Lisp Extraction Mission

Common Lisp, 99

The following 99 bytes solution is a CL version of the nice Scheme answer.

(defun g(l n &optional(o'l))(if(eql n l)o(and(consp l)(or(g(car l)n`(car,o))(g(cdr l)n`(cdr,o))))))

I originally tried to make use of position and position-if, but it turned out to be not as compact as I would have liked (209 bytes):

(lambda(L x &aux(p'l))(labels((f(S &aux e)(cons(or(position x S)(position-if(lambda(y)(if(consp y)(setf e(f y))))S)(return-from f()))e)))(dolist(o(print(f L))p)(dotimes(i o)(setf p`(cdr,p)))(setf p`(car,p)))))

Expanded

(lambda
  (l x &aux (p 'l))
  (labels ((f (s &aux e)
             (cons
              (or (position x s)
                  (position-if
                   (lambda (y)
                     (if (consp y)
                         (setf e (f y))))
                   s)
                  (return-from f nil))
              e)))
    (dolist (o (print (f l)) p)
      (dotimes (i o) (setf p `(cdr ,p)))
      (setf p `(car ,p)))))

Example

(funcall *fun* '(4 5 (1 2 (7) 9 (10 8 14))) 14)

The list is quoted, but if you really want, I can use a macro. The returned value is^[1]:

(CAR (CDR (CDR (CAR (CDR (CDR (CDR (CDR (CAR (CDR (CDR L)))))))))))

For tests, I used to generate a lambda form where l was a variable:

(LAMBDA (#:G854) (CAR (CDR (CDR (CAR (CDR (CDR (CDR (CDR (CAR (CDR (CDR #:G854))))))))))))

Calling this with the original list returns 14.

---

[1] (caddar (cddddr (caddr l))) would be nice too

Retina, 170 142 125 115 114 87 84 83 75 73 70 69 68 67 bytes

Yay, less than 50% of more than 100 bytes off my first attempt. :)

\b(.+)\b.* \1$
(
^.
l
\(
a
+`a *\)|\d


d
+`(.*[l)])(\w)
(c$2r $1)

To run the code from a single file, use the -s flag.

I'm still not convinced this is optimal... I won't have a lot of time over the next few days, I will add an explanation eventually.

Pyth, 62 bytes

JvXz"() ,][")u?qJQG&=J?K}Quu+GHNY<J1)hJtJ++XWK"(cdr "\d\aG\)\l

Try it online: Demonstration or Test Suite

Explanation:

The first bit JvXz"() ,][") replaces the chars "() " with the chars "[]," in the input string, which ends up in a representation of a Python-style list. I evaluate it and store it in J.

Then I reduce the string G = "l" with u...\l. I apply the inner function ... repeatedly to G, until the value of G doesn't change anymore and then print G.

The inner function does the following: If J is already equal to the input number, than don't modify G (?qJQG). Otherwise I'll flatten the the list J[:1] and check if the input number is in that list and save this to the variable K (K}Quu+GHNY<J1)). Notice that Pyth doesn't have a flatten operator, so this is takes quite a few bytes. If K is true, than I update J with J[0], otherwise with J[1:] (=J?KhJtJ). And then I replace G with "(cdr G)" and replace the d the a, if K is true (++XWK"(cdr "\d\aG\)).