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Exocet

a.k.a. jExocet

Full Documentation Coming Soon

Exocet is a pattern that can often occur in very hard puzzles where the candidate density is very high. With few bi-value and bi-location candidates other strategies give up. Exocet takes on three or four candidate sets at a time which is just what is needed in the bottlenecks of extreme puzzles. My first implementation solved 51 out of 123 of the weekly "unsolvables" that were created by David Filmer. We will be replacing any stock known solvable with even harder puzzles.

Phil's concise description is impossible to better: When 2 of the 3 cells in a box-line intersection together contain 3 or 4 candidates, then in each of the two boxes in the same band but in different lines, if there are cells with the same 3 or 4 candidates, any others can be removed.

Order in Strategy List

My instinct is to put this near the end of the Extremes set of strategies but many of the eliminations in Exocet can overlap earlier strategies, I am told. I do not think I have implemented many of those but to test and improve I am putting it at the start of the extremes to give it more exercise. David Bird tells me it can go after the basics but that will have to wait until more variations are in place.

Credits

A number of people explored this strategy and its many variations. I am told the name was coined by forum participant Champagne. The pattern was first discovered by Allan Barker in the "Fata Morgana" puzzle[1],[2]. My main source is the excellent JExocet Compendium written by David P Bird, available in fourteen downloadable documents on the EnjoySudoku Forum and David was kind enough to answer questions as well. I would also like to credit Phil's Sudoku Solver as a source of examples and help. Any other credits not mentioned, please email me.

I don't intend to duplicate David's work but just the parts implemented by the solver and re-expressed in my own way. Terms and references to the original will be supplied.

The Exocet Pattern

Pattern Rule 1
Two Base cells (B) exist in alignment in one box and usually contain three or four candidates *in total*. That could mean {1,2,3,4} + {1,2,3,4} but could also mean {1,2,3}+{2,3,4} or even {1,2,3}+{3,4}. I've not found an instance of two bi-value cells like {1,2}+{3,4} yet. To be explored.

Pattern Rule 2
We then check if there are two Target cells (T) that contain all the digits of the Base cell (plus any extras). The Targets cannot 'see' each other or the Base Cells. They must also be in the same Tier or Stack (group of 3x3 boxes in a row or column). The diagram highlights the top tier in red. There are only three possible cells for each Target given that they must not be 'seen'.
If Bases and Targets are aligned in a Tier (as in these diagrams) we look for three Cross-Lines that descend from the Targets and the cell not occupied by the Bases. These are marked in yellow - columns 3, 4 and 7. We are interested in the six cells outside the Tier (or stack). These cells are called S-Cells.

We will be talking about some of the other cells in the pattern as well, so lets introduce them. Each Target has a Companion cell marked with a C. Each Target has two Mirror Cells that are next to each opposite Target, marked M1 and M2.

Lastly, the asterisks are called Escape cells which can hold the candidates that are found to be false in the base.

Pattern Rule 3
To be a real Exocet pattern the Companion Cells must not contain the Base candidates, not even as clues or givens.

Pattern Rule 4
Now one last complication before we can be certain we have an Exocet. In the diagram I've set the Base candidates to be {1,2,3}. There is hopefully a scattering of these in the S-Cells. We want them to appear no more than twice each. If the Cross-Lines are columns we consider the rows where each Base candidate appears. These are called Cover-Lines and I've drawn all the cover-lines in the three colours of the three numbers.

I've stated that cover-lines are perpendicular to cross-lines. Usually they are, but to get the maximum number of elimination rules we can be flexible. It is possible that one Base candidate appears twice only in one column. If we were strictly perpendicular it would require two cover-lines, but all we want to do is "cover" the candidates, so in the single-column case we can "cover" them vertically. So the cover-line = the cross-line.
This will be useful later.

Pattern Inferences

If an Exocet pattern can be confirmed, then following inferences occur:
• The two Target cells must contain different base digits.
• Mirror cells must contain the same base digits as their 'opposite' Target cells together with one digit that is false in the base cells.
• The two true base digits must each be true in two 'S' cells
As David writes, "These are a rich source of eliminations, some of which can be made immediately, and some that will become available later as the solution progresses. These can also be incorporated into AICs."

Note: When we talk about "true Base digits" we are talking about the final solution and knowing what those cells actually contain. We don't know that at the start but eliminations here might tell us more and we can go back through the Exocet check-list.

Elimination Rule 1

Any candidate in a Target cell that is not one of the Base candidates can be removed. That takes out the 4 in B4 and the 2 and 7 in C7

This example actually contains a whole raft of Exocet eliminations but they rely on another test called the Compatible Digit Check. That I have not implemented so I'm going to ignore those. I will come back to this example in the next update when I've understood this test.

Elimination Rule 3

Bird's definition: 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.

Elimination Rule 4

Bird's definition: A base digit in a target that must be true in the other target is false.

Which I take to mean if eliminations thus far reduce one target to one digit ON then it is strongly linked and eliminates like a single.

Elimination Rule 5

Bird's definition: 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

Elimination Rule 8

Bird documentation states "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."
I couldn't make this work without creating incorrect eliminations. By best implementation that doesn't create incorrect removales goes along the lines of "base digit missing from mirror node or base digits missing in the mirrored target cell".

Exocet Exemplars

These puzzles require the Exocet strategy at some point but are otherwise trivial.
They make good practice puzzles. Klaus Brenner finds.

1EnjoySudoku forum

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... by: Anonymous

Okay, scratch my older comments, I've now perfectly understood the logic of the JExocet.

The aim is to find two 'target' cells that will together contain the digits which are in the base.

Say digits A and B are true in the base. In each cross line, A and B must appear once, therefore in the the three cross lines A and B must each appear three times.

Those 3+3=6 instances cannot be in the escape cells as they're seen by the base, they can't be in the companions either due to the pattern definition. So they can only be in targets or S-cells.

There can be only a maximum of 2 S-cell instances as the possible S-cell instances are confined to two cover houses (two instances can't be true in the same cover house) which consequently means that it should appear in at least one target. It can't be true in both targets though as the other base digit would have no target cell to occupy.

Now, the imprtance of mirror cells: Remember that the corresponding mirror pair for a target cell is actually opposite to it. In the example, B4's mirror is C89. We can say that one of the mirror cells (C89) will eventually contain the base digit which will be in the target (B4). The other will contain any other non-base candidate. This is correct because C89 are the only possibles in the box which will be able to contain the base digit, as all the cells in the box are a) seen by the base cells, b) seen by the target, or c) a companion cell.

Also, the target cells seeing each other is perfectly valid as well. In the example the second target C7 could've been in B7, in that case it's mirror should be C56.

So in conclusion, each true base digit must be true two times in S-cells and one time in a target cell, and a mirror cell pair should contain the base digit which is in its target.

Btw, the term 'Exocet' refers to any kind of pattern which has two targets containing two base digits found in base cells. This pattern is a specific case of that, and is called a jExocet (Junior Exocet). This is the most well known Exocet pattern.

... by: Anonymous

@Steven: The target cells can see each other and it is listed as a separate type in David's documentation. The solver doesn't implement it though.

... by: Steven

047000000500000700680700000700002000008500600900031000005900000400000012000000034

Your solver misses an exocet with base cells A1 A2, targets B4 C7, exactly like the pattern in the first diagram

___
How I understand exocet:

The 3 columns where the S cells are have 2 sections:
1. S cells: has at most 2 of each base candidate
2. not S cells: since the columns require 3 of each base candidate, but the S cells can only supply 2, this section must have at least 1

And there has to be base cells which just happen to eliminate all of the non S cells, *except* the target cells and the companion cells.

But the companion cells don't have base candidates.
So any base digit causes a target cell to have that base digit.
2 base cells, 2 target cells = base cells correspond to target cells

___
I'm not sure why the target cells cannot see each other
It still makes a perfectly valid (base cells correspond to target cells) deduction:

If:
1. 3 lines with 2 sections as described above
--- 1. S cells = not in same tier as base cells and companion cells (max 2 base candidates)
--- 2. not S cells = in same tier as base cells and companion cells (min 1 base candidate)
2. Where base cells limit the second section to 4 cells
3. And 2 of those 4 cells are companion cells (don't have base candidates)

Then:
Base cells correspond to target cells (note that there's no restriction with target cells seeing each other)

___
Or why the target cells must together contain *all* base candidates.

If: All pattern requirements are met, except for a base candidate being missing in the target cells

Then: That base candidate must appear once in the (not S cells) region.

Since that candidate is missing in both the target cells and the companion cells, but still appears in the (not S cells) region, it must see the base cells

So the base cells cannot have that candidate!

So the target cells weren't missing a base candidate after all!

... by: Milten

I haven't studied all the rules, but I can help with Rule 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.

A pair of mirror cells must contain exactly one base digit and one non-base digit. That is because they contain the base digit of the opposite target cell, and then the other base digit is eliminated by the neighbouring target cell. So if one of the mirror cells has only base or non-base candidates, we know the other mirror cell will have the other type (base / non-base). Note that we actually don't need a locked digit to use this logic.

Suppose in the example puzzle that the base digits were determined to be 1 and 8 (alternatively, assume the candidate 5 in C9 were eliminated). Then C9 contains only non-base digits, so we may eliminate {2, 7, 9} from C8.

... by: Milten

@Christopher Eastwood

The mirror cells of F3 are D89, not F12. They contain only {2, 6, 9}, from which it follows that F3 is 2 and D8 is 2.

The elimination follows from the stated Pattern Inference: "Mirror cells must contain the same base digits as their 'opposite' Target cells together with one digit that is false in the base cells." So since 3 and 4 are impossible i the mirror cells, they are eliminated from F3.

I admit that I don't think Bird's Rule 8 gives this. But it sounds like the implementation of Rule 8 more likely correspond to Rules 6 and 7 in the comment below. ("My best implementation that doesn't create incorrect removales goes along the lines of "base digit missing from mirror node or base digits missing in the mirrored target cell".") It sounds like a direct implementation of the Pattern Inference to me.

... by: Christopher Eastwood

OK... not trying to be a pest, but Elimination Rule 8 still isn't making sense in the context of the CtC puzzle/video I posted earlier. (https://www.sudokuwiki.org/sudoku.htm?bd=400000002005082900020000030008010000560090078000060500010000060006150700300000004)

ER8 says "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."

Target Cell F3 contains 234 (which are the three Base candidates), F3's mirror cells (F1 and F2) contain 1279 and 3479 respectively... that's 3 non-base digit values (179), not "only one possible" as ER8 says. So how does ER8 apply?

... by: Anonymous

Let me explain all elimination rules(or at least the ones that I fully understand).
I'm going by the order of which David P Bird listed in his document '02 Junior Exocet Definition'.

Main pattern is explained below in an earlier comment so you could always refer to that.

Rule 1: A base digit with one cover house would have to be true in both target cells which would deny another digit with just two cover houses a target cell to occupy. This digit must therefore be true in two 'escape' cross line cells within the JE band, one in line with the base cells and one in the base cross line, so can't occupy the cell common to both these houses.

I do not understand this fully, but I do have a grasp of what it means. I'll come back to this later.

Rule 2: Any base candidate that isn't capable of being simultaneously true in at least one target cell and its mirror node is false.

First of all let me explain why the mirror cell pairs are important. If a target cell contains digit 'd' for example, 'd' must be true in one of the corresponding mirror cells (remember that the corresponding mirror cells are actually opposite to the target cell!). This is because they are the two cells left in the box which can contain 'd'. Three of the cells are eliminated because of the base digits being able to see them. Another three of the cells are eliminated due to the corresponding target being able to see them. And one more is eliminated because the other target cannot contain 'd'.

Now if 'd' isn't capable of being true in both the target and the corresponding mirror node then it can't be true in the target. If the same happens with the other target as well, d is excluded out of the Exocet.

Rule 3: Any non-base candidate in a target is false.

Basically this is what we have named as 'rule 1'. Explanation can be found in the older comment.

Rule 4: A base digit in a target that must be true in the other target is false.

If a target gets reduced to one possible candidate, it is false in the other target since both base digits cannot be the same.

There is a mistake with the article here, they are not strongly linked but rather weakly linked.

Rule 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.

I do not understand this, I will come back to this later in the future.

Rule 6: Any base candidate that can't be true in the mirror node for a target cell is false in the target cell.

This follows on from the explanation I made earlier about mirror cells. If a base candidate cannot be true in any one of the mirror cells, it cannot be true in its corresponding target.

Rule 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.

At least one of the two mirror cells in a pair must contain a base digit. If one is guaranteed to not contain any base digit, then the other must contain a base digit, hence it is restricted to only base digits.

Rule 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.

Same here, I do not understand this fully, but I have a grasp of what it means.

Rule 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.

I do not understand this. I'll come back later when I've understood it.

Rule 10: A known base digit is false in any cells in full sight of either a) both base cells or b) both target cells.

If we confirm that a digit 'n' is true (which the solver currently cannot do, I think) then it is eliminated in the cells which is in view of both base digits since 'n' should be true in either of the base cells. The same can be said about the target cells.

Rule 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.
Rule 12: For a known base digit, any digit instance that would prevent two of its 'S' cells being true is false.

Do not understand both. I'll come back when I've understood these.

If you have some doubts or if there's anything I missed please do reply, I'm pretty active here and will try to reply to your comment as soon as possible.

And big thanks to Andrew Stuart for extending the documentation! I do suggest adding some explanation similar to the above to the documentation, and would love to see some of these implemented in the solver as well!

Cheers!

... by: Anonymous

Dude there is some chaining in this strategy, check out my last post
Andrew Stuart writes:

True
What I was thinking was some kind of chain hanging off the pattern rather than within the pattern, but didn't come out right.
Certainly going to look closely at the comments when I can get my head into this space again
Cheers

... by: Christopher Eastwood

So, CtC used this rule (kind of) in this video: https://www.youtube.com/watch?v=n2No8hU2OwI

When running it through this solver (as Simon did in the video), it hits on the Exocet rule twice... the first time, is just eliminating the other candidates from F3 and D7, which is fine.

But the second time is causing me some trouble. The solver mentions using "Elimination Rule 4" (to eliminate 2 from D7) and "Elimination Rule 8" (to eliminate 3&4 from F3). The explanation above doesn't make any mention of either of these rules.

I understood the outcome of ER4 by basically just running through the chain of logic for every possible answer for cell E7, which all eliminate 2 from D7. But I don't understand the logic used for ER8... maybe it's a chain that I just don't have the patience to figure out, but I'm thinking that maybe there is some other application of the rules that makes this make more sense in a simpler/more straightforward way?

Anyway... any help is appreciated.
Andrew Stuart writes:

Thanks for this comment. I've posted on the TY video

"I've, hopefully, slightly improved my solver by colouring the cover cells (in yellow) and highlighting the base digits in the cover cells. Might make the Exocet pattern a little more clear. Also, the strategy is visited twice because the second round of eliminations rely on the first Exocet set of eliminations from the target cells (green). P Bird does all eliminations in one step but I insist on separating them since some are knock on effects."

I've also listed the elimination definitions I use and I hope I can expand on them to explain them better. However Exocet documentation is very hard to wade through. There's no chaining as such or knock on chaining.

... by: Anonymous

Replying to kf4abh and Robert MauriÃƒÂ¨s:

I think I know how rule 1 works:

The base usually has four candidates for which digit to go in the base cell. If we prove that the two digits in the base cells will be the same as the two digits in the target cells, then we have essentially proved that Rule 1 works, because a foreign candidate cannot exist in the target cells if the base cells have no such corresponding candidate.

Let's say that the base digits are b1 and b2 (these can be any two candidates found in the base cells). Because of this, all the escape cells cannot contain b1 or b2, which creates a massive AIC which removes foreign candidates from the target cells and only leaves base digits.

For example, let us take the example in the article with base digits 1, 5, 8.
Let's say that b1 is 1 and b2 is 5. Then, this AIC is formed:

1[B4] = 5[B4] - 5[E4] = 5[E3] - 5[J3] = 5[J7] - 5[C7] = 1[C7] - 1[D7] = 1[D3] - 1[J3] = 1[J4] -

where "=" is a weak link and "-" is a strong link.

The companion cells and escape cells not having base candidates make sure that the strong links are formed properly, and Pattern rule 4 takes care of the weak links.

If you do have questions please reply. I hope this was clear to everybody

... by: kf4abh

Following on the first comment by Mist...

Each of the two target cells in the Exemplar 1's Exocet pattern do not each contain all the base pair candidates, although together they do. So maybe the language of Pattern Rule 2 could be modified: "We then check if there are two Target cells (T) that together contain all the digits of the Base cell (plus any extras)."

I've just discovered the Exocet pattern and think it's great, but I sure would like to know why it works.

... by: PseudoFish

Example 1 suggests an Exocet where:
- Base is R3C1 and R3C3
- Target is R1C5 and R2C7

Referring to the samples provided, I get that my Cover Lines would be C257; however, I only get one Cross Link with candidate 8 between the Cross Cover Base and CC1. I cannot make any links to CC2 because it is filled with fixed digits. It looks like I am missing two links for this to work, but somehow SW managed to make this work. What am I missing?

... by: Mist

Hello

I just read the article several times to understand the exocet pattern better.
In the only exercise (not including the example in the article) "Exemplar 1, x1 (score 195)" the first (and only) exocet encountered by the solver, does not follow the rules this article teaches us.

Under "The Exocet Pattern - Pattern 2" it clearly says "We then check if there are two Target cells (T) that contain all the digits of the Base cell". However, they don't. So how are we supposed to know that?

The article is confusing in its current state. It would be nice if you could update the article so we could learn more about this very interesting pattern.

Thanks

... by: Mist

Hello

I tried the puzzle in the given example: 007020004930000600600300000000000050200010008006900400003700900020050001000008000

Later in the same puzzle, the solver references exocet elimination rule 9. It removes candidate 8 from mirror 2 in row B colum 8. However, I could not find anything on any other exocet rules, so it just left me confused. I would be thrilled if you could include rule 9 (and perhaps rules 2-8 since there being a rule 9 implies there must be rules in-between the rule 1 showed here and rule 9 from the solver).

Thank you

... by: Robert Mauriès

Hello,
Excuse my English, I'm French.
In your presentation of the exocet you give the result, but you do not give the demonstration.
Where can one find the demonstration of the results of an exocet?
Article created on 26-October-2016. Views: 28375