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someone_somewhere
Joined: 07 Aug 2005 Posts: 275 Location: Munich

Posted: Fri Aug 26, 2005 4:47 pm Post subject: new advanced 17 nr Sudoku that can be solved with XYwing 


Hi,
For all of you who think that they can do better than Katie. Here an other one:
400000001
080400000
000000000
000700850
001900000
200000000
030006080
600010400
000020000
It can be solved with LOGIC only. It is 17 Nr one. Very, very high candidate score of 319 and if you use XYwing after the "forpaly" of finding the first 33 numbers, you will get to the Gpoint!
have fun, 

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David Bryant
Joined: 29 Jul 2005 Posts: 559 Location: Denver, Colorado

Posted: Sun Aug 28, 2005 12:06 am Post subject: 


All right, someone  I think I can do this one.
After the first 33 moves, all of which are fairly straightforward, I arrive at this grid.
Code: 
4 6 ?5? 2 7/9 8 ?5? *** 1
1 8 *** 4 *** 3/5 2 *** ***
3/5 2 7/9 1 *** 3/5 *** 4 8
9 4 6 7 3 1 8 5 2
3/8 5/7 1 9 5/8 2 *** *** 4
2 5/7 3/8 6 5/8 4 9 1 3/7
7 3 2 5 4 6 1 8 9
6 9 5/8 3/8 1 7 4 2 3/5
5/8 1 4 3/8 2 9 ?5? *** ***

Here I have indicated all the pairs that are as yet unresolved with the "/"  triples or quads are marked ***, except for a few (r1c3, r1c7, and r9c7) that will be important in the ensuing argument.
The next logical move works like this. A "5" can only occupy one of two possible cells in column 7. Similarly, there are only two possibilities for "5" in row 1, and two possibilities for "5" in row 8. Suppose that r9c7 = "5". Then we have the following chain of inference:
r9c7=5 ==> r1c3=5 ==> r8c3=8 ==> r8c9=5
But this can't be right, because it forces two "5"s into the same 3x3 box, in the lower right corner of the grid. So the "5" in column 7 must appear at r1c7.
The next move is also pretty tricky. Start with the {3, 8} pair at r6c3. Since "3" can only appear in two places in row 6, we have the following chain of inference (following across row 6, then down):
r6c3=8 ==> r6c9=3 ==> r8c9=5
But there's another chain to be traced out, because the value "8" can only appear in two places in column 3. Following this chain we see that
r6c3=8 ==> r8c3=5 ==> r8c9=3
But this is impossible  we can't put two different values in the same cell! We conclude that r6c3 is not an "8", and therefore r6c3 = 3.
After these two moves the grid looks like this:
Code: 
4 6 *** 2 7/9 8 5 *** 1
1 8 *** 4 *** 3/5 2 *** ***
3/5 2 7/9 1 *** 3/5 *** 4 8
9 4 6 7 3 1 8 5 2
3/8 5/7 1 9 5/8 2 *** *** 4
2 5/7 3 6 5/8 4 9 1 3/7
7 3 2 5 4 6 1 8 9
6 9 5/8 3/8 1 7 4 2 3/5
5/8 1 4 3/8 2 9 *** *** ***

and it's pretty easy to fill in the rest of the puzzle from there. dcb 

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someone_somewhere
Joined: 07 Aug 2005 Posts: 275 Location: Munich

Posted: Sun Aug 28, 2005 6:22 am Post subject: 


Hi David,
I like the way you think.
It is always easy to find the "complicate solution".
Now let us start with what is the XYwing?
If we have:
XY XZ
YZ *
It can be easily seen that whichever value is in XY, the cell marked with the asterisk cannot be Z.
if XY = X, then XZ = Z, so * cannot be Z
if XY = Y, then YZ = Z, so * cannot be Z
This allows Z to be eliminated from the candidates for the marked cell.
Doesn't it look simple and easy to detect ?
Now lets take a look at the position you have got.
It is the Gpoint!
3 can't be in r6c9, because we have XYWing numbers
X=5 Y=8 in r8c3
X=5 Z=3 in r8c9
Y=8 Z=3 in r6c3
And this is enough. The rest of the numbers are comming by themself before the whisky gets warm and the coffee cold.
It was a pleasure to meet u. This one you killed before Katie.
Would you like an other XYwing? Of course with 17 cells.
see u,
P.S. how long did all the pleasure last for you? 

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alanr555
Joined: 01 Aug 2005 Posts: 195 Location: Bideford Devon EX39

Posted: Tue Sep 06, 2005 1:00 am Post subject: 


The algebraic notation used in explanation can be offputting and so I
have attempted to distil a simple description of when XYWing applies.
Consider a rectangle of four cells (however small or large) using the
four corner cells. If three of the corners from a triplet of values with
two occurrences of each value, the fourth corner CANNOT contain the
digit value that does not occur in its opposite corner.
Example:
18; 16
68; 68
There is a triplet {18,16,68} in three corners (left and top) so that the
bottom right cannot contain 6 (the value not in 18 opposite).
ALSO
There is a triplet {18,16,68} in three corners (RIGHT and top) so that the
bottom LEFT cannot contain 8 (the value not in 16 opposite).
This resolves to
18; 16
6,8
and further progress will depend on nonquoted cells but should be
very much easier.
+++
This rule above applies only when three of the four corners have been
reduced to pairs. Does the rule have wider scope and apply when more
than two digits are in any of the three corners? It would seem that the
number of digits in the fourth (target) corner is not material.
NB: Other readers of this forum should check for themselves that the
suggested rule works before including it in their repertoire! You have been
warned, and only practice will confirm usefulness.
Alan Rayner BS23 2QT 

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geoff h
Joined: 07 Aug 2005 Posts: 58 Location: Sydney

Posted: Tue Sep 06, 2005 11:05 am Post subject: 


Sorry, I only just got around to doing this puzzle. yes, I enjoyed it very much!
Once I found the XY Wing for the 3,5,8, it was easy to put the Nr7 in r6c9
and go on from there.
Thanks for a great puzzle!! 

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