The Euclidean Algorithm (for finding the greatest common divisor)

Question

The Challenge

Write a program or function that takes two input integers, i and j, and outputs their greatest common divisor; calculated by using the Euclidean algorithm (see below).

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Input

Input may be taken as a space-delimited string representation of i and j or as two separate integers. You can assume that integers will be less than or equal to 10,000. You can also assume that the input integers will not be prime to one another.

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Euclidean Breakdown

The larger number between i and j is divided by the smaller as many times as possible. Then, the remainder is added. This process is repeated with the remainder and the previous number, until the remainder becomes 0.

For example, if the input was 1599 650:

1599 = (650 * 2) + 299
 650 = (299 * 2) +  52
 299 =  (52 * 5) +  39
  52 =  (39 * 1) +  13
  39 =  (13 * 3) +   0

The final number, 13, is the greatest common divisor of the two input integers. It can be visualized like this:

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Output

Your output must be the breakdown in the form above, followed by a newline and the GCD. It can be output through any medium.

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Examples

Inputs

18 27
50 20
447 501
9894 2628

Outputs

27 = (18 * 1) + 9
18 =  (9 * 2) + 0
9

50 = (20 * 2) + 10
20 = (10 * 2) +  0
10

501 = (447 * 1) + 54
447 =  (54 * 8) + 15
 54 =  (15 * 3) +  9
 15 =   (9 * 1) +  6
  9 =   (6 * 1) +  3
  6 =   (3 * 2) +  0
3

9894 = (2628 *  3) + 2010
2628 = (2010 *  1) +  618
2010 =  (618 *  3) +  156
 618 =  (156 *  3) +  150
 156 =  (150 *  1) +    6
 150 =    (6 * 25) +    0
6

Note: Outputs do not have to be spaced as they are above. The spacing is only for clarity. Parentheses are required.

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Bonus

If your output is spaced as above, you may add a -10% bonus to your score.

Answer 1

Pyth, 33 bytes

ASQWGs[H\=\(G\*/HG\)\+K%HG)A,KG)H

Try it online: Demonstration or Test Suite

Explanation:

ASQWGs[H\=\(G\*/HG\)\+K%HG)A,KG)H
  Q                                read the two numbers from input
 S                                 sort them
A                                  and assign them to G and H
   WG                              while G != 0:
                      K%HG           assign H mod G to K
     s[H\=\(G\*/HG\)\+K   )          join the following list items and print:
                                        H=(G*(H/G))+K
                           A,KG      assign K, G to G, H
                               )   end while
                                H  print H

Answer 2

CJam, 46 43 39 bytes

q~]$3*~\{N5$"=("3$:G'*3$Gmd")+"\}h]7>NG

Try it online in the CJam interpreter.

How it works

q~]    e# Read all input, evaluate it and wrap the results in an array.
$3*    e# Sort the array and repeat it thrice.
~\     e# Dump the array and swap its last two elements.
{      e# Do:
  N    e#   Push a linefeed.
  5$   e#   Copy the sixth topmost element from the stack.
  "=(" e#   Push that string.
  3$:G e#   Copy the fourth topmost element from the stack. Save it in G.
  '*   e#   Push that character.
  3$   e#   Copy the fourth topmost element from the stack.
  Gmd  e#   Push quotient and remainder of its division by G.
  ")+" e#   Push that string.
  \    e#   Swap the string with the remainder.
}h     e#   If the remainder is positive, repeat the loop.
]7>    e# Wrap the stack in an array and discard its first seven elements.
NG     e# Push a linefeed and G.

Answer 3

Python 2, 70

f=lambda a,b:b and'%d=(%d*%d)+%d\n'%(a,b,a/b,a%b)*(a>=b)+f(b,a%b)or`a`

A recursive function that returns a multiline string. The function creates the first line, then appends it to the recursive result with the next pair of numbers in the Euclidean algorithm. Once the second number is zero, we take the string of the other number as the base case, causing it to be printed on its own line at the end.

The formatting is done via string substitution, using integer division to get the multiplicand.

One hiccup is needing to start with the larger number being taken mod the smaller number. Conveniently, if the numbers are in the wrong order, the first step of the Euclidian algorithm flips them. To prevent this step from being displayed, only add the current line if the first number is at least the second (equality is needed for, say f(9,9)).

Answer 4

awk, 78 77

x=$1{for(x<$2?x+=$2-(y=x):y=$2;t=y;x=t)print x"=("y"*"int(x/y)")+",y=x%y}$0=x

Sorting the input by size takes a lot of bytes :/
It comes down to this:

x=$1;
if(x<$2) x+=$2-(y=x); # add $2 subtract $1 and set y to $1
else y=$2;            # set y to $2

Output

650 1599            (input)
1599=(650*2)+ 299
650=(299*2)+ 52
299=(52*5)+ 39
52=(39*1)+ 13
39=(13*3)+ 0
13

Just for the fun of it I made a properly spaced version too, giving me a score of 233 * 0.9 == 209.7 bytes.

Update I was able to shorten this from 285 bytes and now it works for arbitrarily long numbers if calling gawk4 with the -M option.

x=$1{x<$2?x+=$2-(y=x):y=$2;a=length(x);b=length(y);for(d=length(x%y);t=y;x=t){$++i=x;$++i=y;if(c<l=length($++i=int(x/y)))c=l;$++i=y=x%y}while(j<NF)printf "%"a"d = %"b-length($(j+2))"s(%d * %"c"d) + %"d"d\n",$++j,_,$++j,$++j,$++j}$0=x

But I still got a feeling I ran into some mental block there somewhere...

Output (gawk4 called with awk -M -f code.awk)

6837125332653632513763 18237983363879361  (input)
6837125332653632513763 = (18237983363879361 * 374883) + 15415252446024000
     18237983363879361 = (15415252446024000 *      1) +  2822730917855361
     15415252446024000 =  (2822730917855361 *      5) +  1301597856747195
      2822730917855361 =  (1301597856747195 *      2) +   219535204360971
      1301597856747195 =   (219535204360971 *      5) +   203921834942340
       219535204360971 =   (203921834942340 *      1) +    15613369418631
       203921834942340 =    (15613369418631 *     13) +      948032500137
        15613369418631 =      (948032500137 *     16) +      444849416439
          948032500137 =      (444849416439 *      2) +       58333667259
          444849416439 =       (58333667259 *      7) +       36513745626
           58333667259 =       (36513745626 *      1) +       21819921633
           36513745626 =       (21819921633 *      1) +       14693823993
           21819921633 =       (14693823993 *      1) +        7126097640
           14693823993 =        (7126097640 *      2) +         441628713
            7126097640 =         (441628713 *     16) +          60038232
             441628713 =          (60038232 *      7) +          21361089
              60038232 =          (21361089 *      2) +          17316054
              21361089 =          (17316054 *      1) +           4045035
              17316054 =           (4045035 *      4) +           1135914
               4045035 =           (1135914 *      3) +            637293
               1135914 =            (637293 *      1) +            498621
                637293 =            (498621 *      1) +            138672
                498621 =            (138672 *      3) +             82605
                138672 =             (82605 *      1) +             56067
                 82605 =             (56067 *      1) +             26538
                 56067 =             (26538 *      2) +              2991
                 26538 =              (2991 *      8) +              2610
                  2991 =              (2610 *      1) +               381
                  2610 =               (381 *      6) +               324
                   381 =               (324 *      1) +                57
                   324 =                (57 *      5) +                39
                    57 =                (39 *      1) +                18
                    39 =                (18 *      2) +                 3
                    18 =                 (3 *      6) +                 0
3

Some newlines and tabs added

x=$1{
    x<$2?x+=$2-(y=x):y=$2;
    a=length(x);
    b=length(y);
    for(d=length(x%y);t=y;x=t)
    {
        $++i=x;
        $++i=y;
        if(c<l=length($++i=int(x/y)))c=l;
        $++i=y=x%y
    }
    while(j<NF)
        printf "%"a"d = %"b-length($(j+2))"s(%d * %"c"d) + %"d"d\n",
                                               $++j,_,$++j,$++j,$++j
}$0=x

I can save the lengths of the initial values for x, y and x%y in the beginning, because they can only get shorter each step. But for the factor I have to determine the maximum length before printing anything, because it's length can vary. I use $i as an array here, because it saves two characters compared to using a[i] every time.

Answer 5

C++, GCC compiler, 171 bytes( -10%, so 154 bytes)

okay so my first try..

#include<iostream>
using namespace std;
int main()
{
    int a,b,c;
    cin>>a>>b;
    if(a<b)
    swap(a,b);
    while(b>0)
    {
        c=a;
        cout<<a<<" = ("<<b<<" * "<<a/b<<") + "<<a%b<<endl;
        a=b;
        b=c%b;
    }
    cout<<a;
}

tips to code golf appreciated.

P.S. Is it necessary to count bytes of standard header files and int main while using c++? Excluding it reduces 50 bytes

Answer 6

T-SQL (2012+), 268 bytes

Implemented as an inline table function that executes a recursive CTE. Might be worthwhile trying to put formatting in for the 10% bonus, but that will have to wait.

CREATE FUNCTION E(@ INT,@B INT)RETURNS TABLE RETURN WITH M AS(SELECT IIF(@<@B,@B,@)A,IIF(@>@B,@B,@)B),R AS(SELECT A,B,A/B D,A%B R FROM M UNION ALL SELECT B,R,B/R,B%R FROM R WHERE 0<>R)SELECT CONCAT(A,'=(',B,'*',D,')+',R)R FROM R UNION ALL SELECT STR(B)FROM R WHERE R=0

Explanation and usage:

--Create the function
CREATE FUNCTION E(@ INT,@B INT)RETURNS TABLE RETURN
WITH
    --Order the input correctly
    M AS (
          SELECT IIF(@<@B,@B,@)A,
                 IIF(@>@B,@B,@)B
          )
    --Recursive selection
    ,R AS (
          SELECT A,B,A/B D,A%B R FROM M -- Anchor query
          UNION ALL
          SELECT B,R,B/R,B%R FROM R     -- Recurse until R = 0
          WHERE 0<>R
          )
SELECT CONCAT(A,'=(',B,'*',D,')+',R)R   -- Concat results into output string
FROM R 
UNION ALL                               -- ALL required to maintain order
SELECT STR(B)                           -- Add final number
FROM R WHERE R=0

--Function Usage
SELECT * FROM E(447,501)

R
-----------------------------------------------------
501=(447*1)+54
447=(54*8)+15
54=(15*3)+9
15=(9*1)+6
9=(6*1)+3
6=(3*2)+0
3

Answer 7

PHP, 118 Bytes

for(list(,$n,$m)=$argv,$g=max($n,$m),$l=min($n,$m);$g;$g=$l,$l=$m)
echo$g,$l?"=($l*".($g/$l^0).")+".($m=$g%$l)."
":"";

Try it online!

PHP, 131 Bytes

for(list(,$n,$m)=$argv,$r=[max($n,$m),min($n,$m)];$r[+$i];)echo$g=$r[+$i],($l=$r[++$i])?"=($l*".($g/$l^0).")+".($r[]=$g%$l)."
":"";

Try it online!

-4 Bytes replace list(,$n,$m)=$argv with [,$n,$m]=$argv needs PHP>=7.1

Answer 8

JavaScript (ES6), 74 50 62 61 55 bytes

f=(x,y)=>y?y>x?y:x+`=(${y}*${x/y|0})+${x%=y}
`+f(y,x):x

• Sacrificed 12 bytes allowing the integers to be passed in any order, rather than largest first.

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Try It

f=(x,y)=>y?y>x?y:x+`=(${y}*${x/y|0})+${x%=y}
`+f(y,x):x
o.innerText=f(i.value=683712533265363251376,j.value=18237983363879361)
i.oninput=j.oninput=_=>o.innerText=f(+i.value,+j.value)

<input id=i type=number><input id=j type=number><pre id=o>

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Explanation

f=          :Assign the function to variable f ...
(x,y)=>     :And take the two integer inputs as arguments via parameters x and y.
y?          :If y is greater than 0 then
y>x?        :    If y is greater than x then
f(y,x)      :        Call f again, with the order of the integers reversed.
            :        (This can only happen the first time the function is called.)
:           :    Else
x           :        Start building the string, beginning with the value of x.
+`=(        :        Append "=(".
${y}        :          The value of y.
*           :          "*"
${x/y|0}    :          The floored value of x divided by y
)+          :          ")+"
${x%=y}     :          The remainder of x divided by y, which is assigned to x
            :          (x%=y is the same as x=x%y.)
\n          :          A newline (a literal newline is used in the solution).
`+f(y,x)    :        Append the value f returns when y and the new value of x
            :        are passed as arguments.
:           :Else
x           :    Return the current value of x,
            :    which will be the greatest common divisor of the original two integers.

Answer 9

Japt, 43 42 41 bytes

My newfound Japt "skills" seem to be getting worse, not better!

?U>V?ßVU :[V'='(U'*V/U|0')'+V%=UR,ßVU]¬:V

Try it online

Answer 10

Rev 1: Ruby, 86

Recursive algorithm, thanks to tip from Doorknob.

f=->i,j{j>i&&(i,j=j,i)
0<j ?(print i," = (#{j} * #{i/j}) + #{i%j}
";f[j,i%j]):puts(i)}

Rev 0: Ruby, 93

This really hasn't worked out well at all. The while loop takes up too many characters, especially with the end. I can't see a way of avoiding it. Recursion would require a named function instead of a lambda, which would also eat up a lot of characters.

->i,j{j>i&&(i,j=j,i)
while j>0
print(i," = (#{j} * #{i/j}) + #{i%j}\n")
i,j=j,i%j
end
puts i}

Call it like this:

f=->i,j{j>i&&(i,j=j,i)
while j>0
print(i," = (#{j} * #{i/j}) + #{i%j}\n")
i,j=j,i%j
end
puts i}

I=gets.to_i
J=gets.to_i

f.call(I,J)

Answer 11

PowerShell, 84

A recursive code block, stored in a variable. Invoke it with & $e num1 num2, e.g.:

$e={$s,$b=$args|Sort;if(!$s){$b}else{$r=$b%$s;"$b=($s*$(($b-$r)/$s))+$r";&$e $s $r}}

PS D:\> & $e 9894 2628
9894=(2628*3)+2010
2628=(2010*1)+618
2010=(618*3)+156
618=(156*3)+150
156=(150*1)+6
150=(6*25)+0
6

In a more readable form, it does the following (nb. for clearer code, I've put the full commandlet names, more spaces in the string, and made the pipeline output commands explicit):

function Euclid {
    $small, $big = $args|Sort-Object   #Sort argument list, assign to two vars.

    if (!$small) {                     #Recursion end, emit the last
        Write-Output $big              #number alone, for the last line.

    } else {                           #main recursive code

        $remainder = $big % $small
        Write-Output "$big = ( $small* $(($big-$remainder)/$small)) + $remainder"
        Euclid $small $remainder
    }
}

One annoyance from a codegolf point of view; PoSh has no integer division, 10/3 returns a Double, but cast the result to an integer and it doesn't always round down, it rounds N.5 to the nearest even number - up or down. So [int](99/2) == 50.

That leaves two awkward choices:

$remainder = $x % $y
$quotient = [Math]::Floor($x/$y)

# or, worse

$remainder=$null
$quotient = [Math]::DivRem($x, $y, [ref]$remainder)

Which is why I have to burn some characters doing:

$remainder = $big % $small
($big - $remainder)/$small

Apart from that, it's the number of $$$$ and the lack of ternary operator which really hurts.

---

I also have an iterative version which, rather nicely, is also 84 characters:

{$r=1;while($r){$s,$b=$args|Sort;$r=$b%$s;"$b=($s*$(($b-$r)/$s))+$r";$args=$s,$r}$s}

Completely anonymous codeblock, so run it with

& {*codeblock*} 1599 650

Answer 12

C, 83 bytes

g(x,y,z){y&&(printf("%u=(%u*%u)+%u\n",x,y,x/y,z=x%y),z)?g(y,z,0):printf("%u\n",y);}

test and results

int main()
{g(18,27,0);
 g(50,20,0);
 g(447,501,0);
 g(9894,2628,0);
}

18=(27*0)+18
27=(18*1)+9
18=(9*2)+0
9
50=(20*2)+10
20=(10*2)+0
10
447=(501*0)+447
501=(447*1)+54
447=(54*8)+15
54=(15*3)+9
15=(9*1)+6
9=(6*1)+3
6=(3*2)+0
3
9894=(2628*3)+2010
2628=(2010*1)+618
2010=(618*3)+156
618=(156*3)+150
156=(150*1)+6
150=(6*25)+0
6

Answer 13

JS, 151

a=prompt("g","");b=prompt("l","");c=0;l=[a,b];for(var i=0;i<=1;i++){t=c;o=c+1;r=c+2;n=l[t]%l[o];if(n!==0){l[r]=n;c=c+1;i=0;}else{p=l[o];alert(p);i=2;}}

Answer 14

Java 133

public void z(int i,int j){for(int d=1;d!=0;i=j,j=d){d=i%j;System.out.println(i+"=("+j+"*"+((i-d)/j)+")+"+d);}System.out.println(i);}

It doesn't do the regular Euclidean algorithm. Instead, it uses modulus and then computes the 2nd number to multiply by when it's printed out. You can also make this shorter by turning it into a lambda expression, but I'm not sure how.

public void z(int i, int j)
{
    for(int d=1;d!=0;i=j,j=d)
    {
        d=i%j;
        System.out.println(i+"=("+j+"*"+((i-d)/j)+")+"+d);
    }
    System.out.println(i);
}