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Mogulmeister
Joined: 03 May 2007
Posts: 1151
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 Posted: Mon Nov 30, 2009 2:58 pm Post subject: Sudoku.org.uk 29th Nov |
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This one is designated Diabolical - I'm not sure that it is but unlike many of the Extremes will not explicitly need an ALS or AIC to complete.
(If we disregard for the purposes of this exercise that AIC are a meta language for a lot of the patterns we use to make eliminations) 
| Code: | +---+---+---+
|...|...|2..|
|8.4|...|...|
|..5|.47|.8.|
+---+---+---+
|9..|...|..5|
|.8.|9.5|46.|
|2..|...|..3|
+---+---+---+
|.6.|18.|.5.|
|...|...|6.4|
|..7|.3.|...|
+---+---+---+
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arkietech
Joined: 31 Jul 2008
Posts: 1834
Location: Northwest Arkansas USA
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| Posted: Mon Nov 30, 2009 8:52 pm Post subject: |
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without aic two steps
with one
| Quote: | xyz-wing 379
w-wing 39
or
(3=4)r7c1-(4=9)r7c6-(9)r7c9=(9)r3c9-(9=3)r3c7-(3)r7c7=(3)r8c8
=> r7c7,r8c12<>3
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Mogulmeister
Joined: 03 May 2007
Posts: 1151
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| Posted: Mon Nov 30, 2009 9:29 pm Post subject: |
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After basics I went commando (minus solver) and eventually came up with two steps which were consequential:
| Quote: | xy chain pincers on 1 at r4c8 and r9c7 which means r8c8<>1
Then xy wing on <37> <39> <79> [r28c8 & r7c9] means r12c9<>7 |
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storm_norm
Joined: 18 Oct 2007
Posts: 1741
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| Posted: Tue Dec 01, 2009 12:06 am Post subject: |
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here is another
| Code: | +---------------+--------------------+--------------------+
| 136 137 9 | 3568 56 1368 | 2 4 16(7) |
| 8 1237 4 | 236 269 12369 | 5 (37) 16(7) |
| 136 123 5 | 236 4 7 | 9(3) 8 169 |
+---------------+--------------------+--------------------+
| 9 4 16 | 23678 267 2368 | 178 17 5 |
| 7 8 3 | 9 1 5 | 4 6 2 |
| 2 5 16 | 4678 67 468 | 1789 179 3 |
+---------------+--------------------+--------------------+
| 34 6 2 | 1 8 4-9 | (379) 5 (79) |
| 135 139 8 | 257 2579 29 | 6 13-9 4 |
| 45 19 7 | 456 3 46 | 1-9 2 8 |
+---------------+--------------------+--------------------+ |
loop...ALS[(9)r7c79 = (3)r7c7]r7c79 - (3)r3c7 = (3-7)r2c8 = (7)r12c9 - (7=9)r7c9;
r7c6 <> 9
r8c8 <> 9
r9c7 <> 9
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keith
Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA
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| Posted: Tue Dec 01, 2009 12:12 am Post subject: |
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As I recall, an XYZ-wing -9 revealed an XY-wing -7. Done.
Keith |
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Mogulmeister
Joined: 03 May 2007
Posts: 1151
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| Posted: Tue Dec 01, 2009 1:01 am Post subject: |
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I think with the human eye its 2 steps.
I like the elegance of the ALS but they are not so easy to spot without a solver.
Norm - don't the candidates for elimination (the 9's) have to be able to see all the 9's in the ALS ? Perhaps I've read your diagram wrongly (v possible) but there is a 9 at r3c7 which isn't seen by the candidate 9s at r7c6 and r8c8.
If the locked common is 3 at r37c7 and the two ALS are:
A:r7c79 (2 cells) and B:r12c9, r3c7 (3 cells) then the only 9 that sees all is the one in r9c7. |
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storm_norm
Joined: 18 Oct 2007
Posts: 1741
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| Posted: Tue Dec 01, 2009 1:32 am Post subject: |
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Mogulmeister,
my chain isn't a ALS rule type pattern. its a chain which starts with the Almost locked naked pair {7,9} in r7c79. the naked pair {7,9} would be true if the 3 in r7c7 is false. (this would eliminate the other 9's in box 9 and in r7c6). if the 3 is true (or conversly if the {7,9} pair is false}, then you follow the chain around and you find that the 9 is still true in r7c9. which does the same thing.
hope this helps. |
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keith
Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA
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| Posted: Tue Dec 01, 2009 1:35 am Post subject: |
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There is a simpler one in R37C79:
(3=9) = (16)9 = (9=7) = (39)7 and back to (3=9).
Sorry about the notation, I hope you get the idea.
R7C9 <>9 solves the puzzle.
Keith |
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Mogulmeister
Joined: 03 May 2007
Posts: 1151
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| Posted: Tue Dec 01, 2009 8:39 am Post subject: |
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Thanks Norm and thanks Keith. Nomenclature again!  |
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daj95376
Joined: 23 Aug 2008
Posts: 3854
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| Posted: Tue Dec 01, 2009 6:04 pm Post subject: |
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Using Norm's PM:
*) Strong link on <3> in [c7]
*) Strong link on <7> in [r7]
*) r2c8=<37>
*) => (M-Wing eliminations) r12c9<>7 |
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keith
Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA
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| Posted: Tue Dec 01, 2009 8:35 pm Post subject: |
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| Mogulmeister wrote: | | Nomenclature again! |
 Yes, now that I am starting to see these things, I'll have to learn the notation! (I can read it OK, I just can't write it.)
Keith |
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daj95376
Joined: 23 Aug 2008
Posts: 3854
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| Posted: Wed Dec 02, 2009 2:00 am Post subject: |
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| keith wrote: | I'll have to learn the notation! (I can read it OK, I just can't write it.)
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From my notes file on chains. Simply decide on what you wish to describe and use the notation associated with it.
| Code: | Strong Inference (SI): ~A => B
Weak Inference (WI): A => ~B
(SI) e.g.: ( bilocation (n)a = (n)b ) or ( bivalue cell (m=n)c )
(WI) e.g.: ( peers (n)d - (n)e ) or ( ?-value cell (m-n)f )
bilocation (n)a = (n)b: if [a] is not 'n', then [b] is 'n'
bivalue cell (m=n)c : if [c] is not 'm', then [c] is 'n'
peers (n)d - (n)e: if [d] is 'n', then [e] is not 'n'
?-value cell (m-n)f : if [f] is 'm', then [f] is not 'n'
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The (=) and (-) sign must always alternate when creating a chain.
Regards, Danny |
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Mogulmeister
Joined: 03 May 2007
Posts: 1151
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| Posted: Wed Dec 02, 2009 7:08 pm Post subject: |
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| Quote: | | Simply decide on what you wish to describe and use the notation associated with it. |
Agreed Danny but for us visual types, Norm's picture was mighty helpful. |
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