02-Mar-2019 09:30:30 polyomino_multihedral_example_2x4_test: MATLAB version Set up and solve the 2x4 rectangle polyomino tiling example. Region R: 1 1 1 1 1 1 1 1 Polyomino N: 1 Polyomino O: 1 1 1 Polyomino P: 0 0 1 1 1 1 System matrix A and right hand side B: 1 0 0 0 0 0 0 0 1 0 0 0 0 0 1 0 1 0 1 0 1 0 1 0 0 0 0 0 0 1 1 0 0 0 0 1 1 0 1 1 1 1 0 0 1 0 0 0 0 0 1 1 0 0 1 0 1 1 0 0 1 1 1 0 0 0 1 0 0 0 0 0 1 0 0 0 1 0 1 0 0 0 1 1 0 0 0 0 1 0 0 0 0 0 1 0 1 0 1 0 1 0 0 0 1 0 0 0 0 0 1 0 0 0 0 1 1 1 1 0 1 1 1 0 0 1 0 0 0 0 0 0 1 0 0 0 1 1 1 1 0 0 1 1 1 0 1 0 0 0 0 0 0 0 1 0 0 0 1 0 1 0 0 0 1 0 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 1 1 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 3 3 3 3 4 4 4 4 4 4 4 4 8 Wrote the LP file "2x4.lp" RREF has determinant 4 Row-Reduced Echelon Form system matrix A and right hand side B: 1 0 0 0 0 0 0 0 0-1-1-1 0 0 1 0 1 0 1 0 0 0 1 0 0 0 0 0 0 0 0-1-1 0 0 1 1 0 1 1 1 0 0 0 1 0 0 0 0 0 0 0-1-1 0-1 0 0-1-1 0 0 -1 0 0 0 1 0 0 0 0 0 1 0 0 0 1 0 1 0 0 0 1 1 0 0 0 0 1 0 0 0 0 0 1 0 0-1 0-1 0-1-1-1 0 0 0 0 0 0 1 0 0 0 0 1 1 0 0-1 0 0 0-1-1 0 0 0 0 0 0 0 1 0 0 0 1 1 0 0-1-1 0 0 0-1 0 0 0 0 0 0 0 0 1 0 0 0 1 0 1 0 0 0 1 0 1 1 0 0 0 0 0 0 0 0 1 1 1 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 4 binary solution vectors x: 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 1 0 0 1 0 0 1 0 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 Translate each correct solution into a tiling: Tiling based on solution 1 9 9 9 14 5 14 14 14 Tiling based on solution 2 15 15 15 4 15 12 12 12 Tiling based on solution 3 1 20 20 20 11 11 11 20 Tiling based on solution 4 17 10 10 10 17 17 17 8 polyomino_multihedral_example_2x4_test: Normal end of execution. 02-Mar-2019 09:30:35