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:(derv '(mult (add x 2) (add x 3)) 'x) (ADD (MULT (ADD X 2) (ADD 1 0)) (MULT (ADD X 3) (ADD 1 0)))

There is no intelligence in the DERV function. It should be clear that (MULT (ADD X 2) (ADD 1 0)) can be simplified to (ADD X 2). In fact, there are many similar simplifications that can be performed on polynomials. Here is our list.

Lisp form Simplified form --------- --------------- MULT ? 0 0 MULT 0 ? 0 MULT 1 ? ? MULT ? 1 ? ADD 0 ? ? ADD ? 0 ? SUB ? 0 ? EXP ? 0 1 EXP ? 1 ? EXP 0 ? 0 EXP 1 ? 1 [where ? is any Lisp expression]

Assume we have at our disposal a function called SIMPLIFY which will perform these transformations. Here is some behavior.

:(simplify '(add x 0)) X :(simplify '(exp 0 0)) 1 :(simplify '(mult (mult 0 x) y)) 0 :(setq p (derv '(mult (add x 2) (add x 3)) 'x)) (ADD (MULT (ADD X 2) (ADD 1 0)) (MULT (ADD X 3) (ADD 1 0))) :(simplify p) (ADD (ADD X 2) (ADD X 3))

When we apply the SIMPLIFY function to an expression of the form MULT 1 ?, the result should be the result of applying SIMPLIFY to ?. This means that SIMPLIFY is recursive.

The following function implements the above table:

:(define (simplify1 (lambda (poly) : (cond : ((null poly) nil) : ((atom poly) poly) : ((equal 'mult (function poly)) : (cond : ((equal 0 (firstterm poly)) 0) : ((equal 0 (secondterm poly)) 0) : ((equal 1 (firstterm poly)) : (simplify1 (secondterm poly))) : ((equal 1 (secondterm poly)) : (simplify1 (firstterm poly))) : (t (list 'mult : (simplify1 (firstterm poly)) : (simplify1 (secondterm poly)))) )) : ((equal 'add (function poly)) : (cond ((equal 0 (firstterm poly)) : (simplify1 (secondterm poly))) : ((equal 0 (secondterm poly)) : (simplify1 (firstterm poly))) : (t (list 'add : (simplify1 (firstterm poly)) : (simplify1 (secondterm poly)))) )) : ((equal 'sub (function poly)) : (cond : ((equal 0 (secondterm poly)) : (simplify1 (firstterm poly))) : (t (list 'sub : (simplify1 (firstterm poly)) : (simplify1 (secondterm poly)))) )) : ((equal 'exp (function poly)) : (cond : ((equal 0 (secondterm poly)) 1) : ((equal 1 (secondterm poly)) : (simplify1 (firstterm poly))) : ((equal 0 (firstterm poly)) 0) : ((equal 1 (firstterm poly)) 1) )) : (t poly) )))) SIMPLIFY1

SIMPLIFY1 uses some of the help functions from the last chapter. The structure of the function closely follows the list of semplifications given above. Notice the recursive calls when the terms are not constant.

Let's compare it with our idealized SIMPLIFY.

:(simplify1 '(mult (add x 2) (add 1 0))) (MULT (ADD X 2) 1) :(simplify '(mult (add x 2) (add 1 0))) (ADD X 2)

We have here a discrepancy. What is the source of the problem? Well, when SIMPLIFY1 simplifies (ADD 1 0), it gets 1 as it should. However, the test for multiplication by 1 has already been performed before this. SIMPLIFY1 can't make the second semplification. If we view the polynomial as a tree, then SIMPLIFY1 is moving down the tree and any reduction performed on subtrees can't migrate back to the upper levels. How can we beat this conundrum? What we really want to do is to keep applying SIMPLIFY1 to the polynomial until the application no longer results in any change. Let's write a function which goes around in a loop while continually applying SIMPLIFY1 until a final constant expression is reached.

This is one of those cases, when recursion isn't the easiest way to do things. Thus, let's use a PROG! The following is the PROG for SIMPLIFY:

:(define (simplify (lambda (poly) : (prog (poly1) : loop (setq poly1 (simplify1 poly)) : (cond ((equal poly poly1)) (return poly))) : (setq poly poly1) : (go loop) )))) SIMPLIFY :(simplify '(mult (add x 2) (add 1 0))) (ADD X 2)

This function continues to simplify the polynomial until there is no change between two successive simplifications. It then returns the simplified polynomial.

Contents | Differentiating Polynomials | Efficiency and Elimination of Recursion