## Exocet Example 2 from JExocet Compendium

Post the puzzle or solving technique that's causing you trouble and someone will help

### Re: Exocet Example 2 from JExocet Compendium

This thread gave me an incentive to finally start working through the examples in the Compendium. Pretty cool stuff, and I have to commend David for his work! Regarding those mirror nodes with locked digits, there are some pretty effective examples in the section "08. JE4 Examples with 3 Cross-lines". I have no problem understanding those inferences, but I don't get the Compatibility Check at the end of Example 3 (of that section). Here's my pm before that step:

Code: Select all
`..1....26.6..1....78....1.....3.9..4..8.6.2..3..........798.6.....5.....8....4.5..----------------------.-------------------.-----------------------.| b459  b35     1      | 478  T59     3578 |  78      2      6     ||  59    6     T345    | 278   1      278  | t34      3789   35789 ||  7     8      2      | 46   t359    36   |  1      B349*  B359*  |:----------------------+-------------------+-----------------------:|  126   127   s(5)6   | 3    s2(5)   9    | s(5)78   1678   4     ||  145   45     8      | 17    6      157  |  2       39*    39*   ||  3     1279  s(59)6  | 128  s(4)    128  | s(5)7    167    17    |:----------------------+-------------------+-----------------------:|  1245  345    7      | 9     8      123  |  6       134    123   ||  126   129   s(349)6 | 5    s2(3)7  126  | s(349)   178    1278  ||  8     129   s(39)6  | 126  s2(3)7  4    | s(39)    5      127   |'----------------------'-------------------'-----------------------'(3459)JE4:r1c12,r3c5,r2c7  r3c89,r1c5,r2c3 `

At that point we should apply the previously "identified" incompatibility:

JExocet Compendium wrote:Compatibility check: (39) incompatible in JE2 in r1c12
...
The incompatibility inference can now be used:
(9)r1c1 - (4)JE:r1c1,r2c7 = (3)JE:r2c7,r1c2 -[JE Incompatible]- (9)r1c1 => r1c1 <> 9

I would understand the chain just fine if I could see why (39) is an incompatible base pair in r1c12. There's no UR possibility in its chute, so why are those digits incompatible? Of course that pair is clearly impossible in r3c89 because of the plain UR threat in that chute, but I can't see why it (or any other obvious contradiction) prevents (39) being true in r1c12. Until this example I've understood all the incompatibility checks with the dual UR threats, but here I fail. What am I missing?

Last edited by SpAce on Fri Nov 23, 2018 8:25 am, edited 1 time in total.

SpAce

Posts: 2590
Joined: 22 May 2017

### Re: Exocet Example 2 from JExocet Compendium

I have no idea to both of your questions (rule #8 and this example).
Aren't there any proofs in the compendium (which i don't have) ?
eleven

Posts: 2461
Joined: 10 February 2008

### Re: Exocet Example 2 from JExocet Compendium

eleven wrote:I have no idea to both of your questions (rule #8 and this example).

Thanks, eleven! I wouldn't trust my own judgment with this stuff, so it's good to hear you can't see the logic either. It would be nice to hear Leren's opinion too, as based on the comments on the Compendium thread, he's combed through it before. There's no mention of errata on this particular example, however.

Here's a copy of the full text of this example:
Hidden Text: Show
File: 08. JE4 Examples with 3 Cross-lines

Example 3
Code: Select all
`..1....26.6..1....78....1.....3.9..4..8.6.2..3..........798.6.....5.....8....4.5. 14808 (Morphed) *--------------------------*--------------------------*--------------------------* | 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)
=> r2c3 <> 2, r1c5 <> 7, r3c5 <> 2, r2c7 <> 78 (non-base digits in targets)
=> r3c3 <> 3459, r1c7 <> 3459, r2c4 <> 4, r2c6 <> 35 (base digits seen by all base cells or all target cells)
=> 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)

Compatibility check: (39) incompatible in JE2 in r1c12

Basic moves:
(2)Single:r3c4
(345)HiddenSet:r156c2 => r1c2 <> 9, r5c2 <> 179, r7c2 <> 12
(9)Box/Line:b1c1 => r58c1 <> 9
(39)HiddenSet:r5c89 => r5c8 <> 17, r5c9 <> 157, => r6c789 <> 9
(578)NakedSet: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)

(4=5)r5c2 - (5=6)r4c3 - (6)r9c3 = (6)r9c4 - (6=4)r3c4 => r5c4 <> 4
(4)Single:r6c4
(4)X-Wing:c37\r28 => r2c18 <> 4
(4=5)r5c2 - (5)r5c6 = (5-3)r1c6 = (3)r1c2 => r1c2 <> 4

The incompatibility inference can now be used:
(9)r1c1 - (4)JE:r1c1,r2c7 = (3)JE:r2c7,r1c2 -[JE Incompatible]- (9)r1c1 => r1c1 <> 9

Completion from here is straightforward but lengthy

Aren't there any proofs in the compendium (which i don't have) ?

As you can see, there's no proof for the incompatibility within the example. There is, however, a section (file) dedicated to the topic: 04 Incompatible Base Pair Examples. Some generic statements from that document:

JExocet Compendium wrote:This test identifies when two base digits can't be true together because they would cause the base cells to become part of a deadly UR. This happens when two UR threats involving the base cells exist and it would be impossible to disrupt both of them.

When dual UR threats exist the simple rule is
For a pair of digits to be compatible it must be possible for their their true 'S' cells in the target cross lines to occupy two diagonal boxes.

Sometimes it happens that there is only one way the two digits can occupy diagonal 'S' cells which will then force which way round they would have to be true in the target cells.

When the cover houses for each digit are in different parallel bands this rule will always hold. Experience also shows it holds in other cases but there is a theoretical possibility that a true member digit could exist elsewhere in the unaffected box to eliminate the threat. A formal proof would therefore be required to show that this is never possible which might involve exhaustive case by case analysis.

It also has a couple of Double JE (JE4) examples, which might or might not provide additional insight. I have to go through them again to see if there's some special logic applicable here. (It would have been nice to mention it within the example, though.)

SpAce

Posts: 2590
Joined: 22 May 2017

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