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arkietech
Joined: 31 Jul 2008
Posts: 1834
Location: Northwest Arkansas USA
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 Posted: Sat May 05, 2012 5:56 am Post subject: au tough 5/4/12 |
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| Code: |
*-----------*
|...|7..|...|
|8..|...|.31|
|16.|..2|4..|
|---+---+---|
|...|8..|2..|
|3..|.5.|..6|
|..4|..6|...|
|---+---+---|
|..9|2..|.53|
|67.|...|..9|
|...|..8|...|
*-----------*
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SudoQ
Joined: 02 Aug 2011
Posts: 127
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| Posted: Sat May 05, 2012 12:57 pm Post subject: |
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A controversial solution?
| Code: | | r3c3=5 -> r1c3=3 -> r9c2=35 |
/SudoQ |
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arkietech
Joined: 31 Jul 2008
Posts: 1834
Location: Northwest Arkansas USA
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| Posted: Sat May 05, 2012 1:38 pm Post subject: |
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| SudoQ wrote: | A controversial solution?
| Code: | | r3c3=5 -> r1c3=3 -> r9c2=35 |
/SudoQ |
Wow! 
I get the same thing with a Sue de Cog
| Code: |
*-----------------------------------------------------------------------------*
| 59 3459 *35 | 7 134689 13459 | 569 689 2 |
| 8 2459 27-5 | 4569 469 459 | 5679 3 1 |
| 1 6 7-35 | 359 389 2 | 4 789 78 |
|-------------------------+-------------------------+-------------------------|
| 579 159 6 | 8 13479 13479 | 2 1479 457 |
| 3 1289 128 | 149 5 1479 | 179 14789 6 |
| 579 1589 4 | 19 2 6 | 3 1789 578 |
|-------------------------+-------------------------+-------------------------|
| 4 #18 9 | 2 167 17 | 1678 5 3 |
| 6 7 #*1358 | 1345 134 1345 | 18 2 9 |
| 2 35-1 #*135 | 13569 13679 8 | 167 1467 47 |
*-----------------------------------------------------------------------------*
Sue de Cog (length 4 no extra digit)
X=r189c3{1538}
Y=r7c2,r89c3{1538}
=> r23c3<>35, r9c2<>1; stte
35 in r23c3 is outside of X
1 in r9c2 is outside of Y |
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keith
Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA
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| Posted: Sat May 05, 2012 1:54 pm Post subject: |
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After basics:
| Code: | +----------------------+----------------------+----------------------+
| 59 3459 35 | 7 134689 13459 | 569 689 2 |
| 8 2459 257 | 4569 469 459 | 5679 3 1 |
| 1 6 @357 |@359 389 2 | 4 789 78 |
+----------------------+----------------------+----------------------+
| 579 159 6 | 8 13479 13479 | 2 1479 457 |
| 3 1289 128 | 149 5 1479 | 179 14789 6 |
| 579 1589 4 | 19 2 6 | 3 1789 578 |
+----------------------+----------------------+----------------------+
| 4 18 9 | 2 167 17 | 1678 5 3 |
| 6 7 #1358 |#1345 134 #1345 | 18 2 9 |
| 2 135 135 |13-569 13679 8 | 167 1467 47 |
+----------------------+----------------------+----------------------+ |
Finned X-wing (grouped skyscraper) takes out 5 in R9C4.
| Code: | +----------------------+----------------------+----------------------+
| 59 3459 A35 | 7 134689 13459 | 569 689 2 |
| 8 2459 2-57 | 4569 469 459 | 5679 3 1 |
| 1 6 3-57 | 359 389 2 | 4 789 78 |
+----------------------+----------------------+----------------------+
| 579 159 6 | 8 13479 13479 | 2 1479 457 |
| 3 1289 128 | 149 5 1479 | 179 14789 6 |
| 579 1589 4 | 19 2 6 | 3 1789 578 |
+----------------------+----------------------+----------------------+
| 4 18 9 | 2 167 17 | 1678 5 3 |
| 6 7 138 | 1345 134 1345 | 18 2 9 |
| 2 B135 C135 | 1369 13679 8 | 167 1467 47 |
+----------------------+----------------------+----------------------+ |
M-wing: 3 in A forces 3 in B. AC are pincers on 5.
That solves the puzzle.
Keith |
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SudoQ
Joined: 02 Aug 2011
Posts: 127
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| Posted: Sat May 05, 2012 3:11 pm Post subject: |
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| arkietech wrote: | Wow!
I get the same thing with a Sue de Cog
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Thanks!
Normally, a single candidate contradiction can be rewritten.
In this case, I couldn't do this.
r8c3 or r9c3 must be 3 or 5.
/SudoQ |
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SudoQ
Joined: 02 Aug 2011
Posts: 127
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| Posted: Sat May 05, 2012 7:37 pm Post subject: |
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Here is a solution that only requires singles before and after the step:
| Code: | | r7c2=x -> r5c3=x (x=1/8) => r5c3<>27 |
/SudoQ |
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ronk
Joined: 07 May 2006
Posts: 398
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| Posted: Sun May 06, 2012 3:24 am Post subject: |
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| SudoQ wrote: | Here is a solution that only requires singles before and after the step:
r7c2=x -> r5c3=x (x=1/8 ) => r5c3<>27 |
Nice, you have used an m-ring (see Type C here).
In NL notation: r7c2 -1- r789c3 =1= r5c3 =8= r789c3 -8- r7c2 - continuous loop
In AIC notation: (8=1)r7c2 - (1)r789c3 = (1-8)r5c3 = (8)r789c3 - loop ==> r5c3=18 |
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tlanglet
Joined: 17 Oct 2007
Posts: 2468
Location: Northern California Foothills
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| Posted: Sun May 06, 2012 4:57 am Post subject: |
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After failing to find any productive VH steps, I started looking for anp()s and immediately found an invalid condition.
anp(7=35)r13c3-(35=1=8)r98c3 which would mean that r7c2 is empty. Thus, r3c3=7
Ted |
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aran
Joined: 19 Apr 2010
Posts: 70
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| Posted: Sun May 06, 2012 9:11 am Post subject: |
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Arkietech
Always nice to see a Sue de Coq given its "historical" significance.
SdC being a specific instance of a more general form - Dual-Linked ALS...with the particularity that the specific form is more difficult to define than the general !
Here the general being :
A=3518r189c3
B=18r7c2
=>all the elims
Last edited by aran on Sun May 06, 2012 9:17 am; edited 1 time in total |
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SudoQ
Joined: 02 Aug 2011
Posts: 127
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| Posted: Sun May 06, 2012 9:14 am Post subject: |
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| ronk wrote: | | In NL notation: r7c2 -1- r789c3 =1= r5c3 =8= r789c3 -8- r7c2 - continuous loop |
Hi Ronk!
Is it possible to write your NL notation like this:
r7c2 -1- r89c3 =1= r5c3 =8= r8c3 -8- r7c2 - continuous loop
(since r7c3<>18 and r9c3<>8)?
I don't understand the "continuous loop" part (but you don't need to try to explain it!).
When I look at your link, it seems that this M-wing is also of type D!?
Finally, do you have a suggestion how to write my first example
(r3c3=5 -> r1c3=3 -> r9c2=35) in a nicer way?
/SudoQ |
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