From Hard Drop Tetris Wiki
 
A "checkered" playfield 
Parity is a theoretical concept that explains the exceptional position of T pieces and why zigzag shaped stacks are often suboptimal. If we color the squares (grid cells) of a playfield alternating black and white (as in checkers or chess), then we can make the following statement:
The difference between filled black squares and filled white cells (minos located on black resp. white colored cells) is invariant unless:
 a T piece is played
 a line is cleared
 a garbage line is inserted
In the following we will just drop the word "filled" when we count black and white squares. It is a good idea to balance the number of black and white squares, since this often offers more options where to place a piece.
T Pieces
There are 7 different piece shapes (pieces, "Tetriminoes") in Tetris: I, O, L, J, S, Z and T. I, O, L, J, S and Z pieces will always take 2 black squares and 2 white squares. As a result, the difference between black and white squares will not change, if you place them on a checkered board. T pieces are different: The 3 "arms" lie on the opposite color of the center. As a result, the difference between black and white squares will increase or decrease by 2, if you place a T piece on a checkered board.


The 7 piece shapes. The T piece takes 3 black squares ("arms") and 1 white square ("center"). 
It is important to pay attention to the colors of the T piece centers. Placing the center of 3 successive T pieces on the same color will unbalance your stack. When playing Line Race (Sprint) with Hold feature and 7piece bag randomizer, there is an easy way to ensure that the number of black and white squares are balanced throughout the whole game: Hold the first T piece you get and unhold it after you get another one. Place the 2 T pieces in a way that their "arms" are touching each other.


Examples of how 2 successive T pieces can be placed, so that each pair adds 4 black and 4 white squares to the field. The arms of the darker colored T pieces touch the arms of the brighter ones. 
An unbalanced playfield leaves only a few options to place J, L and O pieces without creating holes. However, it also offers many options for TSpins.



An example of an unbalanced field. Pretty much every overhang you have to create offers an option for a TSpin 
Line Clears
Line clears may also influence the difference between black and white squares. Not every line clear will affect this difference per se. If you clear the top line in your playfield, then 5 black and 5 white squares will disappear which has no effect on the difference at all. On the other hand, line clears will also cause squares above the line clear to fall down. Squares that fall down an odd number of rows will change color (from black to white or vice versa). Note that this effect can be temporarily though. If you clear a line at the very bottom, it will just change the algebraic sign of the difference. If you do this twice, the old sign is restored.




 
Black squares are in the majority.  A row is cleared at the bottom.  White squares are in the majority.  Another row is cleared at the bottom.  Black squares are in the majority again. 
Line clears with some effect
The really interesting case is when you clear a line in the middle of your stack and there is a different number of black and white squares above this line. For example, if you clear the second line from the bottom with an odd number of squares above^{1}, and if you clear the bottom line afterwards, then the difference will be changed permanently.
1: This is equivalent to an odd number of squares in the very bottom line, if you play a mode without garbage lines



Initially, there is an equal number of black and white squares. Then, the 2nd row is cleared with a Z resp. J piece. All squares in the 3rd row fall down an odd number of rows, thus change color. As a result, there are eventually more black squares than white squares.  


 



Initially there are no blocks above the line clear. But this time, the piece adds an odd number of blocks to the lines above the line clear. As a result, there are eventually more black squares than white square.  


 



Clearing 3 lines with no line in between has the same result as clearing 1 line: All blocks above the line clear change color.  



If you clear 2 lines with 1 or 2 lines in between, then all blocks between those 2 cleared lines change color.  



Line clears without any effect
Clearing a Tetris or Skimming 2 lines with an O, S or Z piece will not affect the difference between black and white squares.



On the other hand, clearing 2 lines with no line in between has no effect at all. Blocks above the 2 cleared lines will keep their color, since they fall down an even number of rows.  


 



The same applies for clearing 4 lines at once with an I piece. 
Perfect Clears
Let us assume we want to start a Guideline game with a 4 row (10 pieces) Perfect Clear, but the second T piece shows up very late. So we are forced to complete the perfect clear with exactly 1 T piece used, if we go for the standard PC layout.
Problem is that after we place the first T piece our stack will contain 2 more white squares than black squares (or vice versa). Since there is no second T piece in sight we have to compensate this effect with a special line clear. This usually means that we have to complete the second row before the first row and the first row must contain an odd number of squares when this happens. Another way is to clear the third row before an oddsquared fourth row, but this is barely possible.




 




 




 





Surface
If the stack does not contain any holes (no empty cells with filled cells directly above), then the global difference between black and white squares can be seen directly at the surface. Surface means here the highest filled square for each column. If a column is completely empty, we must take the color of the imaginary row below. We have 10 surface squares and the difference at the surface is double as big as the difference in the stack. If you place a T piece with its center on a white square, then it will decrease the number of white surface squares by 2 and increase the number of black surface squares by 2.
This is also the reason why a blackwhite unbalance causes the stack to be instable. Let us say we have 4 more black squares than white squares in our stack (e.g. by dropping the first 2 T pieces with their center on white squares). Then there is only 1 white square at the surface. Every J, L and O placement must involve this column, if we do not want to create a hole.

global: 12 black, 8 white surface: 9 black, 1 white 12  8 = 4 = (91)/2 
Sidestacked 4 Wide
Most people find it easier to downstack (Combo) a 4wide by having 3 residue (3 minoes in the 4 combo columns). But using an odd number of residue also makes the upstacking part (preparing the combo) easier.
Let us say we use a leftstacked 4 wide. Just consider the surface of the 6 mostleft columns, since we are not supposed to drop any pieces in the 4 mostright columns. When using 3 residue, we start with 4 black squares and 2 white squares. If we place the first T piece on a good spot, we will have 2 black squares and 4 white squares. With another T piece, we are back at 4 black squares and 2 white squares, and so on. So all the time we maintained having at least 2 black and white squares at the surface which is relatively stable in Guideline.
When using 4 residue, we start with 3 black squares and 3 white squares. This is very stable. But when we place the first T piece, we will have 5 black squares and 1 white square (or vice versa). This is not that stable. So when using an even number of residue, we are more often forced to hold and unhold T pieces to avoid those only1whitesquareonsurface situations.
Note, that this is only true, if we start the game with a 4wide. In midgame it may happen that the two described situations are transposed. That is the case if the total number of filled cells is odd (received an odd number of garbage lines).




A 4 wide with 3 residue. At any time, there are at least 2 black and white squares at the surface.  




A 4 wide with 4 residue. If an odd number of T pieces are dropped on the stack, then there is only 1 black or white square at the surface 
Local Unbalances
It is good to have a good balance between black and white squares at the surface. But it also matters, where those squares are located. Imagine the following situation:

There are 5 black and 5 white squares at the surface. All 5 white squares are at the left side and we have a hill in the middle. S, Z and I pieces can be dropped vertically at the left and right side, but this does not affect the color of these columns. J, L and O pieces can only be dropped in those lines where black squares interfere with white squares. This may force us to create holes or form an even bigger hill, if we do not receive a T piece soon.
Such an unbalance can be easier resolved if there is a valley in the middle:



 



 
Initially, the black and white squares at the surface come in cliques. There is just one spot for an J, L or O piece. After dropping an S and a Z piece horizontally, 1 black and 1 white square were separated from their cliques, giving enough space to drop 3 J, L, O pieces simultaneously. At the end the surface is perfectly flat, allowing to drop 5 J, L, O pieces simultaneously. 
Until now we have only paid attention to T pieces, since they change the number of black and white squares at the surface. But we should also pay some attention to J, L pieces as well as horizontal S, Z and I placements, since they change the order of black and white squares. More explicitly:
 J, L: changes the color of 2 adjacent black and white squares
 horizontal I: changes the color of 4 adjacent black and white squares
 horizontal S, Z: changes the color of 2 separated^{2} black and white squares
 horizontal T: changes the color of 2 separated^{2} black squares (resp. white squares)
 vertical T: changes the color of 2 adjacent black squares (resp. white squares)
 O, vertical S, Z, I: no effect
2: Separated means here that there is exactly 1 column in between.


 



From time to time, you may even want to create unbalances, e.g. TSlots (3 adjacent, equally colored squares)



 



