# Multiplying Polynomials for Fun.

## Table of Contents

## Change for a dollar

For some reason, I thought the code to multiply two polynomials would be interesting to implement. To see if I did it right, I opted to solve the "change for a dollar" question this way.

I went with the naive approach of using sequences of all coefficients as the parameters. My first run of this was in the scratch buffer, so copied to here gives us the following.

(defun pmult (as bs) (let ((c (min (length as) (length bs)))) (loop for i from 0 to c collect (apply '+ (loop for j from 0 to i collect (* (or (nth j as) 0) (or (nth (- i j) bs) 0))))))) (defun make-change-seq (d l) (loop for i from 0 to l collect (if (or (eq 0 i) (eq 0 (mod i d))) 1 0))) (defun change-for-amount (a) (nth a (pmult (make-change-seq 50 a) (pmult (make-change-seq 25 a) (pmult (make-change-seq 10 a) (pmult (make-change-seq 5 a) (make-change-seq 1 a))))))) (mapcar 'change-for-amount '(1 5 10 25 50 100 200 300 400 500))

Executing the above will give us the expected answers.

1 | 2 | 4 | 13 | 50 | 292 | 2435 | 9590 | 26517 | 59576 |

I'm new to actually writing elisp, so I suspect I could have done a few things more cleanly. Regardless, I'm happy with this as a rough implementation.

I am curious to know if this is any better than the recursive solution for this question. Just at a glance, this is more work. With how long this takes, I seem to recall that my last attempt at this recursively was a bit faster.

I will note, however, that the closed form solution of this can answer how many ways to make change for a million dollars instantly. So, this seems a much better example for why you should used closed form if you can. (Though, I guess this is not that much less artificial than the typical example of Fibonacci.)