[MS]:
These are some notes on my playing around with [Fraction Math] and the
reference to the nice notes on Continued Fractions at
http://www.inwap.com/pdp10/hbaker/hakmem/cf.html
[http://www.inwap.com/pdp10/hbaker/hakmem/cf.html].

I have started playing with these things and resisted (for now)
looking for other sources. Some of the "errors" I found in the
reference may be due to the informal presentation of results there,
which might be inaccurate or misunderstood by me: ''caveat emptor''.

'''DEFINITIONS:''' a rational approximation p/q to a real number x is "best" iff, for every integer r and s, 
  (s <= q) ==> (|x - p/q| <= |x - r/s|)
It is "best on its side" if
  ((s <= q) & (sgn(x-p/q) = sgn(x-r/s))) ==> (|x - p/q| <= |x - r/s|)
i.e, if no other fraction  on the same side of x with a lesser denominator comes closer.

'''NOTATION:''' let us identify a (positive) real number x with its regular continued fraction representation
  x = {x[0] x[1] x[2] x[3] ...} 
Define the truncation of x after (n+1) terms
  a(x,n) = {x[0] x[1] x[2] x[3] ... x[n]} 
and let
  b(x,n,i) = {x[0] x[1] x[2] x[3] ... x[n-1] i}
be a(x,n) with the last element replaced by 0 < i <= x[[n]]. Note that
  a(x,n) = b(x,n,x[n])


'''NOTE''': [[quotient_rep]] in [Fraction Math] computes the highest-order truncation requiring no integers larger than ''maxint'' in the rational representation p/q.


----

Item 101A (3) clearly says (AFAIU, not all claims reproduced here): 
   A - a(x,n) is "best"
   B - b(x,n,i) is "best" if i>1
   C - b(x,n,1) is never "best" if (x[n] != 1)
Let me provide counterexamples to both B and C, thus showing that they are not true.

The example provided in the text actually contains counterexamples to B. Let x = pi = {3 7 15 1 292 ...}
   . b(x,0,2) = 2/1 = {2} is not best (3/1 is better)
   . b(x,1,2) = 7/2 = {3 2} and b(x,1,3) = 10/3 = {3 3} are not best (3/1 is better) 
   . b(x,2,2) = 47/15 = {3 7 2} is not best (22/7 is better)

For another counterexample to B, easier to follow by hand, consider 
  x = 0.51 = {0 1 1 24 2}
The number
  b(x,3,4) = 5/9 = {0 1 1 4} = 0.555...
is not best, as 1/2 = {0 1 1} = {0 2} is better.

For a counterexample to C, consider x = 7/10 = {0 1 2 3}; now
  b(x,2,1) = 1/2 = {0 1 1} = {0 2} 
is best.

----
So, in light of these counterexamples, one proposition I can hope could be true is:

   D - The set of "best on its side" approximations to 0<x<1 coincides with 
       the set of numbers of the form b(x,n,i)

Remark that a simple corollary would be:
   E - if r is a "best" approximation to x, then r = b(x,n,i) for some (n,i)
as a "best" approx has to be "best on its side".

I think (D) can be proved from ITEM 101C in [http://www.inwap.com/pdp10/hbaker/hakmem/cf.html] (which may be true AFAIK). I'm working on the details, maybe a restriction (i!=1) will be necessary. After a while I'll stop playing and start reading ...

----

The example I provided in [Fraction Math] showing that ''quotient_rep'' does not always provide the best approximation is
   quotient_rep 3.1416305 500 --> 355/113
Now, 3.1416305 = {3 7 16 2 ...},and ''quotient_rep'' produced
   355/113 = {3 7 16} = a(3.1416305,2)
But the fraction
   377/120 = {3 7 16 1} = b(3.1416305,3,1)
is closer - a new counterexample to C.

Remark that the next truncation involves integers that are too large:
   a(3.1416305,3) = {3 7 16 2} = 732/233

----
[Arjen Markus] Some care must be taken, as I discovered myself. I tried to deduce the continued fraction for sqrt(N^2+1)+N (forms such as sqrt(2)+1, sqrt(5)+2, ...):

   (sqrt(N^2+1)+N) * (sqrt(N^2+1)-N = 1

So:

                        1               
   sqrt(N^2+1)+N = -------------
                   sqrt(N^2+1)-N

                             1
                 = ---------------------
                   -2N + (sqrt(N^2+1)+N)

                               1
                 = -------------------------
                                  1
                   -2N + ---------------------
                         -2N + (sqrt(N^2+1)+N)

                 = ...

Now, the approximations are (N=1):

    -0.5, -0.4, ...

How will this ever end up in 2.4141...?

It took me the better part of an evening to realise that I had made a stupid, though understandable, mistake: the continued fraction does not get smaller! Hence my approximations are worthless. They in fact will give 1-sqrt(2).     

What is interesting is that these continued fractions have a natural representation in Tcl, namely lists. Try doing that with plain C!

----

'''Some code to deal with continued fractions'''

a lot of this is ripped off from [KBK]'s code in [Fraction Math]. Some testing of inputs would also be needed.

   #
   # proc cont2frac
   #
   # Returns the fraction corresponding to the list
   # of denominators in a (truncated) regular continued
   # fraction. There is no check for overflow.
   #

   proc cont2frac {lst} {
       foreach {p q p0 q0} {1 0 0 1} break
       foreach a $lst {
	   foreach {p p0} [list [expr {$a*$p+$p0}] $p] break 
	   foreach {q q0} [list [expr {$a*$q+$q0}] $q] break 
      }
      list $p $q
   }


   #
   # proc frac2cont
   #
   # Returns list of denominators in the regular continued
   # expansion of frac. Cannot overflow.
   #

   proc frac2cont {frac} {
       foreach {p0 q0} {1 0} break
       foreach {p q} $frac break
       set clist {}

       while {$q} {
	   set a [expr {$p/$q}]
	   lappend clist $a
	   foreach {p q} [list $q [expr {$p - $q * $a}]] break
       }
       set clist
   }


   # 
   # proc num2cont
   #
   # Returns a list of two lists:
   #   1. the list of denominators in the longest truncated
   #      regular continued fraction expansion of num with both
   #      numerator and denominator <= maxint.
   #   2. the representation of the above list as a fraction
   #
   # Remark that "quotient_rep $num $maxint]" is equivalent to
   # "lindex [num2cont $num $maxint] 1"
   #

    proc num2cont {num { maxint 2147483647 } } {
        foreach {p q p0 q0} {1 0 0 1} break
        set clist {}

        while {1} {
           set a [expr {int($num)}]
           set fract [expr {$num - $a}]

           if {(1.0 * $a * $p + $p0 > $maxint) \
	           || (1.0 * $a * $q + $q0 > $maxint)} {
	       break
           }
           lappend clist $a
           foreach {p p0} [list [expr {$a * $p + $p0}] $p] break 
           foreach {q q0} [list [expr {$a * $q + $q0}] $q] break 

           if {abs($fract * $maxint) < 1} break
           set num [expr {1.0 / $fract}]
        }
        list $clist [list $p $q]
    }