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keith
Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA
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 Posted: Fri Nov 07, 2008 12:25 pm Post subject: LA Times / Free Press - November 7, 2008 |
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| Code: | Puzzle: FP110708
+-------+-------+-------+
| 1 . 8 | . . . | . . . |
| . . . | . 3 . | 9 . . |
| . . . | 1 . 5 | . . 2 |
+-------+-------+-------+
| . . 4 | . 9 . | . 1 6 |
| . 8 . | . . . | . 3 . |
| 9 7 . | . 8 . | 5 . . |
+-------+-------+-------+
| 8 . . | 4 . 3 | . . . |
| 5 . 9 | . 1 . | . . . |
| . . . | . . 9 | 1 . 5 |
+-------+-------+-------+ |
I have not yet looked at it.
Keith |
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nataraj
Joined: 03 Aug 2007
Posts: 1048
Location: near Vienna, Austria
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| Posted: Fri Nov 07, 2008 3:17 pm Post subject: |
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I used:
UR (37) type 4 => r9c2=3
Then, generalized m-wing (2)r2c4=r5c5 (via SL (4) in col 6) => r45c4<>2
x-wing (cols 3,5) and other coloring in (2)
Quite a good puzzle |
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cgordon
Joined: 04 May 2007
Posts: 769
Location: ontario, canada
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| Posted: Fri Nov 07, 2008 4:00 pm Post subject: |
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| Quote: | | UR (37) type 4 => r9c2=3 |
Can't see it. I found an x-wing on 7 and an xy wing for 246. Then it siezed up. In R3C13 I have 347, 37 and in R9C13 - 2347 and 237. Maybe I didn't get far enough with basics? |
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keith
Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA
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| Posted: Fri Nov 07, 2008 4:35 pm Post subject: |
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XY-wing, X-wing, skyscraper. Interestingly, they all eliminate <2>.
After basics:
| Code: | +----------------+----------------+----------------+
| 1 26 8 | 9 7 246 | 46 5 3 |
| 24 2456 25 | 26 3 8 | 9 7 1 |
| 37 9 37 | 1 46 5 | 46 8 2 |
+----------------+----------------+----------------+
| 23 235 4 | 2357 9 27 | 8 1 6 |
| 6 8 125 | 25 24 124 | 7 3 9 |
| 9 7 13 | 36 8 16 | 5 2 4 |
+----------------+----------------+----------------+
| 8 1 6 | 4 5 3 | 2 9 7 |
| 5 24 9 | 267 1 267 | 3 46 8 |
| 2347 234 237 | 8 26 9 | 1 46 5 |
+----------------+----------------+----------------+
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Keith |
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nataraj
Joined: 03 Aug 2007
Posts: 1048
Location: near Vienna, Austria
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| Posted: Fri Nov 07, 2008 7:52 pm Post subject: |
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| cgordon wrote: | | Quote: | | UR (37) type 4 => r9c2=3 |
Can't see it. |
In the grid posted by Keith,
the possible DP is in r39c13. Since one of r9c13 must be 7, none of them can be 3 otherwise DP. |
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cgordon
Joined: 04 May 2007
Posts: 769
Location: ontario, canada
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| Posted: Fri Nov 07, 2008 7:54 pm Post subject: |
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Pls put me out of my misery. I can't see the wood for the trees - and this is only as far as I can get with basics
| Code: |
+-------------+------------+--------+
| 1 2456 8 | 9 2467 2467| 46 75 3|
|247 2456 257| 267 3 8| 9 75 1|
|347 9 37| 1 46 5| 46 8 2|
+-------------+------------+--------+
|23 235 4|2357 9 27| 8 1 6|
| 6 8 125|25 24 124| 7 3 9|
| 9 7 13|36 8 16| 5 2 4|
+-------------+------------+--------+
| 8 1 6| 4 5 3| 2 9 7|
| 5 24 9| 267 1 267| 3 46 8|
| 2347 234 237| 8 26 9| 1 46 5|
+-------------+------------+--------+
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storm_norm
Joined: 18 Oct 2007
Posts: 1741
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| Posted: Fri Nov 07, 2008 7:59 pm Post subject: |
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after basics
| Code: | .------------------.------------------.------------------.
| 1 #26 8 | 9 7 246 |4-6 5 3 |
| 24 2456 25 | 26 3 8 | 9 7 1 |
| 37 9 37 | 1 *46 5 |*46 8 2 |
:------------------+------------------+------------------:
| 23 235 4 | 2357 9 27 | 8 1 6 |
| 6 8 125 | 25 24 124 | 7 3 9 |
| 9 7 13 | 36 8 16 | 5 2 4 |
:------------------+------------------+------------------:
| 8 1 6 | 4 5 3 | 2 9 7 |
| 5 #24 9 | 267 1 267 | 3 46 8 |
|#2347 #234 #237 | 8 #26 9 | 1 46 5 |
'------------------'------------------'------------------' |
(6=2)r1c2-(2)r8c2=(2)r9c123-(2=6)r9c5-(6)r3c5=(6)r3c7; r1c7 <> 6
grouped w-wing plus extension
w-wing on {2,6} the group is in row 9. extension through the 6's
edited for some errors
Last edited by storm_norm on Sat Nov 08, 2008 2:34 am; edited 4 times in total |
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storm_norm
Joined: 18 Oct 2007
Posts: 1741
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| Posted: Fri Nov 07, 2008 8:03 pm Post subject: |
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| Craig, where can the 7 go in col 5? |
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cgordon
Joined: 04 May 2007
Posts: 769
Location: ontario, canada
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| Posted: Fri Nov 07, 2008 8:53 pm Post subject: |
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Norm - actually that didn't help me none cos I made a typo. I actually have an extra 7 in Col 5 at R3
| Code: |
+-------------+------------+--------+
| 1 2456 8 | 9 2467 2467| 46 75 3|
|247 2456 257| 267 3 8| 9 75 1|
|347 9 37| 1 467 5| 46 8 2|
+-------------+------------+--------+
|23 235 4|2357 9 27| 8 1 6|
| 6 8 125|25 24 124| 7 3 9|
| 9 7 13|36 8 16| 5 2 4|
+-------------+------------+--------+
| 8 1 6| 4 5 3| 2 9 7|
| 5 24 9| 267 1 267| 3 46 8|
| 2347 234 237| 8 26 9| 1 46 5|
+-------------+------------+--------+
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keith
Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA
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| Posted: Fri Nov 07, 2008 9:12 pm Post subject: |
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| cgordon wrote: | Norm - actually that didn't help me none cos I made a typo. I actually have an extra 7 in Col 5 at R3
| Code: |
+-------------+------------+--------+
| 1 2456 8 | 9 2467 2467| 46 75 3|
|247 2456 257| 267 3 8| 9 75 1|
|347 9 37| 1 467 5| 46 8 2|
+-------------+------------+--------+
|23 235 4|2357 9 27| 8 1 6|
| 6 8 125|25 24 124| 7 3 9|
| 9 7 13|36 8 16| 5 2 4|
+-------------+------------+--------+
| 8 1 6| 4 5 3| 2 9 7|
| 5 24 9| 267 1 267| 3 46 8|
| 2347 234 237| 8 26 9| 1 46 5|
+-------------+------------+--------+
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Look at <4> in R2 and <7> in C5. Then you should find two pairs in R3, and solve R1C5.
Keith |
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nataraj
Joined: 03 Aug 2007
Posts: 1048
Location: near Vienna, Austria
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| Posted: Fri Nov 07, 2008 9:12 pm Post subject: |
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Craig,
4 in box 1 must be in row 2
7 in box 2 must be in col 5
That should get you going again... |
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cgordon
Joined: 04 May 2007
Posts: 769
Location: ontario, canada
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| Posted: Fri Nov 07, 2008 9:54 pm Post subject: |
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| That is disgraceful. You'd think by now I'd have learnt how to use a rubber. (That's what us Brits call an eraser) |
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keith
Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA
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| Posted: Fri Nov 07, 2008 10:05 pm Post subject: |
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Craig,
I suggest you get a copy of Sudoku Susser. It is great for hints and tips.
I do all my puzzles by hand, with pencil and paper. I use SS to check for errors and to make the grids that I post online.
In the last grid you posted, its hint is: | Quote: | | * Hint: A Row / Block Intersection is your green-light to progress. |
If you need more, it will tell you: | Quote: | | * Hint: A Row / Block Intersection is your green-light to progress. Look at squares with possibility <4>. |
If that's not enough, you can highlight the elimination: | Quote: | Hilighting Intersections - [ESC] to cancel hilighting.
Intersection of row 2 with block 1. The non-intersecting squares of the block cannot contain the value <4>.
Intersection of column 5 with block 2. The non-intersecting squares of the block cannot contain the value <7>. |
Keith |
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Asellus
Joined: 05 Jun 2007
Posts: 865
Location: Sonoma County, CA, USA
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| Posted: Fri Nov 07, 2008 11:33 pm Post subject: |
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I found Norm's elimination, though differently. Either <6> pincer of the otherwise useless 246 XY-Wing in c2 and r8 can be transported. r8c8 can be transported to r3c7 to eliminate <6> in r1c7. Or, r1c2 can be transported to r9c5 to eliminate <6> in r9c7.
Norm,
Nice use of an ER, which is a grouped strong (inference) link. to activate a W-Wing. Your Eureka notation isn't, technically, bidirectional. Writing the ER portion as:
(2)r89c2=(2)r9c123
takes care of that (with the ERI on both sides of the link). Also, the first part should be "(6=2)r1c2" so that the weakly linked <2>s are together and the pincer <6> is evident on the end of the chain. |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Sat Nov 08, 2008 12:11 am Post subject: |
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| I had an XY-Wing on 462, then a flightless XY-Wing on 246. A pincer coloring, or transport, involving four cells finished it off. |
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storm_norm
Joined: 18 Oct 2007
Posts: 1741
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| Posted: Sat Nov 08, 2008 4:43 am Post subject: |
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| Quote: | Writing the ER portion as:
(2)r89c2=(2)r9c123 |
my understanding of the chain says that if r9c5 is not 2, this forces 2 into r9c123, which acts like a box/line reduction... (if all 2's in row 9 lie in box 7 then any other 2's in the box can be eliminated) which says the 2 in r8c2 is not true...
if r9c5 is 2, then that forces 2 out of r9c123, which leaves the remaining 2 in box 7 in r8c2 true...
so I am confused why I should write (2)r89c2=(2)r9c123
oh and I am not even remotely thinking of an ER as part of this chain. |
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Asellus
Joined: 05 Jun 2007
Posts: 865
Location: Sonoma County, CA, USA
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| Posted: Sat Nov 08, 2008 11:14 pm Post subject: |
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| Norm wrote: | | so I am confused why I should write (2)r89c2=(2)r9c123 |
Come to reflect upon it a bit... I guess you don't. There are three ways you could have written it:
(2)r8c2=(2)r9c123
(2)r89c2=(2)r9c13
(2)r89c2=(2)r9c123
depending upon whether you place r9c2 on one side, the other, or both. I guess I just prefer placing it on both since I then don't have to choose (and it gives it a nice symmetry).
| Quote: | | and I am not even remotely thinking of an ER as part of this chain |
All the more reason to point it out, since that's what it is. Any ER is just a grouped strong inference link. Here, the ER in b7 induces a weak link between the <2>s in r1c2 and r9c5, leading to the W-Wing. |
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storm_norm
Joined: 18 Oct 2007
Posts: 1741
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| Posted: Sun Nov 09, 2008 1:30 am Post subject: |
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am I looking into this too deeply??
(2)r89c2 = (2)r9c123
just so I understand what should be going through my head, this simple strong inference says that either there is a 2 in r89c2 or there is a 2 in r9c123.
but if you already said either in r89c2, then how can you include r9c2 in --(2)r9c123,
shouldn't it be (2)r89c2 = (2)r9c13
but going the other way this is still not bidirectional because if you say the 2 in r9c5 is true then none of r9c123 is true which leaves 2 in r8c2
its only bidirectional if (2)r8c2 = (2)r9c123.
 |
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Asellus
Joined: 05 Jun 2007
Posts: 865
Location: Sonoma County, CA, USA
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| Posted: Sun Nov 09, 2008 2:03 am Post subject: |
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| Norm wrote: | | this simple strong inference says that either there is a 2 in r89c2 or there is a 2 in r9c123 |
That's what a strong conjugate link would say. The strong inference link says that either or both of r89c2 and r9c123 contain <2>. If r9c2 is <2>, then both sides contain <2> and both sides are thus true. But, that's okay.
| Quote: | | but going the other way this is still not bidirectional because if you say the 2 in r9c5 is true then none of r9c123 is true which leaves 2 in r8c2 |
(2)r89c2 is true if r8c2 is true. It doesn't matter that r9c2 happens to be false in that case.
The lack of bidirectionality only arises if the ERI cell, r9c2, is true. If you include r9c2 on only one side of the strong link in the notation, then you can only propogate the implications in that direction. By including it on both sides, the implications propogate outward in both directions. In a way, this is a quibble since the victim(s) will be false in any case (which is why I decided it didn't matter after some reflection). |
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storm_norm
Joined: 18 Oct 2007
Posts: 1741
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| Posted: Sun Nov 09, 2008 2:33 am Post subject: |
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| Quote: | | The strong inference link says that either or both of r89c2 |
it does? |
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