This exotic strategy is strongly related to trivial patterns like Naked Pairs and Naked Triples since we are identifying **Locked Sets**. The building blocks are Almost Locked Sets but the most interesting aspect is the alignment of these parts.

To re-cap the terns:

To re-cap the terns:

- A Locked Set is a group of N cells that can see each other which have N candidates - an example being a Naked Pair
- An Almost Locked Set is a group of N cells which are mutually visible and share N+1 candidates in some combination.

- An Almost Almost Locked Set is a group of N cells which are mutually visible and share N+2 candidates in some combination.
- An Almost Almost Almost Locked Set is a group of N cells which are mutually visible and share N+3 candidates in some combination.

Sue-De-Coq, named after the forum handle of the clever chap who identified it, starts with an AALS which must be aligned in a row or column AND be wholly contained within a box. This restricts the size of the AALS group to two or three cells.

In the first example the two yellow cells (N=2) contain {2,3,5,8} which is N+2, or four candidates. The group is contained within box 4. We don't know which of the values {2,3,5,8} will be the solution to D2 and E2 but clearly two of those four will be. Now, if we look along the unit of alignment (column 2) and within the box we can find single cells that contain two of those candidates. B2 contains {2,8} (the green cell) and F3 contains {3,5}. The AALS can see these cells - which is important!

We now know that the solution to the AALS {D2,E2} cannot be 2/8 or 3/5 since it would leave nothing in the cells B2 and F3.

In the first example the two yellow cells (N=2) contain {2,3,5,8} which is N+2, or four candidates. The group is contained within box 4. We don't know which of the values {2,3,5,8} will be the solution to D2 and E2 but clearly two of those four will be. Now, if we look along the unit of alignment (column 2) and within the box we can find single cells that contain two of those candidates. B2 contains {2,8} (the green cell) and F3 contains {3,5}. The AALS can see these cells - which is important!

We now know that the solution to the AALS {D2,E2} cannot be 2/8 or 3/5 since it would leave nothing in the cells B2 and F3.

The logic is as follows. If neither {2/8} can fill the AALS nor {3,5} then some other combination must fill it that leaves a digit free for the single bi-value cells. So effectively {2,3,5,8} must fill all coloured cells. Indeed the group in total contains four cells and there are four candidates, so we have identified the total group as a Locked Set. This means we can remove all candidates X that see all X in the total group. This excludes the 8 in C2 and J2 (aligned in the column) and 2 G2 (also aligned in the columns) and the 3 in E3 (shares the box).

The example above was a relatively easy pattern - we found a four cell **Locked Set**. Sue-De-Coq can be used in more complicated patterns like the second example. In the first example we used bi-value single cells as the 'hooks' to make the 4-cell locked sets. Bi-value cells are by definition **Almost Locked Sets** since they contain two candidates in one cell. Sue-De-Coq can use larger ALSs to make the pattern.

In the second example we combine an AAALS (N+3) in {E7,E8} containing {1,3,6,7,8} with two normal ALSs {D9,F9} and {E2} containing {1,3,7} and {6/8} respectively. The trick is that the total number of cells, 5, equals the total number of candidates in all the cells {1,3,6,7,8}. We get a 5-cell Locked Set.

In the second example we combine an AAALS (N+3) in {E7,E8} containing {1,3,6,7,8} with two normal ALSs {D9,F9} and {E2} containing {1,3,7} and {6/8} respectively. The trick is that the total number of cells, 5, equals the total number of candidates in all the cells {1,3,6,7,8}. We get a 5-cell Locked Set.

To eliminate we look at what candidates OUTSIDE the pattern can see ALL the candidates INSIDE the pattern. These are the 6s and 8s in row E, and the 1s, 3s and 7s in box 6.

The general terms the rule for the pattern is as follows:

- Find a 2-cell or 3-cell group inside a box that is also aligned on a row or column - call it group C
- C contains a set of candidates, V, which must be two or more than the number of cells in C (N+2, N+3 ALS etc).
- We need to find at least one bi-value cell (or larger ALS) in the row or column which only contains candidates from set V, called D
- We need to find at least one bi-value cell (or larger ALS) in the box which only contains candidates from set V, called E
- The candidates in D and E must be different.
- Remove any candidates common to C+D not in the cells covered by C or D in the row or column
- Remove any candidates common to C+E not in the cells covered by C or E in the box

## Comments

Comments Talk## Tuesday 12-Apr-2022

## ... by: Cerberus

Re: Sue de Cog... you say, generally, the candidates in D & E must be different.

... is that, collectively different, or singular different, would 2, 3 in one, be okay, if 3, 6. was in the other, or must all instances of a 2 or a 3 be absent?

## Friday 11-Jun-2021

## ... by: Jonathan Handojo

In the WXYZ-Wing, you are looking for one non-restricted common. But in this strategy, all candidates are restricted commons. Therefore there's a lot more eliminations in it.So my simplest way of understanding this is: Where you have a Locked Set and all of its candidates in it are restricted commons (i.e. where all instances of each and every one of those candidates that can see each other), you can eliminate any other candidates outside that locked set that can see every instance of those candidates within that locked set.

This is why there's a lot more elimination in the Sue-De-Coq rather than the WXYZ Wing.

## Thursday 28-Jun-2018

## ... by: Jan Laloux

When working on my own solver and tackling this strategy I was trying to figure out how it worked. I found it very complicated to be honest. Thinking it trough I realized however that the complexity is not needed and that SdC is nothing more than a (very) special case of a simple locked set.It is not the ordinary locked set, but what I call a "dispersed locked set". In a normal LS all cell involved are bound to one unit. In a DLS there is no such restriction. I have not seen this described anywhere else so far.

You have a DLS of size n when you have n cells anywhere in the grid with in total n digits, with the requirements that for each digit all candidates for that digit are bound to one unit, and that all cells are "connected". The latter condition is a bit more complicated to explain. I'll demonstrate it with a colouring procedure: colour the first cell, then for each digit that is a candidate in that cell colour all other cells in the set that have that digit as candidate, repeat the procedure for all the newly coloured cells. At the end all cells in the unit must be coloured.

In the first example above we have a DLS of size 4 with cells B2, D2, E2 and F3 and 4 digits 2, 3, 5 and 8. Digits 2 and 8 are bound to column 2, digits 3 and 5 to box 4. The eliminations follow the same logic as with an ordinary LS, resulting in the eliminations shown in the SdC example.

There are in the same way also hidden DLS, dispersed almost locked set (DALS) etc. My solver uses DALS wherever an ALS can be used such in an AIC, death blossom, unique rectangle type 3 etc.

## Wednesday 27-Jan-2016

## ... by: Teige

Impressive brain power at work! Great answer!## Saturday 10-May-2014

## ... by: Sherman

There is a restriction that is implied by the general terms: the size of the set of candidates in C+D+E must equal the number of cells in C+D+E. There are a couple of possible extensions if this restriction is made explicit.The first extension is that C can contain candidates not in D and E. This is a possibility if C is N+3 or larger. If C has one or more candidates not in D and E, these candidates can be removed in the cells not covered by C+D+E in C's box and row or column.

The second extension is that D and E can contain candidates not in C, as long as D and E each contain at least one candidate from C. The candidates in D and E must still be different. The rule saying D and E must be a subset of C is not needed. The two removal rules are modified to say that the candidates from D can be removed from the cells not covered by C+D in C's row or column and the candidates from E can be removed from the cells not covered by C+E in C's box.

This works because C+D+E is a locked set and D and E are mutually exclusive. Between C+D+E, all possible locations of their candidates are fixed.

This is not my work. I found it by perusing other Sudoku websites.

## Tuesday 28-Aug-2012

## ... by: SudoNova

In WXYZ-wing comments, Jeff Sandborn asks if there is a VWXYZ-wing.I think the highlighted cells with digits13678 in example 2 are exactly this.

Following my strategy described in WXYZ-wing, the green-highlighted cell

contains the digits WZ (or perhaps VZ here) and the red highlighted digits

68 in row E are the rogue Z's

## Wednesday 27-Apr-2011

## ... by: Prasolov V.

This strategy "Sue_de_cog" of your solver is not full.## Tuesday 7-Apr-2009

## ... by: Herbert Jensen

I use your Sudoku Solver daily as I don't like the bother of filling in all the cues manually. I've probably read 10 or more books on Sudoku strategies, but your site remains the best documented reference work.