Interactive Sudoku Technique
Sue de Coq: When a Line and Box Share the Candidate Budget
Sue de Coq divides a carefully balanced group of candidates across a line, a box, and the cells where those houses intersect. Once the candidates have been partitioned, each companion set locks digits into its own house and removes them from outside cells.
From Two ALSs to Two Sectors
The Next Step After ALS-XZ
ALS-XZ taught us to treat a group of cells as one logical object. Sue de Coq keeps that set-based viewpoint, but replaces the single RCC relationship with an exact division of responsibility between two intersecting houses.
ALS-XZ
Two ALSs share an RCC named X. At least one ALS must provide the shared elimination candidate Z.
Sue de Coq
One intersection and two companion sets divide their candidates into a line family and a box family.
The intersection must absorb exactly what the two companion sets cannot.
Three Participating Regions
The Anatomy of a Sue de Coq
Begin where a row or column crosses a box. The intersection has two surplus candidates. A companion set in the line accounts for one surplus; a companion set in the box accounts for the other.
Intersection
The green region lies in both houses. In the common form it contains N cells with N+2 distinct candidates.
Line Companion
The amber ALS lies in the same row or column, outside the box, and shares one candidate family with the intersection.
Box Companion
The blue ALS lies in the same box, outside the line, and shares a different candidate family with the intersection.
The intersection is not an ordinary ALS in this example. It has two cells and four candidates—N+2 rather than N+1. The two companion ALSs remove those two degrees of freedom.
A Complete Extended Sue de Coq
Column 2 Meets Box 4
Green identifies the two intersection cells. Amber identifies the three-cell column ALS. Blue identifies the two-cell box ALS. Each split red cell is a victim; its non-red half identifies the set responsible for the elimination.
| Role | Cells | Candidate union | What it guarantees |
|---|---|---|---|
| Green intersection | r4c2 {269}r6c2 {259} |
{2,5,6,9} |
One of {2,9} and one of {5,6}. |
| Amber column ALS | r3c2 {37}r7c2 {279}r8c2 {23} |
{2,3,7,9} |
Contains 3, 7, and one of {2,9}. |
| Blue box ALS | r5c1 {45}r5c3 {456} |
{4,5,6} |
Contains 4 and one of {5,6}. |
| Eliminations | r1c2#7r6c1#4,5r6c3#4 |
Four candidates | Remove all four candidates. |
Account for Every Candidate
Six Checks Complete the Deduction
Advance through the actual set accounting. The proof never guesses which candidate occupies either green cell.
- Inventory the green intersection: two cells, four candidates.
- Confirm the amber column ALS: three cells, four candidates.
- Confirm the blue box ALS: two cells, three candidates.
- Partition the green cells between
{2,9}and{5,6}. - Lock
{2,3,7,9}into the marked Column 2 cells. - Lock
{4,5,6}into the marked Box 4 cells.
Begin with the green cells shared by Column 2 and Box 4.
The two sectors are now fully accounted for.
r1c2 ≠ 7r6c1 ≠ 4,5r6c3 ≠ 4
Two Arms, One Intersection
Resolve Each Sector Separately
The line and box deductions do not compete. Each companion ALS leaves one digit for the green intersection, and each arm then becomes complete inside its own house.
Amber Column Arm
The amber ALS has three cells for {2,3,7,9}.
Because green cannot contain 3 or 7, amber must contain both.
Amber takes one of 2 or 9; green takes the other.
Not yet resolved
Blue Box Arm
The blue ALS has two cells for {4,5,6}. Because
green cannot contain 4, blue must contain 4. Blue takes one of
5 or 6; green takes the other.
Not yet resolved
Both companion ALSs have surrendered one candidate to the green intersection. Column 2 and Box 4 now each contain a complete locked family, so every marked red candidate can be removed.
The Doorway to Set Equivalence
Stop Asking Which Cell; Start Balancing Regions
Sue de Coq is still a local pattern, but its proof is already set arithmetic. We know what each companion region must contain, what each one must leave behind, and where that remainder must go.
Set Equivalence Theory expands this same habit beyond one box-line intersection: compare regions known to contain equivalent digit collections, subtract what they share, and reason from the remainder.
A Disciplined Search
How to Look for Sue de Coq
1. Inspect Intersections
Look where a line crosses a box for two cells with four candidates or three cells with five candidates.
2. Split the Families
Find one companion set in the line and another in the box. Their intersection-candidate families must remain disjoint.
3. Audit Both Houses
Determine exactly which digits become locked into the marked cells of each house, then remove only those outside candidates.
Before accepting the eliminations
- The intersection contains at least two cells in both the line and box.
- Each companion group is contained entirely in its respective house.
- The companion groups use disjoint candidate families from the intersection.
- Every extra companion candidate is supported by enough companion cells.
- Each victim lies outside the pattern but inside the house whose digits are locked.
Check the Accounting
Four Quick Sue de Coq Decisions
Count the Intersection
Two green cells contain {2,5,6,9}. How many surplus candidates?
Test the Amber Set
Are three cells with candidate union {2,3,7,9} an ALS?
Resolve the Blue Set
Which digit must occur in the blue ALS because green cannot contain it?
Read a Victim
Which candidate is removed from r6c3 {2349}?
0 of 4 decisions confirmed
Questions Worth Asking
Sue de Coq Frequently Asked Questions
Is Sue de Coq another name for ALS-XZ?
No. Both use set reasoning, but ALS-XZ connects two ALSs through an RCC. Sue de Coq divides candidates between a line, a box, and their intersection.
Why was it originally called Two-Sector Disjoint Subsets?
The pattern operates in two intersecting sectors: one line and one box. The intersection candidates are divided into disjoint families handled by the two companion sets.
Must the companion sets be Bivalue Cells?
No. That is merely the smallest and easiest form to recognize. This example uses a three-cell amber ALS and a two-cell blue ALS, demonstrating an extended Sue de Coq.
Is the green intersection an ALS?
Not here. It contains two cells with four candidates, giving it two surplus candidates. The two companion ALSs account for one surplus apiece.
Why are only some candidates removed from the red cells?
Each arm locks only its proven candidate family. The amber arm removes 7 from r1c2. The blue arm removes 4 and 5 from r6c1 and 4 from r6c3. No other candidate has been proved false.
Next: Generalize the Accounting
From a Local Partition to Set Equivalence Theory
Sue de Coq balances candidates at one box-line intersection. Set Equivalence Theory takes the next step by comparing larger regions that must contain equivalent collections of digits and reasoning from what remains after their overlap is removed.