The New Sudoku Players' Forum

[daj95376]

Which seems to mean that a chain that starts with a weak link isn't properly called an AIC. Does that make it less useful? I've also seen solutions posted that end with a weak link to get a useful elimination. Should we also avoid that?

IIRC, ronk and I had a similar discussion. He convinced me that there wasn't anything wrong with starting and ending with a weak link ... and still calling it an AIC. However, he was talking about a discontinuous loop. The only thing that bothered me was his use of a superflous conclusion. Here's a fictional example: (discontinuous loop does not contain superflous conclusion)

Code: Select all

 AIC w/conclusion:             3r1c1 = 3r3c3 - ... = 4r9c3 - (4=3)r9c9          =>  -3 r9c1

 Discontinuous Loop:   3r9c1 - 3r1c1 = 3r3c3 - ... = 4r9c3 - (4=3)r9c9 - 3r9c1
 conclusion in loop


Multiple eliminations can be handled in the discontinuous loop by changing "3r9c1" to "3r9c1,r1c9" at the start and end of the loop.


Note: There are those who post network expressions and consider them to be AICs because they start and end with a weak link. I don't agree.

[DonM]

...and I'm trying to vary my solutions so that I'm not always starting with a weak link.

Why do it at all?

Because sometimes it's useful for arriving at a solution.

Which seems to mean that a chain that starts with a weak link isn't properly called an AIC. Does that make it less useful? I've also seen solutions posted that end with a weak link to get a useful elimination. Should we also avoid that?

I'm not trying to give you a hard time- I respect anyone trying to come up with their own solutions these days.

While I'm always looking for new, innovative solving methods, I am not familiar with any concept that starting with a weak link allows one to better find an elimination. Also, although discontinuous AICs have, from the beginning, ended with a strong inference, some people here have, inexplicably, started notating the implied weak link that ends in the cell with a discontinuity. I know of no situation where adding a weak link at the end of an AIC gets you a 'useful elimination' not possible without doing so.

This is all about notation and whether one wants to follow the long established rules of AIC notation (including in net-based solutions), the benefit of which means that others can more appreciate one's solutions.

[DonM]

Code: Select all

+--------------+-------------+-----------+
| 2   7    13  | 6   5   139 | 4  8   39 |
| 139 39   8   | 4   2   13  | 7  6   5  |
| 6   5    4   | 39  8   7   | 2  1   39 |
+--------------+-------------+-----------+
| 139 24   7   | 5   19  39  | 8  24  6  |
| 39  6    5   | 2   4   8   | 39 7   1  |
| 8   2349 123 | 379 179 6   | 5  39  24 |
+--------------+-------------+-----------+
| 7   8    29  | 1   3   5   | 6  249 24 |
| 5   23   6   | 79  79  4   | 1  23  8  |
| 4   1    39  | 8   6   2   | 39 5   7  |
+--------------+-------------+-----------+

Fwiw: If the objective is to eliminate (3)r6c8, here's another (rather simple) way to do it:
The AUR (79)r6c45/r8c45 has a simple Type 4 UR pseudocell (13)r6c45 (either 1 or 3 must be true) which along with (123)r6c3 forms a simple 'pseudo' ALS. Thus:

(3=2)r8c8-r8c2=r7c3-pseudo als(2=13)r6c345 -> r6c8<>3 stte

[David P Bird]

An AIC is simply a chain where the links are alternately weak of strong, so strictly speaking it doesn’t matter what type the first one is.

A chain that starts and ends with strong links proves that the first and last nodes can’t both be false and can make multiple eliminations, while one that starts and ends with weak links proves that they can’t both be true and will only make one. It’s therefore sensible to get familiar with using the strong form.

That said, I’m not adverse to using the weak form when the first and last nodes overlap where I think it makes the situation quicker to digest (which I believe to be the prime purpose of a notated string - not to set puzzles). I don’t have an example to hand but say we have a chain that links
(1)r6c13 - (1=pqr)ANS:box4 - [….] – (xyz=1)ANS:row6 - (1)r6c17
Now (1)r6c1 would make the first and last nodes both true so => r6c1 <> 1

DPB

[SteveG48]

I'm not trying to give you a hard time- I respect anyone trying to come up with their own solutions these days.

No problem at all. Solving these puzzles is fun and educational, but it's the occasional discussion like this one that really makes a forum worthwhile.

While I'm always looking for new, innovative solving methods, I am not familiar with any concept that starting with a weak link allows one to better find an elimination.

As Danny has pointed out, I use it quite a bit (more than I like) in networking type solutions. Generally I'll use weak links to two or more different cells to facilitate constructing the chain.