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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
If this question can be reworded to fit the rules in the help center, please edit the question.

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
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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