Vanhegan fiendish November 18, 2012

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Vanhegan fiendish November 18, 2012

Postby ArkieTech » Tue Nov 20, 2012 7:37 am

Code: Select all
 *-----------*
 |1..|9.4|..5|
 |.7.|..6|.4.|
 |..9|.8.|1..|
 |---+---+---|
 |35.|4.7|..6|
 |..8|...|7..|
 |9..|1.8|.34|
 |---+---+---|
 |..5|.4.|6..|
 |.9.|2..|.5.|
 |6..|8.9|..3|
 *-----------*


Play/Print this puzzle online
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Re: Vanhegan fiendish November 18, 2012

Postby tlanglet » Tue Nov 20, 2012 1:08 pm

Code after basics:
Code: Select all
  *--------------------------------------------------*
 | 1    28   6    | 9    7    4    | 3    28   5    |
 | 258  7    23   | 35   1    6    | 9    4    28   |
 | 45   34   9    | 35   8    2    | 1    6    7    |
 |----------------+----------------+----------------|
 | 3    5    12   | 4    9    7    | 28   128  6    |
 | 24   124  8    | 6    3    5    | 7    129  129  |
 | 9    6    7    | 1    2    8    | 5    3    4    |
 |----------------+----------------+----------------|
 | 28   238  5    | 7    4    13   | 6    129  129  |
 | 7    9    34   | 2    6    13   | 48   5    18   |
 | 6    12   124  | 8    5    9    | 24   7    3    |
 *--------------------------------------------------*


An infrequent AUR pattern: all four cells have identical content.

AUR(19)r 57c89 SIS 2r5c89, 2r7c89
(4=2)r5c1-AUR(19)r 57c89[2r5c89=2r7c89]-(28=3)r7c12-(3=4)r3c2 => r3c1,r5c2<>4
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Re: Vanhegan fiendish November 18, 2012

Postby ArkieTech » Tue Nov 20, 2012 2:16 pm

tlanglet wrote:
[/code]
An infrequent AUR pattern: all four cells have identical content.

AUR(19)r 57c89 SIS 2r5c89, 2r7c89
(4=2)r5c1-AUR(19)r 57c89[2r5c89=2r7c89]-(28=3)r7c12-(3=4)r3c2 => r3c1,r5c2<>4


Outstanding! :D

Code: Select all
 *--------------------------------------------------*
 | 1   a28   6    | 9    7    4    | 3    28   5    |
 | 258  7   b23   | 35   1    6    | 9    4    28   |
 | 45   34   9    | 35   8    2    | 1    6    7    |
 |----------------+----------------+----------------|
 | 3    5    12   | 4    9    7    | 28   128  6    |
 | 24   124  8    | 6    3    5    | 7    129  129  |
 | 9    6    7    | 1    2    8    | 5    3    4    |
 |----------------+----------------+----------------|
 | 28  d23-8 5    | 7    4    13   | 6    129  129  |
 | 7    9   c34   | 2    6    13   | 48   5    18   |
 | 6    12   124  | 8    5    9    | 24   7    3    |
 *--------------------------------------------------*
h-wing
(8=2)r1c2-(2=3)r2c3-r8c3=r7c2 => -8r7c2
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Re: Vanhegan fiendish November 18, 2012

Postby David P Bird » Tue Nov 20, 2012 3:16 pm

tlanglet wrote:An infrequent AUR pattern: all four cells have identical content.

AUR(19)r 57c89 SIS 2r5c89, 2r7c89
(4=2)r5c1-AUR(19)r 57c89[2r5c89=2r7c89]-(28=3)r7c12-(3=4)r3c2 => r3c1,r5c2<>4

That's an interesting route. My suggestion for notating it is:

(4=2)r5c1 - (2=19)r5c89 -[UR]- (19=2)r7c89 - (28=3)r7c12 - (3=4)r3c2 => r3c1,r5c2 <> 4

where -[UR]- signifies a justified weak inference with the square brackets containing the justifying pattern. With that understanding I think the chain stands on its own and doesn't need any further explanation.

Alternatively the corresponding strong inference could be justified instead:

(4=2)r5c1 - (2)r5c89 =[(19)UR]= (2)r7c89 - (28=3)r7c12 - (3=4)r3c2 => r3c1,r5c2 <> 4

I find the justified inference notation to be quite versatile. For example, in this variation there is a digit switch from 1 to 2:

(1)r5c2 = (19)r7c89 -[UR]- (19=2)r5c89 - (28=3)r7c12 - (3=4)r3c2 => r5c2 <> 4
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Re: Vanhegan fiendish November 18, 2012

Postby ArkieTech » Tue Nov 20, 2012 3:38 pm

David P Bird wrote:(4=2)r5c1 - (2=19)r5c89 -[UR]- (19=2)r7c89 - (28=3)r7c12 - (3=4)r3c2 => r3c1,r5c2 <> 4


To me it makes more sense this way:

(4=2)r5c1 - (2=19)UR:r57c89 - (19=2)UR:r57c89 - (28=3)r7c12 - (3=4)r3c2 => r3c1,r5c2 <> 4

as it keeps the alternate strong weak strong pattern.

The beautiful thing about Eureka notation is it is like writing. Each person can have and enjoy his own style. Which hopefully others can understand. :D
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Re: Vanhegan fiendish November 18, 2012

Postby Marty R. » Tue Nov 20, 2012 6:54 pm

Code: Select all
  *--------------------------------------------------*
 | 1    28   6    | 9    7    4    | 3    28   5    |
 | 258  7    23   | 35   1    6    | 9    4    28   |
 | 45   34   9    | 35   8    2    | 1    6    7    |
 |----------------+----------------+----------------|
 | 3    5    12   | 4    9    7    | 28   128  6    |
 | 24   124  8    | 6    3    5    | 7    129  129  |
 | 9    6    7    | 1    2    8    | 5    3    4    |
 |----------------+----------------+----------------|
 | 28   238  5    | 7    4    13   | 6    129  129  |
 | 7    9    34   | 2    6    13   | 48   5    18   |
 | 6    12   124  | 8    5    9    | 24   7    3    |
 *--------------------------------------------------*


A non-chain two-stepper is the W-Wing on 12 in boxes 47 followed by a Type 2 UR on 19.

In the one-step chain, r8c7=4-->r2c9=8.

(8=4)r8c7-(4=3)r8c3-(3=2)r3c3-(2=8)r2c9-->r8c9<>8.
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Re: Vanhegan fiendish November 18, 2012

Postby Leren » Tue Nov 20, 2012 9:53 pm

My solution was (2)r4c7 = (4)r9c7 = (1)r9c3 = r4c3 => r4c3<2>

followed by a Kite in 2s r2c39,r9c3,r7c9 => r7c12,r9c7 <2>

Love that H Wing.

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Re: Vanhegan fiendish November 18, 2012

Postby 7b53 » Wed Nov 21, 2012 2:41 am

Marty R. wrote:In the one-step chain, r8c7=4-->r2c9=8.

(8=4)r8c7-(4=3)r8c3-(3=2)r3c3-(2=8)r2c9-->r8c9<>8.

nice about xy-chain is can start the chain at either end. resulting same two end cells .

Leren wrote:Love that H Wing.

indeed. the so called H-wing is very useful and powerful .
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Re: Vanhegan fiendish November 18, 2012

Postby eleven » Wed Nov 21, 2012 1:44 pm

tlanglet wrote:AUR(19)r 57c89 SIS 2r5c89, 2r7c89
(4=2)r5c1-AUR(19)r 57c89[2r5c89=2r7c89]-(28=3)r7c12-(3=4)r3c2 => r3c1,r5c2<>4

Nice,
i used the 19 DP in the same cells, implying that one of r4c8 and r8c9 must be 1 (1r4c8=1r8c9) [and r7c8,r5c9<>1]
Code: Select all
 *--------------------------------------------------*
 | 1    28   6    | 9    7    4    | 3    28   5    |
 | 258  7   b23   | 35   1    6    | 9    4    28   |
 | 45   34   9    | 35   8    2    | 1    6    7    |
 |----------------+----------------+----------------|
 | 3    5   a12   | 4    9    7    | 28  #128  6    |
 | 24   124  8    | 6    3    5    | 7   *129 *129  |
 | 9    6    7    | 1    2    8    | 5    3    4    |
 |----------------+----------------+----------------|
 | 28   238  5    | 7    4    13   | 6   *129 *129  |
 | 7    9  -34    | 2    6   c13   | 48   5   #18   |
 | 6    12  124   | 8    5    9    | 24   7    3    |
 *--------------------------------------------------*

r4c8=1 -> r4c3=2 -> r2c3=3
r8c9=1 -> 3r8c6
=> r8c3<>3
Other notation ?
(3=2)r2c3-(2=1)r4c3-1r4c8[DP19r57c89]=1r8c9-(1=3)r8c6, r8c3<>3
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Re: Vanhegan fiendish November 18, 2012

Postby Leren » Thu Nov 22, 2012 9:57 pm

eleven wrote: i used the 19 DP in the same cells, implying that one of r4c8 and r8c9 must be 1 (1r4c8=1r8c9) [and r7c8,r5c9<>1]


Although it doesn't lead to an immediate solution of this puzzle I think it's worth noting that the UR pattern also => r7c8,r5c9<2> => r7c8,r5c9 = 9

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Re: Vanhegan fiendish November 18, 2012

Postby eleven » Fri Nov 23, 2012 10:50 pm

That's tricky, however you can take the strong link for 2 in box 3 instead of the UR for the same.
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