In a Sue de Coq, two cells where a row or column crosses a box hold four candidates between them. Pair them with a bivalue cell elsewhere in the line and another elsewhere in the box, chosen so the two share no digit and together cover all four. Each outside cell's digits can be removed from the rest of its own house, outside the pattern cells.
Where to look
Every box meets three rows and three columns, and each meeting is three cells long. For this pattern, exactly two of those three cells must be empty, and between them they must hold four different candidates. That is two more candidates than cells, so the pair can't lock on its own.
Next, look along the line, away from the box, for a bivalue cell: a cell with exactly two candidates, both of them intersection digits. Then do the same inside the box, away from the line. The two cells you pick must not share a digit, and together they must cover all four.
Find the intersection
Row nine crosses box seven in columns one to three. Row nine, column one is a given 8, so the intersection is the two empty cells in columns two and three.
Count its candidates
Row nine, column two holds 1, 6 and 9, and row nine, column three holds 1, 2 and 6. Between them they hold 1, 2, 6 and 9. Four candidates in two cells: two more than cells.
Pick a cell in the row
Along row nine, outside box seven, row nine, column four holds 6 and 9. Both are intersection digits, so it can serve as the row cell.
Pick a cell in the box
Inside box seven, outside row nine, row seven, column three holds 1 and 2. It shares no digit with the row cell, and the two cells together cover 1, 2, 6 and 9.
The four cells lock
The two intersection cells take two different digits. They can't be the 6 and the 9, because the row cell sees both and would be left with nothing. They can't be the 1 and the 2 either, because the box cell sees both. So the intersection takes one digit from each pair, and the four cells hold 1, 2, 6 and 9 once each.
Clear the row
Row nine's 6 and 9 now sit in the intersection and the row cell. Any other cell in row nine would lose its 6 or 9. Columns five and eight hold neither, so here the row has nothing to give.
Clear the box
Box seven's 1 and 2 sit in the intersection and the box cell. Row seven, column two holds a 1, so that 1 goes, and the cell is left with 3 and 9.
Why it works
The intersection places two of the four digits, and the outside cells place the rest. The strip shows the two ways that could go wrong, then the only way left.
So row nine's 6 and 9 are confined to three cells of row nine, and box seven's 1 and 2 to three cells of box seven.
In general: the outside cell in the line removes its digits from the rest of the line, and the outside cell in the box removes its digits from the rest of the box. The four pattern cells keep every candidate they have.
Row nine, column five holds 2 and 4. The 4 is not an intersection digit, so this cell can't be the row cell. Row nine, column eight fails the same way, with 1 and 4.
Row nine, column four holds 6 and 9, both intersection digits.
Bigger versions
The pattern grows in two ways. All three cells of the intersection can be empty, holding at least five candidates between them. And either outside cell can become a larger almost locked set, such as two cells with three candidates. The rule stays the same: the line set's digits leave the rest of the line, and the box set's digits leave the rest of the box.
The in-game hint will not show a three-cell intersection, or a set of four or more cells. The game's Sue de Coq check looks only at intersections with two empty cells, with sets of up to three cells beside them.
How it relates to other moves
In its simplest form, Sue de Coq is a restricted common candidate move under another name. Join the intersection to one outside cell and you get an almost locked set: three cells in one house holding the four digits. The other outside cell holds two of those digits, and each copy of either one sees every copy in the set, so both digits are restricted.
Which cell joins the intersection depends on the house you clear. To clear the box, join the row cell. Here that gives row nine, columns two to four, holding 1, 2, 6 and 9. The box cell shares its 1 and 2 with this set. With the 2 as the restricted digit, the rcc rule removes the 1 from any cell that sees every 1 in both sets, which is row seven, column two. The game's rcc check finds this pair and strikes the same 1.
To clear the row, the intersection plus the box cell form an almost locked set in box seven, which links to the row cell through both of its digits, the 6 and the 9. That pairing clears the row, and on this grid the row has nothing to clear.
Longer ALS chains remove this 1 as well.
Common slips
Counting a given as part of the intersection. Row nine, column one is an 8. Only the two empty cells count, and they must hold four candidates.
Picking an outside cell with a digit the intersection lacks. Row nine, column five holds a 4, so it can't be the row cell.
Letting the two outside cells share a digit. Then they cover three of the four digits at most, and nothing locks.
Clearing the wrong house. The box cell's 1 and 2 leave box seven only. Row seven, column three also sits in row seven, but row seven keeps its 1s and 2s, and row seven, column five is the 2 in the solution.
What the in-game hint shows
The in-game hint may show a different or simpler step first. On this grid it opens with a Y-wing. Row five, column eight holds 4 and 6, with wings at row five, column three (1 and 6) and row nine, column eight (1 and 4), so row nine, column three loses its 1. The intersection still holds 1, 2, 6 and 9 between its two cells, so the pattern stands.
The next hint is a 3D Medusa step that strikes the same 1 in row seven, column two.
On the starting grid, the restricted common candidate move removes this 1, and so does an ALS chain. AIC and Nishio remove it as well. The game checks all of these before Sue de Coq, so the hint seldom shows this move. A different route can still be right.
Where a row or column crosses a box at two empty cells holding four candidates, look for a bivalue cell in the line and one in the box that split those four between them. The line cell's digits then leave the rest of the line, and the box cell's digits leave the rest of the box.
Where this sits on the path
This is the sixth technique of the last chapter, after ALS chains. Death blossoms come next and close the chapter.
On a different grid, column three crosses box one at two empty cells that hold 3, 5, 7 and 9 between them. Row four, column three holds 5 and 9, and row three, column two holds 3 and 7. Which of these candidates does the pattern remove?
Each outside cell's digits leave only its own house.
Recap
- Look for
- Two empty cells where a line crosses a box, holding four candidates, plus a bivalue cell in the line and one in the box that split those four between them.
- Scope
- The line and the box that meet at the intersection.
- Result
- The line cell's digits leave the rest of the line, and the box cell's digits leave the rest of the box. The pattern cells keep theirs.
- Tell-tale
- Two bivalue cells on either side of an intersection, with no digit in common.
- If it fails
- If the outside cells share a digit, or one holds a digit the intersection lacks, try another cell in the line or the box.