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Leren, Here's an off-the-cuff response:

The covering rows that Champagne stipulates could be any two covering houses - rows, columns or, in the case of Denis Berther's extension, boxes.

It's very difficult to formulate a precise definition of a pattern, but one essential is that they should be capable of being identified without having to track any logic streams but just by counting and checking the defined elements exist in a qualifying distribution.

We have the concept that an almost pattern is one that just misses qualifying because one required criterion isn't met, and using that terminology you can bring many of your cases into the fold by embedding it in a chain. Compendium file 9 gives some examples of this.

If more than one chain is needed then it becomes a net-based process.

Definitions are required to make our discussions clear and while any route to solving a puzzle is valid, trying to get some methods to appear more acceptable by bending definitions shouldn't be on the agenda. It's easy enough to add an explanatory qualifier – eg 'memory' chains. 'Reasonableness' is a nice concept but experience shows there can never be any agreement about what is and what isn't reasonable within our community.

I'll review how these points are covered in the compendium, check for conflicting terminology, and consider possible modifications later.

David

This tortuous puzzle is from Andrew Stuart's < Unsolvable Series > which contains a single JExocet. It illustrates how repeated use of the JE's derived inferences (starred steps) can allow a purely linear solution.

...4...8..9...1..37...8..1.....5..4...57......3...6..26....21...1..9....3.2.....9 SudokuWiki Unsolvable 198

Code: Select all

 *-------------------------*-------------------------*-------------------------* Cover Rows
 | 125     256     136     | <4>     2367    3579    | 25679   <8>     567     | .57.
 | 2458    <9>     468     | 256     267     <1>     | 24567   2567    <3>     |
 | <7>     2456    346     | 23569   <8>     359     | 24569   <1>     456     | 45..
 *-------------------------*-------------------------*-------------------------*
 | 1289    2678    1678    | 12389   <5>     389     | 3678    <4>     1678    | ..78
 | 12489   2468    <5>     | <7>     1234    3489    | 368     369     168     | 4..8
 | 1489    <3>     1478    | 189     14      <6>     | 578     579     <2>     |
 *-------------------------*-------------------------*-------------------------*
 | <6>     4578    9       | 358     347     <2>     | <1>     357     4578  T |
 | 458 B   <1>     478 B   | 3568    <9>     34578   | 2345678 23567   45678   |
 | <3>     4578    <2>     | 1568    1467    4578  T | 45678   567     <9>     |
 *-------------------------*-------------------------*-------------------------*
           CLb                               CL1                       CL2

(4578)JExocet2:r8c13,r9c6,r7c9
=> r7c8 <> 3, r9c45 <> 6 ((1) is locked as the non-base digit in both mirror nodes)
=> r9c6 <> 48 (base digits absent from mirror node)
. . Single (6)r8c4
. . (23)HPair:r8c78 => r8c7 <> 4578, r8c8<> 57, r8c6 <> 3
Incompatibility checks are inconclusive but (57)target:r9c6 show that (48) can't both be true in the base cells.

* (5)JE:r9c46,r8c1 = (7)r8c3 - (7)r6c3 = (57-9)r6c78 = (239-6)r258c8 = (6)r9c8 => r9c8 <> 5
(5=2)r2c4 - (2)r2c8 = (23-9)r58c8 = (9-5)r6c8 = (5)r6c7 => r2c7 <> 5
* (5=2)r2c4 - (2)r2c8 = (23-9)r58c8 = (59-7)r6c78 = (7)r6c3 - (7=5)JE:r8cb7,r9b8,r7b9 => r2c1,r2c8 <> 5
. . Single (5)r2c4
* JE: => r7c9 <> 5 (absent from mirror node)
* (5)r8c6 = (5)JE:r9c6,r8c1 => r8c9 <> 5
. . (5)LineBox:c9b3 => r13c7 <> 5
(8)r6c13 = (8)r6c4 - (8)r79c4 = (8)r8c6 - (8)r8c13 = (8)r79c2 => r45c2 <> 8
. . (8)LineBox:c2b7 => r8c13,
* JE => r7c9,r9c4 <> 8 (non-base digit in a target and its mirror)
. . (1)Single:r9c4
* (5)r6c7 = (5)r9c7 - (5=7)JE:r7b9,r8b7 - (7)r6c3 = (57)r6c78 => r6c7 <> 8

This is another SodokuWiki unsolvable puzzle that has a couple of notable features

1..4...89......1....9..1.452...6...7..85..4...7.........18..9.......3....6..2..3. SudokuWiki Unsolvable 203

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 *----------------------*----------------------*----------------------*
 | <1>    235    23567  | <4>    357    2567   | 2367   <8>    <9>    |
 | 345678 23458  234567 | 23679  35789  256789 | <1>    267    236    |
 | 3678   238    <9>    | 2367   378    <1>    | 2367   <4>    <5>    |
 *----------------------*----------------------*----------------------*
 | <2>    13459  345    | 139    <6>    489    | 358    159    <7>    | .3..
 | 369    139    <8>    | <5>    1379   279    | <4>    1269   1236   |
 | 34569  <7>    3456   | 1239   13489  2489   | 23568  12569  12368  | 236.
 *----------------------*----------------------*----------------------*
 | 3457   2345   <1>    | <8>    457    4567   | <9>    2567   246    |
 | 45789  24589  2457   | 1679   14579  <3>    | 25678  12567  12468  | 2.67
 | 45789  <6>    457    | 179    <2>    4579   | 578    <3>    148    | ...7
 *----------------------*----------------------*----------------------*
                 CL1      CL2                    CLb

(2367)JE2:r2c89,r1c3,r3c4
=> r1c3 <> 5 (non-base digit in target)
=> r1c2 <> 5, r3c5 <> 8 (1 is non-base digit in these mirror nodes)
=> r1c3 <> 26, r3c4 <> 67 (base digits missing in mirror cells)
=> r2c89 <> 6 (not in either target)
(5)BoxLine:b1r2 => r2c56 <> 5
(8)BoxLine:b2r2 => r2c12 <> 8
(6)BoxLine:b3c7 => r68c7 <> 6
At this point it seems that another (237)JE2 exists with the second pair of base cells either being r1c23 or r3c45 but both of these would need column1 as a cross line and this would break the rule for two cover houses per base digit.
However as (1) is always the third digit in the mini-lines with the base cells and targets, all mini-lines following the / diagonal direction in the JE band will hold the same three digits (a rope pattern).

Code: Select all

 *----------------------*----------------------*----------------------*
 | <1>    23     37     | <4>    357    2567   | 2367   <8>    <9>    |
 | 34567  2345   234567 | 23679  3789   26789  | <1>    27     23     |
 | 3678   238    <9>    | 23     37     <1>    | 2367   <4>    <5>    |
 *----------------------*----------------------*----------------------*
 | <2>    13459  345    | 139    <6>    489    | 358    159    <7>    |
 | 369    139    <8>    | <5>    1379   279    | <4>    1269   1236   |
 | 34569  <7>    3456   | 1239   13489  2489   | 2358   12569  12368  |
 *----------------------*----------------------*----------------------*
 | 3457   2345   <1>    | <8>    457    4567   | <9>    2567   246    |
 | 45789  24589  2457   | 1679   14579  <3>    | 2578   12567  12468  |
 | 45789  <6>    457    | 179    <2>    4579   | 578    <3>    148    |
 *----------------------*----------------------*----------------------*

Tier 1 repeat pattern 
 *----------------------*----------------------*----------------------*
 | <1>    b      a      | <4>    d5     d5     | c      <8>    <9>    |
 | d45    d45    d45    | c9     c89    c89    | <1>    ab     ab     |
 | c8     c8     <9>    | b      a      <1>    | d      <4>    <5>    |
 *----------------------*----------------------*----------------------*

(2)r13c7 = (2)r2c89 - (2)r2c3 = (2)r8c3 => r8c7 <> 2
(3+257=6)r1c2356 - (6+2789=3)r2c45689 => 1c7,r2c34,r3c12 <> 3 (on the same mini-line diagonal)
(3=7)r3c5 - (7)r5c7 = (7-2)r5c6 = (2)r5c89 - (2)r6c7 = (2)r13c7 - (2=3)r2c9
=> r3c7,r1c5,r2c123 <> 3 (same mini-line diagonal)
Single (3)r2c9
JE: => r4c2, r6c1,5 <> 3 (non-'S' cells in cover rows for true base digit)
(7)r5c5 = (7-2)r5c6 = (2)r5c89 - (2)r6c7 = (2)r13c7 - (2=7)r2c8 - (7)r13c7,r2c4 = (7)XWing:r89c47 - (7)r89c13, = (7-3)r7c1 = (3)r5c1 => r5c5 <> 3
Singles (3)r3c5,r1c3 (mirror cell & target)
This breaks the back of the puzzle

The two notable features are
1 Recognising that 3 digits stay together in the mini-lines (a rope pattern)
2 Using the knowledge that if (7) was true in the base cells then there would be an X-Wing in its 'S' cells

(Other eliminations for (3) and (7) before the killer step are available but prove to be insignificant.)

DPB
.
[Edit] typos reported by Gordon F (thanks)

There were serious flaws in File 03 'JE & SK Loop Combinations' because it included some false conjectures which resulted in a flawed solution to the Easter Monster – please accept my apologies for this. Consequently this file and the corresponding example file have been re-written.

Most of the other files have been updated to correct typos and poorly worded sentences, but with no major changes to content. These were reported to me by Gordon Fick for which I am very grateful. My thanks also go to Leren for extracting JE2/SK Loop puzzles for me to analyse during the file 3 re-write.

David P Bird
.

When a base candidate in a potential JE fails the two cover house requirement, if it is a given in another box in the JE band, there may be another way to prove the JE is true. The purpose of the two cover house rule is to ensure that the true candidates in the base cells must also be true in a target cell, but if the digit is a given, this may be forced, and in that case the failed test result can be ignored.

This produces a class of Almost JEs which is missing from the compendium where digits fail the rule but would be forced true in one target. The other target cell must then hold a compliant digit or perhaps another forced one. Although this could be considered to be an extension of the JE pattern specification, the Almost JE treatment is more convenient because why the digit would be forced into a target cell needs to be described to be able to validate the pattern.

Gordon Fick alerted me to the following puzzle from Champagne's collection which nicely illustrates the considerations.

98.7..6..7..........5.8..9.5...7..4..7...9..3..42.6...2.....1...5..2..8....1..3..

Code: Select all

 *------------------------*----------------------*----------------------*
 | <9>    <8>    123      | <7>    1345   12345  | <6>    1235   1245   |
 | <7>    12346  1236     | 34569  134569 12345  | 2458   1235   12458  |
 | 1346 t 12346  <5>      | 346    <8>    1234   | 247    <9>    1247   | CL1
 *------------------------*----------------------*----------------------*
 | <5>    12369  123689 t | 38     <7>    138    | 289    <4>    12689  | CL2
 | 168    <7>    1268     | 458    145    <9>    | 258    1256   <3>    |
 | 138    139    <4>      | <2>    135    <6>    | 5789   157    15789  |
 *------------------------*----------------------*----------------------*
 | <2>    3469 b 36789    | 345689 34569  34578  | <1>    567    45679  |
 | 1346   <5>    13679    | 3469   <2>    347    | 479    <8>    4679   | CLB
 | 468    469  b 6789     | <1>    4569   4578   | <3>    2567   245679 |
 *------------------------*----------------------*----------------------*
                            36            3                      6

(3469)Almost JE2:r79c2, r3c1, r4c3

In this JE2 pattern (4) and (9) are both givens in the JE band and would fail the two cover house rule for their 'S' cells.
Considering (9) first. The mirror cells in box1 are r12c1. As (7)r2c1 is a non-base digit, (9)r1c1 must be the true one.
This would force (9) to be true in the JE and therefore also into the target at r4c3.
It therefore doesn't matter that (9) doesn't comply, it must be true in a target cell when the JE is true.
Turning to (4), if is a member it must be true in the r3c1 target and (4)r6c2 would be the true mirror cell, so it too needn't comply.
Unlike (9) it is possible for (4) to be false in the JE. (4)r6c3 would then become the false mirror cell and (1268)r5c3 the true one.
As (3) and (6) both comply with the two cover house requirement, one of them would be true in target cell r3c1 if (4) was false.
The JE2 pattern is therefore proved. It must contain (9) plus either the forced (4) or one of (23) both of which comply with the rule.
=> r4c3 = 9 (forced), r3c1 <> 1 (non-member digit in a target cell)

A secondary check then shows (3) can't be a member.
=> r3c1,r7c2 <> 3 (missing from mirror cells, eliminated from base and target cells)

But there's more!

Code: Select all

 *----------------------*----------------------*----------------------*
 | <9>    <8>    123  b | <7>    1345   12345  | <6>    1235   1245   |
 | <7>    12346  1236 b | 34569  134569 12345  | 2458   1235   12458  |
 | 46     12346  <5>    | 346    <8>    1234   | 247    <9>    1247   | CLB
 *----------------------*----------------------*----------------------*
 | <5>    1236 t 9      | 38     <7>    138    | 28     <4>    1268   | CL1
 | 168    <7>    1268   | 458    145    <9>    | 258    1256   <3>    |
 | 138    13     <4>    | <2>    135    <6>    | 5789   157    15789  |
 *----------------------*----------------------*----------------------*
 | <2>    469    3678   | 345689 34569  34578  | <1>    567    45679  |
 | 1346 t <5>    1367   | 3469   <2>    347    | 479    <8>    4679   | CL2
 | 468    469    678    | <1>    4569   4578   | <3>    2567   245679 |
 *----------------------*----------------------*----------------------*
                           36           13                     16

(1236)Almost JE2 r12c3,r4c2,r8c1

In this case (136) comply and (2) doesn't but (2)r7c1 is a given and if it is true in the base cells, will be forced to be true in the r4c2 target.
As the other digits (136) all comply with the two cover house rule, the JE2 pattern is sound.
=> r8c1 <> 4 (non-member in target cell)

Secondary check
=> r8c1 <> 6 (missing from mirror cell r6c2 )

It is now possible to check if a third JE2 is true (1368)r56c1, r3c3, r8c3
But (8) does not occupy either target so cannot be a member (note it is a given at r1c2) so the JE2 is not proven.
However (136) all comply with the cover house rule so there is a strong link (8=136)JE2:r56c1 available.
(8)r56c1 = (136)JE2:r56c1 -[a]- (2)r23c2 = (2)r4c2 - (2=8)r4c7 - (8)r6c79 = (8)r6c1 => r5c3 <> 8 ([a]= non-JE Member in target & true mirror cells)

Although this proves this JE2 must be false, the elimination considerably shortens the rest of the solution.

David PB
.

Code: Select all

 +-------+-------+-------+
 | 9 8 . | 7 . . | 6 . . |
 | 7 . . | . . . | . . . |
 | . . 5 | . 8 . | . 9 . |
 +-------+-------+-------+
 | 5 . . | . 7 . | . 4 . |
 | . 7 . | . . 9 | . . 3 |
 | . . 4 | 2 . 6 | . . . |
 +-------+-------+-------+
 | 2 . . | . . . | 1 . . |
 | . 5 . | . 2 . | . 8 . |
 | . . . | 1 . . | 3 . . |
 +-------+-------+-------+


(u in r12c3) where (u in 136) -> (u in r8c1)
| (1 in r12c3) -> (1 in r8c1)
| (3 in r12c3) & (3 in r4c2) -> (3 in r8c1)
| (3 in r12c3) & (3 in r4c46) -> (3 in r3c46) (3 in r7c5) (3 in r8c1)
| (6 in r2c3) & (6 in r4c2) -> (6 in r8c1)
| (6 in r2c3) & (6 in r4c9) -> (6 in r3c4) (6 in r79c5) (6 in r79c8) (6 in r8c1)

So only (one of 136) in r12c3, and (2 in r12c3)

And (2 in r4c2) (r6c2 in 136) (9 in r4c3)

Then basics to the end.

Nice plan, but (6 in r2c3) & (6 in r4c2) still allows 6r9c1.
So you cannot exclude 16 and 36 to be in r12c3 (where 36 is not possible - both would go to r7c5, but to eliminate 16 you need other digits).

My solver found this DBJE

Code: Select all

9    8     123    |7      1345   12345 |6    1235 1245   
7    12346 1236   |34569  134569 12345 |2458 1235 12458 
1346 12346 5      |346    8      1234  |247  9    1247   
--------------------------------------------------------
5    12369 123689 |38     7      138   |289  4    12689 
168  7     1268   |458    145    9     |258  1256 3     
138  139   4      |2      135    6     |5789 157  15789 
--------------------------------------------------------
2    3469  36789  |345689 34569  34578 |1    567  45679 
1346 5     13679  |3469   2      347   |479  8    4679   
468  469   6789   |1      4569   4578  |3    2567 245679

0;4;2316,r1c3 r2c3 r4c2 r6c2
1;4;2316,r1c3 r2c3 r4c2 r8c1
1;4;9634,r7c2 r9c2 r4c3 r3c1

Same 2 JE on my side plus one non JE

EDIT the non JE exocet => 9r4c3

EDIT adding the abi loop exclusions I get
pairs exclusion for r1c3 r2c3 31 26 36
pairs exclusion for r7c2 r9c2 93 63 64 34 => no 3 in base and target

But this does not give evidence of all eliminations shown by yzfwsf (it adds no '3' in the second JE)

eleven, Thanks for pointing out. I was careless.

Code: Select all

 +-------+-------+-------+
 | 9 8 . | 7 . . | 6 . . |
 | 7 . 6 | . . . | . . . |
 | . . 5 | . 8 . | . 9 . |
 +-------+-------+-------+
 | 5 6 . | . 7 . | . 4 . |
 | . 7 . | . . 9 | . . 3 |
 | . . 4 | 2 . 6 | . . . |
 +-------+-------+-------+
 | 2 . . | . . . | 1 . . |
 | . 5 . | . 2 . | . 8 . |
 | . . . | 1 . . | 3 . . |
 +-------+-------+-------+


(6 in r2c3) & (6 in r4c2) & (1 in r1c3) ->
.. (1 in r6c2) (9 in r4c3) (3 in r6c1) (5 in r6c5)
.. (6 in r5c8) (1 in r4c9) (5 in r5c7) (2 in r4c7) (7 in r6c8)
.. (1 in r2c8) (3 in r1c8) (2 in r9c8) (5 in r7c8) (5 in r9c6)
.. (5 in r2c4) (9 in r2c5) (1 in r5c5) (3 in r7c5)

(6 in r2c3) & (6 in r4c2) & (3 in r1c3) -> (3 in r3c46) (3 in r4c46) (3 in r7c5)

i.e. (6 in r2c3) & (6 in r4c2) -> (3 in r7c5)

Then (6 in r3c4) (6 in r9c5) (6 not in r9c1)

Ok, this is a big net now, but it solves a very hard puzzle.

And there is another JExocet.
What i am missing from yzfwsf and champgne, is a readable explanation for the additional eliminations (i found them from the exocets, but maybe there is a better way).

My explanation to eliminate 3r7c2: then 3 neither can be in 3r4c3 (-> 3r3c1, both target cells) nor in r6c1 (-> 3r12c3 => 3r8c1 from 1st exocet), leaving no 3 in box 4.

The DJE is far from solving it, but the ER rating lowered from 10.7 to 8.4.

eleven wrote:Ok, this is a big net now, but it solves a very hard puzzle.

And there is another JExocet.
What i am missing from yzfwsf and champgne, is a readable explanation for the additional eliminations (i found them from the exocets, but maybe there is a better way).

My explanation to eliminate 3r7c2: then 3 neither can be in 3r4c3 (-> 3r3c1, both target cells) nor in r6c1 (-> 3r12c3 => 3r8c1 from 1st exocet), leaving no 3 in box 4.

The DJE is far from solving it, but the ER rating lowered from 10.7 to 8.4.

Hi eleven,

All eliminations given are more or less directly coming out of the exocets.
I don't know how they came in yzfwsf program, but for the main ones on my side

9 r4c3 is forced by the non JE exocet, but also by the abi pairs eliminations.
<3> in the r79c2 base and targets is forced by the abi eliminations

<1> in r3c1 is a side effect of the non JE exocet. If 1 is not in the base r13c3 then 1 must be in r23c2
....

Sorry for becoming OT.
I am not familiar with abi eliminations. When i read the old posts, they are not very manual solver friendly, as non J exocets are. Do you have a link, how they could be found easier than complex nets ?

eleven wrote:Sorry for becoming OT.
I am not familiar with abi eliminations. When i read the old posts, they are not very manual solver friendly, as non J exocets are. Do you have a link, how they could be found easier than complex nets ?

Hi eleven,

First of all, some links to old comments on this topic

http://forum.enjoysudoku.com/exotic-patterns-a-resume-t30508.html
https://gpenet.pagesperso-orange.fr/UX/Sample8FM/FM_fichiers/FMV2.htm

the second link is to fata morgana detailed solution, where the abi loop leads to the valid pair.

The abi loop is a pattern very common in exocets. It uses the UR threat, so it needs the unique solution.

David AFAIK described the effects in terms of "patterns" and this is the way it should be coded in yzfwsf program.

I'll prepare a separate post to remind how it works using this example.

I don't use Davids JE analysis, but the generic rules, so I can not tell where are the corresponding rules in the JExocet Compendium. What is for sure is that the "abi" loop has been described by a manual solver and is something very easy to apply when the exocet has been seen.

In my code, I try to clear pairs "in the mood of the abi loop", just using the 2 digits of the pair, but in some cases, the contradiction can be trivial.

In most cases, if the Exocet has 3 digits, this typical pattern tells which pair is valid. With 4 digits, the best that you can get as in the second Jexocet, is to clear 4 of the 6 pairs.

Hi again eleven,

I have to apologize for the language abuse, here, we have trivial eliminations of pairs
Here is the stack 1 after direct effect of the JE in r79c2

Code: Select all

9    8     123    | 
7    12346 1236   | 
346t 12346 5      |
-------------------
5    12369 369t   | 
168  7     1268   |   
138  139   4      |
-------------------
2    3469b 36789  | 
1346 5     13679  |   
468  469b  6789   |


I reduce it to the exocet digit to make it simpler

Code: Select all

9    -     3+     | 
-    346+  36+    | 
346t 346+  -      |
-------------------
+    369+  369t   | 
6+   -      6+    |   
3+   39+    4     |
-------------------
-    3469b 369+   | 
346+ -     369+   |   
46+  469b  69+    |


my code says
93pair excluded
63pair excluded
64pair excluded
34pair excluded

in fact, for each of these pair, either we are locked in box 4 or we are locked in column 1

if '4' is in the base it is in r3c1 and the other digit is in r4c3
if the other digit is '3' or '6' then it can not be assigned in column 1

Same for the pair 63 for the digit located in r4c3

with the pair 93, we have no room for the digit '3' in box 4

The remaining possibilities are 94 and 96.

The situation for the other exocet is slightly more complex, but again, does not use the abi loop.