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(define (make-sum-list l) (if (= (length l) 2) (list '+ (car l) (cadr l)) (make-sum (car l) (make-sum-list (cdr l))))) (define (make-sum a1 a2) (cond ((=number? a1 0) a2) ((=number? a2 0) a1) ((and (number? a1) (number? a2)) (+ a1 a2)) (else (make-sum-list (list a1 a2))))) (define (make-product-list l) (if (= (length l) 2) (list '* (car l) (cadr l)) (make-product (car l) (make-product-list (cdr l))))) (define (make-product m1 m2) (cond ((or (=number? m1 0) (=number? m2 0)) 0) ((=number? m1 1) m2) ((=number? m2 1) m1) ((and (number? m1) (number? m2)) (* m1 m2)) (else (make-product-list (list m1 m2))))) (define (augend s) (let ((a (cddr s))) (if (= (length a) 1) (car a) (make-sum-list a)))) (define (multiplicand p) (let ((m (cddr p))) (if (= (length m) 1) (car m) (make-product-list m)))) ;; tests (deriv '(* (* x y) (+ x 3)) 'x) ;; (+ (* x y) (* y (+ x 3))) (deriv '(* x y (+ x 3)) 'x) ;; (+ (* x y) (* y (+ x 3)))

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.

See sicp-ex-2.56 for the differentiation program

All that must be done to solve this problem is to change augend and multiplicand so that they return the sum or product, respectively, of the remaining items in the list.

(define (augend s) (accumulate make-sum 0 (cddr s))) (define (multiplicand p) (accumulate make-product 1 (cddr p)))

This works because (accumulate is really awesome.

```
(define (accumulate op initial sequence)
(if (null? sequence)
initial
(op (car sequence)
(accumulate op initial (cdr sequence)))))
```

It recursively applies the make function to add up all the items in the list.

(accumulate make-sum 0 (cddr s))

Is analogous to

(accumulate + 0 (cddr s))

For symbolic data. We must use (cddr s)) to get the rest of the list begining with the 3rd item.

As is mentioned above. All that must be done is to change "augend" and "multiplicand". The above solution is interesting, it use "accumulate" to change the representation of sum and product into the former version. for example: (augend (+ x x x x)) is (+ x (+ x x)), (multiplicand (* x x x x)) is (* x (* x x)). But there is another solution. Here is my code:

;; if there is 4th item, make-sum 3rd item and 4th item (define (augend expr) (if (null? (cdddr expr)) (caddr expr) (make-sum (caddr expr) (cadddr expr)))) ;; if there is 4th item, make-product 3rd item and 4th item (define (multiplicand expr) (let ((first (caddr expr)) (rest (cdddr expr))) (if (null? rest) first (make-product first (cadddr expr)))))

I agree with meteorgan's comment about the use of accumulate. As awesome and simple as it is, it changes the representation away from what we are using. I too came up with similar implementations of "augend" and "multiplicand" with the following changes. Instead of using "make-sum" and "make-product" I simply cons the operator to the beginning of the list.

(define (augend s) (if (null? (cdddr s)) (caddr s) (cons '+ (cddr s)))) (define (multiplicand p) (if (null? (cdddr p)) (caddr p) (cons '* (cddr p)))) ;; using lets might make it easier to understand (define (augend s) (let ((augend-element (cddr s))) (if (null? (cdr augend-element)) (car augend-element) (cons '+ augend-element)))) (define (multiplicand p) (let ((multiplicand-element (cddr p))) (if (null? (cdr multiplicand-element)) (car multiplicand-element) (cons '* multiplicand-element))))

I agree with both AMS and meteorgan's solutions, as mine was quite similar, especially to meteorgan's. My solution is almost exactly the same as meteorgan's, with some slight difference that might make more intuitive sense (but not necessarily better). Instead of using caddr and cadddr to retrieve the 3rd and 4th items of the list, we can use some recursive thinking. Here is my code:

;; if there is no 3rd item, simply return the multiplicand. ;; Otherwise, the multiplicand is the product of the rest of the terms, ie, the product of the multiplier of "cdr p" and the multiplicand of "cdr p" (define (multiplicand p) (if (null? (cdddr p)) (caddr p) (make-product (multiplier (cdr p)) (multiplicand (cdr p))))) (define (augend s) (if (null? (cdddr s)) (caddr s) (make-sum (addend (cdr s)) (augend (cdr s)))))

Note that these two procedures have very similar processes. In the spirit of SICP, we could define a HOP to capture this process.

Here's my attempt at this.

(define (augend s) (if (> (length s) 3) (make-sum (addend (cdr s)) (augend (cdr s))) (caddr s))) (define (multiplicand p) (if (> (length p) 3) (make-product (multiplier (cdr p)) (multiplicand (cdr p))) (caddr p)))

This code accomplishes the same results, but I added simplification to it.

(define (make-sum . l) (let ((lst (if (null? (cdr l)) (car l) l))) ;-> if it's a list inside a list, fix that. (let ((an (map (lambda (x) (if (and (pair? x) (sum? x)) (make-sum (cdr x)) x)) lst))) ;-> this will reduce inner sums. (let ((var-lst (filter (lambda (x) (not (number? x))) an)) ;-> make a list of variable (total (accumulate + 0 (filter number? an)))) ;-> sum all the numbers (cond ((null? var-lst) total) ;-> if there are no variables, than the sum is total ((= total 0) (if (null? (cdr var-lst)) ;-> if total is zero, then return var-lst (car var-lst) ;-> if it has only one element return it (append (list '+) var-lst))) ;-> if it has more, then represent a ;list of the type (+ elements of var-lst) (else (append (list '+) ;-> else, just give a sum var-lst total))))))) (define (augend e) (make-sum (cddr e))) ; and make-product would be the same. (define (make-product . l) (let ((lst (if (null? (cdr l)) (car l) l))) (let ((pn (map (lambda (x) (if (and (pair? x) (product? x)) (make-product (cdr x)) x)) lst))) (let ((var-lst (filter (lambda (x) (not (number? x))) pn)) (prod (accumulate * 1 (filter number? pn)))) (cond ((null? var-lst) prod) ((= prod 1) (if (null? (cdr var-lst)) (car var-lst) (append (list '*) var-lst))) ((= prod 0) 0) (else (append (list '*) var-lst (list prod)))))))) (define (multiplicand p) (make-product (cddr p)))

I have an version will "flatten" nested sum. e.g.

from:

```
(make-sum 'x 5 '(+ y 10) 'z)
```

to:

'(+ 15 x y z)

```
(define (make-sum . s)
(define (sum-iter num-sum symbols seq)
(if (null? seq)
(cond ((null? symbols) num-sum)
((= num-sum 0)
(if (= 1 (length symbols))
(car symbols)
(cons '+ symbols)))
(else (cons '+ (cons num-sum symbols))))
(let ((next (car seq))
(rest (cdr seq)))
(cond ((number? next) (sum-iter (+ num-sum next) symbols rest))
((and (pair? next) (sum? next)) (sum-iter num-sum symbols (append (cdr next) rest)))
(else (sum-iter num-sum (append symbols (list next)) rest))))))
(sum-iter 0 '() s))
```

this is my simple solutin, and this is definitely right.

(define (augend s) (if (> (length s) 3) (cons '+ (cddr s)) (caddr s))) (define (multiplicand p) (if (> (length p) 3) (cons '* (cddr p)) (caddr p)))

Both the solutions above with accumulators and cons symbols are concise and intuitive. In my primary solution, I in fact modified both of the constructor and selectors. But both parts of modifications are small and not difficult to understand.

(define (pure_pair? l) (and (pair? l) (not (sum? l)) (not (product? l)) (not (exponentition? l)))) (define (make-sum a1 a2) (cond ((pure_pair? a2) (make-sum a1 (make-sum (car a2) (cdr a2)))) ((null? a2) a1) ;-> only these two new conditions are added to original make-sum in order to work with list input ((and (number? a1) (= a1 0)) a2) ((and (number? a2) (= a2 0)) a1) ((and (number? a1) (number? a2)) (+ a1 a2)) (else (list '+ a1 a2)))) (define (make-product m1 m2) (cond ((pure_pair? m2) (make-product m1 (make-product (car m2) (cdr m2)))) ((null? m2) m1) ;-> only these two new conditions are added to original make-product in order to work with list input ((and (number? m1) (number? m2)) (* m1 m2)) ((and (number? m1) (= m1 1)) m2) ((and (number? m1) (= m1 0)) 0) ((and (number? m2) (= m2 1)) m1) ((and (number? m2) (= m2 0)) 0) (else (list '* m1 m2)))) (define (augend s) (make-sum (caddr s) (cdddr s))) ;-> augend would be the sum of the rest of the terms (define (multiplicand p) (make-product (caddr p) (cdddr p))) ;-> multiplicand would be the product of the rest of the terms

In my second attempt, I realize that modifications only in selectors should be enough. I end up with new selectors as

(define (augend s) (define (process l) (if (null? (cdr l)) (car l) (make-sum (car l) (process (cdr l))))) (process (cddr s))) (define (multiplicand p) (define (process l) (if (null? (cdr l)) (car l) (make-product (car l) (process (cdr l))))) (process (cddr p)))

When I was doing this, I gradually obtained a strong feeling that recursion should work in this case. I feel I should link the element of list with constructors (make-product / make-sum), together with recursive call of selector it-self (augend / multiplicand). However, I can not do it, because the selectors work with list starting with a sign (+/*). Therefore, an additional function (process) is required to get rid of the sign during recursion. It turns out at the end this method is exactly the same as what accumulate dose. This deepen my understanding that with a higher level abstraction (accumulate), we change our mind of thinking about the subjects at hands and dealing with the problem in a more straightforward way. I still need more practices to get to that level of thinking. This kind of feeling of enlightenment, is just the value of doing SICP.

(define (augend s) (let ((other (cddr s))) (if (= (length other) 1) (car other) (cons '+ other)))) (define (multiplicand p) (let ((other (cddr p))) (if (= (length other) 1) (car other) (cons '* other))))

I reviewed comments up to AThird's. They all use notation like

(+ a (+ b (+ c ...))). But IMHO we can just follow the notation with "arbitrary numbers of (two or more) terms" since the book says "Observe that the later two rules arerecursive in nature" which is done by recursively callingaugendandmultiplicand. So it works by substituting the above(make-sum-list a)with(append '(+) a)(similar for(make-product-list m)).