28 Apr
2003
28 Apr
'03
9:46 a.m.
I think the correspondence works like this. The single node tree is "x". Making a node f2 a child of f1 represents f1^f2. Since (f1^f2)^f3 is just f1^(f2*f3) we can think of it as f1 raised to both f2 and f3, that is, f1 with f2 and f3 as children. So the functions you posted are (sideways) o--o--o--o (f1) o--o--o (f2) \ o o--o--o (f3) \ o o / o--o (f4) \ o o--o (f5) \ o--o Russ