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ZeroAssoluto
Joined: 05 Feb 2017
Posts: 870
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 |
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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