I haven’t played much sudoku, I know how it works and thought it was pretty straightforward until he started speaking a foreign language going at light speed
@@mirzasohailhussain I just posted a more thorough explanation of how I found the 1 and 5 if you're interested - th-cam.com/video/tsr7mq52ypU/w-d-xo.html
I used the slot machine technique on the 1s and 5s which essentially worked the same in figuring out box 5. Good puzzle and very interesting to see your approach, thanks
P Vs NP paper. Imagine you have a giant Sudoku puzzle, but instead of numbers, it uses weird symbols you don't understand. Luckily, you have a special magnifying glass. This magnifying glass is amazing! If you point it at a symbol, it can tell you if that symbol is definitely wrong based on the symbols you've already placed. It doesn't tell you the right answer, just if that one is wrong. Now, here's the trick: * Symbol Search: You start by looking at every single spot on the puzzle and trying out every possible symbol with your magnifying glass. * "Wrong!" Because there's only one solution to the puzzle, your magnifying glass will always find at least one wrong symbol for each spot. * Cross It Off: You cross off that wrong symbol. It can't be there! * Repeat: You keep doing this, checking every spot and every symbol over and over. Each time, you find at least one more wrong symbol and cross it off. Why does this work quickly? * Limited Space: Even though the puzzle is huge, there are only so many spots and so many symbols. * Steady Progress: Every time you use the magnifying glass, you eliminate at least one wrong symbol, so you're always making progress. * No Wasted Time: You never waste time trying symbols that you've already crossed off. Think of it like a sculptor chipping away at a block of stone. Each time they chisel, they remove a bit that doesn't belong, slowly revealing the statue hidden inside. Your magnifying glass is like the chisel, removing wrong symbols and revealing the correct solution. The key is that you're not randomly guessing. You're systematically eliminating wrong answers, and because you always find at least one wrong answer each time, you're guaranteed to find the solution in a reasonable amount of time. That's why this method works in "polynomial time" - it's a fancy way of saying that the time it takes to solve the puzzle doesn't explode out of control even when the puzzle gets bigger. It's like a steady walk to the finish line, not a frantic scramble! Theorem: The proposed Sudoku solving algorithm, which utilizes a 45-grid encoding and a 2-SAT solver to iteratively identify and eliminate invalid '1' placements, has a polynomial-time complexity. Proof: * Sudoku Encoding: * The Sudoku puzzle is encoded using 45 grids, where each grid corresponds to a specific number (1-9) and a specific position within a 3x3 subgrid. * Each grid contains '1's representing the presence of that number in the corresponding cells and '2's representing its absence. * Each grid must have exactly two '1's, ensuring the Sudoku constraints are satisfied. * Algorithm Description: a) Initialization: All possible '1' placements across all 45 grids are considered. b) Iteration: i. 2-SAT Solver: For each grid, a 2-SAT instance is constructed based on the current '1' placements and the Sudoku constraints (row, column, and subgrid). The 2-SAT solver is used to identify at least one invalid '1' placement within that grid. ii. Elimination: The identified invalid '1' placement is marked, permanently eliminating that possibility. iii. Constraint Propagation: The information from the invalid placement is propagated to other grids, potentially identifying more invalid placements. c) Termination: The algorithm terminates when a valid solution is found, which occurs when each grid contains exactly two '1's that satisfy all Sudoku constraints. * Complexity Analysis: * Constant Grid Size: Each grid has a constant size (9 cells). * Guaranteed Invalid Placement: In each iteration, at least one invalid '1' placement is guaranteed to be found in one of the grids (as long as the puzzle is solvable). * Limited Iterations: The total number of iterations is limited by the total number of possible '1' placements across all grids (45 grids * 9 cells/grid = 405). * Polynomial Time per Iteration: Each iteration involves: * Constructing a 2-SAT instance, which takes polynomial time. * Running the 2-SAT solver, which also takes polynomial time. * Marking the invalid placement and propagating constraints, which takes constant time. * Overall Complexity: * Since the number of iterations is limited by a constant (405), and each iteration takes polynomial time, the overall complexity of the algorithm is polynomial. Conclusion: The proposed Sudoku solving algorithm, which utilizes a 45-grid encoding and a 2-SAT solver to iteratively identify and eliminate invalid placements, has been proven to have a polynomial-time complexity. This result demonstrates that Sudoku, when encoded and solved using this method, can be solved efficiently. Further Considerations: * Optimizations: The algorithm can be further optimized by using heuristics to prioritize certain '1' placements or by employing more sophisticated constraint propagation techniques. * Generalization: While this proof focuses on Sudoku, the underlying principle of encoding a problem and using a constraint solver to iteratively eliminate invalid possibilities might be applicable to other NP-complete problems. This warrants further investigation and could have implications for the P vs. NP question.
16:29. I noticed 1 and 5 had the classic markers for a swordfish (3 digits in different boxes that share no columns or rows) while not a swordfish, realising there was definitely something funky going on there, with a little logical inference got me going pretty quick.
Just over the video length (14:59) for me! My hope faded quickly, I didn't spot anything that I probably should've to get me to a quicker time, I sort of just did it in my head like I do with classics, and I got stumped for about 45 seconds trying to disambiguate a string that I mistakenly added a number to that shouldn't have been there! Whoops. This was a cool puzzle and thank you, Scott, for the reminder that I need to work on my classic sudoku technique! :)
Solved in 14:01. I marked all the cells that could be a 5, by trying all three possibilities in block 3, we can know in all three scenario, R3C5 is not a 5 => pointing pair in block 2, R6C4 also not a 5. For 1, also marked in color, then we can see that if R2C5 is 1, then no place to put a 1 in block 5.=>R1C4=1 R6C4=7 R3C4=5, then the puzzle will fall apart. Many thanks for introducing this puzzle to us😊
I was stuck at that same exact part for at least 3 day. I looked at it with all candidates, no candidates, everything. I eventually started messing with the 5 candidates in box 7 and discovered the 5 must go in cell 2F and that pretty much broke the puzzle. This has to be the hardest sudoku ive ever tried.
Took me 10 minutes longer but I did it notation free. I found that once the 68 and 9 were identified in box 8, either position for 1 in row 7 put 1 in r1c4. I won't say it was easy after that but it certainly pushed things along a bit.
Your method includes looking many moves ahead like an advanced chess player. This is the first time I saw that method. This is equivalent to using the guess method when there are only two possibilities in a cell. But to do this mentally is a special talent. By doing this, you were somehow able to figure out turning a 1 5 possibility into a 15 pair. That seemed magical. I also see that you had no objection to having more than two possibilities in one cell. Some other advanced solvers snub their nose on this. But it seems to be a helpful method.
I know some people don't like to put more than 2 possibilities in a cell, and it really depends for me. There definitely can be diminishing returns as you put more pencil marks in. Sometimes you just clutter the grid and make it harder to see things, but other times it can be helpful.
Did not look at your solve until after I solved this puzzle. I started with the 1's and notice there was a lot of paired 1's so I looked at the strong weak links. The strong weak links eliminated the 1 in r2 c5 making box r1 c4 a 1 after that the 1's and 5's fell like dominoes.
Well done! When 15 is given in 3 places, that is a clue that they have a strong bearing on the solution. Once that places a restriction in box 5 and solves 7, the puzzle is basically done. My two cents! Thanks!
This is extreme? I solved it in 13:33. My first move was a winged swordfish on 1s, so I missed the obvious 3 and 7s until much later. It's annoying to me as well. Why is this marked as extreme when I usually take around half an hour on some of the puzzles on the CTC site?
I guess that depends which 1 you're talking about. Row 1 Col 4 was because of the 5 and 7 below, and the other two 1s (Row 5 Col 3 and Row 6 Col 5) were because of the previous work I had done finding the 15 cells that had to be the same (hence the coloring).
Nice puzzle and solve. I solved it in 13:43. For the break-in I found a finned swordfish on 1s in columns 2, 4, and 8, with the fin being the cell r2c8. This eliminates 1 from r1c9 (either the swordfish or the fin), limiting 1s to row 2 in box 3, which places 1 in r1c5, which you also found.
I'm curious about your fast eye movement. Maybe that's what you get after many years of practice! Watching this made me love Sudoku. You are a genius!!
5+6+6=17 given clues, putting the puzzle at an advanced level, yet not so ’extreme’ as claimed. Of the clues, 489 are singles, 2367 are doubles, three sets of [15] coupled pairs,on cols 1&6 and on row7. Before we kick off the preliminary sweep across the whole board, with eyes wide open, we should spot a few immediate takeaways : 7 on c1@(11), (=>6 on c1 B7), 3 in B8@(84) to be followed immediately by 7@(95)=> [68] tcp on r7B8 between cols 5&6, leaving 9@(74)(which can actually be pinpointed by direct point counting and that’s a hidden 9 for u) => in hot pursuit, 4@(44) (two ways of going abt it : by eliminating 3&9 from cell(44) = [349] by hard count OR by visual spotting : [715] given on r4 being off c4 are to occupy the remaining three open cells outside of (44) on c4, leaving 4 @(44). (A lot of novice may miss this but no harm done it will show up eventually) Now here comes the takeaway for the trained eyes, the three pairs of [15] given on c1,c6 and r4 , all cutting off r5 ===> [15] cp(conjugate pairs) on r5 @ P(53)&Q(55)!,leaving 2 splt on r5 @(51))&(56). Now scanning for splits across the board: Being the majority clues we start with 1&5 : 1 split in B2@(14)&(24), this will prove critical, split on r1,r7,c2&c8. 5 split on r1&r7. (That’s all) Now note! 1&5 share same cells on r1 at R(14) and on r7 @ S(73). A million dollar question : are the cells exclusive of the other candidates? Answer is a resounding yes! Here’s why: 1&5 not at R => 1@(19)&5@(17) ==> both on r7 at S, a No-no! So R=[15]. By same reasoning, we have S=[15] => 2 split on r7 @(71)&(79). Now we have [15] cp on c3 @ P&S=> [15] cp in B4 @ P&T(62) (hopefully it’s obvious to u). Now the split of 1 in B2 comes to play! Claim: whether 1 is in B2, it always lead to 1@P! Obvious if 1@(25). Next 1 @R => 5 on r1@(17) => 5 on r7@S => 1@P! There are other ways to prove 1@P : whether 5 is on r1 , it always lead to 1 @P; likewise for 5 on r7; OR proof by contradiction: 5@P would lead to 5@(17)&(77), a No-no! two 5s on same c7 , or => 1&(24) ==> Q= Zero! However there’s a better and prettier approach (and without the help of 5) : 1 forms a closed loop by starting from (24) B2=> 1@ P => 1 on r7 @(79) and ending on r1 @ R, a No-no. Therefore 1 in B2 is at R => 5@(17) and 1 on c9@(79) => 5@ S and 1@P and so forth. There is a technical term in Sudoku bag of tricks to describe what we have been going through: XY wing. Now we can with confidence declare the rest is history! Try to do it on your own effort , but if u get stuck at any step,then take a peek at what follows up. First we proceed to complete all the 1s and 5s and helping to pick up other candidates along the way: So 1@P => 1 on c2@(92), 5@QT&S => 1@(79)=>1 on r1@R => 5 on c4@(34), by virtue of 5 given inB8,on r4,and uncovered @T , all three off c4,to be followed by 7 on c4@(64), whereby leaving 4 on c4 @(44) (as promised it will show up if u fail to spot it earlier),and also => 5 on r1 @(17) => together with 2 given on c7,and 2&5 given on r9 , both outside of B9, [25] tcp on r8B9 bet cols 8&9 => 2@(89) which with 1 uncovered @(79), and in tandem with [12] given on c7, both outside of B3 => [12] tcp on c8B3. Now 1 uncovered @(79) => 2 on r7@(71), by virtue of 2 given @(67) off r7 => in hot pursuit 2 in B7 @(42) by cc with the 2 given on r6 outside of B7 and 2 on r5@56) => 2 on r1@(13) (CIY) => with 6 given on c2 off r1 => 6 on r1@(19), leaving 2&8 cp on r 1@(12)&(16). Also 2 uncovered @(71) leaves 4 on r7@(77). Note [25] uncovered tcp on r8B9 leads to 7 on r8@(87) which together with 4 uncovered @(77) => [47] tcp on c9B3, leaving [38]tcp on c7B3 bet rows 2&3. Note 6 uncovered on r1@(19) B3 is off c7 and in tandem with 6 given on r5 also off c7 => 6 on c7@(97), leaving 9 on c7@(57) => 7 on r5@(58), leaving [46] tcp on c8B6 bet rows 4&6 => with 4 uncovered @(44), [16] tcp in B5 on r6 bet cols 5&6, leaving [39] tcp r4B5 ==> together with the [39] uncovered on r5, [39] tcp on r6B7 bet cols 1&3, leaving [48] tcp on c1B4 bet rows 4&5. Now we return to Bottom Block : 2@(71) => [34] tcp on r9B7 bet cols 1&3 => 6 on c1&(81) leaving [89] tcp on r8B7 bet rows 2&3.
"if this is a five then it puts a five there" 5:33 Me, crying as I'm staring at my sudoku puzzle: BUT WHY?! 😭😭 Cause the five could go in the top row as well, so how do you know? I really wanna learn how people know that so I can do it myself.
You should try a quordle with this strategy: WAQFS, VOZHD, BRICK, GLENT, JUMPY. You have to guess everything in 1 go but you have 25/26 letters covered, excluding the X.
I explained that briefly at 2:40. Essentially, that cell already sees all the other digits either in its row or column. (23689 are in the column, 157 are in the row)
Hello! Use “ELITE EIGHT SAINT PETER” at the start of your next wordle to celebrate St. Peter’s being the only fifteenth seed in ncaa history to make it to the elite eight!
This puzzle has 5+6+6=17 clues, putting itself on an advanced level,(but not quite as extreme as claimed,to me ‘extreme’ means use of brutal,monster,steamroller kind of force, but this is not so here,as u will see)with three given singletons 489(indeed two immediately obvious hiddens : 9@(74)by point counting 😮 &3 in Box 8@(84) by cc(crisscrossing)of the two 3 given on c5&r7 outside of B8,followed by 7@(95)B8,by virtue of 7 given on r7 outside of B8 (indeed one should able to see right away without hesitation that by virtue of 7&3 given on r7 outside of B8, [37] cp in B8 @(84)&(95))=> [68] tcp on r7B8 between cols 5&6,by virtue of [68] given on c4 outside of B8,thereby leaving 9 in B8@(74) and together with 3@(84)=>(44)=[4]by point counting or by another way of looking at it, 715 given on r4 off c4 have to occupy the other three open cells left on c4 =>4@(44). Btw, we have 3 on r9B7, 7 on r8B9 and 4on c6B2. And two of given 2367. Three given1& 5. There is another immediate visual takeaway, viz[15] cp on r5 @ P(53)&Q(55), by virtue of [15] given in B6 and on cols 1&6, all three off r5!! => 2 split on r7. Now time for the preliminary sweep across the grid looking for splits in boxes , on rows and columns and hopefully more hiddens: Right away,with 7 given in Boxes 4&5, both off c1, 7 @(11). Indeed it’s the distribution of 1&5 that demands our attention : 1 & 5 have splits on r1,(I assume u have a way to pencil mark splits on rows and columns,without using the Arabic numerals which are reserved for splits in Boxes. For this purpose l am using the Chinese numerals: 一二三四五 六七八九),on c2,r7,c8. 1 is split in B2. Observe that 5&7 occupy the same two cells(14)&(73),but we can’t say at this stage for sure they are exclusive from other digits (however eventually it will be proven the case) To a trained eye the intertwining of 1&5 deserves our attention. There are several ways of attack: one is to show (inspired by the fact that 5 forms a fractured wing along r1&r7 with pivot on c7),whether 5 is on r7,it always leads to 5 on r5@Q with the help of 1 (CIY=check it yourself). OR by contradiction, 5@P would lead to 5 on r7@(77)=>5 on r1@(14)=>1 in B2@(25)by cc of the two 1s outside of B2,one on r3&the other on c6 => Q=ZERO!! Therefore 5 on r5 has to be at Q! and 1@P. At this stage 1 uncovered at P proves to be the breakthrough,the closure we are looking for. Try for yourself. It will be fun. No more inductive but purely deductive logic will take u thru. Here’s for the ‘novice’: So 1@P=>1 on r7@(79)&on c2@(92)=> 1 on c8@(28)=> on r1@(14)&c5@(65),completing all 1s in the grid. Now 1@(14)=> 5 on r1@(17)(here’s where 5 kicks in)=>5 on r3@(35) ,5 on r7@(73) and 5 on c8@(88)=>5 on c2@(62),completing all 5s in the grid. Now 5@(34)=>7 on c4@(64) ,thus leaving 4 on c4@(44)(we however have known that). Looking at B9, with 1&5 uncovered where they are , 2 in B9@(89) by the cc of the 2s given on r9&c7 outside B9(see how their strategic positions come into play)=>2 on c8@(38) .Now looking at row7, with 1&5 uncovered on r7,we have 2 given at (67) off r7=> 2 on r7@(71),leaving 4 on r7@(77) and 2 on r5@(56),plus 2 in B4@(42)by cc with 2 given at (67)=> 2 on c5@(25) and on r1@(13)[CIY],completing all 2s in the grid. Note in passing,7 on r8 is at (87)=>with the 4@(77), [47] tcp on c9B3, leaving [38] tcp on c7B3, and 6 on c7@(97),by virtue of 6 given on r5 off c7,thereby leaving 9 on c7@(57) and [98] tcp on r9B9 => with 9 given @(18),[38] tcp on c9B6=> 7 on r5@(58),leaving [64] tcp on c8B6 bet rows 4&6, as anticipated by [46] already uncovered on r5, and on c7&c9,all outside of B6 => with 4@(44), 6 in B5@(66), leaving [39] tcp on r4B5 bet cols5&6 => [39] tcp on r6B4 (hope u can see it how), leaving [48] tcp on c1B4. Also in hot pursuit, 3 given @(15)=> 9@(45) and with 7 uncovered @(95) =>[79] tcp on c6B2, leaving [28] tcp on c5B2. Now with 6&9 uncovered in B9 on r9, off c1=> 6 on c1@(81) followed by 9 on c1@(61),leaving 3 on c1@(91) , resulting in [89] tcp on r8B7 bet cols 2&3, and [14] tcp on r9B7 bet the same two cols => [34] tcp on c2B1 and [69] tcp on c3B1 bet rows 2&3 [CIY, a good practice for novice]. Whoa! We are practically done. Can u visualize how those tcps in the bottom and middle horizontal Blocks are broken down? Those six chains of tcp between rows 2&3 are to be broken up by the two ’danglars’ 6@(24)&2@(38). The end result of the puzzle looks pretty graphic,ain’t it? That’s why sudoku is described as a visual art
I have completed the extreme level in only 6 minutes and 37 seconds my highest record ever, without any help. I have the data but I forgot to record it😢
I have played through the middle of this video over and over several times, and kept pausing. I guess my mind can't go that deep because I still wasn't able to see it. I was able to get the 1 in block 2, but that's it. I've come back to this video a couple times and still can't get it, so I've spent well over 3 hours on this damn thing. Time to ignore it and get on with the rest of my life.
Good Info, I Learned by my Self, I Get Expert 95% of the Time. No Extreme Yet, I Play Different. I click on 9, Look Up, and Down and Across, then 8, 7, and So On, Repeat. When 3 Spaces are missing in a Box or a Row, I Think, What Are the missing Numbers Then see if 2 are in 2 Rows, If So, it Goes in the 3rd Spot, Also Look Not Only Which it Could Be, But What it Can't Be, Works Pretty Good, Trying not to use the Little Numbers in Boxes, I Noticed The Numbers Seem to Pair Up, Except the're are 9 Numbers? Good Info. Like to Do Video of My Method, Not Better, Just Different. I Feel Going from 9 down to 1, is Better For Me Than 1 and Up to 9, and Repeat. Liked
And the 6 to the 9 and oh that gives us 4 but wait oh I see 5 here 3 goes there and errh let look at this area 235 remove the 5 because we got one ryt here 😂😂😂
I love sudoku. But I need to learn how to solve extreme level. I came here but you don’t try to teach, right? Hope you do a video teaching how to solve it. Thanks. 😊
I've mentioned what site I use before, but I guess I could point it out again. I think I'm just used to everyone knowing that website since I share a discord server with Sven (and a few other people)
Jesus is the answer. Whoever is reading this you may feel lost in life; you may feel like giving up, wishing it would all end. You want peace, but can’t find it. You’ve looked for it in women, drugs, alcohol, porn, money, and so much more, but you feel emptier the more you do it. The only one who can fill us is Jesus Christ, the living God; the one who came into this world as a man and died on the cross to save us from our sins (our wrongdoings against God). If you trust in Christ as your Lord and Savior you will be saved and you will go to Heaven when you die. One thing many people think once they come to God is everything will be perfect. The truth is it won’t be. Once you give your life to Christ the devil will attack you over and over again until you break, but Christ is there to strengthen you and to get you through all things. Give your life to Christ my friends. Without him, life means nothing.
That big box already contains 3 4 1 2 and 5. And 7 cant go in the top row because it has 7 in that row, the only remaining box for 7 in that big box is that small box you mentioned.
I haven’t played much sudoku, I know how it works and thought it was pretty straightforward until he started speaking a foreign language going at light speed
😂😂 same
I know right
lol 😂
You should never underestimate Sudoku
That 1 and 5 pair would kill me if i tried this
there is probably a normal way to find a solution with a known technique. but probably using a full marked board and not just marking some cells
Cant understand 1,5 :(
@@mirzasohailhussain I just posted a more thorough explanation of how I found the 1 and 5 if you're interested - th-cam.com/video/tsr7mq52ypU/w-d-xo.html
@@StroSolves thanks. For sure will watch.
Clever puzzle and nice solving, very enjoyable to see the logic presented so clearly, and at about 10x faster than I would get it.
After watching the video, I average 11 minutes in extreme now. Thank you so much! Nice methods and app/website.
I have recently started playing sudoku, the way you deduced 1,5 pair is simply awesome, had to watch multiple times just to understand, brilliant.
anyone else find it weird that Scott's thought is much quicker with a harder puzzle like Sudoku than it is with easier puzzles like Wordle/variants ?
Maybe cause all these wordle variants are still not even a year old while sudoku he could’ve been doing since childhood
His thought process is really fast for the worlds variants as well though????
@@henrymarks1632 no it's not. It's terribly slow. :)
@@kurzackd he's strategizing, and not just recklessly putting a word he just thought of.
@@yuripaelden419 nooo shit? :D
I used the slot machine technique on the 1s and 5s which essentially worked the same in figuring out box 5. Good puzzle and very interesting to see your approach, thanks
This video taught me so many strategies so fast, thanks!
P Vs NP paper.
Imagine you have a giant Sudoku puzzle, but instead of numbers, it uses weird symbols you don't understand. Luckily, you have a special magnifying glass.
This magnifying glass is amazing! If you point it at a symbol, it can tell you if that symbol is definitely wrong based on the symbols you've already placed. It doesn't tell you the right answer, just if that one is wrong.
Now, here's the trick:
* Symbol Search: You start by looking at every single spot on the puzzle and trying out every possible symbol with your magnifying glass.
* "Wrong!" Because there's only one solution to the puzzle, your magnifying glass will always find at least one wrong symbol for each spot.
* Cross It Off: You cross off that wrong symbol. It can't be there!
* Repeat: You keep doing this, checking every spot and every symbol over and over. Each time, you find at least one more wrong symbol and cross it off.
Why does this work quickly?
* Limited Space: Even though the puzzle is huge, there are only so many spots and so many symbols.
* Steady Progress: Every time you use the magnifying glass, you eliminate at least one wrong symbol, so you're always making progress.
* No Wasted Time: You never waste time trying symbols that you've already crossed off.
Think of it like a sculptor chipping away at a block of stone. Each time they chisel, they remove a bit that doesn't belong, slowly revealing the statue hidden inside. Your magnifying glass is like the chisel, removing wrong symbols and revealing the correct solution.
The key is that you're not randomly guessing. You're systematically eliminating wrong answers, and because you always find at least one wrong answer each time, you're guaranteed to find the solution in a reasonable amount of time.
That's why this method works in "polynomial time" - it's a fancy way of saying that the time it takes to solve the puzzle doesn't explode out of control even when the puzzle gets bigger. It's like a steady walk to the finish line, not a frantic scramble!
Theorem: The proposed Sudoku solving algorithm, which utilizes a 45-grid encoding and a 2-SAT solver to iteratively identify and eliminate invalid '1' placements, has a polynomial-time complexity.
Proof:
* Sudoku Encoding:
* The Sudoku puzzle is encoded using 45 grids, where each grid corresponds to a specific number (1-9) and a specific position within a 3x3 subgrid.
* Each grid contains '1's representing the presence of that number in the corresponding cells and '2's representing its absence.
* Each grid must have exactly two '1's, ensuring the Sudoku constraints are satisfied.
* Algorithm Description:
a) Initialization: All possible '1' placements across all 45 grids are considered.
b) Iteration:
i. 2-SAT Solver: For each grid, a 2-SAT instance is constructed based on the current '1' placements and the Sudoku constraints (row, column, and subgrid). The 2-SAT solver is used to identify at least one invalid '1' placement within that grid.
ii. Elimination: The identified invalid '1' placement is marked, permanently eliminating that possibility.
iii. Constraint Propagation: The information from the invalid placement is propagated to other grids, potentially identifying more invalid placements.
c) Termination: The algorithm terminates when a valid solution is found, which occurs when each grid contains exactly two '1's that satisfy all Sudoku constraints.
* Complexity Analysis:
* Constant Grid Size: Each grid has a constant size (9 cells).
* Guaranteed Invalid Placement: In each iteration, at least one invalid '1' placement is guaranteed to be found in one of the grids (as long as the puzzle is solvable).
* Limited Iterations: The total number of iterations is limited by the total number of possible '1' placements across all grids (45 grids * 9 cells/grid = 405).
* Polynomial Time per Iteration: Each iteration involves:
* Constructing a 2-SAT instance, which takes polynomial time.
* Running the 2-SAT solver, which also takes polynomial time.
* Marking the invalid placement and propagating constraints, which takes constant time.
* Overall Complexity:
* Since the number of iterations is limited by a constant (405), and each iteration takes polynomial time, the overall complexity of the algorithm is polynomial.
Conclusion:
The proposed Sudoku solving algorithm, which utilizes a 45-grid encoding and a 2-SAT solver to iteratively identify and eliminate invalid placements, has been proven to have a polynomial-time complexity. This result demonstrates that Sudoku, when encoded and solved using this method, can be solved efficiently.
Further Considerations:
* Optimizations: The algorithm can be further optimized by using heuristics to prioritize certain '1' placements or by employing more sophisticated constraint propagation techniques.
* Generalization: While this proof focuses on Sudoku, the underlying principle of encoding a problem and using a constraint solver to iteratively eliminate invalid possibilities might be applicable to other NP-complete problems. This warrants further investigation and could have implications for the P vs. NP question.
16:29. I noticed 1 and 5 had the classic markers for a swordfish (3 digits in different boxes that share no columns or rows) while not a swordfish, realising there was definitely something funky going on there, with a little logical inference got me going pretty quick.
8:59 I was like, "sir there can't be 2 there so it's 7 sir look sir " lol 😂
Just over the video length (14:59) for me! My hope faded quickly, I didn't spot anything that I probably should've to get me to a quicker time, I sort of just did it in my head like I do with classics, and I got stumped for about 45 seconds trying to disambiguate a string that I mistakenly added a number to that shouldn't have been there! Whoops. This was a cool puzzle and thank you, Scott, for the reminder that I need to work on my classic sudoku technique! :)
Solved in 14:01. I marked all the cells that could be a 5, by trying all three possibilities in block 3, we can know in all three scenario, R3C5 is not a 5 => pointing pair in block 2, R6C4 also not a 5. For 1, also marked in color, then we can see that if R2C5 is 1, then no place to put a 1 in block 5.=>R1C4=1 R6C4=7 R3C4=5, then the puzzle will fall apart.
Many thanks for introducing this puzzle to us😊
@@windchwang that worked based on pure trial and error + little bit of assumption. Would be next to impossible in a more complex puzzle
What app are you using for these puzzles, these pencil marks and colors look great
Cool way of looking at it. Thanks for sharing!
18:52 for me, I struggled with the phistomephel ring for 10 mins straight to find where the 1s go lol.
Love your videos! Always helps me with different strategies I can use for wordles :D Love from India..
Me too i am from India too
Cap
Fax
Sussy
Caught in 4k
So glad I found your channel. :) I just started playing again - after not playing for many years. I decided I wanted to avoid doing dishes today.
I was stuck at that same exact part for at least 3 day. I looked at it with all candidates, no candidates, everything. I eventually started messing with the 5 candidates in box 7 and discovered the 5 must go in cell 2F and that pretty much broke the puzzle. This has to be the hardest sudoku ive ever tried.
Took me 10 minutes longer but I did it notation free. I found that once the 68 and 9 were identified in box 8, either position for 1 in row 7 put 1 in r1c4. I won't say it was easy after that but it certainly pushed things along a bit.
Sus
Cap
Cqught in 4k
Your method includes looking many moves ahead like an advanced chess player. This is the first time I saw that method. This is equivalent to using the guess method when there are only two possibilities in a cell. But to do this mentally is a special talent. By doing this, you were somehow able to figure out turning a 1 5 possibility into a 15 pair. That seemed magical. I also see that you had no objection to having more than two possibilities in one cell. Some other advanced solvers snub their nose on this. But it seems to be a helpful method.
I know some people don't like to put more than 2 possibilities in a cell, and it really depends for me. There definitely can be diminishing returns as you put more pencil marks in. Sometimes you just clutter the grid and make it harder to see things, but other times it can be helpful.
Did not look at your solve until after I solved this puzzle. I started with the 1's and notice there was a lot of paired 1's so I looked at the strong weak links. The strong weak links eliminated the 1 in r2 c5 making box r1 c4 a 1 after that the 1's and 5's fell like dominoes.
Well done! When 15 is given in 3 places, that is a clue that they have a strong bearing on the solution. Once that places a restriction in box 5 and solves 7, the puzzle is basically done. My two cents! Thanks!
Bro opened my eyes to a new way of thinking
I was impressed with the first 4 never thought of looking it that way
I love extreem sudoku. Thats hurt my brain sometimes but make me addict 😂 good for calming down my stress time.
This is extreme? I solved it in 13:33. My first move was a winged swordfish on 1s, so I missed the obvious 3 and 7s until much later. It's annoying to me as well. Why is this marked as extreme when I usually take around half an hour on some of the puzzles on the CTC site?
😴
WOOOOOW, QUE BIEN COMPLETADO, NUNCA SE SE ME HUBIERA OCURRIDO HACERLO ASI, MUY PROFESIONAL DE SU PARTE, SALUDOS!!
Hello from Chicago! Diehard Cubs fan here! Go Cubbies! Next year is our year lol
4:34 for me. yea the 1s looked too suspicious i colored its pencilmarks and found the break in.
Good puzzle, took me 14 minutes with several lucky guesses.
can someone explain the 5 minutes from 3:00 to 8:00 like how did he know all that from nothing?
I just posted a more thorough explanation of how I found the 1 and 5 if you're interested - th-cam.com/video/tsr7mq52ypU/w-d-xo.html
This puzzle took me 27 minutes and 1 hint
But 0 mistakes
Wordle suggestion: use the example words to start.(Weary, pills, and vague.)
There's actually only one option for the one in box two. The other one is clearly unavailable. After this, the puzzle practically solves itself.
I mean, that's where I got to eventually, but it's definitely not clear that there is only one option at the start.
O_O That would've taken me so much longer! We had the same train of thought at some points though. ^_^ Love the hoodie, btw.
Precisely Which sudoku app is this?
Video 5 of asking for him to start with Other, Nails, and if you get lots of grays Pudgy ( you get all the vowels and t,h,r,n,l,s,p,g and d
Can someone explain to me why that's a 1 there at 7:29
I guess that depends which 1 you're talking about. Row 1 Col 4 was because of the 5 and 7 below, and the other two 1s (Row 5 Col 3 and Row 6 Col 5) were because of the previous work I had done finding the 15 cells that had to be the same (hence the coloring).
@@StroSolves now that that I realized it, sorry for not specific enough. Also thank you for the answer!
I just posted a more thorough explanation of how I found the 1 and 5 if you're interested - th-cam.com/video/tsr7mq52ypU/w-d-xo.html
Where do you play sudoku and GAPP puzzles
I always played paper sudoku and kept numbers in my head
Nice puzzle and solve. I solved it in 13:43.
For the break-in I found a finned swordfish on 1s in columns 2, 4, and 8, with the fin being the cell r2c8. This eliminates 1 from r1c9 (either the swordfish or the fin), limiting 1s to row 2 in box 3, which places 1 in r1c5, which you also found.
what
@@JDBJake sodoku is crazy man
@@JDBJake my thoughts exactly
Ein interessanter Sword-fish mit einem fehlenden Kandidaten, dafür aber der Finne. Die 1 geht aber nach A4.
What application are you using? I would like to use it as well. I'm unable to shade on the site I play on.
It's the CTC webapp
I not even finished once solving extreme level sudoku
I tried doing extreme level multiple times but couldn't solve it completely
I love ur videos, and for wordle, you should start with audio, fruit, and nymph
I used the Empty Rectangle strategy to eliminate 1 from B5 and thereafter the puzzle collapsed
Start with crane ,shout and dream
I'm curious about your fast eye movement. Maybe that's what you get after many years of practice! Watching this made me love Sudoku. You are a genius!!
5+6+6=17 given clues, putting the puzzle at an advanced level, yet not so ’extreme’ as claimed. Of the clues, 489 are singles, 2367 are doubles, three sets of [15] coupled pairs,on cols 1&6 and on row7.
Before we kick off the preliminary sweep across the whole board, with eyes wide open, we should spot a few immediate takeaways :
7 on c1@(11), (=>6 on c1 B7), 3 in B8@(84) to be followed immediately by 7@(95)=> [68] tcp on r7B8 between cols 5&6, leaving 9@(74)(which can actually be pinpointed by direct point counting and that’s a hidden 9 for u) => in hot pursuit, 4@(44) (two ways of going abt it : by eliminating 3&9 from cell(44) = [349] by hard count OR by visual spotting : [715] given on r4 being off c4 are to occupy the remaining three open cells outside of (44) on c4, leaving 4 @(44). (A lot of novice may miss this but no harm done it will show up eventually)
Now here comes the takeaway for the trained eyes, the three pairs of [15] given on c1,c6 and r4 , all cutting off r5 ===> [15] cp(conjugate pairs) on r5 @ P(53)&Q(55)!,leaving 2 splt on r5 @(51))&(56).
Now scanning for splits across the board:
Being the majority clues we start with 1&5 :
1 split in B2@(14)&(24), this will prove critical, split on r1,r7,c2&c8.
5 split on r1&r7. (That’s all)
Now note! 1&5 share same cells on r1 at R(14) and on r7 @ S(73).
A million dollar question : are the cells exclusive of the other candidates? Answer is a resounding yes!
Here’s why: 1&5 not at R => 1@(19)&5@(17) ==> both on r7 at S, a No-no! So R=[15].
By same reasoning, we have S=[15] => 2 split on r7 @(71)&(79).
Now we have [15] cp on c3 @ P&S=> [15] cp in B4 @ P&T(62) (hopefully it’s obvious to u).
Now the split of 1 in B2 comes to play!
Claim: whether 1 is in B2, it always lead to 1@P!
Obvious if 1@(25).
Next 1 @R => 5 on r1@(17) => 5 on r7@S => 1@P!
There are other ways to prove 1@P : whether 5 is on r1 , it always lead to 1 @P; likewise for 5 on r7; OR proof by contradiction: 5@P would lead to 5@(17)&(77), a No-no! two 5s on same c7 , or => 1&(24) ==> Q= Zero!
However there’s a better and prettier approach (and without the help of 5) :
1 forms a closed loop by
starting from (24) B2=> 1@ P => 1 on r7 @(79) and ending on r1 @ R, a No-no.
Therefore 1 in B2 is at R => 5@(17) and 1 on c9@(79) => 5@ S and 1@P and so forth.
There is a technical term in Sudoku bag of tricks to describe what we have been going through: XY wing.
Now we can with confidence declare the rest is history!
Try to do it on your own effort , but if u get stuck at any step,then take a peek at what follows up.
First we proceed to complete all the 1s and 5s and helping to pick up other candidates along the way:
So 1@P => 1 on c2@(92), 5@QT&S => 1@(79)=>1 on r1@R => 5 on c4@(34), by virtue of 5 given inB8,on r4,and uncovered @T , all three off c4,to be followed by 7 on c4@(64), whereby leaving 4 on c4 @(44) (as promised it will show up if u fail to spot it earlier),and also => 5 on r1 @(17) => together with 2 given on c7,and 2&5 given on r9 , both outside of B9, [25] tcp on r8B9 bet cols 8&9 => 2@(89) which with 1 uncovered @(79), and in tandem with [12] given on c7, both outside of B3 => [12] tcp on c8B3.
Now 1 uncovered @(79) => 2 on r7@(71), by virtue of 2 given @(67) off r7 => in hot pursuit 2 in B7 @(42) by cc with the 2 given on r6 outside of B7 and 2 on r5@56) => 2 on r1@(13) (CIY) => with 6 given on c2 off r1 => 6 on r1@(19), leaving 2&8 cp on r 1@(12)&(16).
Also 2 uncovered @(71) leaves 4 on r7@(77). Note [25] uncovered tcp on r8B9 leads to 7 on r8@(87) which together with 4 uncovered @(77) => [47] tcp on c9B3, leaving [38]tcp on c7B3 bet rows 2&3.
Note 6 uncovered on r1@(19) B3 is off c7 and in tandem with 6 given on r5 also off c7 => 6 on c7@(97), leaving 9 on c7@(57) => 7 on r5@(58), leaving [46] tcp on c8B6 bet rows 4&6 => with 4 uncovered @(44), [16] tcp in B5 on r6 bet cols 5&6, leaving [39] tcp r4B5 ==> together with the [39] uncovered on r5, [39] tcp on r6B7 bet cols 1&3, leaving [48] tcp on c1B4 bet rows 4&5.
Now we return to Bottom Block : 2@(71) => [34] tcp on r9B7 bet cols 1&3 => 6 on c1&(81) leaving [89] tcp on r8B7 bet rows 2&3.
I can see the extremity. However, the 3 spot at 1:09 is easy to see.
"if this is a five then it puts a five there" 5:33
Me, crying as I'm staring at my sudoku puzzle: BUT WHY?! 😭😭
Cause the five could go in the top row as well, so how do you know? I really wanna learn how people know that so I can do it myself.
I just posted a more thorough explanation of how I found the 1 and 5 if you're interested - th-cam.com/video/tsr7mq52ypU/w-d-xo.html
I don’t know if you’ve every tried killer sudoku? My 7yo brother and I really enjoy it
Yes, I've both solved and created quite a few killer sudokus in the past. They are a lot of fun!
Loved that intro Scot!
I couldn't understand how you filled 1 and 5
If you are referring to how he turned the 1 5 possibility into a 15 pair. I believe he did this by looking about 5 moves ahead to deduce this.
I just posted a more thorough explanation of how I found the 1 and 5 if you're interested - th-cam.com/video/tsr7mq52ypU/w-d-xo.html
Great intro! I loved it!
Start with my starting words, TEARS, POUND, CLIMB
What soduko solver app or website is that?
there is link in the description
@@twieg Oh thanks
I find 6:50 difficult to understand. Interesting 13:06 explanation though
I recently made another video going through the explanation a bit more thoroughly - th-cam.com/video/tsr7mq52ypU/w-d-xo.html
I just learned how to play a day ago
I confused again
What? How?
You should try a quordle with this strategy: WAQFS, VOZHD, BRICK, GLENT, JUMPY. You have to guess everything in 1 go but you have 25/26 letters covered, excluding the X.
It makes the game not fun bruh
What software is he using?
sudoku pad, made by sven for ctc
You should start with the 4 words "flame brick podgy and shunt" it removes 22 characters
I solved the same as you, just took longer😂 1 hour☠️ You think so fast mygod. My brain can't handle to move on immediately
I don't understand why a 4 is placed on that spot in the middle square
I explained that briefly at 2:40. Essentially, that cell already sees all the other digits either in its row or column. (23689 are in the column, 157 are in the row)
Hello! Use “ELITE EIGHT SAINT PETER” at the start of your next wordle to celebrate St. Peter’s being the only fifteenth seed in ncaa history to make it to the elite eight!
This puzzle has 5+6+6=17 clues, putting itself on an advanced level,(but not quite as extreme as claimed,to me ‘extreme’ means use of brutal,monster,steamroller kind of force, but this is not so here,as u will see)with three given singletons 489(indeed two immediately obvious hiddens : 9@(74)by point counting 😮 &3 in Box 8@(84) by cc(crisscrossing)of the two 3 given on c5&r7 outside of B8,followed by 7@(95)B8,by virtue of 7 given on r7 outside of B8 (indeed one should able to see right away without hesitation that by virtue of 7&3 given on r7 outside of B8, [37] cp in B8 @(84)&(95))=> [68] tcp on r7B8 between cols 5&6,by virtue of [68] given on c4 outside of B8,thereby leaving 9 in B8@(74) and together with 3@(84)=>(44)=[4]by point counting or by another way of looking at it, 715 given on r4 off c4 have to occupy the other three open cells left on c4 =>4@(44). Btw, we have 3 on r9B7, 7 on r8B9 and 4on c6B2.
And two of given 2367.
Three given1& 5.
There is another immediate visual takeaway, viz[15] cp on r5 @ P(53)&Q(55), by virtue of [15] given in B6 and on cols 1&6, all three off r5!! => 2 split on r7.
Now time for the preliminary sweep across the grid looking for splits in boxes , on rows and columns and hopefully more hiddens:
Right away,with 7 given in Boxes 4&5, both off c1, 7 @(11).
Indeed it’s the distribution of 1&5 that demands our attention :
1 & 5 have splits on r1,(I assume u have a way to pencil mark splits on rows and columns,without using the Arabic numerals which are reserved for splits in Boxes. For this purpose l am using the Chinese numerals: 一二三四五 六七八九),on c2,r7,c8.
1 is split in B2.
Observe that 5&7 occupy the same two cells(14)&(73),but we can’t say at this stage for sure they are exclusive from other digits (however eventually it will be proven the case)
To a trained eye the intertwining of 1&5 deserves our attention.
There are several ways of attack: one is to show (inspired by the fact that 5 forms a fractured wing along r1&r7 with pivot on c7),whether 5 is on r7,it always leads to 5 on r5@Q with the help of 1 (CIY=check it yourself).
OR by contradiction, 5@P would lead to 5 on r7@(77)=>5 on r1@(14)=>1 in B2@(25)by cc of the two 1s outside of B2,one on r3&the other on c6 => Q=ZERO!! Therefore 5 on r5 has to be at Q! and 1@P.
At this stage 1 uncovered at P proves to be the breakthrough,the closure we are looking for.
Try for yourself. It will be fun. No more inductive but purely deductive logic will take u thru.
Here’s for the ‘novice’:
So 1@P=>1 on r7@(79)&on c2@(92)=> 1 on c8@(28)=> on r1@(14)&c5@(65),completing all 1s in the grid.
Now 1@(14)=> 5 on r1@(17)(here’s where 5 kicks in)=>5 on r3@(35) ,5 on r7@(73) and 5 on c8@(88)=>5 on c2@(62),completing all 5s in the grid.
Now 5@(34)=>7 on c4@(64) ,thus leaving 4 on c4@(44)(we however have known that).
Looking at B9, with 1&5 uncovered where they are , 2 in B9@(89) by the cc of the 2s given on r9&c7 outside B9(see how their strategic positions come into play)=>2 on c8@(38) .Now looking at row7, with 1&5 uncovered on r7,we have 2 given at (67) off r7=> 2 on r7@(71),leaving 4 on r7@(77) and 2 on r5@(56),plus 2 in B4@(42)by cc with 2 given at (67)=> 2 on c5@(25) and on r1@(13)[CIY],completing all 2s in the grid.
Note in passing,7 on r8 is at (87)=>with the 4@(77), [47] tcp on c9B3, leaving [38] tcp on c7B3, and 6 on c7@(97),by virtue of 6 given on r5 off c7,thereby leaving 9 on c7@(57) and [98] tcp on r9B9 => with 9 given @(18),[38] tcp on c9B6=> 7 on r5@(58),leaving [64] tcp on c8B6 bet rows 4&6, as anticipated by [46] already uncovered on r5, and on c7&c9,all outside of B6 => with 4@(44), 6 in B5@(66), leaving [39] tcp on r4B5 bet cols5&6 => [39] tcp on r6B4 (hope u can see it how), leaving [48] tcp on c1B4.
Also in hot pursuit, 3 given @(15)=> 9@(45) and with 7 uncovered @(95) =>[79] tcp on c6B2, leaving [28] tcp on c5B2.
Now with 6&9 uncovered in B9 on r9, off c1=> 6 on c1@(81) followed by 9 on c1@(61),leaving 3 on c1@(91) , resulting in [89] tcp on r8B7 bet cols 2&3, and [14] tcp on r9B7 bet the same two cols => [34] tcp on c2B1 and [69] tcp on c3B1 bet rows 2&3 [CIY, a good practice for novice].
Whoa! We are practically done.
Can u visualize how those tcps in the bottom and middle horizontal Blocks are broken down?
Those six chains of tcp between rows 2&3 are to be broken up by the two ’danglars’ 6@(24)&2@(38).
The end result of the puzzle looks pretty graphic,ain’t it? That’s why sudoku is described as a visual art
I just got 6min 15 seconds in extreme sudoku
I use to play actively and easily get uner 11min but now I'm bad at it and it takes me like 16
I have completed the extreme level in only 6 minutes and 37 seconds my highest record ever, without any help.
I have the data but I forgot to record it😢
Thanks sir
Day 6 of me asking you to do TUBES FLING CHAMP WORDY for wordle
You should try squardle
He already did. And not once.
I have played through the middle of this video over and over several times, and kept pausing. I guess my mind can't go that deep because I still wasn't able to see it. I was able to get the 1 in block 2, but that's it. I've come back to this video a couple times and still can't get it, so I've spent well over 3 hours on this damn thing. Time to ignore it and get on with the rest of my life.
I just posted a more thorough explanation of how I found the 1 and 5 if you're interested - th-cam.com/video/tsr7mq52ypU/w-d-xo.html
I’m sorry what
The 29th of March,s worldle is pretty easy ss well it is a well known country
When i use notes i fell like i am cheating.
@@rain4279 It is infact cheating, I think...
very smart
Good Info, I Learned by my Self, I Get Expert 95% of the Time. No Extreme Yet, I Play Different. I click on 9, Look Up, and Down and Across, then 8, 7, and So On, Repeat. When 3 Spaces are missing in a Box or a Row, I Think, What Are the missing Numbers Then see if 2 are in 2 Rows, If So, it Goes in the 3rd Spot, Also Look Not Only Which it Could Be, But What it Can't Be, Works Pretty Good, Trying not to use the Little Numbers in Boxes,
I Noticed The Numbers Seem to Pair Up, Except the're are 9 Numbers?
Good Info.
Like to Do Video of My Method, Not Better, Just Different.
I Feel Going from 9 down to 1, is Better For Me Than 1 and Up to 9, and Repeat.
Liked
He lost me on 689. How did he know a 9 would go in row 7, col 4?
solved it in 44:33 without formulas LOL 🤣
999 IQ dude here
And the 6 to the 9 and oh that gives us 4 but wait oh I see 5 here 3 goes there and errh let look at this area 235 remove the 5 because we got one ryt here 😂😂😂
try an extreme killer sudoku
I've done many killer sudokus in the past (and even created many killer sudokus, some of which might be considered extreme)
But have you tried a completely empty one?
you should play escape room games
My wife and I love doing escape rooms!
@@StroSolves there are some nice games out there, namely escape simulator, i would like to watch you try those
I love sudoku. But I need to learn how to solve extreme level. I came here but you don’t try to teach, right?
Hope you do a video teaching how to solve it. Thanks. 😊
NO WAY YOU USED BOMBER BY RIOT AS YOUR INTRO SONG
I LITERALLY KNOW SOMEONE WITH THAT EXACT SONG AS THEIR INTRO
hi
Great video but at least credit Cracking the Cryptic and their programmer as you use their app while solving???
I've mentioned what site I use before, but I guess I could point it out again. I think I'm just used to everyone knowing that website since I share a discord server with Sven (and a few other people)
Can you explain slowly and more clear please
I just posted a more thorough explanation of how I found the 1 and 5 if you're interested - th-cam.com/video/tsr7mq52ypU/w-d-xo.html
Jesus is the answer. Whoever is reading this you may feel lost in life; you may feel like giving up, wishing it would all end. You want peace, but can’t find it. You’ve looked for it in women, drugs, alcohol, porn, money, and so much more, but you feel emptier the more you do it. The only one who can fill us is Jesus Christ, the living God; the one who came into this world as a man and died on the cross to save us from our sins (our wrongdoings against God). If you trust in Christ as your Lord and Savior you will be saved and you will go to Heaven when you die. One thing many people think once they come to God is everything will be perfect. The truth is it won’t be. Once you give your life to Christ the devil will attack you over and over again until you break, but Christ is there to strengthen you and to get you through all things. Give your life to Christ my friends. Without him, life means nothing.
Also Great video
No
This is relatively simple. Doesn't look like extreme. Maybe medium
Could not get your logic…how a 7 you got in the fifth square… no satisfying reasoning ..sorry
That big box already contains 3 4 1 2 and 5. And 7 cant go in the top row because it has 7 in that row, the only remaining box for 7 in that big box is that small box you mentioned.
6:33 BUT HOW?! Please, I need a slowed down explanation of what's going through your head right now 😭
I just posted a more thorough explanation of how I found the 1 and 5 if you're interested - th-cam.com/video/tsr7mq52ypU/w-d-xo.html
First!
Can you give me a shoutout? Love your videos!
This videos are not helpful. I got a sodoku that was super hard it was missing number 7 and 6. This people don't know
Wordle 282 6/6
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