Returning parameter for Lisp function

Question

I'm trying to get this function to display the literal expr2 and expr1 as they are entered. The data that is entered is of the form (+ x y).

(DEFUN  deriv (expr var)    ;  function name and arguments
  (COND  ( (ATOM  expr)     ;  check for atomic expression (variable or constant)
              (IF  (EQL  expr var)
                  1     
                  0     
              ) )       
        ( (EQL  (FIRST  expr)  '+)  (deriv-add (SECOND expr) (THIRD expr) var) )
        ( (EQL  (FIRST  expr)  '-)  (deriv-minus (SECOND expr) (THIRD expr) var) )
        ( (EQL  (FIRST  expr)  '*)  (deriv-multi (SECOND expr) (THIRD expr) var) )
        ( (EQL  (FIRST  expr)  '/)  (deriv-divide (SECOND expr) (THIRD expr) var) )
        ( T  (ERROR  "UNKNOWN arithmetic operator"))
  )
)

(DEFUN  deriv-multi (expr1 expr2 var)
   (LIST '+ (* (deriv expr1 var) expr2) (* expr1 (deriv expr2 var)))
)

(SETQ e2 '(* (+ x y) (+ y 7)) )
(DERIV e2 'x)

Rainer Joswig · Accepted Answer · 2011-11-17 22:51:23Z

3

This is a good basic introduction to Lisp:

http://www.cs.cmu.edu/~dst/LispBook/

'Common Lisp: A Gentle Introduction to Symbolic Computation' by David S. Touretzky.

Free PDF.

Btw., your function displays nothing. It calls the function 'LIST'. Why are there parentheses around expr1and expr2?

About your code:

it is good coding style in Lisp to use self-explaining code. Since you can use symbols to really name things (and not be abbreviations of concepts), you can use descriptive symbols. Entering longer symbols is then usually done by using symbol completion in the editor. This allows you to get rid of the comments.
don't add additional space. Write compact code.
format your code nice and compact.
use indentation and additional lines where it helps reading the code

Example:

(defun derive (expression variable)
  (if (atom expression)
      (if (eql expression variable) 1 0)
    (funcall (case (first expression)
               (+ #'derive-addition)
               (- #'derive-subtraction)
               (* #'derive-multiplication)
               (/ #'derive-division)
               (otherwise (error "unknown arithmetic operator")))
             (second expression) (third expression) variable)))

(defun derive-multiplication (expression1 expression2 variable)
  (list '+
        (* (derive expression1 variable)
           expression2)
        (* expression1
           (derive expression2 variable))))

Above function still has an error, which is easy to fix. You don't want to execute the multiplication.

(defun test ()
  (assert (equal (derive '(* (+ x y) (+ y 7)) 'x)
                 '(+ (* (+ 1 0)
                        (+ y 7))
                     (* (+ x y)
                        (+ 0 0)))))))

edited Nov 17, 2011 at 22:51

answered Nov 17, 2011 at 21:37

Rainer Joswig

140k11 gold badges230 silver badges357 bronze badges

Sign up to request clarification or add additional context in comments.

8 Comments

Brian Over a year ago

I added more code to better demonstrate what it's trying to be used for.

Rainer Joswig Over a year ago

@Brian: what for are the parentheses around expr1?

Brian Over a year ago

Sorry, that was a mistake. I removed them.

Rainer Joswig Over a year ago

@Brian: you create a list with the + in front. But you execute the multiplication. Do you want to do that?

Brian Over a year ago

I added a test case, and this to my description. Running with (DERIV e2 'x) should result in the output (1 + 0) * (y + 7) + (x + y) * (0 + 0). Though in prefix notation. If that doesn't answer your question let me know.

|

Collectives™ on Stack Overflow

Returning parameter for Lisp function

1 Answer 1

8 Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

8 Comments

Your Answer

Sign up or log in

Post as a guest

Related