How to Solve Bricks Puzzles
Bricks, like Sudoku, is a type of latin square puzzle. All latin square puzzles use the same broad template:
- Start with an NxN square grid.
- Fill in cells with digits from 1 to N.
- Don’t repeat numbers in any row or column.
Beyond that, each variant of a latin square puzzle has its own special requirements to give it a unique feel. For a Bricks puzzle, this involves the 1×2 “bricks” subdividing the grid.
Rules
Fill in the grid with the numbers 1 to N (N being the length of a side), without repeating a digit in any row or column.
- Each 1×2 “brick” contains one odd and one even number.
- If the dimensions of the grid are even (6×6, 8×8, etc), half-bricks at each end of a row are considered one brick.
- Otherwise, half-bricks are simply a digit not used elsewhere in the row.
Basic Techniques to Solve
In Bricks puzzles, the techniques tend to work together in such a way that it’s impractical to give examples without using a whole grid. This time, I’m just going to summarize the tips, and we’ll go straight into solving our example puzzle.
- Prioritize rows and columns with more known numbers.
- Use the even/odd requirement of whole bricks to narrow down candidates. Note that in any odd-sized wall (9×9, 7×7, etc), the half-bricks at the ends will always be odd.
- Cross-reference with multiple rows/columns to find a number position in one row/column.
- Marking candidates is a powerful tool.
Solving the Puzzle
In Sudoku puzzles, you would normally start by scanning along a row or column of a given number to find its position in one of the 3×3 regions. Because Bricks puzzles don’t have these regions, those methods don’t help, and it can be difficult to find where to start. That’s why priority one is always to look for rows and columns that have the most given numbers. Ideally, you want them to be mostly even, or mostly odd.
Breaking In
This cell is part of the purple brick, red column, and blue row. The brick requires an even number, and 2, 4, and 6 are already used. So it must be an 8. 
Now we can solve this brick, which also needs an even number. 2, 4, and 8 are used, so it must contain a 6. 
We have another brick that needs an even number. 4, 6, and 8 are already used around it, so this cell must be a 2.
On the same row, we still need an 8. We have a full brick left, and the 8 in this column forces it onto the left side. 
At this point, we’ve run out of cells with only one option. So we’ll think more broadly. In this column, we’ve used all the even numbers. 
What this means is that in these four bricks, the blue side must be an odd number, and the red side must be even. So let’s look at the possible candidates that result.
Odd and Even
Notice that 1, 5, and 7 are the only options in these three cells. Because of this, you can eliminates them from the green cell, leaving only 9. 
Meanwhile, in column 8, these three even cells, together with the 6, mean that the rest are odd. Therefore, their counterparts in column 9 are odd. 
Look, we found a 6! And again, we know where all the even numbers are, so the remaining cells must be odd.
Now we have a 1-3 pair in two cells, which means we can eliminate them elsewhere in the column. That puts a 5 in the green cell, and a 9 in the purple one. 
Next, this 1-7 pair causes a chain reaction. The green cell must be 5, forcing the purple cell to be a 1, and the red cell is a 3. 
Learning which cells are odd and even is a powerful tool to eliminate candidates. Watch what happens when we mark known odd/even cells in the whole grid.
Each row and column contains 4 even and 5 odd digits. So where all of one type are accounted for, the remaining cells must be the opposite. 
In these rows, the even digit in the brick is the last one, so the half-brick must be odd. Of course, in any odd-sized grid (9×9, 7×7, etc), the half-bricks will all be odd. 
Now we have all 5 odd cells in the first column, so the remaining cells must be even.
Single Candidate Chain Reaction
Well, the grid is certainly colorful now, even if we didn’t figure out all the cells. Let’s see what that did to possible candidates in all these cells. 
Right away, we reveal two cells with only one option. If we hadn’t determined even/odd locations first, these cells would have seemed to have more options. 
Solving that 9 eliminates it from this cell. The resulting chain reaction starts with this 1.
Next, we have one candidate left in these two cells. 
That causes single candidates in these cells. 
We created more single candidates.
And now we found a 1 , 3 and 5. 
Finally, we mark these two cells, ending the chain. 
Here, the 2-4 pair eliminates the 4 in this cell, leaving a 6.
Subtle Tricks
In this row, 3, 5, and 9 are already used, so they can’t be in this brick. But, there is also a 7 in both columns, eliminating it. That means the green cell must be a 7. 
That puts a 1 in the green cell, which forces the purple cell to hold a 7. 
In this cell, we can use a normal Sudoku technique. 1, 2, 3, 4, 5, 7, 8, and 9 are used around it, so it must contain a 6.
This starts another chain reaction, as these cells now have a single candidate. 
This leads to two more cells with one option. 
We have two more solvable cells. The one with no candidates is simply the only cell left in its row.
We end the chain with these two cells. 
In the first column, the 2-8 pair eliminates those options in the green cell. That only leaves a 6. 
We’re nearing the end of the solve. To speed things up, let’s see the candidates in these remaining bricks.
Finishing Up
Right away, we find several cells with single candidates. 
This results in a chain that spreads to these cells. 
That leads to these four solvable cells.
Next, we reveal the final candidates for these three cells. 
That only leaves the final two cells. 
We’ve solved the Bricks puzzle!
