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new advanced 17 nr Sudoku that can be solved with XY-wing

 
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someone_somewhere



Joined: 07 Aug 2005
Posts: 275
Location: Munich

PostPosted: Fri Aug 26, 2005 4:47 pm    Post subject: new advanced 17 nr Sudoku that can be solved with XY-wing Reply with quote

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 XY-wing after the "forpaly" of finding the first 33 numbers, you will get to the G-point!

have fun,
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David Bryant



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

PostPosted: Sun Aug 28, 2005 12:06 am    Post subject: Reply with quote

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

PostPosted: Sun Aug 28, 2005 6:22 am    Post subject: Reply with quote

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 XY-wing?
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 G-point!

3 can't be in r6c9, because we have XY-Wing 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 XY-wing? 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: 193
Location: Bideford Devon EX39

PostPosted: Tue Sep 06, 2005 1:00 am    Post subject: Reply with quote

The algebraic notation used in explanation can be off-putting and so I
have attempted to distil a simple description of when XY-Wing 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 non-quoted 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

PostPosted: Tue Sep 06, 2005 11:05 am    Post subject: Reply with quote

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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