Blokus Puzzles: 2 Colors 12x14 (no I5's) | Home | Blokus Discoveries | Send Your Feedback |

In 2016, Evan O'Dorney proposed filling a 12x14 rectangle with two colors, omitting the I5's. Presented below are different solutions with two colors, with only corner-touching of the same color, and all pieces of each color in one continuous chain, on a 12x14 board, with no empty squares.

All of these were found by computer and are based on symmetric halves. In most cases, the two halves can be exchanged across a central axis to produce another solution.

Because of the symmetry, the two halves can be rotated 90 degrees
each and rejoined along the top or bottom, creating 6x28 or 7x24 shaped solutions,
as shown here:

This example, like some others, can be played in Bernard Tavitian's preferred order:
which is descending by size (pentominoes first, then tetrominoes, and so on).
Here is one possibility for this case:

U, N, T5, Y, W, Z5, F, P, X, V, L5,

L4, Z4, O4, I4, T4, V3, I3, I2, I1

Two sets of solutions are presented, one based on a 7x12 half, the other on 6x14. It is also possible to start with a 4x21 half, resulting in an 8x21 or 4x42 rectangle.

The solutions are presented based on the location of the X pentomino in (x,y) coordinates, with the upper left being (0,0).

It may be possible to find an asymmetric solution, but that would be a very difficult task.

Related pages:

10x18 with all the pieces,
created out of two 10x9 halves.

12x15 with all the pieces,
created out of two 6x15 halves.

13x14 with all the pieces,
a slightly larger rectangle.