Vanhegan extreme December 18, 2012

Post puzzles for others to solve here.

Re: Vanhegan extreme December 18, 2012

Postby David P Bird » Tue Feb 05, 2013 11:59 am

While exploring <this puzzle>, I had a flashback to this exchange on notating ANSs.
Code: Select all
 *----------------------*----------------------*----------------------*
 | <6>    289    279    | 2789   <4>    15789  | 125    129    <3>    |
 | 489    <1>    2349   | 23689  2568   689    | 246    <7>    569    |
 | 479    2349   <5>    | 23679  1267   1679   | <8>    12469  169    |
 *----------------------*----------------------*----------------------*
 | 14789  4689   14679  | <5>    678    <2>    | 13467  134689 16789  |
 | <3>    568    167    | 4678   <9>    4678   | 1567   168    <2>    |
 | 45789  24689  24679  | <1>    678    <3>    | 467    4689   56789  |
 *----------------------*----------------------*----------------------*
 | 145    346    <8>    | 2467   12567  1467   | <9>    23     167    |
 | 149    <7>    13469  | 24689  1268   14689  | 23     <5>    168    |
 | <2>    569    169    | 6789   <3>    156789 | 167    168    <4>    |
 *----------------------*----------------------*----------------------*
In box 3 there is an ANS (12569)r1c78,r23c9

One possible inference is (2569=1)r1c78,r23c9 – (1)r3c8 . . .
If we notated this using tuples as (2569=1256) then it would signify that (6) must be true in the ANS to eliminate it from r2c7 & r3c8 which would be invalid. In fact the link 2569 ~ 1256 isn't strong it's weak! They both can't be true, but they could both be false when the cells contain 1569 (the actual case).

However this puzzle contains an ANS chain which is similar to an SK loop*:
(1569=2)r1c78,r23c9 - (2789=4)r1c23,r23c1 - (1459=6)r78c1,r9c23 - (6)r9c78 = (6)r78c9 - Loop
=> r1c4 <> 2, r46c1 <> 4, r9c46 <> 6, r46c9 <> 6

This can also be notated as a longer AHS chain:
(6)r23c9 = (46-2)r2c7,r8c8 = (2)r1c78 - (2)r1c23 = (23-4)r2c3,r3c2 =
(4)r23c1 - (4)r78c1 = (43-6)r7c2,r8c3 = (6)r9c23 - (6)r9c78 = (6)r78c9 - Loop
=> r2c7,r3c8 <> 19, r1c4 <> 2, r2c3,r3c2 <> 9, r46c1 <> 4, r7c2,r8c3 <> 19, r9c46 <> 6, r46c9 <> 6,

So how did the extra eliminations come about?
The 16 edge cells in the 4 boxes have been proved to be a Multi-Sector Naked Set:
MSNS (6)c9,(159)b3,(2)r1,(789)b1,(4)c1,(159)b7,(6)r9,(178)b9 16 candidates/constrained cells
Any external candidate seen by all internal instances of it in one of the containing houses is therefore false.

When the loop is closed all links become conjugate and so can be notated as either weak or strong. Switching them over, the tuple notation can be used:
(1569-1259)r1c78,r23c9 = (2789-4789)r1c23,r23c1 = (1459-1569)r78c1,r9c23 = (6)r9c78 - (6)r78c9 = Loop

This now makes the extra eliminations. The divisions in each of the ANSs are those needed to provide the necessary ongoing links, and because the chain closes, they're proved to be good.

Somewhere in another thread daj said that he used the tuple notation for ANSs for closed loops but didn't say why. This analysis shows why that practice is legitimate.

*It's not an SK loop because the loop only uses single digit links between terms.
David P Bird
2010 Supporter
 
Posts: 1043
Joined: 16 September 2008
Location: Middle England

Re: Vanhegan extreme December 18, 2012

Postby aran » Tue Feb 05, 2013 2:28 pm

David P Bird
Very interesting loop given the role played by the LS in B9, dual-linking on candidate 6 to two ALS.

Using a personal notation, I would present in this way :
ALS in B3 B1 B7 :
{61592} {27894} {41596}
where Restricted Commons are placed at beginning and end of the terms.

The logic is then straightforward :
anything which would lock any of the ALS, including its RCs, locks all of them, as a result eliminating 6 from the LS : impossible, hence false.
The eliminations can then be written down at a stroke.
aran
 
Posts: 334
Joined: 02 March 2007

Re: Vanhegan extreme December 18, 2012

Postby David P Bird » Tue Feb 05, 2013 3:49 pm

Aran, Xsudo users would probably notate this loop as truth and link sets!

My personal favourite is the MSNS notation I gave which achieves the same thing in one line rather than two. The loop path is given by the order the houses are listed. Working through the 8 houses in order, any instance of a listed candidate in a singly covered cell can then be eliminated.
David P Bird
2010 Supporter
 
Posts: 1043
Joined: 16 September 2008
Location: Middle England

Re: Vanhegan extreme December 18, 2012

Postby ronk » Wed Feb 06, 2013 1:02 am

David P Bird wrote:However this puzzle contains an ANS chain which is similar to an SK loop*:
(1569=2)r1c78,r23c9 - (2789=4)r1c23,r23c1 - (1459=6)r78c1,r9c23 - (6)r9c78 = (6)r78c9 - Loop
=> r1c4 <> 2, r46c1 <> 4, r9c46 <> 6, r46c9 <> 6

This can also be notated as a longer AHS chain:
(6)r23c9 = (46-2)r2c7,r8c8 = (2)r1c78 - (2)r1c23 = (23-4)r2c3,r3c2 =
(4)r23c1 - (4)r78c1 = (43-6)r7c2,r8c3 = (6)r9c23 - (6)r9c78 = (6)r78c9 - Loop
=> r2c7,r3c8 <> 19, r1c4 <> 2, r2c3,r3c2 <> 9, r46c1 <> 4, r7c2,r8c3 <> 19, r9c46 <> 6, r46c9 <> 6,

So how did the extra eliminations come about?

In the first case, you seemingly ignore the fact that the ALSs become locked. In the second case, you are listing exclusions that don't exist, e.g., r2c7, r7c2<>19 in cells which see givens r2c2=1 and r7c7=9.
ronk
2012 Supporter
 
Posts: 4764
Joined: 02 November 2005
Location: Southeastern USA

Re: Vanhegan extreme December 18, 2012

Postby daj95376 » Wed Feb 06, 2013 6:58 am

The n-tuples are typically only needed when a loop is involved. Deriving the proper n-tuples is no great mystery. You simply fill in the blank spaces from writing it as a chain.

Code: Select all
 +--------------------------------------------------------------------------------+
 |  6       289     279     |  2789    4       15789   |  125     129     3       |
 |  489     1       2349    |  23689   2568    689     |  246     7       569     |
 |  479     2349    5       |  23679   1267    1679    |  8       12469   169     |
 |--------------------------+--------------------------+--------------------------|
 |  14789   4689    14679   |  5       678     2       |  13467   134689  16789   |
 |  3       568     167     |  4678    9       4678    |  1567    168     2       |
 |  45789   24689   24679   |  1       678     3       |  467     4689    56789   |
 |--------------------------+--------------------------+--------------------------|
 |  145     346     8       |  2467    12567   1467    |  9       23      167     |
 |  149     7       13469   |  24689   1268    14689   |  23      5       168     |
 |  2       569     169     |  6789    3       156789  |  167     168     4       |
 +--------------------------------------------------------------------------------+
 # 171 eliminations remain

 Chain:

 (6   =   2)r23c9,r1c78 - (2   =   4)r1c23,r23c1 - (4   =   6)r78c1,r9c23 - (6)r9c78 = (6)r78c9

 =>  r46c9    <>6

 Loop:

 (6159=1592)r23c9,r1c78 - (2789=7894)r1c23,r23c1 - (4159=1596)r78c1,r9c23 - (6)r9c78 = (6)r78c9 - loop

 =>  r3c8     <>19   from 159 in [b3]
     r2c3,r3c2<>9    from 789 in [b1]
     r8c3     <>19   from 159 in [b7]
     r1c4     <>2    from first  weak link: <2> ... but not <9>
     r46c1    <>4    from second weak link: <4> ... but not <9>
     r9c46    <>6    from third  weak link: <6>
     r46c9    <>6    from loop   weak link: <6>
daj95376
2014 Supporter
 
Posts: 2624
Joined: 15 May 2006

Re: Vanhegan extreme December 18, 2012

Postby David P Bird » Wed Feb 06, 2013 10:51 am

At school my advanced mathematics teacher was in the habit of writing an opening equation and them jumping several steps at a time to write a couple of intermediate stages before finally writing the final derivation. Being a plodder this put me back considerably because if I forgot a derivation, I couldn't go back to basics and work it out for myself when I needed.

I therefore wanted to produce a definitive piece that showed why using the AIC tuple notation for an ANS was acceptable in a closed loop but not in an open chain. Saying that the 'ANSs become locked' or 'it's no great mystery' is equivalent to what my maths teacher did; it presents a fact but fails to properly provide a supporting proof. This then risks that fact being misremembered later on - as demonstrated on page 2 of this thread:

daj95376 wrote:In fact, the first chain should be:
Code: Select all
(1=3)r6c7 - (3268=2681)r5c3589 => r5c7 <> 1

No doubt someone else has explained this before but I'm not aware of any place I can find it.
David P Bird
2010 Supporter
 
Posts: 1043
Joined: 16 September 2008
Location: Middle England

Re: Vanhegan extreme December 18, 2012

Postby aran » Wed Feb 06, 2013 12:53 pm

David P Bird wrote:Aran, Xsudo users would probably notate this loop as truth and link sets!

Code: Select all
*----------------------*----------------------*----------------------*
 | <6>    289    279    | 2789   <4>    15789  | 125    129    <3>    |
 | 489    <1>    2349   | 23689  2568   689    | 246    <7>    569    |
 | 479    2349   <5>    | 23679  1267   1679   | <8>    12469  169    |
 *----------------------*----------------------*----------------------*
 | 14789  4689   14679  | <5>    678    <2>    | 13467  134689 16789  |
 | <3>    568    167    | 4678   <9>    4678   | 1567   168    <2>    |
 | 45789  24689  24679  | <1>    678    <3>    | 467    4689   56789  |
 *----------------------*----------------------*----------------------*
 | 145    346    <8>    | 2467   12567  1467   | <9>    23     167    |
 | 149    <7>    13469  | 24689  1268   14689  | 23     <5>    168    |
 | <2>    569    169    | 6789   <3>    156789 | 167    168    <4>    |
 *----------------------*----------------------*----------------------*

Presented as a base and cover set solution (aka truth/links)
base set : all 16 of the cells involved ie a total of 16 truths being :
B1 : r1c23 c1r23
B3 : r1c78 r23c9
B7 : r78c1 r9c23
B9 : r78c9 r9c78
These can then be covered as follows
- by rows and columns tracking the restricted commons ie
2r1 4c1 6r9 6c9
- by box cover for the remaining candidates ie
B1 789
B3 159
B7 159
B9 178
that is a total of 4+12=16
so producing a much sought after object, a rank 0 structure i: base-cover = 16-16 = rank 0
Hence "anything in the cover not in the base" is eliminated providing all the eliminations already mentioned.
aran
 
Posts: 334
Joined: 02 March 2007

Previous

Return to Puzzles