递归计算过程（recursive process）：这种类型的计算过程由一个推迟执行的运算链条刻画。
迭代计算过程（iterative process）：其状态可以用固定数目的状态变量描述。与此同时，又存在一套固定规则，描述了计算过程从一个状态到下一个状态转换时，变量的更新方式，还有一个结束检测，描述计算过程应该终止的条件。
范例：换零钱
1 (define (count-change amount) 2 (cc amount 5)) 3 (define (cc amount kinds-of-coins) 4 (cond ((= amount 0) 1) 5 ((or (< amount 0) (= kinds-of-coins 0)) 0) 6 (else (+ (cc amount 7 (- kinds-of-coins 1)) 8 (cc (- amount 9 (first-denomination kinds-of-coins)) 10 kinds-of-coins))))) 11 (define (first-denomination kinds-of-coins) 12 (cond ((= kinds-of-coins 1) 1) 13 (= kinds-of-coins 2) 5) 14 (= kinds-of-coins 3) 10) 15 (= kinds-of-coins 4) 25) 16 (= kinds-of-coins 5) 50)))
部分习题：
exercise 1.9
1 (+ 4 5) 2 (inc (+ 3 5)) 3 (inc (inc (+ 2 5))) 4 (inc (inc (inc (+ 1 5)))) 5 (inc (inc (inc (inc (+ 0 5))))) 6 (inc (inc (inc (inc 5)))) 7 (inc (inc (inc 6))) 8 (inc (inc 7)) 9 (inc 8) 10 9 11 12 (+ 4 5) 13 (+ 3 6) 14 (+ 2 7) 15 (+ 1 8) 16 (+ 0 9) 17 9
exercise 1.11
1 ;recursive process 2 (define (f n) 3 (if (< n 3) 4 n 5 (+ (f (- n 1)) 6 (* 2 7 (f (- n 2))) 8 (* 3 9 (f (- n 3)))))) 10 11 ;iterative process 12 (define (f n) 13 (define (f-iter a b c n) 14 (if (= n 0) 15 a 16 (f-iter b 17 c 18 (+ (* 3 a) 19 (* 2 b) 20 c) 21 (- n 1))) 22 (f-iter 0 1 2 n))
exercise 1.12
需要输入两个参数，代表第几行第几个元素，索引从0开始，则第一个元素表示为 (pascal 0 0)。
1 (define (pascal row col) 2 (cond ((= col 0) 1) 3 ((= col row) 1) 4 ((> col row) (display "col can't larger then row")) 5 (else (+ (pascal (- row 1) 6 (- col 1)) 7 (pascal (- row 1) 8 col)))))
exercise 1.15
a) p将被使用5次，代换模型如下。
1 (sine 12.15) 2 (p (sine 4.05)) 3 (p (p (sine 1.35))) 4 (p (p (p (sine 0.45)))) 5 (p (p (p (p (sine 0.15))))) 6 (p (p (p (p (p (sine 0.05))))))
exercise 1.16
1 (define (fast-expt b n) 2 (define (fast-expt-iter b n a) 3 (cond ((= n 0) a) 4 ((even? n) (fast-expt-iter (square b) 5 (/ n 2) 6 a)) 7 (else (fast-expt-iter b 8 (- n 1) 9 (* a b))))) 10 (fast-expt-iter b n 1))
exercise 1.17
1 (define (double x) 2 (+ x x)) 3 (define (halve x) 4 (/ x 2)) 5 (define (fast-multi a b) 6 (cond ((= b 0) 0) 7 ((even? b) (fast-multi (double a) (halve b))) 8 (else (+ a (fast-multi a (- b 1))))))
exercise 1.18
1 (define (double x) 2 (+ x x)) 3 (define (halve x) 4 (/ x 2)) 5 (define (fast-multi a b) 6 (define (fast-multi-iter a b product) 7 (cond ((= b 0) product) 8 ((even? b) (fast-multi-iter (double a) (halve b) product)) 9 (else (fast-multi-iter a (- b 1) (+ a product))))) 10 (fast-multi-iter a b 0))
exercise 1.19
1 (define (fib n) 2 (fib-iter 1 0 0 1 n)) 3 (define (fib-iter a b p q count) 4 (cond ((= count 0) b) 5 ((even? count) (fib-iter a 6 b 7 (+ (square p) (square q)) 8 (+ (* 2 p q) (square q)) 9 (/ count 2))) 10 (else (fib-iter (+ (* b q) (* a q) (* a p)) 11 (+ (* b p) (* a q)) 12 p 13 q 14 (- count 1)))))
exercise 1.20
1 ;normal order evaluate 2 (gcd 206 40) 3 (if (= 40 0) 4 206 5 (gcd 40 (remainder 206 40))) 6 (gcd 40 (remainder 206 40)) 7 (if (= (remainder 206 40) 0) 8 40 9 (gcd (remainder 206 40) (remainder 40 (remainder 206 40)))) 10 ...... 11 12 13 ;application order evaluate 14 (gcd 206 40) 15 (gcd 40 6) 16 (gcd 6 4) 17 (gcd 4 2) 18 (gcd 2 0) 19 2
