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		| immpy 
 
 
 Joined: 06 May 2017
 Posts: 574
 
 
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				|  Posted: Wed Dec 08, 2021 7:44 pm    Post subject: VH+ 120821 |   |  
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				| Hello all, enjoy the puzzle. 
 
  	  | Code: |  	  | +-------+-------+-------+
 | 8 . 2 | 7 5 . | 4 6 . |
 | . . 1 | . . . | . 7 . |
 | 6 5 . | . . . | . 2 . |
 +-------+-------+-------+
 | 3 . . | . . 4 | 8 1 . |
 | 5 . . | . . 8 | . . . |
 | . . . | 3 1 . | . 4 9 |
 +-------+-------+-------+
 | 9 . . | 6 . . | . . . |
 | 7 2 6 | . . . | . 5 . |
 | . . . | . . 2 | . . 6 |
 +-------+-------+-------+
 
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 Play this puzzle online at the Daily Sudoku site
 
 cheers...immp
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		| dongrave 
 
 
 Joined: 06 Mar 2014
 Posts: 572
 
 
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				|  Posted: Fri Dec 10, 2021 6:27 pm    Post subject: |   |  
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				| After basics: 
  	  | Code: |  	  | +--------------+---------------+-------------+ | 8   39  2    | 7    5    139 | 4   6   13  |
 | 4   39  1    | 289  289  6   | 59  7   358 |
 | 6   5   7    | 1489 489  139 | 19  2   138 |
 +--------------+---------------+-------------+
 | 3   67  9    | 25   267  4   | 8   1   257 |
 | 5   1   4    | 29   2679 8   | 26  3   27  |
 | 2   67  8    | 3    1    57  | 56  4   9   |
 +--------------+---------------+-------------+
 | 9   4   35   | 6    37   57  | 12  8   12  |
 | 7   2   6    | 189  89   19  | 3   5   4   |
 | 1   8   35   | 45   34   2   | 7   9   6   |
 +--------------+---------------+-------------+
 | 
 Then I used the following chain: 2*r4c4-(2=9)r5c4-(29=8*)r2c4-(89=1)r8c4-(1=9)r8c6-r1c6=r1c2-(9=3)r2c2-(38*=5)r2c9-(2*5=7)r4c9-(7=6)r4c2-(67=2)r4c5 contradiction => r4c4 <> 2; stte.
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		| TomC 
 
 
 Joined: 30 Oct 2020
 Posts: 359
 Location: Wales
 
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				|  Posted: Sun Dec 12, 2021 8:58 am    Post subject: |   |  
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				| I had to break this down bit by bit focusing on box 2 
 First, an almost XY wing of 259 pivot r5c7 gives pincers on 9 in r2c7 and r5c4 to remove the 9 in r2c4
 
 Second, (is it a XYZ?) wing of 289 using the <89> in box 8 removes the 8 from r3c5
 
 Third, now that r3c5= <49> using an XY chain if r3c5=9 then 1-2-6-5-7-5-4 gives r9c4 = 4 and no <4> in box 2
 
 So, r3c5 = 4
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		| Clement 
 
 
 Joined: 24 Apr 2006
 Posts: 1113
 Location: Dar es Salaam Tanzania
 
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				|  Posted: Sun Dec 12, 2021 5:22 pm    Post subject: VH  120821 |   |  
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				|  	  | Code: |  	  | +-----------------+------------------------+-------------------+
 | 8     39     2  | 7        5         139 | 4      6      13  |
 | 4     39     1  |b28*9     289       6   | 59     7     c358 |
 | 6     5      7  |c1^489    489      e1^39|f19     2     d1^38|
 +-----------------+------------------------+-------------------+
 | 3     67     9  |b25        267      4   | 8      1     a257 |
 | 5     1      4  |b29        2679     8   | 26     3     a27  |
 | 2     67     8  | 3         1        57  | 56     4      9   |
 +-----------------+------------------------+-------------------+
 | 9     4      35 | 6         37       57  | 2-1    8     a12  |
 | 7     2      6  |b1*89      89       19  | 3      5      4   |
 | 1     8      35 | 45        34       2   | 7      9      6   |
 +-----------------+------------------------+-------------------+
 
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 (1=275)r754c9 – (5=298*1*)r4528c4 – (1^)r3c4,(8)r2c9 = (8-1^3)r3c9 = (3-1^)r3c6 = (1)r3c7 => - 1r7c7; stte
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		| Mogulmeister 
 
 
 Joined: 03 May 2007
 Posts: 1151
 
 
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				|  Posted: Mon Dec 13, 2021 1:59 pm    Post subject: |   |  
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				| Nice solutions all! Very elegant Clement and thanks Immp! 
 
   
 I have an overlapping ALS
 
 Set A = {2,5,6,7} {yellow + pink square}
 
 Set B = {5,6,7} {green + pink square}
 
 Restricted Common is 5
 
 Common Candidate is 6 {red} which is eliminated; stte
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