
| Denote the Sierpinski gasket by S and the Line segment by L. |
| Recall the intersection formula: for almost all placements of
S and L in
|
| dim(S ∩ L) = dim(S) + dim(L) - E |
|   |
| For E = 2 this becomes |
| dim(S ∩ L) = Log(3)/Log(2) + 1 - 2 ≈ 0.585 |
|   |
| For E = 3 this becomes |
| dim(S ∩ L) = Log(3)/Log(2) + 1 - 3 ≈ -0.415 |
Return to Dimension Algebra Exercises.