Vanhegan fiendish November 25, 2012

Post puzzles for others to solve here.

Vanhegan fiendish November 25, 2012

Postby ArkieTech » Mon Nov 26, 2012 7:47 am

Code: Select all
 *-----------*
 |...|.37|4..|
 |...|..6|.1.|
 |72.|...|..5|
 |---+---+---|
 |84.|6..|.7.|
 |...|.2.|...|
 |.5.|..1|.49|
 |---+---+---|
 |5..|...|.38|
 |.1.|4..|...|
 |..2|98.|...|
 *-----------*


Play/Print this puzzle online
dan
User avatar
ArkieTech
 
Posts: 2656
Joined: 29 May 2006
Location: NW Arkansas USA

Re: Vanhegan fiendish November 25, 2012

Postby storm_norm22 » Mon Nov 26, 2012 9:01 am

my one stepper is finned
Last edited by storm_norm22 on Mon Nov 26, 2012 7:20 pm, edited 1 time in total.
Norm
storm_norm22
 
Posts: 59
Joined: 21 November 2012

Re: Vanhegan fiendish November 25, 2012

Postby Leren » Mon Nov 26, 2012 9:04 am

Code: Select all
*--------------------------------------------------------------------------------*
| 16      68      1569     | 25      3       7        | 4      c2-68-9  26       |
| 4       38      35       | 25      9       6        |e-2-378  1      d237      |
| 7       2       369      | 1       4       8        | 369     69      5        |
|--------------------------+--------------------------+--------------------------|
| 8       4       13       | 6       5       9        | 23      7       123      |
| 169     369     7        | 38      2       4        | 3568    568     136      |
| 2       5       36       | 38      7       1        | 368     4       9        |
|--------------------------+--------------------------+--------------------------|
| 5       69      4        | 7       1       2        | 69      3       8        |
| 39      1       8        | 4       6       35       | 2579   b259    a27       |
| 36      7       2        | 9       8       35       | 1       56      4        |
*--------------------------------------------------------------------------------*


S Wing with Strong link

(7=2)r8c9 - r8c8 = r1c8; 7r8c9 - r2c9 = r2c7; (8)r1c8 = r2c7: => r1c8<69>, r2c7<23>

Leren
Leren
 
Posts: 2898
Joined: 03 June 2012

Re: Vanhegan fiendish November 25, 2012

Postby David P Bird » Mon Nov 26, 2012 10:54 am

Leren, your cells produce this continuous AIC loop which gives one more elimination and also makes (2)r4c7 single:

(8)r1c8 = (8-7)r2c7 = (7)r2c9 - (7=2)r8c9 - (2)r8c8 = (2)r1c8 - Loop => r2c7<>23, r8c7 <>2, r1c8 <> 69

For those unfamiliar with AIC loops, they can be broken at any weak link to provide different pairs of strong end links that may provide extra eliminations. Every link in the loop then becomes conjugate.
David P Bird
2010 Supporter
 
Posts: 960
Joined: 16 September 2008
Location: Middle England

Re: Vanhegan fiendish November 25, 2012

Postby ArkieTech » Mon Nov 26, 2012 12:04 pm

Code: Select all
 *-----------------------------------------------------------*
 |a16    8-6   159-6 | 25    3     7     | 4    289-6 d26    |
 | 4     38    35    | 25    9     6     | 2378  1     237   |
 | 7     2     369   | 1     4     8     | 369   69    5     |
 |-------------------+-------------------+-------------------|
 | 8     4     13    | 6     5     9     | 23    7     123   |
 |b169   369   7     | 38    2     4     | 3568  568  c136   |
 | 2     5     36    | 38    7     1     | 368   4     9     |
 |-------------------+-------------------+-------------------|
 | 5     69    4     | 7     1     2     | 69    3     8     |
 | 39    1     8     | 4     6     35    | 2579  259   27    |
 | 36    7     2     | 9     8     35    | 1     56    4     |
 *-----------------------------------------------------------*
m-wing
(6=1)r1c1-r5c1=(1-6)r5c9=6r1c9 => -6r1c238
dan
User avatar
ArkieTech
 
Posts: 2656
Joined: 29 May 2006
Location: NW Arkansas USA

Re: Vanhegan fiendish November 25, 2012

Postby 7b53 » Mon Nov 26, 2012 3:03 pm

Leren wrote:S Wing with Strong link

(7=2)r8c9 - r8c8 = r1c8; 7r8c9 - r2c9 = r2c7; (8)r1c8 = r2c7: => r1c8<69>, r2c7<23>


David P Bird wrote:(8)r1c8 = (8-7)r2c7 = (7)r2c9 - (7=2)r8c9 - (2)r8c8 = (2)r1c8 - Loop => r2c7<>23, r8c7 <>2, r1c8 <> 69


will someone be kind enough to translate these notations to simple words.

for "David P Bird" first notation (8)r1c8. is it a (+8 or -8) for r1c8 ?

please excuse my poor english.
7b53
2012 Supporter
 
Posts: 156
Joined: 01 January 2012
Location: New York

Re: Vanhegan fiendish November 25, 2012

Postby David P Bird » Mon Nov 26, 2012 5:55 pm

7b53, an Alternating Inference Chain (AIC) consists of a series of Boolean nodes (something that is either true or false) that are alternately linked by weak and strong inferences.

A strong link, shown by "=", mean the two nodes cannot both be false
A weak link, shown by "-", mean the two nodes cannot both be true

Code: Select all
*--------------------------------------------------------------------------------*
| 16      68      1569     | 25      3       7        | 4       2689    26       |
| 4       38      35       | 25      9       6        | 2378    1       237      |
| 7       2       369      | 1       4       8        | 369     69      5        |
|--------------------------+--------------------------+--------------------------|
| 8       4       13       | 6       5       9        | 23      7       123      |
| 169     369     7        | 38      2       4        | 3568    568     136      |
| 2       5       36       | 38      7       1        | 368     4       9        |
|--------------------------+--------------------------+--------------------------|
| 5       69      4        | 7       1       2        | 69      3       8        |
| 39      1       8        | 4       6       35       | 2579   259      27       |
| 36      7       2        | 9       8       35       | 1       56      4        |
*--------------------------------------------------------------------------------*

(8)r1c8 = (8-7)r2c7 = (7)r2c9 - (7=2)r8c9 - (2)r8c8 = (2)r1c8 - Loop => r2c7<>23, r8c7 <>2, r1c8 <> 69

(8)r1c8 = (8)r2c7 these nodes cannot both be false (because box 3 wouldn't hold an (8))
(8)r2c7 – (7)r2c7 these nodes cannot both be true (because r2c7 would hold two digits)
(7)r2c7 = (7)r2c9 these nodes cannot both be false (because row 2 wouldn't hold a (7))

The outcome of this part of the chain is that (8)r1c8 and (7)r2c9 cannot both be false (but they could both be true).
Confirm this theorem for yourself first by checking that if (8)r1c1 is false then (7)r2c9 must be true, and then reading the links in the opposite direction, that if (7)r2c9 is false then (8)r1c8 must be true.

We can therefore shorten this part chain to (8)r1c8 = (7)r2c9. Now, by the same logic, the whole length of the chain can be shortened to (8)r1c8 = (2)r1c8, which proves that r1c8 cannot hold any other digit. (See my previous post for the other eliminations.)

For this to work the strong and weak links MUST alternate.

We don't have to assume if any of the terms are true or false to be able to make AIC eliminations.
David P Bird
2010 Supporter
 
Posts: 960
Joined: 16 September 2008
Location: Middle England

Re: Vanhegan fiendish November 25, 2012

Postby Leren » Mon Nov 26, 2012 10:10 pm

ArkieTech wrote: m-wing (6=1)r1c1-r5c1=(1-6)r5c9=6r1c9 => -6r1c238


That M Wing is also an M Ring with an additional elimination r5c9<3>

Leren
Leren
 
Posts: 2898
Joined: 03 June 2012

Re: Vanhegan fiendish November 25, 2012

Postby Marty R. » Tue Nov 27, 2012 12:25 am

Some sort of loop/chain which eliminates 6s in r1. Not sure about the 3rd notation term.

Code: Select all
*--------------------------------------------------------------------------------*
| 16      68      1569     | 25      3       7        | 4       2689    26       |
| 4       38      35       | 25      9       6        | 2378    1       237      |
| 7       2       369      | 1       4       8        | 369     69      5        |
|--------------------------+--------------------------+--------------------------|
| 8       4       13       | 6       5       9        | 23      7       123      |
| 169     369     7        | 38      2       4        | 3568    568     136      |
| 2       5       36       | 38      7       1        | 368     4       9        |
|--------------------------+--------------------------+--------------------------|
| 5       69      4        | 7       1       2        | 69      3       8        |
| 39      1       8        | 4       6       35       | 2579   259      27       |
| 36      7       2        | 9       8       35       | 1       56      4        |
*--------------------------------------------------------------------------------*


(6=2)r1c9-(2=7)r8c9-(27=3)r2c9-(3=8)r2c2-(8=6)r1c2=>r1c138<>6
Marty R.
 
Posts: 1420
Joined: 23 October 2012
Location: Rochester, New York, USA

Re: Vanhegan fiendish November 25, 2012

Postby 7b53 » Tue Nov 27, 2012 5:04 am

appreciate your time and patience. David P Bird

my understanding on Leren's notation, eliminations was base on r8c9 .


rather for your eliminations seems to come directly from the strong linked of digit 8 @ box 3.
if i could translate your notation correctly, its like saying...
if r2c7=8 => r1c8=2
if r2c7<>8 then r1c8 must be an 8 . therefore r1c8 can only be a {2 or 8} resulting r1c8<>69.

am i getting there ? thanks
Last edited by 7b53 on Sat Dec 22, 2012 3:19 am, edited 2 times in total.
7b53
2012 Supporter
 
Posts: 156
Joined: 01 January 2012
Location: New York

Re: Vanhegan fiendish November 25, 2012

Postby David P Bird » Tue Nov 27, 2012 8:25 am

7b53, some solvers look for particular patterns of candidates in cells that they know will give eliminations. To justify these eliminations they must then identify the pattern name and those candidates and cells that fit that pattern. I believe this is the way Leren works.

Perhaps Leren can give you a link to where "S Wings" are defined. I can't as they should have been called "S-Wings" for the forum search engine to be able to find it.

Large families of patterns can be expressed as AICs that usually have 5 or 7 links. So, I (and others) don’t look for any particular one of these but look for chains instead. In my case I won't even know the names of the patterns I find. Of course some chains can get much longer than 7 links but can often be seen as two or more patterns linked together.

You seem to be getting there with AICs but missed the point I made in my earlier response to Leren. Because this chain makes a continuous loop (there is a weak link between the first and last terms) it can be broken at different weak links to give multiple eliminations like this:

(8)r1c8 = (8-7)r2c7 = (7)r2c9 - (7=2)r8c9 - (2)r8c8 = (2)r1c8 => r1c8 <> 69

(7)r2c7 = (7)r2c9 - (7=2)r8c9 - (2)r8c8 = (2-8)r1c8 = (8)r2c7 => r2c7<>23

(2)r8c8 = (2-8)r1c8 = (8-7)r2c7 = (7)r2c9 - (7=2)r8c9 => r8c7 <>2

This is identified by adding " – Loop " at the end of the chain.
David P Bird
2010 Supporter
 
Posts: 960
Joined: 16 September 2008
Location: Middle England


Return to Puzzles