prime boxes of (31) 18.02.93 / 01.06.93 / 07.08.93 Torsten Sillke 1 1 31 complete 2 1 31/32 complete // 024.06.94 2x5x 2p, 4, 6, 8, 10, 12, 13p, ... 12..13+2n 4x5x 2, 4, 5p, ... 4..5+2n 5x5x 4p, 6p, 8, 9p, 10, 11p, ... 8..11+4n 6x5x 2, 4, 5p, ... 4..5+2n 7x5x 4, 6, 7p, 8, 9p, ... 6..9+4n 2x10x 2, 4, 5, ... 4..5+2n 3x10x 4p, 6p, 7p, 8, 9p, ... 6..9+4n 2x15x 2, 4, 6, 7p, ... 6..7+2n 3x15x 4p, 6p, 8, 9p, 10, 11p, ... 8..11+4n 2x20x 2, 4, 5, ... 4..5+2n 3x20x 4, 6, 7, 8, 9, ... 6..9+4n 2x25x 2, 4, 6, 7, ... 6..7+2n 3x25x 4, 6, 7p, 8, 9, ... 6..9+4n 2x2x5x 1p, ... n 2x3x5x 2, 3p, ... 2..3+2n 3x3x5x 2p, 3p, ... 2..3+2n 2xZx 2, 3p, ... 2..3+2n 3xZx 2p, 4, 5p, ... 4..5+2n Impossible: 2x3xN (dies out) 3x3xZ 2x5x{3,5,7,9,11} 3x5xN (dies out after 40 steps) 5x5x5, 5x5x7, 3x7x15 The impossible proofs use the dist2 coloration: * . * . * . . . . . * . * . * Polycubes, which are subsets of the octacubes: 2 2 2 2 2 2 or 2 2 can hit one * only. A longer explanation is made for pentacube (61). prime boxes of (31/32) with reflection. 2x5x 2p, 4, 5p, ... 4..5+2n 3x5x 8p, 12p, ... {8,12}+8n 4x5x 2, 4, 5, ... 4..5+2n 5x5x 4p, 6p, 8, 9p, 10, 11p, ... 8..11+4n 2x10x 2, 4, 5, ... 4..5+2n 3x10x 4p, 6p, 7p, 8, 9p, ... 6..9+4n 2x15x 2, 4, 5, ... 4..5+2n 3x15x 4p, 6p, 8, 9p, 10, 11p, ... 8..11+4n 2x20x 2, 4, 5, ... 4..5+2n 3x20x 4, 6, 7, 8, 9, ... 6..9+4n 2x25x 2, 4, 6, 7, ... 6..7+2n 3x25x 4, 6, 7p, 8, 9, ... 6..9+4n 2x2x5x 1p, ... n 2x3x5x 2, 3p, ... 2..3+2n 3x3x5x 2p, 3p, ... 2..3+2n 2xZx 2, 3p, ... 2..3+2n 3xZx 2p, 4, 5p, ... 4..5+2n Impossible: 2x3*N (dies out) 3x3xZ 3x5xu (with u odd) 3x5xk (k = 2 (modulo 4)) // computation at the end of this text 3x5x4 5x5x5, 5x5x7, 3x7x15 Annotations: The 3x5xe (with e even) have zero defect for the dist two coloration, (x,y,z) with x, y in 2Z, 31/32 can hit this coloring at most two times. There are point-symmetric tilings (31/32) for 3x5x8, 3x5x12. 4 4 4 4 4 8 8 8 8 8 give you 3x5x8 and 3x5x12. 3 3 4 5 5 7 7 8 9 9 4 4 4 4 4 8 8 8 8 8 The 2x2x5x Box is unique: 2 3 two times 3 2 The 2x5x13 Box is unique: 32 32 29 29 27 27 21 21 17 17 13 13 10 33 31 31 27 27 23 22 21 17 19 13 11 10 33 33 31 28 26 23 22 22 18 16 14 11 11 34 35 35 28 28 24 25 22 18 18 14 15 11 35 35 30 30 28 25 25 20 20 18 15 15 12 33 32 32 29 29 23 23 19 19 17 17 13 13 33 34 32 27 29 23 24 21 19 19 14 14 10 34 34 31 28 26 24 24 21 18 16 14 10 10 34 30 31 26 26 24 20 22 16 16 12 12 11 35 30 30 26 25 25 20 20 16 15 15 12 12 The 3x4x10 Box (2 solutions): 33 33 28 28 25 25 two times 34 32 28 25 25 34 35 35 29 29 35 35 29 29 . 31 33 33 28 28 32 32 33 25 26 34 34 30 24 29 35 30 30 27 27 31 31 26 26 . 32 31 31 26 26 32 34 27 24 24 30 30 27 27 24 24 The 4x5x5 Box: 17 17 13 11 11 18 18 13 14 . 18 17 13 13 11 18 17 14 14 11 18 15 14 13 10 19 17 14 12 11 19 15 15 12 10 19 16 16 12 10 19 19 15 12 12 16 16 15 10 10 16 The 3x6x10 Box: 43 38 38 34 34 27 27 17 17 15 43 43 34 34 27 27 21 17 15 15 44 43 35 30 30 25 21 21 12 12 44 44 35 35 28 23 23 21 13 12 45 44 39 39 31 31 24 24 13 13 45 39 39 31 31 24 24 20 20 13 43 37 38 38 34 22 19 19 17 17 41 40 35 38 27 29 19 15 15 10 41 41 35 30 25 25 23 18 11 10 45 45 33 30 28 28 23 21 13 12 45 44 33 36 26 31 20 16 16 12 42 39 36 36 32 24 20 20 14 14 41 37 37 29 29 22 22 19 19 11 41 40 37 37 29 29 22 22 11 11 40 40 33 33 25 28 23 18 11 10 40 42 33 30 25 28 18 18 10 10 42 42 32 32 26 26 18 14 16 16 42 36 36 32 32 26 26 14 14 16 you can tile a L3x2x5: unique a b b c c a a c c f - b b d d c - bent - e a a f f - b e e d d e e f f d one L3*2*5 and three 2x2x5 give the 2x3x3x5. a solution of 2x3xZ: a a b b . . a c d e . . f f e e c c a a b b . f c c d d b . f f e e d d ----------------------------------------------------------- (31\32) Impossibility of the 3x5xk box for k = 2 (modulo 4) Try to fill a 3x5xk Brick. What are the borders of a partial packing? Catalogue this borders with 6+12n pieces. There are three possible savings: 1) Use symmetry (factor 4) 2) Use the coloration (a,b,c) in {0..2}x{0..4}xN is back iff a and b are even. For packings of the 3x5xk (k even) there mustn't be any defect, each piece must hit two black ones. So partial packings with defect are uninteresting. 3) Only extendable partial packings are interesting. If there is a packing of the 3x5xk box it is of course extendable ad infinitum, as you can iterate it. (I look for an enlargment of 20 pieces.) There are 3 different borders at the beginning according to these reductions. pieces | new borders 6 3 18 24 30 7 42 0 As none of this borders is a right plane, you can't fill a 3x5xk (k=2 (mod 4)). The borderline is in each case only two levels wide. This is encoded with the Schroeppel code for packings (binary coding of each pile). The one-line representation "abcdefghijklmno" expands to a d g j m b e h k n c f i l o Table of different borders: 101311131320101 111011131311101 131020131311101 --------------- 101320131113101 101320131311101 111110131113101 111110131311101 131110101113111 131110101311111 131110101323101 131110111110131 131110111113101 131110111311101 131110131011111 131110131020131 131110131023101 131110131110111 131110131320101 131311101020131 131311101023101 131311101320101 131311111011101 131311111110101 131320101011131 131320101110131 131320101113101 131320131110101 --------------- 101311111311101 101323101311101 101323131110101 111311101113101 111311101311101 131320101311101 131320131011101 --------------- nothing new