This method may look hard, but is actually easy to spot and very common. We look at the intersection
of a square with either a row or a column. The intersection consists of 3 cells in a square which are aligned
either in a row or a column. If a given candidate appears only in the intersection with regards
to either the square or the row/column, we can eliminate it from the other house.

Since we look at a square versus row or column and the eliminations can appear in both, we can
differentiate between four cases (each is illustrated with an example).

- Square x Row - elimination in the square
- Square x Row - elimination in the row
- Square x Column - elimination in the square
- Square x Column - elimination in the column

Consider the intersection of row **A** with the top-middle square. The intersecting cells are
highlighted in dark blue. Looking at candidate **9**, we see that it appears only in the intersecting cells in
row **A**. Hence, the **9** will be the value of either **A4** or **A5**, but we do not know where
exactly. What we do know is that both of these are in the top-middle square and so no other **9** can be
in the square. We can eliminate the candidate **9** from **C5**. To see why the method works, let
us consider placing the **9** on **C5**. This forces its elimination from **A4** and **A5**,
and there will be no place for the **9** in row **A**.

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In the next example, we are faced with a similar situation - an intersection between
a row and a square, but this time the elimination appears in the row.
Consider the intersection of row **H** with the bottom-middle square. In this square the candidate
**2** appears only in **H5** and **H6**. Hence the value **2** will be in one of these, but
cannot be anywhere else in row **H**. Again, if we think of placing **2** in, say, **H7**,
this would leave the square without a cell for the **2**.

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The previous two examples were looking at an intersection between a row and a square.
Symmetrically, we can search for an intersection between a column and a square.
Considering row **1** and the bottom-left square, we see that candidate **2** appears in
the column only at the intersection. Hence, we can remove all other **2**s in the square.

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Finally, when the intersection is between a column and a square, we can eliminate from the
column. The candidate **5** is eliminated from column **6**, by using the same logic.

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