Minkowski question-mark function

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Revision as of 00:16, 27 February 2022 by Rdm (talk | contribs) (J: trivia)
Task
Minkowski question-mark function
You are encouraged to solve this task according to the task description, using any language you may know.

The Minkowski question-mark function converts the continued fraction representation [a0; a1, a2, a3, ...] of a number into a binary decimal representation in which the integer part a0 is unchanged and the a1, a2, ... become alternating runs of binary zeroes and ones of those lengths. The decimal point takes the place of the first zero.

Thus, ?(31/7) = 71/16 because 31/7 has the continued fraction representation [4;2,3] giving the binary expansion 4 + 0.01112.

Among its interesting properties is that it maps roots of quadratic equations, which have repeating continued fractions, to rational numbers, which have repeating binary digits.

The question-mark function is continuous and monotonically increasing, so it has an inverse.

  • Produce a function for ?(x).   Be careful: rational numbers have two possible continued fraction representations:
  •   [a0;a1,... an−1,an]     and
  •   [a0;a1,... an−1,an−1,1]
  • Choose one of the above that will give a binary expansion ending with a   1.
  • Produce the inverse function ?-1(x)
  • Verify that ?(φ) = 5/3, where φ is the Greek golden ratio.
  • Verify that ?-1(-5/9) = (√13 - 7)/6
  • Verify that the two functions are inverses of each other by showing that ?-1(?(x))=x and ?(?-1(y))=y for x, y of your choice


Don't worry about precision error in the last few digits.


See also



11l

Translation of: Python

<lang 11l>-V MAXITER = 151

F minkowski(x) -> Float

  I x > 1 | x < 0
     R floor(x) + minkowski(x - floor(x))
  V p = Int(x)
  V q = 1
  V r = p + 1
  V s = 1
  V d = 1.0
  V y = Float(p)
  L
     d /= 2
     I y + d == y
        L.break
     V m = p + r
     I m < 0 | p < 0
        L.break
     V n = q + s
     I n < 0
        L.break
     I x < Float(m) / n
        r = m
        s = n
     E
        y += d
        p = m
        q = n
  R y + d

F minkowski_inv(=x) -> Float

  I x > 1 | x < 0
     R floor(x) + minkowski_inv(x - floor(x))
  I x == 1 | x == 0
     R x
  V cont_frac = [0]
  V current = 0
  V count = 1
  V i = 0
  L
     x *= 2
     I current == 0
        I x < 1
           count++
        E
           cont_frac.append(0)
           cont_frac[i] = count
           i++
           count = 1
           current = 1
           x--
     E
        I x > 1
           count++
           x--
        E
           cont_frac.append(0)
           cont_frac[i] = count
           i++
           count = 1
           current = 0
     I x == floor(x)
        cont_frac[i] = count
        L.break
     I i == :MAXITER
        L.break
  V ret = 1.0 / cont_frac[i]
  L(j) (i-1 .. 0).step(-1)
     ret = cont_frac[j] + 1.0 / ret
  R 1.0 / ret

print(‘#2.16 #2.16’.format(minkowski(0.5 * (1 + sqrt(5))), 5.0 / 3.0)) print(‘#2.16 #2.16’.format(minkowski_inv(-5.0 / 9.0), (sqrt(13) - 7) / 6)) print(‘#2.16 #2.16’.format(minkowski(minkowski_inv(0.718281828)), minkowski_inv(minkowski(0.1213141516171819))))</lang>

Output:
 1.6666666666696983  1.6666666666666667
-0.5657414540893350 -0.5657414540893351
 0.7182818280000091  0.1213141516171819

Factor

Works with: Factor version 0.99 2020-08-14

<lang factor>USING: formatting kernel make math math.constants math.continued-fractions math.functions math.parser math.statistics sequences sequences.extras splitting.monotonic vectors ;

CONSTANT: max-iter 151

>continued-fraction ( x -- seq )
   0 swap 1vector
   [ dup last integer? pick max-iter > or ]
   [ dup next-approx [ 1 + ] dip ] until nip
   dup last integer? [ but-last-slice ] unless ;
? ( x -- y )
   >continued-fraction unclip swap cum-sum
   [ max-iter < ] take-while
   [ even? 1 -1 kernel:? swap 2^ / ] map-index
   sum 2 * + >float ;
(float>bin) ( x -- y )
   [ dup 0 > ]
   [ 2 * dup >integer # dup 1 >= [ 1 - ] when ] while ;
float>bin ( x -- n str )
   >float dup >integer [ - ] keep swap abs
   [ 0 # (float>bin) ] "" make nip ;
?⁻¹ ( x -- y )
   dup float>bin [ = ] monotonic-split
   [ length ] map swap prefix >ratio swap copysign ;
compare ( x y -- ) "%-25u%-25u\n" printf ;

phi ? 5 3 /f compare -5/9 ?⁻¹ 13 sqrt 7 - 6 /f compare 0.718281828 ?⁻¹ ? 0.1213141516171819 ? ?⁻¹ compare</lang>

Output:
1.666666666668335        1.666666666666667        
-0.5657414540893351      -0.5657414540893352      
0.718281828000002        0.1213141516171819       

FreeBASIC

<lang freebasic>#define MAXITER 151

function minkowski( x as double ) as double

   if x>1 or x<0 then return int(x)+minkowski(x-int(x))
   dim as ulongint p = int(x)
   dim as ulongint q = 1, r = p + 1, s = 1, m, n
   dim as double d = 1, y = p
   while true 
       d = d / 2.0
       if y + d = y then exit while
       m = p + r
       if m < 0 or p < 0 then exit while
       n = q + s
       if n < 0 then exit while
       if x < cast(double,m) / n then
           r = m
           s = n
       else
           y = y + d
           p = m
           q = n
       end if
   wend
   return y + d

end function

function minkowski_inv( byval x as double ) as double

   if x>1 or x<0 then return int(x)+minkowski_inv(x-int(x))
   if x=1 or x=0 then return x
   redim as uinteger contfrac(0 to 0)
   dim as uinteger curr=0, count=1, i = 0
   do
       x *= 2
       if curr = 0 then
           if x<1 then
               count += 1
           else
               i += 1
               redim preserve contfrac(0 to i)
               contfrac(i-1)=count
               count = 1
               curr = 1
               x=x-1
           endif
       else
           if x>1 then
               count += 1
               x=x-1
           else
               i += 1
               redim preserve contfrac(0 to i)
               contfrac(i-1)=count
               count = 1
               curr = 0
           endif
       end if
       if x = int(x) then
           contfrac(i)=count
           exit do
       end if
   loop until i = MAXITER
   dim as double ret = 1.0/contfrac(i)
   for j as integer = i-1 to 0 step -1
       ret = contfrac(j) + 1.0/ret
   next j
   return 1./ret

end function

print minkowski( 0.5*(1+sqr(5)) ), 5./3 print minkowski_inv( -5./9 ), (sqr(13)-7)/6 print minkowski(minkowski_inv(0.718281828)), minkowski_inv(minkowski(0.1213141516171819)) </lang>

Output:
 1.666666666669698           1.666666666666667
-0.5657414540893351         -0.5657414540893352
 0.7182818280000092          0.1213141516171819

Go

Translation of: FreeBASIC

<lang go>package main

import (

   "fmt"
   "math"

)

const MAXITER = 151

func minkowski(x float64) float64 {

   if x > 1 || x < 0 {
       return math.Floor(x) + minkowski(x-math.Floor(x))
   }
   p := uint64(x)
   q := uint64(1)
   r := p + 1
   s := uint64(1)
   d := 1.0
   y := float64(p)
   for {
       d = d / 2
       if y+d == y {
           break
       }
       m := p + r
       if m < 0 || p < 0 {
           break
       }
       n := q + s
       if n < 0 {
           break
       }
       if x < float64(m)/float64(n) {
           r = m
           s = n
       } else {
           y = y + d
           p = m
           q = n
       }
   }
   return y + d

}

func minkowskiInv(x float64) float64 {

   if x > 1 || x < 0 {
       return math.Floor(x) + minkowskiInv(x-math.Floor(x))
   }
   if x == 1 || x == 0 {
       return x
   }
   contFrac := []uint32{0}
   curr := uint32(0)
   count := uint32(1)
   i := 0
   for {
       x *= 2
       if curr == 0 {
           if x < 1 {
               count++
           } else {
               i++
               t := contFrac
               contFrac = make([]uint32, i+1)
               copy(contFrac, t)
               contFrac[i-1] = count
               count = 1
               curr = 1
               x--
           }
       } else {
           if x > 1 {
               count++
               x--
           } else {
               i++
               t := contFrac
               contFrac = make([]uint32, i+1)
               copy(contFrac, t)
               contFrac[i-1] = count
               count = 1
               curr = 0
           }
       }
       if x == math.Floor(x) {
           contFrac[i] = count
           break
       }
       if i == MAXITER {
           break
       }
   }
   ret := 1.0 / float64(contFrac[i])
   for j := i - 1; j >= 0; j-- {
       ret = float64(contFrac[j]) + 1.0/ret
   }
   return 1.0 / ret

}

func main() {

   fmt.Printf("%19.16f %19.16f\n", minkowski(0.5*(1+math.Sqrt(5))), 5.0/3.0)
   fmt.Printf("%19.16f %19.16f\n", minkowskiInv(-5.0/9.0), (math.Sqrt(13)-7)/6)
   fmt.Printf("%19.16f %19.16f\n", minkowski(minkowskiInv(0.718281828)),
       minkowskiInv(minkowski(0.1213141516171819)))

}</lang>

Output:
 1.6666666666696983  1.6666666666666667
-0.5657414540893351 -0.5657414540893352
 0.7182818280000092  0.1213141516171819

Haskell

In a lazy functional language Minkowski question mark function can be implemented using one of it's basic properties:

?(p+r)/(q+s) = 1/2 * ( ?(p/q) + ?(r/s) ), ?(0) = 0, ?(1) = 1.

where p/q and r/s are fractions, such that |ps - rq| = 1.

This recursive definition can be implemented as lazy corecursion, i.e. by generating two infinite binary trees: mediant-based Stern-Brocot tree, containing all rationals, and mean-based tree with corresponding values of Minkowsky ?-function. There is one-to-one correspondence between these two trees so both ?(x) and ?-1(x) may be implemented as mapping between them. For details see the paper [[1]] (in Russian).

<lang haskell>import Data.Tree import Data.Ratio import Data.List

intervalTree :: (a -> a -> a) -> (a, a) -> Tree a intervalTree node = unfoldTree $

 \(a, b) -> let m = node a b in (m, [(a,m), (m,b)])

Node a _ ==> Node b [] = const b Node a [] ==> Node b _ = const b Node a [l1, r1] ==> Node b [l2, r2] =

 \x -> case x `compare` a of
         LT -> (l1 ==> l2) x
         EQ -> b
         GT -> (r1 ==> r2) x

mirror :: Num a => Tree a -> Tree a mirror t = Node 0 [reflect (negate <$> t), t]

 where
   reflect (Node a [l,r]) = Node a [reflect r, reflect l]

sternBrocot :: Tree Rational sternBrocot = toRatio <$> intervalTree mediant ((0,1), (1,0))

 where
   mediant (p, q) (r, s) = (p + r, q + s)

toRatio (p, q) = p % q

minkowski :: Tree Rational minkowski = toRatio <$> intervalTree mean ((0,1), (1,0))

mean (p, q) (1, 0) = (p+1, q) mean (p, q) (r, s) = (p*s + q*r, 2*q*s)


questionMark, invQuestionMark :: Rational -> Rational questionMark = mirror sternBrocot ==> mirror minkowski invQuestionMark = mirror minkowski ==> mirror sternBrocot


-- Floating point trees and functions

sternBrocotF :: Tree Double sternBrocotF = mirror $ fromRational <$> sternBrocot

minkowskiF :: Tree Double minkowskiF = mirror $ intervalTree mean (0, 1/0)

 where
   mean a b | isInfinite b = a + 1
            | otherwise = (a + b) / 2

questionMarkF, invQuestionMarkF :: Double -> Double questionMarkF = sternBrocotF ==> minkowskiF invQuestionMarkF = minkowskiF ==> sternBrocotF</lang>

λ> mapM_ print $ take 4 $ levels farey
[1 % 2]
[1 % 3,2 % 3]
[1 % 4,2 % 5,3 % 5,3 % 4]
[1 % 5,2 % 7,3 % 8,3 % 7,4 % 7,5 % 8,5 % 7,4 % 5]

λ> mapM_ print $ take 4 $ levels minkowski
[1 % 2]
[1 % 4,3 % 4]
[1 % 8,3 % 8,5 % 8,7 % 8]
[1 % 16,3 % 16,5 % 16,7 % 16,9 % 16,11 % 16,13 % 16,15 % 16]

λ> questionMark (1/2) 1 % 2 λ> questionMark (2/7) 3 % 16 λ> questionMark (-22/7) (-193) % 64 λ> invQuestionMark (3/16) 2 % 7 λ> invQuestionMark (13/256)

5 % 27

λ> questionMark $ (sqrt 5 + 1) / 2
1.6666666666678793
λ> 5/3
1.6666666666666667
λ> invQuestionMark (-5/9)
-0.5657414540893351
λ> (sqrt 13 - 7)/6
-0.5657414540893352

J

Implementation:

<lang J>ITERCOUNT=: 52

minkowski=: Template:F=. 1

invmink=: Template:F=. 1</lang>

That said, note that this algorithm introduces significant numeric instability for √7 divided by 3:

<lang J> (minkowski@invmink - invmink@minkowski) (p:%%:)3 1.10713e_6</lang>

I see this same instability using the python implementation and appending:

<lang python> print(

       "{:19.16f} {:19.16f}".format(
           minkowski(minkowski_inv(4.04145188432738056)),
           minkowski_inv(minkowski(4.04145188432738056)),
       )
   )</lang>

Using an exact fraction for 4.04145188432738056 and bumping the iteration count from 52 up to 200 changes that difference to 1.43622e_12.

Julia

Translation of: FreeBASIC

<lang julia>function minkowski(x)

   p = Int(floor(x))
   (x > 1 || x < 0) && return p + minkowski(x)
   q, r, s, m, n = 1, p + 1, 1, 0, 0
   d, y = 1.0, Float64(p)
   while true
       d /= 2.0
       y + d == y && break
       m = p + r
       (m < 0 || p < 0) && break
       n = q + s
       n < 0 && break
       if x < (m / n)
           r, s = m, n
       else
           y, p, q = y + d, m, n
       end
   end
   return y + d

end

function minkowski_inv(x, maxiter=151)

   p = Int(floor(x))
   (x > 1 || x < 0) && return p + minkowski_inv(x - p, maxiter)
   (x == 1 || x == 0) && return x
   contfrac = [0]
   curr, coun, i = 0, 1, 0
   while i < maxiter
       x *= 2
       if curr == 0
           if x < 1
               coun += 1
           else
               i += 1
               push!(contfrac, 0)
               contfrac[i] = coun
               coun = 1
               curr = 1
               x -= 1
           end
       else
           if x > 1
               coun += 1
               x -= 1
           else
               i += 1
               push!(contfrac, 0)
               contfrac[i] = coun
               coun = 1
               curr = 0
           end
       end
       if x == Int(floor(x))
           contfrac[i + 1] = coun
           break
       end
   end
   ret = 1.0 / contfrac[i + 1]
   for j in i:-1:1
       ret = contfrac[j] + 1.0 / ret
   end
   return 1.0 / ret

end

println(" ", minkowski((1 + sqrt(5)) / 2), " ", 5 / 3) println(minkowski_inv(-5/9), " ", (sqrt(13) - 7) / 6) println(" ", minkowski(minkowski_inv(0.718281828)), " ",

   minkowski_inv(minkowski(0.1213141516171819)))

</lang>

Output:
 1.6666666666696983   1.6666666666666667
-0.5657414540893351   -0.5657414540893352
 0.7182818280000092   0.12131415161718191

Mathematica / Wolfram Language

<lang Mathematica>ClearAll[InverseMinkowskiQuestionMark] InverseMinkowskiQuestionMark[val_] := Module[{x}, (x /. FindRoot[MinkowskiQuestionMark[x] == val, {x, Floor[val], Ceiling[val]}])] MinkowskiQuestionMark[GoldenRatio] InverseMinkowskiQuestionMark[-5/9] // RootApproximant MinkowskiQuestionMark[InverseMinkowskiQuestionMark[0.1213141516171819]] InverseMinkowskiQuestionMark[MinkowskiQuestionMark[0.1213141516171819]]</lang>

Output:
5/3
1/6 (-7+Sqrt[13])
0.121314
0.121314

Nim

Translation of: Go

<lang Nim>import math, strformat

const MaxIter = 151


func minkowski(x: float): float =

 if x notin 0.0..1.0:
   return floor(x) + minkowski(x - floor(x))
 var
   p = x.uint64
   r = p + 1
   q, s = 1u64
   d = 1.0
   y = p.float
 while true:
   d /= 2
   if y + d == y: break
   let m = p + r
   if m < 0 or p < 0: break
   let n = q + s
   if n < 0: break
   if x < m.float / n.float:
     r = m
     s = n
   else:
     y += d
     p = m
     q = n
 result = y + d


func minkowskiInv(x: float): float =

 if x notin 0.0..1.0:
   return floor(x) + minkowskiInv(x - floor(x))
 if x == 1 or x == 0:
   return x
 var
   contFrac: seq[uint32]
   curr = 0u32
   count = 1u32
   i = 0
   x = x
 while true:
   x *= 2
   if curr == 0:
     if x < 1:
       inc count
     else:
       inc i
       contFrac.setLen(i + 1)
       contFrac[i - 1] = count
       count = 1
       curr = 1
       x -= 1
   else:
     if x > 1:
       inc count
       x -= 1
     else:
       inc i
       contFrac.setLen(i + 1)
       contFrac[i - 1] = count
       count = 1
       curr = 0
   if x == floor(x):
     contFrac[i] = count
     break
   if i == MaxIter:
     break
 var ret = 1 / contFrac[i].float
 for j in countdown(i - 1, 0):
   ret = contFrac[j].float + 1 / ret
 result = 1 / ret


echo &"{minkowski(0.5*(1+sqrt(5.0))):19.16f}, {5/3:19.16f}" echo &"{minkowskiInv(-5/9):19.16f}, {(sqrt(13.0)-7)/6:19.16f}" echo &"{minkowski(minkowskiInv(0.718281828)):19.16f}, " &

    &"{minkowskiInv(minkowski(0.1213141516171819)):19.16f}"</lang>
Output:
 1.6666666666696983,  1.6666666666666667
-0.5657414540893351, -0.5657414540893352
 0.7182818280000092,  0.1213141516171819

Perl

Translation of: Raku

<lang perl>use strict; use warnings; use feature 'say'; use POSIX qw(floor);

my $MAXITER = 50;

sub minkowski {

   my($x) = @_;
   return floor($x) + minkowski( $x - floor($x) ) if $x > 1 || $x < 0 ;
   my $y = my $p = floor($x);
   my ($q,$s,$d) = (1,1,1);
   my $r = $p + 1;
   while () {
       last if ( $y + ($d /= 2)  == $y ) or
               ( my $m = $p + $r) <  0   or
               ( my $n = $q + $s) <  0;
       $x < $m/$n ? ($r,$s) = ($m, $n) : ($y += $d and ($p,$q) = ($m, $n) );
   }
   return $y + $d

}

sub minkowskiInv {

   my($x) = @_;
   return floor($x) + minkowskiInv($x - floor($x)) if $x > 1 || $x < 0;
   return $x if $x == 1 || $x == 0 ;
   my @contFrac = 0;
   my $i = my $curr = 0 ; my $count = 1;
   while () {
       $x *= 2;
       if ($curr == 0) {
           if ($x < 1) {
               $count++
           } else {
               $i++;
               push @contFrac, 0;
               $contFrac[$i-1] = $count;
               ($count,$curr) = (1,1);
               $x--;
           }
       } else {
           if ($x > 1) {
               $count++;
               $x--;
           } else {
               $i++;
               push @contFrac, 0;
               @contFrac[$i-1] = $count;
               ($count,$curr) = (1,0);
           }
       }
       if ($x == floor($x)) { @contFrac[$i] = $count; last }
       last if $i == $MAXITER;
   }
   my $ret = 1 / $contFrac[$i];
   for (my $j = $i - 1; $j >= 0; $j--) { $ret = $contFrac[$j] + 1/$ret }
   return 1 / $ret

}

printf "%19.16f %19.16f\n", minkowski(0.5*(1 + sqrt(5))), 5/3; printf "%19.16f %19.16f\n", minkowskiInv(-5/9), (sqrt(13)-7)/6; printf "%19.16f %19.16f\n", minkowski(minkowskiInv(0.718281828)), minkowskiInv(minkowski(0.1213141516171819));</lang>

Output:
 1.6666666666696983  1.6666666666666667
-0.5657414540893351 -0.5657414540893352
 0.7182818280000092  0.1213141516171819

Phix

Translation of: FreeBASIC
with javascript_semantics
constant MAXITER = 151
 
function minkowski(atom x)
    atom p = floor(x)
    if x>1 or x<0 then return p+minkowski(x-p) end if
    atom q = 1, r = p + 1, s = 1, m, n, d = 1, y = p
    while true do
        d = d/2
        if y + d = y then exit end if
        m = p + r
        if m < 0 or p < 0 then exit end if
        n = q + s
        if n < 0 then exit end if
        if x < m/n then
            r = m
            s = n
        else
            y = y + d
            p = m
            q = n
        end if
    end while
    return y + d
end function
 
function minkowski_inv(atom x)
    if x>1 or x<0 then return floor(x)+minkowski_inv(x-floor(x)) end if
    if x=1 or x=0 then return x end if
    sequence contfrac = {}
    integer curr = 0, count = 1
    while true do
        x *= 2
        if curr = 0 then
            if x<1 then
                count += 1
            else
                contfrac &= count
                count = 1
                curr = 1
                x -= 1
            end if
        else
            if x>1 then
                count += 1
                x -= 1
            else
                contfrac &= count
                count = 1
                curr = 0
            end if
        end if
        if x = floor(x) then
            contfrac &= count
            exit
        end if
        if length(contfrac)=MAXITER then exit end if
    end while
    atom ret = 1/contfrac[$]
    for i = length(contfrac)-1 to 1 by -1 do
        ret = contfrac[i] + 1.0/ret
    end for
    return 1/ret
end function
 
printf(1,"%20.16f %20.16f\n",{minkowski(0.5*(1+sqrt(5))), 5/3})
printf(1,"%20.16f %20.16f\n",{minkowski_inv(-5/9), (sqrt(13)-7)/6})
printf(1,"%20.16f %20.16f\n",{minkowski(minkowski_inv(0.718281828)), 
                              minkowski_inv(minkowski(0.1213141516171819))})
Output:
  1.6666666666696983   1.6666666666666668
 -0.5657414540893351  -0.5657414540893352
  0.7182818280000092   0.1213141516171819

Python

Translation of: Go

<lang python>import math

MAXITER = 151


def minkowski(x):

   if x > 1 or x < 0:
       return math.floor(x) + minkowski(x - math.floor(x))
   p = int(x)
   q = 1
   r = p + 1
   s = 1
   d = 1.0
   y = float(p)
   while True:
       d /= 2
       if y + d == y:
           break
       m = p + r
       if m < 0 or p < 0:
           break
       n = q + s
       if n < 0:
           break
       if x < m / n:
           r = m
           s = n
       else:
           y += d
           p = m
           q = n
   return y + d


def minkowski_inv(x):

   if x > 1 or x < 0:
       return math.floor(x) + minkowski_inv(x - math.floor(x))
   if x == 1 or x == 0:
       return x
   cont_frac = [0]
   current = 0
   count = 1
   i = 0
   while True:
       x *= 2
       if current == 0:
           if x < 1:
               count += 1
           else:
               cont_frac.append(0)
               cont_frac[i] = count
               i += 1
               count = 1
               current = 1
               x -= 1
       else:
           if x > 1:
               count += 1
               x -= 1
           else:
               cont_frac.append(0)
               cont_frac[i] = count
               i += 1
               count = 1
               current = 0
       if x == math.floor(x):
           cont_frac[i] = count
           break
       if i == MAXITER:
           break
   ret = 1.0 / cont_frac[i]
   for j in range(i - 1, -1, -1):
       ret = cont_frac[j] + 1.0 / ret
   return 1.0 / ret


if __name__ == "__main__":

   print(
       "{:19.16f} {:19.16f}".format(
           minkowski(0.5 * (1 + math.sqrt(5))),
           5.0 / 3.0,
       )
   )
   print(
       "{:19.16f} {:19.16f}".format(
           minkowski_inv(-5.0 / 9.0),
           (math.sqrt(13) - 7) / 6,
       )
   )
   print(
       "{:19.16f} {:19.16f}".format(
           minkowski(minkowski_inv(0.718281828)),
           minkowski_inv(minkowski(0.1213141516171819)),
       )
   )

</lang>

Output:
 1.6666666666696983  1.6666666666666667
-0.5657414540893351 -0.5657414540893352
 0.7182818280000092  0.1213141516171819

Raku

Translation of: Go

<lang perl6># 20201120 Raku programming solution

my \MAXITER = 151;

sub minkowski(\x) {

  return x.floor + minkowski( x - x.floor ) if x > 1 || x < 0 ;
  my $y = my $p = x.floor;
  my ($q,$s,$d) = 1 xx 3;
  my $r = $p + 1;
  loop {
     last if ( $y + ($d /= 2)  == $y )        ||
             ( my $m = $p + $r) <  0 | $p < 0 ||
             ( my $n = $q + $s) <  0           ;
     x < $m/$n ?? ( ($r,$s) = ($m, $n) ) !! ( $y += $d; ($p,$q) = ($m, $n) );
  }
  return $y + $d

}

sub minkowskiInv($x is copy) {

  return $x.floor + minkowskiInv($x - $x.floor) if  $x > 1 || $x < 0 ;
  return $x if $x == 1 || $x == 0 ;
  my @contFrac = 0;
  my $i = my $curr = 0 ; my $count = 1;
  loop {
     $x *= 2;
     if $curr == 0 {
        if $x < 1 {
           $count++
        } else {
           $i++;
           @contFrac.append: 0;
           @contFrac[$i-1] = $count;
           ($count,$curr) = 1,1;
           $x--;
        }
     } else {
        if $x > 1 {
           $count++;
           $x--;
        } else {
           $i++;
           @contFrac.append: 0;
           @contFrac[$i-1] = $count;
           ($count,$curr) = 1,0;
        }
     }
     if $x == $x.floor { @contFrac[$i] = $count ; last }
     last if $i == MAXITER;
   }
   my $ret = 1 / @contFrac[$i];
   loop (my $j = $i - 1; $j ≥ 0; $j--) { $ret = @contFrac[$j] + 1/$ret }
   return 1 / $ret

}

printf "%19.16f %19.16f\n", minkowski(0.5*(1 + 5.sqrt)), 5/3; printf "%19.16f %19.16f\n", minkowskiInv(-5/9), (13.sqrt-7)/6; printf "%19.16f %19.16f\n", minkowski(minkowskiInv(0.718281828)),

  minkowskiInv(minkowski(0.1213141516171819))</lang>
Output:
 1.6666666666696983  1.6666666666666667
-0.5657414540893351 -0.5657414540893352
 0.7182818280000092  0.1213141516171819

REXX

Translation of: FreeBASIC
Translation of: Phix

<lang rexx>/*REXX program uses the Minkowski question─mark function to convert a continued fraction*/ numeric digits 40 /*use enough dec. digits for precision.*/ say fmt( mink( 0.5 * (1+sqrt(5) ) ) ) fmt( 5/3 ) say fmt( minkI(-5/9) ) fmt( (sqrt(13) - 7) / 6) say fmt( mink( minkI(0.718281828) ) ) fmt( mink( minkI(.1213141516171819) ) ) exit 0 /*stick a fork in it, we're all done. */ /*──────────────────────────────────────────────────────────────────────────────────────*/ floor: procedure; parse arg x; t= trunc(x); return t - (x<0) * (x\=t) fmt: procedure: parse arg a; d= digits(); return right( format(a, , d-2, 0), d+5) /*──────────────────────────────────────────────────────────────────────────────────────*/ mink: procedure: parse arg x; p= x % 1; if x>1 | x<0 then return p + mink(x-p)

      q= 1;    s= 1;    m= 0;    n= 0;    d= 1;   y= p
      r= p + 1
                   do forever;   d= d * 0.5;      if y+d=y | d=0  then leave   /*d= d÷2*/
                   m= p + r;                      if m<0   | p<0  then leave
                   n= q + s;                      if n<0          then leave
                   if x<m/n      then do;   r= m;       s= n;           end
                                 else do;   y= y + d;   p= m;   q= n;   end
                   end   /*forever*/
      return y + d

/*──────────────────────────────────────────────────────────────────────────────────────*/ minkI: procedure; parse arg x; p= floor(x); if x>1 | x<0 then return p + minkI(x-p)

                                               if x=1 | x=0  then return x
      cur= 0;                limit= 200;       $=               /*limit: max iterations*/
      #= 1                                                      /*#:  is the count.    */
                 do  until #==limit | words($)==limit;                        x= x * 2
                 if cur==0  then if x<1  then      #= # + 1
                                         else do;  $= $ #;  #= 1;   cur= 1;  x= x-1;  end
                            else if x>1  then do;           #= # + 1;        x= x-1;  end
                                         else do;  $= $ #;  #= 1;   cur= 0;           end
                 if x==floor(x)          then do;           $= $ #;  leave;           end
                 end   /*until*/
      z= words($)
      ret= 1 / word($, z)
                              do j=z  for z  by -1;    ret= word($, j)    +    1 / ret
                              end   /*j*/
      return 1 / ret

/*──────────────────────────────────────────────────────────────────────────────────────*/ sqrt: procedure; parse arg x; if x=0 then return 0; d=digits(); numeric digits; h=d+6

     numeric form; m.=9; parse value format(x,2,1,,0) 'E0' with g "E" _ .; g=g *.5'e'_ %2
       do j=0  while h>9;     m.j= h;             h= h % 2 + 1;      end  /*j*/
       do k=j+5  to 0  by -1; numeric digits m.k; g= (g + x/g) * .5; end  /*k*/; return g</lang>
output   when using the internal default inputs:
     1.66666666666666666666666666673007566392      1.66666666666666666666666666666666666667
    -0.56574145408933511781346312208825067563     -0.56574145408933511781346312208825067562
     0.71828182799999999999999999999999992890      0.12131415161718190000000000000000000833

Wren

Translation of: FreeBASIC
Library: Wren-fmt

<lang ecmascript>import "/fmt" for Fmt

var MAXITER = 151

var minkowski // predeclare as recursive minkowski = Fn.new { |x|

   if (x > 1 || x < 0) return x.floor + minkowski.call(x - x.floor)
   var p = x.floor
   var q = 1
   var r = p + 1
   var s = 1
   var d = 1
   var y = p
   while (true) {
       d = d / 2
       if (y + d == y) break
       var m = p + r
       if (m < 0 || p < 0 ) break
       var n = q + s
       if (n < 0) break
       if (x < m/n) {
           r = m
           s = n
       } else {
           y = y + d
           p = m
           q  = n
       }
   }
   return y + d

}

var minkowskiInv minkowskiInv = Fn.new { |x|

   if (x > 1 || x < 0) return x.floor + minkowskiInv.call(x - x.floor)
   if (x == 1 || x == 0) return x
   var contFrac = [0]
   var curr = 0
   var count = 1
   var i = 0
   while (true) {
       x = x * 2
       if (curr == 0) {
           if (x < 1) {
               count = count + 1
           } else {
               i = i + 1
               var t = contFrac
               contFrac = List.filled(i + 1, 0)
               for (j in 0...t.count) contFrac[j] = t[j]
               contFrac[i-1] = count
               count = 1
               curr = 1
               x = x - 1
           }
       } else {
           if (x > 1) {
               count = count + 1
               x = x - 1
           } else {
               i = i + 1
               var t = contFrac
               contFrac = List.filled(i + 1, 0)
               for (j in 0...t.count) contFrac[j] = t[j]
               contFrac[i-1] = count
               count = 1
               curr = 0
           }
       }
       if (x == x.floor) {
           contFrac[i] = count
           break
       }
       if (i == MAXITER) break
   }
   var ret = 1/contFrac[i]
   for (j in i-1..0) ret = contFrac[j] + 1/ret
   return 1/ret

}

Fmt.print("$17.16f $17.14f", minkowski.call(0.5 * (1 + 5.sqrt)), 5/3) Fmt.print("$17.14f $17.14f", minkowskiInv.call(-5/9), (13.sqrt - 7)/6) Fmt.print("$17.14f $17.14f", minkowski.call(minkowskiInv.call(0.718281828)),

                            minkowskiInv.call(minkowski.call(0.1213141516171819)))</lang>
Output:
 1.66666666666970  1.66666666666667
-0.56574145408934 -0.56574145408934
 0.71828182800001  0.12131415161718