# weima learns to program

my attempt to do the exercises in sicp.

## Monday, July 21, 2008

### sicp exercise 2.57

;; Exercise 2.57.  Extend the differentiation program to handle sums and products of arbitrary numbers of (two or more) terms. Then the last example above could be expressed as

;; (deriv '(* x y (+ x 3)) 'x)

;; Try to do this by changing only the representation for sums and products, without changing the deriv procedure at all. For example, the addend of a sum would be the first term, and the augend would be the sum of the rest of the terms.

(define (variable? exp) (symbol? exp))
(define (same-variable? v1 v2)
(and (variable? v1) (variable? v2) (eq? v1 v2)))

(define (sum? exp) (and (pair? exp) (eq? '+ (car exp))))
(define (augend exp) (if (null? (cdddr exp))

(define (product? exp) (and (pair? exp) (eq? '* (car exp))))
(define (multiplicand exp) (if (null? (cdddr exp))

(define (exponentiation? exp) (and (pair? exp) (eq? '** (car exp))))

(define (make-sum x y)
(cond ((and (number? x) (= x 0)) y)
((and (number? y) (= y 0)) x)
((and (number? x) (number? y)) (+ x y))
(else (list '+ x y))))

(define (make-product x y)
(cond ((and (number? x) (= x 1)) y)
((and (number? y) (= y 1)) x)
((and (number? x) (= x 0)) 0)
((and (number? y) (= y 0)) 0)
((and (number? x) (number? y)) (* x y))
(else (list '* x y))))

(define (make-exponentiation base exponent)
(cond ((and (number? exponent) (= exponent 0)) 1)
((and (number? exponent) (= exponent 1)) base)
(else (list '** base exponent))))

(define (deriv exp var)
(cond ((number? exp) 0)
((variable? exp) (if (same-variable? exp var) 1 0))
((sum? exp)
(deriv (augend exp) var)))
((product? exp)
(make-sum
(make-product (multiplicand exp)
(deriv (multiplier exp) var))
(make-product (multiplier exp)
(deriv (multiplicand exp) var))))
((exponentiation? exp)
(make-product (exponent exp)
(make-product
(make-exponentiation (base exp) (- (exponent exp) 1))
(deriv (base exp) var))))
(else (error "ERROR-" ))))

(display (deriv '(+ x 3) 'x)) (newline)
(display (deriv '(* (* x y) (+ x 3)) 'x)) (newline)
(display (deriv '(** x 4) 'x)) (newline)

(display (deriv (deriv '(** x 4) 'x) 'x)) (newline)
;; (* 4 (* 3 (** x 2))) ;; needs improvement. should be (* 12 (** x 2))

(display (deriv '(* x y (+ x 3)) 'x)) (newline)