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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.)

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closed as off-topic by Per Alexandersson, Ryan Budney, Felipe Voloch, Stefan Waldmann, Yemon Choi Nov 17 '14 at 19:47

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question does not appear to be about research level mathematics within the scope defined in the help center." – Per Alexandersson, Ryan Budney, Felipe Voloch, Stefan Waldmann
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Why is the count in this particular case mathematically interesting? – Douglas Zare Feb 1 '10 at 8:10
Further to Douglas' question, what's special about 5? – Yemon Choi Feb 1 '10 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 '10 at 11:51

1 Answer 1

up vote 6 down vote accepted

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 Source code is also available.

I think a proof just involves writing out the cases.

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