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

Phil's concise description is impossible to better:

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.

Lets start with the 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'.

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.

We then check if there are two Target cells (T) that contain

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 M_{1} and M_{2}.

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.

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 M

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

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

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

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.

If an Exocet pattern can be confirmed, then following inferences occur:*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.

- 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

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.

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.

They make good practice puzzles. Klaus Brenner finds.

- Exemplar 1, x1 (score 195)
*More as we find them...*

## Comments

Comments Talk## Thursday 4-Feb-2021

## ... by: Anonymous

Replying to kf4abh and Robert MauriÃƒÂ¨s:I think I know how rule 1 works:

The base

usuallyhas 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

## Monday 23-Mar-2020

## ... 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.

## Monday 9-Dec-2019

## ... 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?

## Wednesday 2-May-2018

## ... by: Mist

HelloI 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

## Wednesday 2-May-2018

## ... by: Mist

HelloI 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

## Thursday 2-Feb-2017

## ... 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?