car and cdr in Scheme are driving me crazy ...

Question

Hi Im facing a problem with the car and cdr functions

for example:

first I defined a list called it x

(define x (a (bc) d ( (ef) g ) ))

so x now is equal to (a (bc) d ( (ef) g ) )

now for example I need to get the g from this list using only car and cdr (!! noshortcuts as caddr cddr !!) the correct answer is:

(car(cdr(car(cdr(cdr(cdr x))))))

BUT how ? :-( I work according to the rules (the car gives the head of list and cdr gives the tail)

and instead of getting the answer above I keep reaching wrong answers. Can any one help me in understanding this ... give me step or a way to solve it step by step

Thanks in advance. I'm really sick of Scheme.

Answer 1

Have you tried using a REPL (read-eval-print-loop) such as csi? That way you can work on this interactively, which will make it easier for you to learn and work through this (and other) problems using Scheme.

Answer 2

(cdr x) = ((bc) d ( (ef) g ) )
(cdr(cdr x)) = (d ( (ef) g ) )
(cdr(cdr(cdr x))) = (( (ef) g ) )
(car(cdr(cdr(cdr x)))) = ( (ef) g )
(cdr(car(cdr(cdr(cdr x))))) = (g)
(car(cdr(car(cdr(cdr(cdr x)))))) = g

Answer 3

Try to do it step by step:

• cdr yields the list without the first element
  

• car yields the first element of a list

                                x         is  (a (bc) d ( (ef) g ))
                           (cdr x)        is  (  (bc) d ( (ef) g ))
                      (cdr (cdr x))       is  (       d ( (ef) g ))
                 (cdr (cdr (cdr x)))      is  (         ( (ef) g ))
            (car (cdr (cdr (cdr x))))     is            ( (ef) g )
       (cdr (car (cdr (cdr (cdr x)))))    is            (      g )
  (car (cdr (car (cdr (cdr (cdr x))))))   is                   g
  

Answer 4

do it iteratively. Also, realize that scheme always looks backwards.

(cdr x) = ( (b c) ...)
(cdr (cdr (cdr x))) = (( (ef) g))
(car (cdr (cdr (cdr x)))) = ((ef) g)
(cdr (car (cdr (cdr (cdr x))))) = (g)
(car (cdr (car (cdr (cdr (cdr x)))))) = 'g

hope that helps

Answer 5

do the transforms one at a time. cdr gives you a list without the first element, car gives you the first element.

(cdr (a (bc) d ( (ef) g ) )) -> ( (bc) d ( (ef) g ) )
(cdr ( (bc) d ( (ef) g ) ))  -> ( d ( (ef) g ) )
(cdr ( d ( (ef) g ) ))       -> ( ( (ef) g ) )
(car ( ( (ef) g ) ))         -> ( (ef) g )  <- pulls the first element out, which happens to be a list itself
(cdr ( (ef) g ))             -> (g)
(car (g))                    -> 'g

Answer 6

this is easy/compact way to get the value of the list.

(cadr (cadddr x))

by combining repeating functions, you get elegant easy to read statement.