Question

Nested lists in Common Lisp really confused me. Here is the problem:

By using recursion, let (nested-list 'b '(a (b c) d)) return t
if the first argument appears in the second argument (which could be
a nested list), and nil otherwise.

I tried find, but it only works if the first argument is '(b c).
I turned my eyes on lambda expressions. I want to flatten the second
argument first, and then use eq to compare the arguments.

(defun nested-list (x y)
  (cond     
    ((null y) ())    
    (t (append (lambda (flatten) (first y))

Then I got stuck. Even though I read a lot of stuff about lambda
expessions, it still confused me. I do not know how to recall it when
I need, I knew the funcall function, but you know I just cannot get
it. I just learnt Common Lisp for 5 days, so I hope you can give me a
hint. Thanks a lot!

Answer 1

First of all unless you mistyped if instead of iff the problem is quite trivial, just return t and you’re done 🙂

Seriously speaking instead when you need to solve a problem using recursion the idea often is simply:

1. if we’re in a trivial case just return the answer
2. otherwise the answer is the same answer we’d get by solving a problem that it’s just a little bit simpler that this, and we call ourself to solve this simplified version.

In the specific consider:

1. If the second argument is an empty list the answer is NIL
2. If the first argument is equal to the first element of the second argument then just return T instead
3. Otherwise if the first element of the second list is a list (therefore also possibly a nested list) and the element is contained in this multilist then return true (to check this case the function is calling itself)
4. Otherwise just check the same problem, but first dropping the first element of the second argument, because it has been checked (this also calls recursively the same function)

So basically 1 and 2 are the trivial cases; 3 and 4 are the cases in which you solve a simpler version of the problem.