A classic sudoku with a host of powerful techniques !
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- Опубликовано: 9 сен 2024
- rockratzero's hard classic sudoku is a tough nut to crack. The start is complicated but the video explains all tricks required to unravel it. Enjoy !
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Rules :
Place a digit from 1 to 9 in each empty cell so that each digit appears exactly once in each row, column and 3x3 box.
Hello Everyone ! Welcome to this RUclips channel :
"Unshackling Sudokus & Puzzles "
#puzzles #sudokus #brainteasers # Classic
I am Kishore Kumar Sridharan and I have been part of the team representing India in the World Sudoku Championships 2015-2019,2022. I have also won the Classic Sudoku Master Event held during the Indian Sudoku Championship for a streak of 4 successive times. Solving sudokus and puzzles is my passion.
Online solving Link : tinyurl.com/ytd...
I solved this using standard techniques (though not easy to spot).
First I spotted a skyscraper on 4s in columns 4 and 8 which limited the candidates in R8C5 to 2/3.
The digit in R4C4 is also from 2/3 and the content of column 6 in box 2 says that R8C5 and R4C4 cannot contain the same digits, which gives the value of R4C5 (I think this a W-wing on both 2 and 3).
A great load of deductions followed and I had to spot another skyscraper later (on 3s this time), after which it was a smooth ride to the end.
Very enjoyable on the whole.
Thanks Sudokanard for your feedback !
I spent 12 mins marking Snyder notation and then ALL candidates. Saw the host of 23s or 38s around and knew I needed a chain. I think it was a 7, either box 5 or 6, but I got the 7 ruled out of r4c9. From there I determined R1C8 couldn't be an 8 from the chain in R1&R4. the rest collapsed from basic logic, no tricks required. 17mins+ but definitely fun.
Thank you @Frie_Jemi !
Quite tough puzzle. Solved it in 24:41
At 10:00 my break-in looked at cell r4c9, which at that point can be 236. If it is 6, that puts 4 in r6c9 and 7 in r6c7, while if it is not 6, then we have a 23 pair in row 4, placing 7 into r4c5. Either way, 7 cannot be in r6c5, so it must be in r4c5 in box 5. This gives a 69 pair in r4c12, making r4c9 a 23, and we get a 238 triple in column 9, placing digits to r1c9=4 and r3c9=7 and r6c9=6. This then gives a lot of digits, and I finished using the same skyscraper and coloring of 23 pairs.
Well done !
Very nice puzzle, but only after the initial deduction, which I didn't particularly enjoy. Always a pleasure to watch your videos though!
The truth is that whatever technique we use to solve a Sudoku puzzle that cannot be solved with the sole use of counting and/or crosshatching, we are always bifurcating, because we analyze two different outcomes trying to find a deduction based on logic which would eliminate a candidate or candidates from cell or cells where the two paths intersect. Bifurcation is not to be confused with guessing where you blindly put a digit and hope for a positive outcome. At 9:35 I analyzed both outcomes for 7 in box 6 and came up with a logical deduction that 7 must go in r4c5 because if you put 7 in r5c7 a 58 Naked Pair in r5c12 forms a 23 Naked Pair in box 5
I tried, filled no's in all cells, only find WXYZ wing 2349 in box 8&9, but no improvement. Then i saw the vedio it is bifurcation method. So i don't want to go further.
You took 46 (r6c9) to get 6 in r6c9. The explanation not understand properly, i saw many more times that particular stage, the right no is 7 in r4c5. After saw comments MILAS also mentioned the same point.
If r6c9(46) 4 or 6 we get 7 in r4c5.
* If -4- then r6c7 is 7, 7 eliminated in r6c5 and 7 spotted on r4c5.
* If -6- then 23 pairs in row 4 then 7 spotted in r4c5. We get also 6 in r6c9.
If we also take 47 (r6c7) it's 4 or 7 r4c5 mustbe 7.
If -4- r6c9 is 6, 23 pairs in row 4, then 7 in r4c5.
If -7- r6c5 7 eliminated and spotted 7 in r4c5.
Nice puzzle.
Guys please solve this sudoku named as everest by arto inkala. It is not on your channel. I already checked. This sudoku is different.
It's not possible to solve the sudoku you refer without heavy bifurcations