¹Definition: A Sue de Coq ("SDC") consists of:

1. A base of:
   • 2-3 cells in the same lane and block
   • with a total number of different candidates equal to two more than the number of cells. (i.e. 4 unique candidates in 2 cells, or 5 unique candidates in 3 cells).
2. Two BVC's:
   • A BVC within the base block, containing two of the base's unique candidates.
   • A BVC outside of the block., along the base lane, containing two of the other base's candidates.

²Look For: A lane within a block in which there are 4 different candidates in 2 cells or 5 different candidates in 3 cells (the "base" [yellow cells in the diagram above]). Then see if you can find a BVC in the block with two of the base candidates, and a BVC outside the block, along the same lane, with two of the other base candidates (blue cells in the diagram above).

³Consequence: Within the block, the candidates of the in-block BVC (2 and 3 in our example) can be erased (other than in the SDC cells).
Outside the block, along the lane, the candidates of the out-block BVC (1 and 4 in our example) can be erased (other than in the SDC cells).

So, the erasures take place in the red cells in the diagram above. The darkest red cell (E3) is in both the block and the lane, so all four of the SDC candidates can be erased from it.

⁴Why it Works: The 2-base-cell SDC behaves as a Naked Quad that is bent, split, or "disjointed." In this way it is similar to a WXYZ-Wing, which is also a bent Naked Quad. The differences are: each candidate in the SDC can see all of its matching candidates, and only one cell of the SDC lies outside the block.

⁵If the in-block BVC were located at E3 instead of F3 we would have a Naked Quad in column E, and, as we already know, this would eliminate all of the Quad's candidates from column E outside of the Quad cells themselves.

⁶Conversely, if the out-block cell at E5 were in the block, say at D3 for instance, then the Naked Quad would appear in block 2, and, as we already know, this would eliminate all of the Quad's candidates from block 2 outside of the Quad cells themselves.

⁷Our "bent quad" is not a lane-quad nor a block-quad. It is, however, an overlapping-house [of the base-cells] Quad; the pertinent cells of which are all of the colored cells in the diagram above.

⁸Since one of the Quad's cells lies outside the block (E5), and one lies outside the lane (F3), the effects of the Naked Quad are also split in two: eliminating just the (2,3) from the block, and the (1,4) from the lane.

⁹To prove this, think what would happen if, for instance a 2 were to appear in block 2 outside of the SDC (i.e. not in E2 or F3). This would force E2 to be a 4 and F3 to be a 3, which in turn would force E1 to be a 1. But this would leave no possible candidate for E5; the 1 and 4 already being used in column E (at E1 and E2).

¹⁰Likewise, if a 4 were to appear in column E outside of the SDC (i.e. not in E2 or E4) this would force E5 to be a 1 and E2 to be a 2, forcing E1 to be a 3, leaving no possible candidate for F3; the 2 and 3 having already been spoken for in its house (at E2 and E1).

¹¹You could continue this proof with all of the SDC candidates, and it would lead to the same conclusion: In Block 2 the candidates 2 and 3 can only appear in the SDC cells, and in Column E candidates 1 and 4 can only appear in the SDC cells.

¹²Working through the example on the board, we find the base SDC cells at H4 and I4 for the four candidates 1, 3, 7, and 9. The in-block BVC is I6 (candidates 3 and 9).The out-block BVC, along row 4 is E4 (candidates 1 and 7).

¹³This means that candidates 3 and 9 can be erased from block 6 (outside of the SDC cells). There are no 3's, but we find a 9 in H6 which can be erased.
We can also erase candidates 1 and 7 from row 4 (outside of the SDC cells). We don't find any 1's, but the 7 in B4 can be erased.

¹⁴

Three-Cell Based SDC's

A three-cell based SDC is similar to the two-cell version we've been focusing on up till now. But with a 3-cell base (in a lane within a block) we need five different candidates instead of just four.
We still need an in-block BVC (with two of the base-candidates), as well as an out-block BVC along the lane (with two of the other base-cell candidates). In effect, the 3-cell SDC is a "bent Naked Quint" (just as the 2-cell SDC is a "bent Naked Quad.")

¹⁵We can still erase candidates 2 and 3 from block 2 (other than in the SDC cells), and erase candidates 1 and 4 from column E (outside the SDC cells). We have lost the ability to erase all the candidates from the single cell (E3) that resided in both the block and the lane; it is now one of the SDC cells, and so is untouchable. In recompense, however, we have gained the potential to make even more erasures, as we will next see!

¹⁶There is a candidate in the base cells that is not in either of the BVC cells (F3 and E5): candidate 5. We will call this the "remainder candidate." And here's the great thing about remainder candidates: they can be erased from both the block and the lane (outside of the SDC cells)!So, here are all the potential erasures, shown in red.

¹⁷Why does the remainder candidate give us so many potential erasures? Imagine if one of the red 5's were true. No matter which one you choose it would cross-out the 5 in E3. That would leave five SDC cells with only four possible candidates; one of the SDC cells would be empty, so the 5 cannot occur outside the SDC cells, and so can be erased from them.

¹⁸Looking at a 3-base-cell SDC real puzzle example, we find the three base cells at H4-6. Our five candidates are 1,4,5,6, and 7. The in-block BVC is at G4 with candidates 5 and 6. The out-block BVC is at H8 with candidates 1 and 7.

¹⁹This means that we can erase candidates 5, 6, and 4 (the remainder candidate) from block 6, and candidates 1, 7, and 4 from column H.
There are no 6's to be erased in the block, but we can erase a 4 and three 5's.
There are no 1's to be erased in column H, but we can erase two 4's and 7's.
A total of 8 eliminations just from applying one technique!