The following theorems are easy (n>1 assumed in them all): 1: f(n+1) >= f(n) 2: f(2*n-1) >= 2*f(n) 3: f(n+5) >= 3+f(n) 4: f(n+2) >= 1+f(n) PROOFS: 1: just add a new direction and don't use it. 2: take an n-direction configuration twice, and rotate the 2nd configuration around until it shares a point with the 1st. 3: Like Wechsler's proof f(8)>=4 (which is a special case of this); essentially you adjoin |> |> |> to an n-configuration and make the two furthest apart points of the adjoined thing, coincide with old points. 4: Adjoin |> but make one of the new points coincide with an old point. Q.E.D. Application: My cruddy search hack found f(27)>=17 and then applying (3) we deduce f(32)>=20. CONJECTURE: |2*f(n) - 3*n| remains bounded by some constant independent of n. This conjecture arises by counting degrees of freedom D and counting constraining equations E, and hoping that a solution never exists if E>=D+SuitableConstant, but always exists if D>=E+SuitableConstant. Here is the latest table of bounds (where * suffix is improvement): N=3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 L=1 1 2 2 3 4* 4 5 5 6 7* 7 8 9 10*10 11 11 13*13 14 14 15 16* 17* 17 18 18 19 20* U=1 1 2 2*3*8 12 13 17 20 26 28 35 37 44 48 57 60 70 73 83 88 100 104 117 121 134 140 155 160