The New Sudoku Players' Forum

[Serenity's Embrace]

Can't figure out what technique to use next. Where do I go from here?

[JasonLion]

I see a Hidden Unique Rectangle and a Sashimi Swordfish right off, though neither of them looks like they will get you very far. You will almost certainly need chains to make real progress.


2.658....83...46.5.5.36....59..23168.238.65496.89...7...2.38.56.65......38.6..7..

[Pat]

Serenity's Embrace, you have somehow excluded 9 at r3c3,
this is wrong and you will be unable to solve the puzzle

[nedBlake]

The puzzle does have a single solution, but it's a tough one to find. I couldn't see anyway to begin to solve it, and so exported it to www.sudokuwiki.org. If you step through the solution at that site, you will see it solved by applying about a dozen strategies. The first step I have drawn out below:

Extreme50 puzzle.png (34.1 KiB) Viewed 1221 times

Assuming that 4 is the solution at row 7 col 2, a chain of implications shows that it cannot be a solution since it would imply that the 4 in the cell above would also be one. Therefore 4 at (7,2) can be removed:

+4(7,2) -7(7,2) +7(1,2) -7(2,3 & 3,3) +7(4,3) +7(4,3) -4(4,3) +4(6,2)

What makes this chain work is that although there are three 7's in column 3, there is a strong link between two of them and the third one below the two. This is true because when there are n possibles in a unit, and n-1 of them are the same color, then the one with the single color must be the solution.

Most of the other strategies used in this puzzle are digit, or cell, or unit forcing strategies. I am still looking at them.

[pjb]

Agree with Jason. After the sashimi swordfish elimination of 7 from r7c4 and the 1 from r9c5 due to the hidden UR, it only requires simple AICs. (I will list them if you want).
Phil

[Kristoforus]

HI Lets continue what started Nedblake by his chain. I help you make 2 steps forward. Starting cell is G1here are 4 candindates lets check 7 If 7G1 is true then: 7E1 must be false then 7E5 must be true then 7D4 must be false then 4 D4 must be true then 4G4 must be false and beacouse ned proved that 4G2 is false now false 4G4 mean that in cells G2 and G4 we have same candidates 1and 7 so naked par . This mean 7 G1 must be false we have contradiction 7 out from G1. Next chain will be harder but profitable . Start from C2 here are 2 candidates one is true second not. Lets try that 7 is false. So if -7G2->7H2->-7B3->{7B4 and 7B5 locked candidates}->{-7A6 and -7C6}->7H6->-7H1-> and......we have not any cell for 7 values in that square if we remove 7 from G2 chain iimplies that must be. Its not end yorurs problems here so good luck;)