Vanhegan Fiendish December 3, 2012

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Vanhegan Fiendish December 3, 2012

Postby ArkieTech » Tue Dec 04, 2012 8:25 am

Code: Select all
 *-----------*
 |.8.|.3.|2..|
 |.3.|..2|...|
 |.27|.84|.1.|
 |---+---+---|
 |67.|...|..8|
 |39.|...|.74|
 |8..|...|.92|
 |---+---+---|
 |.1.|76.|48.|
 |...|4..|.3.|
 |..5|.1.|.2.|
 *-----------*


Play/Print this puzzle online
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Re: Vanhegan Fiendish December 3, 2012

Postby Leren » Tue Dec 04, 2012 9:05 am

Code: Select all
*--------------------------------------------------------------------------------*
| 14      8       69       | 159     3       17       | 2       46      57       |
| 14      3       69       | 159     79      2        | 8       46      57       |
| 5       2       7        | 6       8       4        | 9       1       3        |
|--------------------------+--------------------------+--------------------------|
| 6       7       2-4      |c29     a49      13       | 13      5       8        |
| 3       9       12       |c28      5       68       | 16      7       4        |
| 8       5      b14       |c13      -47     67       | 136     9       2        |
|--------------------------+--------------------------+--------------------------|
| 2       1       3        | 7       6       5        | 4       8       9        |
| 7       6       8        | 4       2       9        | 5       3       1        |
| 9       4       5        |c38      1       38       | 7       2       6        |
*--------------------------------------------------------------------------------*


ALS XY Wing (4=9)r4c5 - r4c4 = 1r6c4 (ALS r4569c4) -(1=4)r6c3 => r4c3, r6c5 <4>

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Re: Vanhegan Fiendish December 3, 2012

Postby Leren » Tue Dec 04, 2012 10:43 am

Code: Select all
*--------------------------------------------------------------------------------*
| 14      8       69       | 159     3      a17       | 2       46      57       |
| 14      3       69       | 159     79      2        | 8       46      57       |
| 5       2       7        | 6       8       4        | 9       1       3        |
|--------------------------+--------------------------+--------------------------|
| 6       7       24       | 29      49     b13       |c13      5       8        |
| 3       9       12       | 28      5      e68       |d16      7       4        |
| 8       5       14       | 13      47     f6-7      | 136     9       2        |
|--------------------------+--------------------------+--------------------------|
| 2       1       3        | 7       6       5        | 4       8       9        |
| 7       6       8        | 4       2       9        | 5       3       1        |
| 9       4       5        | 38      1       38       | 7       2       6        |
*--------------------------------------------------------------------------------*


Another one stepper: Extended S Wing (7=1) r1c6 - r4c6 = r4c7 - (1=6) r5c7 - r5c6 = r6c6 => r6c6<7>

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Re: Vanhegan Fiendish December 3, 2012

Postby ArkieTech » Tue Dec 04, 2012 12:44 pm

Code: Select all
 *--------------------------------------------------*
 | 14   8    69   | 159  3    17   | 2    46   57   |
 | 14   3    69   | 159  79   2    | 8    46   57   |
 | 5    2    7    | 6    8    4    | 9    1    3    |
 |----------------+----------------+----------------|
 | 6    7    24   | 29   49   13   | 13   5    8    |
 | 3    9   a12   |b28   5    68   | 16   7    4    |
 | 8    5    4-1  |d13   47   67   | 136  9    2    |
 |----------------+----------------+----------------|
 | 2    1    3    | 7    6    5    | 4    8    9    |
 | 7    6    8    | 4    2    9    | 5    3    1    |
 | 9    4    5    |c38   1    38   | 7    2    6    |
 *--------------------------------------------------*
xy-chain
(1=2)r5c3-(2=8)r5c4-(8=3)r9c4-(3=1)r6c4 => -1r6c3; stte


and a short chain.... :D
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Re: Vanhegan Fiendish December 3, 2012

Postby Marty R. » Tue Dec 04, 2012 5:37 pm

xy-chain


Also XY-Chain; I noticed a few different ones.
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Re: Vanhegan Fiendish December 3, 2012

Postby tlanglet » Wed Dec 05, 2012 1:58 am

Hi all. Been traveling and had a great time but it is great to be home working on sudoku.

Almost MUG (1469)r12c1348 with external SIS r1c6=1 & r4c4=9
2r4c3=(2-1)r5c3=r5c7-r4c7=r4c6-AMUG(1469)r12c1348[1r1c6=9r4c4] => r4c4<>2
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Re: Vanhegan Fiendish December 3, 2012

Postby David P Bird » Wed Dec 05, 2012 1:03 pm

tlanglet wrote:Hi all. Been traveling and had a great time but it is great to be home working on sudoku.

Almost MUG (1469)r12c1348 with external SIS r1c6=1 & r4c4=9
2r4c3=(2-1)r5c3=r5c7-r4c7=r4c6-AMUG(1469)r12c1348[1r1c6=9r4c4] => r4c4<>2

Firstly it seems you omitted a final – (9=4)r4c5 in your chain.

Secondly the threatened MUG (1469)r12c1348, like < Marty's Almost UR >, poses no threat because (5)r12c4 must be true. Any inference available by looking at other ways to avoid it will always be available much more simply, without needing to check out an 8 cell pattern!

Following your chain up to the extended MUG, the way I see it concluding is:

(2)r4c3 = (2-1)r5c3 = (1)r5c7 – (1)r4c7 = (1)r4c6 – (1=7)r1c6 – (7=9)r2c5 - (9=4)r4c5 => r4c3 <> 4

[Edit] final elimination corrected following Leren's response.
Last edited by David P Bird on Thu Dec 06, 2012 1:03 am, edited 1 time in total.
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Re: Vanhegan Fiendish December 3, 2012

Postby Leren » Thu Dec 06, 2012 12:28 am

David P Bird wrote: (2)r4c3 = (2-1)r5c3 = (1)r5c7 – (1)r4c7 = (1)r4c6 – (1=7)r1c6 – (7=9)r2c5 - (9=4)r4c5 => r4c3 <> 2


David, I think you meant to end your move with something like .... - (9=4)r4c5 = (9-2) r4c4 => r4c4 <> 2

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Re: Vanhegan Fiendish December 3, 2012

Postby David P Bird » Thu Dec 06, 2012 1:01 am

Leren wrote:
David P Bird wrote: (2)r4c3 = (2-1)r5c3 = (1)r5c7 – (1)r4c7 = (1)r4c6 – (1=7)r1c6 – (7=9)r2c5 - (9=4)r4c5 => r4c3 <> 2

David, I think you meant to end your move with something like .... - (9=4)r4c5 = (9-2) r4c4 => r4c4 <> 2

Thanks for pointing out my error, but my typo was the elimination. The one I had in mind which was r4c3 <> 4 which achieves the same effect.
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Re: Vanhegan Fiendish December 3, 2012

Postby tlanglet » Fri Dec 07, 2012 12:14 am

David, thanks for your feedback. I always appreciate the effort by others to add to my knowledge base.

I do not follow your comment about a missing node in my AIC. My leading statement is that r4c3=2 and I end with r4c4=9 from which I conclude that r4c4<>2. To me, your additional node simply changes the deletion to r3c4<>4. Is it possible that you misread my deletion notation? Also,I did not end my AIC with the adjustment noted by Leren since it is my simple-minded understanding that an AIC starts and ends with a strong link even though I think the alteration adds clarity.

I agree with your second comment about an alternate solution but it is my understanding that my solution is valid. I initially noticed the potential MUG and when I found a solution, I posted it which is my tendency even if alternate, maybe simpler solutions are possible.
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Re: Vanhegan Fiendish December 3, 2012

Postby Luke » Fri Dec 07, 2012 1:55 am

Ted, your pattern is a BUG-Lite, not a MUG. No matter what it's called, the only outs worth considering are the 5s locked in box 2. There's nothing more this pattern can tell you other than one 5 is true, and that's something already known.
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Re: Vanhegan Fiendish December 3, 2012

Postby David P Bird » Sat Dec 08, 2012 12:12 am

tlanglet wrote:David, thanks for your feedback. I always appreciate the effort by others to add to my knowledge base.

I do not follow your comment about a missing node in my AIC. My leading statement is that r4c3=2 and I end with r4c4=9 from which I conclude that r4c4<>2. To me, your additional node simply changes the deletion to r3c4<>4. Is it possible that you misread my deletion notation?

Sorry. yes I'm afraid I did – I was comparing different solutions at the time and got my wires crossed.
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