Tetris Piece Rotation Algorithm

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)?

Solution 1:

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 ]

Solution 2:

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.

1. 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.

2. 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).

3. 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.

4. 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) .

5. 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.

Solution 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.