Comments on: SICP section 1.2.1

By: Ivo

Ivo — Mon, 07 Sep 2009 19:36:24 +0000

Note that the Ackermann function as defined in SICP is different from the Ackermann function in both links. They state:
A(0, y) = y + 1
A(1, y) = y + 2
A(2, y) = 2y + 3
A(3, y) = 2^(y+3) – 3
A(4. y) explodes

By: eliben

eliben — Fri, 28 Aug 2009 04:26:51 +0000

OK, I’ve fixed the computation of (A 3 3). Thanks for noticing.

By: zok

zok — Mon, 24 Aug 2009 20:56:05 +0000

There is an error in the calculation of (A 3 3)
You wrote:
; we saw that (A 2 2) is 16
But (A 2 2) is 4 !!!
Right sequence is:
(A 3 3)
(A 2 (A 3 2))
(A 2 (A 2 (A 3 1)))
(A 2 (A 2 2))
(A 2 (A 1 (A 2 1)))
(A 2 (A 1 2))
(A 2 4)
(A 1 (A 2 3))
(A 1 (A 1 A(2 2)))
(A 1 (A 1 (A 1 (A 2 1))))
(A 1 (A 1 (A 1 2)))
So,
(A 3 3) = 2^(2^(2^2))=2^(2^4)=2^16 = 65536

By: Joe Snikeris

Joe Snikeris — Fri, 03 Apr 2009 23:27:05 +0000

@eli,

Cool, thanks for pointing that out.

A recursive definition of h: h(n) = 2^{h(n-1)}

By: gwern

gwern — Mon, 30 Mar 2009 02:23:30 +0000

Looks like some of your formatting is messed up: I see ‘; (add-iter 1 8)’ as ‘add-iter 1 smiley-face’!

By: eliben

eliben — Tue, 10 Mar 2009 17:54:27 +0000

@Joe, Exponentation is right-associative. Hence: 2^2^2^2 = 2^(2^(2^2)) = 2^(2^4) = 2^16 = 65536

By: Joe Snikeris

Joe Snikeris — Mon, 09 Mar 2009 15:49:02 +0000

For (h n) your definition is wrong. It falls apart at (h 4).

(h 0) = 0
(h 1) = 2
(h 2) = 2^2 = 4
(h 3) = 2^2^2 = 16
(h 4) = 2^2^2^2 = 256 <> 65536

h(0) = 0, h(1) = 2, h(n) = 2^{h(n-1)} is more accurate.

By: poq

poq — Wed, 04 Feb 2009 02:22:49 +0000

I think the answer to Exercise 1.10 which calculates A(3,3) is not right

; (A 3 3)
; (A 2 (A 3 2))
; (A 2 (A 2 (A 3 1)))
; (A 2 (A 2 2))
; we saw that (A 2 2) is 16
; So, (A 3 3) = 2^16 = 65536 //here,A(1,16)=65536,A(2,16) >> 65536

By: mike

mike — Wed, 28 Jan 2009 22:36:21 +0000

Oops need to modify my previous post(should have previewed)
(define (newAh1 n)
(define (iter p c)
(if (= c n)
p
(iter (expt 2 p) (+ c 1))))
(iter 2 1))

By: mike

mike — Fri, 16 Jan 2009 20:26:41 +0000

(h n) can also be written by the following iterative process
(define (h n) (define (iter p c) (if (> (+ c 1) p (expt 2 (iter p (+ c 1))))) (iter 2 1))