dailysudoku.com

[ZeroAssoluto]

Hi everyone,

[Ajò Dimonios]

If R1C8=2-->delete 2 in R3C7 and 3 inR3C5-->two 1 in line 3 (contradiction)-->delete 2 in R1C8-->solution basic strategies.

[hughwill]

Gianni wrote:


so many solutions this time

Quote:
XY-Wing 2,7,1 r2c6, r3c47;
XY-Wing 1,7,3 r2c16, r3c5;
XY-Wing 1,2,3 r1c8, r3c57;
XY-Chain r1c4, r1c5, r3c5, r3c7 and -2 in r1c8,r3c4;
UR 3,4 r39c12.


I don't think the first XY wing solves it but 2 and 3 do. I haven't the attention span to work out whether the UR works!

Hugh

[Ajò Dimonios]

Gianni wrote:

[the_lock_man]

[Ajò Dimonios]

XY-Chain R2C6, R3C4,R3C7-->delete 1 in R3C5-->solution with basic strategies

[ZeroAssoluto]

Hi Hugh,
I try to answer you but keep in mind that I don’t speak English very well.

With the first XY-Wing you can delete number 1 from r3c5, r2c89.
This allows you to solve box 2 and 3 and then the rest of the scheme.
It seems to me that you can get to the solution; If you seem stuck at a certain point we can always look at it.

As for the UR with numbers 3,4 in cells r39c12 there are three possible arguments that can be made.

At the bottom of everything there is the idea that you can’t have 4 pairs equal to the tops of a rectangle (in two box) to prevent the schema from having more than one solution.
There is not much argument for row 9; is sure that there will be number 3 in a cell and the number 4 in the other.
Then we need to think about r3c12 cells.
I will not repeat cell references continually, I always talk about these.

You can not have 3 AND 4 here at the same time.

The first argument that can be made is that by observing box 1 you can see that number 4 is only in one of the two cells in question; then it’s presence in one of the two cells will be compulsory; this prevents the simultaneous presence of number 3, which can be eliminated from both cells.
Now in row 3, number 3 remains only in column 5, and we are in a situation similar to the first XY-Wing.

The second argument we can do is that since there can’t be 3 AND 4 simultaneously, the alternatives are r3c1 = 7 OR r3c2 = 2; these two numbers are the same present in r3c4; as a result, you create a kind of “naked pair” that allows you to delete all the other numbers 2 and 7, possibly in line 3; in this case r3c7 -2 = 1, and you get to the schema solution.

The third argument is an extension of the second: the alternatives r3c1 = 7 OR r3c2 = 2 connected to the cells r3c4 (2,7) and r3c7 (1,2) show you that you can delete number 1 from r3c5; r3c5 -1 = 3, and you get to the solution.

Alla similar considerations that lead to the same solution.
I hope to have been clear, otherwise we can discuss again.
Ciao Gianni

[hughwill]

[Marty R.]

[czerwiec]

OMG, that's too hard for me!