How Hard Is It To Add Up To 10?

[@PuzzleQodec]

It's really remarkable how close this whole thought process that I went through to set this puzzle is to this solve. Like... I've built some clues in there to ease up some deductions that I thought were a little too hard, but you didn't even need them. What a mesmerizing and amazing solve, I'm so so impressed, wow!

[@martysears]

that was sensational. And absolutely mental. I wouldn't have solved that in a million years. Really intrigued to know what your starting point / core concept was for creating this puzzle? Was it the idea of being economical with low digits in those 3 green rows? Either way, this central idea was handled with such care and mastery that this video deserves a standing ovation. Genius stuff, and big props to Simon for figuring it all out

[@davidrattner9]

Again..just absolutely astonishing from you. Huge respect for how your mind works in constructing!

[@PuzzleQodec]

@@martysears Yeah this shortage of low digits was the key idea. And thanks!

[@waldolala2]

I believe Simon shouldn’t have abandoned thinking about the maximum of green. By realising there is a single 5 in at least a three cell sum, the maximum total drops to 110. This makes blue total to 25. It also means that the pyramid needs to contribute 20 to green, as one of the bigger ones only contributes 40. Then yellow needs to be on the bigger one with the single 5, otherwise both bigger ones could only contribute 40.

[@PuzzleQodec]

@@waldolala2 To be fair, there was a lot to think about, several leads to follow. It's not easy to predict which one is the most promising in terms of advancing the solve.

[@MichaelGreen831]

“Never talk to me at a party”, Simon says to several thousand fans who are willing to watch him noodle out a puzzle for an hour and 40 plus minutes. Sorry Simon, you get us at a party with you, and you will be swarmed by fans. You are a class act all day long.

[@longwaytotipperary]

Amen X 10!!

[@davidrattner9]

So couldn't agree with you any more!!!

[@bethelliot7140]

👍👍

[@Anne_Mahoney]

Let's all fly to Britain next week and have that party! 😼💜

[@longwaytotipperary]

@@Anne_Mahoney are you paying??? 🤪

[@amysteele2488]

Simon... apologies, I meant to post this as soon as the video ended, but did my usual and fell asleep! Thank you so much for the birthday wishes, and your kind words - it really made my day! I feel highly honoured to be so mentioned. My thanks too to David Rattner for letting you know - that made such a lovely birthday present. I can confirm that there was a plenteous supply of cake, though owing to a medical condition, any small amount of chocolate involved was white... And what a stunning puzzle it was today! Again, feeling highly honoured. Thank you, thank you, thank you

[@PuzzleQodec]

Happy birthday!!

[@Mr_Top_Hat_Jones]

I hope you had a great birthday!

[@emhi1078]

happy birthday! and many happy returns

[@emilywilliams3237]

Happy Birthday, Amy!

[@johnedwards7473]

The fact that the blues add up to 25 was a little easier than Simon made it. The large loops have 10 cells and one of them (at least) has a 5 grouped with something other than a 5. So three cells. That leaves at most 7 other cells, so at most 3 tens. So the two big loops have at most 9 tens. Little loop has at most 2 tens. So green adds up to at most 110. So blues, whose sum ends in 5 are at least 25. So have to be 25.

[@ashenfang]

Yeah, he started off strong when he noted if they were all sets of 2, that his max would be 120, which is the only way the blues could be 15. The moment you can't pair the third 5 with a fourth 5 in the three rows, means at least it needs two low digits, therefore there's at least a set of 3 numbers, meaning you can no longer reach that 120 needed for the blues to be 15.

[@samlight1387]

I saw this as well, was shouting at the screen! That being said I love seeing something that Simon doesn’t, it doesn’t happen very often!

[@nortonmalcontent8778]

It was shocking to me how he came up with the deduction about the 10s only being able to add up to 110 or 120, 120 only being achievable by having the longer lines with 5 groups of 10 and the small line with 2 groups of 10, and then failed to use it to clean up some of the things he then puzzled over for over 1/2 hour. As you said, as soon as he realized that to have the maximum number of groups of 10 would require four 5s in those 3 rows, you can quickly deduce that there can only be eleven groups of 10, four on one long line, five on the other, and two in the triangle, meaning that the blue cells were 7-8-9, and the triangle could not be 1-2-3-4 (which he also took a much more convoluted path to determine). The logic paths that he finds consistently amaze me, so I'm still not disappointed. It is kind of like being surprised when Federer would hit a simple forehand into the net.

[@jasonclough9380]

I did something similar. 1) Can't have a 5 on the blue dominos, otherwise it breaks. 2) The black loops must have 3 5s in them. 3) The combination of sum for the loops can either be (a) 50-50-20 (no triples) with blue = 15, (b) 50-40-20 (2 triples or 1 quad) with blue = 25, or (c) 50-50-10 (1 quad) with blue = 25. 4) Since we have 3 5's, there has to be at least a triple or a quad. 5) Therefore, the blue dominos had to add to 25.

[@tman9641]

That moment when you think you're sure you know a digit that Simon missed........then 45 minutes later he proves you know nothing. brilliant.

[@Playmaker6174]

Yeah... my comment about "how intricate this puzzle is" is still valid after this video. Certainly one of the most intricate puzzles I've seen from Qodec, but also among the most brilliant ones. I guess I'll binge watching this movie first just to experience this crazy solve path once again x)

[@stangerrits6712]

These colouring puzzles always have this tipping point where the colours are not really adding to the solve anymore. This is where I would clean up and stop colouring everything, but after watching many many CtC videos I learned to accept that whenever Simon starts colouring, he doesn’t reckon his solve completed before every single cell has been coloured. Also, since Simon looked at the lines in the grid and saw it resembled a spectacled man looking back at him, the polychromatic chaos he produced might very well resemble what’s going on in the man’s mind 😊

[@ShakalDraconis]

A much cleaner way to identify that the three starting blue digits had to add to 25 was to continue on just slightly longer on the 10-pair count from the start. At most 12 digit-pairs giving a max of 120 in green, thus requiring at least 15 in the blue digits. BUT, then realizing you have 3 5's in green, which can form ONE pair but require at LEAST ONE 10 to be made from more then 2 digits, drops the maximum sum of green down to 110, thus demanding blue add up to it's maximum possible of 25.

[@inspiringsand123]

Rules: 03:47 Let's Get Cracking: 05:39 Simon's time: 1h34m45s Puzzle Solved: 1:40:24 What about this video's Top Tier Simarkisms?! Bobbins: 6x (22:57, 22:57, 31:14, 1:05:51, 1:32:42, 1:32:42) The Secret: 5x (07:12, 07:12, 07:21, 07:27, 1:05:15) Knowledge Bomb: 1x (09:37) Three In the Corner: 1x (1:38:28) Maverick: 1x (1:01:30) And how about this video's Simarkisms?! Hang On: 22x (10:05, 15:08, 15:13, 16:45, 16:55, 17:35, 19:56, 21:58, 22:55, 43:51, 1:02:13, 1:04:19, 1:08:38, 1:12:02, 1:12:09, 1:20:16, 1:23:31, 1:23:31, 1:27:09, 1:27:21, 1:33:19, 1:36:31) Ah: 22x (12:23, 14:15, 23:02, 28:29, 28:32, 39:05, 39:16, 45:01, 45:40, 1:04:14, 1:04:16, 1:04:28, 1:08:47, 1:10:03, 1:13:55, 1:14:33, 1:30:00, 1:34:52, 1:35:24, 1:36:46, 1:38:13, 1:40:03) Sorry: 17x (09:59, 18:30, 24:45, 27:51, 32:12, 38:36, 44:38, 45:07, 51:59, 57:23, 57:23, 57:23, 1:01:32, 1:09:53, 1:14:36, 1:30:21, 1:37:38) By Sudoku: 14x (34:12, 39:41, 46:19, 53:01, 1:09:43, 1:19:46, 1:20:34, 1:25:12, 1:26:26, 1:28:40, 1:33:58, 1:34:33, 1:34:37, 1:37:09) Obviously: 10x (01:15, 02:43, 09:04, 10:38, 12:39, 12:39, 25:21, 55:39, 55:49, 58:55) Clever: 8x (22:45, 22:48, 41:26, 50:12, 1:01:43, 1:13:19, 1:40:43, 1:41:34) Beautiful: 5x (23:09, 23:52, 30:28, 1:04:28, 1:31:23) Wow: 5x (38:56, 49:46, 1:04:36, 1:30:36, 1:40:03) Magnificent: 4x (00:26, 01:18, 1:35:48, 1:41:42) Unbelievable: 4x (1:41:24, 1:41:27, 1:41:30, 1:41:30) Brilliant: 3x (02:12, 02:23, 1:40:34) I've Got It!: 3x (22:43, 30:28, 1:33:47) Thingy Thing: 3x (29:01, 29:01, 29:05) Good Grief: 2x (50:40, 1:04:28) What on Earth: 2x (12:59, 1:17:03) Useless: 2x (08:13, 1:22:13) Goodness: 2x (50:38, 1:01:28) I Have no Clue: 2x (44:22) Lovely: 2x (03:02, 1:38:25) Come on Simon: 2x (1:06:33, 1:28:52) What Does This Mean?: 2x (04:16, 1:37:25) Pencil Mark/mark: 2x (26:28, 29:14) Cake!: 2x (02:24, 02:41) Naked Single: 1x (1:39:23) Out of Nowhere: 1x (1:27:28) Recalcitrant: 1x (1:35:37) Naughty: 1x (54:22) Bingo: 1x (1:27:41) Horrible Feeling: 1x (05:42) Extraordinary: 1x (1:41:06) Deadly Pattern: 1x (1:38:57) Hypothecate: 1x (45:36) Discombobulating: 1x (36:32) Bonkers: 1x (1:08:29) Shouting: 1x (45:04) Facetious: 1x (1:12:13) Surely: 1x (18:18) In Fact: 1x (1:23:20) That's Huge: 1x (45:44) Nature: 1x (1:23:08) Symmetry: 1x (42:25) Most popular number(>9), digit and colour this video: Ten (98 mentions) Two (125 mentions) Green (80 mentions) Antithesis Battles: Low (75) - High (59) Even (14) - Odd (6) Row (29) - Column (10) FAQ: Q1: You missed something! A1: That could very well be the case! Human speech can be hard to understand for computers like me! Point out the ones that I missed and maybe I'll learn! Q2: Can you do this for another channel? A2: I've been thinking about that and wrote some code to make that possible. Let me know which channel you think would be a good fit!

[@jamesshearer3936]

You are amazing. At one point early I figured out a restriction on the fives in the top right box before you spotted it. I was quite impressed with myself because I couldn't have gotten anywhere at all with this puzzle without your genius guiding the way. Hat's Off to Simon!

[@jonnybolton4659]

I think an easier start is to consider Column 1. 5 cannot go on any of the 2 cell lines nor the 4 cell line, so it has to go in R3or4, i.e. in the green area. Then when Simon says around 10:30 that the left spectacle could add up to a maximum of 50, this maximum is actually reduced to 40 because the 5 must be in a 3 cell sequence. Then the maximums for all three lines are 40, 20 & 50 and the blue cells have to sum to 25. And then thinking about where the 5s can be placed in the green area (in particular where it is limited on the right spectacle) actually places 5 in C1 and reveals there must be a double 5 on one of the spectacles. All without having to do the more complicated logic around the count of high and low cells in green.

[@mikkotoivo]

If there was no 5-5 pair on either spectacle, those would add to maximum of 80. Then the blue cells would need to add up to at least 35.

[@davidrattner9]

Happy birthday Amy!! Nice that you get a fabulous Qodec puzzle!!

[@PuzzleQodec]

Happy happy birthday Amy! (And many thanks to you David for notifying Simon!)

[@davidrattner9]

@@PuzzleQodec can't wait to see this puzzle solved of yours!! Your setting skills is amazing !!

[@PuzzleQodec]

@@davidrattner9 Thanks David! I try :)

[@longwaytotipperary]

Happy birthday Amy!!!

[@amysteele2488]

Thank you all - especially to David for notifying Simon. That was a very kind thing to do, and really made my day

[@markbennet9058]

Very good to see Qodec being featured - there are many more good puzzles out there

[@june4976]

A full feature film from Cracking the Cryptic! Great puzzle, great solve!

[@vishvananda]

I definitely needed some help on this one. Simon made some amazing deductions. I did a lot more coloring before getting digits. I was surprised he didn’t look at column 1 more. I saw pretty quickly that the L in the bottom can’t be 1234 because you can’t fit two other 2 digit 10-sums in the column, which means the two digits on the eyeglasses in column 1 are 5 and the digit in r9c2. That helped me color all of the boxes on the left and determine that the light green/red pair was the same as orange.

[@emilywilliams3237]

I am always, always happy to see a Qodec puzzle on the channel - really interesting how the puzzle worked, and I loved watching you solve it, Simon. What a great video. I thought it was quite brave of you starting a Qodec puzzle with those sorts of statistics so late in the day, and I'm very glad you got through it and posted it on time! Unfortunately I could not watch it when posted, but I knew I would come back to finish it because, well, Qodec. Thanks, Simon.

[@pouletbelette]

Same here! Thanks for your comments, Emily. The kindness they contain matches the regularity with which you post them (including on Mark's videos).

[@emilywilliams3237]

@@pouletbelette and your kindness makes me very happy - thank you!

[@maxgreve7856]

Amazing, Amazing puzzle! Made good progress at first and then I got REALLY bogged down in the geometry. Over two hours for me. Great puzzle by Qodec and great solve by Simon. Sorry, I know I say it a lot, but I kind of like these longer puzzles. I really hope we continue to see these kinds of solves on the channel even if they do take longer.

[@aidarosullivan5269]

Please don't apologize for Maverick, I'm looking forward to it making a cameo in every video

[@flobiish]

So much time trying to put 1234 on the triangle, so little time remembering the high vs. low digit math he's done so many times. Keep it up, Simon. You'll get there before I finish the video, I'm certain of it. Edit: And once again, I make the comment, hit play and then he does it. LOL I have to say, I'm having fun with this video. Kinda feel bad for him having a rough time though.

[@emagrant1485]

I tend to do the solves alongside as I am still just learning - it is fun sometimes when i remember him saying something that lets me go ahead while he explains it again, but this one was slightly painful listening to a long debate after I had already put in yellow and resolved my 5s

[@davidhughes7174]

Astonishing, amazing and beautiful. Thank you Qodec for testing Simon. Simon, another A , thank you

[@studgerbil9081]

Amazing. About once a week, CtC solve a seemingly impossible puzzle and I'm certain it is the most difficult they have ever solved. This one holds up that standard.

[@eveengineer4147]

33:50 I figured out that the middle triangle had to sum to 20 because the blue cells summed to 25. Based on the low digit allocation one spectacle must sum to 50 (5 different 2 cell 10 sums) and the other one must sum to 40 (either a 4 cell sum including the 5 plus 3 different 2 cell 10 sums, or a 3 cell sum including the 5 plus a 3 cell 10 sum and 3 different 2 cell 10 sums). Thus the middle triangle must sum to 20 to make the math for the 3 rows work. Knowing that the triangle is 20 means the yellow digit must be on a spectacle and as Simon points out later it must be part of a 3 cell sum, meaning it has to be on the left spectacle with the 3 cell sum including the 5.

[@olivier2553]

As soon as Simon mentioned the three 5 in green, one 5 had to be in a group of 3, there where only 11 possible groups of sum 10 (total of 110) in green. So the blue cells sum to 25 There were only 24 green cells left. The only combination that would work was a double 5, 8 groups of 2, 1 group of 5+2 digits, and 1 group of 3 digits. The possibility of 5 in a group of 4 cells was not even remotely possible :) On the different ways the human brain works, it is quite amazing. I would not be able to start that puzzle, I would not be able to guess about using the 3 rows, but as soon as the 5 came in the picture, I could see the above logic.

[@RichSmith77]

​@@olivier2553 Why do you say 5 in a group of four digits wasn't remotely possible? There could have been a repeated low digit, so either 1+1+3+5 or 1+2+2+5 was possible (at least initially).

[@olivier2553]

@@RichSmith77 You are right, the left spectacle could have been 5 in a group of 4 and 3 groups of 2.

[@imblackmagic1209]

absolutely monstrous, congratz Simon! thanks for the solve!

[@HunterJE]

Video comes back at 20:15, doesn't seem to miss much but Simon pondering

[@CrackingTheCryptic]

That is a weird glitch. I checked the source video and you don't miss much: I say "I’ve got to pair up two 5s and the 5 that’s left over is going to consume at least two low digits… I think it could take three with something like that" and then it continues...!

[@kathyjohnson2043]

I thought it must have been my internet stream

[@Coyotek4]

I worked on this, took a break, worked again on this, realized I errored after about 75 minutes, backtracked, worked again, ended up bifurcating on r5c3 (I'm not proud) ... and eventually clocked in at a time of 2:12:21. Insanity, thy name is ...

[@eytanz]

What a great puzzle! I did things in a slightly different order - I first figured out that the 5 in column 1 has to be on the 10-cell line, and that then constrained it in such a way that when I got to looking at rows 3-5 I moved through them a bit quicker. And then I got stuck for a long time at the end because I totally blanked on the diamond being able to have a double digit.

[@CreativityCurve]

Really happy I managed to solve this on my own, even if it took a while. Definitely one of the most colorful puzzles I've done!

[@DrGonzoSokkaJenius]

Well done Simon. Love the way you start with realising there must be 11 ten cells in the middle then completely forgetting that by trying to make the triangle a 4 cell 10. Hilarious 😂😂😂

[@jetranger57b]

Had the same thought - much more consumed with the number of high and low digits as to how to divide up the required 11 groupings of 10

[@Emmibean77]

A phenomenal puzzle, beautifully solved by Simon. Well done, truly! Very impressive 🎉

[@redteardrop]

remarkable solve on that one. I am very surprised I didn't see you do more with column 1. you can immediately rule out the bottom line being 1234 as that would not leave enough low digits to finish both the 10 cell pairs in r1,c1 or r5c1 meaning it has to have a pair of 10s. combine that with the only place a 5 could be (r4c1 or r5c1), the other of which would have to be the partner to r9c1 for the column to add up to 45 since there would now be 3 other 10 sums already used (r12, r45, r78 + the 5 in one of 2 places). It all works out in the end and there's no way I would have been able to do this puzzle in it's entirety, but I am quite proud of myself for seeing that very early on. I'm just glad it was proven to be true later on :)

[@isabellns]

YAY, another happy birthday!! Thank you so much, Mariana and Simon!! ❤

[@davidrattner9]

Hope u had a wonderful bday !!

[@PuzzleQodec]

Happy birthday!

[@longwaytotipperary]

Happy Birthday

[@sanabas1]

Did this one when it appeared on LMD, it's a 10-star puzzle. Absolutely fantastic. Skimming the video, it's weird just how different Simon's solve path is to what mine was.

[@phillipcarlroberts4640]

"Light green is orange or purple, but I don't know which"....Vintage stuff, great solve BTW.

[@flobiish]

@45:56 "I've got to put yellow in one of those two squares, and either of these squares is really interesting. Yep, never talk to me at a party." I think anyone watching these videos would love to talk to you at a party.

[@steve470]

But only because we're Simon's favourite people.

[@schoeMful]

My thought process was: If the green region adds to a maximum of 120, with all 10-sums being in dominos, there can only be a maximum of two 10-sums that aren't dominos to sum to 110. This rules out the 1234-triangle, because there wouldn't be a match for the third 5, and there was some deduction earlier that seemed easier to me as well. But I might have missed something there.

[@mstmar]

Great solve as always. i had a slightly different break in that involved Some really beautiful logic. In col 1, both of the lined dominoes contain a low digit, so you can only have at most 2 of them in the L, meaning the L is 2 dominoes. everything in the first column is a domino that adds to 10, so where does the 5 go? on the spectacles. in particular, in places where it can't be a 55 domino (since both see their neighbors), and so it's in a 3+ cell region. now if we count like simon did at the beginning, we can only put 40 on the left, 50 on the right, 20 in the middle which leaves at least 25 in the 3 blue cells. this also tells us that the right and center are all dominoes, which simon took a while to notice in other ways. It does miss out on knowing that there needs to be a 55 domino somewhere on the spectacles tho.

[@Raven-Creations]

This is another fiendish masterpiece from Qodec. The use of geometry and maths to ensure a unique, discoverable solution is excellent. It would be quite easy to make a mistake in this, because the use of diagonals means that there are often more possibilities than initially spring to mind. I had to check myself a few times, where I nearly jumped to a wrong conclusion. I used the line tool to overlay the segments, which helped a lot in spotting where paired digits were restricted. I started with a similar train of thought to you, but at the point where you got to 5s, you veered off into deep undergrowth. The path was much easier. The crucial thing you missed was that one of the large lines could have double-5s, but the third 5 must need at least a 3-cell run, which reduces its line's maximum to 40, and bumps up the extra three cells to 25. We can't afford to lose any more 2-cell sequences because that would reduce the total further, and bump up the total of the three extra cells to 35, which is impossible. The three extra cells must therefore sum to 25 and be 889 or 799. The triangle must have two 2-digit sequences. There must be either: nine 2-cell segments (including the double-5) and two 3-cell segments, or there are ten 2-cell segments and a 4-cell segment. Apart from the double-5, all 2-cell segments use a low and a high digit. If the single 5 is in a 4-cell sequence, it's 5 + three lows. If there are two 3-cell sequences, one is 5 + two lows, and the other is a high + two lows. This ensures that we have the right count of highs and lows. Whichever digit is doubled in the three extra cells, it is excluded from the triangle, so its doubles partner must also be excluded. E.g. if they are 799, there is no 9 on the triangle, so neither is 1. There is a 9 at one end of R5, so there is a 1 on one end of R6. This means there must be a 1 in R4C4/6. The geometry means this can't be 19, so it's part of a 3 or 4-cell sequence. It's definitely on the side where the single 5 goes. The same logic applies if the extra cells are 889. In C1, the 5 can only go in R3 (R4 would prevent any double-5). This must be a single 5, and it cannot reach the 12 which must be in R4C4, so there must be two 3-cell segments on the line. The double-5 cannot go between boxes 4 & 5, because R4C4 needs to be the 12 which is part of the second 3-cell segment. Therefore the double-5 is between boxes 5 & 6. We now know there are two 3-cell sequences, and two 2-cell sequences on the left, and five 2-cell sequences on the right, and the double-5 tells us how they are positioned. I wish I had found some smart way to finish the puzzle. I ended up colouring the four 2-cell combinations like you, but also flashing one of each pair with light grey, and one with dark grey - effectively giving me light-blue/dark-blue, light-green/dark-green, etc. This style of colouring has proven very effective before, and it was here too. It enabled me to progress the colouring to an extent where I could prove the diamond had to be a single sequence, but it couldn't contain 1-4, so had to include 5 (where you had the option of 6, my colouring ruled that out), and it must have a repeated digit. This was enough to complete the puzzle, without requiring it all to be coloured. Considering you missed the deduction caused by the single 5, reducing the large lines' sum to 90, it was a pretty good solve from you. I would strongly recommend you try my flashing method to identify the ends of paired digits; it's effective, economical with colours, and easy to scan. Because the greys are numerically lower than the colours, it means the flashes are consistently placed below the colour, so you can scan across the top half of a row to easily spot missing colours.

[@spoilers9723]

At the 18-20 minute mark it feels like the articulate way to say it is that you have 3 5's, but since you can't put them both in 5-5 pairs, therefore you can't have the maximum number of 10's... Like we have either 11, 12, or 13 tens to leave behind 25, 15, or 5 in blue, then we realize that each of them being a 2-cell pair would mean you have 12, but because you need 3 5's they can't all be in a 2-cell pair, meaning that there must be 11 10's, and therefore 25 sum in the blue cells.

[@RiseofPerfection]

Yeah, my way to figure this was, if one 5 has to pair with 2 low digits, on one line there have to be 2 three cell 10s. Meaning, one of them has a max of 4 10s

[@yaboyJJJJJ]

its weird that he literally proved this like 10 minutes before that even, then just refused to use the logic

[@randomjunk1977]

I think there's a much simpler way to demonstarte very quickly that the three extra cells in the "face" have to add to 25 without going through all the low digit allocation etc. We know the eyes and nose have to add to a multiple of 10. So the first question for me is, what's the largest possible number? There are 24 squares forming sets of 10, and obviously every set of ten must have at least 2 squares. So the maximum number of sets is 12, which gives us 120. The exact number we need if the extras added to 15. So 15 and 25 are the only options for the extras. Except we can show almost instantly that 120 for the face is impossible. For the reasons Simon explained, it's fairly obvious we can't put a 5 on any of the stub lines. Well, the only way to get to 120 is if all three of the 5s are able to exist as part of 2-cell sets of 10, which is obviously impossible. As soon as even a single set of any kind requires three digits we can't get to 120 and the three extra cells now must total more than 15 and the only valid number left is 25.

[@pascalm1305]

This puzzle is brutal! It's just such an amazing puzzle and solve! Well done both.

[@KSilverlode]

Thanks for the birthday shout-out. I'm happy if my email has made your day; watching your videos has so often made mine! 😊

[@JohnSmith-zq9mo]

The deduction to get that yellow is equal to 1 was great.

[@ymiros0953]

To be honest I do not quite understand the conclusion at 1:25:00 that the 6 has to go into c4r5, c5r5 or c6r5 just because either c7r6 or c8r6 is either a 4 or a 6. The 4 having to go there is quite obvious if you just look at the row and see that blue can't be a 4...... Well now that I types it out I feel stupid, the 6 obviously has to go with the 4 since there needs to be a 10-pair on the triangle, but I'll post it anyway in case anyone else is confused 😂

[@tedg1278]

I really needed the help finding the first place to look and then a bit to get over the hurdle of how to use that information. Getting the green and blue groupings up and the mention of 5s lead to some very simple ways of seeing part of it. The "spectacles" could have at most 10 groups for a total of 100. The triangle can't have a 5. There are three 5s in the green region meaning at least one set containing a 5 with at least 2 other cells. A 3 cell group leaves 17 cells to pair for a maximum of 8 pairings with another 3 cell group. Maximum total for the "spectacles" is now 90. Maximum total of the blue cells(25) requires the green cells to total at least 110. Maximum total of the green triangle is 20. "Spectacle" maximum of 90 plus triangle maximum of 20 is required to reach the maximum for the blue cells. Therefore the triangle must be formed with 2 pairs, the spectacles must have a double 5 pair and the remaining 5 in that area groups with as a 3 or 4 cell cluster.

[@ymiros0953]

I think it's really interesting how many different ways there are to figure out that the 3 digits Simon marked in blue very quickly must add up to 25 (The way Simon did it in the end of course is the most complicated I've seen so far :P)

[@chrissolnordal9421]

This is one of the best puzzles I have done in months! Thank you so much to Qodec and of course to Simon and Mark for highlighting these puzzles and showcasing them in such an entertaining manner. As seems to be becoming a usual behaviour for me, I got fixated on some erroneous logic that held me up for hours. I eventually took a peek at the video and saw Simon say "so this diamond must be a set of four" (something I had already deduced), "I still don't know if it is a 1-2-3-4 or has a repeated digit". Oops! There was my error. I'd assumed it must be 1-2-3-4 if it was a set of four. With that sorted, I got to the solution quite quickly. Lovely logic!

[@RichSmith77]

I think the easier way to get the initial break-in is to think about how many units of 10 you have to have in green. The most you could have would be if every 10 was a simple domino, giving 5 on each spectacle and 2 on the middle triangle, for 120 of the required 135 total of the three rows. But that wouldn't work, because it's not possible to allocate the three 5s across dominos alone. At least one of the 5s has to be in a three or four cell string. But you cannot reduce the number of 10 strings in green below 11, since that would leave blue having to add up to 35 or more. So there are EXACTLY 11 '10’ strings in green , summing to 110, and blue sums to 25. Also, 5s cannot appear on the triangle, so one of the spectacles contains a non-domino with a 5, and will therefore contain at most four '10’ strings. So the other spectacle and the central triangle will have to max out at five and two '10' dominos respectively. I haven't finished watching, or read any other comments yet, so I may be repeating what someone else has said. Sorry. Took me over three hours to solve. 😂

[@Paolo_De_Leva]

I agree 😏👍 By analyzing *column 1,* it's easy to find that there must be a *5* along the long line on the left, which forces it to be a 4-segment line, adding up to *40...*

[@MrDaDisha]

The birds chirping in the background turn sudoku into zen-like experience!

[@ronjohnson6916]

Thar accidental blue in R4C4 was driving me crazy. Glad he caught it before it became a problem.

[@bena-z5k]

at 54:34 why can't purple and orange be the same pair, but flipped?

[@PuzzleQodec]

That's because r4c7 sees both r3c7 and r4c8, so has to be different from both.

[@bena-z5k]

@@PuzzleQodec ah you're right, so easy. Thanks

[@Gonzalo_Garcia_]

55:36 for me, although not a proper solve, as I completely broke the puzzle at one point and needed help from the video to fix it. Not happy with this one, but it is what it is. Very beautiful puzzle anyways!

[@piarittersporn]

A great puzzle and a tremendous challenge. Fortunately, after several restarts and interruptions, I was finally able to solve it.

[@ianoz1]

An absolute beast, but a lot of fun once you get your colours happening.

[@PathOfShrines]

Very cool puzzle. I had dismissed early in my solve that sums would be helpful, but when I revisited that and figured out that the right structure had to have five pairs, that was a really nice moment. I spent quite some time staring at that structure and trying to directly prove that arranging it orthogonally wasn't going to work before I finally found the line of logic proving that there had to be 5-5 pair and it couldn't be on the left structure. The reasoning from there was... still not easy; just a fun time from start to finish. 98:16 for me.

[@highpath4776]

Simon needs to play "green grow the rushes - o" tomorrow. (I'll give you one-oh, green grow the rushes (etc) Gringo.)

[@ericklugaro730]

I went and noticed that I couldn't get enough grouping of tens from the green rows, with the 5s i could only get to 120 and that set me on to the double digit path but also made it easier to limit the triangle as 1234

[@ericpraline1302]

I found this monstrously hard, seemingly making endless deductions which got me nowhere. I feel like I laboured to a conclusion even though I never bifurcated. Now to see the litany of things I missed and which Simon spotted in a few minutes...

[@Faktor13]

Me reading the top right corner: Oh Qodec...oh my god, Qodec *panic mode* I fear his puzzles more than Phistomephel...

[@srwapo]

86:45, I'm shocked I was able to beat this. I needed a tiny bit of help, though. I was concentrating on Box 5, wasn't getting anywhere, and I skimmed enough to see that I had to think about the pairs in Box 6. I found disambiguating the long lines on the left side of the grid the hardest, but eventually came up with some neat logic.

[@shaunholt]

I want to see Mark pencilsimoning numbers in a Sudoku puzzle.

[@sushirunner]

Shorter chain of reasoning for about 20:00 : think where 5 needs to go in column 1, this 5 can't go with another one. therefore the sum of the "10 segments" can be a maximum of 110.

[@michaelkaster5058]

did go down the path of 1234 in the central box to failure of backing up to there, but solved it in 90, without talking to the camera lol

[@BramMertens]

Remarkable how the line colours for the ten sums in boxes 1 and 3 end up being correct :)

[@HunterJE]

35:00 Wouldn't using up four lows to put 1234 on the triangle cause problems w.r.t. high digits in green as discussed before?

[@CarterPatterson1228]

Yeah, 4 lows on the triangle plus at least 9 paired with the highs would be 13 lows in 3 rows

[@megoesmo0]

The easiest way to figure out the triangle can’t be 1234 is, putting 1234 on the triangle means the green cells only add up to 100 (4 groups on the left squiggly line, 5 groups on the right squiggly line, and the triangle. That means the blue cells would need to be 35 which is impossible.

[@Baritocity]

Early in the puzzle, I thought I found pretty good logic how by sudoku a repeat digit on the second three-cell 10 sum would break. Played myself.

[@bluerizlagirl]

I was shouting at the screen for a good half hour there! There was only ever one way to partition the digits in R3-R5. If you tried to put 120 on the ten-lines, you could use up two of the fives in a 5-5 sum, but the third would have to go in R3C5/R5C1/R5C9 where it can't. So you have to put 110 (= 12 low digits, three fives and nine high digits) on the ten-lines and 25 (= three high digits) in the remaining cells. That leftover five will have to go on a ten-line somewhere, which cannot be only two cells long. We can make 90 in 18 cells with double five and eight high-low pairs. Now we have 20 in the form of a five, a high digit and four low digits to fill the remaining six cells; either as two three-cell sums, or another high-low pair and a four-cell sum which would have to include a repeated digit. All Simon would have had to do was think about how the ten-lines were made up. He skated close to the edge of it quite a few times ..... Still, I suppose everything's obvious once you've seen it ..... and I can't say for certain I would have spotted the trick if I'd had a camera on me .....

[@shteevuk]

Superb puzzle! VERY Tough tho, I had to restart completely a few times, kept falling into some evil pitfalls!

[@oneeyedman4431]

Lost me at 48.00 brilliant puzzle, brilliant solve. Thanks for my nightly entertainment CTC. Get very excited when I see the solve times are long……🍿🍿🍿

[@logiciananimal]

Given the first digits deduced seem to "have to be" 5, is the title based on _Hawaii 5-O_?

[@flobiish]

I'm about a half hour into the video wondering when Simon's going to ask where 5 goes in column 1. Edit: apparently it's right after I post this comment, lol.

[@AngelWedge]

I was thinking the same :)

[@flobiish]

@@AngelWedge I’d been thinking about it since the puzzle started.

[@tnkurse]

I could only get to the point where 3 cells would sum up to 25 and detected 5s in C1, I figured the left group has 40 sum which I dont have a clue what to do next then I gave up lol Magnificent puzzle indeed.

[@bait6652]

Its crazy which opening determines how fast u can solve a puzzle. Choosing the 5 pairs is slower then the special digit...but choosing the 2-bifurc(the digit overlap in the leftover cells) after the 1st special digit is faster then playing w the lines

[@richardglover314]

That solve was magic and like all good magic you stare in disbelief how it was done. There were two ways to make the leftover 25, double 8 + a 9 or double 9 + a 7 ; so two surplus 2s + an odd 1 or two surplus 1s + an odd 3. I am watching Simon solve thinking this does not add up not seeing that take out two 8 from the three rows leaves only one 8 to go with the two 2s or take out two 9s leaving just one 9 to go with the two 1s. In the end the odd 3 went with a 2 that "belonged" to an 8 that made up it's ten by claiming the two surplus 1s. WOW

[@dinane]

I had to redo my coloring 4 separate times. I kept having more than 9 colors and getting confused… I did really enjoy the puzzle though.

[@AndrewHartshorn]

Love the puzzle - no way I could do this myself and haven't finished the video yet but the triangle can't be 1234 because it uses up too many low digits - leaving only 8 for the 9 remaining high digits and the lonely 5.

[@AndrewHartshorn]

60 seconds later in the video Simon gets there - sorry for shouting Simon!

[@jdferreira]

28:38 Simon spots an L-shaped line in the corner of the grid 28:43 Simon says «this reversed-Italy-boot here» 😆

[@Nevir202]

Anyone else seeing this and going, *You feel your sins crawling on your back*...

[@HunterJE]

53:10 Can remove the 8 pencil mark from the tip of the triangle - that cell must be the non-repeated digit in green, and the non repeat is never an 8 (8-8-9 or 9-9-7 being the only two ways to hit 25)

[@asbjrnfossmo1589]

Not that I was able to finish the puzzle, but it is funny to see Simon struggle so much with the very first deduction I made: The center triangle CANNOT be 1234 because of sheer low digit shortage. 9 high digits in green needs at least 9 low digits, 4 wasted in the centre would make 13 low digits in 3 rows before factoring in the last 5, which would take us to 15 or 16 low digits in just 3 rows...

[@ukdavepianoman]

Extraordinary puzzle - very, very difficult. Simon's first half was slow - forgettng deductions made e.g. the triangle in box 5 not being 1234 and forgetting what the two "eyes" must add up to. Second half he was brilliant especially determining that yellow was one. Why did one have to continue to have two colours? Just use dark green for 19.

[@Rach881101]

94:33 for me. Brutal puzzle!

[@roryjackson1524]

the green area is always a multiple of ten so the 3 pink digits are a multiple of 5 because of the secret applied to the 3 rows as it is a total of 135. you can't put 5 in the pink cells as they are on 10 lines and so you need to put 3 x 5s in the green region. that means there is a maximum sum on the green region of 110, 2 x10 regions on the triangle, 4x 10 regions on the line with the 2 fives (as if you have one 3 cell region on the line you need to have 2 3 cell regions) and 5 x 10 regions on the line with one 5 (which has a repeated 5 on it.) If you were to reduce the number of 10 regions you would reduce the 110 total by 10 every time which wont work as you can't make 35 with 3 sudoku cells. hence the pink numbers are 997. I did a bit more than that and then messed up the sudoku, cheers.

[@titusadduxas]

Why couldn’t they be 988?

[@roryjackson1524]

Good point, I think it could have been so it doesn't solve like that. I got lucky. Ffs

[@roryjackson1524]

I suppose you would still get 9or8 at the top and then and 897 in the other two and it might work

[@roryjackson1524]

Yeh it works like that also I just missed 988 as an option, you'd end up with what Simon has in the pink squares

[@titusadduxas]

I haven’t got that far yet (after 65 mins)

[@AFT3RDAY5]

In the end I couldn’t do it without some slight hint from Simon. But it is very funny that I made a lot of the deductions completely different and yet they all come together the same way 🤩👍🏻

[@bait6652]

Wow one ridiculously hard puzzle even tho the opening wasnt. completed 2 10 dominos with relative ease but struggled on the other 2 Cant wait to watch the video to see a nonbifur solve. best part of P&P solve is when u strip all the nonsymmetric parts , solve the opening only to realize you mirror-flip the grid cuz you chose the wrong openning(10 count was the wrong choice better to use the special XVsum digit). Cuz i chose 10-counting ended up with 2-bifurc, but the wrong path took few steps beform elim. The right choice quickly place 2 sets of dominos.

[@daveturner5305]

Strangely my first thought was c1 - where can the 5 go? r3 or r4! box 5 can't have a 5 in r6. So 5 must be in box 6 r5. Am I wrong?

[@daveturner5305]

Was going to delete the back half of the comment 'cos I realised I got confused by the colouring and forgot about r6.

[@13vatra]

Man, you know it's a brutal puzzle when Simon only has three digits an hour in. And to clarify, that is not bad mouthing Simon, despite how self deprecating he's been in this video.

[@BakingFiend-rl7mo]

Oh man you mustve finished solving this at around dinner time judging by the videi length. Hope you had a great dinner afterward!

[@WeeTurtles7]

It's amazing to me that Simon was able to deduce that he only had 1 low digit to play with but somehow concluded at one point that 1234 was possible in the middle triangle, which would use 4 low digits, going well past the 1 digit leeway he had earlier deduced. 🙄

[@AFT3RDAY5]

I started on the leftmost column which made the start way easier. Not that the rest was any easy 😅

[@rustypotatoes]

tenno?

[@Robert_H.]

*General* First, High (H) = {6,7,8,9}, Low (L) = {1,2,3,4}. To add to 10, there are several possibilities: (01) H + L (02) H + L + L (03) H + L + L + L [ only {6,1,1,2} ] (04) 5 + 5 (05) 5 + L + L [ only {5,1,4} and {5,2,3} ] (06) 5 + L + L + L [ only {5,1,1,3} and {5,1,2,2} ] (07) 5 + L + L + L + L [ only {5,1,1,1,2} ] (08) L + L + L (09) L + L + L + L (10) L + L + L + L + L (11) L + L + L + L + L + L [ only {1,1,1,2,2,3} ] *Step 1* If we now look at rows 3, 4 and 5, we know that there is 12 x H, 12 x L and 3 x 5. We can divide the cells into Green (whole groups of 10) and Blue (half groups of 10) [8:05]. It must apply: Sum(Green) = {110, 120, 130} as well as Sum(Blue) = {5, 15, 25}, since Sum(Green) + Sum(Blue) = 3 * 45 = 135. Moreover, there cannot be a 5 in the Blue. If there is a 5 in Blue, that 10-sum line located in it cannot be made possible because then the other cell on the 10-sum line would also have to be a 5. This is not possible because they are in the same box/column. *Step 2* If all Highs are in Green (and no High in Blue), they must be together with at least one Low. Thus all Highs and Lows would be used up and in Blue the 5s would have to be, which does not work (Step 1). Accordingly, not all Highs are in Green. ------------- If 11 of 12 Highs are in Green (and one High in Blue), then all 11 Highs need at least one Low. So only one Low and three 5s are left to be placed in the two areas. So we get: Green = {11 x (H,L), (5,5)} Blue = {H, L, 5} OR Green = {11 x (H,L), (5,L)} Blue = {H, 5, 5} These cannot occur because a 5 is in Blue (Step 1). Accordingly, no 11 of 12 Highs are in Green. ------------- If 10 of 12 Highs are in Green (and two Highs in Blue), then all 10 Highs need at least one Low. So only two Lows and three 5s are left to be placed in the two areas. So we would get: Green = {10 x (H,L), (5,5), (5,L)} Blue = {H, H, L} OR Green1 = {8 x (H,L), 2 x (H,L,L), (5,5)} Green2 = {9 x (H,L), (H,L,L,L), (5,5)} Blue = {H, H, 5} The first case cannot occur because (5,L) in Green is not a possible 10-Sum. The second case cannot occur because a 5 is in Blue (Step 1). Accordingly, no 10 of 12 Highs are in Green. ------------- If 9 of 12 Highs are in Green (and three Highs in Blue), then all 9 Highs need at least one Low. So only 3 Lows and three 5s are left to be placed in Green. There are two possibilities here: Green1 = {8 x (H,L), (H,L,L), (5,5), (5,L,L)} Green2 = {9 x (H,L), (5,5), (5,L,L,L)} Blue = {H,H,H} They have to add up to: Sum(Green) = 110, Sum(Blue) = 25. *Step 3* So in the end we have the following configuration: Green1 = {8 x (H,L), (H,L,L), (5,5), (5,L,L)} Green2 = {9 x (H,L), (5,5), (5,L,L,L)} Blue = {H,H,H} Sum(Green) = 110, Sum(Blue) = 25. To get Blue to 25, only the following possibilities exist: {7,9,9} OR {8,8,9}. Therefore there has to be a digit twice (8 or 9), which has to be in r3c5. The other blue cells have to be 7, 8 or 9. No 6! *Step 4*

[@RichSmith77]

Shouldn't that be green 2: {(5,5), 9*(H,L), (5,L,L,L)}? [Edit: You changed it as I was typing 🙂] I feel all the High/Low analysis over-complicates the break-in. It's simpler just to think of how many '10’ strings you can have in green. 12 is the most, but this doesn't work because you cannot distribute the three 5s amongst only domino strings. It cannot be fewer than 11 because then blue adds to 35 or more. So exactly 11 strings, one of which has to contain a 5 in a three or four cell string. (And once you realise you need a 5 in r3 or r4 in column 1, then that pins down the left spectacle as being four strings, with the right spectacle being five domino strings and the triangle two domino strings.)

[@Robert_H.]

@@RichSmith77 That's pretty darn smart. I didn't understand Simon's explanation of why 3 High Digits should be in blue. So I just tried listing all the possible configurations.

[@mortalcoilerinokripperino110]

Me: L Simon, an intellectual: Reverse Italy boot

[@Ardalambdion]

A 10 sum isn't supposed to be so brutal.

[@blabla-rg7ky]

I would have never solved this one, because - even though Simon figured the break-in - I still don't get it despite of having been explained. Yeah, brutal puzzle, harder than others that have taken Simon longer to solve...

[@patrickelliott2169]

Was waiting for some else to comment on this, but... if "contiguous" means sharing an edge, then the center triangle only has a single group of cells that have to add to 10, and many other fills violate this rule. Also, the clues explicitly said that such groups, "cannot overlap". While I have no clue how to solve this thing myself, it seems like the 828 pattern that spans box 3 and 6 also violates the, "cannot overlap" rule. Now I could be utterly wrong, but a huge flaw in the system, unless set by someone else with the correct answer, is that the grid software only cares if you filled in a grid with numbers that fit the 1-9 in each box, line, and column rule. So... as I understand the rules, you got a valid solution, but may have broken the rules to get there. Am I utterly wrong somehow?

[@theblinkingbrownie4654]

I doubt most puzzles with unique rules have software specialized for them, I think they just store the solution set by the setter. I assume your confusion about the ruleset was cleared by watching the video.

[@minamagdy4126]

For early on in the puzzle, roughly 23 minutes in, after what you deduced about a pair of 5's being required for a ten instance in green, what about the third 5 being in blue? That would let all 12 high digits pair one-to-one with all 12 low digits, completing the sum as desired. EDIT: Simon himself showed that 5 cannot be in blue, as all blue cells are part of vertical dominoes adding up to 10.

[@SzanyiAtti]

5 can't be in blue, because it would require another 5 to sum to 10, and they would see eachother.

[@minamagdy4126]

@Szanyi Atti not true, blue sums up to a number that is 5 mod 10 (here, 5, 15 or 25). That was a corollary of Simon's first discovery. Also there are 3 blue cells. Also, I just realized why 5 cannot be in blue, that was Simon's second discovery.

[@BryanLu0]

​@@minamagdy4126 It is true, a blue 5 would require another 5 above or below it, which see it

[@iceberg54321]

54:42, SImon got lucky, The orange and purple pair could have been the same. and the grey pair could have been a different pair, repeating at the bottom of box six.

[@PuzzleQodec]

r4c7 sees both digits of the other pair so orange and purple must be different.

[@davejacob5208]

i dont understand the last part of the rules (and therefore dont want to start or get a spoiler): "digits may repeat freely (...) within sums" what sums? what is that supposed to mean? if it means the 10s, then the rule that says these "contiguous groups" mustnt overlap forbids these digits from somehow serving purposes for more than one sum of 10... !?

[@JASPACB750RR]

Am I wrong in thinking that the triangle in the center can’t be a 1234 quad? So for the lines we need to place 9 highs, 12 lows, and 3 mids (5). So if the center triangle is a 1234 it takes 4 of the 12 lows. leaving 8 low digits to deal with the 9 highs and the two needed for the 5. So we’re out of low digits with that method. EDIT: Never-mind, Simon finally got it 20 minutes after I made my post.

[@AngelWedge]

Took me 11:04 to place the first digit, then got completely stuck. Maybe colouring everything wasn't the best idea…