The New Sudoku Players' Forum

[daj95376]

Note: I'll be away for a few days, but I'll check on this thread when I return.

Actually, I think we're about done, but the discussion is fun so let's summarize. I started with:

(5=694)b7p167 - (4=194)r5c159 - (94=65)r4c25 - (5=86)r7c27 => -6 r7c1

After discussion I would change it to this:

(6=94)b7p67 - (4=9*)r5c1 - (9=14)r5c59 - (4*9=65)r4c25 - (5=86)r7c27 => -6 r7c1

The only point remaining is the highlighted term, the "parallel move". I seems to me that if both 4 and 9 are true in r5c15, then they must both be false in r4c25, so while I called it 2 separate weak links, it's still a proper weak link between two sets, albeit two houses, b4 and b5, are involved. (Naturally, the fact that the 9 is in b4 and the 4 is in b5 is critical, but there's nothing unusual about using the known locations of candidates in a set.) Would we agree, then, that the chain as modified is a proper AIC, given the star notation?

Whew!!! This discussion has taken on a life of its own. I'm going to forego getting wrapped up in the subsequent discussions to this post.

To answer your question first. No, I do not consider it a proper AIC. Back when everyone started collapsing their notation to be as short as possible, several liberties were taken. The primary liberty was presenting a network in chain-style notation -- with the inclusion of asterisk (*). That's still your situation!

In my initial post on your notation, I presented the following, expanded, interpretation of how your notation should have read:

Code: Select all

(5=6)r7c1 - (6=9)r8c3 - (9=4)r9c1 - (4=9*)r5c1 - (9=1)r5c9 - (1=4)r5c5 - (4=6)r4c5 - (*96=5)r4c2  =>  -5 r7c2


To maintain your elimination, it would be written as:

Code: Select all

(6=9)r8c3 - (9=4)r9c1 - (4=9*)r5c1 - (9=1)r5c9 - (1=4)r5c5 - (4=6)r4c5 - (*96=5)r4c2 - (5=68)r7c27  =>  -6 r7c1


At this point, I added that it might be acceptable to collapse it to:

Code: Select all

(6=94)b7p67 - (4=9*)r5c1 - (9=4)r5c95 - (*94=65)r4c25 - (5=86)r7c27 => -6 r7c1


This is equivalent to your second network above. However, I'm not a fan of collapsing the notation this way. I would prefer:

Code: Select all

(6=94)b7p67 - (4=9*)r5c1 - (9=4)r5c95 - (4=6)r4c5 - (*96=5)r4c2 - (5=68)r7c27  =>  -6 r7c1


_

[SteveG48]

This is equivalent to your second network above. However, I'm not a fan of collapsing the notation this way. I would prefer:

Code: Select all

(6=94)b7p67 - (4=9*)r5c1 - (9=4)r5c95 - (4=6)r4c5 - (*96=5)r4c2 - (5=68)r7c27  =>  -6 r7c1


_

Fair enough. Thanks.

[David P Bird]

One more if you don't mind, David.

We frequently use constructs like (12)r1c12 - (12=34)r1c89. Would it be more technically correct to write (1&2)r1c12 - (1|2=3&4)r1c89 ?

The quick answer is yes as by itself the link (12=34)r1c89 isn't strong.

However your use of the '&' symbols will only help absolute novices and normally can be considered redundant (a point DonM makes). Generally I'm in favour of some redundancy because a) it makes it easier to interpret chains that contain typos, & b) it can make it easier to pick up a chain again having checked the grid, but in this case neither of those considerations applies.

[SteveG48]

Thanks, David.