View previous topic :: View next topic |
Author |
Message |
daj95376
Joined: 23 Aug 2008 Posts: 3854
|
Posted: Sat Jul 09, 2011 9:55 pm Post subject: Puzzle 11/07/09: ~ XY |
|
|
Code: | +-----------------------+
| . 7 . | 9 . . | . . . |
| 9 6 . | 5 4 . | . 1 . |
| . . 5 | 7 . 2 | 9 . 6 |
|-------+-------+-------|
| 7 3 4 | 8 5 9 | . 6 . |
| . 1 . | 3 . 4 | . . . |
| . . 6 | 2 7 . | 3 . 4 |
|-------+-------+-------|
| . . 3 | . . 7 | 6 . . |
| . 5 . | 4 . . | . 7 1 |
| . . 7 | . . 5 | . 3 9 |
+-----------------------+
|
Play this puzzle online at the Daily Sudoku site |
|
Back to top |
|
|
peterj
Joined: 26 Mar 2010 Posts: 974 Location: London, UK
|
Posted: Sun Jul 10, 2011 8:28 am Post subject: |
|
|
Without going "extreme" this one didn't give in easy (for me)!
Quote: | w-wing(28) (2=8)r2c3 - r2c6=r8c6 - (8=2)r8c7 ; r2c7<>2
remote-pair(48) r7c1, r1c1, r3c2, r3c8 ; r7c8<>48
x-wing(8) r3, r6 ; r79c2<>8, r15c8<>8
xy-wing(28-5) r8c7 ; r5c8<>5 |
Danny, also let me echo Ted's comments - thanks for the puzzles! |
|
Back to top |
|
|
tlanglet
Joined: 17 Oct 2007 Posts: 2468 Location: Northern California Foothills
|
Posted: Sun Jul 10, 2011 2:03 pm Post subject: |
|
|
Like Peter, I also tried more standard moves but could not resist the anp() third step.
#1: Flightless xy-wing with pseudocell pincer plus transport; r1c1<>2
(2)r1c8=(2)r12c7-r8c7=r8c5-(2=8)r9c5-r8c6=r2c6-(8=2)r2c3;
#2: RP (48)r7c1=>r3c8; r7c8<>48
Code: | *-----------------------------------------------------------*
| 48 7 1 | 9 38 6 | 2458 2458 358 |
| 9 6 2 | 5 4 38 | 78 1 378 |
| 3 48 5 | 7 1 2 | 9 48 6 |
|-------------------+-------------------+-------------------|
| 7 3 4 | 8 5 9 | 1 6 2 |
| 2 1 89 | 3 6 4 | 578 589 578 |
| 5 89 6 | 2 7 1 | 3 89 4 |
|-------------------+-------------------+-------------------|
| 48 2489 3 | 1 289 7 | 6 25 58 |
| 6 5 89 | 4 2389 38 | 28 7 1 |
| 1 248 7 | 6 28 5 | 248 3 9 |
*-----------------------------------------------------------* |
#3: Two options:
either
anp(89=5)r65c8*-(5=2)r7c8-(2=8)r8c7-r8c3=(8)r5c3 => r5c79<>58=7; invalid result thus r5c8<>5
(As always, any suggestions on a better/proper description of this step is appreciated.)
or
anp(5=78)r5c79-(8)r5c3=r8c3-(8=2)r8c7-(2=5)r7c8; r5c8<>5
Ted |
|
Back to top |
|
|
daj95376
Joined: 23 Aug 2008 Posts: 3854
|
Posted: Sun Jul 10, 2011 4:02 pm Post subject: |
|
|
tlanglet wrote: | Code: | *-----------------------------------------------------------*
| 48 7 1 | 9 38 6 | 2458 2458 358 |
| 9 6 2 | 5 4 38 | 78 1 378 |
| 3 48 5 | 7 1 2 | 9 48 6 |
|-------------------+-------------------+-------------------|
| 7 3 4 | 8 5 9 | 1 6 2 |
| 2 1 89 | 3 6 4 | 578 589 578 |
| 5 89 6 | 2 7 1 | 3 89 4 |
|-------------------+-------------------+-------------------|
| 48 2489 3 | 1 289 7 | 6 25 58 |
| 6 5 89 | 4 2389 38 | 28 7 1 |
| 1 248 7 | 6 28 5 | 248 3 9 |
*-----------------------------------------------------------* |
anp(89=5)r65c8*-(5=2)r7c8-(2=8)r8c7-r8c3=(8)r5c3 => r5c79<>58=7; invalid result thus r5c8<>5
(As always, any suggestions on a better/proper description of this step is appreciated.)
|
Ted, I'd use:
Code: | (5=2)r7c8-(2=8)r8c7-r8c3=r5c3-r5c789=(8-9)r6c8=(9)r5c8 => r5c8<>5
|
If there had been a <5> in r6c8 as well, then I'd use =(8-9)r6c8* and deduce r56c8<>5.
Regards, Danny |
|
Back to top |
|
|
ronk
Joined: 07 May 2006 Posts: 398
|
Posted: Sun Jul 10, 2011 4:51 pm Post subject: |
|
|
tlanglet wrote: | #3: Two options:
either
anp(89=5)r65c8*-(5=2)r7c8-(2=8)r8c7-r8c3=(8)r5c3 => r5c79<>58=7; invalid result thus r5c8<>5
(As always, any suggestions on a better/proper description of this step is appreciated.)
or
anp(5=78)r5c79-(8)r5c3=r8c3-(8=2)r8c7-(2=5)r7c8; r5c8<>5 |
Or do both with an ALS (anp) on each end of one chain.
(57=8)r5c79 - (8)r5c3 = (8)r8c3 - (8=2)r8c7 - (2=5)r7c8 - (5=89)r56c8 ==> r5c79<>8, r56c8<>5 potentially |
|
Back to top |
|
|
tlanglet
Joined: 17 Oct 2007 Posts: 2468 Location: Northern California Foothills
|
Posted: Sun Jul 10, 2011 7:23 pm Post subject: |
|
|
Danny, I played around with various possibilities including the first portion of your suggested chain. I simply missed the obvious two final steps, marked in red.
(5=2)r7c8-(2=8)r8c7-r8c3=r5c3-r5c789=(8-9)r6c8=(9)r5c8 => r5c8<>5
Ron, I like your suggestion also. Once I have found a path, I often fail to seek another variation, such as your use of (57=8) rather than my (5=78). Hopefully I will get better in a few more years......
Thanks to both of you,
Ted |
|
Back to top |
|
|
susume
Joined: 13 May 2011 Posts: 36 Location: Southeastern US
|
Posted: Tue Jul 12, 2011 10:20 pm Post subject: |
|
|
A different ANP approach:
(4)r3c8=r3c2-r1c1=(4)r7c1 => r7c8<>4, r9c7=4
W wing (2=8)r2c3-r2c6=r8c6-(8=2)r8c7 => r3c7<>2
anp (38=5)r1c59-(5=8)r7c9-(8=2)r8c7-r12c7=(2-4)r1c8=(4)r1c1 => r1c18<>8 |
|
Back to top |
|
|
Marty R.
Joined: 12 Feb 2006 Posts: 5770 Location: Rochester, NY, USA
|
Posted: Wed Jul 13, 2011 4:42 am Post subject: |
|
|
W-Wing (28), SL8 in c6; r2c7<>2
Kite; r1c7<>4
X-Wing; r1c8, r5c8, r7c28, r9c2<>8
BUG+2 |
|
Back to top |
|
|
|