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exercise1.10.scm
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exercise1.10.scm
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(define (A x y)
(cond ((= y 0) 0)
((= x 0) (* 2 y))
((= y 1) 2)
(else (A (- x 1)
(A x (- y 1))))))
; basically 1 << 10
(A 1 10)
(A 0 (A 1 9))
(A 0
(A 0
( A 1 8 )))
(A 0
(A 0
( A 0
( A 0
(A 0
(A 0
(A 0
(A 0
(A 0
(A 1 1))))))))))
(A 0
(A 0
( A 0
( A 0
(A 0
(A 0
(A 0
(A 0
(A 0 2)))))))))
(A 0
(A 0
( A 0
( A 0
(A 0
(A 0
(A 0
(A 0 4))))))))
(A 0 512)
1024
; second one
(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)))
(A 1
(A 1
(A 0
(A 1 1))))
(A 1
(A 1
(A 0 2)))
(A 1
(A 1 4))
(A 1
(A 0
(A 1 3)))
(A 1
(A 0
(A 0
(A 1 2))))
(A 1
(A 0
(A 0
(A 0
(A 1 1)))))
(A 1
(A 0
(A 0
(A 0 2))))
(A 1
(A 0
(A 0 4)))
(A 1
(A 0 8))
(A 1 16)
; at this point it's just the original
65536
; third one
(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
(A 1 2))
(A 2
(A 0
(A 1 1)))
(A 2
(A 0 2))
(A 2 4)
; we know what (A 2 4) is already
65536
; 2n
(define (f n) (A 0 n))
; 1 << n or 2^n
(define (g n) (A 1 n))
(g 4)
(g 12)
; written in png; h(n) = 2^h(n -1) except when n = 0
(define (h n) (A 2 n))
; 0
(h 0)
; 2
(h 1)
; 4
(h 2)
; 16
(h 3)
; 65536
(h 4)
; call stack too deep
(h 5)
(define (h2
(cond ((= y 0) 0)
((= y 1) 2)
(else (A 1
(A 2 (- y 1))))))