17
\$\begingroup\$

Write a program or a function that takes two non-empty lists of the same length as input and does the following:

  • uses elements of first list to get numerators,
  • uses elements of the second list to get denominators,
  • displays resulting fractions after simplification (2/4=>1/2), separated by "+"s,
  • displays "=" and result of addition after last fraction.

Example:

Input

[1, 2, 3, 3, 6]
[2, 9, 3, 2, 4]

Output

1/2+2/9+1+3/2+3/2=85/18

About rules

  • elements of lists will be positive integers,
  • elements can be separated by spaces, eg: 1/2 + 2/9 + 1 + 3/2 + 3/2 = 85/18 is ok,
  • trailing newline is allowed,
  • lists can be taken in other formats than above, eg.: (1 2 3 3 6) or {1;2;3;3;6}, etc.,
  • 1 can be expressed as 1/1,
  • instead of printing you can return appropriate string,
  • you do not need to handle wrong input,
  • shortest code wins.
\$\endgroup\$
18
  • \$\begingroup\$ What range of values does it have to support? \$\endgroup\$ Commented May 19, 2017 at 12:26
  • \$\begingroup\$ @BradGilbertb2gills I would say at least -30 000 to 30 000, but then I don't know whether it would be extra problem for some languages. So maybe just standard integer range of your language of choice. \$\endgroup\$
    – user65167
    Commented May 19, 2017 at 12:36
  • \$\begingroup\$ @PrzemysławP saying "standard integer range of your language of choice" is not a good idea, some languages have standard integer as booleans \$\endgroup\$ Commented May 19, 2017 at 12:37
  • \$\begingroup\$ Thank you! @BradGilbertb2gills Then at least -30 000 to 30 000. \$\endgroup\$
    – user65167
    Commented May 19, 2017 at 12:40
  • \$\begingroup\$ Can we get fractions as [1, 2] [2, 9] [3, 3] ... instead? \$\endgroup\$ Commented May 19, 2017 at 12:42

26 Answers 26

7
\$\begingroup\$

Ruby 2.4, 54 53 characters

->n,d{a=n.zip(d).map{|n,d|n.to_r/d};a*?++"=#{a.sum}"}

Thanks to:

  • Value Ink for the Ruby 2.4 specific version (-3 characters)
  • Value Ink for optimizing the Rational initialization (-1 character)

Ruby, 58 57 56 characters

->n,d{t=0;n.zip(d).map{|n,d|t+=r=n.to_r/d;r}*?++"=#{t}"}

Sample run:

irb(main):001:0> puts ->n,d{t=0;n.zip(d).map{|n,d|t+=r=n.to_r/d;r}*?++"=#{t}"}[[1, 2, 3, 3, 6], [2, 9, 3, 2, 4]]
1/2+2/9+1/1+3/2+3/2=85/18

Try it online!

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4
  • 1
    \$\begingroup\$ a=n.zip(d).map{|f|(f*?/).to_r};a*?++"=#{a.sum}" in Ruby 2.4 saves you 3 bytes. \$\endgroup\$
    – Value Ink
    Commented May 19, 2017 at 18:46
  • \$\begingroup\$ Thanks @ValueInk. I suspected that may be possible, just had no 2.4 neither locally nor on TIO. \$\endgroup\$
    – manatwork
    Commented May 20, 2017 at 16:13
  • 1
    \$\begingroup\$ Yeah, I installed 2.4 specifically so I could test out solutions with sum haha. Also I just remembered that .map{|i,j|i.to_r/j} is shorter by 1 byte \$\endgroup\$
    – Value Ink
    Commented May 22, 2017 at 20:25
  • \$\begingroup\$ Doh. I tried various approaches through .to_f and division, but didn't thought to dividing Rational with Fixnum. Thanks again, @ValueInk. \$\endgroup\$
    – manatwork
    Commented May 24, 2017 at 19:54
6
\$\begingroup\$

Mathematica, 33 bytes

Row@{Row[#/#2,"+"],"=",Tr[#/#2]}&

input

[{1, 2, 3, 3, 6}, {2, 9, 3, 2, 4}]

\$\endgroup\$
3
  • \$\begingroup\$ Isn't Row@@{#/#2,"+"} the same as Row[#/#2,"+"]? \$\endgroup\$
    – feersum
    Commented May 19, 2017 at 13:57
  • \$\begingroup\$ yes! you are right! \$\endgroup\$
    – ZaMoC
    Commented May 19, 2017 at 14:04
  • 1
    \$\begingroup\$ Fantastic! I didn't realize Row was so convenient for things like this :) \$\endgroup\$ Commented May 19, 2017 at 18:08
3
\$\begingroup\$

M, 12 11 bytes

÷µFj”+;”=;S

This is a dyadic link. Due to a bug, it doesn't work as a full program. F is also required due to a bug.

Try it online!

How it works

÷µFj”+;”=;S  Dyadic link. Left argument: N. Right argument: D

÷            Perform vectorized division, yielding an array of fractions (R).
 µ           Begin a new, monadic chain. Argument: R
  F          Flatten R. R is already flat, but j is buggy and has side effects.
   j”+       Join R, separating by '+'.
      ;”=    Append '='.
         ;S  Append the sum of R.
\$\endgroup\$
1
  • \$\begingroup\$ I like how more than a quarter of the program is to append the '='. :) \$\endgroup\$ Commented Jul 19, 2017 at 10:19
3
\$\begingroup\$

Python 3, 104 bytes

9 bytes thanks to Felipe Nardi Batista.

from fractions import*
def f(*t):c=[Fraction(*t)for t in zip(*t)];print('+'.join(map(str,c)),'=',sum(c))

Try it online!

\$\endgroup\$
6
3
\$\begingroup\$

Perl 6,  77  73 bytes

{join('+',($/=[@^a Z/ @^b]).map:{.nude.join('/')})~"="~$/.sum.nude.join('/')}

Try it

{($/=[@^a Z/@^b])».nude».join('/').join('+')~'='~$/.sum.nude.join('/')}

Try it

Expanded:

{  # bare block lambda with two placeholder params 「@a」 and 「@b」

  (
    $/ = [              # store as an array in 「$/」 for later use

      @^a Z/ @^b        # zip divide the two inputs (declares them)

    ]

  )».nude\              # get list of NUmerators and DEnominators
  ».join('/')           # join the numerators and denominators with 「/」

  .join('+')            # join that list with 「+」

  ~
  '='                   # concat with 「=」
  ~

  $/.sum.nude.join('/') # get the result and represent as fraction
}
\$\endgroup\$
3
\$\begingroup\$

Clojure, 71 bytes

#(let[S(map / % %2)](apply str(concat(interpose '+ S)['=(apply + S)])))

Yay for built-in fractions!

\$\endgroup\$
2
\$\begingroup\$

Mathematica, 61 bytes

t=#~ToString~InputForm&;Riffle[t/@#,"+"]<>"="<>t@Tr@#&[#/#2]&
\$\endgroup\$
2
\$\begingroup\$

JavaScript (ES6), 111 bytes

Takes the lists in currying syntax (a)(b).

let f =

a=>b=>a.map((v,i)=>F(A=v,B=b[i],N=N*B+v*D,D*=B),N=0,D=1,F=(a,b)=>b?F(b,a%b):A/a+'/'+B/a).join`+`+'='+F(A=N,B=D)

console.log(f([1, 2, 3, 3, 6])([2, 9, 3, 2, 4]))

\$\endgroup\$
2
\$\begingroup\$

Java, 225 bytes

int c(int a,int b){return b>0?c(b,a%b):a;}
(N,D)->{int n=0,d=1,i=0,g;String s="";for(;i<N.length;g=g(N[i],D[i]),N[i]/=g,D[i]/=g,s+=(i>0?"+":"")+N[i]+"/"+D[i],n=n*D[i]+N[i]*d,d*=D[i++],g=g(n,d),n/=g,d/=g);return s+"="+n+"/"+d;}

N and D are both int[], contextualized.

I reused Kevin Cruijssen's GCD function.

See it and test it online!

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2
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Julia v0.4+, 66 53 bytes

-13 bytes thanks to Dennis

a^b=replace(join(a.//b,"+")"=$(sum(a.//b))","//","/")

Try it Online!

Alternately, if fractions can be displayed using // rather than /, the following works for 35 bytes:

a^b=join(a.//b,'+')"=$(sum(a.//b))"
\$\endgroup\$
0
2
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setlX, 103 bytes

f:=procedure(a,b){i:=1;l:=[];while(i<=#a){l:=l+[a[i]/b[i]];i+=1;}s:=join(l,"+");return s+"="+eval(s);};

Creates a function called f where your insert two lists.

ungolfed:

f := procedure (a,b) {
    i:=1;
    l:=[];
    while(i<=#a){
        l:=l+[a[i]/b[i]];
        i+=1;
    }
    s:=join(l,"+");
    return s+"="+eval(s);
};

with named variables and annotations:
setlX doesn't provide a comment feature so let's just pretend that we can comment with %

f := procedure(firstList,secondList) {
    count := 1;
    list := []; 
    while(count <= #firstList) {
        % calculate every fraction and save it as a list
        list := list + [firstList[count]/secondList[count]];
        count += 1;
    }
    % Seperate every list entry with a plus ([2/3,1] -> "2/3+1")
    str := join(list, "+");
    % eval executes the string above and thus calculates our sum
    return str + "=" + eval(str);
};

\$\endgroup\$
3
  • \$\begingroup\$ What if #firstList is different from #secondList? \$\endgroup\$
    – user58988
    Commented May 21, 2017 at 19:05
  • \$\begingroup\$ you mean differnet size? The question states that first list is used by enumerator and wrong input can be ingored \$\endgroup\$
    – BlueWizard
    Commented May 21, 2017 at 20:04
  • \$\begingroup\$ but other than that: if the second list is longer, the remaining entries will be ignored. If the list is shorter a runtime error will occur. \$\endgroup\$
    – BlueWizard
    Commented May 21, 2017 at 20:05
1
\$\begingroup\$

05AB1E, 35 34 bytes

‚øvyy¿÷'/ý'+}¨'=¹.¿©¹÷²*O®‚D¿÷'/ýJ

Try it online!

\$\endgroup\$
1
\$\begingroup\$

Jelly, 31 bytes

,:gj”/
ṙJ$ṖP×Sç⁸P¤,ç@"j”+$¥Ṛj”=

Try it online!

\$\endgroup\$
1
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PHP>=7.1, 190 Bytes

<?function f($x,$y){for($t=1+$x;$y%--$t||$x%$t;);return$x/$t."/".$y/$t;}[$n,$d]=$_GET;for($p=array_product($d);$x=$n[+$k];$e+=$x*$p/$y)$r[]=f($x,$y=$d[+$k++]);echo join("+",$r)."=".f($e,$p);

Online Version

+14 Bytes for replacement return$x/$t."/".$y/$t; with return$y/$t>1?$x/$t."/".$y/$t:$x/$t; to output n instead of n/1

\$\endgroup\$
1
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F#, 244 241 239 bytes

let rec g a=function|0->abs a|b->g b (a%b)
let h a b=g a b|>fun x->a/x,b/x
let s,n,d=List.fold2(fun(s,n,d)N D->
 let(N,D),(n,d)=h N D,h(n*D+N*d)(d*D)
 s@[sprintf"%i/%i"N D],n,d)([],0,1)nom div
printf"%s=%i/%i"(System.String.Join("+",s))n d

Try it online!

\$\endgroup\$
1
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setlX, 62 bytes

[a,b]|->join([x/y:[x,y]in a><b],"+")+"="++/[x/y:[x,y]in a><b];

ungolfed:

[a,b]|->                  define a function with parameters a and b
  join(                 
    [ x/y :               using set comprehension, make a list of fractions 
      [x,y] in a><b       from the lists zipped together
    ],
    "+"
  )                       join them with "+"
  + "="                   concat with an equals sign
  +                       concat with
  +/[x/y:[x,y]in a><b]    the sum of the list
;

interpreter session

\$\endgroup\$
1
\$\begingroup\$

Haskell (Lambdabot), 94 91 86 bytes

t=tail.((\[n,_,d]->'+':n++'/':d).words.show=<<)
a#b|f<-zipWith(%)a b=t f++'=':t[sum f]

Try it online!

Thanks @Laikoni for -8 bytes!

Ungolfed

-- convert each fraction to a string (has format "n%d", replace '%' with '/' and prepend '+' ("+n/d"), keep the tail (dropping the leading '+')
t = tail.((\[n,_,d]->'+':n++'/':d).words.show=<<)
-- build a list of fractions
a # b = let f = zipWith(%) a b in
-- stringify that list, append "=" and the stringified sum of these fractions
  t f ++ "=" ++ t [sum f]
\$\endgroup\$
8
  • \$\begingroup\$ Your missing a import Data.Ratio for % which is not in Prelude. \$\endgroup\$
    – Laikoni
    Commented Jul 19, 2017 at 14:16
  • 1
    \$\begingroup\$ You can save some bytes by replacing "?"++ with '?':. \$\endgroup\$
    – Laikoni
    Commented Jul 19, 2017 at 14:21
  • 1
    \$\begingroup\$ The shortening also works for "/"++d and "="++. \$\endgroup\$
    – Laikoni
    Commented Jul 19, 2017 at 14:25
  • 1
    \$\begingroup\$ Rearranging saves some more bytes: tail(f>>=t)++'=':(tail.t.sum)f \$\endgroup\$
    – Laikoni
    Commented Jul 19, 2017 at 14:46
  • 1
    \$\begingroup\$ Putting tail and =<< into t saves some more: Try it online! \$\endgroup\$
    – Laikoni
    Commented Jul 19, 2017 at 14:55
1
\$\begingroup\$

Google Sheets, 83 81 bytes

=ArrayFormula(JOIN("+",TRIM(TEXT(A:A/B:B,"?/???")))&"="&TEXT(sum(A:A/B:B),"?/???"

Saved 2 bytes thanks to Taylor Scott

Sheets will automatically add 2 closing parentheses to the end of the formula.

The two arrays are input as the entirety of columns A and B. Empty rows below the inputs will throws errors.

\$\endgroup\$
1
  • \$\begingroup\$ you should be able to drop 2 bytes by dropping the terminal )) \$\endgroup\$ Commented Sep 5, 2017 at 16:17
0
\$\begingroup\$

Perl 6, 72 bytes 65 bytes

Native and automatic rationals should make this easy, but default stringification is still as decimal, so we have to .nude (numerator and denominator) which kills our score and makes the 1 ugly :(

my \n = 1,2,3,3,6; my \d = 2,9,3,2,4;
(n Z/d)».nude».join("/").join("+")~"="~([+] n Z/d).nude.join("/")

Update: Removed unneeded brackets, kill more space and use smarter map. Saves characters over Brad's solution at the cost of no not being a lambda sub.

\$\endgroup\$
1
  • \$\begingroup\$ Welcome to the site! Nice first answer! \$\endgroup\$
    – user58826
    Commented May 19, 2017 at 13:13
0
\$\begingroup\$

R, 109 bytes

f=MASS::fractions;a=attributes
g=function(n,d)paste(paste(a(f(n/d))$f,collapse='+'),a(f(sum(n/d)))$f,sep='=')

requires the MASS library (for its fractions class). the function g returns the required output as a string.

Try it online! (R-fiddle link)

\$\endgroup\$
0
\$\begingroup\$

TI-BASIC, 100 bytes

:L₁⁄L₂                                              //Creates a fraction from Lists 1 & 2, 7 bytes
:toString(Ans→Str1                                  //Create string from list, 7 bytes
:inString(Ans,",→A                                  //Look for commas, 9 bytes
:While A                                            //Begin while loop, 3 bytes
:Str1                                               //Store Str1 to Ans, 3 bytes
:sub(Ans,1,A-1)+"+"+sub(Ans,A+1,length(Ans)-A→Str1  //Replace "," with "+", 33 bytes
:inString(Ans,",→A                                  //Check for more commas, 9 bytes
:End                                                //End while loop, 2 bytes
:Str1                                               //Store Str1 to Ans, 3 bytes
:sub(Ans,2,length(Ans)-2                            //Remove opening and closing brackets, 13 bytes
:Ans+"="+toString(expr(Ans                          //Add "=" and answer, 11 bytes

Note the at the beginning, different from /. This makes the fractions hold their forms. It does work with negative fractions.

Sigh. TI-BASIC is horrible with strings. If all we had to do was print the fractions, and then their sum, the code would be:

TI-BASIC, 12 bytes

:L₁⁄L₂    //Create fractions from lists, 7 bytes
:Disp Ans //Display the above fractions, 3 bytes
:sum(Ans  //Display the answer, 2 bytes

That means that 88 bytes of my code is spent just formatting the answer! Hmph.

\$\endgroup\$
0
\$\begingroup\$

C, 171 bytes

Try Online

i;t;x;y;f(int s,int*a,int*b){x=*a;y=*b;while(++i<s)x=(y-b[i])?(x*b[i])+(a[i]*y):(x+a[i]),y=(y-b[i])?(b[i]*y):y;for(i=1;i<=(x<y?x:y);++i)t=(x%i==0&&y%i==00)?i:t;x/=t;y/=t;}
\$\endgroup\$
0
\$\begingroup\$

Axiom, 212 bytes

C==>concat;S==>String;g(a)==C[numerator(a)::S,"/",denom(a)::S];h(a:List PI,b:List PI):S==(r:="";s:=0/1;#a~=#b or #a=0=>"-1";for i in 1..#a repeat(v:=a.i/b.i;s:=s+v;r:=C[r,g(v),if i=#a then C("=",g(s))else"+"]);r)

test

(5) -> h([1,3,4,4,5,6], [2,9,5,5,6,7])
   (5)  "1/2+1/3+4/5+4/5+5/6+6/7=433/105"
                                                             Type: String
(6) -> h([1,3,4,4], [2,9,5,5,6,7])
   (6)  "-1"
                                                             Type: String
\$\endgroup\$
0
\$\begingroup\$

Casio Basic, 161 Bytes

Dim(List 1)->A
for 1->I to A step 1
3*I-2->B
List 1[I]->C
List 2[I]->D
locate 1,B,C
locate 1,B+1,"/"
locate 1,B+2,D
C/D+E->E
next
locate 1,B+3,"="
locate 1,B+4,E

Explanation:

  • Number of input is saved in A
  • A iterations
  • B acts as a counter for correct displaying
  • I'th item of List 1 and 2 saved in C and D
  • Displaying of Variable C / Variable D
  • save C/D+E in E
  • After last number locate = and E
\$\endgroup\$
0
\$\begingroup\$

Vyxal , 12 bytes

/₌≬vS\+j∑\=$

Try it Online!

10 bytes of this is just getting the formatting right... /:

Explained

/₌≬vS\+j∑\=$
/           # divide the lists, vectorising
 ₌≬vS\+j     # join each fraction on "+" (there's a bug where they get converted to decimal without vS :/)
       ∑    # get the sum of the fractions. 
        \=$ # place a "=" between the two
            # the Ṫ flag prints the entire stack without spaces. If it wasn't for the aforementioned joining bug, the last $ could be a j and the flag wouldn't be needed. 
\$\endgroup\$
0
\$\begingroup\$

MATL, 32 bytes

/YQv'%i/%i+'wYD3L)61yUYQVwV47b&h

Try at MATL online!

Explanation

Consider [1, 2, 3, 3, 6], [2, 9, 3, 2, 4] as input.

/         % Implicit inout. Divide element-wise
          % STACK: [0.5 0.222 1 1.5 1.5]
YQ        % Rational approximation (with default tolerance)
          % STACK: [1 2 1 3 3], [2 9 1 2 2]
v         % Conctenate all stack elements vertically
          % STACK: [1 2; 2 9; 1 2; 3 2; 3 2]
'%i/%i+'  % Push this string (sprintf format specifier)
          % STACK: [1 2; 2 9; 1 2; 3 2; 3 2], '%i/%i+'
wYD       % Swap, sprintf
          % STACK: '1/2+2/9+1/1+3/2+3/2+'
3L)       % Remove last entry
          % STACK: '1/2+2/9+1/1+3/2+3/2'
61        % Push 61 (ASCII for '=')
          % STACK: '1/2+2/9+1/1+3/2+3/2', 61
y         % Duplicate from below
          % STACK: '1/2+2/9+1/1+3/2+3/2', 61, '1/2+2/9+1/1+3/2+3/2'
U         % Evaluste string into a number
          % STACK: '1/2+2/9+1/1+3/2+3/2', 61, 4.722
YQ        % Rational approximation 
          % STACK: '1/2+2/9+1/1+3/2+3/2', 61, 85, 18
VwV       % Convert to string, swap, convert to string
          % STACK: '1/2+2/9+1/1+3/2+3/2', 61, '18', '85'
47        % Push 47 (ASCII for '/')
          % STACK: '1/2+2/9+1/1+3/2+3/2', 61, '18', '85', 47
b         % Bubble up in stack
          % STACK: '1/2+2/9+1/1+3/2+3/2', 61, '85', 47, '18'
&h        % Concatenate all stack elements horizontally. Implicitly display
          % STACK: '1/2+2/9+1/1+3/2+3/2=85/18'
\$\endgroup\$

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