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VH++ 101323

 
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immpy



Joined: 06 May 2017
Posts: 571

PostPosted: Fri Oct 13, 2023 6:21 pm    Post subject: VH++ 101323 Reply with quote

Hello all, enjoy the puzzle.

Code:

+-------+-------+-------+
| 3 2 . | . . . | . . 4 |
| 6 4 . | . . 8 | 7 5 . |
| . . 7 | . . . | 3 . . |
+-------+-------+-------+
| . . . | 6 . 5 | . . . |
| . . . | 1 2 9 | 5 . 6 |
| . . . | 7 . 4 | . . . |
+-------+-------+-------+
| . . 4 | . . . | 9 . . |
| 7 5 6 | . . 3 | 1 2 8 |
| 9 8 . | . . . | . . 5 |
+-------+-------+-------+

Play this puzzle online at the Daily Sudoku site

cheers...immp
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dongrave



Joined: 06 Mar 2014
Posts: 567

PostPosted: Sat Oct 14, 2023 2:23 pm    Post subject: Reply with quote

Thanks for the puzzle immp!

After basics:
Code:
+-----------------+---------------+-----------------+
| 3    2   c58    | d59  1679 167 | 68  19     4    |
| 6    4   f19    |  3  e19   8   | 7   5      2    |
| 58  g19   7     |  45  46   2   | 3   68     19   |
+-----------------+---------------+-----------------+
| 148  379  2389  |  6   38   5   | 248 134789 1379 |
| 48   37  b38    |  1   2    9   | 5   3478   6    |
| 158  6    23589 |  7   38   4   | 28  1389   139  |
+-----------------+---------------+-----------------+
| 2    3-1  4     |  8   5    167 | 9   367    37   |
| 7    5    6     |  49  49   3   | 1   2      8    |
| 9    8   a13    |  2   167  167 | 46  3467   5    |
+-----------------+---------------+-----------------+

Then I used the chain (1=3)r9c3-(3=8)r5c3-(8=5)r1c3-(5=9)r1c4-(9=1)r2c5-r2c3=r3c2 => r7c2 <> 1; stte.
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Clement



Joined: 24 Apr 2006
Posts: 1110
Location: Dar es Salaam Tanzania

PostPosted: Sun Oct 15, 2023 3:25 pm    Post subject: VH++101323 Reply with quote

Code:

+------------------+-----------------+---------------------+
| 3     2    a58   |a59   1679   167 |a68    19       4    |
| 6     4    c19   | 3   b19     8   | 7     5        2    |
| 58   d19    7    | 45   46     2   | 3     68       19   |
+------------------+-----------------+---------------------+
| 148   379   2389 | 6    38     5   | 248   134789   1379 |
| 48    37    38   | 1    2      9   | 5     3478     6    |
| 158   6     259  | 7    38     4   | 28    1389     139  |
+------------------+-----------------+---------------------+
| 2    e13    4    | 8    5      167 | 9    f367     f37   |
| 7     5     6    | 49   49     3   | 1     2        8    |
| 9     8     13   | 2    167    167 |h4-6  g3467     5    |
+------------------+-----------------+---------------------+

(6=859)r1c734 - (9=1)r2c5 - r2c3 = r3c2 - (1=3)r7c2 - r7c89 = (3-4)r9c8 = (4)r9c7 => -6r9c7; stte
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glesco



Joined: 12 May 2022
Posts: 36

PostPosted: Thu Oct 19, 2023 5:47 pm    Post subject: Reply with quote

I used a colouring technique.

In r1c3, I set the 5 to Green & the 8 to Blue. As 8 is a strong link in r1 so the 8 in r1c7 is Green with the 6 being Blue. In c2 the 5 is also a strong link so in r6c3 the 5 is set to Blue. There are a lot more cells that can be coloured but it turns out not to be necessary to find the solution.

If Green is the solution and looking at the 28 bivalued cell in r6c7, the 2 would also be Green as the 8 sees a Green 8. That 2 is also, however, part of a strong link in r6 which would imply 2 in r6c3 is Blue. But r6c3 already has a Blue 5! So Green can not be the solution and thus solves for all the cells coloured Blue.

Trust this explanation is relatively clear!
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