# Learning LISP

I first thought about learning LISP when I was still an undergrad, but I was stymied by Real Life and a lack of material to learn from. Shortly before I submitted my MSc thesis I picked up two LISP books - LISP and On LISP: Advanced Techniques for Common LISP - but my MSc was taking up all my time, so I put them on a shelf and forgot about them. About a month ago, I read Recursive Functions of Symbolic Expressions and their Computation by Machine (Part I), the original paper about LISP. It’s very clearly written, and explains the design of LISP so well (in only 34 pages) that someone could make a reasonable attempt at implementing LISP based solely on reading it. Inspired by the paper, I dug out my books and started learning LISP; I’ve now reached a point where the solutions to some exercises are interesting enough to post.

Problem 5-3: Now write a pair of procedures KEEP-FIRST-N-CLEVERLY and
KEEP-FIRST-N-CLEVERLY-AUX, that together make a list of the first *n* elements
in a list. Be sure that KEEP-FIRST-N-CLEVERLY-AUX is tail recursive.

My solution:

```
(defun keep-first-n-cleverly (n alist)
(keep-first-n-cleverly-aux n alist nil)
)
(defun keep-first-n-cleverly-aux (n alist newlist)
(if (zerop n)
newlist
(keep-first-n-cleverly-aux
(- n 1)
(rest alist)
(append newlist (list (first alist)))
)
)
)
```

I like tail recursion: lots of problems are simpler to solve recursively, and knowing that a tail recursive call will be optimised to a goto satisfies the part of my mind that thinks “What if my function is run on a list with 1000 elements? Would I be better writing it iteratively, so that it doesn’t run out of stack space?”.

Problem 5-9: Define SQUASH, a procedure that takes an expression as its argument and returns a non-nested list of all atoms found in the expression. Here is an example:

```
* (squash '(a (a (a (a b))) (((a b) b) b) b))
(A A A A B A B B B B)
```

Essentially, this procedure explores the fringe of the tree represented by the list given as its argument, and returns a list of all the leaves.

My solution:

```
(defun squash (alist)
(cond
((null alist) nil)
((atom alist) (list alist))
(t (append
(squash (first alist))
(squash (rest alist))
)
)
)
)
```

Problem 5-12: The version of Fibonacci we have already exhibited is inefficient beyond comparison. Many computations are repeated. Write a version with optional parameters that does not have this flaw. Think of working forward from the first month rather than backward from the nth month.

My solution:

```
(defun fib (n &optional (count 2) (fibn-2 0) (fibn-1 1))
(case n
(0 0)
(1 1)
(otherwise
(if
(equal n count)
(+ fibn-2 fibn-1)
(fib n (+ count 1) fibn-1 (+ fibn-2 fibn-1))
)
)
)
)
```

The point of this exercise was to use optional parameters; if I was writing
`fib()`

for real, I would use an auxiliary procedure, like this:

```
(defun fib (n)
(case n
(0 0)
(1 1)
(otherwise (fib-aux n 2 0 1))
)
)
(defun fib-aux (n num-calculated fibn-2 fibn-1)
(if (equal n num-calculated)
(+ fibn-2 fibn-1)
(fib-aux n (+ num-calculated 1) fibn-1 (+ fibn-2 fibn-1))
)
)
```

My first inclination when writing a Fibonacci function is to use Memoization; if I was writing it in Perl I would use the standard module Memoize, where Fibonacci is presented as an example in the documentation. I don’t know yet how hard it would be to do this in LISP, but I expect that closures should be easy enough.