View previous topic :: View next topic |
Author |
Message |
daj95376
Joined: 23 Aug 2008 Posts: 3854
|
Posted: Fri Apr 08, 2011 6:11 am Post subject: Puzzle 11/04/08: ~ Difficult++ |
|
|
Code: | +-----------------------+
| . . . | . . . | . 2 . |
| . . 8 | . . 3 | . 7 . |
| . 6 1 | 2 . . | . 8 3 |
|-------+-------+-------|
| . . 3 | 4 . 5 | . . 7 |
| . . . | . . 7 | . 5 . |
| . 7 . | 6 2 . | . 3 9 |
|-------+-------+-------|
| . . . | . . . | 3 . . |
| 5 3 6 | . 9 2 | . . . |
| . . 7 | 3 . 6 | . . 8 |
+-----------------------+
|
Play this puzzle online at the Daily Sudoku site |
|
Back to top |
|
|
JC Van Hay
Joined: 13 Jun 2010 Posts: 494 Location: Charleroi, Belgium
|
Posted: Fri Apr 08, 2011 10:31 am Post subject: |
|
|
Quote: | The bivalues in C3, C7\NP(19), R3, B6, B9 form a huge boolean variable. By colouring these bivalues, it is observed that for one of the colour the DP(18)R57C12 is obtained. The other colour solves the puzzle. |
|
|
Back to top |
|
|
peterj
Joined: 26 Mar 2010 Posts: 974 Location: London, UK
|
Posted: Fri Apr 08, 2011 10:46 am Post subject: |
|
|
Two steps - could be combined into an 'almost'..
Quote: | AIC ; (9=4)r1c3 - r5c3=r5c9 - r6c7=r3c7 - (4=5)r3c5 - (5=1)r2c4 - (1=9)r2c7 ; r1c7<>9
w-wing(29) (2=9)r4c2 - r1c2=r1c3 - (9=2)r7c3 ; r5c3<>2, r79c2<>2 |
|
|
Back to top |
|
|
JC Van Hay
Joined: 13 Jun 2010 Posts: 494 Location: Charleroi, Belgium
|
Posted: Fri Apr 08, 2011 10:48 am Post subject: |
|
|
Eureka notation (or "solution" for some)
Quote: | 6r5c1=2r7c123(because of UR subset)-2r7c9=2r5c9-(2=68)r45c7 => -6r5c7 |
|
|
Back to top |
|
|
peterj
Joined: 26 Mar 2010 Posts: 974 Location: London, UK
|
Posted: Fri Apr 08, 2011 11:38 am Post subject: |
|
|
JC Van Hay wrote: | Eureka notation (or "solution" for some) |
JC, nice UR move. Just out of interest why the als at the end?
I read your move from the description as (in a longer hand)...
Code: | ur(18)r57c12[(6)r5c1=qnp(29)r7c123] - (2)r7c9=r5c9 - (2=6)r4c7 ; r4c1<>6 |
|
|
Back to top |
|
|
Marty R.
Joined: 12 Feb 2006 Posts: 5770 Location: Rochester, NY, USA
|
Posted: Sun Apr 10, 2011 12:33 am Post subject: |
|
|
XY-Wing (249); r12c2<>9
X-Wing (5); r127c5<>5
UR (18), boxes 47. R5c1=6 or r7c12 forms pseudo cell 29; common outcome, r5c7=8 |
|
Back to top |
|
|
|