, I really loved the logic. I'm really enjoying all of Sam Cappleman-Lynes's 'uncertainty series' puzzles; some of them are brutally difficult, but once you get the initial break-in, unlocking the rest, for me, is just the right mix of arithmetic and sudoku logic. Brilliant!

[@jimpassaro]

There's actually a more simple way of getting your first few numbers. I needed help (which I got watching the first few seconds of his solve) with realizing that each row/column adds up to 45, which, by comparing that sum to the sum of the two clues outside a row/column, informs if there's either an overlapping x-sum or which, if any, digits separate the two sums. This is how Simon placed the initial pencil marks for 1s and 2s in rows7&9 and columns7&9. They need to be in the central three positions of the row/column because you need at least three digits on the outside for each sum, as they were all greater than 17. What Simon missed is that, the placement of those 1s and 2s within those central three positions necessitates an X-sum on at least one side of it, and, the leading digit of that/those X-sums. So, if you place, for example, a 1 in r7c5, the center position of row7, you need a leading 4 on both ends of row7 for their x-sums. Therefore you can't have a 1 in r7c5, or a single separating digit in row9 or columns 7&9. Therefore, row7 demands a 5 on one of its ends (r7c1 or r7c9). Similarly row9 needs a 5 on either end (r9c1 or r9c9), column7 and column9 need 5s on their ends. The intersection of these requirements gives you a 5 placed in r9c9, which resolves position of the 5 in row7 and column7, and the position of the four separating digits (1s and 2s) in rows7&9 and columns7&9. Those 5s also restrict possible X-sums on row5 and column5, giving you a 1 in r5c5, among other digits. How elegant!

[@JaggerG]

Unfortunate Simon missed that break in. A bit more elegant than the one he found.

[@nemoyatpeace]

This was a beautiful puzzle. It took me a while to get in, but I solved it without having to watch the video at all. Now I'm watching to compare how you solved with me. I printed all 10 of his puzzles and am slowly making my way through them all. This was one I actually completed first time without having to erase and back up. First thing I saw was that many of the clues totaled 45 and the rest are near 45. I noted that a sum of 44 means a 1 somewhere in the middle. Then I knew that this (r7) required the clues to be a total of 8 cells, this meant that they were a 5 clue and a 3 clue. So I knew that r7c1 or r7c9 had to be a 5, same applies to C7, C9, R9. And this gave me a 5 in r9c9. This got me going.

[@dkalambokis78]

I'm watching your tactics / solving process for more than 3 hours. Never apologize for your solving time!

[@donnaml8776]

I think I’ll join the children’s hunt, that’s more my speed. 🥴

[@Afterthoughtbtw]

Pretty pleased with that - took me 40 odd minutes. The difficulty is finding a way into it I think. I never got stuck for an extended period after that, there was always a square that showed itself as the one to tackle next.

[@no-feetmcgee5577]

Love, love, love this puzzle! Very intriguing logic needed. Took me just under an hour.

[@ExoKiller4]

Quite a puzzle. Looking at the starting position, the fact it even has a unique solution is mind-blowing, let alone the fact that it can be discovered by a human.

[@aidarosullivan5269]

Petition to name this type of sudoku a "geometric sudoku" or geodoku, as it's amazing how solution unfolds from boundaries to the center

[@richardfarrer5616]

A slightly different start for me. In columns 7 and 9 we can't have the extra digit being in the fifth row or else both both the top and bottom sum would be four digits long, requiring 4 repeated in the column. Therefore the extra digit is in the fourth or sixth row, giving one number of length 5 in each of those columns. That gives an X-wing on 5s in columns 7 and 9. Symmetry gives the same in rows 7 and 9 so we can place the 5s immediately. This means that the X-Sum in row and column 5 can't be five digits so it must be four, which means that both the column and the row need two digits between the sums, showing that they are a 12 pair and a 13 pair. Additionally, in top left block we know both 21s are frame clues so the total value in the box is 6 + 21 + 21 + unfilled cells in row/ column 2 - r3c3. The only way that works is if r3c3 is 9 and the unfilled cells sum to 6. A similar but less restrictive conclusion works in the bottom right. After that my solve path was much the same, just slower :-(. 45:44 eventually.

[@sasopalma4993]

I liked this puzzle. At first I thought I'm not gonna make it, but it went pretty good. Let me just explain my break-in into the puzzle, which I think is much better than Simon's. First I realized, that 43 and 44 row/col sums means one number between two sets. As Simon pointed out, it must be in the middle three positions, but you can rule out the middle one, because then you would have 4s on both sides! Ergo, there must be a 5 on either end of the column/row (to get to divider number with X-sum). You apply this to row7 and row9 and to column 7 and column9. This immediately sets a 5 in the bottom right box on r9c9 and thereafter 5 in r1c7 and r7c1 (and of course the 1s and 2s in proper places). And now for r1c1: from Simon's 456 possibility, 5 is ruled out because of 5s in row and column. The 4 is ruled out because 23 cannot be made with 3 numbers to form Frame clue, and it cannot be X-sum, because then it would have to be a 5 on the other end to make another X-sum, which is again ruled out by the 5s already placed. Therefore, c1r1 is 6. And then continue... So no difficult sums and look aheads. What do you think?

[@barrydavies2977]

It's interesting to see, Simon, that your normal scanning direction is up and down and so you are slower to pick up pencil marks in the rows. Where as Mark's normal is across and the reverse is true. Slower is relative, btw. You are both hyper speed compared to me. Really enjoying both of you puzzle solving.

[@dliessmgg]

Just out of curiosity if I have spotted it correctly, at 35:00, do the 8's in columns 6 and 8 form an x-wing? Can anybody confirm?

[@batdowg]

Yes, and that resolves the 68 in box 4

[@Vishnu_00]

Wow.. I cant believe... Did it in 36:05...🤘

[@rmath101]

34:20. I'm pretty pleased that I managed to solve it!

[@TegukiSix]

It's always a good day when I can solve the puzzle without help. This one flowed beautifully (though I did make two errors and had to start over).

[@04LightningFan]

I started before dinner and did not finish till after four in the morning. Most of that time trying to figure out an 'in' to the puzzle. Even when I completed a valid sudoku, I had to double check the I used the clues correctly. I didn't spot the same thing that Simon did. I knew there were columns and rows where both clues could not be x-sums. Eliminated one of those clues from being a frame clue. Once I figured out those two clues, I could make deductions about the other clues and putting numbers in went pretty quickly.

[@remigregoire2219]

Ah! that's so sweet, you think I did this puzzle!

[@broski89]

Wow I actually found a first number before Simon. I found that 6 in r1 c1 probably 7 minutes before him and I was yelling at my screen for 5 minutes that it was a 6

[@thelostpiranha]

It was a 9/10 difficulty just watching this! Simply stunning work once again

[@OyvindSOyvindS]

I am awestruck of this puzzle, and also the solve!

[@codywest8199]

The 22/23 pairs and 21/24 pairs both sum to 45, so you know the X-sum needs to be of size six. This immediately gives you three 6s in the grid (r1c1, r3c9, r9c3)

[@bluefloyd1]

at 23:25, I don't quite follow. Could the x-sums add to 20 with 4682, leaving the 9 ambiguous?

[@lasttobi1219]

No it cant. I will try to explain it. If you add both of the given outside numbers of that column (23+20) you get 43. The total of one column always is 45. Therefor it leaves us with the rest of 2. This rest (2) is neither part of one of the two sums (23 or 20) of this column. So it cant be included in the possible x-wing that you are trying to build. I hope that was understandable.

[@bluefloyd1]

@@lasttobi1219 Oh of course! sorry, I forgot, in my comment, that the 2 was the separating number, and not a number that can be included in the sum. Totally right, I withdraw my objection! EDIT: I will say, though, that you can still get through the puzzle using the logic from the top-right box, which is where I went. Definitely not as efficient as finding that 9. damn!

[@lasttobi1219]

@@bluefloyd1 i was checking my Numbers every now and then to see If did a mistake and have to admit, that Like 80% of the time i was filling in Others numbers than He did. Well took me way longer so i guess His solvingpattern was more efficient than Mine. But ya know, many ways lead to rome

[@bluefloyd1]

Jeez, 1:54:38, a long one, with that blunder!

[@lasttobi1219]

@@bluefloyd1 1:22:39 Here with around 10 Minutes of afking. Took a while ^^

[@CuratedPile]

About 33 minutes. Very clever puzzle. Just watching your solve now, the 45s broke this open early for me.

[@ronnymanfred]

Awesome puzzle, way to hard for me. Maybe you could do some more Frame or X-Sum puzzles in the future :) (just one of both rules in a puzzle for a start ;) )

[@leppyr64]

I'm happy to see this puzzle. I started it yesterday and deduced that the 23s in the top left cannot be Frames... I intend to struggle more but happy there will be a spoiler if I stay stuck.

[@ArisKatsaris]

I started solving this by figuring which of the rows/column needed a "5" in one end, because a single "1" or "2" had to be squeezed between the clues -- marking these potential 5s (7th column, 9th column, 7th row, 9th row) one sees that the 7th column needs a "5" at the top because a "5" at the bottom of the 7th row would have blocked both possibilities for the 9th row. Similar with the 7th row and the 9th column. Those were the first definite digits I got.

[@bristolrovers27]

Fantastic puzzle I’m afraid a strictly sit and watch and admire puzzle

[@CCHdennis1]

54:37 But after I saw Simon video, I discover my logic problem. For R1 & C1, I forget that there could be 2 X-sum with 4 or 5. And I directly set a 6 on it. Fortunately, I can get the answer still. However, the other idea is right and solving path is totally different. coz I spot that R1R3C1C3 they are all sum to 45, and so we can immediately get all pencil mark on box 1 and find out 6 on R9C3 & C9R3. Afterward it's totally different.

[@jdpjamesp]

Really enjoyed the Junior Puzzle Hunt. That's about my level. lol

[@mdesonie]

I actually figured out how to start this one right away, but it took Simon awhile.

[@DevilishlyDutch]

I just settled on backseat puzzle solving for this one, I tried to get into it but had no idea how to begin at all hahah

[@JackDeHearts]

It really broke when you found that 2 in the middle box but you didn't utilise it properly, missed it placing a three in C1. Amazing solve anyway :D

[@Saintsdevotee]

I solved in an hour without needing a clue from the video. My proudest solve ever!

[@RoderickEtheria]

Solved it in 38:23 once I verified I understood the rules. I got stuck at a point where I'd added up both the center row and column to the same number, but after rechecking realized they differed by 1. I also immediately realized that the rows that summed 43 or 44 had to use combinations of 5 and 3 digits. Had they used combinations of 6 and 3, there wouldn't be any leftover, and had one of the combinations been of size 4, you would have to have the other combination also of size 4, which would break the row or column. This puts a 5 on either end of these lines, to correspond to line length, and, thus, you have two competing x-wings of 5s, in the lower box, which forces the 5 in the only space of that box that sees both x-wings, and then 5s in the opposite corners of the x-wings. This eliminates any other possibilities of 5 length columns or rows. There were two of each of the rows that totalled 43 or 44 both vertically and horizontally, which instantly meant that the 45 totals had to consist of combinations of size 6 and 3 length, as there was no chance of a combination of 5 length. This creates two x-wings of 6s where the only spot on both of these x-wings is the top left corner, and there can be a 6 placed in the opposite corner of each of the x-wings, eliminating any other possible combinations of length 6 around the grid. The middle row/column had to then be consisting of combinations of 4 and 3 length, as both the 6 and the 5 length combinations had been used. The totals of 42 and 41 in the middle immediately then identified the center cell of the grid as a 1. This wasn't actually all that hard of a puzzle, especially with what we've come to expect from the setter. It isn't a monster; it's just misunderstood, and I'm referencing both "monster"s. And, yes, the best place to start is this framework that could quickly be identified. And I almost immediately grasped it. I don't like how you resorted to guessing the numbers to start off when as I have presented, there was a much more logical approach to this puzzle.

[@MichaelJPartyka]

I always wonder how much making a fill-in attempt crosses the line into bifurcation. I got as far as the 456 in the top left square, but I couldn't visualize a solution past that, and it felt like, "Let's try this number!" and then playing it out as far as it might go would be bifurcating, so I gave up. I wasn't that far away from figuring it out, I think -- once you get that first couple of steps in, there aren't that many more hard battles to fight.

[@Namknat]

Wow. I did this one. It took me 1h30 I think (the timer is at 3h but I take a pause lasting ~1h30). Once you understand 1 2 and 3 can't be in the middle of the 44-43-42 pairs, it became less difficult. And once I add the 6 on the borders of the grid, it became almost obvious (just have to did the math in my head, the most difficult part at the end). This sudoku was really difficult, but damn I loved it.

[@RealCadde]

It is not the length of the video that's bothering me, it's that you keep on missing some basic stuff. But i still come back occasionally to watch you solve puzzles because your spotting abilities are excellent and i learn a lot from that.

[@killroy42]

Oh man, I'd love to back-seat-puzzle on a live stream :D

Great solve! I've had no idea...

[@mdesonie]

First puzzle I've solved without ANY help from Simon!

[@alangbaker]

Did anyone notice that the divider for the X-sums can never be in the centre square of a row or column, because then both sides would have to begin with 4?

[@micheltordoir7734]

Finally 01:32:04 to do it, but had to restart it once after being quite far in the puzzle. So I probably spent 4 hours on it! But I cracked it!

[@draconicdusk5911]

1 hour exactly great puzzle.

[@Laserstormelpies]

amazing puzzle, really entertaining solve

[@brettaspivey]

Honestly this one was quite a bit easier than some of the recent monsters

[@DJejbarros]

I started with the 45 sums. Maybe I'm wrong, but it would force one of them to be 6 and the other one to be a 3cell sum... so I got a lot of 6s

[@Zarrow1]

While I love doing sudokus, I quite often still make easy mistakes in basic puzzles. I do sudokus on paper with a pen so I don't have luxury of making so many notes and I have to track everything manually. Do you think I could improve on these manual sudokus by doing harder ones on pc and actually pencil marking everything down? I've noticed these past months my logic has improved quite a lot and I'd love to improve even more.

[@neilramsey1368]

Great puzzle too good for me

[@fussyboy2000]

The software needs the ability to mark off the clues outside the grid.

[@UofMFan628]

stared at the puzzle awhile, tapped out and decided to just watch Simon do it. first one I've tapped out of in a long time at least but the uncertainty really messed with me and I just could not find a break in point.

[@aidefcon1]

Anybody please correct me if I’m wrong but wasn’t the whole logic chain for deducing that the top left corner a bit over complicated? He had already deduced that both top left corners where X-sums clues and the opposite numbers frame clues. Then it had to be a 6 because the total had to be 45

[@Keyboardje]

YaY 46m 29s ! ! ! It's really weird, but I seem to have more problems with easier ones O.o Is that posssible? Or might it depend on the sort of logic behind them? Or the symmetry maybe? Sorry, I'm only doing puzzles like this for about 3 weeks so I have no idea yet.

[@JaggerG]

I feel like a lot of easier puzzles require lots of possible paths and searching for eliminations, and harder ones have some point pulling your attention to them. On top of that, if you’re expecting hard stuff, you might not be looking for easy stuff. Simon and Mark miss easy things alllll the time.

[@JonGorman]

There is a point, at which, a puzzle is so monstrously difficult, it is no longer interesting. For me, at least, this puzzle is there.

[@RoderickEtheria]

It's not monstrously difficult. It just looks that way on first glance.

[@lasttobi1219]

Tbh i wasnt able to solve some of the other puzzles that were introduced on the channel before. And some of these got a lower difficulty. Guess it depends on how your brain works and how you trying to solve different kind of puzzles. Some find it easier, some harder.

[@DerKarlKopp]

Very fun puzzle, solved it in exactly 33 minutes

[@philipgill4024]

my first instinct would be the 21 24 row, as they total 45, putting 6 as the first number in the row, and 789 in some order in the last box

[@SailSmBi]

I solved it assuming the opposite pairs weren't both x-sums from the beginning so it create a completely different set of logic. Lots of 5 and 6 x-wings in the corners and a different set of symmetry. In the end I guess I got lucky

[@michaelvos3839]

When are the puzzlehunt solution videos on or are they for patreon members only?

[@Tahgtahv]

There has not been mention there will ever be a solution video for any of the puzzlehunts.

[@CrackingTheCryptic]

At some stage we may do videos of the hunt solutions but we want to let them run for a while first.

[@Rangar1997]

Wow, that start . . .

[@RoderickEtheria]

could have been much smoother had he reasoned with the 44 and 43 rows and columns rather than the 45 rows and columns.

[@DarkChasm]

beautiful!

[@schnitipuff]

is this in a different key?

[@ezapi]

Was completely lost until I watched the video. I was under the assumption that the X-sums and frames in the same row or column could have overlapped.

[@kevinb1743]

They were allowed to overlap but as it turned out the values in the frames weren't large enough to actually let them. Like a 21 and a 23 only add up to 44 and since a row or column must total to 45 there must be a 1 left in between. The numbers across from each other would have had to add up to at least 46 for them to overlap.

[@masaakunokouchi]

Normally I give them a go, at least with a little start, this one i just said, no way, let your brain melt

[@TheBlackZodiacGhost]

I see a 30+ minute video, I instantly press like! :D EDIT: Amazing puzzle!

[@Fernando_Cabanillas]

can someone explain to me the logic he used to clear the 3 from column 9 row 5 in minute 24:53? i did clear that 3 too but I used a different method and I am having a hard time figuring out what he did =S

[@s9ecera]

The frame has to be 20, and if he uses a a three in c9r5 he needs a 9 i either c7r5 or c8r5. Neither is possible because of the 9s in c8r1 and c7r7.

[@Fernando_Cabanillas]

@@s9ecera ah gotcha, thanks =)

[@BlueCyann]

I liked this. It was hard but not grindy. Wish I had a time, but too many interruptions, probably about an hour.

[@OidhcheMhath]

2:22 skip intro

[@iosifoulis]

44.38 after burning half of my brain cells, i've managed to solve it. ty for the puzzle

[@tiemen9095]

29:50!

[@zzzhishamzz]

Took a long time but its pretty solvable... 1hr 30 min :)

[@msscoventry]

I understand it all when demonstrated and watch appreciatively, but this is too difficult for me to get a good rate of enjoyment when solving myself.

[@kazmurf8218]

Took me 01:43:44. But it felt rewarding

[@RichieDcik]

Won’t do a better puzzle this month... because it’ll take me all month to solve!

[@brendanbush2174]

I was wondering why I couldnt hear him over the munching of my food

[@RealCadde]

20:00 And since the bottom three of row 7 is a frame clue, there can't be a one in it. Witch means either R7C2 or R7C3 is a one. No matter how good you are at doing advanced stuff, please don't forget the basics. Check what everything you do gives you in form of clues.

[@Wecoc1]

This puzzle is beautiful but sadly I'm too bad with this type.

[@coryr305]

You can deduce from the start that a row with outside clues summing to 45 must have a 6 at one of its ends, since the X-Sum + Frame covers the whole row. Similarly, a row summing to 43 or 44 must have a 5 at one end, with a 1 or 2 in the 5th or 7th cell. This gives you X wings on 5s and 6s that you can use to make early progress.

[@chompyzilla]

That’s not true. If the sum was 45, you could have have them both be x sums with 4 and 5 at both ends. You made a lucky guess though cause this puzzle’s solution doesn’t have that arrangement.

[@RoderickEtheria]

@@chompyzilla You can however be completely justified in the realization that any sum of 43 or 44 around the board must be a combination of 5 and 3 distance combinations if the numbers on opposite sides of the line are both greater than 17. If there are two of such row or column(as in the same row or column, not just one row and one column) in evidence, then you can surmise that any totals of 45 in the corresponding of rows or columns must be a combination of 6 and 3 length combinations if both sides of the rows/columns totaling to 45 are greater than 17. Upon further consideration, the limit of 17 is too specific, and as long as both sides of a row/column are greater than 11 this must be the case.

[@coryr305]

@chompyzilla thanks, I forgot to consider that. However, if David is right, you can still deduce that about the 5s, getting an X wing from r7c1 to r9c9. This rules out the 45 possibility in rows 1 and 3, making the X sum a 6 and thus giving a 6 X wing from r1c1 to r3c9. So I guess my conclusion was right, just got a little lucky without realizing the complete logic.

[@xyzzyx7812]

This was the first time I solved a puzzle faster than the length of the video. So I start the video and realized that I assumed all rows are Frame+Xsum. Luckily the real solution ended up no two Xsums in a line :D

[@tachyon1132]

I could have done this all by myself, but I am bad at seeing what the sums could be, like when there is unique set available. Rest, I would say I did all by myself without help from Simon cause he helped me reach here.

[@RandomBurfness]

I LOVED doing this one a week ago.

[@efa666]

It's like a nonogram lol

[@Coyotek4]

Here is how I started (in case there was a flaw in my logic and I just got lucky): First, I reasoned that ANY X-sum in this puzzle has to start with 3, 4, 5, or 6. Then I looked at column 9, noting that the sum of the two numbers was 44. The numbers in that column (21 and 23) cannot both be Frame clues, as that would leave a sum of 1 for the r456 in c9. Now, IF the two clues were both X-sums, then r1c9 and r9c9 would have to sum to 8 (the one cell not being part of either having to be a 1). They couldn't be 4 and 4, so they'd have to be 5 and 3. Meanwhile, if exactly one clue was X-sum, the 'Frame' clue would consist of exactly 3 cells, meaning that the X-sum clue begins with 5. The point: either r1c9 or r9c9 must be a 5! And the same holds true for column 7 (with the sum of the two numbers being 43, leaving only 2)! So either r1c7 or r9c7 must be a 5 as well! Voila: I can pencil-mark 5s into those 4 cells. So now neither r1c5 nor r9c5 can contain a 5, but there's a similar situation brewing with the sum of the two clues (20 and 22) being 42. This now forces at least one of those clues to be an X-sum starting with 4, the other clue using exactly 3 cells (either Frame or X-sum starting with 3), and the two middle numbers (r5c5 and either the cell above or below it) being a 1-2 pair. I would go on, but I think I've said enough.

[@farmersix4434]

I thought I had it cracked but nope 😣

[@salvadorsanchez5057]

when solving the 45 sums on opposite clues i dont understand why you assumed that they cant overlap. for example as you said in 12:19 "if this is 7 then these 2 squares [on the other side] would have to add up to 22". however, i dont see why the sums cant overlap? could be for example be 7 in one side then a number bigger than 2 on the other. maybe the overlappnig sum breaks down when filling the digits but i cant see why its excluded out of the gate.

[@ewanjacksonmusic]

2:14 to skip the intro

[@RealCadde]

21:10 And since those two cells are 67, there's only two other cells where a one can go.

[@RealCadde]

21:24 And there's symmetry in the puzzle. If R9C1 can't be a nine then R1C9 can't be a nine either.

[@mcwbyu82]

32 min.

[@RealCadde]

The sound quality is a bit off in this one. Could you perhaps look into getting a lapel microphone?