📕Get my FREE Solving Guide that will help you solve over 80% of all Sudoku puzzles🧩to include NYT Hard👉👉www.buymeacoffee.com/timberlakeB/e/125822 Timestamps 0:00 Intro 00:18 It’s Solving Time 00:29 Puzzle Story 00:56 Finding Snyder Restrictions 02:56 BONUS Tip #1 06:45 How To Find Extreme Strategy 07:45 Extreme Strategy Revealed 11:17 Next Puzzle Steps 13:49 BONUS Tip #2
"Mark solved this with bifurcation, but we're going to try something different..." Proceeds to solve through bifurcation and incorrectly refers to a broken puzzle as a deadly pattern.
Took just a minute and half shy of an hour, but I finally got there. Not too sure what the difference between bifurcation and a logic chain is, but I'm going to say that I used a logic chain to solve without resorting to using deadly pattern (that I 100% noticed in the first place... absolutely) once I reached your 7:00 min mark (my like, 20-30min mark lol) I focused on box 2 and box 8. If you make the middle cell in box 8 (r8c5) a 6, it forces a 6 in r1c6 which forces a 9 in r1c3 (note that this 9 would make a 68 pair in box 4 r4c3 & r5c3), The 9 in r1c3 forces a 9 in r3c8 causing r1c8 to be a 45. The 45 in r1c8 would then pair up with the 45 in r8c8. This pair would then result in 68 pair in c8 of box 6 (r4c8 & r5c8). Then you look back over at the 68 pair you made in box 4 - whoa it's an x-wing (that also counts as a deadly pattern now that I think about it)! But... what's that? The x-wing on the 68 pairs you just made in r4 and r5 would force box 5 to have two 3's in column 4. And, I may not be the best at sudoku, but even I know that is not allowed. Tripple, quadruple, quintuple check the logic and you can confidently conclude that the middle cell in box 8 (r8c5) can never be a 6. Which means it must be an 8 :) After that murderous process, it solves pretty quickly.
Nice deduction. Thank you for sharing. Mark Goodliffe would say that a logic chain is not bifurcation if you can see the whole chain to make your deduction, whereas bifurcation is just picking 1 of 2 choices and seeing what happens. How much have you watched Cracking The Cryptic?
10:27 I was always taught that using uniqueness to solve a puzzle is a no no. It doesn't prove that a puzzle has a unique solution, just makes it possible to solve it one way.
That is a great point. What I did here did not rely on the puzzle being unique. It started with a premise of what could be the value of 2 of the cells in the puzzle. Without a 5 in one of those cells, there is no way for the 6's to solve themselves. This is a similar approach used when solving Sudoku Oddagons. Hope that helps.
Super insane sudoku. My time was around 13min, but I only could solve after watch the part "extreme strategy revealed". I didn't remember this one at all! Thanks for the tips! 🎉
You are welcome. Great job getting this in 13 minutes, even with a little help. I’m featuring popular CTC Classic videos I haven’t before. I have to go back to 2020 and 2019 to get puzzles I haven’t covered. Do you have any recommendations?
@@SmartHobbies I haven't sorry. After so many years following them, I did and watch so many puzzles (plus Sleuth and Bremster), that I can solve sudokus from last year as a brand new one 😅
7:00 This took me forever. My bottom band and center-cell 4 were identical to the video's position here. I had a lot more corner marks, including the green 37 restricted to block 6 in column 7. I followed my usual practice of centermarking the blocks, and that may be what got me lost in the middle of nowhere. I really don't want to describe what I went through to solve this puzzle.
Tough puzzle, tried and stuck at 7:00, no strategy was helped, saw vedio , very nice soft and easy solution and explanation. I tried after the same little change... r3c8 -9 is eliminated, because, If r3c8 is 9, then 45,45,45,245 UR in Col 7&8 row 1&8. It gives 2,4,5 in box 9 middle. r8c8 4 make r1c8 to 5, and it gives 68 pairs in box 6. Now X wing 8s in Col 3 & 8, gives 36 pairs in box 5, and it makes r8c5 is 6. Now X wing 6s in Col 4 & 8., eliminate 6s in box 4., and it makes r1c3 is 6. Now r1c3 is 6 & r8c5 is 6s, it makes the same problem no place for 6 in box 2. Because 9 eliminated from r3c8 and spotted 9 in r1c8, puzzle solved. Nice Tough puzzle thanks.
Great question. After you place the 9 in R7C4, you can look at R7C3 and see what digit can go there. 3 is the only digit possible because of all other digits see that cell either in the row, column, or block. Does that help?
@@SmartHobbies not really, I'm referring to R1C5, you put 3 there right after putting 2 to R1C6 (around min 13:10) - no idea why only 3 is left as an option there
Thank you John. Another way to look at it is that the 89 from row 1 could only go in row 2 in Block 2, leaving 2 spots left for 2 and 3, and there is a 3 already in column 6
As usual, there is no way in the world that I could have solved this without virtual chaining. I solved it half a dozen times, but always with at least one chain. Here is perhaps my simplest way through, using a two-way chain (75 = row 7 column 5; second write-in): 99, 75, 76, 78, 98, 91, 74, 82, 92, 73, 55 (all easy up to here, but then…), 59 (forced, irrespective of the order of the 6/8 pair in box 8)*, 69, 89, 87, 88, 18 (restricted 5s in box 6, irrespective of the order of the 6/8 pair in box 8), 13 (naked single), 35, 84, 85, 26, 25, 56, 31, 43, 53, 44, 54, 15, 16, 95, 96, 66, 41, 58, 23, 42, 71, 72, 47, 57, 45, 65, 52, 48, 11, 37, 38, 67, 68, 27, 32, 21, 22, 61, 62, 37. * R8C5 = 6 → R1C6 = 6 → R3C1 = 6 → R6C8 = 6 → R6C7 = 1 → R5C9 = 2. R8C5 = 8 → R2C6 = 8 → R6C7/C8 = 1/8 pair → R5C9 = 2. Some of my other solutions included uniqueness arguments, but Mark would certainly disapprove of those. Afterthought: Well, my reasoning at the crunch point was very different, but the result was much the same. I'm quite happy with that, but your 'virtual' UR (Type 4?) was far more elegant. Interestingly, the HoDoKu rating you give is rather lower than it might be. My version reports Extreme 4492 or 4872, depending on whether uniqueness arguments are allowed. Edited to add the afterthought.
Right from the start, I noticed that 1,5,7 couldn't be placed on the first 4 cells of row 7. That observation, typical for The New York Times Hard Sudoku made it very easy to sove the last 3 blocks to the point where you then have to use advanced strategies.
I cld do only the bott horizontal chute with those pairs and a triple left , after that a BIG NO NO... But that is very difficult , u magically pick up the 68 pair in col 8 of box 6 and start ur investigation in its vicinity to find an error . I guess the SIA gave u some clues to start ur investigation in that area of the grid. BTW what is SIA? Sudoku intelligence agency. I'll again watch the oddagon strategy, if I find time. Thanks for the extreme version !
Thank you for asking. It is like an Oddagon. I explain the concept in depth on this tutorial: ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-Vf-HMsB00AI.htmlsi=IBXXHG-i-QtPGNc-
Hi, Thank you for the post! I have a question. When there are 8s in col 3, 4 and 8 on rows 4 and 5, do you still have an X wing? When I saw that, I just assumed the 8 in col 8 were no longer valid. Was I correct? Thank you!
Great question Kenneth. A true X-Wing covers 4 different blocks. When you have 8s in rows 4 and 5, keeping the 4 cells in 2 blocks, then that is commonly referred to as a "mini X-wing" which is useful for finding Claiming Pairs but is not a true X-Wing. Does that help?
@@SmartHobbies Thank you for your reply. I think I understand what an X wing is, but in this case, there are 6 eights in a rectangle. Are you saying that since there are two additional 8s, it’s no longer an X wing? Thank you!
(5,5)=4 cuat(3,6,8,9) en 4a col ent (7,4)=9 (8,2)=9 (9,1)=8; par (6,8) en 8a fila dentro del bloque 8 y tercia(2,4,5) dentro del bloque 9 ent (9,2)=6; par (2,4) en 7a fila dentro del bloque 7 ent (7,3)=3; cuat (6,7,8,9) en 3a col; cuat(1,2,5,9 en 9a col ent (9,9)=1 (9,8)=3 (7,8)=7; par(1,5) en 7a fila dentro del bloque 8 ent (7,5)=1 (7,6)=5; par(2,7) en 9a fila dentro del bloque 8; hasta aquí queda atorado el Sudoku, trato de desbloquearlo haciendo un Tanteo por Combinaciones entre las casillas (5,9)=(2,5,9) con (6,9)=(5,9). Primero la combinación (6,9)=5 con (5,9)=2 se llega a un Error! en las casilla (8,9)=(2,5) no habría lugar para cualquier número, quedaría vacia; luego la combinación (6,9)=5 con (5,9)=5 obviamente es un Error! se está duplicando el #5 en 9a col y dentro del bloque 6 y finalmente la combinación (6,9)=5 con (5,9)=9 no se ve tan obvio pero desarrollando un poco se llega a un Error! se duplica el #7 en 6a fila, por lo tanto en la casilla (6,9)=9...
