package Maths;

/**
 * Amicable numbers are two different numbers so related
 * that the sum of the proper divisors of each is equal to the other number.
 * (A proper divisor of a number is a positive factor of that number other than the number itself.
 * For example, the proper divisors of 6 are 1, 2, and 3.)
 * A pair of amicable numbers constitutes an aliquot sequence of period 2.
 * It is unknown if there are infinitely many pairs of amicable numbers.
 * *
 * <p>
 * * link: https://en.wikipedia.org/wiki/Amicable_numbers
 * * </p>
 * <p>
 * Simple Example : (220,284) 220 is divisible by {1,2,4,5,10,11,20,22,44,55,110 } <- Sum = 284
 * 284 is divisible by -> 1,2,4,71,142 and the Sum of that is. Yes right you probably expected it 220
 */

public class AmicableNumber {

 public static void main(String[] args) {

 AmicableNumber.findAllInRange(1,3000);
 /* Res -> Int Range of 1 till 3000there are 3Amicable_numbers These are 1: = ( 220,284)	2: = ( 1184,1210)
 3: = ( 2620,2924) So it worked	*/

 }

 /**
 * @param startValue
 * @param stopValue
 * @return
 */
 static void findAllInRange(int startValue, int stopValue) {

 /* the 2 for loops are to avoid to double check tuple. For example (200,100) and (100,200) is the same calculation
 * also to avoid is to check the number with it self. a number with itself is always a AmicableNumber
 * */
 StringBuilder res = new StringBuilder();
 int countofRes = 0;

 for (int i = startValue; i < stopValue; i++) {
 for (int j = i + 1; j <= stopValue; j++) {
 if (isAmicableNumber(i, j)) {
 countofRes++;
 res.append("" + countofRes + ": = ( " + i + "," + j + ")" + "\t");
 }
 }
 }
 res.insert(0, "Int Range of " + startValue + " till " + stopValue + " there are " + countofRes + " Amicable_numbers.These are \n ");
 System.out.println(res.toString());
 }

 /**
 * Check if {@code numberOne and numberTwo } are AmicableNumbers or not
 *
 * @param numberOne numberTwo
 * @return {@code true} if {@code numberOne numberTwo} isAmicableNumbers otherwise false
 */
 static boolean isAmicableNumber(int numberOne, int numberTwo) {

 return ((recursiveCalcOfDividerSum(numberOne, numberOne) == numberTwo && numberOne == recursiveCalcOfDividerSum(numberTwo, numberTwo)));
 }

 /**
 * calculated in recursive calls the Sum of all the Dividers beside it self
 *
 * @param number div = the next to test dividely by using the modulo operator
 * @return sum of all the dividers
 */
 static int recursiveCalcOfDividerSum(int number, int div) {

 if (div == 1) {
 return 0;
 } else if (number % --div == 0) {
 return recursiveCalcOfDividerSum(number, div) + div;
 } else {
 return recursiveCalcOfDividerSum(number, div);
 }
 }






}