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keith
Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA
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 Posted: Sat May 01, 2010 4:14 pm Post subject: Free Press Apr 30, 2010 |
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| Code: | Puzzle: FP043010
+-------+-------+-------+
| . . 3 | . 5 . | . . . |
| 7 . 5 | 2 . . | 3 . . |
| . . . | . . 8 | . . 1 |
+-------+-------+-------+
| 8 . 1 | . . . | 5 . . |
| . 9 . | 8 . 5 | . 2 . |
| . . 6 | 9 . . | 1 . 3 |
+-------+-------+-------+
| 5 . . | 1 . . | . . . |
| . . 8 | . . 7 | 6 . . |
| . . . | . 3 . | 2 . . |
+-------+-------+-------+ |
Enjoy!
Keith |
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keith
Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA
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| Posted: Sat May 01, 2010 7:49 pm Post subject: |
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After basics: | Code: | +-------------------+-------------------+-------------------+
| 1469 146 3 | 7 5 49 | 8 469 2 |
| 7 8 5 | 2 469 1 | 3 469 49 |
| 2469 246 249 | 3 469 8 | 7 5 1 |
+-------------------+-------------------+-------------------+
| 8 24 1 | 6 24 3 | 5 79 79 |
| 3 9 7 | 8 1 5 | 4 2 6 |
| 24 5 6 | 9 7 24 | 1 8 3 |
+-------------------+-------------------+-------------------+
| 5 23467 24 | 1 8 26 | 9 347 47 |
| 129 123 8 | 4 29 7 | 6 13 5 |
| 1469 1467 49 | 5 3 69 | 2 147 8 |
+-------------------+-------------------+-------------------+ |
Do you see the XY cycle? (4 cells.)
Keith |
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Mogulmeister
Joined: 03 May 2007
Posts: 1151
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| Posted: Sat May 01, 2010 11:23 pm Post subject: |
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| I can see a skyscraper on 9s and that is in 4 cells. |
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tlanglet
Joined: 17 Oct 2007
Posts: 2468
Location: Northern California Foothills
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| Posted: Sun May 02, 2010 2:12 am Post subject: |
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Do you mean the one in r79c36 It was fun but not sufficient.
| Quote: | Then I used a m-wing 24 to delete (2) in r8c1, and a xy-wing 16-9 with vertex 16 in r9c1 to complete te puzzle.
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Ted |
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keith
Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA
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| Posted: Sun May 02, 2010 2:12 am Post subject: |
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| Code: | +-------------------+-------------------+-------------------+
| 1469 146 3 | 7 5 49 | 8 469 2 |
| 7 8 5 | 2 469 1 | 3 469 49 |
| 2469 246 2-49 | 3 469 8 | 7 5 1 |
+-------------------+-------------------+-------------------+
| 8 24 1 | 6 24 3 | 5 79 79 |
| 3 9 7 | 8 1 5 | 4 2 6 |
| 24 5 6 | 9 7 24 | 1 8 3 |
+-------------------+-------------------+-------------------+
| 5 -23-467 24a | 1 8 26b | 9 347 47 |
| 129 123 8 | 4 29 7 | 6 13 5 |
|1-46-9 1-467 49d | 5 3 69c | 2 147 8 |
+-------------------+-------------------+-------------------+ |
Keith |
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storm_norm
Joined: 18 Oct 2007
Posts: 1741
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| Posted: Sun May 02, 2010 7:15 am Post subject: |
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if the 2 is false in r8c2 then this kite is true on 2's
r8c1 = r8c5 - r7c6 = r6c6
| Code: | .---------------------.---------------------.---------------------.
| 1469 146 3 | 7 5 49 | 8 469 2 |
| 7 8 5 | 2 469 1 | 3 469 49 |
| 2469 246 249 | 3 469 8 | 7 5 1 |
:---------------------+---------------------+---------------------:
| 8 24 1 | 6 24 3 | 5 79 79 |
| 3 9 7 | 8 1 5 | 4 2 6 |
| 24 5 6 | 9 7 24 | 1 8 3 |
:---------------------+---------------------+---------------------:
| 5 23467 24 | 1 8 26 | 9 347 47 |
| 129 123 8 | 4 29 7 | 6 13 5 |
| 1469 1467 49 | 5 3 69 | 2 147 8 |
'---------------------'---------------------'---------------------' |
if the kite is false then the 2 is true in r8c2...
(2-3)r8c2 = (3-6)r7c2 = (6-2)r7c6 = (2)r6c6
proving that the 2 in r6c1 can be eliminated because either the kite is true (which eliminates the 2 in r6c1) or the 2 is true in r8c2( leads to 2 true in r6c6)
or as a diagram, at least one of the 2's in row 8 must be true.
(2)r8c1
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(2-3)r8c2 = (3-6)r7c2 = (6-2)r7c6 = (2)r6c6
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(2)r8c5 - (2)r7c6 = (2)r6c6 |
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arkietech
Joined: 31 Jul 2008
Posts: 1834
Location: Northwest Arkansas USA
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| Posted: Sun May 02, 2010 12:08 pm Post subject: |
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| storm_norm wrote: | (2)r8c1
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(2-3)r8c2 = (3-6)r7c2 = (6-2)r7c6 = (2)r6c6
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(2)r8c5 - (2)r7c6 = (2)r6c6 |
Some people see beauty in a flower or sunset.
I like sudoku! |
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daj95376
Joined: 23 Aug 2008
Posts: 3854
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| Posted: Sun May 02, 2010 2:08 pm Post subject: |
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I'm sorry, but I'm not in favor of complexity in the name of step reduction.
(2)r8c5 = (2-6)r7c6 = (6-3)r7c2 = (3)r8c2 => r8c2<>2
followed by the Kite.
-or-
| Code: | r8c3 2-String Kite <> 9 r3c5
r6c5 2-String Kite <> 2 r8c1
r48 X-Wing <> 2 r37c2
<16+9> XYZ-Wing r9c1/r8c1+r9c6 <> 9 r9c3
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keith
Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA
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| Posted: Sun May 02, 2010 2:25 pm Post subject: |
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I think this is simpler: | Code: | +-------------------+-------------------+-------------------+
| 1469 146 3 | 7 5 49 | 8 469 2 |
| 7 8 5 | 2 469 1 | 3 469 49 |
| 2469 -246a 249 | 3 469 8 | 7 5 1 |
+-------------------+-------------------+-------------------+
| 8 24b 1 | 6 24c 3 | 5 79 79 |
| 3 9 7 | 8 1 5 | 4 2 6 |
| 24 5 6 | 9 7 24d | 1 8 3 |
+-------------------+-------------------+-------------------+
| 5 -234-67 24 | 1 8 26e | 9 347 47 |
|1-29 123 8 | 4 29 7 | 6 13 5 |
| 1469 1467 49 | 5 3 69 | 2 147 8 |
+-------------------+-------------------+-------------------+ |
Two steps:
1. Coloring on 2 removes 2 in three cells as shown.
2. An extended XY-wing 24-6 takes out 6 in R7C2.
Keith |
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daj95376
Joined: 23 Aug 2008
Posts: 3854
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| Posted: Sun May 02, 2010 3:30 pm Post subject: |
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| keith wrote: | Coloring on 2 removes 2 in three cells as shown.
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I'll agree that the three eliminations exist, but I don't think your markings account for all of them at once. |
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keith
Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA
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| Posted: Sun May 02, 2010 4:06 pm Post subject: |
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| daj95376 wrote: | | keith wrote: | Coloring on 2 removes 2 in three cells as shown.
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I'll agree that the three eliminations exist, but I don't think your markings account for all of them at once. |
That is correct. Initially, the coloring takes out 2 in R7C2 and R8C1. Then it can be extended to take out R3C2.
Keith |
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arkietech
Joined: 31 Jul 2008
Posts: 1834
Location: Northwest Arkansas USA
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| Posted: Sun May 02, 2010 7:00 pm Post subject: |
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| daj95376 wrote: | | I'm sorry, but I'm not in favor of complexity in the name of step reduction |
Complexity is in the eye of the beholder.  |
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peterj
Joined: 26 Mar 2010
Posts: 974
Location: London, UK
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| Posted: Sun May 02, 2010 8:35 pm Post subject: |
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| daj95376 wrote: | | keith wrote: | Coloring on 2 removes 2 in three cells as shown.
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I'll agree that the three eliminations exist, but I don't think your markings account for all of them at once. |
X-Colours (ref Sudopedia) would give you the eliminations in one named step - if you think such an algortihmic method can be treated as an atomic step. 
I've tried using it occasionally but cant ever recall it doing anything more useful then colouring and then colouring again.... |
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daj95376
Joined: 23 Aug 2008
Posts: 3854
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| Posted: Sun May 02, 2010 9:32 pm Post subject: |
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[Original response withdrawn]
Yes, the Sudopedia definition of X-Colors is more powerful than the originally way I learned to do X-Colors. It would work in one step for all three eliminations. Congratulations!
Last edited by daj95376 on Sun May 02, 2010 10:27 pm; edited 2 times in total |
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keith
Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA
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| Posted: Sun May 02, 2010 9:54 pm Post subject: |
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What I have done is simple coloring. You can view it as one or two steps.
It is NOT X-coloring. It may well be a subset of X-coloring, I don't know.
I have been down the road about X-coloring once before, and I do not intend to go there again.
Keith |
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