"I don't know what to do next." Proceeds to solve the next number. "Bobbins! I have no idea what I'm doing." Proceeds to solve the next few numbers. "How am I supposed to solve this?" Proceeds to solve the next numbers. Etc.
To be fair, that's how I solve pretty much every Sudoku on this channel. "This grid is impossible and I don't think I can make a single logical deduction...Except I guess that this digit can't be a 5?...I guess that means that this one can't be a 4." Repeat for anywhere from 30 to 300 minutes until the puzzle is solved.
The "extra power" of the givens is pretty clear by the end. For example, in a normal sudoku any given digit would only be able to "see" 20 cells (8 cells in it's box, 6 more cells in its column and another 6 in its row). However, in this puzzle the 4 in c7r4 can "see" 29 other cells because of the law of leftovers.
Let's Get Cracking: 07:37 Simon's time: 39m43s Puzzle Solved: 47:20 What about this video's Top Tier Simarkisms?! Bobbins: 1x (43:19) Maverick: 1x (06:42) And how about this video's Simarkisms?! Pencil Mark/mark: 16x (13:33, 15:14, 15:17, 15:41, 16:19, 19:30, 21:01, 29:31, 29:34, 34:51, 35:36, 37:35, 39:52, 41:09, 42:50, 42:55) Ah: 7x (08:06, 13:58, 24:54, 25:00, 30:14, 35:27, 46:07) Brilliant: 6x (02:03, 03:35, 06:13, 06:13, 47:28, 47:33) By Sudoku: 5x (19:20, 35:54, 39:13, 43:31, 45:31) Cake!: 5x (04:28, 05:20, 05:23, 05:54, 05:59) Sorry: 4x (12:21, 32:32, 34:59, 40:37) In Fact: 4x (25:38, 27:09, 27:09, 33:02) What on Earth: 3x (18:43, 25:03) Beautiful: 3x (02:10, 34:59, 37:03) Weird: 3x (25:44, 45:19, 47:53) The Answer is: 2x (09:26, 32:41) Clever: 2x (43:39, 43:41) I Have no Clue: 2x (32:52, 36:49) Ridiculous: 2x (20:16, 47:12) Hang On: 2x (10:52, 40:13) Whoopsie: 2x (34:16, 37:21) What Does This Mean?: 2x (26:31, 28:19) That's Huge: 2x (35:08) Unique: 2x (00:33, 01:05) Useless: 1x (24:51) Apologies: 1x (04:01) Naughty: 1x (39:43) Extraordinary: 1x (01:55) Gorgeous: 1x (33:59) Come on Simon: 1x (41:03) Epiphany: 1x (41:27) Masterpiece: 1x (03:28) Magnificent: 1x (48:14) Surely: 1x (35:49) Obviously: 1x (31:17) Progress: 1x (40:08) Wow: 1x (21:22) Let's Take Stock: 1x (35:49) Next Trick: 1x (19:47) Have a Think: 1x (13:22) Nature: 1x (02:34) Symmetry: 1x (45:17) Most popular number(>9), digit and colour this video: Fourteen (7 mentions) One (66 mentions) Green (22 mentions) Antithesis Battles: Even (6) - Odd (0) Outside (2) - Inside (0) Column (36) - Row (25) 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!
If anyone is curious about why this can have an unique solution with only 14 digits, when a regular sudoku needs a minimum of 17, here's some ideas to think about -not a proof by any means, but perhaps some things that can help form an intuition: In a regular sudoku, there is a certain amount of 'information' that's encoded from the limitations of the rules (1-9 in each row/column/box, and no repeating digits). So if we take a stack of 3 boxes, the digits visible in the boxes above and below, serves to restrict the permutations of digits that a box below can have. But if we shift one of the boxes over by one cell. Because of the requirement of boxes needing 1-9, only 2 columns of overlap are needed to define what the remaining cells in those 3 boxes (that aren't in the 2 overlapping columns) will consist of. When you compare to a stack of 3 boxes in a regular sudoku, you realize that once you fully define 2 of the columns, you know what the 3 digits are for the 3 remaining cells, in each box, and you don't require 'column logic' for that conclusion. In contrast, in stack with one shifted box, you get a more visibly disjoint set of 9 cells that have 1-9 becuase the column logic was never required for that - basically what we refer to as the law of leftovers. And in this configuration, the shifted box can provide additional information to the next stack, without sacrificing the integrity of the information it can squeeze from it's own stack. In that way, a regular arrangement of 9x9 grids has both limitations on how much information can be transferred across it's different 'constraining sets' (rows/columns/boxes) as well as redundant limitations provided by it's structure, that provide multiple unique logis paths of restricting the value of a cell to a unique digit. One way to think about how a cell can be restricted is to think about the last digit you fill in a sudoku grid. That can be determined by seeing four of a digit pointing into that box and going in the only empty space (hidden single logic) OR you could look at the box and see which number is missing, or the box (naked single logic). Essentially, in a regular sudoku, the redundant limitations on cells that's provided by rows and stacks of boxes, means that you end up with a 'choppy' information horizon, where it's difficult to design a solve path for a puzzle where there isn't consistently too many, or too few deductions that can be made. That leads you to require an 'inefficient' number of given digits, to weed out all the deadly patterns that remain. Once you decouple box logic and column logic, where the box logic can finalize the horizontal poistion of digits within its box, which allows that to transfer over column logic to another stack etc. (or decoupled rows of boxes providing row logic information to other box-rows), you reduce the information redundancy among a stack, and provide additional column information to other stacks that is otherwise 'lost information efficiency'. You can use the example of placing the last digit in this puzzle to see how 'less pieces of logic' COULD be available to solve the last digit of this puzzle (depending on its location) e.g., if r1c1 was the last digit placed, you would only have 'naked single logic' from the box available, rather than four seperate logic paths that all lead to the same conclusion. This sort of 'decoupling' of box, row and column logic, allows the information distribution along the solve path to be more efficient, and require fewer given digits to arrive at an unique solution. One thing to note, is the ability to preserve the information density of a stack, comes from shifting only one box by one column from fully aligned (and similarly for rows). A 10x10 grid ensures that this will be true for every stack and box-row. The proof of why is left as an exercise to the reader! 🙂
One thing i noticed while watching is that the given 4 in the dark blue box interacts with 8/9 boxes, whereas in standard sudoku it would only interact with 5
One thing I noticed on the starting grid, was there were 16 guaranteed cells in (as-opposed to only 9 on an 11x11). That's a huge disparity. That's why (that 16 as opposed to 9) I think had something to do with it. What's the minimum of digits on an "11x11" of this nature? Very good though I guess, but what's the answer? (To the "11x11") Ciao ;) 😉 😂
@@stevesebzda570 There's actually 36 guaranteed cells, before considering any givens - each box has a 4x4 area it can exist within, and the center 2x2 area for each of those have to be a part of that box.
@@SourabhDas95 Oh, you're right. Because of those middle twos, you can extend the corners and the edges out one. That's 36 to that 9 (on the 11x11 grid now). Even more of a disparity. I wonder if you can extend something on that "9" from the "11x11" now?! Huge disparity now though (36 - 9). Wow, thanks. What about the minimum digits on that 11x11 though? Thanks again.
@stevesebzda570 I'm not sure how a 11x11 would fare. The initial digits would have to play more of a role in either keeping boxes aligned close enough to communicate enough information through rows and columns, or more directly providing extra digits to narrow down the set of possible solutions to 1. I think the bigger point of appreciation is that the constructor noticed that this was an inherent property of 10x10 grids that could be exploited, and then went on to make a puzzle that showcased it's power. To answer your question is significantly more difficult! The proof for a regular sudoku needing a minimum of 17 digits is complicated enough on it's own and took a team of mathematicians, several months, and a lot of programming and computer power to prove. This puzzle doesn't prove that the minimum for a 10x10 deconstruction puzzle is 14 given digits - just that it's 14 or lower. So giving you a conclusive number for 11x11 is a little outside my scope 😄 But since you have raised the question, I'll answer to the extent that my intuition carries me: My gut says that it might be possible for a 11x11 deconstruction to still have it's minimum required digits be under 17, but would almost definitely require more than the minimum for 10x10. Additionally, the final layout of boxes in an 11x11 decomposition sudoku that maximizes the information efficiency of given digits, would probably fit into a 10x10 area anyway. This is just based on some micro-level ideas - like two of the same digit, diagonally across, can do a lot of work in positioning 2 whole boxes AND providing a lot of row/column info, if placed in the right parts of the grid, and would likely outperform a layout with boxes that are more spread out. But that's about as far as I can take it without spending significantly more time, effort and resources behind researching this!
Rhyming Rules, for those who enjoy them: Lepton's cheeky puzzle grid exploded! (If you're sneaky, it can be decoded...) First, look for the logic which is trapping cells in boxes with no overlapping. Make each box measure in at 3 cells square, then place at leisure 1 to 9 in there. Outside the boxes, digits cannot go; naught may repeat in column or in row. Nine boxes drawn and digits tightly placed, Fourteen's conundrum will be rightly faced; and when these rules are purely satisfied, the grid (and boxes) surely ratified.
Lovely poem! If im not misunderstanding you, the only line that is not true is "cells in boxes with no overlapping". I am interpreting this as the boxes not being overlapping and I see 7/9 boxes overlapping with other boxes. But, again lovely poem! Edit: perhaps you mean that no box is on-top of each other as overlapping, and, in that case, you're correct!
@@ugglorimossen They're described in the original puzzle rules as "non-overlapping boxes" so "boxes with no overlapping" feels correct - I have noticed I completely failed to mention that there must be _nine_ 3x3 boxes though, which a sneaky edit is about to correct... 🤪
this is one of those puzzles that one should solve as a way to "cement" their knowledge of this type of sudoku. it's not necessarily the "big boss" that would spawn a 2 hour video, but very effective at reminding the solver the "axioms" following this ruleset. bravo to lepton!
Thanks for solving this one Simon! I was one of the people who wrote in about it because I found it to be an incredible discovery. Nice work Lepton & Simon!
Wow! I made a puzzle with the same rules a while ago (it's called "Herding Cats" on Logic Masters, if anyone wants to try it), but I needed 31 given digits to make it work. I'm amazed that you can do it with just 14!
The reason this puzzle can have fewer digits than a regular Sudoku is because it has additional box interactions that normal sudoku doesn't. If you look at how box 3 is disjointed and shifted down one row from being in parity with box 1 and 2 it allows the bottom row of box 3 to interact with the top row of box 4. Something which would be impossible in normal sudoku. This same general rule applies across the puzzle anywhere when one box in a row or column is misaligned with the rest of its row or column. Meanwhile all of the normal row and column interactions that would be in a normal sudoku are still maintained through "the law of leftovers" and the extra regions that Simon mentiones frequently throughout the puzzle. So ultimimately all original sudoku relationships are maintained, even if in a slightly disjointed way, and then a handful of extra interactions are added from boxes that normally would never interact directly.
Wow! I can’t believe I got this. 80:04 but I got it! I worked ahead then got stuck a bit after placing box 4 (it just struck me how appropriate the constraint was that finished the placement) then watched until Simon got there (+one 7, -a few others). Unlike simon, I pencilmarked all remaining digits. It wasn’t super helpful yet. I was faulting out on when numbers could be in a r/c vs must be. The explanation of law of leftovers clarifies that, but also the “only if there are 9 cells across/down must a digit be there” which seems obvious but wasn’t for me until I heard Simon’s phrasing. With those 2 bits of logic updated, I kicked along with 72514s until I got stuck again. I watched again until Simon fixed “A” (by ruling it out box 4 options. Again a useful box, 4! ) Once I got the “A” placed in boxes 2,3 with Simon’s example, I finished up. My eyes started pulsing the colors between black & red, making the conflict checking take 4X as long as if I could see, but yay! Go me! I solved a 14 digit free-form sudoku with only 3 digits of help & some logic boosts!
the offset means that for example the red box effectively looks at 4 whole boxes and 3 partial boxes which is more than the 4 whole boxes a box can look at in a normal sudoku.
I find it incredible to find that logic about the extra region for the columns, and perpetuate it on all columns, and yet not use it on rows until extreme exhaustion of the column logic when there was very simple digits to be gotten with the row extra regions
Brilliant puzzle. The offsetting of a box pulled into play additional constraints from boxes that were outside the original 3 boxes but aligned with the spaces of the offset box.
i like pausing right before youre about to explain what youve done to find it out and sometimes I get it and it is so super satisfying hearing you explain my exact thought process
35:31 for me. This is a wonderful puzzle - in addition to it being really cool and weird that it exists, it's enjoyable to solve, and challenging but not brutally difficult.
This video introduced me to sudoku over a month ago and now that I have a good few solves under my belt, I decided to give this puzzle a go. I promise I didn't remember the actual solve, but I did remember the strategies you used. It took me 122 minutes, but I did it! It seriously feels like magic, making digits appear out of what feels like thin air.
I was wondering, if you could cut out rows/columns of a 3x3 and rearange them, so that you get a 9x9, and if that means that you can make a regular sudoku with only 14 given digits. But you loose information by doing this, because of the interactions between the boxes. e.g. transpose the top row of box 7 into the bottom row of the 10x10 grid. That would not change the set of the boxes 7-9, but you lose the information about the relationship of box 7 with boxes 5 and 6. What a careful construction!
What a remarkable puzzle! I often get frustrated with "chaos construction" puzzles, but I found this one somewhat more open to standard logic, plus SET. Lovely logic throughout. My time today was 35:36, solver number 458.
I spent ninety minutes on this, _and_ I had to look at the video for a hint. I would never have figured out the extra region thing on my own. Thanks to you and thanks to Lepton, I'm proud that I stuck through the puzzle regardless of that small bit of cheating I did!
The leftovers effect from the sticky-out bits causes digits in certain positions to make extra eliminations compared to a standard classical Sudoku, and thus creates a similar effect to a king's or knight's move restriction. This in turn means fewer given digits are required to lead to a unique solution.
The best part about this puzzle is that it doesn’t feel like it’s the hardest this kind of puzzle can be. It feels relatively easy when you get past the first stages. I wonder about mixing it with other sudoku mechanics as well. Wonderful puzzle
@16:15 in the video (haven't finished it yet) I Believe 6 is mistakenly (maybe correctly?) Assigned to row 6 when it could just as easily be in row 4, Simon forgot he was dealing with 10 rows not 9
The reason you have fewer clues required is that instead of the puzzle having the usual 9 row and 9 column uniqueness constraints, it has 10 row and 10 column constraints, for a total of 29 uniqueness constraints vs 27 for a normal sudoku.
I've never solved a sudoku puzzle for myself before, but I like to watch a lot of videos about topics I've had little exposure to. This was a very interesting exercise in logic!
I'd say it seems like whenever you add more rules, you add limitations, so the possible solutions drop at any amount of numbers drops as well, so the amount of numbers you need to minimise that to 1 solution lowers as well. Thus why you've also had sudokus with no numbers placed initially. I would also say this doesn't lower the minimum for standard sudokus. The reason you need less numbers for this one is precisely because of the symmetries you found. More symmetries means more limited solutions Instead of having 9 numbers that relate in a row, you now have 12 numbers that relate (of which there's 2x3 that match), and then that happens every 2 rows basically. So you have more ins to the puzzle essentially.
I actually spent over a week trying to solve this puzzle (using a spreadsheet), my approach was completely different, I started by working out where the boxes went an fixed 8 of them, then chose one of two positions for the 9th (the one that created the most symmetry, the intention was that it led to a complete contadiction I would try again with the other) I skip the mistakes and the useless things I did here: instead of marking where a number could go I marked where it couldn't go with highlights (one grid for each number) and eliminated what I could, making a few deductions along the way, then started to look at the 3 columns and rows with only 6 numbers and made assumptions (looking at what digits could be in them before looking at the order for each set) the first set all led to conflicts allowing for a few deductions, the second set I tried had a lot of different ordering but various deductions about "what ifs" help eliminate all of them quicker. On the third one I checked after eliminating some orderings that automatically led to a conflict, I checked one of the remaining ones and managed toget lucky and put in everything
59:54 for me, just barely within one hour. This puzzle made me realize that after all the hype around set theory following Phistomefel’s Ring puzzle, I internalized all consequences of set theory in a regular 9x9 sudoku, without actually applying the set theory. Took me an embarrassingly long time to see i could use set theory here to “shift” odd boxes by one to align with the two boxes in a row. Definitely a good reminder to look at set theory more often! For real tho, what an amazing puzzle, really enjoyable, and idk if it was intended, but the logic I used to figure out the exact position of box 4 using only the cornermarks for 5s and 7s in row 4 was incredibly satisfying!
9:10 You can assign a box position to the indicated cells; since you can't put two boxes in the five rows above it and can't put two boxes in the four cells below it (and a 10x10 grid is not big enough for a given column of boxes to not be stacked on top of each other) it must be box 6.
Like Simon, I drew teh boxes very quickly. The law of leftovers gave me the last box. To make things easier I coloured the "leftover" regions. This made things a lot easier. I also discovered the duplicate duo across the central part of the grid a bit sooner than Simon. That was best bit and unravelled the rest. I really enjoyed that one, basic rule sets are definitely preferred!
Wow! This was my first time solving something more difficult and involved than a regular or killer sudoku. After listening to your pointers and explanations about the leftovers I was able to solve it all myself in 56 minutes. Thanks for a great video!
Noticed there were some interesting cells that I want to call ghost cells that existed outside the boxes and didn't see 9 cells both horizontally or vertically (r1c10, r4c4, r4c10, r10c1 and r10c4) and I also noticed that there was only 1 number each of them could hold. Curious to find out if ghost cells seeing 8 of the numbers from 1-9 is characteristic of this type of puzzle or if there are variations that don't have this. Edit: after looking at it a bit more I see that the values that can go in r1c10 and r10c1 are also values that are a part of 2 extra regions, which would explain why they are what they are. The value in r4c4 is 1 (the same as in r1c1) but I can't tell why and I'm also unsure about c4r10 and c10r4.
Fantastic in so many ways - and in particular your solve, Simon. Thank you for showing us, again, what a genius you are with regard to the region-defining kinds of puzzles. I love watching you do this.
I'm missing something at 38:54, how is that 5 determined? I thought 9 digits in a row, but they aren't in a row. What am I missing? I get why it isn't a 5 from the placed 5 and quadruple, but why is that cell a 5? How is its value determined? (Why not a possible cell below?) edit: I think the fact Simon is using no repeats in any row, I find it confusing to split that across two rows. Maybe 5's were knock out of the two right grey cells. edit2: Same with 40:20. I think I missed the logic of the extra region knocking out the total of 3 sets of digits. edit3: I've realized, which Simon may have said is that any extra section outside of overlapping rows/groups must have the digits 1-9. Brilliant, clearly not something I would deduce.
The way in which that solved was absolutely stunning (and with the bonus of 14 being my favourite number). As for the quality of setting, it seems that every year produces an even greater number of incredible puzzles than the last one. This one’s been crazy so far!! 😅
solving this alongside you is neat because i can pause the video while i'm working on it and then keep watching when i get stuck. i'm taking things in a bit of a different order than you and making some discoveries of my own along the way so i have more information than you in some areas but in others i have no idea where you're going with it and need to wait a bit longer for you to explain. it's really fun!
So it is indeed the power of the geometry of this puzzle, which is actually making it ambiguous from a normal 9x9 sudoku, but still in some or the other way, it is making it more powerful than a normal one. It is ridiculous how it can be both at the same time! Like we know that we need at least 17 digits in a sudoku grid to be able to solve it, and absurdly, we actually needed this puzzle's scaled up geometry to help us get those extra 3 initial digits to obtain the minimum number of digits (17) to be able to solve a sudoku puzzle! And adding more to the weirdness, all of the help in getting those first 3 digits was solely from this geometry, how bizarre! It is just unbelievable how someone could actually discover this! Take a bow, Lepton!
This is one of those puzzles that seems counterintuitive at first, but makes sense while you're solving it. The Law of Leftovers is quite powerful 🙂 Which raises the question if a similar idea in an 11x11 grid could work with even fewer digits.
I haven't finished solving, but the "law of leftovers" trick seems to be critical in creating additional "9" regions. By applying it twice, you can prove which of the two possible locations for box 4 are correct, since you end up with an impossible overlap, with conflicting 1s and 4s for one of the positions.
23:35 I think maybe the easier way to explain this is that light green and red both need a 4, but only column 3 is available. Columns 2 and 4 already have 4s. Thus the puzzle breaks.
As someone who almost never does sudoku but gets these videos recommended to me occasionally, I maybe shouldn't have attempted this, but I was happy to have done it in just over two hours with only a couple hints (I had never heard of this Law of Leftovers thing and it seems pretty critical, but it made sense once I saw it explained, and I couldn't find the 9 at 30:40)
I am amazed at some of the leaps of logic you make, seriously if I was doing this I'm not sure I could have figured out the last box, and yet I spend 15 minutes yelling at my screen about 3 or 4 digits that seem obvious to me,
As usual, I gave this puzzle a go and got completely stuck, then started the video to see how long it would take for Simon to "catch up to me"... Well, that was about halfway into the video. You then started marking those A's and B's so I paused the video, marked my A's and B's too and continued from there, relatively quickly resolved the A as 9 but was left with B being a 2 or an 8... So I then marked its complement 2 or 8 as C. After having finished the puzzle, I continued the video and saw Simon follow kind of the same path without marking B's complement but he came to the same path to disambiguate B as I did so I don't think my C's helped me. Anyway... Brilliant puzzle.
In terms of minimum digits, I count five extra ‘virtual’ clues. The 2 in r4 is a clue for the virtual r1 part of box 3. The 5 in c6 is a virtual clue in c1. The 6 and 1 in c7 are clues for the virtual c10 part of box 3. And that same 6 (r7) is also a clue for the virtual r10 part of box 7. Which gives a total of 19 given digits. 17 for the classic sudoku, and two extra to define the boxes. Very impressive.
Simon, due to set theory Box 7 and 3 mirror one of their rows and columns into row 1, column 9 and vice versa. In short the given 4 in box 3 is mirrored twice and the given 5 in box 7 is morrored twice and they therefore count for 2 or 3 givens. If you don't see it, concentrate on box 3 of the final grid. The 468 in the bottom row of it could be written into row 1 above it and complete the first row, the column 294 of box 3 could be written into the last column right of box 3. Therefore the given 4 in box 3 is worth 2 given 4s in these places and the same applies to the given 5 in box 7 due to symmetry. That makes it 18 givens in 14 givens.
Every cells has information provided by at least one 12 long row or column, narrowing the posibilities. Explaining the 14 digits minimum cause every starting digits spread its information over more cells. So clever!!
I suppose when you have the "extra region" tech the digits can do double duty. For instance the 4 in the dark blue box sees both the rightmost column in light blue, as well as the column in pink and yellow.
Pencilmarking *9* together with *A,* In the same cells... ...immediately after discovering that *A = 9❗* 🤦♂ Anyway, thanks for featuring such a magnificent puzzle.
That was mad genius! Loved it! And, if you spot the weird geometry early on it's not that difficult. I will have to agree with @Babinzo, this should be in CTC tutorials playlist as a textbook example of the law of leftovers!
At around 22:00 does it indicate in the ruleset that the left hand side of the dark blue box and the right hand side of the light blue box are part of the same column? If not, why can't the latter have a 4 in it?
It's incredible how loose boxes provides more restriction than normal Sudoku boxes. Took me 90 mins to solve by myself, what a wonderful puzzle this has been ❤.
Definitely one of the most enjoyable puzzles I’ve solved. I have no clue how you’re supposed to come up with something like this. I wonder if the digit limit could be lowered further
20:08 for me. that was fun. i use to do badly at new constraints (like the dam previous *puzzle*), but this gave me the oportunity to give the last laugh
The 9th 3x3 box can be determined after getting the first two digits. With a 1 in columns 1 and 10, if box 4 extended into column 4 then there would be a 1 in each of colums 2-9 as well for a total of 10 1s and the puzzle would be broken.
I think the "extra power" comes down roughly to breaking of symmetries inherently meaning fewer possibilities to still arrive at something valid. *Any* nontrivial symmetry is going to allow you to find situations where more than one solution is possible until you give enough constraints to break that symmetry. 17 is the number of constraints you need at a minimum to ensure breaking all the symmetries in regular Sudoku. But this slightly less symmetric situation must be causing three more symmetries to disappear somehow, giving you a need of just 14 filled-in constraints. And I think two of those are quite obvious: You are no longer connecting up full columns nor full rows. Perhaps the third is that one of the diagonals is also broken up? Not sure...
@@galoomba5559 on an empty Sudoku (without any further constraints but basic Sudoku rules), you have complete freedom of choice. you can pick any one out of 81 spots for any one out of 9 digits: 729 possible choices for that first digit alone. As soon as you fill in one digit, you automatically constrain the puzzle in all kinds of ways: That particular digit can only go again into 60 remaining spot (none of the 9 in the same box, and 2x6 additional removed choices from the column and row) - it's therefore a much less symmetric space. You can also only pick it 8 more times while the other digits still are available 9 times. This next bit I'm not looking up so I may well make some reasoning error here, but as far as I can tell: For a *filled* grid, without extra constraints beyond the basic Sudoku ones, you can freely swap labels to arrive at essentially the same puzzle, giving you 9! = 362880 "identical" puzzles up to the labelling symmetry. You also can freely rotate or reflect the puzzle, giving you, I *think* 8 more ways puzzles might be considered "identical" - those are just the symmetries of a basic square. Additionally, for row or column of larger boxes, you may reorder the three rows or columns within them freely, giving you 2x3x2x3 = 36 choices (2 for whether you are dealing with columns or rows, 3 for which big column or row you are dealing with, and 2x3 for the 6 ways you may rearrange stuff in those columns and rows) And you may *also* freely swap the big rows and columns (2x2x3 = 12 for whether cols or rows, and each possible order of those) Altogether, that gives you 9! * 8 * 36 * 12 = 1254113280 Sudoukus that are effectively "the same" under its basic symmetries. At least those are all the symmetries I can think of. Maybe there are more. Probably something about how subset that have all nine digits in it can be related or something. Not sure how you'd count that though, and I'm not sure whether that still would count as "essentially the same". But at any rate, at a minimum, to define a *specific* Sudoku, you need to fill in enough values, or add enough extra constraints, to break all the freedoms those symmetries give you. 1.25 billion might seem a lot, but of course, every choice you make very quickly reduces this spaces down to something much more manageable. All of this math assumes the 9x9 grid of course. And that is a highly symmetric playing field. This slightly *larger* field is less symmetric immediately, because not every place gets a digit. You still work with the same number of digits after all. You also get constraints on the box shapes. The original 9x9 grid fits into this larger grid in 4 ways of course. However, you can completely take out most of the symmetries regarding that by simply putting a single number "off grid" (relative to an original 9x9; the exact condition is that every row and every column has at least 1 digit in it somewhere), thereby forcing the utilization of all 10 rows and columns. And despite the extra space, the remaining possibilities are probably more constrained than you'd think, as all the most symmetric ones (just working with the original 9x9) are instantly gone. Doing the math there is gonna be much trickier though, as you'd have to carefully keep track of all the ways boxes may distort and what not.
The extra regions don't add any extra power. They are just like the the third row (or column) of those three boxes.... But wait! There is this part of three digit sticking out and in a row(or column) of six that is looking at another three squares that would normally not be "looked at".
The colors are quite good for color blind people. 5 (purple) and 9 (blue) are a bit similar but all other’s distinguish quite well from each other. If you permanently replace blue by dark blue it would be perfect for me (red-green blind).
I'm neither a Sudoku nor math person. But I can give a bit of intuition for why this setup should allow for a smaller number. And that is because certain boxes are twice constrained. Let's take row 4 from top. The blue box from row 4 is constrained because it has to form a 1-9 with row 1. But at the same time it can't have the same numbers as the lime box in the same row 4. That way you reduce the number of independent variables and therefor you reduce the minimum number needed to give a unique Sudoku. I have never seen letters being used in solving a Sudoku. That's a first for me, but it makes sense somehow, especially in such a strange Sudoku such as this one.
A more general statement about 10×10 grids is that you automatically know 4 cells for each cage. No matter how you giggle the boxes, they must contain a 2×2 square that's 1 cell away from the perimeter and each other. But in this puzzle your method is far more direct.
Looking at the opening grid in the link; I think, that "14-digit conundrum," is because 16 cells are given to be in (whereas, on a 11x11 grid, only 9 cells are given to be in at the start) I had to look at it twice (to make sure those inner ones were two -- and 2x2 in the center). That might be a way to view this as less "counter-intuitive," (on an "11x11 grid," only 9 cells of the boxes are given to be in at the start). (In other words, on an "11x11 grid," you're truly "finding'' the boxes more) Don't know. What's the minimum of digits on an 11x11 grid of this nature? Very good (very gouda) though (
16 cells given to be in at the start (on a 10x10) - versus 9 (on an 11x11). That's a huge disparity. That's why I think those "16" (as-opposed to 9) has something to do with it. Don't know, though.. Very good though again. "Snow.?!" (It's "outside," literally) 😂 ;) 😉
PS: Because of those middle twos on the 10x10, you can extend out the corners and edges by one. (Counting, that's 36) 36 (on this 10x10) to that "9" (on an 11x11). A huge, HUGE disparity now. (36 - 9) Wow. Something's up there (maybe)? 😂
FInding the boxes is easy, but I struggled badly with the sudoku bit. Naturally, it took Simon about 2 minutes to spot something that evaded me for half an hour, but I'm not bitter and twisted about it at all.
I was worried for a minute when I noticed that the conflict checker was not working for boxes, only r/c. But I’m guessing that Sven in his brilliance did that purposely to foil any cheating through guessing… 54:09
One thing that might be relevant as to why 14 digits was possible for this one, is the fact that horizontal striping occurs in the top 6 boxes.I wonder if there is a theory, that can prove that with such low count of given digits some striping must occur?
1:32 “So apparently this is solvable-by a human-and I do *loosely* qualify as such…” really caught me off guard there 😂
Well when you’re a demigod like Simon… 😊
"I don't know what to do next."
Proceeds to solve the next number.
"Bobbins! I have no idea what I'm doing."
Proceeds to solve the next few numbers.
"How am I supposed to solve this?"
Proceeds to solve the next numbers.
Etc.
To be fair, that's how I solve pretty much every Sudoku on this channel.
"This grid is impossible and I don't think I can make a single logical deduction...Except I guess that this digit can't be a 5?...I guess that means that this one can't be a 4."
Repeat for anywhere from 30 to 300 minutes until the puzzle is solved.
This video should get filed away into a CTC Tutorials playlist as a textbook example of the Law of Leftovers
I agree.
Would love a playlist like that, just one or two videos showcasing the best and approachable examples of all different variants 🤩
The "extra power" of the givens is pretty clear by the end. For example, in a normal sudoku any given digit would only be able to "see" 20 cells (8 cells in it's box, 6 more cells in its column and another 6 in its row). However, in this puzzle the 4 in c7r4 can "see" 29 other cells because of the law of leftovers.
Thanks for the explanation!
Let's Get Cracking: 07:37
Simon's time: 39m43s
Puzzle Solved: 47:20
What about this video's Top Tier Simarkisms?!
Bobbins: 1x (43:19)
Maverick: 1x (06:42)
And how about this video's Simarkisms?!
Pencil Mark/mark: 16x (13:33, 15:14, 15:17, 15:41, 16:19, 19:30, 21:01, 29:31, 29:34, 34:51, 35:36, 37:35, 39:52, 41:09, 42:50, 42:55)
Ah: 7x (08:06, 13:58, 24:54, 25:00, 30:14, 35:27, 46:07)
Brilliant: 6x (02:03, 03:35, 06:13, 06:13, 47:28, 47:33)
By Sudoku: 5x (19:20, 35:54, 39:13, 43:31, 45:31)
Cake!: 5x (04:28, 05:20, 05:23, 05:54, 05:59)
Sorry: 4x (12:21, 32:32, 34:59, 40:37)
In Fact: 4x (25:38, 27:09, 27:09, 33:02)
What on Earth: 3x (18:43, 25:03)
Beautiful: 3x (02:10, 34:59, 37:03)
Weird: 3x (25:44, 45:19, 47:53)
The Answer is: 2x (09:26, 32:41)
Clever: 2x (43:39, 43:41)
I Have no Clue: 2x (32:52, 36:49)
Ridiculous: 2x (20:16, 47:12)
Hang On: 2x (10:52, 40:13)
Whoopsie: 2x (34:16, 37:21)
What Does This Mean?: 2x (26:31, 28:19)
That's Huge: 2x (35:08)
Unique: 2x (00:33, 01:05)
Useless: 1x (24:51)
Apologies: 1x (04:01)
Naughty: 1x (39:43)
Extraordinary: 1x (01:55)
Gorgeous: 1x (33:59)
Come on Simon: 1x (41:03)
Epiphany: 1x (41:27)
Masterpiece: 1x (03:28)
Magnificent: 1x (48:14)
Surely: 1x (35:49)
Obviously: 1x (31:17)
Progress: 1x (40:08)
Wow: 1x (21:22)
Let's Take Stock: 1x (35:49)
Next Trick: 1x (19:47)
Have a Think: 1x (13:22)
Nature: 1x (02:34)
Symmetry: 1x (45:17)
Most popular number(>9), digit and colour this video:
Fourteen (7 mentions)
One (66 mentions)
Green (22 mentions)
Antithesis Battles:
Even (6) - Odd (0)
Outside (2) - Inside (0)
Column (36) - Row (25)
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cool bot!
14:35 he said by sodoku
If anyone is curious about why this can have an unique solution with only 14 digits, when a regular sudoku needs a minimum of 17, here's some ideas to think about -not a proof by any means, but perhaps some things that can help form an intuition:
In a regular sudoku, there is a certain amount of 'information' that's encoded from the limitations of the rules (1-9 in each row/column/box, and no repeating digits). So if we take a stack of 3 boxes, the digits visible in the boxes above and below, serves to restrict the permutations of digits that a box below can have. But if we shift one of the boxes over by one cell. Because of the requirement of boxes needing 1-9, only 2 columns of overlap are needed to define what the remaining cells in those 3 boxes (that aren't in the 2 overlapping columns) will consist of. When you compare to a stack of 3 boxes in a regular sudoku, you realize that once you fully define 2 of the columns, you know what the 3 digits are for the 3 remaining cells, in each box, and you don't require 'column logic' for that conclusion.
In contrast, in stack with one shifted box, you get a more visibly disjoint set of 9 cells that have 1-9 becuase the column logic was never required for that - basically what we refer to as the law of leftovers. And in this configuration, the shifted box can provide additional information to the next stack, without sacrificing the integrity of the information it can squeeze from it's own stack.
In that way, a regular arrangement of 9x9 grids has both limitations on how much information can be transferred across it's different 'constraining sets' (rows/columns/boxes) as well as redundant limitations provided by it's structure, that provide multiple unique logis paths of restricting the value of a cell to a unique digit. One way to think about how a cell can be restricted is to think about the last digit you fill in a sudoku grid. That can be determined by seeing four of a digit pointing into that box and going in the only empty space (hidden single logic) OR you could look at the box and see which number is missing, or the box (naked single logic).
Essentially, in a regular sudoku, the redundant limitations on cells that's provided by rows and stacks of boxes, means that you end up with a 'choppy' information horizon, where it's difficult to design a solve path for a puzzle where there isn't consistently too many, or too few deductions that can be made. That leads you to require an 'inefficient' number of given digits, to weed out all the deadly patterns that remain.
Once you decouple box logic and column logic, where the box logic can finalize the horizontal poistion of digits within its box, which allows that to transfer over column logic to another stack etc. (or decoupled rows of boxes providing row logic information to other box-rows), you reduce the information redundancy among a stack, and provide additional column information to other stacks that is otherwise 'lost information efficiency'.
You can use the example of placing the last digit in this puzzle to see how 'less pieces of logic' COULD be available to solve the last digit of this puzzle (depending on its location) e.g., if r1c1 was the last digit placed, you would only have 'naked single logic' from the box available, rather than four seperate logic paths that all lead to the same conclusion.
This sort of 'decoupling' of box, row and column logic, allows the information distribution along the solve path to be more efficient, and require fewer given digits to arrive at an unique solution.
One thing to note, is the ability to preserve the information density of a stack, comes from shifting only one box by one column from fully aligned (and similarly for rows). A 10x10 grid ensures that this will be true for every stack and box-row. The proof of why is left as an exercise to the reader! 🙂
One thing i noticed while watching is that the given 4 in the dark blue box interacts with 8/9 boxes, whereas in standard sudoku it would only interact with 5
One thing I noticed on the starting grid, was there were 16 guaranteed cells in (as-opposed to only 9 on an 11x11).
That's a huge disparity.
That's why (that 16 as opposed to 9) I think had something to do with it.
What's the minimum of digits on an "11x11" of this nature?
Very good though I guess, but what's the answer?
(To the "11x11")
Ciao ;) 😉 😂
@@stevesebzda570 There's actually 36 guaranteed cells, before considering any givens - each box has a 4x4 area it can exist within, and the center 2x2 area for each of those have to be a part of that box.
@@SourabhDas95 Oh, you're right.
Because of those middle twos, you can extend the corners and the edges out one.
That's 36 to that 9 (on the 11x11 grid now).
Even more of a disparity.
I wonder if you can extend something on that "9" from the "11x11" now?!
Huge disparity now though (36 - 9).
Wow, thanks.
What about the minimum digits on that 11x11 though?
Thanks again.
@stevesebzda570 I'm not sure how a 11x11 would fare. The initial digits would have to play more of a role in either keeping boxes aligned close enough to communicate enough information through rows and columns, or more directly providing extra digits to narrow down the set of possible solutions to 1. I think the bigger point of appreciation is that the constructor noticed that this was an inherent property of 10x10 grids that could be exploited, and then went on to make a puzzle that showcased it's power.
To answer your question is significantly more difficult! The proof for a regular sudoku needing a minimum of 17 digits is complicated enough on it's own and took a team of mathematicians, several months, and a lot of programming and computer power to prove. This puzzle doesn't prove that the minimum for a 10x10 deconstruction puzzle is 14 given digits - just that it's 14 or lower. So giving you a conclusive number for 11x11 is a little outside my scope 😄
But since you have raised the question, I'll answer to the extent that my intuition carries me:
My gut says that it might be possible for a 11x11 deconstruction to still have it's minimum required digits be under 17, but would almost definitely require more than the minimum for 10x10.
Additionally, the final layout of boxes in an 11x11 decomposition sudoku that maximizes the information efficiency of given digits, would probably fit into a 10x10 area anyway.
This is just based on some micro-level ideas - like two of the same digit, diagonally across, can do a lot of work in positioning 2 whole boxes AND providing a lot of row/column info, if placed in the right parts of the grid, and would likely outperform a layout with boxes that are more spread out.
But that's about as far as I can take it without spending significantly more time, effort and resources behind researching this!
Rhyming Rules, for those who enjoy them:
Lepton's cheeky puzzle grid exploded!
(If you're sneaky, it can be decoded...)
First, look for the logic which is trapping
cells in boxes with no overlapping.
Make each box measure in at 3 cells square,
then place at leisure 1 to 9 in there.
Outside the boxes, digits cannot go;
naught may repeat in column or in row.
Nine boxes drawn and digits tightly placed,
Fourteen's conundrum will be rightly faced;
and when these rules are purely satisfied,
the grid (and boxes) surely ratified.
Lovely poem! If im not misunderstanding you, the only line that is not true is "cells in boxes with no overlapping". I am interpreting this as the boxes not being overlapping and I see 7/9 boxes overlapping with other boxes. But, again lovely poem!
Edit: perhaps you mean that no box is on-top of each other as overlapping, and, in that case, you're correct!
@@ugglorimossen They're described in the original puzzle rules as "non-overlapping boxes" so "boxes with no overlapping" feels correct - I have noticed I completely failed to mention that there must be _nine_ 3x3 boxes though, which a sneaky edit is about to correct... 🤪
Wonderful as always! 👏🏻👏🏻👏🏻👏🏻👏🏻👏🏻
These are always excellent. Bravo!
Brilliant again!
this is one of those puzzles that one should solve as a way to "cement" their knowledge of this type of sudoku. it's not necessarily the "big boss" that would spawn a 2 hour video, but very effective at reminding the solver the "axioms" following this ruleset. bravo to lepton!
Simon works like an archeologist, carefully unearthing sudoku digits. Amazing to watch!
Thanks for solving this one Simon! I was one of the people who wrote in about it because I found it to be an incredible discovery. Nice work Lepton & Simon!
Wow! I made a puzzle with the same rules a while ago (it's called "Herding Cats" on Logic Masters, if anyone wants to try it), but I needed 31 given digits to make it work. I'm amazed that you can do it with just 14!
u can solve it with less than 87178291200 digits actually
Wow, this video is truly captivating! t’s clear a lot of effort and creativity went into making this sudoku. Great job to Lepton!
The reason this puzzle can have fewer digits than a regular Sudoku is because it has additional box interactions that normal sudoku doesn't.
If you look at how box 3 is disjointed and shifted down one row from being in parity with box 1 and 2 it allows the bottom row of box 3 to interact with the top row of box 4. Something which would be impossible in normal sudoku. This same general rule applies across the puzzle anywhere when one box in a row or column is misaligned with the rest of its row or column.
Meanwhile all of the normal row and column interactions that would be in a normal sudoku are still maintained through "the law of leftovers" and the extra regions that Simon mentiones frequently throughout the puzzle. So ultimimately all original sudoku relationships are maintained, even if in a slightly disjointed way, and then a handful of extra interactions are added from boxes that normally would never interact directly.
Wow! I can’t believe I got this. 80:04 but I got it!
I worked ahead then got stuck a bit after placing box 4 (it just struck me how appropriate the constraint was that finished the placement) then watched until Simon got there (+one 7, -a few others). Unlike simon, I pencilmarked all remaining digits. It wasn’t super helpful yet.
I was faulting out on when numbers could be in a r/c vs must be. The explanation of law of leftovers clarifies that, but also the “only if there are 9 cells across/down must a digit be there” which seems obvious but wasn’t for me until I heard Simon’s phrasing. With those 2 bits of logic updated, I kicked along with 72514s until I got stuck again. I watched again until Simon fixed “A” (by ruling it out box 4 options. Again a useful box, 4! ) Once I got the “A” placed in boxes 2,3 with Simon’s example, I finished up.
My eyes started pulsing the colors between black & red, making the conflict checking take 4X as long as if I could see, but yay! Go me! I solved a 14 digit free-form sudoku with only 3 digits of help & some logic boosts!
29:55 "Can nearly stop A being 2" - the pencil marks in box 4 brings that "nearly" a lot nearer...
the offset means that for example the red box effectively looks at 4 whole boxes and 3 partial boxes which is more than the 4 whole boxes a box can look at in a normal sudoku.
I find it incredible to find that logic about the extra region for the columns, and perpetuate it on all columns, and yet not use it on rows until extreme exhaustion of the column logic when there was very simple digits to be gotten with the row extra regions
Brilliant puzzle. The offsetting of a box pulled into play additional constraints from boxes that were outside the original 3 boxes but aligned with the spaces of the offset box.
i like pausing right before youre about to explain what youve done to find it out and sometimes I get it and it is so super satisfying hearing you explain my exact thought process
The idea of the left over was honestly a stroke of genius well played sir
35:31 for me. This is a wonderful puzzle - in addition to it being really cool and weird that it exists, it's enjoyable to solve, and challenging but not brutally difficult.
This video introduced me to sudoku over a month ago and now that I have a good few solves under my belt, I decided to give this puzzle a go. I promise I didn't remember the actual solve, but I did remember the strategies you used. It took me 122 minutes, but I did it! It seriously feels like magic, making digits appear out of what feels like thin air.
I was wondering, if you could cut out rows/columns of a 3x3 and rearange them, so that you get a 9x9, and if that means that you can make a regular sudoku with only 14 given digits. But you loose information by doing this, because of the interactions between the boxes.
e.g. transpose the top row of box 7 into the bottom row of the 10x10 grid. That would not change the set of the boxes 7-9, but you lose the information about the relationship of box 7 with boxes 5 and 6.
What a careful construction!
What a remarkable puzzle! I often get frustrated with "chaos construction" puzzles, but I found this one somewhat more open to standard logic, plus SET. Lovely logic throughout. My time today was 35:36, solver number 458.
This puzzle could have been called "OffSets"
Just under an hour. Happy with that. Very fun puzzle and very interesting. It's easy to forget about the extra sets of digits that see each other.
I spent ninety minutes on this, _and_ I had to look at the video for a hint. I would never have figured out the extra region thing on my own. Thanks to you and thanks to Lepton, I'm proud that I stuck through the puzzle regardless of that small bit of cheating I did!
FWIW, TopAutism on the CTC community discord has already lowered this to 12 givens… in an 11x11 grid. 😅
Ah, I should have read the comments before asking that very same question. Of course TopAutism jumped right on it, I could've known 🙂
@@Pulsar77 It's a slick puzzle, too, I hope it gets published :)
Man that autism emoji looks way better on discord
I'm glad I checked the comments, LOL
How far can we push this concept?
41:43 "I've got loads of 2s in the grid!" (5 of them)
41:59 "How many 1s have we got? Not many..." (5 of them)
The leftovers effect from the sticky-out bits causes digits in certain positions to make extra eliminations compared to a standard classical Sudoku, and thus creates a similar effect to a king's or knight's move restriction.
This in turn means fewer given digits are required to lead to a unique solution.
Well said/explained!
Absolute genius. Nothing else I can say. Been enjoying your videos for years. Thanks.
What a brilliant discovery in Sudoku history!
The best part about this puzzle is that it doesn’t feel like it’s the hardest this kind of puzzle can be. It feels relatively easy when you get past the first stages. I wonder about mixing it with other sudoku mechanics as well.
Wonderful puzzle
@16:15 in the video (haven't finished it yet) I Believe 6 is mistakenly (maybe correctly?) Assigned to row 6 when it could just as easily be in row 4, Simon forgot he was dealing with 10 rows not 9
Amazing puzzle! My solve was completely different. The 9 you placed at 30:42 was one of the last digits i placed.
The reason you have fewer clues required is that instead of the puzzle having the usual 9 row and 9 column uniqueness constraints, it has 10 row and 10 column constraints, for a total of 29 uniqueness constraints vs 27 for a normal sudoku.
Simon: But is it a useful pencil mark?
Me: no
Simon: *demonstrates that it was actually a useful pencil mark"
I've never solved a sudoku puzzle for myself before, but I like to watch a lot of videos about topics I've had little exposure to. This was a very interesting exercise in logic!
I'd say it seems like whenever you add more rules, you add limitations, so the possible solutions drop at any amount of numbers drops as well, so the amount of numbers you need to minimise that to 1 solution lowers as well.
Thus why you've also had sudokus with no numbers placed initially.
I would also say this doesn't lower the minimum for standard sudokus.
The reason you need less numbers for this one is precisely because of the symmetries you found. More symmetries means more limited solutions
Instead of having 9 numbers that relate in a row, you now have 12 numbers that relate (of which there's 2x3 that match), and then that happens every 2 rows basically. So you have more ins to the puzzle essentially.
At the 16-minute mark, Simon goes over the law of leftover. Then totally forget it 4 minutes later and miss the 7 in box 9.
Was going to say that myself, it's 90⁰ rotationally symmetrical.
What part of the video did he miss at?
@@chipsounder4633 He really misses scanning rows more than columns, a lot - at 40 min he could have resolved B to be 2 ...
I actually spent over a week trying to solve this puzzle (using a spreadsheet), my approach was completely different, I started by working out where the boxes went an fixed 8 of them, then chose one of two positions for the 9th (the one that created the most symmetry, the intention was that it led to a complete contadiction I would try again with the other)
I skip the mistakes and the useless things I did here: instead of marking where a number could go I marked where it couldn't go with highlights (one grid for each number) and eliminated what I could, making a few deductions along the way, then started to look at the 3 columns and rows with only 6 numbers and made assumptions (looking at what digits could be in them before looking at the order for each set) the first set all led to conflicts allowing for a few deductions, the second set I tried had a lot of different ordering but various deductions about "what ifs" help eliminate all of them quicker. On the third one I checked after eliminating some orderings that automatically led to a conflict, I checked one of the remaining ones and managed toget lucky and put in everything
59:54 for me, just barely within one hour. This puzzle made me realize that after all the hype around set theory following Phistomefel’s Ring puzzle, I internalized all consequences of set theory in a regular 9x9 sudoku, without actually applying the set theory. Took me an embarrassingly long time to see i could use set theory here to “shift” odd boxes by one to align with the two boxes in a row. Definitely a good reminder to look at set theory more often!
For real tho, what an amazing puzzle, really enjoyable, and idk if it was intended, but the logic I used to figure out the exact position of box 4 using only the cornermarks for 5s and 7s in row 4 was incredibly satisfying!
00:46:53 for me. That was a great puzzle with some fun unique logic! Kind comment.
9:10 You can assign a box position to the indicated cells; since you can't put two boxes in the five rows above it and can't put two boxes in the four cells below it (and a 10x10 grid is not big enough for a given column of boxes to not be stacked on top of each other) it must be box 6.
Like Simon, I drew teh boxes very quickly. The law of leftovers gave me the last box. To make things easier I coloured the "leftover" regions. This made things a lot easier. I also discovered the duplicate duo across the central part of the grid a bit sooner than Simon. That was best bit and unravelled the rest. I really enjoyed that one, basic rule sets are definitely preferred!
the puzzle symmetry with the horizontal line is gorgeous
Delightful to watch. Such a clever puzzle and a wonderful solve.
Wow! This was my first time solving something more difficult and involved than a regular or killer sudoku. After listening to your pointers and explanations about the leftovers I was able to solve it all myself in 56 minutes. Thanks for a great video!
17:48 Wow, absolutely beautfiul reasoning, love it!
Noticed there were some interesting cells that I want to call ghost cells that existed outside the boxes and didn't see 9 cells both horizontally or vertically (r1c10, r4c4, r4c10, r10c1 and r10c4) and I also noticed that there was only 1 number each of them could hold.
Curious to find out if ghost cells seeing 8 of the numbers from 1-9 is characteristic of this type of puzzle or if there are variations that don't have this.
Edit: after looking at it a bit more I see that the values that can go in r1c10 and r10c1 are also values that are a part of 2 extra regions, which would explain why they are what they are.
The value in r4c4 is 1 (the same as in r1c1) but I can't tell why and I'm also unsure about c4r10 and c10r4.
Fantastic in so many ways - and in particular your solve, Simon. Thank you for showing us, again, what a genius you are with regard to the region-defining kinds of puzzles. I love watching you do this.
I did this one before watching, I got half of it done and then had to brute force the 8s and 9s.
the beginning of the puzzle was absolutely brilliant
I'm missing something at 38:54, how is that 5 determined? I thought 9 digits in a row, but they aren't in a row. What am I missing?
I get why it isn't a 5 from the placed 5 and quadruple, but why is that cell a 5? How is its value determined? (Why not a possible cell below?)
edit: I think the fact Simon is using no repeats in any row, I find it confusing to split that across two rows. Maybe 5's were knock out of the two right grey cells.
edit2: Same with 40:20. I think I missed the logic of the extra region knocking out the total of 3 sets of digits.
edit3: I've realized, which Simon may have said is that any extra section outside of overlapping rows/groups must have the digits 1-9. Brilliant, clearly not something I would deduce.
The way in which that solved was absolutely stunning (and with the bonus of 14 being my favourite number). As for the quality of setting, it seems that every year produces an even greater number of incredible puzzles than the last one. This one’s been crazy so far!! 😅
solving this alongside you is neat because i can pause the video while i'm working on it and then keep watching when i get stuck. i'm taking things in a bit of a different order than you and making some discoveries of my own along the way so i have more information than you in some areas but in others i have no idea where you're going with it and need to wait a bit longer for you to explain. it's really fun!
So it is indeed the power of the geometry of this puzzle, which is actually making it ambiguous from a normal 9x9 sudoku, but still in some or the other way, it is making it more powerful than a normal one. It is ridiculous how it can be both at the same time! Like we know that we need at least 17 digits in a sudoku grid to be able to solve it, and absurdly, we actually needed this puzzle's scaled up geometry to help us get those extra 3 initial digits to obtain the minimum number of digits (17) to be able to solve a sudoku puzzle! And adding more to the weirdness, all of the help in getting those first 3 digits was solely from this geometry, how bizarre!
It is just unbelievable how someone could actually discover this! Take a bow, Lepton!
With every video I learn something new, I didn't know the law of leftovers was a thing!
That was like a cross breed between regular sudoku and irregular sudoku.. the two zets of logic let to some insane symmetry. Top 3 boxes had roping ❤
This is one of those puzzles that seems counterintuitive at first, but makes sense while you're solving it. The Law of Leftovers is quite powerful 🙂 Which raises the question if a similar idea in an 11x11 grid could work with even fewer digits.
I haven't finished solving, but the "law of leftovers" trick seems to be critical in creating additional "9" regions. By applying it twice, you can prove which of the two possible locations for box 4 are correct, since you end up with an impossible overlap, with conflicting 1s and 4s for one of the positions.
23:35 I think maybe the easier way to explain this is that light green and red both need a 4, but only column 3 is available. Columns 2 and 4 already have 4s. Thus the puzzle breaks.
As someone who almost never does sudoku but gets these videos recommended to me occasionally, I maybe shouldn't have attempted this, but I was happy to have done it in just over two hours with only a couple hints (I had never heard of this Law of Leftovers thing and it seems pretty critical, but it made sense once I saw it explained, and I couldn't find the 9 at 30:40)
Was having a night in, saw this in my suggested. Might have to give sudoku a try!
I am amazed at some of the leaps of logic you make, seriously if I was doing this I'm not sure I could have figured out the last box, and yet I spend 15 minutes yelling at my screen about 3 or 4 digits that seem obvious to me,
As usual, I gave this puzzle a go and got completely stuck, then started the video to see how long it would take for Simon to "catch up to me"... Well, that was about halfway into the video. You then started marking those A's and B's so I paused the video, marked my A's and B's too and continued from there, relatively quickly resolved the A as 9 but was left with B being a 2 or an 8... So I then marked its complement 2 or 8 as C. After having finished the puzzle, I continued the video and saw Simon follow kind of the same path without marking B's complement but he came to the same path to disambiguate B as I did so I don't think my C's helped me. Anyway... Brilliant puzzle.
In terms of minimum digits, I count five extra ‘virtual’ clues. The 2 in r4 is a clue for the virtual r1 part of box 3. The 5 in c6 is a virtual clue in c1. The 6 and 1 in c7 are clues for the virtual c10 part of box 3. And that same 6 (r7) is also a clue for the virtual r10 part of box 7.
Which gives a total of 19 given digits. 17 for the classic sudoku, and two extra to define the boxes. Very impressive.
31:00 B can be placed in column 1, as box 4 already has 2 and 8
Amazing. The logic is just mad, but this is quite extraordinary - and convoluted
Simon, due to set theory Box 7 and 3 mirror one of their rows and columns into row 1, column 9 and vice versa. In short the given 4 in box 3 is mirrored twice and the given 5 in box 7 is morrored twice and they therefore count for 2 or 3 givens.
If you don't see it, concentrate on box 3 of the final grid. The 468 in the bottom row of it could be written into row 1 above it and complete the first row, the column 294 of box 3 could be written into the last column right of box 3. Therefore the given 4 in box 3 is worth 2 given 4s in these places and the same applies to the given 5 in box 7 due to symmetry. That makes it 18 givens in 14 givens.
Wonderful explanation!
Every cells has information provided by at least one 12 long row or column, narrowing the posibilities. Explaining the 14 digits minimum cause every starting digits spread its information over more cells. So clever!!
*How he chaptered the video*
Chapter 1: _Introduction_
- Fancy music
- TH-cam username
- Credits and professions
- Video title
- Intriguing question
- Surprising background
Chapter 2: _Streaming_
- Exciting poster
- Promotion
Chapter 3: _Something_
- Idk what this is tbh
Chapter 4: _Birthdays_
- Congratulate people
- Happy slides
- Birthday present
- Funny photos
Chapter 5: _Official rules_
- Explanation
and finally Chapter 6: _Solving_
- The last 40 minutes of the 48-minute video 💀
- Quickly ends
I suppose when you have the "extra region" tech the digits can do double duty. For instance the 4 in the dark blue box sees both the rightmost column in light blue, as well as the column in pink and yellow.
It's outrageous, letting me do sudoku 14 minutes in a sudoku ... named 'Fourteen'.
it's interesting how the shifted boxes make it so the digits in some parts give info on two colums or two rows. brilliant setting
Pencilmarking *9* together with *A,* In the same cells...
...immediately after discovering that *A = 9❗* 🤦♂
Anyway, thanks for featuring such a magnificent puzzle.
Simon does that a lot, also keeping various colourings and stuff when they're no longer needed
That was mad genius! Loved it! And, if you spot the weird geometry early on it's not that difficult. I will have to agree with @Babinzo, this should be in CTC tutorials playlist as a textbook example of the law of leftovers!
2:09:18 - I got the boxes very quickly but sorting out the digits was a real challenge for me. Happy to have finished it though.
Perfect puzzle for a Law of Leftovers explanation.
At around 22:00 does it indicate in the ruleset that the left hand side of the dark blue box and the right hand side of the light blue box are part of the same column? If not, why can't the latter have a 4 in it?
It's incredible how loose boxes provides more restriction than normal Sudoku boxes. Took me 90 mins to solve by myself, what a wonderful puzzle this has been ❤.
Definitely one of the most enjoyable puzzles I’ve solved. I have no clue how you’re supposed to come up with something like this. I wonder if the digit limit could be lowered further
20:08 for me. that was fun. i use to do badly at new constraints (like the dam previous *puzzle*), but this gave me the oportunity to give the last laugh
The 9th 3x3 box can be determined after getting the first two digits. With a 1 in columns 1 and 10, if box 4 extended into column 4 then there would be a 1 in each of colums 2-9 as well for a total of 10 1s and the puzzle would be broken.
I think the "extra power" comes down roughly to breaking of symmetries inherently meaning fewer possibilities to still arrive at something valid.
*Any* nontrivial symmetry is going to allow you to find situations where more than one solution is possible until you give enough constraints to break that symmetry.
17 is the number of constraints you need at a minimum to ensure breaking all the symmetries in regular Sudoku.
But this slightly less symmetric situation must be causing three more symmetries to disappear somehow, giving you a need of just 14 filled-in constraints. And I think two of those are quite obvious: You are no longer connecting up full columns nor full rows.
Perhaps the third is that one of the diagonals is also broken up? Not sure...
I don't think given digits correspond one-to-one with symmetries
@@galoomba5559 on an empty Sudoku (without any further constraints but basic Sudoku rules), you have complete freedom of choice. you can pick any one out of 81 spots for any one out of 9 digits: 729 possible choices for that first digit alone.
As soon as you fill in one digit, you automatically constrain the puzzle in all kinds of ways: That particular digit can only go again into 60 remaining spot (none of the 9 in the same box, and 2x6 additional removed choices from the column and row) - it's therefore a much less symmetric space. You can also only pick it 8 more times while the other digits still are available 9 times.
This next bit I'm not looking up so I may well make some reasoning error here, but as far as I can tell:
For a *filled* grid, without extra constraints beyond the basic Sudoku ones, you can freely swap labels to arrive at essentially the same puzzle, giving you 9! = 362880 "identical" puzzles up to the labelling symmetry.
You also can freely rotate or reflect the puzzle, giving you, I *think* 8 more ways puzzles might be considered "identical" - those are just the symmetries of a basic square.
Additionally, for row or column of larger boxes, you may reorder the three rows or columns within them freely, giving you 2x3x2x3 = 36 choices (2 for whether you are dealing with columns or rows, 3 for which big column or row you are dealing with, and 2x3 for the 6 ways you may rearrange stuff in those columns and rows)
And you may *also* freely swap the big rows and columns (2x2x3 = 12 for whether cols or rows, and each possible order of those)
Altogether, that gives you
9! * 8 * 36 * 12 = 1254113280 Sudoukus that are effectively "the same" under its basic symmetries.
At least those are all the symmetries I can think of. Maybe there are more. Probably something about how subset that have all nine digits in it can be related or something. Not sure how you'd count that though, and I'm not sure whether that still would count as "essentially the same".
But at any rate, at a minimum, to define a *specific* Sudoku, you need to fill in enough values, or add enough extra constraints, to break all the freedoms those symmetries give you.
1.25 billion might seem a lot, but of course, every choice you make very quickly reduces this spaces down to something much more manageable.
All of this math assumes the 9x9 grid of course. And that is a highly symmetric playing field.
This slightly *larger* field is less symmetric immediately, because not every place gets a digit. You still work with the same number of digits after all. You also get constraints on the box shapes.
The original 9x9 grid fits into this larger grid in 4 ways of course. However, you can completely take out most of the symmetries regarding that by simply putting a single number "off grid" (relative to an original 9x9; the exact condition is that every row and every column has at least 1 digit in it somewhere), thereby forcing the utilization of all 10 rows and columns.
And despite the extra space, the remaining possibilities are probably more constrained than you'd think, as all the most symmetric ones (just working with the original 9x9) are instantly gone. Doing the math there is gonna be much trickier though, as you'd have to carefully keep track of all the ways boxes may distort and what not.
How in the world did you stumble in to the exact logic you needed that was like 5 logical steps deep right away at 30:00
The extra regions don't add any extra power. They are just like the the third row (or column) of those three boxes.... But wait! There is this part of three digit sticking out and in a row(or column) of six that is looking at another three squares that would normally not be "looked at".
The colors are quite good for color blind people. 5 (purple) and 9 (blue) are a bit similar but all other’s distinguish quite well from each other.
If you permanently replace blue by dark blue it would be perfect for me (red-green blind).
I'm neither a Sudoku nor math person. But I can give a bit of intuition for why this setup should allow for a smaller number. And that is because certain boxes are twice constrained. Let's take row 4 from top. The blue box from row 4 is constrained because it has to form a 1-9 with row 1. But at the same time it can't have the same numbers as the lime box in the same row 4. That way you reduce the number of independent variables and therefor you reduce the minimum number needed to give a unique Sudoku.
I have never seen letters being used in solving a Sudoku. That's a first for me, but it makes sense somehow, especially in such a strange Sudoku such as this one.
A more general statement about 10×10 grids is that you automatically know 4 cells for each cage. No matter how you giggle the boxes, they must contain a 2×2 square that's 1 cell away from the perimeter and each other.
But in this puzzle your method is far more direct.
you know shit's about to get real if letters even managed to to introduced in sudoku 😭🤣💀
The law of leftovers makes it more powerful, you already said it.
Haven't liked a video in a week or so but this was magnificent
The geometry of the whole thing was fascinating and I just kind of stumbled my way through it.
Looking at the opening grid in the link;
I think, that "14-digit conundrum," is because 16 cells are given to be in (whereas, on a 11x11 grid, only 9 cells are given to be in at the start)
I had to look at it twice (to make sure those inner ones were two -- and 2x2 in the center).
That might be a way to view this as less "counter-intuitive," (on an "11x11 grid," only 9 cells of the boxes are given to be in at the start).
(In other words, on an "11x11 grid," you're truly "finding'' the boxes more)
Don't know.
What's the minimum of digits on an 11x11 grid of this nature?
Very good (very gouda) though (
16 cells given to be in at the start (on a 10x10) - versus 9 (on an 11x11).
That's a huge disparity.
That's why I think those "16" (as-opposed to 9) has something to do with it.
Don't know, though..
Very good though again.
"Snow.?!"
(It's "outside," literally)
😂 ;) 😉
PS: Because of those middle twos on the 10x10, you can extend out the corners and edges by one.
(Counting, that's 36)
36 (on this 10x10) to that "9" (on an 11x11).
A huge, HUGE disparity now.
(36 - 9)
Wow.
Something's up there (maybe)? 😂
FInding the boxes is easy, but I struggled badly with the sudoku bit. Naturally, it took Simon about 2 minutes to spot something that evaded me for half an hour, but I'm not bitter and twisted about it at all.
I was worried for a minute when I noticed that the conflict checker was not working for boxes, only r/c.
But I’m guessing that Sven in his brilliance did that purposely to foil any cheating through guessing… 54:09
The geometric "power" you are looking for is probably for example
the 294 triplet in dark blue square "participating" in 2 columns instead of 1
Wow what a cool concept! Thanks Simon it was a pleasure to watch.
One thing that might be relevant as to why 14 digits was possible for this one, is the fact that horizontal striping occurs in the top 6 boxes.I wonder if there is a theory, that can prove that with such low count of given digits some striping must occur?
I love how this puzzle ie easier of you look for normal sudoku. At least, for pencil marking at the beginning.
Fascinating puzzle and discovery! Loved it