${\displaystyle {\text{of }}\,\{30,\ 24\,\},\,}$

${\displaystyle 2\,|\,30\,{\text{ and }}\,2\,|\,24.\,}$

${\displaystyle {\text{of }}\,\{30,\ 24\,\}:\,}$

${\displaystyle \,{\text{GCD}}(30,\ 24)=6.\,}$

${\displaystyle {\text{And GCD}}(0,\ 0)=0\ }$

${\displaystyle {\text{is }}\,\leqslant .\,}$

${\displaystyle {\text{Denoting by }}\,C\,}$ the set of common divisors ${\displaystyle \,{\text{of }}\,\{30,\ 24\,\},\,}$ ${\displaystyle \,30\,{\text{ and }}\,24\,}$ are multiples of any element ${\displaystyle d\,{\text{of }}\,C.\,}$ ${\displaystyle {\text{Therefore }}\,30-24=6\,}$ is also a multiple ${\displaystyle {\text{of }}\,d.\,}$ Conversely, every common divisor ${\displaystyle \,{\text{of }}\,\{24,\ 6\,\}\,}$ is also a divisor ${\displaystyle \,{\text{of }}\,24\,{\text{ and }}\,6+24=30.\,}$ In other words, ${\displaystyle \,C\,{\text{ is}}\,}$ also the set of common divisors ${\displaystyle {\text{of }}\,\{24,\ 6\,\}.}$

Under the algorithm of the image, ${\displaystyle \{30,\ 24\,\}\,}$ is replaced ${\displaystyle {\text{with }}\,\{24,\ 6\,\},\,}$ and then by repeating the process replaced ${\displaystyle {\text{with }}\,\{18,\ 6\,\},\,}$ ${\displaystyle {\text{because }}\,24-6=18.\,}$ And then replaced ${\displaystyle \,{\text{with }}\,\{12,\ 6\,\},\,}$ ${\displaystyle \,{\text{and with }}\,\{6,\ 6\,\}.\,}$ Finally the step‑by‑step process replaces the initial pair ${\displaystyle {\text{with }}\,\{6,\ 0\,\},\,}$ goes out of the loop of subtractions, and affirms: ${\displaystyle {\text{6 is the GCD of }}\,30\,{\text{ and }}\,24.}$

Instead of replacing ${\displaystyle \{24,\ 6\,\}\,}$ with the four successive pairs: ${\displaystyle \{18,\ 6\,\},\ \{12,\ 6\,\},\ \{6,\ 6\,\},\ \{0,\ 6\,\},\,}$ we could get directly the last pair by the Euclidean division ${\displaystyle {\text{of }}\,24\,{\text{ by }}\,6.\,}$ It does not matter there are 4 successive subtractions of 6, we can ignore 4: the quotient value of this Euclidean division. Actually, every sequence of subtractions of a same number can be replaced with an Euclidean division by this number. Thus we discover the more known algorithm of GCD through successive Euclidean divisions, by improving the present algorithm of subtractions.

A novice in coding can copy and paste in a window dedicated to JavaScript one of the following comparisons, and then command the execution: 30 < 24; or 6 == 0; or 6 < 24. Within the image, six symbols look like assignments in OCaml. Here is an example of assignment in JavaScript: r = 30; commanding to store into ${\displaystyle {\text{variable }}\,r\,}$ the object of type Number and value 30. Everyone can download Firefox, and then open a Firefox window dedicated to JavaScript code. In this special window, we can copy and paste the following translation from the flow chart into JavaScript.

/* To open a Firefox window
 dedicated to JavaScript code:  Shift + F4  */

d = r = k = 182; p = 238;  // example of input values,
// that we can replace with two other natural numbers
if( s = p){  // if the common value of s and p is not zero
while(r){  // while the value of r is not zero
if(r < s){  // in this case, reverse the values of r and s
d = s; s = r; r = d }
r = r-s }  // end of the loop 'while(r)'
d = s }  // end of the block that begins with 'if( s = p)'
"   GCD("+ k +", "+ p +") = "+ d;  // output: a String object

// Keyboard shortcut in Firefox to execute the code:  Ctrl + L

On the image top,  ${\displaystyle r\in \mathbb {N} \ }$ means that  ${\displaystyle r\,{\text{ is a member}}}$ ${\displaystyle {\text{of set }}\mathbb {N} }$ of natural numbers. The previous JavaScript code works only if the two input values are natural numbers. Here is an improvement of the previous code, where the input values assigned ${\displaystyle {\text{to }}\,k\,{\text{ and }}\,p\,}$ are verified.

try{ //  in case of error in this block,
//  execution failure of this code block, go to 'catch'
d = r = k = 408; p = 255;  // example of input values
var b;  // global scope declaration
s = function(n){
// to test the value of parameter n: is it a natural number?
b = n.constructor == Number; // Boolean value
if( !b // first incorrect case
  || n < 0 || n != Math.floor(n)  // other incorrect cases
) throw n
//  in one of the previous cases, n is thrown as error
};  // end of assignment to variable s
s(k); s(p);  // verifications
if( s = p){  // if the common value of s and p is not zero
while(r){
if(r < s){d = s; s = r; r = d} r = r-s } d = s }
"  GCD("+ k +", "+ p +") = "+ d
}catch(e){  // in case of error (if e is thrown)
"  "+( b ? e +"  is not a natural number.":
"  Incorrect code.")
}

${\displaystyle {\text{For example, GCD}}(408,\ 255)=51,\,}$

${\displaystyle {\frac {\,408\,}{51}}=8\ {\text{ and }}\ {\frac {\,255\,}{51}}=5,\,}$

${\displaystyle {\text{so }}\,{\frac {\,8\,}{5}}\,}$

${\displaystyle {\text{of }}\,{\frac {\,408\,}{255}}.}$