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VH++ 112923

 
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immpy



Joined: 06 May 2017
Posts: 569

PostPosted: Wed Nov 29, 2023 8:03 pm    Post subject: VH++ 112923 Reply with quote

Hello all, enjoy the puzzle.

Code:

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

Play this puzzle online at the Daily Sudoku site

cheers...immp
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dongrave



Joined: 06 Mar 2014
Posts: 560

PostPosted: Fri Dec 01, 2023 5:40 pm    Post subject: Reply with quote

Thanks for the puzzle immp! I couldn't find a single step solution so I settled for 2 (very convoluted) chains.

After basics:
Code:
+--------------------+------------------+-------------------+
|  k2478 2348    6   | 2489  23479 3789 |  5    1      249  |
|  j2478 9      e2@8 | 12458 2457  178  |  3    6     f24*  |
|   1    234     5   | 249   23469 369  |  8    249    7    |
+--------------------+------------------+-------------------+
|   5    126     4   | 19    8     169  |  7    29     3    |
|   289  128     7   | 1459  3459  139  |  6    24589  2459 |
|   3    68     d89  | 7     4569  2    | b49   4589   1    |
+--------------------+------------------+-------------------+
|   2469 24      3   | 289   279   789  |  1    459    4569 |
| hL2489 7    dhm289 | 6     1     5    | c2*49 3     g489  |
|  i68   5       1   | 3     29    4    | a29*  7      68   |
+--------------------+------------------+-------------------+

Step 1: 9*r9c7-(9=4)r6c7-(49=2*)r8c7-(2=8|9)r68c3-(8=2@)r2c3-(2=4*)r2c9-(49*=8)r8c9-r8c13=r9c1-(2@4*8=7)r2c1-(2@78=4)r1c1-(2*48=9)r8c1-(2*9=8)r8c3 contradiction (i.e. 8r8c39) => r9c7 != 9.

Code:
+---------------+-------------------+------------------+
| 47   238  6   | 2489   2347  3789 |  5   1      249  |
| 47   9   g28  | 12458  2457  178  |  3   6     i24   |
| 1   h23*  5   | 249    2346 k369  |  8  j249    7    |
+---------------+-------------------+------------------+
| 5    126  4   | 19     8    f169  |  7  b2*9    3    |
| 289  128  7   | 1459   345   139  |  6   24589  2459 |
| 3   d68  c89  | 7     e456   2    | a49  4589   1    |
+---------------+-------------------+------------------+
| 69   4    3   | 28     27    78   |  1   59     569  |
| 289  7    289 | 6      1     5    |  49  3      489  |
| 68   5    1   | 3      9     4    |  2   7      68   |
+---------------+-------------------+------------------+

Step 2: 9r6c7-[(9=2*)r4c8 -(9=8)r6c3]-[(8=6)r6c2-r6c5=r4c6 -(8=2)r2c3]-[(2=3*)r3c2 -(2=4)r2c9]-(2*4=9)r3c8-(3*9=6)r3c6 contradiction (i.e. 6r34c6) => r6c7 != 9; stte.
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immpy



Joined: 06 May 2017
Posts: 569

PostPosted: Fri Dec 01, 2023 10:42 pm    Post subject: Reply with quote

Yes, I can understand that dongrave. This one was closer to an "extreme" or "unfair" level. I needed many steps to whittle it down, including chains. Maybe a rating of +++ would have been more accurate.

cheers...immp
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glesco



Joined: 12 May 2022
Posts: 35

PostPosted: Thu Dec 07, 2023 10:47 pm    Post subject: Reply with quote

I used a candidate colouring method that keys off bivalued cells though the solution is similar to dongraves. The steps I took:

- The 9s in c37 eliminate 9 in r8c9 in a finned x-wing thus creating a 48 bivalued cell in r8c9.
- Then I starting colouring candidates in Green or Blue ("G" or "B") if there is a strong link. Picked r9c1 to start making 6B & 8G. That makes in r7c1, 6G as it is a strong link in box 7.
- In r9c9 is 6G & 8B. In r8c9 the 8 is G as it is a strong in box 9 and thus the 4 becomes B.
- Now I looked for bivalued cells that see a colour with the same #, like the 24 in r2c9 that sees a 4B in r8c9. If B is the solution the 2 would be B.
- That 2B sees the 28 cell in r3c3 thus that 8 would B.
- We can eliminate 8 from r8c3 it sees a 8G in r8c9 as well as the 8B from above. This highlights an XY-wing
- That elimination creates a 29 bivalued cell in r8c3.
- Continuing on down c3 with the what if B is the solution scenario, in r6c3 the 9 becomes a B so in r8c3, the 2 becomes a B.
- But the 6 in r7c1 is G so r7c1 can not be 2!
- Continuing in box 7 with the same B is the solution scenario, in r8c3 the 2 is B so the 4 in r7c2 is B.
- But the 6 in r7c1 is G so r7c1 can not be 4!

Hope this is somewhat clear.

Eliminating 2&4 from r7c1 seems to be the key in this puzzle.

I used different approach from a lot of replies in this forum, let me know what you think!
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immpy



Joined: 06 May 2017
Posts: 569

PostPosted: Sat Dec 09, 2023 5:19 pm    Post subject: Reply with quote

Nice going glesco, that works.

cheers...immp
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