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Earl
Joined: 30 May 2007
Posts: 677
Location: Victoria, KS
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 Posted: Sat Apr 11, 2009 2:17 pm Post subject: April 11 DB |
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I chased this Easter Bunny down many dead ends, till I found this two route solution. Any single shots out there?
A Solution: Either solution of the (83)UR nets a <1> in R2C1. Then an xy-chain eliminates <3> in R2C3 to finish it off.
Earl
| Code: |
+-------+-------+-------+
| . 5 . | 4 . . | . . 1 |
| . 8 . | . 9 . | . 6 . |
| 4 . 9 | . 1 . | . . . |
+-------+-------+-------+
| . . 8 | . . 2 | . . . |
| 7 . . | 5 . 9 | . . 3 |
| . . . | 3 . . | 7 . . |
+-------+-------+-------+
| . . . | . 4 . | 5 . 9 |
| . 7 . | . 3 . | . 2 . |
| 2 . . | . . 6 | . 1 . |
+-------+-------+-------+
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Play this puzzle online at the Daily Sudoku site |
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wapati
Joined: 10 Jun 2008
Posts: 472
Location: Brampton, Ontario, Canada.
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| Posted: Sat Apr 11, 2009 3:22 pm Post subject: Re: April 11 DB |
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| Earl wrote: | | I chased this Easter Bunny down many dead ends, till I found this two route solution. Any single shots out there? |
Use the UR to make a 25 pair and xy-wing 25-3 to set Box1.  |
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Louise56
Joined: 21 Sep 2005
Posts: 94
Location: El Cajon, California USA
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| Posted: Sat Apr 11, 2009 4:23 pm Post subject: |
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| I did the basics on this puzzle and have a 38 pair in r1c67. In row 3 the 38s have a 5 with them in c6 and a 2 with them in c7. Could someone explain to me how the UR works in this case? Thanks! |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Sat Apr 11, 2009 4:52 pm Post subject: |
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| I messed around with three Hidden URs, an X- and XYZ-Wing until an XY-Loop finished things off. A fun puzzle. |
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wapati
Joined: 10 Jun 2008
Posts: 472
Location: Brampton, Ontario, Canada.
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| Posted: Sat Apr 11, 2009 4:54 pm Post subject: |
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| Louise56 wrote: | | I did the basics on this puzzle and have a 38 pair in r1c67. In row 3 the 38s have a 5 with them in c6 and a 2 with them in c7. Could someone explain to me how the UR works in this case? Thanks! |
You have observed correctly that to avoid a double solution at least one of those two cells must be 2 or 5.
You can use that to build an xy-wing but that isn't really your question.
| Code: | .---------------.-------------.-------------.
| 6 5 7 | 4 2 #38 |*38 9 1 |
| 13 8 123 | 7 9 35 | 234 6 45 |
| 4 23 9 | 6 1 *358 |@28-3 7 58 |
:---------------+-------------+-------------:
| 35 34 8 | 1 7 2 | 9 45 6 |
| 7 246 246 | 5 68 9 | 1 48 3 |
| 159 169 156 | 3 68 4 | 7 58 2 |
:---------------+-------------+-------------:
| 8 16 16 | 2 4 7 | 5 3 9 |
| 59 7 45 | 89 3 1 | 6 2 48 |
| 2 349 34 | 89 5 6 | 48 1 7 |
'---------------'-------------'-------------' |
The cell marked "#" has a strong link on 8s to the two cells marked "*".
If # is 3 than the "*" cells will be 8 and "@" cannot be 3 or the UR wrecks the puzzle. If # is 8 then r1c7 is 3 and "@" cannot be 3.
This is the elimination possible with this 38 UR. |
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daj95376
Joined: 23 Aug 2008
Posts: 3854
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| Posted: Sat Apr 11, 2009 6:28 pm Post subject: |
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| Louise56 wrote: | | I did the basics on this puzzle and have a 38 pair in r1c67. In row 3 the 38s have a 5 with them in c6 and a 2 with them in c7. Could someone explain to me how the UR works in this case? Thanks! |
There are a number of ways this UR can be used.
| Code: | after basics
+-----------------------------------------------------+
| 6 5 7 | 4 2 *38 | *38 9 1 |
| 13 8 123 | 7 9 35 | 234 6 45 |
| 4 23 9 | 6 1 *38+5 | *38+2 7 58 |
|-----------------+-----------------+-----------------|
| 35 34 8 | 1 7 2 | 9 45 6 |
| 7 246 246 | 5 68 9 | 1 48 3 |
| 159 169 156 | 3 68 4 | 7 58 2 |
|-----------------+-----------------+-----------------|
| 8 16 16 | 2 4 7 | 5 3 9 |
| 59 7 45 | 89 3 1 | 6 2 48 |
| 2 349 34 | 89 5 6 | 48 1 7 |
+-----------------------------------------------------+
# 43 eliminations remain
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Method #1:
Use the fact that [r3c6]=5 or [r3c7]=2 must be true, and look for common eliminations resulting from the assumption that both are true.
| Code: | [r3c6]=5 => [r2c6]=3 => [r2c13]<>3
[r3c7]=2 => [r3c2]=3 => [r2c13]<>3
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I believe that Earl did this but made the mistake of fixating on [r2c1]<>3 alone. For sure, this is equivalent to wapati's pseudo-cell in XY-Wing 25-3.
Method #2:
Use the fact that at least one of <38> can not be in cells [r3c67]. This forces [r3c2]=3 or [r3c9]=8 to be true.
| Code: | [r3c2]=3 => [r2c13]<>3
[r3c9]=8 => [r2c9]=5 => [r2c6]=3 => [r2c13]<>3
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Not a better choice than Method #1 in this puzzle, but sometimes a better choice in other puzzles.
Method #3:
Take advantage of bivalue cells and/or strong links on one or both UR candidates. This is what wapati mentioned above, and for which I'll give a slightly different explanation.
The UR cells in [r1] are bivalue cells and there is a strong link on <8> in [c6]. This means that <3> can be deleted from the fourth UR cell; i.e., [r3c7]<>3. To verify how this works, set [r3c7]=3 and see what follows:
| Code: | [r3c7]=3 => bivalue [r1c7]=8 => bivalue [r1c6]=3 => strong link [r3c6]=8 => <38> UR [r13c67]
____________________________________________________________________________________________
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There are other methods, but these three methods should be a good starting point for understanding what can be done with this UR. |
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wapati
Joined: 10 Jun 2008
Posts: 472
Location: Brampton, Ontario, Canada.
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| Posted: Sat Apr 11, 2009 7:37 pm Post subject: |
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| Keith, that doesn't solve the puzzle. ( IMHO ). |
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Louise56
Joined: 21 Sep 2005
Posts: 94
Location: El Cajon, California USA
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| Posted: Sat Apr 11, 2009 7:38 pm Post subject: |
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| Thank-you both Wapati and daj95376 for your great explanations! I understand how the 3 is removed from r3c7. I don't see how to solve the puzzle from there. There is a similar UR set up with 16's and if I use the same strategy I can remove the 6 from r6c3. Then I get a 153 wing that solves the puzzle. I'll hang in there each week and hopefully be able to do these faster. |
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keith
Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA
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| Posted: Sat Apr 11, 2009 8:15 pm Post subject: |
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| wapati wrote: | | Keith, that doesn't solve the puzzle. ( IMHO ). |
Wapati is referring to a post I deleted because it was incorrect. Here you go:
| Code: | +-------------+-------------+-------------+
| 6 5 7 | 4 2 38 | 38 9 1 |
| 1-3 8 123 | 7 9 35# | 234 6 45d |
| 4 23 9 | 6 1 358 | 238 7 58 |
+-------------+-------------+-------------+
| 35# 34 8 | 1 7 2 | 9 45 6 |
| 7 246 246 | 5 68 9 | 1 48 3 |
| 159 169 156 | 3 68 4 | 7 58 2 |
+-------------+-------------+-------------+
| 8 16 16 | 2 4 7 | 5 3 9 |
| 59a 7 45 | 89b 3 1 | 6 2 48c |
| 2 349 34 | 89 5 6 | 48 1 7 |
+-------------+-------------+-------------+ |
abcd is an XY-chain with pincers <5>. With the cells # it makes a W-wing, eliminating <3> in R2C1. This does not solve the puzzle, but I am still looking.
Edit: There is then a 5-cell chain that takes out <3> in R2C3. I don't see anything more elegant than that.
Keith |
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daj95376
Joined: 23 Aug 2008
Posts: 3854
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| Posted: Sat Apr 11, 2009 9:52 pm Post subject: |
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| Louise56 wrote: | | Thank-you both Wapati and daj95376 for your great explanations! I understand how the 3 is removed from r3c7. I don't see how to solve the puzzle from there. There is a similar UR set up with 16's and if I use the same strategy I can remove the 6 from r6c3. Then I get a 153 wing that solves the puzzle. I'll hang in there each week and hopefully be able to do these faster. |
I think we're talking apples-'n-oranges here. The elimination [r3c7]<>3 was not meant to answer Earl's request for a single step solution. It was part of the answer to your query about what could be done with this UR. The elimination [r2c13]<>3 answers your query and cracks the puzzle in one step. |
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Kdelle
Joined: 20 Mar 2008
Posts: 59
Location: Hudson, NH
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| Posted: Sun Apr 12, 2009 4:05 pm Post subject: |
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| A little UR trick I learned from Norm (I think) is that to avoid the DP 38, both r3c6 and r3c7 can't be false. If r3c7 is NOT 2, the r3c6 can not be 5, so r3c7 mut be 2.....which solves the puzzle. |
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Asellus
Joined: 05 Jun 2007
Posts: 865
Location: Sonoma County, CA, USA
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| Posted: Sun Apr 12, 2009 7:05 pm Post subject: |
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For those interested, Kdelle's solution is an AIC loop with strong inference discontinuity:
38UR[(2)r3c7=(5)r3c6] - (5=3)r2c6 - ALS[(3)r2c13=(2)r2c3] - (2)r2c7=(2)r3c7; r3c7=2 |
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Louise56
Joined: 21 Sep 2005
Posts: 94
Location: El Cajon, California USA
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| Posted: Mon Apr 13, 2009 12:16 am Post subject: |
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| daj95376 wrote: |
I think we're talking apples-'n-oranges here. The elimination [r3c7]<>3 was not meant to answer Earl's request for a single step solution. It was part of the answer to your query about what could be done with this UR. The elimination [r2c13]<>3 answers your query and cracks the puzzle in one step. |
That's true, I was trying to figure out how to resolve the UR which you helped me with. Sadly it wasn't the way to solve the puzzle though. Thanks for your help. |
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storm_norm
Joined: 18 Oct 2007
Posts: 1741
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| Posted: Mon Apr 13, 2009 4:40 pm Post subject: |
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Kdelle,
very nice.
a little bit simpler maybe
UR38[(2)r3c7 = (5)r3c6] - (5=3)r2c6 - (3)r2c13 = (3)r3c2
which eliminates 2 in r3c2. |
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