I happened to observe this commit from today by Neil Toronto: <a href="http://git.racket-lang.org/plt/commitdiff/47fcdd4916a2d33ee5c28eb833397ce1d2a515e2">http://git.racket-lang.org/plt/commitdiff/47fcdd4916a2d33ee5c28eb833397ce1d2a515e2</a> I may have some useful input on this, having dealt with similar problems myself. The problem: Given b and x, you want to find the largest integer n such that b^n <= x (or, for the ceiling problem, the smallest integer n such that b^n ≥ x). This can profitably be viewed as a case of a more general problem: given a monotonically increasing but otherwise opaque function f, find the largest integer n such that f(n) ≤ x. The previous algorithm to solve this can now be viewed as a linear search, taking n steps. This took too long, and Neil Toronto replaced it with an algorithm that depended on floating-point numbers; but I hate to see that happen, because floats will screw up when the numbers get too large; careful reasoning may prove that floats will be sufficiently accurate for a given application, but it would be nice to have an exact algorithm so that one didn't have to do that. I come here to present such an algorithm. Simply put: Use a binary search rather than a linear search. How do we do binary search on "the nonnegative integers"? Well, start with 0 and 1 as your bounds, and keep doubling the 1 until you get something that's actually too high; then you have a real lower and upper bound. This will take log(n) steps to find the upper bound, and then a further log(n) steps to tighten the bounds to a single integer. Thus, this takes roughly 2*log(n) steps, where each step involves calculating (expt b n) plus a comparison. It might be possible to do slightly better by not treating b^n as a black box (e.g. by using old results of b^n to compute new ones rather than calling "expt" from scratch, or by using "integer-length" plus some arithmetic to get some good initial bounds), but I suspect this is good enough and further complexity is not worth it. Note that this algorithm should work perfectly with any nonnegative rational arguments [other than 0 for b or x, and 1 for b], and should give as good an answer as any with floating-point arguments. Also, "integer-finverse", as I called it, might be useful for many other "floor"-type computations with exact numbers (I have so used it myself in an ad hoc Arc math library). ;Finds n such that n >= 0 and f(n) <= x < f(n+1) in about 2*log(n) steps. ;Assumes f is monotonically increasing. (define (integer-finverse f x) (define upper-bound (let loop ((n 1)) (if (> (f n) x) n (loop (* n 2))))) (define (search a b) ;b is too big, a is not too big (let ((d (- b a))) (if (= d 1) a (let ((m (+ a (quotient d 2)))) (if (> (f m) x) (search a m) (search m b)))))) (search (quotient upper-bound 2) upper-bound)) (define (floor-log/base b x) (cond (( ((= b 1) (error "Base shouldn't be 1.")) ((< x 1) (- (ceiling-log/base b (/ x)))) (else (integer-finverse (lambda (n) (expt b n)) x)))) (define (ceiling-log/base b x) (cond (( ((= b 1) (error "Base shouldn't be 1.")) ((< x 1) (- (floor-log/base b (/ x)))) (else (let ((u (floor-log/base b x))) (if (= x (expt b u)) u (+ u 1)))))) Testing: > (floor-log/base 10 3) 0 > (floor-log/base 3 10) 2 > (ceiling-log/base 3 10) 3 > (ceiling-log/base 1/3 10) -2 > (floor-log/base 1/3 10) -3 > (floor-log/base 2/3 10) -6 > (ceiling-log/base 2/3 10) -5 > (exact->inexact (expt 3/2 6)) 11.390625 > (exact->inexact (expt 3/2 5)) 7.59375 I might add, by the way, that I'm inclined to expect the base to be a second, optional argument (perhaps defaulting to 10) to a function called simply "floor-log" or "ceiling-log". I don't know if that fits with desired conventions, though. --John Boyle<div></div><div>Science is what we understand well enough to explain to a computer. Art is everything else we do. --Knuth</div>