The New Sudoku Players' Forum

by **David P Bird** » Sat May 07, 2016 12:30 pm

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

by **David P Bird** » Tue May 10, 2016 8:54 am

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

Continuation: Show

by **David P Bird** » Wed Jun 08, 2016 9:07 am

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

Code: Select all: *----------------------*----------------------*----------------------* | <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)

by **David P Bird** » Sat Mar 18, 2017 11:19 pm

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
.

by **David P Bird** » Wed Mar 28, 2018 11:23 am

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
.

by **numpl_npm** » Tue Apr 28, 2020 2:07 pm

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.

by **eleven** » Tue Apr 28, 2020 11:12 pm

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).

by **yzfwsf** » Tue Apr 28, 2020 11:43 pm

My solver found this DBJE

: DBJE.png (48.09 KiB) Viewed 2513 times

by **champagne** » Wed Apr 29, 2020 5:50 am

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)

by **numpl_npm** » Wed Apr 29, 2020 4:09 pm

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)

by **eleven** » Wed Apr 29, 2020 7:13 pm

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.

by **champagne** » Wed Apr 29, 2020 9:00 pm

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
....

by **eleven** » Wed Apr 29, 2020 9:31 pm

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 ?

by **champagne** » Thu Apr 30, 2020 2:43 am

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.

by **champagne** » Thu Apr 30, 2020 4:44 am

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.

The New Sudoku Players' Forum

JExocet Compendium

Re: JExocet Compendium

Later Use of JE Inferences

Re: JExocet Compendium

Re: JExocet Compendium Corrections

Extra Class of Almost JE

Maybe this solution is common to JExocet.

Re: JExocet Compendium

Re: JExocet Compendium

Re: JExocet Compendium

Re: JExocet Compendium

Re: JExocet Compendium

Re: JExocet Compendium

Re: JExocet Compendium

Re: JExocet Compendium

Re: JExocet Compendium