Tetris Piece Rotation Algorithm - algorithm

Question

What are the best algorithms (and explanations) for representing and rotating the pieces of a tetris game? I always find the piece rotation and representation schemes confusing.
Most tetris games seem to use a naive "remake the array of blocks" at each rotation:
http://www.codeplex.com/Project/ProjectDirectory.aspx?ProjectSearchText=tetris
However, some use pre-built encoded numbers and bit shifting to represent each piece:
http://www.codeplex.com/wintris
Is there a method to do this using mathematics (not sure that would work on a cell based board)?

Answer 1

When I was trying to figure out how rotations would work for my tetris game, this was the first question that I found on stack overflow. Even though this question is old, I think my input will help others trying to figure this out algorithmically. First, I disagree that hard coding each piece and rotation will be easier. Gamecat's answer is correct, but I wanted to elaborate on it. Here are the steps I used to solve the rotation problem in Java.
For each shape, determine where its origin will be. I used the points on the diagram from this page to assign my origin points. Keep in mind that, depending on your implementation, you may have to modify the origin every time the piece is moved by the user.
Rotation assumes the origin is located at point (0,0), so you will have to translate each block before it can be rotated. For example, suppose your origin is currently at point (4, 5). This means that before the shape can be rotated, each block must be translated -4 in the x-coordinate and -5 in the y-coordinate to be relative to (0,0).
In Java, a typical coordinate plane starts with point (0,0) in the upper left most corner and then increases to the right and down. To compensate for this in my implementation, I multiplied each point by -1 before rotation.
Here are the formulae I used to figure out the new x and y coordinate after a counter-clockwise rotation. For more information on this, I would check out the Wikipedia page on Rotation Matrix. x' and y' are the new coordinates:
x' = x * cos(PI/2) - y * sin(PI/2) and y' = x * sin(PI/2) + y * cos(PI/2)
.
For the last step, I just went through steps 2 and 3 in reverse order. So I multiplied my results by -1 again and then translated the blocks back to their original coordinates.
Here is the code that worked for me (in Java) to get an idea of how to do it in your language:
public synchronized void rotateLeft(){
Point[] rotatedCoordinates = new Point[MAX_COORDINATES];
for(int i = 0; i < MAX_COORDINATES; i++){
// Translates current coordinate to be relative to (0,0)
Point translationCoordinate = new Point(coordinates[i].x - origin.x, coordinates[i].y - origin.y);
// Java coordinates start at 0 and increase as a point moves down, so
// multiply by -1 to reverse
translationCoordinate.y *= -1;
// Clone coordinates, so I can use translation coordinates
// in upcoming calculation
rotatedCoordinates[i] = (Point)translationCoordinate.clone();
// May need to round results after rotation
rotatedCoordinates[i].x = (int)Math.round(translationCoordinate.x * Math.cos(Math.PI/2) - translationCoordinate.y * Math.sin(Math.PI/2));
rotatedCoordinates[i].y = (int)Math.round(translationCoordinate.x * Math.sin(Math.PI/2) + translationCoordinate.y * Math.cos(Math.PI/2));
// Multiply y-coordinate by -1 again
rotatedCoordinates[i].y *= -1;
// Translate to get new coordinates relative to
// original origin
rotatedCoordinates[i].x += origin.x;
rotatedCoordinates[i].y += origin.y;
// Erase the old coordinates by making them black
matrix.fillCell(coordinates[i].x, coordinates[i].y, Color.black);
}
// Set new coordinates to be drawn on screen
setCoordinates(rotatedCoordinates.clone());
}
This method is all that is needed to rotate your shape to the left, which turns out to be much smaller (depending on your language) than defining each rotation for every shape.

Answer 2

There is a limited amount of shapes, so I would use a fixed table and no calculation. That saves time.
But there are rotation algorithms.
Chose a centerpoint and rotate pi/2.
If a block of a piece starts at (1,2) it moves clockwise to (2,-1) and (-1,-2) and (-1, 2).
Apply this for each block and the piece is rotated.
Each x is the previous y and each y - the previous x. Which gives the following matrix:
[ 0 1 ]
[ -1 0 ]
For counterclockwise rotation, use:
[ 0 -1 ]
[ 1 0 ]

Answer 3

This is how I did it recently in a jQuery/CSS based tetris game.
Work out the centre of the block (to be used as a pivot point), i.e. the centre of the block shape.
Call that (px, py).
Each brick that makes up the block shape will rotate around that point.
For each brick, you can apply the following calculation...
Where each brick's width and height is q, the brick's current location (of the upper left corner) is (x1, y1) and the new brick location is (x2, y2):
x2 = (y1 + px - py)
y2 = (px + py - x1 - q)
To rotate the opposite direction:
x2 = (px + py - y1 - q)
y2 = (x1 + py - px)
This calculation is based on a 2D affine matrix transformation.
If you are interested in how I got to this let me know.

Answer 4

Personally I've always just represented the rotations by hand - with very few shapes, it's easy to code that way. Basically I had (as pseudo-code)
class Shape
{
Color color;
ShapeRotation[] rotations;
}
class ShapeRotation
{
Point[4] points;
}
class Point
{
int x, y;
}
At least conceptually - a multi-dimensional array of points directly in shape would do the trick too :)

Answer 5

You can rotate a matrix only by applying mathematical operations to it. If you have a matrix, say:
Mat A = [1,1,1]
[0,0,1]
[0,0,0]
To rotate it, multiply it by its transpose and then by this matrix ([I]dentity [H]orizontaly [M]irrored):
IHM(A) = [0,0,1]
[0,1,0]
[1,0,0]
Then you'll have:
Mat Rotation = Trn(A)*IHM(A) = [1,0,0]*[0,0,1] = [0,0,1]
[1,0,0] [0,1,0] = [0,0,1]
[1,1,0] [1,0,0] = [0,1,1]
Note: Center of rotation will be the center of the matrix, in this case at (2,2).

Answer 6

Representation
Represent each piece in the minimum matrix where 1's represent spaces occupied by the tetriminoe and 0's represent empty space. Example:
originalMatrix =
[0, 0, 1]
[1, 1, 1]
Rotation Formula
clockwise90DegreesRotatedMatrix = reverseTheOrderOfColumns(Transpose(originalMatrix))
anticlockwise90DegreesRotatedMatrix = reverseTheOrderOfRows(Transpose(originalMatrix))
Illustration
originalMatrix =
x y z
a[0, 0, 1]
b[1, 1, 1]
transposed = transpose(originalMatrix)
a b
x[0, 1]
y[0, 1]
z[1, 1]
counterClockwise90DegreesRotated = reverseTheOrderOfRows(transposed)
a b
z[1, 1]
y[0, 1]
x[0, 1]
clockwise90DegreesRotated = reverseTheOrderOfColumns(transposed)
b a
x[1, 0]
y[1, 0]
z[1, 1]

Answer 7

Since there are only 4 possible orientations for each shape, why not use an array of states for the shape and rotating CW or CCW simply increments or decrements the index of the shape state (with wraparound for the index)? I would think that might be quicker than performing rotation calculations and whatnot.

Answer 8

I derived a rotation algorithm from matrix rotations here. To sum it up: If you have a list of coordinates for all cells that make up the block, e.g. [(0, 1), (1, 1), (2, 1), (3, 1)] or [(1, 0), (0, 1), (1, 1), (2, 1)]:
0123 012
0.... 0.#.
1#### or 1###
2.... 2...
3....
you can calculate the new coordinates using
x_new = y_old
y_new = 1 - (x_old - (me - 2))
for clockwise rotation and
x_new = 1 - (y_old - (me - 2))
y_new = x_old
for counter-clockwise rotation. me is the maximum extent of the block, i.e. 4 for I-blocks, 2 for O-blocks and 3 for all other blocks.

Answer 9

If you're doing this in python, cell-based instead of coordinate pairs it's very simple to rotate a nested list.
rotate = lambda tetrad: zip(*tetrad[::-1])
# S Tetrad
tetrad = rotate([[0,0,0,0], [0,0,0,0], [0,1,1,0], [1,1,0,0]])

Answer 10

If we assume that the central square of the tetromino has coordinates (x0, y0) which remains unchanged then the rotation of the other 3 squares in Java will look like this:
private void rotateClockwise()
{
if(rotatable > 0) //We don't rotate tetromino O. It doesn't have central square.
{
int i = y1 - y0;
y1 = (y0 + x1) - x0;
x1 = x0 - i;
i = y2 - y0;
y2 = (y0 + x2) - x0;
x2 = x0 - i;
i = y3 - y0;
y3 = (y0 + x3) - x0;
x3 = x0 - i;
}
}
private void rotateCounterClockwise()
{
if(rotatable > 0)
{
int i = y1 - y0;
y1 = (y0 - x1) + x0;
x1 = x0 + i;
i = y2 - y0;
y2 = (y0 - x2) + x0;
x2 = x0 + i;
i = y3 - y0;
y3 = (y0 - x3) + x0;
x3 = x0 + i;
}
}

Answer 11

for 3x3 sized tetris pieces
flip x and y of your piece
then swap the outer columns
that's what I figured out some time

Answer 12

I have used a shape position and set of four coordinates for the four points in all the shapes. Since it's in 2D space, you can easy apply a 2D rotational matrice to the points.
The points are divs so their css class is turned from off to on. (this is after clearing the css class of where they were last turn.)

Answer 13

If array size is 3*3 ,than the simplest way to rotate it for example in anti-clockwise direction is:
oldShapeMap[3][3] = {{1,1,0},
{0,1,0},
{0,1,1}};
bool newShapeMap[3][3] = {0};
int gridSize = 3;
for(int i=0;i<gridSize;i++)
for(int j=0;j<gridSize;j++)
newShapeMap[i][j] = oldShapeMap[j][(gridSize-1) - i];
/*newShapeMap now contain:
{{0,0,1},
{1,1,1},
{1,0,0}};
*/

Answer 14

Python:
pieces = [
[(0,0),(0,1),(0,2),(0,3)],
[(0,0),(0,1),(1,0),(1,1)],
[(1,0),(0,1),(1,1),(1,2)],
[(0,0),(0,1),(1,0),(2,0)],
[(0,0),(0,1),(1,1),(2,1)],
[(0,1),(1,0),(1,1),(2,0)]
]
def get_piece_dimensions(piece):
max_r = max_c = 0
for point in piece:
max_r = max(max_r, point[0])
max_c = max(max_c, point[1])
return max_r, max_c
def rotate_piece(piece):
max_r, max_c = get_piece_dimensions(piece)
new_piece = []
for r in range(max_r+1):
for c in range(max_c+1):
if (r,c) in piece:
new_piece.append((c, max_r-r))
return new_piece

Answer 15

In Ruby, at least, you can actually use matrices. Represent your piece shapes as nested arrays of arrays like [[0,1],[0,2],[0,3]]
require 'matrix'
shape = shape.map{|arr|(Matrix[arr] * Matrix[[0,-1],[1,0]]).to_a.flatten}
However, I agree that hard-coding the shapes is feasible since there are 7 shapes and 4 states for each = 28 lines and it will never be any more than that.
For more on this see my blog post at
https://content.pivotal.io/blog/the-simplest-thing-that-could-possibly-work-in-tetris and a completely working implementation (with minor bugs) at https://github.com/andrewfader/Tetronimo

Answer 16

In Java:
private static char[][] rotateMatrix(char[][] m) {
final int h = m.length;
final int w = m[0].length;
final char[][] t = new char[h][w];
for(int y = 0; y < h; y++) {
for(int x = 0; x < w; x++) {
t[w - x - 1][y] = m[y][x];
}
}
return t;
}
A simple Tetris implementation as a single-page application in Java:
https://github.com/vadimv/rsp-tetris