Convert decimal number to rational

Convert decimal number to rational is a draft programming task. It is not yet considered ready to be promoted as a complete task, for reasons that should be found in its talk page.

The task is to write a program to transform a decimal number into fraction in lowest terms.

For transform a decimal to a fraction, previously we have to know what type is the decimal number, here are some examples

'Examples of types decimal numbers:

67 / 74 = 0.9054054 >>> Mixed decimal number

42 / 81 = 0.518518 >>> Pure decimal number

3/4 = 0.75 >>> Exact decimal number

Option >>>

4527027/ 5000000 = 0.9054054

259259/ 500000 = 0.518518

3/ 4 = 0.75

BASIC

<lang qbasic> 'Program to transform a decimal to a fraction 'LRCVS 11.06.11

'program to transform a decimal number to fraction declare sub exacto (a$) declare sub puro (a$, b$()) declare sub mixto (a$, b$()) declare function factor (j , k ) as integer

dim as integer l, r, s, t, k, w1, i, m, x, ll, pp, ps, u, v, j

dim as string a, c, d, a2

dim y () as string dim w2 () as string

cls input "Decimal number = ";a$ a2$ = a$ print if instr(a$,".") = 0 then print "It's not a decimal number " : goto 100 cls l = len(a$)

for r = 1 to l for s = 1 to l if s + r = l + 2 then exit for k = k + 1 next s next r

w1 = k redim y$(w1) redim b$(w1)

k = 0 for r = 1 to l for s = 1 to l c$ = mid$(a$,r,s) if s + r = l + 2 then exit for if len(c$) <= int(l/2) then k = k + 1 : y$(k) = c$ next s next r t = 0

for r = 1 to k i = 0 f = 0 x = 0 m = 0 if i = 0 then i = instr(a$,y$(r)):x = 1 for s = 1 to len(a$) if x = 1 then f = instr(s,a$,y$(r)) if x = 1 and f > m then m = f next s

h = 0 k = 0 for n = i to m step len(y$(r)) if h = 0 and mid$(a$,n,len(y$(r))) = y$(r) then k = k + 1 else h = 1 next n if k > 1 then t = t + 1 :b$(t) = y$(r) next r

for r = 1 to w1 for s = r + 1 to w1 if b$(r) = b$(s) then b$(s) = "" next s next r

print "decimal number = ";a$ print

ll = len(a$) pp = instr(a$,".") d$ = mid$(a$,pp+1,ll) ps = instr(d$,b$(1)) if ps = 0 then

               print "Decimal number exact"
               print
               call exacto (a$)
               end if

if ps = 1 then

               print "Decimal number pure"
               print
               call puro (a$, b$())
               end if

if ps > 1 then

                print "Decimal number mix"
                print
                call mixto (a$, b$())
                end if

100: print print "End" sleep end

':::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: sub exacto (a as string) dim as integer b, c, d, g, may, j, k, l, r, s, u, v, w, f dim as string z, h, g1 b = len(a$) c = instr(a$,".") d = b - c g = int(val(a$)) h$ = right$(a$, b - c)

may = 0 j = 10^d k = val(h$) for n = 9 to 1 step - 1 if j mod (1*(10^n)) = 0 and k mod (1*(10^n)) = 0 then j = j/(1*(10^n)) : k = k/(1*(10^n)) :exit for next n l = factor (j,k) u = k/l v = j/l w = (g * v) + u print print w;"/";v ;" = " ;w/v

end sub

':::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: sub puro (a as string, b() as string) dim as integer b2, c, d, g, may, j, k, l, u, v, w, f, lr,x5 dim as string z, h, g1, x, a3

z$ = b$(1) x5 = int(val(a$)) lr = len (z$) b2 = len (a$) c = instr (a$,".") g = int (val(a$)) b2 = len(z$) + 1 + len(str$(g)) a3$ = str$(g) + "." + z$ h$ = right$(a3$, b2 - c)

may = 0 x$ = "" for n = 1 to lr x$ = x$ + "9" next n

j = val(x$) k = val(h$)

for n = 9 to 1 step - 1 if j mod (1*(10^n)) = 0 and k mod (1*(10^n)) = 0 then j = j/(1*(10^n)) : k = k/(1*(10^n)) :exit for next n l = factor (j,k) u = k/l v = j/l w = (g * v) + u print w;"/";v ;" = ";w/v print print "Option >>> " call exacto (a$)

end sub

':::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: sub mixto (a as string, b() as string)

dim as integer b3, c, d, g, j, k, l, u, v, w, f, lr, lz, ly, x5 dim as string z, h, g4, g7, x, y

z$ = b$(1) x5 = int(val(a$)) w = instr(a$, z$) v = instr(a$,".") y$ = mid$(a$,v+1,w-v-1) lz = len(z$) ly = len(y$) b3 = (val(y$)*(9*(10^ly))) + ((1*(10^ly))* (val(z$))) c = (9*(10^ly))*(1*(10^ly)) j = b3 k = c for n = 9 to 1 step - 1 if j mod (1*(10^n)) = 0 and k mod (1*(10^n)) = 0 then j = j/(1*(10^n)) : k = k/(1*(10^n)): exit for next n l = factor (b3,c) u = k/l v = j/l if x5 <> 0 then print (x5*v)+ u;"/";u ;" = ";((x5*v)+ u)/u else print v;"/";u;" = "; v/u print print "Option >>> " call exacto (a$) end sub ':::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: function factor (j as integer, k as integer) as integer dim as integer may, n, s, r, l5, j5, k5 may = 0 l5 = 1 j5 = j k5 = k for n = 9 to 1 step - 1 if j5 mod (1*(10^n)) = 0 and k5 mod (1*(10^n)) = 0 then j5 = j5/(1*(10^n)) : k5 = k5/(1*(10^n)): exit for next n if j5 > k5 then may = j5 else may = k5 for n = may to 1 step - 1

  r = (j5 mod n)
  s = (k5 mod n)
  if r = 0 and s = 0  then l5 = n :exit for

next n factor = l5 end function end</lang>

C

Since the intention of the task giver is entirely unclear, here's another version of best rational approximation of a floating point number. It's somewhat more rigorous than the Perl version below, but is still not quite complete.<lang C>#include <stdio.h>

1. include <stdlib.h>
2. include <math.h>
3. include <stdint.h>

/* f : number to convert.

* num, denom: returned parts of the rational.
* md: max denominator value.  Note that machine floating point number
*     has a finite resolution (10e-16 ish for 64 bit double), so specifying
*     a "best match with minimal error" is often wrong, because one can
*     always just retrieve the significand and return that divided by 
*     2**52, which is in a sense accurate, but generally not very useful:
*     1.0/7.0 would be "2573485501354569/18014398509481984", for example.
*/

void rat_approx(double f, int64_t md, int64_t *num, int64_t *denom) { /* a: continued fraction coefficients. */ int64_t a, h[3] = { 0, 1, 0 }, k[3] = { 1, 0, 0 }; int64_t x, d, n = 1; int i, neg = 0;

if (md <= 1) { *denom = 1; *num = (int64_t) f; return; }

if (f < 0) { neg = 1; f = -f; }

while (f != floor(f)) { n <<= 1; f *= 2; } d = f;

/* continued fraction and check denominator each step */ for (i = 0; i < 64; i++) { a = n ? d / n : 0; if (i && !a) break;

x = d; d = n; n = x % n;

x = a; if (k[1] * a + k[0] >= md) { x = (md - k[0]) / k[1]; if (x * 2 >= a || k[1] >= md) i = 65; else break; }

h[2] = x * h[1] + h[0]; h[0] = h[1]; h[1] = h[2]; k[2] = x * k[1] + k[0]; k[0] = k[1]; k[1] = k[2]; } *denom = k[1]; *num = neg ? -h[1] : h[1]; }

int main() { int i; int64_t d, n; double f;

printf("f = %16.14f\n", f = 1.0/7); for (i = 1; i <= 20000000; i *= 16) { printf("denom <= %d: ", i); rat_approx(f, i, &n, &d); printf("%lld/%lld\n", n, d); }

printf("\nf = %16.14f\n", f = atan2(1,1) * 4); for (i = 1; i <= 20000000; i *= 16) { printf("denom <= %d: ", i); rat_approx(f, i, &n, &d); printf("%lld/%lld\n", n, d); }

return 0; }</lang>Output:<lang>f = 0.14285714285714 denom <= 1: 0/1 denom <= 16: 1/7 denom <= 256: 1/7 denom <= 4096: 1/7 denom <= 65536: 1/7 denom <= 1048576: 1/7 denom <= 16777216: 1/7

f = 3.14159265358979 denom <= 1: 3/1 denom <= 16: 22/7 denom <= 256: 355/113 denom <= 4096: 355/113 denom <= 65536: 104348/33215 denom <= 1048576: 3126535/995207 denom <= 16777216: 47627751/15160384</lang>

J

J's x: built in will find a rational number which "best matches" a floating point number.

<lang j> x: 0.9054054 0.518518 0.75 NB. find "exact" rational representation 127424481939351r140737488355328 866492568306r1671094481399 3r4</lang>

Note that the concept of "best" has to do with the expected precision of the argument: <lang j> x: 0.9 0.5 9r10 1r2

  x: 0.9054 0.5185

4527r5000 1037r2000

  x: 0.9054054 0.5185185

127424481939351r140737488355328 1037037r2000000

  x: 0.9054054054 0.5185185185

5358191125333r5918002138463 6073341499873r11712872893031

  x: 0.9054054054054 0.5185185185185

67r74 14r27

  x: 0.9054054054054054 0.5185185185185185

67r74 14r27</lang>

Note that J allows us to specify an epsilon, for the purpose of recognizing a best fit, but the allowed values must be rather small. In J version 6, the value 5e_11 was nearly the largest epsilon allowed:

<lang j> x:(!. 5e_11) 0.9054054054 0.5185185185 67r74 14r27</lang>

Here are some other alternatives for dealing with decimals and fractions:

<lang j> 0j10": x:inv x: 0.9054054 0.518518 0.75 NB. invertable (shown to 10 decimal places) 0.9054054000 0.5185180000 0.7500000000

  0j10": x:inv 67r74 42r81 3r4             NB. decimal represenation (shown to 10 decimal places)

0.9054054054 0.5185185185 0.7500000000

  x: x:inv 67r74 42r81 3r4                 NB. invertable

67r74 14r27 3r4</lang>

Perl

Note: the following is considerably more complicated than what was specified, because the specification is not, well, specific. Three methods are provided with different interpretation of what "conversion" means: keeping the string representation the same, keeping machine representation the same, or find best approximation with denominator in a reasonable range. None of them takes integer overflow seriously (though the best_approx is not as badly subject to it), so not ready for real use. <lang perl>sub gcd {

       my ($m, $n) = @_;
       ($m, $n) = ($n, $m % $n) while $n;
       return $m

}

sub rat_machine {

       my $n = shift;
       my $denom = 1;
       while ($n != int $n) {
               # assuming the machine format is base 2, and multiplying
               # by 2 doesn't change the mantissa
               $n *= 2;

               # multiply denom by 2, ignoring (very) possible overflow
               $denom <<= 1;
       }
       if ($n) {
               my $g = gcd($n, $denom);
               $n /= $g;
               $denom /= $g;
       }
       return $n, $denom;

}

1. helper, make continued fraction back into normal fraction

sub get_denom {

       my ($num, $denom) = (1, pop @_);
       for (reverse @_) {
               ($num, $denom) = ($denom, $_ * $denom + $num);
       }
       wantarray ? ($num, $denom) : $denom

}

sub best_approx {

       my ($n, $limit) = @_;
       my ($denom, $neg);
       if ($n < 0) {
               $neg = 1;
               $n = -$n;
       }

       my $int = int($n);
       my ($num, $denom, @coef) = (1, $n - $int);

       # continued fraction, sort of
       while (1) {
               # make sure it terminates
               last if $limit * $denom < 1;
               my $i = int($num / $denom);

               # not the right way to get limit, but it works
               push @coef, $i;

               if (get_denom(@coef) > $limit) {
                       pop @coef;
                       last;
               }

               # we lose precision here, but c'est la vie
               ($num, $denom) = ($denom, $num - $i * $denom);
       }

       ($num, $denom) = get_denom @coef;
       $num += $denom * $int;

       return $neg ? -$num : $num, $denom;

}

sub rat_string {

       my $n = shift;
       my $denom = 1;
       my $neg;

       # trival xyz.0000 ... case
       $n =~ s/\.0+$//;
       return $n, 1 unless $n =~ /\./;

       if ($n =~ /^-/) {
               $neg = 1;
               $n =~ s/^-//;
       }

       # shift decimal point to the right till it's gone
       $denom *= 10    while $n =~ s/\.(\d)/$1\./;
       $n =~ s/\.$//;

       # removing leading zeros lest it looks like octal
       $n =~ s/^0*//;
       if ($n) {
               my $g = gcd($n, $denom);
               $n /= $g;
               $denom /= $g;
       }
       return $neg ? -$n : $n, $denom;

}

my $limit = 1e8; my $x = 3/8; print "3/8 = $x:\n"; printf "machine: %d/%d\n", rat_machine $x; printf "string: %d/%d\n", rat_string $x; printf "approx below $limit: %d/%d\n", best_approx $x, $limit;

$x = 137/4291; print "\n137/4291 = $x:\n"; printf "machine: %d/%d\n", rat_machine $x; printf "string: %d/%d\n", rat_string $x; printf "approx below $limit: %d/%d\n", best_approx $x, $limit;

$x = sqrt(1/2); print "\n1/sqrt(2) = $x\n"; printf "machine: %d/%d\n", rat_machine $x; printf "string: %d/%d\n", rat_string $x; printf "approx below 10: %d/%d\n", best_approx $x, 10; printf "approx below 100: %d/%d\n", best_approx $x, 100; printf "approx below 1000: %d/%d\n", best_approx $x, 1000; printf "approx below 10000: %d/%d\n", best_approx $x, 10000; printf "approx below 100000: %d/%d\n", best_approx $x, 100000; printf "approx below $limit: %d/%d\n", best_approx $x, $limit;

$x = -4 * atan2(1,1); print "\n-Pi = $x\n"; printf "machine: %d/%d\n", rat_machine $x; printf "string: %d/%d\n", rat_string $x;

for (map { 10 ** $_ } 1 .. 10) {

       printf "approx below %g: %d / %d\n", $_, best_approx($x, $_)

}</lang>

Output:

3/8 = 0.375:
machine: 3/8
string:  3/8
approx below 100000000:  3/8

137/4291 = 0.0319272896760662:
machine: 2300603678209305/72057594037927936
string:  159636448380331/5000000000000000
approx below 100000000:  137/4291

1/sqrt(2) = 0.707106781186548
machine: 6369051672525773/9007199254740992
string:  176776695296637/250000000000000
approx below 10:  5/7
approx below 100:  29/41
approx below 1000:  408/577
approx below 10000:  5741/8119
approx below 100000:  33461/47321
approx below 100000000:  38613965/54608393

-Pi = -3.14159265358979
machine: -884279719003555/281474976710656
string:  -314159265358979/100000000000000
approx below 10: -22 / 7
approx below 100: -22 / 7
approx below 1000: -355 / 113
approx below 10000: -355 / 113
approx below 100000: -208341 / 66317
approx below 1e+06: -1146408 / 364913
approx below 1e+07: -5419351 / 1725033
approx below 1e+08: -245850922 / 78256779
approx below 1e+09: -1881244168 / 598818617
approx below 1e+10: -9978066541 / 3176117225

Perl 6

<lang perl6>sub decimal_to_fraction ( Str $n, Int $rep_digits = 0 ) returns Str {

   my ( $int, $dec ) = ( $n ~~ /^ (\d+) \. (\d+) $/ )».Str or die;

   my ( $numer, $denom ) = ( $dec, 10 ** $dec.bytes );
   if $rep_digits {
       my $to_move = $dec.bytes - $rep_digits;
       $numer -= $dec.substr(0, $to_move);
       $denom -= 10 ** $to_move;
   }

   my $rat = Rat.new( $numer.Int, $denom.Int ).perl;
   return $int ?? "$int $rat" !! $rat;

}

my @a = ['0.9054', 3], ['0.518', 3], ['0.75', 0], (^4).map({['12.34567', $_]}); for @a -> [ $n, $d ] {

   say "$n with $d repeating digits = ", decimal_to_fraction( $n, $d );

}</lang>

Output:

0.9054 with 3 repeating digits = 67/74
0.518 with 3 repeating digits = 14/27
0.75 with 0 repeating digits = 3/4
12.34567 with 0 repeating digits = 12 34567/100000
12.34567 with 1 repeating digits = 12 31111/90000
12.34567 with 2 repeating digits = 12 17111/49500
12.34567 with 3 repeating digits = 12 1279/3700

Python

Note that the decimal values of the task description are truncated in some cases.

The first loop limits the size of the denominator the second uses the Decimal module to convert an accurate Decimal representation of the given values to a fraction. <lang python>>>> from fractions import Fraction >>> for d in (0.9054054, 0.518518, 0.75): print(d, Fraction(d).limit_denominator(100))

0.9054054 67/74 0.518518 14/27 0.75 3/4 >>> from decimal import Decimal >>> for d in '0.9054054 0.518518 0.75'.split(): print(d, Fraction(Decimal(d)))

0.9054054 4527027/5000000 0.518518 259259/500000 0.75 3/4 >>> </lang>