+;;; This algorithm for RATIONALIZE, due to Bruno Haible, is included
+;;; with permission.
+;;;
+;;; Algorithm (recursively presented):
+;;; If x is a rational number, return x.
+;;; If x = 0.0, return 0.
+;;; If x < 0.0, return (- (rationalize (- x))).
+;;; If x > 0.0:
+;;; Call (integer-decode-float x). It returns a m,e,s=1 (mantissa,
+;;; exponent, sign).
+;;; If m = 0 or e >= 0: return x = m*2^e.
+;;; Search a rational number between a = (m-1/2)*2^e and b = (m+1/2)*2^e
+;;; with smallest possible numerator and denominator.
+;;; Note 1: If m is a power of 2, we ought to take a = (m-1/4)*2^e.
+;;; But in this case the result will be x itself anyway, regardless of
+;;; the choice of a. Therefore we can simply ignore this case.
+;;; Note 2: At first, we need to consider the closed interval [a,b].
+;;; but since a and b have the denominator 2^(|e|+1) whereas x itself
+;;; has a denominator <= 2^|e|, we can restrict the seach to the open
+;;; interval (a,b).
+;;; So, for given a and b (0 < a < b) we are searching a rational number
+;;; y with a <= y <= b.
+;;; Recursive algorithm fraction_between(a,b):
+;;; c := (ceiling a)
+;;; if c < b
+;;; then return c ; because a <= c < b, c integer
+;;; else
+;;; ; a is not integer (otherwise we would have had c = a < b)
+;;; k := c-1 ; k = floor(a), k < a < b <= k+1
+;;; return y = k + 1/fraction_between(1/(b-k), 1/(a-k))
+;;; ; note 1 <= 1/(b-k) < 1/(a-k)
+;;;
+;;; You can see that we are actually computing a continued fraction expansion.
+;;;
+;;; Algorithm (iterative):
+;;; If x is rational, return x.
+;;; Call (integer-decode-float x). It returns a m,e,s (mantissa,
+;;; exponent, sign).
+;;; If m = 0 or e >= 0, return m*2^e*s. (This includes the case x = 0.0.)
+;;; Create rational numbers a := (2*m-1)*2^(e-1) and b := (2*m+1)*2^(e-1)
+;;; (positive and already in lowest terms because the denominator is a
+;;; power of two and the numerator is odd).
+;;; Start a continued fraction expansion
+;;; p[-1] := 0, p[0] := 1, q[-1] := 1, q[0] := 0, i := 0.
+;;; Loop
+;;; c := (ceiling a)
+;;; if c >= b
+;;; then k := c-1, partial_quotient(k), (a,b) := (1/(b-k),1/(a-k)),
+;;; goto Loop
+;;; finally partial_quotient(c).
+;;; Here partial_quotient(c) denotes the iteration
+;;; i := i+1, p[i] := c*p[i-1]+p[i-2], q[i] := c*q[i-1]+q[i-2].
+;;; At the end, return s * (p[i]/q[i]).
+;;; This rational number is already in lowest terms because
+;;; p[i]*q[i-1]-p[i-1]*q[i] = (-1)^i.
+;;;
+;;; See also
+;;; Hardy, Wright: An introduction to number theory
+;;; and/or
+;;; <http://modular.fas.harvard.edu/edu/Fall2001/124/lectures/lecture17/lecture17/>
+;;; <http://modular.fas.harvard.edu/edu/Fall2001/124/lectures/lecture17/lecture18/>
+