| View previous topic :: View next topic |
| Author |
Message |
keith
Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA
|
 Posted: Fri Mar 07, 2008 10:48 am Post subject: LA Times / Freep Fri Mar 7 |
 |
|
Interesting, this one:
| Code: | Puzzle: FP030708
+-------+-------+-------+
| . . . | 4 . . | . 9 6 |
| . . 1 | 9 . 8 | . . . |
| . . . | . 1 . | 7 8 . |
+-------+-------+-------+
| . . . | . 5 . | . 4 . |
| 6 2 . | . . . | . 3 9 |
| . 8 . | . 9 . | . . . |
+-------+-------+-------+
| . 1 5 | . 2 . | . . . |
| . . . | 5 . 3 | 2 . . |
| 2 4 . | . . 7 | . . . |
+-------+-------+-------+ |
Keith |
|
| Back to top |
|
 |
Johan
Joined: 25 Jun 2007
Posts: 206
Location: Bornem Belgium
|
| Posted: Fri Mar 07, 2008 11:47 am Post subject: |
|
|
| UR [37] in R12C15 opens up an xyz-wing, that solves the puzzle. |
|
| Back to top |
|
|
Earl
Joined: 30 May 2007
Posts: 677
Location: Victoria, KS
|
| Posted: Fri Mar 07, 2008 2:12 pm Post subject: La |
|
|
How does the 37 UR (my grid has two unknowns) work?
A three-step XY-chain eliminates the 7 in R2C1 and opens the puzzle.
Earl |
|
| Back to top |
|
|
ravel
Joined: 21 Apr 2006
Posts: 536
|
| Posted: Fri Mar 07, 2008 2:23 pm Post subject: |
|
|
There are at least 2 w-wings, which lead to the same xyz-wing. This also can be replaced by an xy-wing.
| Earl wrote: | | How does the 37 UR (my grid has two unknowns) work? |
Strong link for 3 in row 1, for 7 in row 2:
r1c1=3 => r1c5=7 => r2c5=3 => r2c1=7
makes a deadly pattern |
|
| Back to top |
|
|
Steve R
Joined: 24 Oct 2005
Posts: 289
Location: Birmingham, England
|
| Posted: Fri Mar 07, 2008 4:17 pm Post subject: |
|
|
Here is the (37) rectangle in r12c15:
| Code: | +------------------------------------------+
| 37+8 357 2378 | 4 37+ 25 | 1 9 6 |
| 37+4 6 1 | 9 37+ 8 | 34 2 5 |
| 349 359 239 | 26 1 256 | 7 8 34 |
--------------------------------------------
| 1 379 379 | 236 5 26 | 8 4 27 |
| 6 2 4 | 7 8 1 | 5 3 9 |
| 5 8 37 | 23 9 4 | 6 17 127 |
--------------------------------------------
| 37 1 5 | 8 2 9 | 34 6 347 |
| 789 79 6 | 5 4 3 | 2 17 178 |
| 2 4 38 | 1 6 7 | 9 5 38 |
+------------------------------------------+ |
The two lower cells of the rectangle are conjugate with respect to 7 so, if r2c1 contains 4, r2c5 contains 7 and r1c5 contains 3. This would eliminate 3 from r1c1. However, if r2c1 does not contain 4, r1c1 must contain 8 to avoid the deadly pattern and again it cannot contain 3.
At the same time there is a (34) bent six:
| Code: | +------------------------------------------+
| 378 357 2378 | 4 37 25 | 1 9 6 |
| 34+7 6 1 | 9 37 8 | 34+ 2 5 |
| 34+9 359 239 | 26 1 256 | 7 8 34+ |
--------------------------------------------
| 1 379 379 | 236 5 26 | 8 4 27 |
| 6 2 4 | 7 8 1 | 5 3 9 |
| 5 8 37 | 23 9 4 | 6 17 127 |
--------------------------------------------
| 37 1 5 | 8 2 9 | 34+ 6 34+7 |
| 789 79 6 | 5 4 3 | 2 17 178 |
| 2 4 38 | 1 6 7 | 9 5 38 |
+------------------------------------------+ |
Using the almost locked sets r7c1 and {r1c1, r7c1, r8c1} eliminates 3 from r2c1 and r3c1 as well.
These two patterns alone solve the puzzle.
Steve |
|
| Back to top |
|
|
keith
Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA
|
| Posted: Fri Mar 07, 2008 9:04 pm Post subject: |
|
|
What about the Type 6 UR that takes out <6> in R3C6 and R4C4?
Keith |
|
| Back to top |
|
|
storm_norm
Joined: 18 Oct 2007
Posts: 1741
|
| Posted: Fri Mar 07, 2008 9:10 pm Post subject: |
|
|
xy-chain starting in r7c1 37-34-34-37 removes 7 in r2c1
xy-wing {378} pivot in r7c1 removes 8 in r8c1
-------
just as a side note
another UR exists in boxes 6 and 9 on {1,7} type 4 i believe. |
|
| Back to top |
|
|
Asellus
Joined: 05 Jun 2007
Posts: 865
Location: Sonoma County, CA, USA
|
| Posted: Sat Mar 08, 2008 12:28 am Post subject: |
|
|
Lots of Deadly Pattern action here. Interesting indeed.
Regarding Steve's nice bent six, the r178c1 ALS isn't necessary due to the strongly linked <4>s in r23c1. (In fact, I'm not sure this two ALS approach would work without those strongly linked <4>s. I suspect that only the <3> at r3c1 could be eliminated in that case without resorting to additional ALSs in b9 and r8. However that may be...) Those <4>s and the {37} bivalue at r7c1 are all that are required to eliminate the <3>s in r23c1. Pretty nifty! |
|
| Back to top |
|
|
Steve R
Joined: 24 Oct 2005
Posts: 289
Location: Birmingham, England
|
| Posted: Sat Mar 08, 2008 1:26 am Post subject: |
|
|
Yes, the argument using the conjugates with respect to 4 is a much neater way of using the bent six … but I missed it.
When I wrote before I was pretty sure the als argument did the job as well. Now, a bottle of wine later, I am less certain. For what it’s worth my thinking was:
(1) The bent six must contain 7 or 9.
(2) 7 can be placed only in r2c1 orr7c9. Either puts 3 in r7c1 and eliminates it from r23c1.
(3) If there is no 7, r3c1 contains 9 (not 3). But then r1378c1 becomes a locked (3789) set so 3 may be eliminated from r2c1 as well.
Perhaps it is the wine talking!
Steve |
|
| Back to top |
|
|
ravel
Joined: 21 Apr 2006
Posts: 536
|
| Posted: Sat Mar 08, 2008 1:44 am Post subject: |
|
|
Nice catch, Steve R, i understood it this way:
r2c1=7 or r7c9=7 => r7c1=3 => r123c3<>3
r3c1=9 => r178c1=378 => r23c1<>3 |
|
| Back to top |
|
|
|
FAQ
Search
Memberlist
Usergroups
Register
Profile
Log in to check your private messages
Log in
