[To avoid any confusion, note that it would not be possible to generate these particular boards during an actual Tetris game.]
Note that in the example above, the dimensions have reversed. The old grid was 7 rows by 10 columns, whereas the new grid is 10 rows by 7 columns. Our number of rows and number of columns swap!
We want to make sure that our piece remains centered in the same location as before (to avoid having pieces 'drift'); that means that the piece's location may need to change. To compute where the new location is, we must calculate where the center of the old piece was, set it equal to the center of the new piece, and then work backwards to calculate the new piece's location.
Observe that the center row of the falling piece can be computed as the sum of its top row plus half its height in rows. That is:
centerRow = fallingPieceRow + fallingPieceRows//2
We actually have two pieces: the old falling piece (before the rotation), and the new falling piece (after the rotation). We can compute each of their centers as such:
oldCenterRow = oldRow + oldNumRows//2
newCenterRow = newRow + newNumRows//2
Note that oldNumRows and newNumRows are not the same, since the dimensions change when the piece is rotated. In any case, to keep the old and new centers the same, we just set oldCenterRow equal to newCenterRow in the equations above.
oldRow + oldNumRows//2 = oldCenterRow = newCenterRow = newRow + newNumRows//2
Then we solve for newRow in this new equation, which gives us:
newRow = oldRow + oldNumRows//2 - newNumRows//2
We then repeat for the column, and we're done!Finally, we need to update the piece itself. This is done in two parts: finding the new row, and finding the new column.
Column: Consider what happens as we move in the old grid from red to green (that is, moving downward with rows increasing from 0 to 6). This maps in the new grid to moving across with columns increasing from 0 to 6. Thus, our new column is equal to our old row.
Row: Consider the other dimension, as we move in the old grid from red to white (that is, moving across with columns increasing from 0 to 9. This maps in the new grid to moving up with rows decreasing from 9 to 0. Thus, our new row is equal to (9 minus our old column). More generally, we replace "9" with "one less than the number of old columns".
Now, to write rotateFallingPiece, we take the following steps. If you aren't sure how to implement any of these steps, re-read the explanation above!
- Store the data associated with the old piece (its dimensions, location, and the piece itself) into temporary local variables. We'll need these if we need to undo the action.
- Compute the number of new rows and new columns according to the old dimensions.
- Compute the new location of the piece according to the old location.
- Generate a new 2D list based on the new dimensions, and fill it with None values.
- Iterate through each of the cells in the original piece, and move each value to its new location in the new piece.
- Set fallingPiece and the other variables equal to their new values.
- Check whether the new piece is legal. If it isn't, restore the values based on the temporary variables we created in Step 1.
