## Vanhegan fiendish November 18, 2012

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### Vanhegan fiendish November 18, 2012

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| *-----------*`

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dan

ArkieTech

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Location: NW Arkansas USA

### Re: Vanhegan fiendish November 18, 2012

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
tlanglet
2010 Supporter

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### Re: Vanhegan fiendish November 18, 2012

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!

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`
dan

ArkieTech

Posts: 2956
Joined: 29 May 2006
Location: NW Arkansas USA

### Re: Vanhegan fiendish November 18, 2012

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
David P Bird
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Location: Middle England

### Re: Vanhegan fiendish November 18, 2012

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.
dan

ArkieTech

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Joined: 29 May 2006
Location: NW Arkansas USA

### Re: Vanhegan fiendish November 18, 2012

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.
Marty R.

Posts: 1508
Joined: 23 October 2012
Location: Rochester, New York, USA

### Re: Vanhegan fiendish November 18, 2012

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.

Leren
Leren

Posts: 3313
Joined: 03 June 2012

### Re: Vanhegan fiendish November 18, 2012

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 .
7b53
2012 Supporter

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Location: New York

### Re: Vanhegan fiendish November 18, 2012

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
eleven

Posts: 1865
Joined: 10 February 2008

### Re: Vanhegan fiendish November 18, 2012

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

Leren
Leren

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Joined: 03 June 2012

### Re: Vanhegan fiendish November 18, 2012

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

Posts: 1865
Joined: 10 February 2008