## JExocet Compendium

Advanced methods and approaches for solving Sudoku puzzles

### Re: JExocet Compendium

sultan vinegar wrote:With my (slightly more SK-loop like cover sets in b45) I miss out on the elimination of 4r4c8 though. As you say the eliminations are very interesting. Here it seems that there is the SK-loop like AMSLS in boxes 4,5,7,8 and the ALS in r12c7 which get stitched together and magically the truths balance because of the cells in r47c7. Is this just a freaky coincidence or is there a reason for it, hidden in the puzzle symmetries somewhere?

Platinum Blonde contains an almost-sk-loop logic set with at least four equivalents. The equivalents aren't as neat and pretty as with a true sk-loop, but the same principles hold.
Code: Select all
`After hidden single r9c3=5000000012000000003002300400001800005060070800000009000008500000900040500475006000  Platinum Blonde unmorphed Plat [22,251] 58 Candidates, 0-rank all-cell truths     19 Truths = {12N7 4N12567 56N34 7N12567 8N34 9N4}     19 Links = {23r4 123r7 3c3 12c4 679c7 479b4 46b5 6b7 79b8}     0-rank almost-sk-loop truths     18 Truths = {4679R47 4679C34 12N7}     18 Links = {679c7 12n3 12n4 47n8 7n9 479b4 46b5 6b7 79b8}0-rank almost-all-column truths     18 Truths = {4679C3489 58B3}     18 Links = {49r5 467r6 67r8 9r9 12n3 12n4 2347n8 37n9}0-rank almost-all-row truths     21 Truths = {4R1247 679R12347 58B3}     21 Links = {4c26 6c157 7c167 9c257 12n34 2347n8 37n9}0-rank almost-all-box truths     18 Truths = {123C7 1B578 23B4578 5B45 8B48}     18 Links = {23r4 123r7 3c3 12c4 5n16 6n1257 8n26 9n57}Each of the above has the same 17 exclusions     17 Eliminations --> r6c57<>6, r6c17<>7, r9c57<>9, r47c8<>23, r2c4<>12, r1c3<>3,      r5c6<>4, r6c2<>4, r7c9<>1, r8c6<>7`

Note that r4c8<>4 is not listed as an exclusion.
ronk
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### Re: JExocet Compendium

SV Congratulations on your discovery (or in the light of Ronk's revelations, should I say rediscovery) of the Almost SK Loop. It certainly passed me by.

Applying the eliminations arising from eliminating (6) as a base digit first, this is the Multi-Sector Locked Set equivalent

Code: Select all
`    *--------------------------*--------------------------*--------------------------*    | 3568    3458    4679-3   | 4679    568     458      | 79      <1>     <2>      |     | 1568    1458    4679     | 4679-12 12568   12458    | 79      568     <3>      |     | 15678   1589    <2>      | <3>     15689   1578     | <4>     568     68       |     *--------------------------*--------------------------*--------------------------* 23 | 237     2349    <1>      | <8>     236     234      | 236     79      <5>      | b4 679     | 235     <6>     349      | 124     <7>     1235-4   | <8>     2349    149      | b5 46    | 2358    2358-4  347      | 1246    1235-6  <9>      | 123-6   23467   1467     |     *--------------------------*--------------------------*--------------------------*123 | 1236    123     <8>      | <5>     1239    1237     | 1236    479     479      | b7 6     | <9>     123     36       | 127     <4>     1238     | <5>     2368    1678     | b8 79    | <4>     <7>     5        | 129     1238    <6>      | 123     238     189      |     *--------------------------*--------------------------*--------------------------*                      3          12                         6     MS-NS:(23)r4,(123)r7,(3)c3,(12)c4,(6)c7,(479)b4,(46)b5,(6)b7,(79)b8  (17 Digit Instances/Cells)=> 7 Elims: 3r1c3, 12r2c4, 4r5c6, 4r6c2, 6r6c5, 6r6c7 (6 Cells)`

So the net gain is 7 more eliminations.
As (79):LockedPair:r12c7 now exists the covers are reduced by 2. This de-mystifies your observations about r47c7 somewhat.

The eliminations in box5 look particularly promising as a means of completing the puzzle just using AICs with supporting patterns.

I've explored Almost SK Loops in combination with JExocets in the past. In some cases it's possible to get a rank 0 pattern by adding a cover set for the base digits through the box containing the base cells (as you did in c7 in this case) but when that failed I hit a brick wall. I would get N potential eliminations containing a single invalid one but no way to isolate it to any sub-group.

DPB
David P Bird
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### Re: 8. JE4 Examples

David P Bird wrote:8. JE4 Examples
Example 3

..1....26.6..1....78....1.....3.9..4..8.6.2..3..........798.6.....5.....8....4.5. 14808 (Morphed)
Code: Select all
` *--------------------------*--------------------------*--------------------------* | 459 b   3459 b  <1>      | 478     34579 T 3578     | 345789  <2>     <6>      |  | 2459    <6>     23459 T  | 2478    <1>     23578    | 345789  34789   35789    |  | <7>     <8>     23459    | 246     23459 t 2356     | <1>     349 B   359 B    |  *--------------------------*--------------------------*--------------------------* | 1256    1257    256      | <3>     257     <9>      | 578     1678    <4>      |  5  | 1459    14579   <8>      | 147     <6>     157      | <2>     1379    13579    |  | <3>     124579  24569    | 12478   2457    12578    | 579     1679    1579     | 45  *--------------------------*--------------------------*--------------------------* | 1245    12345   <7>      | <9>     <8>     123      | <6>     134     123      |  | 12469   12349   23469    | <5>     237     12367    | 34789   134789  123789   |34   | <8>     1239    2369     | 1267    237     <4>      | 379     <5>     12379    |3   *--------------------------*--------------------------*--------------------------*                    9                                     9                   CLb                CL1                CLB`

(3459)JE4:r1c12,r3c5,r2c7 r3c89,r1c5,r2c3 (digit (9) covered by c3 & c7)
=> r3c3 <> 2459, r1c7 <> 3459, r2c4 <> 4, r2c6 <> 35 (base digits seen by all base cells or all target cells)
=> r2c3 <> 2, r1c5 <> 7, r3c5 <> 2, r2c7 <> 78 (non-base digits in targets)
=> r8c2689,r9c29 <> 3, r6c24,r8c128 <> 4, r4c12,r6c269 <> 5, r2c3,r2c7 <> 9 (base digits in non-'S' cells)
=> r1c5 <> 3 (target node base digit missing in mirror cells)
=> r2c1 <> 2, r3c46 <> 2 (the non-base digits in these mirror nodes must be 6)
After basic moves there are two more JExocet eliminations that further simplify the rest of the solution
JE => r1c5 <> 3 (target node base digit missing in mirror cells)
JE => r3c6 <> 5 (mirror node base digit missing in target cell)

I am unable to confirm your last exclusion r3c6<>5. A counter-example is shown below. Note that the targets reflect their respective base, and columns 3, 5, and 7 each have placements for digits <345>. Would you please revisit this and tell me where I've gone astray.

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`+-----------------+----------------+------------------+|b4    b5     1   | 78   T9   378  | 78   2      6    || 29    6    T3   | 278   1   278  |t5    478    78   || 7     8     2   | 26   t4  *5    | 1   B3     B9    |+-----------------+----------------+------------------+| 126   127  *5   | 3     27  9    | 78   1678   4    || 19    179   8   | 147   6   17   | 2    179    1357 || 3     1279 *4   | 1278 *5   1278 | 789  16789  178  |+-----------------+----------------+------------------+| 125   1234  7   | 9     8   12   | 6    1      12   || 1269  129   269 | 5    *3   1267 |*4    1789   1278 || 8     129   269 | 1267  27  4    |*3    5      127  |+-----------------+----------------+------------------+`
ronk
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### Re: JExocet Compendium

RonK wrote:I am unable to confirm your last exclusion r3c6<>5. A counter-example is shown below. Note that the targets reflect their respective base, and columns 3, 5, and 7 each have placements for digits <345>. Would you please revisit this and tell me where I've gone astray.

Thanks for picking up on my typo, and I also note that I'd repeated the (3)r1c5 eliminaton (ugh!).
There are so many eliminations to make here they become tripping hazards. Although many of the JE eliminations can be made more simply, I wanted to show that they would be possible in other circumstances too.

Here is the draft edit, which I hope I've got right now:

(3459)JE4:r1c12,r3c5,r2c7 r3c89,r1c5,r2c3 (digit (9) covered by c3 & c7)
=> r3c3 <> 3459, r1c7 <> 3459, r2c4 <> 4, r2c6 <> 35 (base digits seen by all base cells or all target cells)
=> r2c3 <> 2, r1c5 <> 7, r3c5 <> 2, r2c7 <> 78 (non-base digits in targets)
=> r8c2689,r9c29 <> 3, r6c24,r8c128 <> 4, r4c12,r6c269 <> 5, r2c3,r2c7 <> 9 (base digits in non-'S' cells)
=> r1c5 <> 3 (target node base digit missing in mirror cells)
=> r2c1 <> 2, r3c46 <> 2 (the non-base digits in these mirror nodes must be 6)

Basic moves:
(2)Single:r3c4
(345)HiddenSet:r156c2 => r1c2 <> 9, r5c2 <> 179, r7c2 <> 12
(9)Box/Line:b1c1 => r58c1 <> 9
(39)HiddenSet:r5c89 => r3c8 <> 17, r5c9 <> 157, => r6c789 <> 9
(578)HiddenSet:r289c7 => r2c7 <> 5*, r8c7 <> 78, r9c7 <> 7
(7)Line/Box:r5b5 => r4c5,r6c456 <> 7
(7)Line/Box:c5b8 => r8c6,r9c4 <> 7
(9)Line/Box:c7b9 => r8c89,r9c9 <> 9

These reveal one further JExocet elimination:
JE => r3c6 <> 5 (mirror node base digit missing in target cell)

The (578)HiddenSet:r289c7 eliminates (5)in the r2c7 target, so it can also be eliminated in its mirror node r3c46.

It appears you either missed the hidden triples or didn't consider them as being basic moves. (In your grid r7c2 is the only cell capable of holding either (3) or (4) in r7, c2 & b7.)
David P Bird
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### Re: JExocet Compendium

David P Bird wrote:Here is the draft edit, which I hope I've got right now:

(3459)JE4:r1c12,r3c5,r2c7 r3c89,r1c5,r2c3 (digit (9) covered by c3 & c7)
=> r3c3 <> 3459, r1c7 <> 3459, r2c4 <> 4, r2c6 <> 35 (base digits seen by all base cells or all target cells)
=> r2c3 <> 2, r1c5 <> 7, r3c5 <> 2, r2c7 <> 78 (non-base digits in targets)
=> r8c2689,r9c29 <> 3, r6c24,r8c128 <> 4, r4c12,r6c269 <> 5, r2c3,r2c7 <> 9 (base digits in non-'S' cells)
=> r1c5 <> 3 (target node base digit missing in mirror cells)
=> r2c1 <> 2, r3c46 <> 2 (the non-base digits in these mirror nodes must be 6)

Basic moves:
(2)Single:r3c4
(345)HiddenSet:r156c2 => r1c2 <> 9, r5c2 <> 179, r7c2 <> 12
(9)Box/Line:b1c1 => r58c1 <> 9
(39)HiddenSet:r5c89 => r3c8 <> 17, r5c9 <> 157, => r6c789 <> 9
(578)HiddenSet:r289c7 => r2c7 <> 5*, r8c7 <> 78, r9c7 <> 7
(7)Line/Box:r5b5 => r4c5,r6c456 <> 7
(7)Line/Box:c5b8 => r8c6,r9c4 <> 7
(9)Line/Box:c7b9 => r8c89,r9c9 <> 9

These reveal one further JExocet elimination:
JE => r3c6 <> 5 (mirror node base digit missing in target cell)

The (578)HiddenSet:r289c7 eliminates (5)in the r2c7 target, so it can also be eliminated in its mirror node r3c46.

It appears you either missed the hidden triples or didn't consider them as being basic moves. (In your grid r7c2 is the only cell capable of holding either (3) or (4) in r7, c2 & b7.)

Typos continue to haunt you. As best as I can determine, here is your grid after basics:

Code: Select all
` +-----------------------------------------------------------------------+ | B459   B345    1      |  478   q459    3578   |  78     2      6      | | q459    6     r345    |  278    1      278    | Q34     34789  35789  | |  7      8      2      |  46    R3459   356    |  1     b349   b359    | |-----------------------+-----------------------+-----------------------| |  126    127    56     |  3      25     9      |  578    1678   4      | |  145    45     8      |  147    6      157    |  2      39     39     | |  3      1279   4569   |  128    245    128    |  57     167    17     | |-----------------------+-----------------------+-----------------------| |  1245   345    7      |  9      8      123    |  6      134    123    | |  126    129    3469   |  5      237    126    |  349    178    1278   | |  8      129    369    |  126    237    4      |  39     5      127    | +-----------------------------------------------------------------------+ # 115 eliminations remain`

Since '5' must be true in R(r3c5) or b(r3c9), we get r3c6<>5 . Similar logic would apply for '9' ... if it had any candidates to eliminate.

_
daj95376
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### Re: JExocet Compendium

Hi Danny. On the subject of typos shouldn't you remove the q from r2c1 in your PM ?

Leren
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### Re: JExocet Compendium

David P Bird wrote:
Example 2 Having a Cross-Line as a Cover House

..2.8...76...........67..........7...2...6.54..5....12..1...4....8.5.....573..2.8
Code: Select all
` *-----------------------*-----------------------*-----------------------* | 1359 b 139 b  <2>     | 149    <8>    349     | 359    6      <7>     |  | <6>    7      349     | 1259 t 239    2359    | 8      24     139     |  | 13589  1389   349     | <6>    <7>    239     | 359 t  24     139     |  *-----------------------*-----------------------*-----------------------* | 139    1349   6       | 259    2349   2359    | <7>    8      39      |   5   | 78     <2>    39      | 78     1      <6>     | 39     <5>    <4>     |  3 9 | 3789   3489   <5>     | 789    349    3789    | 6      <1>    <2>     |   *-----------------------*-----------------------*-----------------------* | 239    6      <1>     | 2789   29     2789    | <4>    37     5       | 1 | 239    39     <8>     | 2479   <5>    2479    | 1      37     6       | 1 | 4      <5>    <7>     | <3>    6      1       | <2>    9      <8>     |  3  *-----------------------*-----------------------*-----------------------*                           9                        CLb       CL1                     CL2     `

(1359)JE2:r1c12.r2c4,r3c7 (cover houses for (9) are r5 & c4 ) (this could be easily be missed)
=> r2c4 <> 2 (non-base digit in target cell)
=> r2c4 <> 5 (base digit missing from mirror node)
=> r2c4 <> 9 (target cell is a non-'S' cell for (9))

This puzzle now demonstrates how this last derived inference for (9)r2c4 operates.
"Derived Inference 5: A base digit candidate that has a cross-line as an 'S' cell cover house is false in a target cell in that cross-line."
If (9)r2c4 was true only one of its cover houses would remain (at row 5) so it would also have to be true in in the other target which would make the pattern false and produce an immediate contradiction.

After the follow-on eliminations a (39) repeating pair chain: (x)r5c3 = (x)r5c7 - (x)r4c9 = (x)r2c9 => r2c3 <> 39 now demolishes the puzzle.

HI David, This is from your JE2 examples file. Everything is fine except that you appear to have missed the other half of the mirror node inference -39 r3c9, so only one basic move plus a Skyscraper are required to finish the puzzle.

I know this is not the main point of the example but it would be nice to have this cleared up for the sake of easy reading.

Leren
Leren

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### Re: JExocet Compendium

=> r3c12 <> 1 ((1) must be true in the base cells)
=> r3c9 = 1 (as (1) is true in target:r2c4, it must be true in the r3c89 mirror node and this can only be in r3c9)

I have to be careful with the wording here as if (1)r3c8 exisited as a candidate, boldly eliminating instances of the other base candidates in the mirror node would be unsound.

However I still favour the (39) repeating pair chain as for a manual solver it should stand out a mile.

Keep them coming!
David P Bird
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### Re: JExocet Compendium

Leren wrote:Hi Danny. On the subject of typos shouldn't you remove the q from r2c1 in your PM ?

No. The value in r1c5 must be true in r3c89 but not in r2c3. Where else in r2c123 is it possible to place this value?

My solver sees it as a secondary equivalence:

Code: Select all
` +--------------------------------------------------------------------------------+ |  459     3459    1       |  478     34579   3578    |  345789  2       6       | |  2459    6       23459   |  2478    1       23578   |  345789  34789   35789   | |  7       8       23459   |  246     23459   2356    |  1       349     359     | |--------------------------+--------------------------+--------------------------| |  1256    1257    256     |  3       257     9       |  578     1678    4       | |  1459    14579   8       |  147     6       157     |  2       1379    13579   | |  3       124579  24569   |  12478   2457    12578   |  579     1679    1579    | |--------------------------+--------------------------+--------------------------| |  1245    12345   7       |  9       8       123     |  6       134     123     | |  12469   12349   23469   |  5       237     12367   |  34789   134789  123789  | |  8       1239    2369    |  1267    237     4       |  379     5       12379   | +--------------------------------------------------------------------------------+ # 186 eliminations remain ### -3459- QExocet   Base = r1c12 <34>   Target = r2c7,r3c5 ### -3459- QExocet   Base = r3c89 <59>   Target = r1c5==r2c1,r2c3 *** double QExocet`

What caught me by surprise was DPB's deduction "-9 r2c37" from the double-exocet.

_
daj95376
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### Re: JExocet Compendium

Hi Danny, I now understand why you put q in r2c1. However from your last post you don't seem to mention the dual secondary equivalence for the first Exocet r2c7==r3c4/r3c6.

Following your notation convention I would have thought you would have put Q in r3c46.

Leren
Leren

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Joined: 03 June 2012

### Re: JExocet Compendium

Leren wrote:Hi Danny, I now understand why you put q in r2c1. However from your last post you don't seem to mention the dual secondary equivalence for the first Exocet r2c7==r3c4/r3c6.

Following your notation convention I would have thought you would have put Q in r3c46.

I'm unaware of anything logically noteworthy coming from dual cells as a secondary equivalence. My solver only marks when a single cell is present as a secondary equivalence.

_
daj95376
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### Re: JExocet Compendium

daj95376 wrote : I'm unaware of anything logically noteworthy coming from dual cells as a secondary equivalence.

What I'm calling here a dual secondary equivalence is where the two cells in the mini-line opposite a target cell are linked by a strong link on a non=base digit.

For the r1c12 Exocet a dual secondary equivalence occurs in r3c46 opposite the target cell in r2c7 with a strong link on the non-base digit 6. So one of these 2 cells must be 6 and the other must eventually hold the same digit as r2c7. The nett result of all this is that you can eliminate all non-common digits in r3c46 and r2c7 (but not of course the 6 in r3c46). In this puzzle the direct result is r3c46 <> 2, which solves r3c3 = 2. In David's solution he notes this as "r3c46 <> 2 (the non-base digits in these mirror nodes must be 6)". So as far as I can see David refers to both single and dual secondary equivalences as "mirror node inferences".

How common are dual secondary equivalences? Surprisingly, they are quite common, particularly when double Exocets can be found.

Leren
Leren

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### Re: JExocet Compendium

David P Bird Wrote :

4 JE2 Examples

Example 1 JE2 with 3 Base Digits

..5....36....5....89....5..4..........9.8.6.....4.1.2..7.....1....2....7..3.6.9..

After basic eliminations including Swordfish for (3) & (6)
Code: Select all
` *--------------------------*--------------------------*--------------------------* | 158 b   158 b   <7>      | 1568    <2>     1569     | 1358    1389    <4>      |  | <9>     <3>     12458    | 1458 t  478     1457     | <6>     1278    257      |  | <6>     1458    12458    | <3>     4789    14579    | 12578 t 12789   2579     |  *--------------------------*--------------------------*--------------------------* | 13478   14789   1489     | 248     34678   247      | 127     <5>     69       | 1 8  | <2>     4579    459      | 456     <1>     34567    | 37      69      <8>      |  5  | 13578   1578    <6>      | <9>     378     257      | <4>     127     237      |  *--------------------------*--------------------------*--------------------------* | 1458    14568   <3>      | <7>     46      12       | <9>     248     256      |  | 478     <2>     489      | 46      <5>     39       | 378     34678   <1>      |   8 | 1457    145679  1459     | 12      39      <8>      | 257     247     23567    | 15  *--------------------------*--------------------------*--------------------------*                   CLb        CL1                        CL2`

Hi David, found another typo in your file 05 JE2 examples. You'll notice in the above, that the puzzle doesn't match the PM

I think the correct puzzle should be ..7.2...493....6..6..3............5.2...1...8..69..4....37..9...2..5...1.....8...

Leren
Leren

Posts: 3319
Joined: 03 June 2012

### Re: JExocet Compendium

David P Bird wrote :

Example 3 Using JE Group Nodes in AICs

..514.....7.....3.3..9.......2....89..74....11...6...2..13..72.5.32..916......... (Morphed)
Code: Select all
` *-----------------------*-----------------------*-----------------------* | 268    68     <5>     | <1>    <4>    3       | 268    9      7       |  | 24689  <7>    4689    | 568    258    2568    | 1      <3>    458     |  | <3>    1      468     | <9>    2578   25678   | 2568   456    458     |  *-----------------------*-----------------------*-----------------------* | 46     3456   <2>     | 7      135    15      | 3456   <8>    <9>     | 2 | 689    35689  <7>     | <4>    23589  2589    | 356    56     <1>     |  | <1>    34589  489     | 58     <6>    589     | 345    7      <2>     |   8 *-----------------------*-----------------------*-----------------------* | 4689   4689   <1>     | <3>    589    45689   | <7>    <2>    458     |  | <5>    48     <3>     | <2>    78     478     | <9>    <1>    <6>     | 2 | 7      2      4689    | 568    1589   145689  | 58     45     3       |  68 *-----------------------*-----------------------*-----------------------*                                                    6                     CLb      CL1                     CL2 `

(268)JE2:r1c12,r2c4,r3c7 (Cover houses for (6) = r9 & c4 )
=> r2c4 <> 5, r3c7 <> 5 (Non-base digits in target cells)
=> r3c7 <> 6 (Target cell in an 'S' cell cover house)

There are two more eliminations for digit 6, which has orthogonal lines cover houses : r2c1, r9c6 <> 6

Leren
Leren

Posts: 3319
Joined: 03 June 2012

### Re: JExocet Compendium

Leren wrote:
David P Bird wrote :

Example 3 Using JE Group Nodes in AICs

..514.....7.....3.3..9.......2....89..74....11...6...2..13..72.5.32..916......... (Morphed)
Code: Select all
` *-----------------------*-----------------------*-----------------------* | 268    68     <5>     | <1>    <4>    3       | 268    9      7       |  | 24689  <7>    4689    | 568    258    2568    | 1      <3>    458     |  | <3>    1      468     | <9>    2578   25678   | 2568   456    458     |  *-----------------------*-----------------------*-----------------------* | 46     3456   <2>     | 7      135    15      | 3456   <8>    <9>     | 2 | 689    35689  <7>     | <4>    23589  2589    | 356    56     <1>     |  | <1>    34589  489     | 58     <6>    589     | 345    7      <2>     |   8 *-----------------------*-----------------------*-----------------------* | 4689   4689   <1>     | <3>    589    45689   | <7>    <2>    458     |  | <5>    48     <3>     | <2>    78     478     | <9>    <1>    <6>     | 2 | 7      2      4689    | 568    1589   145689  | 58     45     3       |  68 *-----------------------*-----------------------*-----------------------*                                                    6                     CLb      CL1                     CL2 `

(268)JE2:r1c12,r2c4,r3c7 (Cover houses for (6) = r9 & c4 )
=> r2c4 <> 5, r3c7 <> 5 (Non-base digits in target cells)
=> r3c7 <> 6 (Target cell in an 'S' cell cover house)

There are two more eliminations for digit 6, which has orthogonal lines cover houses : r2c1, r9c6 <> 6

Leren
Leren thank you. Your post reveals a point that I don't remember making in the write-up and which I'll need to check. However this example demonstrates it well and is a good place to reinforce it.

This is my draft:
"Normally non-'S' cell eliminations can only be made for digits that must be true in the base cells. But when a target cell is a non-'S' cell for a digit it produces an automatic contradiction; it can only be true in that target if it is a true base digit, but if that were the case it would be eliminated as occupying a non-'S' cell. The (6)r3c7 elimination can therefore be made immediately but other (6) eliminations can only be made when it is known to be true in the base cells."

That states what I believe is the generic case for JE patterns but perhaps you have found an immediate follow-on in this puzzle that I can't see.

DPB
David P Bird
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