Hi David,

David P Bird wrote:Perhaps my wording could be improved but it's no so far out IMO

It isn't all that far out, but it's misleading. The MSLS doesn't appear until the 4's in the S cells in c4 have been cleared, and only one digit other than the base digits can go in those cells.

Maybe my point will be clear if you look at things this way: rather than taking the long route and coming up with the single for 4r7c4 in b8, you can note that just after the exocet eliminations, 8r7c1 is a column single. Placing that, eliminates 8r7c4. At that point, what you said is still true: each column has a single digit locked in the S cells -- 6 in c1, 8 in c4, 5 in c7. Also, at that point, I don't think you could rule out the possiblity that only five S cells contain a 1,2 or 3 -- with c4 containing one digit from 123, and a 4 and an 8 in the other two cells.

There are 9 S cells to fill in total.

In each column 1 cell must contain the digit that's locked in these cells which leaves 6 cells still to populate.

But in each of the 3 rows there are only two digits available to do this – ie each of which is a member of the JE pattern.

Therefore the 3 cells in each row must contain a digit that is locked in the column and two members out of the three in the JE pattern.

The reason that the line in blue is true, is because like you say, the only remaining candidates are for 1, 2 or 3.

This doesn't have anything to do with the fact that a digit was locked in the column, but instead, that there just weren't any other digits with candidates in those cells.

In the 9 cells, we know that

no more than 6 digits from 123, can appear in the S cells, since for each digit, there are only two rows that it can go in. The thing to show, is that it's 6 exactly, rather than something smaller. For that, what you need (in this case anyway), is that there's only one other digit that could go in each column (in the S cells), and it can only go in one cell. The part about "can only go in one cell", indicates that it's a cover set that we're talking about, rather than a base set. For the cover set, it wouldn't matter if the digit was locked or not.

The difference in our viewpoints here is that I'm thinking in truths and you're thinking in weak and strong cover sets.

(...)

You're right that I'm thinking in terms of a base/cover setup.

You're right that there's more than one way to skin a cat.

The "Alternatively ..." note in my earlier post, was refering to the fact that there at least two.

The two possiblities that I had in mind, were these:

1) base sets are the 9 "cell truths", cover sets are the 6 "row links" for digits 1,2,3 (in bands 1 & 2), and the 3 "column links" for the digits other than 123, that can appear in those cells.

2) base sets are the 9 "column truths" for every digit besides the ones used as column links in (1), cover sets are the 6 row links for 123 (again), and "cell links" for every open cell in the base band portion of the S columns.

In the first case, the column links are cover sets, and it doesn't matter if the digit is

locked in the S cells or not. It's the fact that there's only one digit to be "covered", that's important -- only 3 additional cover sets are required, to cover all of the candidates in the S cells.

In the 2nd case, the digits in question, don't even enter into the picture.

One final thing to bitch about: After 48+ pages about exocets and JExocets, it seems that you're now wanting people to use the term "member digits", for what we've been calling "base digits", and to use "base digits" to refer to something else. I have a hard time with that. It may have been a good idea long ago, but at this late date, it seems like it could only lead to confusion. Coming up with another term for digits that are known to occupy one of the base cells, seems like a better idea.

No, two things to bitch about: There is no such thing as a "strong cover set" -- base sets are "strong", cover sets are "weak", period. In your 'abc' example (assuming those are the only cells that contain 'bc' candidates), the way to get that the a's are weakly linked, using the 'bc' digits, is with 'bc' strong links as

base sets and (weak)

cell links as cover sets. With that, if one 'a' was forced, the 'bc' would be a locked pair in the other two cells, and the other 'a' would be eliminated via the relevant cell link. You know

why the 'a's are weakly linked, of course ... an AHS for 'bc' comes to mind here ... I'm just refering to the terminology. Sorry I didn't continue this in the other thread, but it seemed like it would be out of place.

Regards,

Blue.