Question

I had a question about a program I’ve been trying to get running.
Encrypt takes a message, public-key, and private-key, and returns the message with the letters from the message in the public-key changed to the letters from the private-key.

For ex, (encrypt “abcd” “abcd” “efgh”) should return “efgh”
and (encrypt “abcl” “abcd” “efgh”) should return “efgl” (the letter from message that was not in the public-key will stay the same).

I’ve written a few helper programs to solve this but I keep getting the error “exception in car, __ is not a pair” when I try to run it.. but I’m not sure what’s amiss. If anyone has any pointers, let me know. Thanks!

(define encrypt
  (lambda (message public-key private-key)
    (cond
      [(list->string (encrypt-helper (string->list message)
      (string->list public-key) (string->list private-key)))])))

(define encrypt-helper
  (lambda (msg-ls public-ls private-ls)
    (cond
      [(null? public-ls) '()]
      [(null? private-ls) '()]
      [(and (null? public-ls) (null? private-ls)) msg-ls]
      [else (cons (encrypt-key (car msg-ls) (car public-ls) (car private-ls))
        (encrypt-helper (cdr msg-ls) (cdr public-ls) (cdr private-ls)))])))

;should encrypt all letters in msg-ls. not working correctly

(define encrypt-key
  (lambda (char pub-key priv-key)
    (cond
      [(null? pub-key) char]
      [(equal? char (car pub-key)) (car priv-key)]
      [else (encrypt-key char (cdr pub-key) (cdr priv-key))])))

;encrypts just one letter, ex: (encrypt-key 'a '(a) '(b)) => b
;works correctly

Answer 1

The problem is that inside encrypt-helper, you’re calling

[else (cons (encrypt-key (car msg-ls) (car public-ls) (car private-ls)...

but (car public-ls) (and (car private-ls)) is an atom, while inside encrypt-key you also perform

[(equal? char (car pub-key) ...

and you can’t car pub-key here because car only works on a list, while pub-key is an atom.

In the example you give that works, i.e.

(encrypt-key 'a '(a) '(b)) => b

you’ll notice that '(a) and '(b) are specified as lists, for exactly this reason. Hint:

>(cons 'a ())
(a)
> 

I’ll leave it there 🙂