The New Sudoku Players' Forum

[Leren]

Code: Select all

*--------------------------------------------------------------*
| 9     8     34     | 7     36    1234   | 126   125   12356  |
| 12    127   6      | 138   5     1238   | 4     12789 1239   |
| 124   5     347    | 13468 368   9      | 1267  1278  1236   |
|--------------------+--------------------+--------------------|
| 8     127   479    | 14569 679   147    | 1269  3     12569  |
| 1246  3     49     | 14568 689   148    | 1269  12459 7      |
| 146   17    5      | 13469 2     1347   | 8     149   169    |
|--------------------+--------------------+--------------------|
| 5     6     2      | 38    1     378    | 39    79    4      |
| 37    4     8      | 39    79    6      | 5     12    12     |
| 37    9     1      | 2     4     5      | 37    6     8      |
*--------------------------------------------------------------*

Hi Danny,

That's where I get to after applying the JE3 and JE4 eliminations (same as yours as far as I can see) + basics - maybe I'm missing something obvious from there.

The point I was making is that if you treat this position as a 2 * JE4 (3789) then one of 8 eliminations is invalid. If you treat it as a 2 * JE3 (379) then it's OK.

I can use other moves such as finned fish and franken fish but then the puzzle solves without needing Exocet moves at all.

Leren

[champagne]

Code: Select all

*--------------------------------------------------------------*
| 9     8     34     | 7     36    1234   | 126   125   12356  |
| 12    127   6      | 138   5     1238   | 4     12789 1239   |
| 124   5     347    | 13468 368   9      | 1267  1278  1236   |
|--------------------+--------------------+--------------------|
| 8     127   479    | 14569 679   147    | 1269  3     12569  |
| 1246  3     49     | 14568 689   148    | 1269  12459 7      |
| 146   17    5      | 13469 2     1347   | 8     149   169    |
|--------------------+--------------------+--------------------|
| 5     6     2      | 38    1     378    | 39    79    4      |
| 37    4     8      | 39    79    6      | 5     12    12     |
| 37    9     1      | 2     4     5      | 37    6     8      |
*--------------------------------------------------------------*

Hi Danny,

That's where I get to after applying the JE3 and JE4 eliminations (same as yours as far as I can see) + basics - maybe I'm missing something obvious from there.

The point I was making is that if you treat this position as a 2 * JE4 (3789) then one of 8 eliminations is invalid. If you treat it as a 2 * JE3 (379) then it's OK.

I can use other moves such as finned fish and franken fish but then the puzzle solves without needing Exocet moves at all.

Leren

Hi leren,

I checked the status starting from Danny's PM (Danny raises another interesting point, but i'll see that later to-day).

For me it is a double exocet (JE3+JE4) and the eliminations coming out of the double exocet logic are correct.

What I don't see is "how do you come to r3c8 <>8"

can you give more details

[Leren]

Code: Select all

*--------------------------------------------------------------------------------*
| 9       8       34       | 7       36      124-3    | 126     125     1256-3   |
| 12      127     6        | 138     5       1238     | 4       12789   1239     |
| 124     5       347      | 1468-3  368     9        | 1267    128-7   126-3    |
|--------------------------+--------------------------+--------------------------|
| 8       12-7    479      | 1456-9  679     14-7     | 1269    3       1256-9   |
| 1246    3       49       | 14568-9 689     148      | 1269    1245-9  7        |
| 146     17      5        | 13469   2       1347     | 8       149     169      |
|--------------------------+--------------------------+--------------------------|
| 5       6       2        | 38      1       378      | 39      79      4        |
| 37      4       8        | 39      79      6        | 5       12      12       |
| 37      9       1        | 2       4       5        | 37      6       8        |
*--------------------------------------------------------------------------------*

My apologies, I was incorrectly applying fin cell eliminations to digit 8 in rows 5.
As 8 does not appear in the S cells in Column 3 these were not valid.

The above PM shows what I now believe to be the correct double JE4 fin cell eliminations for this puzzle.

I still don't see how the puzzle can be solved by basics from this position.

Leren

[David P Bird]

Leren, your example is an excellent illustration of what happens when there are known cells in the mini-lines containing diagonal targets - thank you.

As DAJ says, only the single JE is required

(379)JE2D:r9c12, r8c5, r7c7 => r8c5 <> 8, r7c7 <> 2
Because this has diagonal targets the true base digits must occupy r8b8 & r7b9 both of which have non-base givens
[JED](379#2)r8c45 => r8c4 <> 8
[JED](379#2)r7c78 => r7c8 <> 2
HiddenSingles (2)r7c3,(8)r8c3
HiddenPair(12)r8c89 => r8c8 <> 79, r8c9 <> 39
Hidden Single(9)r9c2
Naked pair(37)r89c1
Because (9)r9c2 is a base cell the (9) fin cells can be eliminated => r4c249,r5c48 <> 9
(9) Must now be true in one of the target cells
(9)r7c7 =[JE]= (9-7)r8c5 = (7)r7c6 => r7c7 <> 7
(7)SwordFish:r268c268 => r3c8,r4c26 <> 7
(9)r7c7 =[JE]= (9-3)r8c5 = (3)r9c7 - (3)r9c1 = (3)r8c1 => r8c5 <> 3
(3)box.line c5b2 = > r12c6,r23c4 <> 3
sste

If you opt to use the double Exocet points that should become obvious with experience are:
There are 4 target cells so therefore 4 true base digits
Hence fin cell eliminations can be made immediately
Only one taget cell holds 8 so that can be made true
r7c3 sees all base cells so reduces to 2

David

[champagne]

Champagne, I assure you this exchange is just as frustrating to me as it is to you.

Sure the example I posted only had 3 candidates in each pair of base cells but it would have worked the same if we had a few extra candidates in those cells as well, as I said. Then it wouldn't be clear which digits were true in the four base cells but we would know there would be three of them not four, and the one that was common to both pairs of base cells would be true in r8c2.

Every bit of information helps with tough puzzles.

David

sorry David, but now, I am completely lost.
In the logic of an exocet, the target is defined in reference to the base. I can't see any room for "extra digits" in it.

What I think, but I have no proof, is that in a valid sudoku we will never find 2 JEs in the same band, JE3 and JE4 with targets sharing one cell.

[Leren]

David P Bird wrote: As DAJ says, only the single JE is required

You're quite right David, after some bug fixes the puzzle solved via the 379 JE and basics as Danny said. In fact the fin
cell eliminations you mentioned weren't even required, although I've taken that on board for the future. I think you'll
find that the 7 Swordfish was r267 c268.

BTW do you have an example of a JE with an orthogonal lines digit that is opposite a target cell, I'm keen to try out that
move but haven't found one yet.

Leren

[ronk]

Because (9)r9c2 is a base cell the (9) fin cells can be eliminated => r4c249,r5c48 <> 9

I see nothing in the pattern that justifies the use of the "fin cell" term for these cells. Would someone please clarify?

[David P Bird]

sorry David, but now, I am completely lost.
In the logic of an exocet, the target is defined in reference to the base. I can't see any room for "extra digits" in it.

Champagne, Oh dear! It seems that you can't see the wood for the trees.

Clearly only two digits will be true in the base cells, but they will also contain a number of other candidate digits which must be proved to be false. Those are the 'extra digits' I was referring to.

Two examples for you:

Code: Select all

*-------------*-------------*--------------*   
| abc abd \   | .   \   .   | .    .   .   |   
| .   .   abx | .   acy .   | bcz  .   .   |   
| .   .   .   | .   \   .   | \    acd abd | 
*-------------*-------------*--------------*

Here we have
(abcd)JEC:r1c12,r3c5,r3c7 => r3c5 <> y, r3c7 <> z
(abcd)JEC:r3c89,r2c3,r2c7 => r3c3 <> x, r3c5 <> y

The two JEs have three target cells the one at r3c5 being common target to both. As it contains (ac), one of these digits must be true in both r1c12 and r3c89. This in turn means one of the four digits must be false in both pairs of base cells.

Now I've made it easy by making sure that digit (d) doesn't appear in any target so that is the one that’s false and can be eliminated from both sets of base cells.

Therefore the theorem that for double JEs every target cell must hold a different digit is potentially very useful.

Now note that if (d) didn't appear in r1c12 you would be calling that a JE3 and the one in r3c89 a JE4 which doesn’t tell us anything

Code: Select all

*-------------*-------------*--------------*   
| abc ab  \   | .   .  \    | .    .   .   |   
| .   .   abx | ady .  acy  | bcz  .   .   |   
| .   .   .   | \   .   .   | \    acd abd | 
*-------------*-------------*--------------*

Here we have a similar situation
(abc)JEC:r1c12, r3c4,r3c7 => r3c4 <> y, r3c7 <> z
(abcd)JEC:r3c89, r3c3, r3c6 => r3c3 <> x, r3c4 <> y

There are 4 targets so the base cells must hold different digits. (d) must therefore be true in r3c89 and also & r3c4 as that's the only target where it occurs. Note if r3c89 didn't contain (a) this would still work and (a)r3c3 could be eliminated.

So a double JE can resolve to either having different digits in the 4 base cells or can have one digit unique to each base set and one common to both.

That's why I'm saying that when we categorise JEs, the number of different digits that must be true in the base cells is more useful than having the candidate count. Note we get the candidate count easily by checking the candidate list at the start the way I'm notating them.

If you still don't get it then I give up!

David

[champagne]

Here are the individual JExocet eliminations (as I see them):

Code: Select all

 +--------------------------------------------------------------------------------+
 |  9       8       234     |  7       36      1234    |  1236    125     12356   |
 |  1237    127     6       |  138     5       1238    |  4       12789   1239    |
 |  12347   5       2347    |  13468   368     9       |  12367   1278    1236    |
 |--------------------------+--------------------------+--------------------------|
 |  8       1279    2479    |  14569   679     147     |  1269    3       12569   |
 |  1246    3       249     |  145689  689     148     |  1269    12459   7       |
 |  1467    179     5       |  13469   2       1347    |  8       149     169     |
 |--------------------------+--------------------------+--------------------------|
 |  5       6       23789   |  389    Q1       378     | R39-27  q79-2    4       |
 |  237     4       23789   | r39-8   Q79-38   6       | R5       1279    1239    |
 | B37     B79      1       |  2       4       5       |  379     6       8       |
 +--------------------------------------------------------------------------------+
 

Hi Danny,

In my view, this is more than just direct eliminations from a JE, but you are using the word Qexocet.

In fact, looking for all exocets in band for that puzzle, my solver finds

r7c4 r7c6 r8c1 r8c3
r7c4 r7c6 r8c3 r9c7
r9c1 r9c2 r7c7 r7c8
r9c1 r9c2 r7c7 r8c5
r9c1 r9c2 r8c4 r8c5

and we have the properties you apply. (r7c8==r8c5 r7c7==r8c4)
Eliminations made are very close to what can be done using the double exocet.

With the JE4, we have also r8c1 == r9c7

[David P Bird]

Because (9)r9c2 is a base cell the (9) fin cells can be eliminated => r4c249,r5c48 <> 9

I see nothing in the pattern that justifies the use of the "fin cell" term for these cells. Would someone please clarify?

< Here > I called the S cells in the JExocet definition a "Partial Fish" where two cover sets are sufficient to cover all the instances of a subject digit in the 3 part columns. If only row cover sets are used the pattern is 2/3rds if a Swordfish. We can define a partial fish fin to be any cell which occupied by the subject digit would prevent the PF from holding a requisite number of truths. This is in keeping with the use of the term in respect to other fish forms.

The concept of considering the S cells this way isn't strange to you as witnessed by your post <here> So it appears when you say you see nothing in the pattern that justifies the use of the "fin cell" you are back to your old tricks again.

If you want to help rather than criticise, what term would you suggest instead that would keep our descriptions succinct?

AFAIAC you've had your three strikes with me several times over and I don't welcome such non-constructive contributions from you.

[David P Bird]

BTW do you have an example of a JE with an orthogonal lines digit that is opposite a target cell, I'm keen to try out that move but haven't found one yet.

Sorry I'm afraid not.

David

[ronk]

Because (9)r9c2 is a base cell the (9) fin cells can be eliminated => r4c249,r5c48 <> 9

I see nothing in the pattern that justifies the use of the "fin cell" term for these cells. Would someone please clarify?

< Here > I called the S cells in the JExocet definition a "Partial Fish" where two cover sets are sufficient to cover all the instances of a subject digit in the 3 part columns. If only row cover sets are used the pattern is 2/3rds if a Swordfish. We can define a partial fish fin to be any cell which occupied by the subject digit would prevent the PF from holding a requisite number of truths. This is in keeping with the use of the term in respect to other fish forms.

Other uses of the term have fin cells in strong inference sets. You are using the term for cells in weak inference sets outside their intersection with strong inference sets. Big difference IMO.

The concept of considering the S cells this way isn't strange to you as witnessed by your post <here> So it appears when you say you see nothing in the pattern that justifies the use of the "fin cell" you are back to your old tricks again.

If you want to help rather than criticise, what term would you suggest instead that would keep our descriptions succinct?

A misunderstanding on your part does not equate to a "trick" on my part. Take another look at the post I referenced and you will see the strong/weak inferences as I noted above. But as you hint, these eliminations have no name AFAIK. Only partially in jest, maybe we should call them barnacles.

AFAIAC you've had your three strikes with me several times over and I don't welcome such non-constructive contributions from you.

Ad hominem AFAIAC.

[blue]

Hi David,

[ I was writing this, while ronk made his last post -- I see it on the preview page, but haven't reviewed in detail it yet. ]

Because (9)r9c2 is a base cell the (9) fin cells can be eliminated => r4c249,r5c48 <> 9

I see nothing in the pattern that justifies the use of the "fin cell" term for these cells. Would someone please clarify?

< Here > I called the S cells in the JExocet definition a "Partial Fish" where two cover sets are sufficient to cover all the instances of a subject digit in the 3 part columns. If only row cover sets are used the pattern is 2/3rds if a Swordfish. We can define a partial fish fin to be any cell which occupied by the subject digit would prevent the PF from holding a requisite number of truths. This is in keeping with the use of the term in respect to other fish forms.

The concept of considering the S cells this way isn't strange to you as witnessed by your post <here> So it appears when you say you see nothing in the pattern that justifies the use of the "fin cell" you are back to your old tricks again.

If you want to help rather than criticise, what term would you suggest instead that would keep our descriptions succinct?

AFAIAC you've had your three strikes with me several times over and I don't welcome such non-constructive contributions from you.

I agree very much with ronk.

From what I remember, the "Ultimate Fish Guide" term for the candidates for digit 9, in r45, outside of c357, is "PE candidate(s)" -- PE standing for "potential elimination". The term "Fin candidate" (my term?), or "Fin cell" (UFG), always refers to a candidate a base set (the base sets being the "column truths" in this case). Candidates in cover sets, are "PE" candidates.
These aren't "fish" in the usual sense, and so what to call them is debatable, but for another term, how about like "PE candidates in the covering rows (for digit 9)". It's kind of long to write, and something shorter would be nice. The word "fin" should not be used, however ... and not just "IMO".

Hello ronk,
I, for one, am very happy to see you contributing again.
Please don't be put off by David's comments.

For David: Cut ronk some slack.
I know I've found myself in a situations where I just don't bother posting in a thread, because I don't like the feeling that I'm always pointing out an error or inconsistency in someone else's post. I don't like it. I'm sure that ronk was only trying to keep things on a "sane" level.

Best Regards,
Blue.

[blue]

Hi David,

Code: Select all

+--------------------------------------------------------------------------------+
|  9       8       234     |  7       36      1234    |  1236    125     12356   |
|  1237    127     6       |  138     5       1238    |  4       12789   1239    |
|  12347   5       2347    |  13468   368     9       |  12367   1278    1236    |
|--------------------------+--------------------------+--------------------------|
|  8       1279    2479    |  14569   679     147     |  1269    3       12569   |
|  1246    3       249     |  145689  689     148     |  1269    12459   7       |
|  1467    179     5       |  13469   2       1347    |  8       149     169     |
|--------------------------+--------------------------+--------------------------|
|  5       6       23789   |  389     1       378     |  2379    279     4       |
|  237     4       23789   |  389     3789    6       |  5       1279    1239    |
|  37      79      1       |  2       4       5       |  379     6       8       |
+--------------------------------------------------------------------------------+

(379)JE2D:r9c12, r8c5, r7c7 => r8c5 <> 8, r7c7 <> 2
Because this has diagonal targets the true base digits must occupy r8b8 & r7b9 both of which have non-base givens
[JED](379#2)r8c45 => r8c4 <> 8
[JED](379#2)r7c78 => r7c8 <> 2
HiddenSingles (2)r7c3,(8)r8c3
HiddenPair(12)r8c89 => r8c8 <> 79, r8c9 <> 39
Hidden Single(9)r9c2
Naked pair(37)r89c1
Because (9)r9c2 is a base cell the (9) fin cells can be eliminated => r4c249,r5c48 <> 9
(9) Must now be true in one of the target cells
(9)r7c7 =[JE]= (9-7)r8c5 = (7)r7c6 => r7c7 <> 7
(7)SwordFish:r268c268 => r3c8,r4c26 <> 7
(9)r7c7 =[JE]= (9-3)r8c5 = (3)r9c7 - (3)r9c1 = (3)r8c1 => r8c5 <> 3
(3)box.line c5b2 = > r12c6,r23c4 <> 3
sste

There's an error near the end.

(9)r7c7 =[JE]= (9-3)r8c5 = (3)r9c7 - (3)r9c1 = (3)r8c1 => r8c5 <> 3
--> should be
(9)r8c5 =[JE]= (9-3)r7c7 = (3)r9c7 - (3)r9c1 = (3)r8c1 => r8c5 <> 3

Also, for these two lines:

Because (9)r9c2 is a base cell the (9) fin cells can be eliminated => r4c249,r5c48 <> 9
(9) Must now be true in one of the target cells

I think the order is wrong. Since 9r9c2 is/was a base cell, and it has been filled with a '9', we know that it must occur in one of the target cells. Only with that, can we eliminate the 9's in r45, outside the S columns.

Danny said, though, that the puzzle could be solved using the JE3 and "simple techniques".
I wouldn't put these lines in that category:

(9)r7c7 =[JE]= (9-7)r8c5 = (7)r7c6 => r7c7 <> 7
(9)r8c5 =[JE]= (9-3)r7c7 = (3)r9c7 - (3)r9c1 = (3)r8c1 => r8c5 <> 3

Regards,
Blue.

Added: the elimnation for the 9's, is also in the not in the category of a "simple technique", but (on the other hand, it) can be seen as following from the JExocet . To go with the last line in the next post, it's something like this:

XWing:9r45c35 = 9r8c5 - (3 or 7)r8c5 =[JE*9]= (3 or 7)r7c7 - 9r7c7 = XWing:9r45c57 => r4c49<>9,r5c48<>9

Here, "=[JE*9]=" (maybe not the best symbology) is refering to the fact that we have a (one time) 379 JExocet, where we know that one of the base cells must (and in this case, does in fact) hold a 9, and the other must hold a 3 or a 7. The elimination for the 9's doesn't actually follow from the partial fish knowledge and the (9r8c5 = 9r7c7) SIS alone, but requires the full knowledge from the JExocet.

Note: If we didn't have the 9r9c2 as the digit in a filled base cell, and instead, only knew that one base cell must contain a 9, we would need to use that knowledge to eliminate the 9's in r89c3, before the first strong link would be valid.

[blue]

These aren't "fish" in the usual sense, and so what to call them is debatable, but for another term, how about like "PE candidates in the covering rows (for digit 9)". It's kind of long to write, and something shorter would be nice. The word "fin" should not be used, however ... and not just "IMO".

What about just calling them PE candidates ?
-- with the idea being that they are "potential eliminations" for the JExocet pattern (rather than for a fish pattern).
The idea would be that as soon as you can prove that the target cells must contain a particular digit, it's PE candidates can be eliminated.

Blue.

Added: After making the addition to the previous post.
The "last line" that I refered to in that post, is the last line above.

To clarify the situtation: refering to the previous post ... it isn't so much that we know that "at least one" target cell, must contain a particular digit, but instead that we know that a base cell must contain a particular digit. In a way, knowledge of one, implies knowledge of the other, and so maybe I'm more concerned with the details than I should be.

The point I wanted to make, was this, though: When we have a JExocet (or an exocet, "in general"), then we (can say that) we know that candidates in the target cells for the same digit, are "weakly linked". The logic behind that (in brief), is that if a base digit candidate were true in one target, then via a long deduction, it would need to be true in a base cell as well. Then since the base cells can see each other, and so, can't contain the same digit (in a solution), the other base cell (in any solution) would contain a different digit, and it would be forced (for the JExocet case, via the partial fish logic) into the second target cell ... eliminatiing the candidate for the original digit, in that cell.

Summarizing the paragraph above: candidates for the same base digit, in the target cells, are weakly linked (regardless of what we know about the presence of that digit in a base cell).

In order to make the kind of X-wing chain from the previous post, "work", though -- we need to know that the digit, definitely must appear in one of the base cells. At that point, we can eliminate the candidates for that digit, in the other cells in the box that contains the base cells -- in particular, in the cells in the S column. It's (only) at that point, that the X-wing chain (from the previous post) becomes valid.

So, to summarize a bit farther: what I said in the first part of this post -- that we can eliminate the PE candidates, once we know that a base digit, must occupy one of the target cells -- is a bit of a stretch. More correct, is that once we know that a particular digit must occupy one of the base cells -- and given that the candidates for that digit in the target cells, are weakly linked -- then we can eliminate the PE candidates, via the kind of X-wing chain that I presented. Again, however, once we know one fact (base or target), the other is implied.

A question then: has anyone ever used the fact that candidates for a base digit, in the target cells, are weakly linked )regardless of what we know about whether the digit must occur in a base cell) ... to produce a "chain-based" elimination ?