This is a very subtle yet beautiful extension of logic. We're looking at formations that could potentially be X-Wings but have a corner that's not quite right. In an X-Wing we are looking for four cells in a rectangle which contain a candidate N that exists just twice in either the two rows that form the rectangle or the two columns. Our Finned version is still a rectangle but it has extra candidates of the number in question that prevent one of the two pairs we need from existing.

Let’s look at the distribution of 8 candidates on the board in Figure 1. We have a potential X-Wing marked with the green X. The two blue lines show the top pair of 8s and the potential pair of 8s on the bottom row, F. It’s not a real X-Wing because two cells have gotten in the way. These are the green +8s marked in F8 and F9. If these cells didn’t exist, we’d be able to eliminate 8s in columns 1 and 7 (marked with a green strike-out line).

These +8 cells are the “fin”. The “fin” or “fillet” rule goes as follows:

It's important to remember we can only have one fin at a time!

In our example, the -8 is the only cell that shares a box with the +8 cells. It would have been eliminated anyway if the X-Wing were real. However, none of the other X-Wing eliminations are valid.

In our example, the -8 is the only cell that shares a box with the +8 cells. It would have been eliminated anyway if the X-Wing were real. However, none of the other X-Wing eliminations are valid.

Turning to a real example, consider the potential X-Wing on 7 marked in yellow in Figure 2. We would dearly like to remove all the 7s marked with a green circle. However, the are extra 7s in box 9, marked in green. These are the fin cells. But the fin rule allows us to remove the 7 on H9 at least (red circle).

Now there is more to the idea of Finned X-Wings as I demonstrate in this example.

It so happens, that when using the "fin" or “filleting” rule, it is permissible for the X-Wing to be missing a corner in the finned box. The logic can still be applied! It's going to be fun to explain how and why it works, but first lets look at the example on the right.

We are looking at candidate 4. The fin is again marked in green but the corner of the X-Wing missing. There is no 4 at D6 - which so happens to be a clue, and therefore was never a candidate 4 there at any time! But it doesn't matter, we can remove the 4s from E6 and F6 because they are in the same box as the fin and the potential X-Wing eliminates in the columns in this example.

Where the Finned X-Wing is missing the candidate in the finned box, the type is called a Sashimi Finned X-Wing.

It so happens, that when using the "fin" or “filleting” rule, it is permissible for the X-Wing to be missing a corner in the finned box. The logic can still be applied! It's going to be fun to explain how and why it works, but first lets look at the example on the right.

We are looking at candidate 4. The fin is again marked in green but the corner of the X-Wing missing. There is no 4 at D6 - which so happens to be a clue, and therefore was never a candidate 4 there at any time! But it doesn't matter, we can remove the 4s from E6 and F6 because they are in the same box as the fin and the potential X-Wing eliminates in the columns in this example.

Where the Finned X-Wing is missing the candidate in the finned box, the type is called a Sashimi Finned X-Wing.

It is possible to consider this example in another way. Either there will be a 4 in D9 or there will be a 4 in one of the two cells {D4,D5}. If the later then the 4s in E6/F6 must go (same box situation) or the 4 in D9 forces a chain giving us 6 in H9 and 4 in H6 - which also eliminates the 4s in E6/F6.

## Comments

Comments Talk## Thursday 2-Apr-2015

## ... by: Ate no basil

Is it *required* that the fin be immediately adjacent to a corner of the x-wing?For example, in the 4s example, could the fin be in D3 and D4? (rather than D4 and D5)

If that is indeed the case, then in my example we could still eliminate 4s in E6 and F6 (but not if the fin were shifted further left completely into box 4)

## Wednesday 26-Dec-2012

## ... by: Banjo

In your example of Finned Exwings, why is it that neither row D or H complete the Exwing with row B, instead of relying upon row F? In fact, the combination of rows B, D, and H would seem to give a 2 x 3 situation would seem cannot exist.## Tuesday 27-Mar-2012

## ... by: sean t

David,In the example, the 7 in J7 and J8 cannot be eliminated because it is possible that G6 contains a 7. If that's the case, then C9 does as well, but none of G7, G8, G9 or H9 do. That leaves only J7 and J8 as candidates for the 7 in box 9. Therefore, the 7 in J7 and J8 cannot be eliminated.

In general, for a finned x-wing, only those cells that can see all the fin cells AND are cells that would be eligible for elimination if it were just a regular x-wing can be eliminated. In this example, without the fin cells present, J7 and J8 would not be eligible for elimination via the regular x-wing.

## Wednesday 1-Feb-2012

## ... by: Anton Delprado

I agree with Roland that Finned X-Wings are just a specific form of Grouped X-Cycles. Also they are underneath Grouped X-Cycles in the solver. So why does the solver find them at all? Shouldn't it find them as Grouped X-Cycles first? or does the solver deliberately avoid classing them as Grouped X-Cycles?There are no restrictions on Grouped X-Cycles or any other strategy, especially in terms of 'letting through' to any other strategy. However, what I am searching for may be limited to how I search and the generalization I've made. I'll look again at the correspondance to X-Cycles and see if there is something the solver is missing.

## Wednesday 28-Dec-2011

## ... by: David

In the "real example" given following the Finned X-Wing explanation, the rule that is stated is: "If you can form an X-Wing by ignoring the fin cells, then you can keep your elimination of any cell that shares the same unit as all the cells in the fin." Why, then is only the red-circled 7 eliminated as sharing the same unit, but not the 7s at cells J7 and J8, which appear also to share the same unit (i.e. the box being the unit in which the two yelleo fin cells are located). I thought I understood the fin strategy, but if the 7s at J7 and J8 cannot be eliminated, the theory is completely lost on me! Someone please explain why only the red-circled 7 is elinated but not that other two at J7 and J8. Thx.## Thursday 28-Apr-2011

## ... by: SueinOz

I think in this example that the fin is green not yellow, and also the E5/F6 reference at the end should be E6/F6.This site is so amazing! I'm learning so much. A few typos should be expected when your mind is full of all these incredible strategies. My head hurts just reading them!

## Saturday 26-Feb-2011

## ... by: John Wilcox

Above, under Sashimi Finned X-Wings, 5th paragraph, 2nd line, I think the reference to E5/F6 should be E6/F6. Same correction at the very end of the sentence.## Tuesday 3-Nov-2009

## ... by: suneet

Despite removing tick mark solver does use this strategy. which normally it should not use in that case. I request you to remove this anomalythanks, love your site

## Saturday 18-Apr-2009

## ... by: Roland Zito-Wolf

It's worth mentioning that the finned x-wings can be explained as grouped x-cycles of the rule-3 nice loop variety, the kind with 2 weak links in a row.

For example 1, we have C6=C9-H9-[G7,G8,G9]=G6-C6, allowing 7 at H9 to be removed. This clarifies why the 7 at H9 can be removed but not the ones at J7 and J8: only H9 has a link to C9.

The sashimi variation works because it uses exactly the same kind of grouped x-cycle: H9=H6-E4-[D4,D5]=D9-H9 and likewise for F5.

That raises the question of whether the Swordfish and Jellyfish can be related to more general strategies as well...

love your site!