# Empty Rectangle

## Difficulty

## Description

**Empty Rectangle** is a technique, where we look at groups of identical candidates within a box.

The candidates involved form a rectangle, hence the name.

This is a (very reduced) chaining strategy, so **weak links** and **strong links** will play a role. **Check their definitions** on the page about Simple Coloring.

## How does it work?

### Elimination candidate's point of view

The Empty Rectangle technique requires:

**a strong link**between two candidates (If one of the two 6s is correct, then the other is incorrect. If one of the two is incorrect, then the other is correct.)**a box that contains the corner of a rectangle**in the broadest sense (6)**a candidate**in the fourth corner of the rectangle**that can be eliminated**(6)

If 6 **were** correct, then it would eliminate all 6s in its line.

Also it would eliminate the 6 to which it is connected, which then would make the other 6 the solution due to the strong link.

That solution would then eliminate all 6s in *its* line.

Since we're in an invalid state where all 6s of the box have been eliminated, we know that our initial assumption was wrong, and thus we can eliminate the top left **6**.

### Group box's point of view

Another way of showing it is this:

Either one of those 6s will be the solution (directly eliminating the 6)...

... or one of these 6s.

In this case, the bottom right 6 will be false, the top right 6 will be the solution. The consequence is, again, that 6 can be eliminated.

## What is not allowed

It is easy to see that this reasoning **does not work**, if there are additional candidates in the box that **do not line up** with the other corners of the rectangle, like in this example.

What is also **important** is that the link to the right is a **strong link**. If there was an additional candidate in that column, making it a weak link, then we cannot use the "if 6 is false, then the other 6 must be true" logic.

## What is allowed

We can **omit candidates** as long as two groups remain that go in orthogonal directions, so this example is also completely **valid**.

The two groups **don't need to meet in the center of the box**. It can certainly be something like this.