SICP Exercise 2.58 modified sum and product infix Part b

Exercise 2.58.  Suppose we want to modify the differentiation program so that it works with ordinary mathematical notation, in which + and * are infix rather than prefix operators. Since the differentiation program is defined in terms of abstract data, we can modify it to work with different representations of expressions solely by changing the predicates, selectors, and constructors that define the representation of the algebraic expressions on which the differentiator is to operate.

a. Show how to do this in order to differentiate algebraic expressions presented in infix form, such as (x + (3 * (x + (y + 2)))). To simplify the task, assume that + and * always take two arguments and that expressions are fully parenthesized.

b. The problem becomes substantially harder if we allow standard algebraic notation, such as (x + 3 * (x + y + 2)), which drops unnecessary parentheses and assumes that multiplication is done before addition. Can you design appropriate predicates, selectors, and constructors for this notation such that our derivative program still works?

SOLUTION

 The code and tests are here.

 Allow standard algebraic notation, such as (x + 3 * (x + y + 2)), which does not use unnecessary parentheses and assumes that multiplication is done before addition.

Note: I have added a new procedure: remove-superfluous-parens that helps simplify expressions.

Let's take this example: (x * 3 + (x + y + 2) + (5 * x ** 2))
We work our way from the left to the right assuming that any expression no matter how complex it is, is a sum of other expressions. So the logic is: look for the first occurrence of the + sign. Everything to its left will be the addend and everything to its right will be the augend. 

 Then we apply the sum rule of differentiation on the addend and the augend just like in the previous exercise. When we do this recursively, all the sum rules will get applied correctly regardless of how many terms are '+'ed together in the overall expression. If we do not find any + operators as we scan the list, we re-scan the list for * operators and apply product-rule. And if we do not find any * operators either, then we look for ** operators.

 This order of evaluation ensures that multiplication is done before addition and exponentiation is done before multiplication. This logic also allows us to treat +, * and ** as if these operators work on only two operands.

Comments

Post a Comment

Popular posts from this blog

SICP Exercise 2.56 differentiation rule

Image
Exercise 2.56.    Show how to extend the basic differentiator to handle more kinds of expressions. For instance, implement the differentiation rule by adding a new clause to the  deriv  program and defining appropriate procedures  exponentiation? ,  base ,  exponent , and  make-exponentiation . (You may use the symbol  **  to denote exponentiation.) Build in the rules that anything raised to the power 0 is 1 and anything raised to the power 1 is the thing itself. SOLUTION The code and tests are here .
Read more

SICP Exercise 1.28 (Miller-Rabin Test)

Exercise 1.28.    One variant of the Fermat test that cannot be fooled is called the  Miller-Rabin test  (Miller 1976; Rabin 1980). This starts from  an alternate form of Fermat's Little Theorem, which states that if  n  is a prime number and  a  is any positive integer less than  n , then  a  raised to the ( n  - 1)st power is congruent to 1 modulo  n . To test the primality of a number  n  by the Miller-Rabin test, we pick a random number  a < n  and raise  a  to the ( n  - 1)st power modulo  n  using the  expmod  procedure. However, whenever we perform the squaring step in  expmod , we check to see if we have discovered a ``nontrivial square root of 1 modulo  n ,'' that is, a number not equal to 1 or  n  - 1 whose square is equal to 1 modulo  n . It is possible to prove that if such a nontrivial square root of 1 exists, then  n  is not prime. It is also possible to prove that if  n  is an odd number that is not prime, then, for at least half the numbers a < n , co
Read more

SICP Exercise 4.18 a alternative strategy for interpreting internal definitions

Exercise 4.18.   Consider an alternative strategy for scanning out definitions that translates the example in the text to (lambda < vars >   (let ((u '*unassigned*)         (v '*unassigned*))     (let ((a < e1 >)           (b < e2 >))       (set! u a)       (set! v b))     < e3 >)) Here  a  and  b  are meant to represent new variable names, created by the interpreter, that do not appear in the user's program. Consider the  solve  procedure from section  3.5.4 : (define (solve f y0 dt)   (define y (integral (delay dy) y0 dt))   (define dy (stream-map f y))   y) Will this procedure work if internal definitions are scanned out as shown in this exercise? What if they are scanned out as shown in the text? Explain. SOLUTION & DISCUSSION The streams based procedure 'solve' in this exercise exposed several issues in my metacircular evaluator implementation. I discovered that streams are not at all handled properly by my evaluator. Moreover, the &
Read more