Empty Rectangle
The Empty Rectangle is a single-digit solving technique which uses the absence of candidates to perform an elimination.
An alternative term is hinge, which was actually coined before the term Empty Rectangle, but to a smaller audience. The acronym ER is used by many players.
How it works
Take a look at this diagram:
.-------.-------.-------. | . . . | . . . | . . . | | - A - | - - - | - B - | | . . . | . . . | . . . | :-------+-------+-------: | . . . | . . . | . . . | | . . . | . . . | . . . | | . . . | . . . | . . . | :-------+-------+-------: | . . . | . . . | - X - | | . * . | . . . | X . X | | . . . | . . . | - X - | '-------'-------'-------'
Row 2 has a strong linked pair of candidates in cells A and B. Call that candidate x. Box 9 has all candidates x confined to 1 boxcol and 1 boxrow. The 4 cells without candidate x form the Empty Rectangle (marked with a dash). When x is true in A, x is false in r8c2. When x is false in A, it is true in B, and hence the remaining candidates x in box 9 are confined to row 8, causing x to be false in r8c2. Thus x can never be true in r8c2, and so can be eliminated from r8c2.
This technique is a special case of Grouped Turbot Fish.
DOUBLE EMPTY RECTANGLE
A special case of the simple ER shown above is the double ER in which each of the conjugate pairs is aligned with the non-ER row(column) of an ER box. This can result in additional cell eliminations as illustrated in the partially worked puzzle shown below. Here the double ER consists of the 9 conjugate pair in row 1 and the ER boxes 4 and 5. If r1c2=9 then r5c3 must also be 9 and if r1c5=9 then r5c6 must also be 9. Thus either r5c3 or r5c6 must be 9 and 9 can be eliminated from r5c58.
Double ER Example 1
.-------------------+-------------------+---------------------. | 4 289 3 | 7 289 6 | 28 1 5 | | 6 1 89 | 3 5 289 | 278 79 4 | | 27 2789 5 | 4 1 289 | 3 6 89 | |-------------------+-------------------+---------------------| | 8 79 6 | 5 279 3 | 4 279 1 | | 3 4 79 | 1 2789 289 | 5 279 6 | | 1 5 2 | 6 79 4 | 78 3 89 | |-------------------+-------------------+---------------------| | 5 6 1 | 2 4 7 | 9 8 3 | | 279 278 4 | 89 3 1 | 6 5 27 | | 279 3 78 | 89 6 5 | 1 4 27 | '-------------------+-------------------+---------------------'
The original puzzle is:
003000000 010050000 000400060 800003001 040000500 002600000 000007980 000001000 000000040
In the second Double ER example, the 2 ER boxes are not in the same row(column). It consists of the 4 conjugate pair in row 6 and ER boxes 3 and 7. If r6c9=4 then r2c8=4, and if r6c3=4, then r9c2=4 and also r2c1=4. Therefore either r2c1 or r2c8 must be 4 and it can be eliminated from the four cells. r2c3569. This example requires correction: The second Double ER example in Sudopdia is incorrect, there cannot be a resolved 1 in H7 as indicated in the grid, and there is a candidate 4 in A7 which does not permit box c to hold an ER. The original puzzle is: 003000000 010050000 000400060 800003001 040000500 002600000 000007980 000001000 000000040 The problem is in box j. The original string shown above shows a given 4 in J8, rather than the given 9 in the numeric grid of the example, also the numeric grid in the example has a resolved 1 in H7 immediately next to a resolved 1 in H6 in box j. When these problem are corrected as above, box c holds an ER in 4, but box g holds only a single 4 in H3. Using the 4-strong link in row F this forces F9 to 4 and removes 4’s from F3, E7, E9, D7, D8, D9, C9, B9, and A9. This forces A7 to 4 and removes 4’s from A5, A3, and A2, this forces B1 to 4 which removes 4’s from B3, C3, B5, B6, D1 and E1.
Double ER Example 2
.-------------------+-------------------+---------------------. | 6 347 347 | 8 12347 9 | 13 5 12347 | | 348 1 3457 | 2356 23467 3456 | 9 27 2347 | | 9 2 3467 | 1347 1347 345 | 8 6 1347 | |-------------------+-------------------+---------------------| | 2348 3489 16 | 356 346 3456 | 13 1247 123479 | | 234 349 16 | 7 346 8 | 135 124 123459 | | 7 5 34 | 1 9 2 | 6 8 34 | |-------------------+-------------------+---------------------| | 238 38 9 | 4 2368 7 | 15 1 1568 | | 5 6 2478 | 9 28 1 | 27 3 78 | | 1 378 2378 | 236 5 36 | 27 4 678 | '-------------------+-------------------+---------------------'
EXTENDED FORM of the EMPTY RECTANGLE
The pattern shown in the diagram below is an extension of the ER technique which can result in 1 or 2 additional candidate cell eliminations. In order to use this extended technique there must be an ER pattern and at least one additional conjugate pair with one cell which is a peer either of the cell A in the ER pattern or the two cells in row 8 box 9. This example is the former with two conjugate pairs CE and CD. Note that the ER pattern in the diagram is the same as in the original example.
.-------.-------.-------. | . . . | . . . | . . . | | - A - | - - - | - B - | | . . . | . . . | . . . | :-------+-------+-------: | - - E | . . . | . . . | | - C - | - D - | - - - | | - - - | . . . | . . . | :-------+-------+-------: | . . . | . . . | - X - | | . * * | . * . | X . X | | . . . | . . . | - X - | '-------'-------'-------'
How it works. If B is X then X is in row 8 of box 9 and r8c235 cannot be X. If B is not X, it's conjugate A is X and C which is a peer of A must be not X. Therfore both of C's conjugates D and E must also be X and therefore r8c235 cannot be X. Note that this extension technique can also be used with the 2-string kite
Here is a second example using the same ER pattern. In this case a single additional conjugate pair CD is in column 5 and C is the peer cell.
.-------.-------.-------. | . . . | . - . | . . . | | - A - | - - - | - B - | | . . . | . - . | . . . | :-------+-------+-------: | . . . | . - . | . . . | | . * . | . D . | . . . | | . . . | . - . | . . . | :-------+-------+-------: | . . . | . - . | - X - | | . * . | . C . | X . X | | . . . | . - . | - X - | '-------'-------'-------'
