JExocet Pattern Definition

Advanced methods and approaches for solving Sudoku puzzles

Re: JExocet Pattern Defintion

Postby champagne » Thu Jun 13, 2013 8:19 am

Leren wrote:Hi Champagne,

I'm going through your latest list of puzzles - so far this list looks well behaved - I'm on holidays so I only have time for Sudoku in short spells at the moment.

One difference I found was in the following puzzle.

98.7.....6...5......4..86..4....69.....3...7.....2...5.5...18....85....2..1....3.; ;1;1;match type;0;r1c3 r2c3 r4c2 r7c1 237;;;;23;;

I found a double: r8c1 r9c1 r3c2 r4c2 237 / r1c3 r2c3 r4c2 r7c1 237 - this may indicate a bug in your code (or mine!)

I'll try to run thru the full list and report any interesting findings when I have time.

It's hot in Darwin !!

Leren


sorry leren. I took the wrong file. I added the good one (keeping the false) in my post.

regarding that file, it's likely an interim situation in some test runs. I'll look in it.

BTW what is the moisture level in Darwin now
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Re: JExocet Pattern Defintion

Postby David P Bird » Thu Jun 13, 2013 9:48 am

When I started this thread my intention was to concentrate on patterns that could just about be recognised by a determined manual player. However to my dismay the discussions have ranged much wider than that, and I got rather overwhelmed trying to respond to various posts.

In an exchange of PMs with Blue he brought me to realise that I've was hindering what could be a useful discussion here by insisting on the pattern being recognisable. For example using weak and strong cover sets may be more powerful than simple truth counting methods in exploring ways of extending the concept, and the 'absent knowns' only approach, although practical, can be limiting. I therefore leave this thread open to follow whatever direction contributors want to follow.

I took a break, firstly because responding here was frustrating, leaving me with little time for what needed to be done, and secondly because I have little interest in patterns that need solving programs to find. However I have now completed a summary of the various inferences available from the JExocet pattern in the different forms it can take (but I think there may still be gaps) which will be in a separate post following this.

I also have made some slight headway in finding examples illustrating different scenarios, but working manually this is time consuming. As a result of this work I've modified my search method. I still start by looking for a pattern where all the member digits aren't known to occupy any cell in a band, but now when I find a pattern, I widen the search to check for a double JE in the same band that includes other digits.

I'm still hopeful that Champagne will post the results of his screened search runs on his project page to help us.
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Re: JE Pattern Eliminations and Inferences

Postby David P Bird » Thu Jun 13, 2013 9:51 am

JE Pattern Eliminations and Inferences

Terms
Member digit: A digit that appears as a candidate in the base cells
Base digit: A digit that is true in one of the base cells.

Spot Candidate: Any instance of a of a member digit that would prevent it from being able to occupy two S cells (similar to a fin for a fish pattern).


Target Cells in different boxes

Note: The eliminations and inferences detailed here apply to the original JE specification which requires that the base and object cell pairs all should occupy different boxes in the JE band. However cases are possible where the two target cells can also occur in the same box. At the time of writing. (June 2013) these cases haven't been studied to investigate whether the pattern will produce any inferences that aren't subsumed by simpler methods.

Common To All Pattern Forms
Code: Select all
*-----------*-----------*----------*
| #ab #ab . |  .  .  .  | .  .  .  | base cells = (ab)r1c12
|  .   .  . | #a  .  .  | Q  .  .  | target 1 = (a)r2c4
|  .   .  . |  \  .  .  | Q  ax ay | Q = object cells = r23c7 
*-----------*-----------*----------* non-member digits = x,y

This pattern is common to all two target box forms of JE.
1) The JE theorem proves that a digit in a target cell must also occur in the base cells which eliminates instances of it in sight of either.
2) The theorem also proves that object cells pairs will contain different base digits
3) A digit that is true in a target cell will also be true in the base cells and one of the two non-object cells in the remaining unoccupied line in the third box.

Eliminations
1. Any digit that has been eliminated from either the base or target cell pair can be eliminated from the other cell pair.
2. A digit found to be true in a target cell can also be eliminated from the other target cell and the cells in sight of the base cells
3. Any member digit that isn't common to one target cell and the non-object cells in the diagonal mini-line in the other target box can be eliminated from these three cells.

Inferences
1. A weak inference exists between instances of the same member digit in the object cell pairs* - it can't be true in both
2. For a true base digit, a conjugate inference exists between the two pairs of object cells* - it must be true in just one of them
3. An equivalence inference between a digit's truth in the base and the four object cells* - it can't be true in one and false in the other
4. An equivalence inference between instances of a member digit in a target cell and the non object-cells* in the diagonal mini-line in the other target box.
5. Weak inferences between a member digit in the base or target cells and all the spot candidates for the same digit.

* Usually the inference is confined to the target cells, but this covers cases when a target could be either object cell.

Other inferences are now specific to the different forms the JE pattern can take


The Diagonal JE form
Code: Select all
*-----------*-----------*-----------*
| #ab #ab . |  .  .  .  |  .  .  .  | base cells = (ab)r1c12
|  .   .  . | #a  bx bx |  \  .  .  | target 1 = (a)r2c4
|  .   .  . |  \  .  .  | #b  ay ay | target 2 = (b)r3c7 
*-----------*----------*------------* non-base digits = x,y

Here the target cells are on diagonal mini-rows within JE band.
1) From the previous proof, each mini-line containing a target will also contain the true base digit that will occupy the other target.
2) The diagonal target mini-lines can only contain one non-member digit at most.

Inferences
1. If one non-member digit is locked in a target mini-line, all other non-member digits can be eliminated from that mini-line


The Collinear JE form
Code: Select all
*-----------*-----------*------------*
| #ab #ab . |  .  .  .  |  .  .  .   | base cells = (ab)r1c12
|  .   .  . | #a  .  .  | #b  .  .   | target 1 = (a)r2c4
|  .   .  . |  \  bx bx |  \  ay ay  | target 2 = (b)r2c7 
*-----------*-----------*------------* non-base digits = x,y

Now the target cells are on the same line.
1) The target and companion cell mini-lines will each contain one true base digit.

Inferences
1. If two non-member digits are locked in a companion mini-line, all other non-member digits can be eliminated from that mini-line


The Twin JE Form
Code: Select all
*-----------*-----------*--------------* base cells = (ab)r1c12
| #ab #ab . |  .  .  .  |  .    .  .   | target 1 = (a)r2c4
|  .   .  . | #a  .  .  | #bz-x .  .   | target/companion 2 = (b)r23c7
|  .   .  . |  \  .  .  | #bz-y ax ay  | locked non-base digit = (z)r23c7 
*-----------*-----------*--------------* non-base digits = x,y

The twin JE form departs from the others in that a pair of object cells contains a locked non-member digit, and both object cells may contain a mix of member and non-member digits.

1) One cell must eventually hold the locked non-member digit leaving just one capable of holding a member digit, and so complies with the JE pattern requirements.

Eliminations
1. For a twinned object cell pair all non-member digits apart from the one that is locked can be eliminated from the object cells.

Note 1. The object cell pair can be seen as an Almost Hidden Pair consisting of the locked non-base digit and one base digit.
Note 2 There are other AHS sub-patterns that will satisfy the "one member, one non-member" JE pattern requirement, but they are comparatively rare and difficult to spot, and aren't covered here.


Double JEs with 3 Common Cross-Lines

Double JEs occur when two sets of base cells with corresponding target cells in the same band of boxes. All instances seen have the use the same three cross-lines.
Code: Select all
*-------*-------*-------*   *-------*-------*-------*  B = base cell 1st JE
| B B \ | . t . | . . . |   | B B \ | . \ . | . . . |  T = target cells 1st JE
| . . t | . \ . | T . . |   | . . t | . # . | T . . |  b = base cells 2nd JE
| . . . | . T . | \ b b |   | . . . | . \ . | \ b b |  t = target cells 2nd JE   
*-------*-------*-------*   *-------*-------*-------*  # = common target cell   
       4 Target Cells              3 Target Cells

Two JEs can co-reside in the same band, and sharing the same three cross lines as these diagrams illustrate. On the left there are 4 target cells but on the right one target cell is common to both patterns and there are only 3. In the two base boxes the target and companion cells may be in either order so any mix of diagonal and collinear forms is possible. Twin forms are also possible provided that any shared target cell is not involved.

1) As each component JE can taken independently, the target cells must hold true base digits in their respective base cell pairs.
2) The target cells in the same boxes as one 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 are two target cells in the third box they must hold different true digits in their respective base cells, otherwise a single target cell must hold a digit that is true in both pairs of base cells.
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.

Eliminations
1.Any member digit in sight of both base pairs
2 Any member digit in sight of all the target cells

Note 1: As soon it can be shown that a member digit must be true in one or other base set, the spot candidates for that digit can be eliminated. This usually happens as soon as a double JE is found.

Double JEs With 4 or more Cross lines ???

I'm not sure that we have any examples.


Spot Cell Eliminations

To check how many instances of a digit can occur in the S cells the minimum number of rows and/or columns that are necessary to cover all the S cells where it occurs as a candidate. To comply with the pattern requirements this must be no more than two.

External spot cells are those cells in the covering lines that are not S cells, and when one of the covering houses is a cross-line, these will lie in the JE band itself.
Internal spot cells are any S-cell at the intersection of two covering lines.
Code: Select all
B B . | . x . | . . .   B = Base cells 
. . . | . # . | R . .   R = Object cells
. . . | . # . | R . .   # = Spot object cells
------+-------+------     
. . / | . O . | / . .   
x x O | x X x | O x x <   
. . / | . O . | / . .   O = Occupied (maximum)
------+-------+------   / = Excluded S-cell
. . / | . O . | / . .   x = External spot cells
. . / | . O . | / . .   X = Internal spot cell   
. . / | . O . | / . .
Example with covering lines row 5 & column 5.

1) The two target cells within each pair of object cells are the only cells available in the cross lines capable of holding a true base digit in the JE band.
2) When both true base digits are confined to two instances in the S cells, within the 3 cross lines, each digit must occur once in the JE band and twice in the other bands, in the S cells.
3) A member digit confined to one instance in the S cells will therefore invalidate the pattern if it's true in the base cells or reduce the set of members if it's false.
4) If a spot cell for a digit were true, the S cells would be reduced to holding one instance and the JE band would have to hold two, at least one of which would be in sight of the base cells so it could not also be true in the base cells.
5) When a spot cells for a digit is also an object cell, then if it were true it would create a contradiction as it would also have to be a true base digit but there would be insufficient cells in the cross lines to hold the its required 3 instances.
6) If two covering houses intersect, their intersection cell (r5c5 above) is also an internal spot cell as if it were true no other S cell could also be true.

Eliminations
1, Any member digit that has a spot cell covering a target cell can be eliminated from that target cell

Inferences (repeated)
1. Weak inferences between a member digit in the base or target cells and all the spot candidates for the same digit.
2. Any member digit that is restricted to one instance in the S cells won't invalidate the pattern provided it can be shown to be false in the base cells by other means. Until this can be proved the pattern must be reduced to an 'Almost' status.


Note: With the targets in different boxes, it is only necessary to count the minimum number of rows and columns needed to cover the instances of a digit in the S cells. Should the targets lie in the same box, this check would need to be extended to include box covers as well. It may then be possible to select two covering houses in more than one way when more than one internal spot cell may be identifiable.[/quote]

Edit 15th June 2013: Spot Cell Eliminations section: Theory point 3) and corresponding changes under Eliminations & Inferences
Last edited by David P Bird on Sat Jun 15, 2013 3:06 pm, edited 1 time in total.
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Re: JExocet Pattern Defintion

Postby Leren » Thu Jun 13, 2013 12:20 pm

Champagne wrote; I added the good one (keeping the false) in my post.

Hi Champagne, I've quickly run through your latest Exocet puzzles and my results are very close to yours.
Only a few actually use double Exocet methods in the solution, so maybe their rating is a bit low. There are some
minor discrepancies that I'll look into when I have more time.

Darwin has 2 kinds of weather (1) hot and humid or (2) hotter and more humid :)

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Re: JExocet Pattern Defintion

Postby champagne » Thu Jun 13, 2013 4:50 pm

David P Bird wrote:I'm still hopeful that Champagne will post the results of his screened search runs on his project page to help us.


I am working on that, watching among other things what comes out of the first lot.

I have a strong feeling that the most interesting fraction will come from the "grey area". We have seen many puzzles with low ratings where the solving power of exocets was poor.

But the consequence is that the file(s) must be first rated. I am running some rating batches using skmpp, significantly faster than skfr, but I am just at the middle of the first batches. As I said, I could be in a position to restart the extraction of interesting puzzles at the end of June.

Then, I'll publish sub files per areas of rating. BTW, this will ease the uploading work.
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Re: JExocet Pattern Defintion

Postby ronk » Thu Jun 13, 2013 10:20 pm

Leren wrote:
ronk wrote: There is a similar explanation for 6b4 being a 0-rank link yielding r6c1<>6

In terms of David's Exemplar diagram, for JExocet digits with Orthogonal line cover and r1c7 not having the digit
then it may be eliminated from positions marked # and x whether or not it eventually occupies a Base cell.

Code: Select all
Single JExocet

B B . | . . . | \ . .   B = base cells 
x x . | . \ . | T . .   T = target cell
x x . | . # . | \ . .   # = exclusion target cell
------+-------+------     
. . \ | . O . | \ . .   
x x O | x x x | O x x <   
. . \ | . O . | \ . .   O = occupied (maximum)
------+-------+------   \ = does not contain digit of interest
. . \ | . O . | \ . .   x = exclusion cells
. . \ | . O . | \ . .     
. . \ | . O . | \ . .

Not surprisingly it's a continuous loop: (a)r5c3 = (a)r123c3 - (a)r1c12 ; (a)r2c7 = (a)r5c7 - loop
    where " - (a)r1c12 ; (a)r2c7 = " means:
      If r1c12<>a then r2c7<>a and
      if r2c7=a then r1c12=a
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Re: JExocet Pattern Defintion

Postby daj95376 » Fri Jun 14, 2013 4:50 am

Maybe it's time to expand the Eureka notation and add equivalence:

Code: Select all
X: r5c3 = r123c3 - r1c12==r2c7 = r5c7 - loop

In this case, the equivalence is forced by the JExocet being present.
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Re: JExocet Pattern Defintion

Postby David P Bird » Fri Jun 14, 2013 7:57 am

I would write (a)r5c3 = (a)r123c3 - (a#2)JE:r1c12,r2c7 = (a)r5c7 - Loop

The inclusion of JE into the node provides the justification for considering these apparently unconnected cells together, and the #2 makes it plain that the three cells must hold two instances of (a), not one as could otherwise be implied.
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Re: JExocet Pattern Defintion

Postby Leren » Fri Jun 14, 2013 12:40 pm

Hi all , I'd realised what was covered by those last 3 posts but haven't had the time to mention it. I'm happy to follow any agreed convention on equivalence.

BTW Danny, the 1248 Exocet you mentioned a few posts back looks odd. If r9c1 = 1 then the Base does not contain 1 but 1 is forced into Target cell r8c4 by the Group Strong link in Box 8.

Gotta go.

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Re: JExocet Pattern Defintion

Postby daj95376 » Fri Jun 14, 2013 4:05 pm

Leren wrote:BTW Danny, the 1248 Exocet you mentioned a few posts back looks odd. If r9c1 = 1 then the Base does not contain 1 but 1 is forced into Target cell r8c4 by the Group Strong link in Box 8.

You forgot to consider the implications of the Exocet:

Code: Select all
E:(1a=bc)r9c23==(b)r8c4 - (1)r8c4 = (1)r9c46 - loop  =>  r9c1<>1

Congratulations, you've found another elimination associated with the Exocet.

Personal note: The implications of the Exocet being folded into finding eliminations is giving me a headache !!! Especially on trying to decide how to notate the implications.


[Edit: Inadvertantly used JExocet when I should have been using Exocet.]
Last edited by daj95376 on Sat Jun 15, 2013 2:02 pm, edited 1 time in total.
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Re: JExocet Pattern Defintion

Postby daj95376 » Fri Jun 14, 2013 4:42 pm

Now, to consider something even more disturbing about this grid.

Code: Select all
 +-----------------------------------------------------------------------+
 |  9      8      124    |  7      125    1235   |  6      345    12345  |
 |  7      3      124    |  8      1256   1256   |  9      45     1245   |
 |  12     5      6      |  9      4      123    |  1278   378    12378  |
 |-----------------------+-----------------------+-----------------------|
 |  8      124    127    |  5      126    12467  |  3      9      1467   |
 |  1234   6      1237   |  124    8      9      |  147    457    1457   |
 |  14     9      5      |  14     3      67     |  78     2      678    |
 |-----------------------+-----------------------+-----------------------|
 |  2345   7      238    | Q6      25     245    | R248    1      9      |
 |  1246   124    9      | Q1234   7      8      | R5      346    234    |
 |  12456 B124   B128    |  1234   9      1245   |  2478   34678  23478  |
 +-----------------------------------------------------------------------+
 # 118 eliminations remain

We have givens in r7c89 and r8c56. If we consider r7c8=1 and r8c6=8 to be candidates in their respective cells, then we can apply secondary equivalences to get r8c4=1, r7c7=8 -->> r9c3=8, and r9c2=1.

I'm sure there is a flaw in this logic, and I'll work on it just as soon as I get over this headache ! _ :lol: _
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Re: JExocet Pattern Defintion

Postby Leren » Sat Jun 15, 2013 12:21 pm

daj 95376 wrote:
You forgot to consider the implications of the JExocet:

Code: Select allJE:(1a=bc)r9c23==(b)r8c4 - (1)r8c4 = (1)r9c46 - loop => r9c1<>1


Maybe I've gone troppo, but I just don't understand what you are saying. You seem to be assuming that
1 is a valid Jexocet digit but I don't see it. There are 1's in 3 rows of the S columns so the way I'd proceed
is to show that (1) if either of r9c23 = 1 then 1 appears in exactly 1 of r8c4 or r7c7 and (2) if neither of r9c23 = 1 then 1 does not appear in either of
r8c4 or r7c7. This seems to be contingent on r9c1 <> 1.

Can somebody help me out here ?

Leren


Danny,

After a long day on the bus the fog has now cleared and I now see the way to truth and justice. Having proven that a base digit either satisfies S cell rules (for 2 and 4 in your example) or one of a number of simple forcing chain patterns (for 1 and 8 in your example) proving that if the digit is in one of the Base cells then it is in exactly one of the Target cells, then an Exocet is proven to exist.! Full Stop. Exclamation mark! So any base candidate in a cell that can see all instances of that digit in both Base cells and which forces that digit to occupy a Target cell is in conflict with the proven Exocet and can therefore be eliminated. r9c1 <> 1. I still can't understand the first part of your loop notation but that is beside the point, your 1248 Exocet find and r9c1 <> 1 elimination are correct. Regarding your other post about secondary equivalences being driven by solved or given cells in the opposing mini-line, I find that it happens quite often and I've had no incorrect solutions, so as far as I'm concerned the logic is fine and you can put the Panadol away.

Yours humbly, Leren
Last edited by Leren on Sun Jun 16, 2013 9:35 am, edited 1 time in total.
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Re: JExocet Pattern Defintion

Postby David P Bird » Sat Jun 15, 2013 3:19 pm

Leren, I don't pretend to understand DAJ's POV, but checking out the puzzle, it proved to be a counter-example for a misconception I had.

I considered that if the S cells for a member digit could only hold one truth, it could safely be deleted from the base cells. There's a (2346)r8c89 potential base set where (6) is a case in point. However, it's true in the base cells, which invalidates the JE inferences.

I've edited the JE Pattern Inferences post accordingly.
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Twin JExocet Example

Postby David P Bird » Sun Jun 16, 2013 7:51 am

Twin JE Example

This puzzle is an example of using locked non-member digits in the target cell boxes to derive further inferences.

98.7..6..5..9..4....7.3....4..5..8...7.....2...8..1....6..597.....6...48........5
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    |
 *----------------------------*----------------------------*----------------------------*
 | 1238     <6>      1234     | 12348    <5>      <9>      | <7>      # 13     # 123    |
 | # 1237   1235-9   1235-9   | <6>      127      237      | 1239     <4>      <8>      |
 | # 1237-8 12349    12349    | #123-48  1278-4   2378-4   | 1239     6        <5>      |
 *----------------------------*----------------------------*----------------------------*

As (7) is locked in r89c1 there is a Twin pattern in tier 3
(123)JET:r8c89,r8c4,(7)r89c1
=> r9c4 <> 48 (non-member digits in the simple target cell)
=> r9c1 <> 8 (non-member in twin cells not equal to the locked digit)
=> r8c23 <> 9 (as (5) is locked and one cell must hold a member)
=> r9c56 <> 4 (if (7)r8c1 then (78) become locked, if (7)r9c1 then (8) locked + member digit)
The targets will either be collinear or diagonal

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. This allows the spot candidates for each member digit to be eliminated.
Multi-sector Locked Set:(12)r3,(13)r5,(23)r6,(6)c1,(8)c4,(5)c7
=> r3c3689 <> 12, r5c369 <> 13, r6c2589 <> 23
Singles and tuples now finish the solution
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Re: Twin JExocet Example

Postby daj95376 » Sun Jun 16, 2013 11:51 am

[Edit: Withdrawn. Erroneously deduced that r9c1<>7 from the 4-value Exocet. It was late and I was tired !!!]
Last edited by daj95376 on Sun Jun 16, 2013 3:44 pm, edited 1 time in total.
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