## Exocet Example 2 from JExocet Compendium

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

### Exocet Example 2 from JExocet Compendium

The following is taken from David Bird's "JExocet Compendium" -- "Example 2 Having a Cross-Line as a Cover House".
(It has a couple additional candidate removals using other strategies before resorting to Exocet but is essentially the same situation.)

Code: Select all
`002080067670000800000670000006000780020016054005000612061000405008050106457361298+----------------+----------------+-------------+| 1359  139  2   | 149  8    349  | 359 6   7   || 6     7    349 | 1259 239  2359 | 8   24  139 || 13589 1389 349 | 6    7    239  | 359 24  139 |+----------------+----------------+-------------+| 139   1349 6   | 259  2349 2359 | 7   8   39  || 78    2    39  | 78   1    6    | 39  5   4   || 3789  3489 5   | 79   349  3789 | 6   1   2   |+----------------+----------------+-------------+| 239   6    1   | 2789 29   2789 | 4   37  5   || 239   39   8   | 247  5    2479 | 1   37  6   || 4     5    7   | 3    6    1    | 2   9   8   |+----------------+----------------+-------------+`

It gives these eliminations:
(1359)JE2:r1c12.r2c4,r3c7 (cover houses for (9) are r5 & c4 )
=> 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))
=> r3c9 <> 39, (base digits missing in the mirrored target cell)

I am able to understand why 2,5,9 can be eliminated from cell (2,4).
But I don't understand why base candidates (3 and 9) missing in the opposite target cell (2,4) cannot be true in the mirror cell (3,9).

I've been trying to do a proof-by-contradiction but never arrive at a contradiction.
So to to be fair, the explanation should NOT USE the inferences made in parallel or unneeded assumptions of the pattern:
- Do not use that 1 must be in (2,4), so it can not be in (2,9), so (3,9) must be 1.
- Do not use that 3/9 are a locked-pair in row 5, just that there is (at least) a weak-link there.

So in my reasoning... let's say (for contradiction sake) that (3,9) is 3, then (5,7) is 3, then (5,3) is not 3.
Also because 3 is in neither target cells, then 3 is not in (1,1) or in (1,2), so (2,3) must be 3 and (1,6) must be 3.
After these types of naive eliminations there doesn't seem to be an obvious contradiction for the candidate 3s:
Code: Select all
`+--------------+----------------+---------+| 159  19    2 | 149  8    3    | 59 6  7 || 6     7    3 | 1259 29   259  | 8  24 1 || 1589 189   4 | 6    7    29   | 59 24 3 |+--------------+----------------+---------+| 139   1349 6 | 259  2349 259  | 7  8  9 || 78    2    9 | 78   1    6    | 3  5  4 || 3789  3489 5 | 79   349  789  | 6  1  2 |+--------------+----------------+---------+| 239   6    1 | 2789 29   2789 | 4  37 5 || 239   39   8 | 247  5    2479 | 1  37 6 || 4     5    7 | 3    6    1    | 2  9  8 |+--------------+----------------+---------+`

Obviously, I will later see contradictions, but those are too indirect. So there must be something else I'm missing.
DougCube

Posts: 3
Joined: 17 November 2018

### Re: Exocet Example 2 from JExocet Compendium

DougCube wrote:
I am able to understand why 2,5,9 can be eliminated from cell (2,4).
But I don't understand why base candidates (3 and 9) missing in the opposite target cell (2,4) cannot be true in the mirror cell (3,9).

Hi DougCube and welcome here,

You made 99% of the job

I am not responsible for the original post, but

I am able to understand why 2,5,9 can be eliminated from cell (2,4).

if this is ok, then

r2c4=1
r3c9=1

and the rest is trivial
champagne
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### Re: Exocet Example 2 from JExocet Compendium

DougCube wrote:So to to be fair, the explanation should NOT USE the inferences made in parallel or unneeded assumptions of the pattern:
- Do not use that 1 must be in (2,4), so it can not be in (2,9), so (3,9) must be 1..

Being an exocet amateur (forever) i just do:
Each of 359 in r1c12 would go to r3c7 too.
So not 2 of them can be in r1c12.
Thus 1 must be there, and in r2c4 and in r3c9.
eleven

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Joined: 10 February 2008

### Re: Exocet Example 2 from JExocet Compendium

=> 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))
=> r3c9 <> 39, (base digits missing in the mirrored target cell)

The way I understand it, the elimination rules can be applied in parallel, so the reasoning given by champagne is not what I'm looking for.

I kind of understand eleven's explanation but it is relying on something else not in the Exocet pattern -- that there are 3 base-candidates that need to go in r3c789 (forcing a locked-triple).
However, if say 5 was not yet eliminated from r2c4 (or if the pattern was slightly different and it could not be eliminated there), then why can we still eliminated 3 and 9 from r3c9???
My approach is to follow a proof-by-contradiction: Place a 3 in r3c9 and look for the contradiction but using only the pattern requirements -- and then again with 9.
I spent more than a few hours yesterday on this but and still stuck so I came here for help.

champagne wrote:r2c4=1
r3c9=1

and the rest is trivial

This is exactly what I said not to use because this depends on a parallel elimination rule or something not always true in the pattern.

The general elimination rule I'm trying to understand (i.e. prove to myself) is that "base-candidates that cannot be true in a Target cell, cannot be true in the (opposite) Mirror cells".
I don't know if this was spelled out explicitly anywhere (there are so many elimination rules), but the solver on sudokuwiki.org is giving it as 'Exocet Rule #9'.
(I contacted the person for that solver yesterday and am hoping for a reply from him.)
DougCube

Posts: 3
Joined: 17 November 2018

### Re: Exocet Example 2 from JExocet Compendium

DougCube wrote:
=> 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))
=> r3c9 <> 39, (base digits missing in the mirrored target cell)

The way I understand it, the elimination rules can be applied in parallel, so the reasoning given by champagne is not what I'm looking for.

Are you sure you're not assuming too much about the parallelism? It seems to me that, as stated, half of the last elimination step depends on first eliminating 9r2c4, which kind of implies that those rules are not totally independent of each other. If that's so, then wouldn't it make sense that the last rule as a whole is just a logical consequence of the application of the previous rules, and you might not even be able to prove it independently? (Caveat: I know very little about JExocets, so this is just an observation with very limited value.)

I don't know if this was spelled out explicitly anywhere (there are so many elimination rules), but the solver on sudokuwiki.org is giving it as 'Exocet Rule #9'.
(I contacted the person for that solver yesterday and am hoping for a reply from him.)

While at it, could you ask him to list all of those elimination rules with their explanations? Stating just "Rule #X" is not very helpful if those numbers aren't explained anywhere. I don't think the numbering matches with David's list:

JExocet Compendium wrote:5. Eliminations

1. A base candidate that is restricted to only one 'S' cell cover house is invalid and is false in the base mini-line and target cells.
2. Any base candidate that isn't capable of being simultaneously true in at least one target cell and its mirror node is false.
3. Any non-base candidate in a target is false.
4. A base digit in a target that must be true in the other target is false.
5. A base candidate that has a cross-line as an 'S' cell cover house must be false in the target cell in that cross-line which may make other simple colouring eliminations available.
6. Any base candidate that can't be true in the mirror node for a target cell is false in the target cell.
7. If one mirror node cell can only contain non-base digits, the second one will be restricted to the base digits in the opposite object cells.
8. If a mirror node contains only one possible non-base digit value, it is true in that node and false in the cells in sight of it.
9. If a mirror node contains a locked digit, any other digits it contains of the same type (known base or non-base) are false.
10. A known base digit is false in any cells in full sight of either a) both base cells or b) both target cells.
11. A known base digit, or one that can only occur once in the 'escape' cells in the cross-lines, is false in the non-'S' cells in its cover houses.
12. For a known base digit, any digit instance that would prevent two of its 'S' cells being true is false.

SpAce

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### Re: Exocet Example 2 from JExocet Compendium

DougCube wrote:
champagne wrote:r2c4=1
r3c9=1

and the rest is trivial

This is exactly what I said not to use because this depends on a parallel elimination rule or something not always true in the pattern.

It's not a direct exocet effect, but so evident that I would likely have considered it as an exocet derived elimination as well.
And even in the full list of eliminations mentioned by SpAce, some are basic, and some "derived"

The basic rule is very simple
if a digit is valid in the base, it is valid in one of the target cells (by definition, and must have been checked)
All the rest is derived, mainly the 2 basic consequences

The target cells can not contain extra digits
for any digit valid in the base, cells seeing both targets can not contain this digit.
Last edited by champagne on Tue Nov 20, 2018 5:26 am, edited 1 time in total.
champagne
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### Re: Exocet Example 2 from JExocet Compendium

Just to clarify (sorry being Off Topic for you)
DougCube wrote:I kind of understand eleven's explanation but it is relying on something else not in the Exocet pattern -- that there are 3 base-candidates that need to go in r3c789 (forcing a locked-triple).

I don't know, what you mean.
My way is following, needing no JExocet at all.
5r1c12 -> 5r3c7
3r1c12 -> r5c3 -> r4c9 -> 3r3c7 (can't be in r1)
9r1c12 -> r5c3 -> r4c9 -> 9r3c7 (can't be in r1)
Since each of 359 in r1c12 would go to r3c7, only one can be true in r1c12, leaving 1 as the only second digit there.

[Added:] The relation to the JExocet is, that the 4th digit 1 is going to another exocet target, so that you have 4 digits constrained to 2 targets. But it is not needed at all for the eliminations, no 2nd target, no S-lines, no mirrored cells, no rule #9 ...
eleven

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### Re: Exocet Example 2 from JExocet Compendium

champagne wrote:The target/mirror cells can not contain extra digits
for any digit valid in the base, cells seeing both targets can not contain this digit.

What is meant by "extra digits"?
On Mirror cells r2c56 there can be a non-base candidate 2 there. And r3c8 has no base-candidates, only 2 and 4.
Also, r3c9 doesn't see both Targets.
I can understand that if 5 was in r2c4, then 5 cannot be in r3c7 according to the Exocet. But inferences about the mirror cells I don't understand at all.

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     `
DougCube

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### Re: Exocet Example 2 from JExocet Compendium

Hi Dougcube,

I suppressed the word mirror in my post, not sure that it is the same as "target", my assumption reading your first post.

The exocet itself says only that '2r2c4' extra digit not in the base is not in the targets
Telling that 5r2c4 and 9r5c4 are not valid is already specific to the puzzle and not an exocet property, but I agree that it can be a JExocet specidic pattern effect. I did not study all JE extensions.

I can understand that if 5 was in r2c4, then 5 cannot be in r3c7 according to the Exocet.
first and simplest combined effect puzzle basic logic plus exocet for me, The second, similar, using "eleven"'s chain is for 9r2c4

Also, r3c9 doesn't see both Targets.
Right, but as the three previous eliminations leave 1r2c4 alone, 1r2c4 is valid. And then, r3c9=1.

And "eleven" is right, here, the main facts can be established easily without the exocet property. A player has big chances to solve the puzzle in the same way as him.

EDIT and eleven has also the fact that "1" is valid in the base -> r2c4=1
EDIT2 In some ways, the JE patterns establishes the Exocet property. What does eleven is part of the work needed to establish that we have an exocet.
champagne
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### Re: Exocet Example 2 from JExocet Compendium

DougCube wrote:
champagne wrote:The target/mirror cells can not contain extra digits
for any digit valid in the base, cells seeing both targets can not contain this digit.

What is meant by "extra digits"?
On Mirror cells r2c56 there can be a non-base candidate 2 there. And r3c8 has no base-candidates, only 2 and 4.
Also, r3c9 doesn't see both Targets.
I can understand that if 5 was in r2c4, then 5 cannot be in r3c7 according to the Exocet. But inferences about the mirror cells I don't understand at all.

Here's my take on that, with the same disclaimer as before. There's a crucial difference between a mirror node and a mirror cell. Both adjacent cells on the mini-line next to a target cell form a mirror node (for the other target), but only one of them is the actual mirror cell (we don't necessarily know which one). For example, r2c56 is the mirror node for t2 (r3c7), but we don't yet know which one of those two is the mirror cell because both can hold base digits. In box 3, r3c89 is the mirror node for t1 (r2c4), but of those r3c8 only has non-base digits so it can't be the mirror cell -> therefore r3c9 must be. Knowing that, we can eliminate 3 and 9 (base digits missing in t1) from it according to rule 7 (assuming that 9 has already been cleared from t1 using another rule), and we could also clear any non-base digits from it if it had any:

6. Any base candidate that can't be true in the mirror node for a target cell is false in the target cell.
7. If one mirror node cell can only contain non-base digits, the second one will be restricted to the base digits in the opposite object cells.
8. If a mirror node contains only one possible non-base digit value, it is true in that node and false in the cells in sight of it.
9. If a mirror node contains a locked digit, any other digits it contains of the same type (known base or non-base) are false.

SpAce

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### Re: Exocet Example 2 from JExocet Compendium

DougCube,

the proof is quite simple.

Either 3 is in r3c7 and can't be in r3c9,

or two of the other digits occupy r1c12 and r2c4 and r3c7.
The one in r1c12 and r2c4 now can't be in r12c789, not in r3c7 (there is the other one), and not in r3c8, which contains no base digits.
So it it must be in r3c9, and 3 can't be there.

Same for 9 (after elimination from r2c4).
eleven

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Joined: 10 February 2008

### Re: Exocet Example 2 from JExocet Compendium

I have a question about those mirror node inferences too:

6. Any base candidate that can't be true in the mirror node for a target cell is false in the target cell.
7. If one mirror node cell can only contain non-base digits, the second one will be restricted to the base digits in the opposite object cells.
8. If a mirror node contains only one possible non-base digit value, it is true in that node and false in the cells in sight of it.
9. If a mirror node contains a locked digit, any other digits it contains of the same type (known base or non-base) are false.

How exactly is rule #8 to be interpreted? In the example puzzle we have "only one possible non-base digit value" (2) in the mirror node r2c56. However, it will not be true in that node. Instead, there will be one false base digit (3) in addition to the mirrored true base digit (5), i.e. zero non-base digits in that mirror node. So, what does that rule mean and when is it applicable?

SpAce

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### Re: Exocet Example 2 from JExocet Compendium

DougCube wrote : But I don't understand why base candidates (3 and 9) missing in the opposite target cell (2,4) cannot be true in the mirror cell (3,9).

I might be coming into this discussion a bit late but I've been involved in a huge job elsewhere on the site (see my recent post) and this old piece of Exocet theory seems like a bit of light amusement after that.

It's obvious that the contents of r2c4 and r3c9 must eventually be the same base digit, as shown in this PM.

Code: Select all
`*------------------------------------------------*| a     b    2     | 149  8    349  |-a   6   7  || 6     7      349 |*a    239  2359 | 8  -a  -a  || 13589 1389   349 | 6    7    239  | b   /  *a  ||------------------+----------------+------------|| 139   1349   6   | 259  2349 2359 | 7   8   39 || 78    2      39  | 78   1    6    | 39  5   4  || 3789  3489   5   | 789  349  3789 | 6   1   2  ||------------------+----------------+------------|| 239   6      1   | 2789 29   2789 | 4   37  5  || 239   39     8   | 2479 5    2479 | 1   37  6  || 4     5      7   | 3    6    1    | 2   9   8  |*------------------------------------------------*`

I've shown the Exocet Base cells solving to ab and the Target cells solving to a and b. So, in Box 3 a can't go in Row 1 (Exocet Base cell solution) or Row 2 (Exocet Target cell solution). In Row 3 it can't go in r3c7 (Target cell must be b) and crucially it can't go in r3c8 (cell in Row 3 Taarget cell mini-line that doesn't hold a base digit, which I've indicated by /). So, by a process of elimination, in Box 3, there is only one cell that a can possibly go, r3c9.

So even if you don't know what a is, it's obvious that non-common digits can be eliminated from r2c4 and r3c9. In the old days we used to write this as r2c4==r3c9 meaning that these cells must eventually hold the same base digit, so non-common digits can be eliminated from these two cells. We, or at least I, used to call this a secondary elimination.

So the only requirement for a secondary elimination opportunity is that in r3c89 there is one cell that doesn't hold a base digit. In this case it's r3c8.

An extra spin on this pattern occurs when both cells in r3c89 hold base digits but there is a non-base digit in them that is the only two instances if it in Row 3 (call it, say, e), so one of r3c89 must eventually hold e (a non-base digit) and the other cell must then hold the same base digit as r2c4 (This extra spin doesn't actually happen here, but hopefully you can see what I'm getting at). If this had been the case, you could have eliminated non-common base digits in both r3c89 but e stays in both cells for the moment. To cater for this extra spin case David coined the term Mirror Node. I had figured this out years before the Compendium was written, but I forgot to tell anyone about it, so at least David and I independently discovered this. Originally I thought this was probably quite rare, but when I did some testing on a collection of Exocet puzzles I found that it was surprisingly common. It's very common when there is an associated double Exocet pattern. I have no idea why this is the case but it is.

Hopefully this is a simple explanation for all the fancy words in the Compendium.

Leren
Leren

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### Re: Exocet Example 2 from JExocet Compendium

I'll put my money where my mouth is and illustrate an extra spin example on a purported Double Exocet puzzle, the second I tried in an old Champagne collection, although it looks like eleven was the discoverer :

.23.....94.....1...9..3..4.2..81...4.....78..9...4...23...9...1.6..........5.....;264;elev;266;r1c7 r1c8 r3c1 r2c5 5678;r2c2 r2c3 r1c5 r3c9 5678

This was what my solver did:

Code: Select all
`*--------------------------------------------------------------------------------*| 15678   2       3        | 1467    5678    14568    |B567    B5678    9        || 4       578     5678     | 2679   T5678-2  25689    | 1       235678  35678    ||T5678-1  9       15678    | 1267    3       12568    | 2567    4       5678     ||--------------------------+--------------------------+--------------------------|| 2       357     567      | 8       1       3569     | 35679   35679   4        || 156     1345    1456     | 2369    256     7        | 8       13569   356      || 9       13578   15678    | 36      4       356      | 3567    13567   2        ||--------------------------+--------------------------+--------------------------|| 3       4578    24578    | 2467    9       2468     | 24567   25678   1        || 1578    6       1245789  | 12347   278     12348    | 234579  235789  3578     || 178     1478    124789   | 5       2678    123468   | 234679  236789  3678     |*--------------------------------------------------------------------------------*`

Exocet 1: r1c7 r1c8 r3c1 r2c5 5678 : Eliminate non base candidates in Target cells.

Code: Select all
`*--------------------------------------------------------------------------------*| 15678   2       3        | 1467    5678    14568    |B567    B5678    9        || 4       578     5678     | 2679   T5678    25689    | 1       235678  35678    ||T5678    9       5678-1   | 1267    3       12568    | 2567    4       5678     ||--------------------------+--------------------------+--------------------------|| 2       357     567      | 8       1       3569     | 35679   35679   4        || 156     1345    1456     | 2369    256     7        | 8       13569   356      || 9       13578   15678    | 36      4       356      | 3567    13567   2        ||--------------------------+--------------------------+--------------------------|| 3       4578    24578    | 2467    9       2468     | 24567   25678   1        || 1578    6       1245789  | 12347   278     12348    | 234579  235789  3578     || 178     1478    124789   | 5       2678    123468   | 234679  236789  3678     |*--------------------------------------------------------------------------------*`

Exocet 1: r1c7 r1c8 r3c1 r2c5 5678 ; r3c3==r2c5 ; Eliminate non-common candidates.

and here is the Killer move :

Code: Select all
`*--------------------------------------------------------------------------------*| 1       2       3        | 467     5678    4568     |B567    B5678    9        || 4       578     5678     | 67+9-2  T5678   568+9-2  | 1       235678  35678    ||T5678    9       5678     | 1267    3       12568    | 2567    4       5678     ||--------------------------+--------------------------+--------------------------|| 2       357     567      | 8       1       3569     | 35679   35679   4        || 56      1345    1456     | 2369    256     7        | 8       13569   356      || 9       13578   15678    | 36      4       356      | 3567    13567   2        ||--------------------------+--------------------------+--------------------------|| 3       4578    24578    | 2467    9       2468     | 24567   25678   1        || 578     6       1245789  | 12347   278     12348    | 234579  235789  3578     || 78      1478    124789   | 5       2678    123468   | 234679  236789  3678     |*--------------------------------------------------------------------------------*`

Exocet 1: r1c7 r1c8 r3c1 r2c5 5678 ; Eliminate cands in r2c4 & r2c6 that are not in r3c1 and are not SIS cand 9.

This is an example of the extra spin situation I mentioned in my previous post. The puzzle solves with basics from there. Without this move you might have to use the second Exocet and consequent common eliminations.

Leren
Leren

Posts: 3925
Joined: 03 June 2012

### Re: Exocet Example 2 from JExocet Compendium

Leren wrote:An extra spin on this pattern occurs when both cells in r3c89 hold base digits but there is a non-base digit in them that is the only two instances if it in Row 3 (call it, say, e), so one of r3c89 must eventually hold e (a non-base digit) and the other cell must then hold the same base digit as r2c4 (This extra spin doesn't actually happen here, but hopefully you can see what I'm getting at). If this had been the case, you could have eliminated non-common base digits in both r3c89 but e stays in both cells for the moment.

As far as I understand, this case with the locked digit in the mirror node is covered by Elimination Rule #9 and seems pretty obvious to me. However, can you explain what Rule #8 means, as I previously asked? It doesn't mention locked digits.

SpAce

Posts: 2550
Joined: 22 May 2017

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