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Guest Guest

Posted: Mon Sep 19, 2005 10:13 pm Post subject: Times Sunday 8 ....Next move please ...and why 


Have got this far and just cannot work out what to do next. I would appreciate an explanation (in simple language) from a helpful person please.
5 . 2367. 67. 139. 13689. 1268. 134789. 1348. 1278.
4. 23. 1. 39. 7. 28. 389. 5. 6.
9. 2367. 8. 4. 136. 5. 137. 13. 127.
2. 1. 67. 57. 456. 3 . 478. 9. 578.
38. 78. 5. 2. 14. 9. 6. 134. 17.
36. 4. 9. 8. 156. 167. 137. 2. 157
18. 89. 2. 6. 189. 4. 5. 7. 3.
7. 59. 3. 159. 2. 18. 18. 6. 4.
168. 568. 4. 1357. 1358. 178. 2. 18. 9[/img][/list] 

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chobans
Joined: 21 Aug 2005 Posts: 39

Posted: Tue Sep 20, 2005 2:35 am Post subject: 


You need XWing for this.
By looking at the grid, 8 has a XWing at r2c6,r2c7,r8c6 and r8c7. So you can eliminate 8 at r1c6, r1c7, r4c7 and r9c6.
See if you can solve it after those eliminations. 

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Guest Guest

Posted: Tue Sep 20, 2005 7:22 am Post subject: 


Thankyou. Would you mind explaining X Wing as I do not understand it. 

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chobans
Joined: 21 Aug 2005 Posts: 39


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alanr555
Joined: 01 Aug 2005 Posts: 193 Location: Bideford Devon EX39

Posted: Tue Sep 20, 2005 10:35 pm Post subject: 


I do not see that XYwing is necessary to solve this one.
The key area is in column 2
(2367,23,2367;1,378,4;89,59,568)
The first three cells in the column include an embedded triplet
{236,23,236} so that the digit 7 may be removed from rows 1
and 3 in this column  leaving row 5 as the only possibility for
the location of the digit 7.
Having set r5c2 to '7' the rest of the puzzle resolves quite easily.
I do not know if this is a valid method or whether I have just made
a lucky guess. Perhaps someone else can elaborate the rule?
Where THREE cells within a row, column or box contain THREE digits
forming a triplet PLUS other digits, can any conclusion be drawn?
A suggestion has been made theat XYwing applies. OK, if it does, we
need to know which cells are involved. I looked for one as a pattern
of the cells containing just two values but did not find one. The search
was for three cells (each containing only a pair of values) which form
a rightangled pattern with only three unique values within them
(eg 12,23,13). It would then be possible to learn something about the
cell in the opposite corner of the rectangle formed by the three cells  but
in this case I found no such pattern.
Two questions:
Is my elimination of 7 from r1c2 and r3c2 (described above) correct?
What cells are involved in the XYwing?
Alan Rayner BS23 2QT 

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chobans
Joined: 21 Aug 2005 Posts: 39

Posted: Wed Sep 21, 2005 4:15 am Post subject: 


alanr555 wrote:  Perhaps someone else can elaborate the rule?
Two questions:
Is my elimination of 7 from r1c2 and r3c2 (described above) correct?
What cells are involved in the XYwing? 
Well, 6 is a possibility at r9c2 so you can't say that it forms a naked triple at those 3 cells.
8 forms the XYwing at r2c6, r2c7, r8c6 and r8c7. The rows of 2 and 8 only has 2 possibilities for number 8, and they happen to fall on same columns on both rows. Thus XYwing. 

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Guest Guest

Posted: Wed Sep 21, 2005 8:24 am Post subject: 


Thankyou for your efforts! I have in fact solved this puzzle but am not satsified with my method. I took one pair 8,3 (R5,C1) ...selected one of them and followed through from there. With the 3 it did not work, with the 8 it all came out fine. Not a very satisfying method, there surely must be a more logical one.
"The first three cells in the column include an embedded triplet
{236,23,236} so that the digit 7 may be removed from rows 1
and 3 in this column  leaving row 5 as the only possibility for
the location of the digit 7."
I understood that to be a triplet there must be three digits ONLY in three cells.
"A suggestion has been made that XYwing applies. OK, if it does, we
need to know which cells are involved. I looked for one as a pattern
of the cells containing just two values but did not find one. The search
was for three cells (each containing only a pair of values) which form
a rightangled pattern with only three unique values within them
(eg 12,23,13). It would then be possible to learn something about the
cell in the opposite corner of the rectangle formed by the three cells"
I have found the cells referred to by Chobans: 2,8 (R2,C6); 1,8 (R8,C7); 1,8 (R8,C and 3,8,9 (R2,C7) but still cannot understand what this tells me. Please explain!!
Thanks 

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Guest Guest

Posted: Wed Sep 21, 2005 8:27 am Post subject: 


Oops! The smiley face above is supposed to be C6! 

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chobans
Joined: 21 Aug 2005 Posts: 39

Posted: Wed Sep 21, 2005 3:46 pm Post subject: 


Guest wrote:  I have found the cells referred to by Chobans: 2,8 ( R2,C6 ); 1,8 ( R8,C7 ); 1,8 ( R8,C8 ) and 3,8,9 ( R2,C7 ) but still cannot understand what this tells me. Please explain!! 
Actually it's r8c6 and r8c7, not r8c7 and r8c8.
Anyway, if you put 8 in r2c6 then you know it's not a 8 at r2c7/r8c6. So r8c7 MUST be an 8.
But if you put 8 at r2c7 then you know it's not a 8 at r2c6/r8c7. So r8c6 MUST be an 8.
They basically form a "X" where if one corner is certain number then the other corner becomes certain number. But you know this number will only fall in these rows and columns. So you can eliminate this number from other cells that falls out of these criteria.
But it only can show up twice in two unique same direction row or column. That's the rule.
The link I posted above has a guy explaining it with examples. You probably can understand it better by reading that. 

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

Posted: Wed Sep 21, 2005 4:28 pm Post subject: It's XWing, _not_ XYWing 


chobans wrote: 
8 forms the XYwing at r2c6, r2c7, r8c6 and r8c7. The rows of 2 and 8 only has 2 possibilities for number 8, and they happen to fall on same columns on both rows. Thus XYwing. 
Some confusion arose here. You identified this formation correctly as "XWing" in the first place, Chobans. It got confused when AlanR called it an "XYWing".
Here's another explanation for our guest. In the matrix you exhibited, there are only two places an "8" can appear in row 2  it must go in r2c6 or in r2c7. Similarly, there are only two places an "8" can appear in row 8  it must go in r8c6 or in r8c7.
Since these two rowwise unique possibilities appear in the same two columns (columns 6 and 7), we can be certain that the only way an "8" can fit in column 6 is either at r2c6, or at r8c6. Similarly, the only way an "8" can fit in column 7 is either at r2c7, or at r8c7. Therefore we can eliminate the possibility of an "8" at r1c6, r1c7, r4c7, and r9c6.
This leaves r4c9 as the ONLY cell in which "8" can appear in row 4, which in turn says that "5" must appear at r6c9. The rest of the puzzle should be simple enough from there. dcb 

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Guest Guest

Posted: Wed Sep 21, 2005 11:17 pm Post subject: 


Thankyou both David and Chobans. Got it!! 

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