I was wondering if this is the fastest possible version of this function.
(defun foo (x y)
(cond
;if x = 0, return y+1
((zp x) (+ 1 y))
;if y = 0, return foo on decrement x and 1
((zp y) (foo (- x 1) 1))
;else run foo on decrement x and y = (foo x (- y 1))
(t (foo (- x 1) (foo x (- y 1))))))
When I run this, I usually get stack overflow error, so I am trying to figure out a way to compute something like (foo 3 1000000) without using the computer.
From analyzing the function I think it is embedded foo in the recursive case that causes the overflow in (foo 3 1000000). But since you are decrementing y would the number of steps just equal y?
edit: removed lie from comments
12 years ago I wrote this:
(defun ackermann (m n)
(declare (fixnum m n) (optimize (speed 3) (safety 0)))
(let ((memo (make-hash-table :test #'equal))
(ncal 0) (nhit 0))
(labels ((ack (aa bb)
(incf ncal)
(cond ((zerop aa) (1+ bb))
((= 1 aa) (+ 2 bb))
((= 2 aa) (+ 3 (* 2 bb)))
((= 3 aa) (- (ash 1 (+ 3 bb)) 3))
((let* ((key (cons aa bb))
(val (gethash key memo)))
(cond (val (incf nhit) val)
(t (setq val (if (zerop bb)
(ack (1- aa) 1)
(ack (1- aa) (ack aa (1- bb)))))
(setf (gethash key memo) val)
val)))))))
(let ((ret (ack m n)))
(format t "A(~d,~d)=~:d (~:d calls, ~:d cache hits)~%"
m n ret ncal nhit)
(values ret memo)))))
As you can see, I am using an explicit formula for small a and memoization for larger a.
Note, however, that this function grows so fast that it makes little sense to try to compute the actual values; you will run out of atoms in the universe faster - memoization or not.