[math-fun] sl(2,z)-unique convext lattice polygons
hihi, all - it occurs to me that in the few cases that had more than one polygon left, after applying rigid motions and SL(2,Z), i should provide the polygons for n=7, area 6.5 longest edge 3, perimeter 12 (0,0) (1,0) (3,1) (3,2) (2,3) (1,3) (0,1) longest edge 3, perimeter 12 (0,0) (1,0) (2,1) (2,2) (1,3) (-1,2) (-1,1) for n=11, area 21.5 longest edge 4, perimeter 22 (0,0) (1,0) (4,1) (5,2) (5,3) (4,4) (2,5) (1,5) (0,4) (-1,2) (-1,1) longest edge 3, perimeter 22 (0,0) (1,0) (3,1) (4,2) (4,3) (3,5) (2,6) (1,6) (0,5) (-1,3) (-1,2) for n=15, area 51.5 longest edge = 5, perimeter = 36 (0,0) (1,0) (4,1) (6,2) (7,3) (8,5) (8,6) (7,8) (6,9) (5,9) (3,8) (2,7) (0,4) (-1,2) (-1,1) longest edge = 4, perimeter = 34 (0,0) (1,0) (4,1) (6,2) (7,3) (7,4) (6,6) (5,7) (3,8) (2,8) (0,7) (-1,6) (-2,4) (-2,3) (-1,1) for n=20, area 121.0 longest edge = 5, perimeter = 54 (0,0) (1,0) (4,1) (6,2) (9,4) (10,5) (11,7) (11,8) (10,10) (9,11) (7,12) (6,12) (3,11) (1,10) (-2,8) (-3,7) (-4,5) (-4,4) (-3,2) (-2,1) longest edge = 4, perimeter = 52 (0,0) (1,0) (4,1) (6,2) (7,3) (8,5) (8,6) (7,8) (6,9) (4,10) (1,11) (0,11) (-3,10) (-5,9) (-6,8) (-7,6) (-7,5) (-6,3) (-5,2) (-3,1) for n=22, area 164.0 longest edge = 5, perimeter = 62 (0,0) (1,0) (4,1) (6,2) (9,4) (10,5) (11,7) (12,10) (12,11) (11,13) (10,14) (8,15) (7,15) (4,14) (2,13) (-1,11) (-2,10) (-3,8) (-4,5) (-4,4) (-3,2) (-2,1) longest edge = 5, perimeter = 62 (0,0) (1,0) (4,1) (6,2) (9,4) (10,5) (11,7) (11,8) (10,10) (9,11) (7,12) (4,13) (3,13) (0,12) (-2,11) (-5,9) (-6,8) (-7,6) (-7,5) (-6,3) (-5,2) (-3,1) all of these polygons and the programs will eventually be on the web site, but they need explaining and it needs reorganizing (next week probly) more soon, cal Chris Landauer Aerospace Integration Science Center The Aerospace Corporation cal@aero.org
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Chris Landauer