Mixed Number to an Improper Fraction

Question

Mixed Number to an Improper Fraction

In this challenge you will be converting a mixed number to an improper fraction.

Because improper fractions use fewer numbers, your code will need to be as short as possible.

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Examples

4 1/2
9/2

12 2/4
50/4

0 0/2
0/2

11 23/44
507/44

Specification

You may assume the denominator of the input will never be 0. The input will always be in the format x y/z where x,y,z are arbitrary nonnegative integers. You do not need to simplify the output.

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This is code-golf so shortest code in bytes wins.

Answer 1

Japt, 10 bytes

Woohoo, currently beating CJam!

U*W+V+'/+W

Try it online!

How it works

       // Implicit: [U,V,W] = eval(input). This automatically discards the slash.
U*W+V  // Calculate the new numerator: (whole part * denominator) + numerator.
+'/+W  // Add the slash and the denominator.
       // Implicit: output last expression

Answer 2

LabVIEW, 29 LabVIEW Primitives

Answer 3

CJam, 16 15 14 bytes

l'/']er~:Xb'/X

or

l'/']er~_@b'/@

Test it here.

Explanation

l      e# Read input.
'/']er e# Replace the "/" with a "]".
~      e# Evaluate the string as code. If the input was "i n/d", this pushes [i n] d.
:X     e# Store the denominator in X.
b      e# Treat [i n] as base-d digits. This effectively computes i*d + n.
'/     e# Push a slash.
X      e# Push the denominator.

The other version avoids using a variable by using a bit more stack shifting.

Answer 4

Mathematica, 58 bytes

ToExpression@StringReplace[#," "->"+"]~ToString~InputForm&

This returns the simplified result. If outputting a rational number instead of a string is fine, we can save 19 bytes:

ToExpression@StringReplace[#," "->"+"]&

Answer 5

PowerShell, 47 44 42 Bytes

_{Crossed out 44 is still regular 44 ;(}

$l,$n,$d=$args-split'\D';"$(+$l*$d+$n)/$d"

Golfed a couple bytes by using regex -split. Golfed a couple more thanks to TessellatingHeckler by swapping the regex.

The $args-split'\D' takes our input argument and splits on non-digit characters. Here it performs two splits, one on whitespace, the other on the / character. The results are then stored in the three variables using a simultaneous assignment. We then formulate the string output as (the $left number times the $denominator plus the $numerator) executed as a code block, a / slash, and then the $denominator again.

Answer 6

Java with Ten Foot Laser Pole 1.03, 79+25 (import) = 104 bytes

Requires import sj224.tflp.math.*;

String m(String[]a){return ""+new BigRational(a[0]).add(new BigRational(a[1]));}

This will almost certainly work with 1.04 as well, but so far I've only tested it with 1.03 because I already happened to have a java project set up with 1.03 in the build path.

Answer 7

JavaScript (ES6), 44 41 bytes

m=>([x,y,z]=m.match(/\d+/g),+y+x*z+"/"+z)

Saved 3 bytes thanks to @ETHproductions!

Explanation

Very simple.

m=>
  ([x,y,z]=m.match(/\d+/g), // x, y and z = numbers from input
    +y+x*z                  // numerator
    +"/"+z                  // denominator
  )

Test

Test is without destructuring assignment to work in most browsers.

var solution = m=>+(p=m.match(/\d+/g))[1]+p[0]*p[2]+"/"+p[2]

<input type="text" id="input" value="11 23/44" />
<button onclick="result.textContent=solution(input.value)">Go</button>
<pre id="result"></pre>

Answer 8

Haskell, 74 67 63 bytes

r=read
f x|(a,(c,s:d):_)<-lex<$>lex x!!0=show(r a*r d+r c)++s:d

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Explanation

As H.PWiz figured out we can use Haskell's lexer here to break up the string into it's parts. (Earlier I was using span(>'/')) And Laikoni pointed out that <$> works just like mapSnd from Data.Tuple.

The pattern guard breaks up our code into the three numbers we want using lex. lex invokes haskell's lexer to break off the first token. It returns a list with each element representing a possible way to parse the string. These elements are tuples with the first element being the first token and the rest of the string being the second element. Now since the input format is very regular we are only ever going to have exactly one parse, so we can always take the first one. The first thing we do is invoke lex on the input

lex x

Then we unwrap it from it's list giving us a 2-tuple

lex x!!0

The first token will be the whole part of the mixed fraction leaving the fraction prepended by a space to still parse. Then since tuples are Functors we can use (<$>) an alias for fmap to apply lex to the second element of the tuple.

lex<$>lex x!!0

This eats through the space and breaks off the next token, the numerator of our fraction. Now we bind this to a pattern match using <-. Our pattern is

(a,(c,s:d):_)

a grabs the whole part of the fraction, our first token. :_ unwraps the list resulting from our second lex. c grabs the second token we lexed, that is the numerator of the fraction. Everything that remains is bound to s:d which splits it into its first character, guaranteed by the format to be a / and the remainder which will be the denominator.

Now that we have parsed the input we do the actual computation:

show(r a*r d+r c)++s:d

Where r is the read function we bound earlier.

It is important to note that lex returns a list empty if it fails and non-empty if it succeeds. Why this is not a Maybe I do not know.

Answer 9

Bash + coreutils, 28

dc<<<${@/\// }sarla*+n47Plap

$@ expands to all command-line parameters, so ${@/\// } expands to all command-line parameters with / replaced with , which is put on dc's stack. The rest is simple stack manipulation and arithmetic.

Answer 10

Julia, 58 50 bytes

s->eval(parse((r=replace)(r(s," ","+"),"/","//")))

This is an anonymous function that accepts a string and returns a Rational type object. To call it, give it a name, e.g. f=s->....

We can take advantage of the fact that the input can be manipulated slightly to be an expression that evaluates to a rational. In particular, an integer plus a rational is a rational, and rationals are denoted with double slashes. So if we turn 4 1/2 into 4+1//2, the evaluated result will be 9//2.

Ungolfed:

function f(s::AbstractString)
    # Replace the space in the input with a plus
    r1 = replace(s, " ", "+")

    # Replace the / with //
    r2 = replace(r1, "/", "//")

    # Parse the resulting expression as a rational
    return eval(parse(r2))
end

Answer 11

Smalltalk – 76 characters

The input exactly matches the array delimiter and inherent fraction representation of Smalltalk. If it just weren't so verbose, it could have been a serious contender!

Compiler evaluate:'|p|p:=0.#(',FileStream stdin nextLine,')do:[:q|p:=p+q].p'

It's too bad simplification wasn't a requirement, Smalltalk does it automatically!

Answer 12

05AB1E, 17 15 bytes

#`'/¡R`Š©*+®'/ý

-2 bytes thanks to @MagicOctopusUrn.

Try it online or verify all test cases.

Explanation:

#`         # Split input by spaces and push all items to the stack
           #  i.e. "4 1/2" → "4" and "1/2"
  '/¡      # Push the second item by "/"
           #  i.e. "1/2" → [1,2]
     R`    # Revert the list, and also push all items to the stack
           #  i.e. [1,2] → [2,1] → 2 and 1
Š          # Triple-swap the stack
           #  [4,2,1] → [1,4,2]
 ©         # Store the 2 in the register
  *        # Multiple the top two items
           #  4 and 2 → 8
   +       # Add the top two items
           #  1 and 8 → 9
®          # Push the 2 from the register to the stack again
 '/ý       # Join the two items by "/"
           #  9 and 2 → "9/2"

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With flexible input- and output-format, taking the integers in the order x,z,y and outputting the nominator and denominator on separated lines it would be 4 bytes (which is why I added the parsing-tag to the challenge..):

*+²»

Try it online or verify all test cases.

Explanation:

*        # Multiply the first two inputs (x and z)
         #  i.e. 4 and 2 → 8
 +       # Add the third input (y)
         #  i.e. 8 and 1 → 9
  ²      # Take the second input again (z)
   »     # Join the stack by newlines and implicitly print it

Answer 13

ARBLE, 13 bytes

a*c+b.."/"..c

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Answer 14

Perl 5 with -la -Mfeature=say, 32 bytes 25 bytes

m|/|;say$_*$'+$F[1],"/$'"

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(-7 bytes thanks to Dom Hastings)

$_ is the whole input x y/z, which evaluates the value of x in numeric contexts (like the * here). $' is the regex post-match string, which here contains whatever comes after / - so, z. To get the y value, we use the -a flag which splits the input on spaces and places them in the @F array. So here, @F = ("x", "y/z"), which means $F[1]="y/z" which evaluates in y in numeric contexts (since y is the initial contiguous sequence of digits with $F[1]).

Answer 15

Python 3, 78 76 bytes

def f(s):a,b=s.split();b,c=b.split('/');return'%s/'%(int(a)*int(c)+int(b))+c

Try it online!

Answer 16

Stax, 1 byte

+

Run and debug it (although there's not much to debug)

The challenge specification says "You do not need to simplify the output." Assuming it's allowed to simplify, then there's a built-in instruction in stax to do this. The input is implicitly interpreted as an integer and a rational number. The + instruction widens both to rationals, adds, and simplifies. The result is implicitly printed.

Answer 17

Perl, 82 61 38 bytes

#!perl -paF/\D/
$_=$F[0]*$F[2]+$F[1]."/$F[2]"

This can probably be golfed more.

Changes

• Saved 16 bytes by using a regex in split, and 5 by using <> instead of <STDIN>.
• Saved another 16 bytes thanks to Dennis.

Answer 18

Java, 159 148 142 120 110 bytes

String m(String[]a){Long b=new Long(a[0]),d=new Long((a=a[1].split("/"))[1]);return b*d+new Long(a[0])+"/"+d;}

Saved a bunch of bytes thanks to FlagAsSpam.

Answer 19

Javascript ES6, 62 bytes

p=prompt;b=p(a=+p()).split`/`;alert((+b[1]*a+ +b[0])+"/"+b[1])

Answer 20

Mathematica, 51 bytes

Interpreter["ComputedNumber"]@#~ToString~InputForm&

Interestingly, Mathematica supports this with a built-in. If outputting a number is allowed, than we only need 28 bytes:

Interpreter@"ComputedNumber"

Answer 21

Check, 120 bytes

>]+>:>32r#v
#d@0+\)  ##:>4;:>5'=:>48-\R-?
dd)R>32-#v
#>15-#v  #?
47r@>@   #v
#dd#v #?
r@>@     #v
    #\d@\: @*@+pd"/"op

Try it online!

I might be able to save some bytes by not trying to reuse the parsing loop (the second line). That way I could make the loop more specific, avoid the huge mess of conditionals, and I could use the register for other things.

Answer 22

Ruby, 23 bytes

->x{eval x.split*?++?r}

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Answer 23

C#, 112 bytes

s=>{string[]a=s.Split(),b=a[1].Split('/');int o=int.Parse(b[1]);return int.Parse(a[0])*o+int.Parse(b[0])+"/"+o;}

Full/Formatted Version:

namespace System.Linq
{
    class P
    {
        static void Main()
        {
            Func<string, string> f = s =>
            {
                string[] a = s.Split(), b = a[1].Split('/');
                int o = int.Parse(b[1]);
                return int.Parse(a[0]) * o + int.Parse(b[0]) + "/" + o;
            };

            Console.WriteLine(f("4 1/2"));
            Console.WriteLine(f("12 2/4"));
            Console.WriteLine(f("0 0/2"));
            Console.WriteLine(f("11 23/44"));

            Console.ReadLine();
        }
    }
}

Answer 24

APL (Dyalog Unicode), 31 bytes

∊∘(⍕¨⊃,⍨'/',⍨⊃⊥1∘↓)¯1⌽#⍎¨∊∘⎕D⊆⊢

Try it online!

Thanks to ngn for the ⊃⊥1∘↓ trick

Answer 25

PHP, 65 Bytes

Try it online

Code

<?=(($a=preg_split("/[\s,\/]/",$argv))[0]*$a[2]+$a[1])."/".$a[2];

Explanation

$a=preg_split("/[\s,\/]/",$argv); # Split the string on "/" and " "
(($a)[0]*$a[2]+$a[1]) # as always denominator*whole number + numerator 
."/"                  # echo an slash
.$a[2];               # echo de denominator

Answer 26

Java 10, 87 bytes

A lambda from String to String.

s->{var p=s.split(" |/");return new Long(p[0])*new Long(p[2])+new Long(p[1])+"/"+p[2];}

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Answer 27

Ouroboros, 16 bytes

rr\r.@*@+n47on1(

Try it here!

Explanation

The reason why Ouroboros is great for this challenge: its integer-reading primitive, r, skips over any non-digit characters until it comes to a number. So no parsing is required; we just read three numbers and process them.

rr                Read the first two numbers
                   Stack: 11 23
  \r              Swap and read the third one
                   Stack: 44 23 11
    .             Duplicate
                   Stack: 44 44 23 11
     @            Rotate (moving third item to top of stack)
                   Stack: 23 44 44 11
      *           Multiply
                   Stack: 1012 44 11
       @+         Rotate and add
                   Stack: 1023 44
         n        Output as number
                   Stack: 44  Output: 1023
          47o     Push 47 and output as character
                   Stack: 44  Output: 1023/
             n    Output as number
                   Output: 1023/44
              1(  Terminate

Answer 28

C, 64

main(i,d,n){scanf("%d %d/%d",&i,&n,&d);printf("%d/%d",i*d+n,d);}

Reads input from STDIN. Fairly self-explanatory, I think.

Answer 29

Lua, 123 Bytes

m=io.read()z=string x=z.find c=z.sub b=tonumber s=x(m," ")f=x(m,"/")d=c(m,f+1)print(b(c(m,1,s))*b(d)+b(c(m,s,f-1)).."/"..d)

Answer 30

Milky Way 1.6.0, 31 bytes

'" "\="/"\=>;<A;A;:>*;A+"/"+<+!

Ended up being much longer than I thought it would be.

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Explanation

'" "\="/"\=>;<A;A;:>*;A+"/"+<+!

'                                # read input from the command line
 " "  "/"               "/"      # push a string to the stack
    \    \                       # split the STOS at the TOS
     =    =                      # dump the TOS to the stack
           >       >             # rotate the stack rightward
            ;  ; ;   ;           # swap the TOS and STOS
             <              <    # rotate the stack leftward
              A A     A          # push the integer representation of the TOS
                  :              # duplicate the TOS
                    *            # multiply the STOS by the TOS
                       +   + +   # add the TOS and STOS
                              !  # output the TOS

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Usage

./mw <path-to-code> -i <input>

Answer 31

Python 2.7, 88 Bytes

a=input().split('/');print int(a[-1])*int(a[0].split()[0])+int(a[0].split()[1]),'/',a[1]

Try it online!

You have to type the input in quotes.

Probably not the best ...