¹Definition: An Empty Rectangle (ER) consists of a digit which occurs only along one row and one column of a block (the remaining cells forming an otherwise irrelevant "empty rectangle"), where one of these lanes points to one end of a perpendicular conjugate pair (CP) of the same digit. The intersection of the row and column forms the Empty Rectangle Intersection (ERI). If the same digit appears in a cell that can see both the ERI and the other end of the conjugate pair, it can be erased.

²Look For: Any configuration of two to five of the same digit within a block which can be seen (after imagining any missing cells along the paths) as forming one horizontal and one vertical line. A test of the correct configuration is if the remaining cells form an empty rectangle (which need not be contiguous.) Here are numerous examples of valid patterns, with their ERI's highlighted in gray (note that some can have two conceivable ERI's depending on which way you imagine the lines going):

• Example 1.
• Example 2.
• Example 3.
• Example 4.
• Example 5. Two ERI's since each 5 could represent the horizonal or the vertical line.
• Example 6.
• Example 7.
• Example 8.
• Example 9. Two ERI's since each 5 could represent the horizonal or the vertical line.
• Example 10. Two ERI's since each 5 could represent the horizonal or the vertical line.

³You then need to see if one end of a CP is pointed to by a line running from the ERI. The CP must run perpedicular to the ER line, and must span at least two blocks.

⁴Consequence: If the digit exists as a candidate in the cell that can see both the ERI and the other end-point of the CP, the candidate can be erased from that cell. Note: since there are so many variations of the ER pattern, and they can be hard to spot, you may find it easier to backtrack to find them: find a CP first, and see if one ends points to an eraseable cell, and the other end points to an ERI of an ER.

⁵Why it Works: The key is the CP and the fact that the digit must be placed along one of the ER lines in the ER block. In our example, we have an Empty Rectangle in Block 1 for digit 2, with the ERI at C3. The CP is at C6-F6. Since the candidate 2 at F3 can see both the opposite end of the CP as well as the ERI, we can erase candidate 2 at F3.

⁶But why? Well, let's start with the CP: we know that either C6 or F6 will resolve to a 2.

• If F6 resolves to a 2, then obviously the 2 at F3 can be erased.
• If C6 resolves to a 2, that eliminates all the 2's in column C, so in block 1 only row 3 can resolve to a 2 (either A3 or B3) making them pointing Locked Candidates, so again the 2 at F3 can be erased.

Either way, the 2 at F3 is a goner.

⁷Your Turn!
Find the Empty Rectangle in this puzzle along with the CP it points to. Then find the cell that can see the ERI as well as the opposite end of the CP.Erase the candidate that the ER technique eliminates.

Hint