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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.)

edited Feb 1 2010 at 8:23

Yemon Choi
8,465●1●19●45

asked Feb 1 2010 at 8:04

David
9●1

3

Why is the count in this particular case mathematically interesting? – Douglas Zare Feb 1 2010 at 8:10

Further to Douglas' question, what's special about 5? – Yemon Choi Feb 1 2010 at 8:21

I've heard of someone attempting to do this in 4-d, of course 3d projections of the 4d surface... apparently it was so strange that it was unplayable – Michael Hoffman Feb 1 2010 at 11:51

TonyK · Accepted Answer · 2010-02-01 11:14:37Z UTC

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.

Tetris in 3D with 5 units

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