**Background:** There are 7 "bricks" used in the game of Tetris. These are the 7 *unique* combinations of 4 unit squares in which every square shares at least one edge with another square. ("unique" in this case refers to the idea that no brick can be rotated in 2-D space to become another brick.)


**Question:** Using 5 unit cubes, how many *unique* "bricks" could be formed in which each cube shares at least one face with another cube? (Please provide a proof to this in your answer if you can find one.)



Tetris in 3D with 5 units
There are 29 distinct 5-cube bricks (counting mirror images as distinct). Together with one 1-cube brick, one 2-cube brick, two 3-cube bricks, and eight 4-cube bricks, these constitute the brick set for the highly addictive 3-D Tetris game Blockout II, available at <http://www.blockout.net/blockout2/>. Source code is also available.


I think a proof just involves writing out the cases.


