(defun numeral-value (numeral)
  (cond ((char-equal numeral #\M) 1000)
	((char-equal numeral #\D) 500)
	((char-equal numeral #\C) 100)
	((char-equal numeral #\L) 50)
	((char-equal numeral #\X) 10)
	((char-equal numeral #\V) 5)
	((char-equal numeral #\I) 1)
	(t 0)))

(defun numeral= (a b)
  (= (numeral-value a) (numeral-value b)))

(defun numeral< (a b)
  (< (numeral-value a) (numeral-value b)))

(defun numeral> (a b)
  (> (numeral-value a) (numeral-value b)))

(defun numeral<= (a b)
  (<= (numeral-value a) (numeral-value b)))

(defun numeral>= (a b)
  (>= (numeral-value a) (numeral-value b)))

(defun parse-roman-numeral (numeral &key (current-value 0) (subtract 0))
  (cond ((= (length numeral) 0)
	 ;; in this case the whole input string has been consumed
	 (- current-value subtract))

	((= (length numeral) 1)
	 ;; in this case the input string is only one character long,
	 ;; so just return its value plus whatever value has been
	 ;; accumulated so far
	 (- (+ (numeral-value (char numeral 0)) current-value) subtract))

	((numeral>= (char numeral 0) (char numeral 1))
	 ;; in this case the first numeral in the input string is
	 ;; either equal to or greater than the second and so is
	 ;; assumed to be taken literally
	 (parse-roman-numeral (subseq numeral 1)
			      :current-value (- (+ current-value (numeral-value (char numeral 0))) subtract)))

	((numeral< (char numeral 0) (char numeral 1))
	 ;; in this case the first numeral in the input string is less
	 ;; than the second and so is assumed to be a subtracter
	 (parse-roman-numeral (subseq numeral 1)
			      :current-value current-value
			      :subtract (+ subtract (numeral-value (char numeral 0)))))))