Euler_Problem-033.b93

v
>110p120p"cc" v
vp09p08       <     >80g55+/90g55+/-#v_v                           
>90g80g`!#v_>80g55+/|                  >80g90g55+/*90g80g55+/*-#v_v
            v       <                <                          <
                    >80g55+/90g55+%-#v_v
            >80g55+/|                  >80g90g55+/*90g80g55+%*-#v_v
 @v\p05:<   v       <                <                          <  
 .>50g%:|: <        >80g55+%90g55+/-#v_v
 ^/\g02$<  g>80g55+%|                  >80g90g55+%*90g80g55+/*-#v_v
           0v       <                <                          <
           2        >80g55+%90g55+%-#v_v
           g>80g55+%|                  >80g90g55+%*90g80g55+%*-#v_v
           0v       <                <                          <
           1>                                v
           ^ $_v#!-9p08:-1g08<_v#!-9p09:-1g09<p02*g09g02p01*g08g01<
^       p0 9"c"<
^         >#                 ^#<



[1,0] numerator
[2,0] denominator

[5,0] euclid (temp)

[8,0] A
[9,0] B


// Iterate through every possible fraction

          v         >#                                    #< v#     p09"c"<
                          
"cc"80p90p>90g80g`!#^_     80g."/",90g.55+,    90g1-:90p9-#^_>80g1-:80p9-#^_$  @


// isDCF

        >80g55+/90g55+/-#v_v
>80g55+/|                  >80g90g55+/*90g80g55+/*-#v_>  @ true
v       <                <                          <  > @ false
        >80g55+/90g55+%-#v_v
>80g55+/|                  >80g90g55+/*90g80g55+%*-#v_^
v       <                <                          <
        >80g55+%90g55+/-#v_v
>80g55+%|                  >80g90g55+%*90g80g55+/*-#v_^
v       <                <                          <
        >80g55+%90g55+%-#v_v
>80g55+%|                  >80g90g55+%*90g80g55+%*-#v_^
v       <                <                          <
>                                                      ^


// GCD (stack,stack -> stack) # uses 50 as temp

   >:50p\v
> :|:%g05<
   >$    @


---------------------------------------

This one was relaxing. You can simple iterate through all 8100 cases and test each one in short time.  
Three notable things:

- You can test two fraction `a/b` and `c/d` simple of equality with the formula `a*d == b*c`.
- This is my compact [GCD algorithm](http://en.wikipedia.org/wiki/Euclidean_algorithm) in befunge (stack -> stack)
- This is one of the few *real* Befunge-93 programs in this series. (It doesn't violate the [80x25](https://github.com/catseye/Befunge-93/blob/master/doc/Befunge-93.markdown) size rule)