Tiling a Rectangle with L-tetrominoes

Find a necessary and sufficient condition that an a×b rectangle can be exactly covered (completely, and without overlaps) with L-tetrominoes.

The problem has been posted by G. W. Golomb in 1962 and the solution below is due to D. A. Klarner, Humboldt State College, Arcata, California (The American Mathematical Monthly, Vol. 70, No. 7 (Aug. - Sep., 1963), pp. 760-761)

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Solutions

Find a necessary and sufficient condition that an a×b rectangle can be exactly covered (completely, and without overlaps) with L-tetrominoes.

Solution

Since 4 divides ab, a may be taken even. Let a/2 alternate rows of b squares each be colored black in the rectangle.

Impossibility of tiling of a chessboard with 15 L- and 1 square tetromino

Then every L-tetromino in the covering must cover three squares of one color and one square of the other. If m L-tetrominoes cover three black squares and n L-tetrominoes cover three white squares, then 3m + n = ab/2 = 3n + m; hence m = n. This means that the covering must use an even number of L-tetrominoes, and hence that 8 divides ab. Except for the 1×8k rectangle, every rectangle of area 8k can be partitioned into exhaustive, disjoint rectangles of dimensions 2×4 and/or 3×8, but both the 2×4 and 3×8 rectangle can be packed with L-tetrominoes in an obvious way. Hence, the necessary and sufficient conditions are that a and b be greater than 1 and ab = 8k.


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  • Golomb's inductive proof of a tromino theorem
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  • Covering a Chessboard with a Hole with L-Trominoes
  • Tromino Puzzle: Deficient Squares
  • Tiling a Square with Tetrominoes Fault-Free
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  • Tiling a 12x12 Square with Straight Trominoes
  • Bicubal Domino
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