The New Sudoku Players' Forum

[Leren]

David P Bird wrote: Twin JE Example

Some comments and questions on your example:

=> r8c23 <> 9 (as (5) is locked and one cell must hold a member

I discovered this move myself during the currency of this thread. I call it (for want of a better term) a Tertiary Equivalence - that's a Secondary Equivalence in 2 cells of the same mini-row as a Target cell linked by a non-base SIS candidate. Here 8r2c3==r9c4 (the opposite Target cell) or 5, so you can eliminate all non-common candidates in all 3 cells (except for the SIS candidate 5).

Continuing your post I am with you all the way up to :

if (7)r9c1 then (8) locked + member digit

You seem to be saying that if 8 is in one if r9c56 (I'm OK up to there) then the other cell must hold a member digit ie not 4. Can't see it. Could you explain your reasoning ?

I managed to achieve a similar outcome for the 4's using another Secondary Equivalence move that I also discovered during this thread (perhaps others already know about this but I didn't).

Code: Select all

*--------------------------------------------------------------*
| 9     8     1234   | 7     124   245    | 6     135   123    |
| 5     123   1236   | 9     1268  268    | 4     1378  1237   |
| 126   124   7      | 1248  3     24568  | 125   1589  129    |
|--------------------+--------------------+--------------------|
| 4     1239  12369  | 5     2679  2367   | 8     1379  13679  |
| 136   7     13569  | 348   4689  3468   | 135   2     13469  |
| 236   2359  8      | 234   24679 1      | 35    3579  34679  |
|--------------------+--------------------+--------------------|
| 8     6     123-4  | 1234  5     9      | 7     13    123    |
| 1237  1235  1235   | 6    B127  B237    | 9     4     8      |
|T1237  12349 12349  | 123   12478 23478  |t123   6     5      |
*--------------------------------------------------------------*

At this point I discovered a 3rd Exocet r8c5 r8c6 r9c1 r9c7 1237.

This is a 2 row Exocet that wouldn't allow for "normal" Secondary Equivalences.
However, in r7c123 there are 2 cells that are given or solved as non-base digits. A little thought will reveal that
the third cell in that mini-row must == the Target cell in the other Target box, ie r7c3==r9c7 => r7c3 <> 4 exposing the same single 4 that you did.

From there one of the twin JETs and the 3rd Exocet allowed for Double Exocet fin cell eliminations outside of the Exocet chute followed by stte.

Leren

[David P Bird]

You seem to be saying that if 8 is in one if r9c56 (I'm OK up to there) then the other cell must hold a member digit ie not 4. Can't see it. Could you explain your reasoning ?

The targets are either collinear, when (7) is in r8c1 or diagonal when it's in r9c1. Go back to my post "JE Pattern Eliminations and Inferences" on the previous page and see the section on the diagonal form (you've used this inference yourself before now).

I can see your point talking about secondary and tertiary inferences but I don't see the necessity for that terminology which I consider would present an unnecessary hurdle for newcomers to the pattern to master, and there's enough of those already. I won't comment at length on notation because there is a general unwillingness by contributors to conform to any standard – (yet more hurdles for newcomers). The only thing I would say is that, I use (3)r1c1 =[(12)UR:r12c14]= (4)r2c4 to indicate a strong inference between UR disrupting digits, so using "==" to indicate equivalence would conflict with that, particularly when it's necessary to justify why the equivalence exists.

I pounced on the Multi-Sector Locked set because I was pleased to find it and it provided multiple eliminations to wrap everything up in one step.

David

[blue]

Twin JE Example

(...)

One of the singles created by these eliminations is (4)r7c4 eliminating (4)r36c4
Now this puzzle is an oddity because the S cells for (123) all lie in 3 rows and in each column a single digit is locked in the 3 cells. Consequently the 9 cells must hold 6 instances of the base digits and 3 instances of the locked column digits are therefore a multi-sector locked set.

I think you aren't presenting the right argument for why it's a MSLS.
I think you understand what's going on, but present it in a confusing way.
It isn't because there's a digit locked in the 3 S cells in the column.
It isn't because there's a single digit locked in the 3 S cells in the column.
Locked or not, it's because only one digit other than 123 can go in the 3 cells.

[ Alternatively, it's because the other digits (if any) that need to be placed in the column, are locked, but locked in the cells in the base band, leaving room for only one digit from 123 to be in the base band, and forcing at least two of them into the S cells. ]

Regards,
Blue.

[David P Bird]

Now this puzzle is an oddity because the S cells for (123) all lie in 3 rows and in each column a single digit is locked in the 3 cells. Consequently the 9 cells must hold 6 instances of the base digits and 3 instances of the locked column digits are therefore a multi-sector locked set.

I think you aren't presenting the right argument for why it's a MSLS.
I think you understand what's going on, but present it in a confusing way.
It isn't because there's a digit locked in the 3 S cells in the column.
It isn't because there's a single digit locked in the 3 S cells in the column.
Locked or not, it's because only one digit other than 123 can go in the 3 cells.

[ Alternatively, it's because the other digits (if any) that need to be placed in the column, are locked, but locked in the cells in the base band, leaving room for only one digit from 123 to be in the base band, and forcing at least two of them into the S cells. ]

Perhaps my wording could be improved but it's no so far out IMO

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.
This makes it impossible for any of these member digits to occupy a spot cell.

All this results from the fact that only 3 rows are needed to cover 2 instances of each of the member digits, and the only way that is possible is for each row to cover two of them. That's why I called it an oddity.

The difference in our viewpoints here is that I'm thinking in truths and you're thinking in weak and strong cover sets. A weak cover set in a house yields exactly the same inferences as a strong cover set for the complementary digits. As a trivial example, 3 cells containing (ab) (bc) (ac) can be covered by a weak set for (a) or a strong set for (bc). One tells us only one instance of (a) can be true, and the other tells us (b) and (c) must each be true once, hence (a) can only occupy one of these cells. Therefore, provided no individual cells sets are needed, it's possible to express any rank 0 pattern in terms of strong cover sets for different houses. (This crops up again in the Exotic Patterns thread which is the best place for any continuation of this discussion).

[Leren]

David P Bird Wrote: The targets are either collinear, when (7) is in r8c1 or diagonal when it's in r9c1. Go back to my post "JE Pattern Eliminations and Inferences" on the previous page and see the section on the diagonal form (you've used this inference yourself before now).

OK David thanks, I've got it now - it's the same type of inference as for r8c23 <> 9.

So in your solution you could have said:
...
=> r9c1 <> 8 (non-member in twin cells not equal to the locked digit)
=> r8c23 <> 9 (as (5) is locked and one cell must hold the same digit as r9c4)
=> r9c56 <> 4 (as (8) is locked and one cell must hold the same digit as r8c1)
...

Leren

[blue]

Hi David,

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.

[daj95376]

[Withdrawn: abandoning ship as well.]

_

[David P Bird]

Blue, thanks for the tablets carved out of stone! I now realise that there is another important difference between our approaches, I'm reacting to the state of the grid as it exists at any time, and you're relating to it as it was when the pattern was located.

Regarding what is a cover set, I learnt that fom the early descriptions of fish logic where the rows and columns for fish were described as primary and secondary cover sets or something similar. The essence of the term in my mind was that only the cells that contained the focus digit were covered, or included, in the set. I feel others also have the same impression.

Coming onto the difference between a base digit and a member digit, it was only when I tried to write that summary that I realised that it was necessary to distinguish between candidates in the base cells when the pattern is identified and those that are eventually found to be true in them, and the term 'base digit' has been used ambiguously. However throughout the piece I used 'member digit' and 'true base digit' because I realised I was instigating a change, and omitting the 'true' qualifier would create confusion.

I'll reflect on the points you've made.

David

[David P Bird]

DAJ, Regarding how far the inference set from JE patterns should be extended to cover special cases I considered the points made by Denis and also how I would go about coding them if I were ever to write a solver. I concluded that it was best to include
a) The equivalence checks that can be made between the diagonal mini-lines in the object cell boxes as these can always be made when the pattern is found.
b) The differences between row and column cover set 'spot' cells as I must now call them.

When non-member digits are locked in the diagonal mini-lines we get a bonus which I thought would be easy to check for either by a player or by a program. I therefore took the view they should be considered as part of the equivalence checks.

However when a candidate has been eliminated in the JE band later on, these equivalence checks can be re-run, when I think they should be considered as subsequent deductions.

Should the eliminations r9c56<>4 be considered part of the JE pattern, or as a subsequent deduction?

That's a good question, and IMO it’s a subsequent deduction because two rival cases have to be examined for a common outcome, and notating it that way allows the reader to realise that. We shouldn’t be required to remember what happens in every freak circumstance.

So I've been making judgment calls which I think I can defend. However Leren and Ronk's work on the spot candidates that can be eliminated regardless of whether a member digit is true or not, lies on the border-line. The case is comparatively rare I think, but another factor is that I doubt if these eliminations will be significant, and they will come out in the wash later on anyway.

I'll weigh up any comments that others may have on this subject, but I remind you I don't own JExocet, so what I write amounts to no more than personal opinion.

[Leren]

David P Bird wrote:

1. The base cells, must hold two of the base digits (ab) which will makes them false in JExocet band,crossline1
2. By selection crosslines 2 & 3 can each only hold one of the base digits in the JExocet band - in the target cells
3. This forces the parallel bands,crosslines123 to hold at least 4 truths for (a) & (b)
4. If these cells are restricted to a single truth for (a) they must therefore hold three truths for (b).
5. When no selection for (b) allows this, the base cells can't hold (a) and it can be eliminated from them.

Hi David, I've moved the Exocet discussion from the Dobrichev puzzle thread to here, as I think it was getting in the way of his original intention.

I've produced the following exemplar to indicate what I think is really going on for the case of a JExocet with a digit with a single Truth in the cross rows.

Code: Select all

. . a | . A  . | A  . .   .   May contain a and or b but not important to the proof
. . \ | . \  . | \  . .   \   Must not contain a or b
. . B | . b  . | b  . .   a   Must contain a and is True  if r7c1 = a
------+--------+-------   b   Must contain b and is True  if r7c2 = b
. . \ | . \  . | \  . .   A   May  contain a and is False if r7c1 = a
. . b | . B  . | B  . .   B   May  contain b and is False if r7c2 = b
. . \ | . \  . | \  . .
------+--------+-------
a b . | . AB . | AB . .
. . . | . aB . | \  . .     
. . . | . \  . | aB . .

The JExocet has Base cells r7c12, Target cells r8c5 & r9c7 and cross columns 3, 5 & 7. a has one Truth and b has 2 Truths in the cross columns.

If r7c1 = a and r7c2 = b then a is forced into both Target cells, but more importantly, b is forced into both r3c5 and r3c7 - a contradiction. If either or both of r3c5 and r3c7 does not contain b then the relevant cross column would be void of b - again a contradiction. So the contradiction may be more succinctly stated as : at least one of the two right hand cross columns would be void of b. All of this would also be the case if r7c1 = b and r7c1 = a.

The roles of the 2 b Truth rows may be reversed without affecting the conclusion. Also the 1 a Truth row may coincide with one of the b Truth rows.

For a 3 digit JExocet there is a third digit C with similar properties to B and for a 4 digit JExocet there would be digits C and D with similar properties to B.

Thus a may be removed from r7c12.

A bit further back in this thread you said there was a counterexample to this situation, but I don't think that case conformed to the above exemplar (the digit of interest was not a candidate in both Target cells).

Leren

<edit> clarified wording of contradiction in digit b.

[daj95376]

[Withdrawn: abandoning ship as well.]

_

[Leren]

daj95376 wrote: Since one of <ab> must be true in the base cells, then it must also be true in one of the target cells. This prevents the base cells -- and subsequently the target cells -- from containing <c>.

Sorry to be such a pain Danny but I definitely don't see it this way - the underlined words do not follow.

My wording would be something like :

Since one of <ab> must be true in the base cells, and c must be true in both target cells then <ab> must be void in at least one of the two righthand cross columns ie there is no valid solution to the puzzle. This prevents the base cells -- and subsequently the target cells -- from actually containing <c>.

Also in your bottom diagram only the lefthand S must contain <c>. If it didn't then either the lefthand cross column would be void of c, or there would be a singleton <c> or pointing pair or triple <c> in r123c3 you would already have eliminated <c> from the base cells via basics. The two righthand Ss may contain <c>.

Your corollary reads OK but begs the question of what would happen for a 4 digit JExocet was found with 2 digits, say <c> and <d>, with the degenerate pattern. I haven't checked this but I think the answer is that both <c> and <d> would be forced into both target cells ! - a contradiction - so they may both be removed from the base and Target cells. It may be that such a beast is exceedingly rare or non-existant for some reason.

Leren

[daj95376]

[Withdrawn: abandoning ship as well.

_

[Leren]

daj95376 wrote:

Okay, let's consider my statement case-by-case:

Case #1: if a is true in one base cell and Q is a.
Case #2: if a is true in one base cell and Q is not a, then strong link aQ = aR is applied ... and R=a follows.
Case #3: if a is true in one base cell and R is a.
Case #4: if a is true in one base cell and R is not a, then strong link aQ = aR is applied ... and Q=a follows.

OK let's consider your statement case-by-case. When the Traditional pattern digit a is accompanied by another Traditional pattern digit, say b, in the base cells then all your cases are valid.

However let's see that happens when a is accompanied by a Degenerate pattern digit, c, in the base cells.

I've recast the exemplar diagram from my earlier post so that the digit labels line up with your allocations. Also I've labelled the Target cells as Q and R to line up with your conventions.

Code: Select all

. . c | . C  . | C  . .   <a> = Traditional pattern digit
. . \ | . \  . | \  . .   <c> = Degenerate  pattern digit
. . A | . a  . | a  . .
------+--------+-------
. . \ | . \  . | \  . .
. . a | . A  . | A  . .
. . \ | . \  . | \  . .
------+--------+-------
c a . | . CA . | CA . .
. . . | . c  . | \  . .    r8c5 = Q
. . . | . \  . | c  . .    r9c7 = R

So c is forced into both target cells Q and R and we have a contradiction that a occupies two cells in Row 3 or alternatively it is void in either Column 5 or Column 7.

OK let's consider Case#1 ie for some reason Q = a

Code: Select all

. . c | . C  . | C  . .   <a> = Traditional pattern digit
. . \ | . \  . | \  . .   <c> = Degenerate  pattern digit
. . A | . A  . | a  . .
------+--------+-------
. . \ | . \  . | \  . .
. . a | . A  . | A  . .
. . \ | . \  . | \  . .
------+--------+-------
c a . | . CA . | CA . .
. . . | . a  . | \  . .    r8c5 = Q
. . . | . \  . | c  . .    r9c7 = R

The problem here is that there are no c's in Column 3. Also if r3c7 does not have a then there are no a's in Column 7. Either way there is a contradiction, so Case#1 can't happen when a is accompanied by a Degenerate pattern digit in the base cells.

Now let's consider Case#2 ie Q <> a. If Q = c then we are back to the first exemplar diagram with a contradiction in digit a.

Now let's suppose Q = b ie some other base cell candidate;

Code: Select all

. . c | . C  . | C  . .   <a> = Traditional pattern digit
. . \ | . \  . | \  . .   <c> = Degenerate  pattern digit
. . A | . a  . | a  . .
------+--------+-------
. . \ | . \  . | \  . .
. . a | . A  . | A  . .
. . \ | . \  . | \  . .
------+--------+-------
c a . | . CA . | CA . .
. . . | . b  . | \  . .    r8c5 = Q
. . . | . \  . | c  . .    r9c7 = R

In this case there are no c's in Column 3 and no a's in either Column 3 or Column 5. so Case#2 can't happen when a is accompanied by a Degenerate pattern digit in the base cells.

Cases#3 and #4 follow similar reasoning. Thus in summary :

Code: Select all

Case #1: if a is true in one base cell and Q is a.                                                               Can't happen when a is accompanied by a Degenerate pattern digit c in the base cells.
Case #2: if a is true in one base cell and Q is not a, then strong link aQ = aR is applied ... and R=a follows.  Can't happen when a is accompanied by a Degenerate pattern digit c in the base cells.
Case #3: if a is true in one base cell and R is a.                                                               Can't happen when a is accompanied by a Degenerate pattern digit c in the base cells.
Case #4: if a is true in one base cell and R is not a, then strong link aQ = aR is applied ... and Q=a follows.  Can't happen when a is accompanied by a Degenerate pattern digit c in the base cells.

More importantly I'm trying to fully understand the requirements of the Degenerate pattern. For example can S in r8c3 of your third diagram be a given or solved cell ? I Think so but I'm not sure. Also <c> must be a candidate in both target cells and be forced into both Target cells if c is assumed to be True in a base cell. Again I think so but I'm not sure.

Leren

[daj95376]

[Withdrawn: covering old territory.]

_