View previous topic :: View next topic |
Author |
Message |
immpy
Joined: 06 May 2017 Posts: 570
|
Posted: Wed Nov 29, 2023 8:03 pm Post subject: VH++ 112923 |
|
|
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 |
|
Back to top |
|
![](templates/subSilver/images/spacer.gif) |
dongrave
Joined: 06 Mar 2014 Posts: 564
|
Posted: Fri Dec 01, 2023 5:40 pm Post subject: |
|
|
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. |
|
Back to top |
|
![](templates/subSilver/images/spacer.gif) |
immpy
Joined: 06 May 2017 Posts: 570
|
Posted: Fri Dec 01, 2023 10:42 pm Post subject: |
|
|
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 |
|
Back to top |
|
![](templates/subSilver/images/spacer.gif) |
glesco
Joined: 12 May 2022 Posts: 36
|
Posted: Thu Dec 07, 2023 10:47 pm Post subject: |
|
|
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! |
|
Back to top |
|
![](templates/subSilver/images/spacer.gif) |
immpy
Joined: 06 May 2017 Posts: 570
|
Posted: Sat Dec 09, 2023 5:19 pm Post subject: |
|
|
Nice going glesco, that works.
cheers...immp |
|
Back to top |
|
![](templates/subSilver/images/spacer.gif) |
|