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		| ZeroAssoluto 
 
 
 Joined: 05 Feb 2017
 Posts: 1019
 Location: Rimini, Italy
 
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				|  Posted: Fri Aug 18, 2017 10:14 pm    Post subject: Aug 19 VH |   |  
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				| Hi everyone, 
 
  	  | Code: |  	  | +-------------+-----------+--------+
 | 259 59  125 | 7  19  4  | 8  3 6 |
 | 4   6   8   | 2  3   5  | 7  9 1 |
 | 39  79  137 | 69 169 8  | 4  2 5 |
 +-------------+-----------+--------+
 | 6   579 57  | 3  2   79 | 1  4 8 |
 | 29  8   27  | 4  79  1  | 6  5 3 |
 | 1   3   4   | 5  8   6  | 9  7 2 |
 +-------------+-----------+--------+
 | 8   4   35  | 69 569 29 | 23 1 7 |
 | 35  1   9   | 8  57  27 | 23 6 4 |
 | 7   2   6   | 1  4   3  | 5  8 9 |
 +-------------+-----------+--------+
 
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 Play this puzzle online at the Daily Sudoku site
 
 some options
 
 
  	  | Quote: |  	  | XYZ-Wing 2,5,9 in r5c1, r1c12 and -9 in r3c1 or
 finned X-Wing with number 9 in r15c15 +r1c2 and -9 in r3c1
 or
 Empty Rectangle with number 9 in sector 2 r3c5, r5c5, r5c1 and again -9 in r3c1
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 Ciao Gianni
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		| Ajò Dimonios 
 
 
 Joined: 01 May 2017
 Posts: 339
 Location: Sassari Italy
 
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				|  Posted: Sat Aug 19, 2017 10:11 am    Post subject: |   |  
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				| Hi everyone 
 Other solutions
 
 [(5)R1C1-(5)R1C2=(5)R4C2-(5=7)R4C3 and (5-2)R1C1=(2)R5C1-(2=7)R5C3]=> contradiction two singles 7 in box 4 or column 3=> -(5)R1C1=>stte.
 
 [(5)R7C3-(5)R4C3=(5)R4C2-(5=9)R1C2 and (5-3)R7C3=(3)R3C3-(3=9)R3C1]=> contradiction two singles 9 in box 1=>-5 R7C3=>stte.
 
 Another solution -3R8C1=>stte.
 
 R8C1=3 is biunivoca with R7C3=5.
 
 Ciao a Tutti
 Paolo
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		| Earl 
 
 
 Joined: 30 May 2007
 Posts: 677
 Location: Victoria, KS
 
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				|  Posted: Sat Aug 19, 2017 11:22 am    Post subject: Aug 19 VH |   |  
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				| The old reliable, less sophisticated xywing! 
 359 xywing @ R3C1, -5R1C1
 
 Early Earl
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		| Ajò Dimonios 
 
 
 Joined: 01 May 2017
 Posts: 339
 Location: Sassari Italy
 
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				|  Posted: Sat Aug 19, 2017 12:10 pm    Post subject: |   |  
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				| Hi Earl 
 
 Of course it is true.
 
 However, I believe that the AIC for the xy chain (xywing)
 (5 = 9) R1C2- (9 = 3) R3C1- (3 = 5) R8C1- (5) R1C1 => -5R1C1 has a similar difficulty of the demonstration that I given with two small chains. It's always a contradiction.
 
 Ciao a Tutti
 
 Paolo
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