SICP Exercise 1.10
> (A 1 10)
1024
> (A 2 4)
65536
> (A 3 3)
65536
(f n) computes 2n.
The next two problems I had trouble on. I am no math wizard. I initially came up with:
(g n) computes to 2(A(1, n - 1))
and figured that I had the wrong notation and that the answer wasn't really completed because A(1, n -1) makes another recursive call. What I didn't intuitively realize was that this is really 2^n.
(h n)
Same thing with this one the answer is 2^h(n-1).
I found the answer to these both at: http://community.schemewiki.org/?sicp-ex-1.10
Although I am trying to go through these myself I had trouble grasping these ones, but now get it. I also got to read the Ackermann's function Wikipedia entry.
