View previous topic :: View next topic |
Author |
Message |
daj95376
Joined: 23 Aug 2008 Posts: 3854
|
Posted: Wed May 06, 2009 6:16 pm Post subject: Set XY_02 Puzzle 105 |
|
|
Code: | +-----------------------+
| 4 2 . | 1 . . | . . . |
| 5 . . | 2 4 9 | . 8 1 |
| . . 9 | . 5 . | . . . |
|-------+-------+-------|
| 8 7 . | . 9 . | . 1 3 |
| . 3 4 | 8 2 . | . . 5 |
| . 5 . | . . 1 | 8 . 4 |
|-------+-------+-------|
| . . . | . . 3 | 4 . . |
| . 8 . | 4 . . | . . . |
| . 4 . | 9 7 2 | . . . |
+-----------------------+
|
Play this puzzle online at the Daily Sudoku site |
|
Back to top |
|
|
storm_norm
Joined: 18 Oct 2007 Posts: 1741
|
Posted: Wed May 06, 2009 9:34 pm Post subject: |
|
|
Danny,
I am going to try and take a new approach to this puzzle.
the chain that results is very very long.
in this puzzle, if you remove the 9 from r1c8, you are left with {3,7}.
that {3,7} cell completes this w-wing
(3=7)A - (7)B = (7)C - (7=3)D which would eliminate the 3 in r1c5.
can this partial chain be made?
w-wing{3,7} = (9)A...
Code: | .---------------------.---------------------.---------------------.
| 4 2 8 | 1 -36 67 | 5 A379 E79 |
| 5 6 37 | 2 4 9 | 37 8 1 |
| 37 1 9 |D37 5 8 |H26 4 G26 |
:---------------------+---------------------+---------------------:
| 8 7 J26 | 5 9 4 |I26 1 3 |
| 1 3 4 | 8 2 67 | 79 679 5 |
| 9 5 K26 |C37 L36 1 | 8 B267 4 |
:---------------------+---------------------+---------------------:
| 27 9 1 | 6 8 3 | 4 5 F27 |
| 2367 8 37 | 4 1 5 | 379 23679 2679 |
| 36 4 5 | 9 7 2 | 1 36 8 |
'---------------------'---------------------'---------------------' |
what that partial chain means is that if the 9 is true, then the w-wing is false and vice versa.
with that in mind, this huge chain can be formed.
{w-wing37[(3=7)A - (7)B = (7)C - (7=3)D]} = (9)A - (9=7)E - (7=2)F - (2)G = (2)H - (2)I = (2)J - (2=6)K - (6=3)L
this huge chain means that either the w-wing is true or the 3 in L is true.
either way the 3 in r1c5 can't exist.
solves the puzzle.
[daj edit] |
|
Back to top |
|
|
daj95376
Joined: 23 Aug 2008 Posts: 3854
|
Posted: Thu May 07, 2009 12:07 am Post subject: |
|
|
Norm,
Your chain may be perfectly valid, but I'm not the one who would know for sure. I've always been uncomfortable with embedded sub-patterns, even though I've seen some that've knocked my socks off once I convinced myself that they worked.
I agree with your elimination, but that's because I see it as two chains with the strong link on <3> in [r1] resulting in contradictions when [r1c8]=7|9.
Specifically:
[r1c8]=7 should force [r1c5]=3, but instead forces [r3c4]=3 by following the W-Wing cells. This results in a contradiction by adding (3)r3c4-(3)r1c5=(3)r1c8.
[r1c8]=9 should force [r1c5]=3, but your =(9)A chain instead forces [r6c5]=3. This results in a contradiction by adding (3)r6c5-(3)r1c5=(3)r1c8.
This then leaves [r1c8]=3 to crack the puzzle.
BTW: As individual chains, [r1c8]<>7|9 can be had with shorter chains. |
|
Back to top |
|
|
tlanglet
Joined: 17 Oct 2007 Posts: 2468 Location: Northern California Foothills
|
Posted: Thu May 07, 2009 1:47 pm Post subject: |
|
|
Three steps: Quote: | x-wing on 3,
coloring on 7,
w-wing 26 with pincers in r3c9 & r8c1 and strong link on 2 in row7 deletes 6 in r8c9. Then transport pincer r8c1 to r9c8 via r9c1 and pincer r3c9 to r4c7 via r3c7 to delete 6 in r56c8 and complete the puzzle. |
Ted |
|
Back to top |
|
|
Marty R.
Joined: 12 Feb 2006 Posts: 5770 Location: Rochester, NY, USA
|
Posted: Sat May 09, 2009 5:23 pm Post subject: |
|
|
I used simple coloring on 7. Then in the rightmost stack there was an XY-Wing Chain, 27-26-26-26-67. It was flightless but a pincer transport removed the 7 from r5c6 and finished the puzzle. |
|
Back to top |
|
|
|
|
You cannot post new topics in this forum You cannot reply to topics in this forum You cannot edit your posts in this forum You cannot delete your posts in this forum You cannot vote in polls in this forum
|
Powered by phpBB © 2001, 2005 phpBB Group
|