The New Sudoku Players' Forum

[champagne]

While it's gone quiet here...

I've been exploring the possible configurations of the target cells in different types of single and double JEs with paired target cells in different boxes.

Code: Select all


*-------*-------*-------*  *-------*-------*-------*   *-------*-------*-------* 
| B B \ | . t . | . . . |  | B B t | . t . | . . . |   | B B t | . t . | . . . |
| . . t | . \ . | T . . |  | . . \ | . \ . | T . . |   | . . \ | . \ . | \ . . |
| . . . | . T . | \ b b |  | . . . | . T . | \ b b |   | . . . | . T . | T b b |
*-------*-------*-------*  *-------*-------*-------*   *-------*-------*-------*
JE4DD                       JE4DC                       JE4CC       
...

Hi David,

as often, we have slightly different views on the same facts.

I have to revise seriously my code establishing the double exocet pattern, so I studied yours patterns.

To have a wording very close to what I can put in code, I would write that in the following way.

A) JE4 + JE 4 (same four digits, same band, 2 different boxes for the base)

- if one base see the 2 targets of the other JE, then the double exocet is established
this cover all your patterns except JE4CC and JE4DD

-if all targets share the same unit, then the double exocet is established (JE4CC)
-if each base see a target of the other exocet and the 2 other targets share the same unit then the double exocet is established (JE4DD)

B) JE3 + JE3

here, none base can see the 2 other targets, the puzzle would have no solution
and for the same reason, the four targets can not share the same unit.
At the end, I strongly believe that one target must be common to both exocets. That target ins in the thir line and in the third row as in you diagrams.

- it is then enough that one base sees one target of the other exocet to have the double exocet established.

In that case, the 2 exocets share one digit and have one of the 2 others digits as second digit in base
All your diagrams fill that condition.

I see a small window with the 3 target in a mini row but I don't know if it is realistic.


C) JE4 + JE3

we have seen that in a former post.
The rules to apply to the 3 digits base are the same as for a 2*JE4 ;
I'll study several cases to have the relevant rules

[David P Bird]

Hi Champagne,

With these double JEs there will be four base cells with either 3 or 4 target cells between them.
When there are three target cells we know that one digit will be true in both base sets so there are three true digits in total and the common one will occupy the shared target cell.
When there are four target cells, there will be four true digits.

In my view it is far better to identify the number of true digits in the 4 base cells than the number of candidates in the two base sets.

A general proof that no two target cells can hold the same digit for the JE3 and JE4 patterns:
1) It has already been proved that the target cells belonging to a base set must hold different digits
2) The target cells in the same boxes as a base set must hold a true base digit in the other base set.
3) The second true member in each base set must therefore be true in a target cell in the third box.
4) If there is a single target cell in this box, it must hold a digit that is common to both base sets, otherwise two targets must hold different digits that are true in their different base sets.
5) Each target cell must therefore hold a different digit, and together these cells will hold the same combination of digits as the full set of base cells.

This means that as soon as a digit is known to be true in one or other of the base sets it can be eliminated from the cells in sight of all the base cells or all the target cells, and can also be eliminated from the fin cells for its partial fish.

So, I think the way you are accepting double JEs even if there are no eliminations in the target cells is therefore right - these other eliminations might still exist.

I too was interested in the target cells for set A that could be seen by the base cells for Set B. This is why I had to show three different ways each of the JE4 patterns could have their targets positioned. From that I produced the general proof, which shows that it's not important!

Your points don't seem to agree with my findings, but maybe that’s because of language issues. If you work though my proof and then go back to your findings you may see what I mean.

David

[champagne]

A general proof that no two target cells can hold the same digit for the JE3 and JE4 patterns:
1) It has already been proved that the target cells belonging to a base set must hold different digits
2) The target cells in the same boxes as a base set must hold a true base digit in the other base set.
3) The second true member in each base set must therefore be true in a target cell in the third box.
4) If there is a single target cell in this box, it must hold a digit that is common to both base sets, otherwise two targets must hold different digits that are true in their different base sets.
5) Each target cell must therefore hold a different digit, and together these cells will hold the same combination of digits as the full set of base cells.

The proof works well for JEs, and this is why I had no problem in the solving part of my program with JE's seen as double, although I made no checking.
The point 3) of your proof is not granted for exocets "not JEs" in the same band and that was part of my concern.

BTW, there is no contradiction between having a list of patterns or a list of rules. But I accept your point that with JEs it always work.

What about twin JEs?? I did not yet look at that pattern. Did you make a similar analysis

(Normally, I can apply the same rules but...)

meantime I thought a little more about JE3 + JE4. It works exactly as 2*JE4.

[ronk]

meantime I thought a little more about JE3 + JE4. It works exactly as 2*JE4.

Hopefully you apply the exclusions of the JE3 before seeing if the JE3 + JE4 still exists.

[champagne]

meantime I thought a little more about JE3 + JE4. It works exactly as 2*JE4.

Hopefully you apply the exclusions of the JE3 before seeing if the JE3 + JE4 still exists.

good remark, but examples are welcome.

JE3 + JE4 has normally 4 target cells. This does not fit with a double JE3 requiring a common cell.

So we can think (but nothing can be granted in that field) that with such a pattern the JE3 will not destroy the JE4.

[David P Bird]

Champagne, to take your points in a different order.

What about twin JEs?? I did not yet look at that pattern. Did you make a similar analysis
(Normally, I can apply the same rules but...)

I already tried to explain that with twin JEs we don't know which cell in an object pair is the target and which is the companion cell. When that is determined it will turn into one of the regular categories. Until then some of the extra inferences won't be available which are easy to identify. I believe we should know how many target cells there will be though. I suggested that T should be used in place of the unknown C or D suffix in the suggested classification system eg JE4DT.

The point 3) of your proof is not granted for exocets "not JEs" in the same band and that was part of my concern.

BTW, there is no contradiction between having a list of patterns or a list of rules. But I accept your point that with JEs it always work.

Yes, I understand what you say. I don't think sorting out the C, D, and T suffix categories is important when filtering the files. I therefore think all the double Exocets could be filtered into one collection with labels giving the number of targets cells. This label could be extended to identify a) your Exocets with targets outside the JE band, and optionally b) if any of the targets were twinned. You could therefore code according to your rules and I can classify them according to mine.

If you want though those that contained twins could be made into a separate collection (which could include both single and double JEs)

meantime I thought a little more about JE3 + JE4. It works exactly as 2*JE4.

If you mean double Exocets with 3 and 4 base digit sets respectively, just count their target cells to find out how many digits will be true in the two sets of base cells – either 3 with a common digit or 4 when they contain different digits when they will be classed either as JE3 or JE4 under the proposed system.

[champagne]

What about twin JEs?? I did not yet look at that pattern. Did you make a similar analysis
(Normally, I can apply the same rules but...)

I already tried to explain that with twin JEs we don't know which cell in an object pair is the target and which is the companion cell. When that is determined it will turn into one of the regular categories. Until then some of the extra inferences won't be available which are easy to identify.

I think the logic gives room for action with double exocets, but it's my duty to come back with an example

meantime I thought a little more about JE3 + JE4. It works exactly as 2*JE4.

If you mean double Exocets with 3 and 4 base digit sets respectively, just count their target cells to find out how many digits will be true in the two sets of base cells – either 3 with a common digit or 4 when they contain different digits when they will be classed either as JE3 or JE4 under the proposed system.

I said implicitly more. let us assume we would have a common cell in that case, then we have 3 target to fill four digits. this is not any more a double exocet, that name assuming (for action) that all digits are locked in the 2 bases (and in the 2 targets).

The only chance in that case, and this can be the implicit idea of ronk, is that we can prove that this is a 2*JE3 through extra eliminations of the digit in excess,.

[ronk]

this can be the implicit idea of ronk, is that we can prove that this is a 2*JE3 through extra eliminations of the digit in excess,.

No, I was speaking of applying the exclusions of the single JE3 (and the unmentioned single JE4 plus basic techniques, of course) before seeing if a JE3 + JE4 double exists. I do not have an example.

[David P Bird]

Champagne,

Under the new system a JE3 pattern indicates there are 3 target cells (and JE4 would indicate 4 target cells) between 2 pairs of base digits. This doesn't provide the number of digits that exist as candidates in the two base sets – just the number that must be true which is far more important.

If that is not what you are talking about, then I simply don’t understand your point. The only way it makes sense to me is if there are three sets of base cells in different mini-lines. If that case ever arose they could be considered in pairs to find the digits common between set1 and set2, set2 and set3 and set3 and set1 which would show the target cells they would have to occupy. I'll wait until a real example is found (perhaps you already have one ) to study the flood of inferences that would then arise.

If you had anything else in mind please post a hypothetical PM band showing the conditions you want to cover.

David

[champagne]

Champagne,

Under the new system a JE3 pattern indicates there are 3 target cells (and JE4 would indicate 4 target cells) between 2 pairs of base digits. This doesn't provide the number of digits that exist as candidates in the two base sets – just the number that must be true which is far more important.

If that is not what you are talking about, then I simply don’t understand your point. The only way it makes sense to me is if there are three sets of base cells in different mini-lines. If that case ever arose they could be considered in pairs to find the digits common between set1 and set2, set2 and set3 and set3 and set1 which would show the target cells they would have to occupy. I'll wait until a real example is found (perhaps you already have one ) to study the flood of inferences that would then arise.

If you had anything else in mind please post a hypothetical PM band showing the conditions you want to cover.

David

Just to be clear, on my side, the "double exocet" qualifier should be kept for a couple of exocets having all digits locked in the 2 bases.

Then, and only with that condition, you have immediate elimination of the base digits in
- all cells seeing the 2 bases
- all cells seeing all the targets.

If only 3 digits out of four are proven in the bases, the solving power is much lower (close to nil).

[David P Bird]

If only 3 digits out of four are proven in the bases, the solving power is much lower (close to nil).

Not necessarily, look at this example (it's the best I can find quickly):

98.7..6..5...4......3..9...4......5..6.2..7....9..3....1.....67...9...8.....281..

Code: Select all

 *----------------------*----------------------*----------------------*
 | <9>    <8>    124    | <7>    135    125    | <6>    1234   12345  |
 | <5>    27     1267   | 138    <4>    126    | 238    12379  12389  |
 | 1267   247    <3>    | 158    1568   <9>    | 2458   1247   12458  |
 *----------------------*----------------------*----------------------*
 | <4>    237    1278   | 18     16789  167    | 238    <5>    123689 |
 | 138    <6>    158    | <2>    1589   145    | <7>    1349   13489  |
 | 1278   257    <9>    | 1458   15678  <3>    | 248    124    12468  |
 *----------------------*----------------------*----------------------*
 | 28     <1>    28     | 345 t  35 #   45 #   | 9      <6>    <7>    |
 | 36     345 tT 456    | <9>    17     17     | 345-2T <8>    2345   |
 | 37     9      457    | 6      <2>    <8>    | <1>    34 #   345 #  |
 *----------------------*----------------------*----------------------*
           S               S                      S

This has two single JExocets with 3 target cells between them as r8c2 is common to both
(345)r7c56, r8c2, r8c7
(345)r9c89, r8c2, r7c4
These would then be classed as JE3CD and everything I maintained is true and would hold even if the base cells held extra candidates.

[champagne]

If only 3 digits out of four are proven in the bases, the solving power is much lower (close to nil).

Not necessarily, look at this example (it's the best I can find quickly):

98.7..6..5...4......3..9...4......5..6.2..7....9..3....1.....67...9...8.....281..

Code: Select all

 *----------------------*----------------------*----------------------*
 | <9>    <8>    124    | <7>    135    125    | <6>    1234   12345  |
 | <5>    27     1267   | 138    <4>    126    | 238    12379  12389  |
 | 1267   247    <3>    | 158    1568   <9>    | 2458   1247   12458  |
 *----------------------*----------------------*----------------------*
 | <4>    237    1278   | 18     16789  167    | 238    <5>    123689 |
 | 138    <6>    158    | <2>    1589   145    | <7>    1349   13489  |
 | 1278   257    <9>    | 1458   15678  <3>    | 248    124    12468  |
 *----------------------*----------------------*----------------------*
 | 28     <1>    28     | 345 t  35 #   45 #   | 9      <6>    <7>    |
 | 36     345 tT 456    | <9>    17     17     | 345-2T <8>    2345   |
 | 37     9      457    | 6      <2>    <8>    | <1>    34 #   345 #  |
 *----------------------*----------------------*----------------------*
           S               S                      S

This has two single JExocets with 3 target cells between them as r8c2 is common to both
(345)r7c56, r8c2, r8c7
(345)r9c89, r8c2, r7c4
These would then be classed as JE3CD and everything I maintained is true and would hold even if the base cells held extra candidates.

Is it a language problem ???

I write 3 out of 4
your example is 3 out of 3 !!!!

for sure, with 3 out of 3 it works

[David P Bird]

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

[Leren]

Champagne wrote: meantime I thought a little more about JE3 + JE4. It works exactly as 2*JE4.

Hi Champagne,

Try this puzzle from your grey area database:

98.7.......6.5.4.......9...8......3..3......7..5.2.8..5...1...4.4...65....12...6.; ;2;0;match type;40;r7c4 r7c6 r8c3 r9c7 3789;r9c1 r9c2 r8c5 r7c7 379;;;;; ' Double JExocet J3/J4 example

When I treat this as a double JE3 the puzzle solves OK, however if I treat it as a double JE4 I get an incorrect fin cell elimination r3c8 <> 8.

Leren

[daj95376]

I'm taking a stab at ronk's point using Leren's recent post.

Hopefully you apply the exclusions of the JE3 before seeing if the JE3 + JE4 still exists.

No, I was speaking of applying the exclusions of the single JE3 (and the unmentioned single JE4 plus basic techniques, of course) before seeing if a JE3 + JE4 double exists. I do not have an example.

Try this puzzle from your grey area database:

98.7.......6.5.4.......9...8......3..3......7..5.2.8..5...1...4.4...65....12...6.; ;2;0;match type;40;r7c4 r7c6 r8c3 r9c7 3789;r9c1 r9c2 r8c5 r7c7 379;;;;; ' Double JExocet J3/J4 example

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       |
 +--------------------------------------------------------------------------------+
 # 147 eliminations remain

 ### -379- QExocet   Base = r9c12   Target = r8c5==r7c8,r7c7==r8c4


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   | B389     1      B378     |  2379    279     4       |
 | r37-2    4      Q3789-2  |  389     3789    6       |  5       1279    1239    |
 |  37      79      1       |  2       4       5       | R37-9    6       8       |
 +--------------------------------------------------------------------------------+
 # 147 eliminations remain

 ### -3789- QExocet   Base = r7c46   Target = r8c3,r9c7==r8c1


If you apply the eliminations from the JE3 and reduce the grid before searching for a JE3+JE4 double JExocet, then you'll never find the double JExocet because the JE3 eliminations crack the puzzle ... along with subsequent basics.