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GCD(int1, int2[, showWork])

Last updated November 8, 2001

author

Shakti Shrivastava

Version: 1 | Requires: CF5 | Library: MathLib

Description:
Calculates the GCD (greatest common factor [divisor]) of two positive integers using the Euclidean Algorithm. Optionally displays all of the steps in the calculation.

Return Values:
Returns a numeric value. Optionally outputs a string.

Example:

<CFSET x=13>
<CFSET y=130>

<cfoutput>
Given:<BR>
x=#x#<BR>
y=#y#<BR>
G.C.D. = #GCD(x,y,"yes")#
</cfoutput>

Parameters:

Name Description Required
int1 Positive integer. Yes
int2 Positive integer. Yes
showWork Boolean. Yes or No. Specifies whether to display work. Default is No. No

Full UDF Source:

/**
 * Calculates the GCD (greatest common factor [divisor]) of two positive integers using the Euclidean Algorithm.
 * 
 * @param int1 	 Positive integer. 
 * @param int2 	 Positive integer. 
 * @param showWork 	 Boolean.  Yes or No.  Specifies whether to display work.  Default is No. 
 * @return Returns a numeric value.  Optionally outputs a string. 
 * @author Shakti Shrivastava (shakti@colony1.net) 
 * @version 1, November 8, 2001 
 */
function GCD(int1, int2)
{
  Var ShowWork=False;
  If (ArrayLen(Arguments) eq 3 AND Arguments[3] eq "yes"){
    ShowWork=True;
  }
   
  // if both numbers are 0 return undefined
  if (int1 eq 0 and int2 eq 0) return 'Undefined';

  // if both numbers are equal or either one of them is equal to 0 
  // then return any 1 of those numbers appropriately
  if (int1 eq int2 or int2 eq 0) return int1;
  if (int1 eq 0) return int2;

  // if int2 divides int1 "properly" then we have reached our final 
  //   step. So we output the last step and exit the function.
  if (int1 mod int2 eq 0) {
    if(ShowWork EQ True) {
      WriteOutput("<br>"&int1&"= "&fix(int1/int2)&" X "&int2&" + "&int1 mod int2); 
    }
    // this line outputs the last iteration
    if (ShowWork EQ True) {
      return "<p>GCD = "&int2; //print the GCD
    }
    else{
      return int2;
    }
  }

  // if int2 does not divides int1 "properly" then we recurse using
  // int1 equal to int2 and int2 equal to int1 mod int2
  else {
    if(ShowWork EQ True)
      // this line outputs calculations from each iteration. you can safely
      // delete/comment out this line if you dont need to display the steps.
      WriteOutput("<br>"&int1&"= "&fix(int1/int2)&" X "&int2&" + "&int1 mod int2); 
  }
  //since we still have not reached the last step we recall the function 		
  //(recurse)
  GCD(int2, int1 mod int2, ShowWork);
}
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