88 lines
2.9 KiB
Java
88 lines
2.9 KiB
Java
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package Maths;
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/**
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* Amicable numbers are two different numbers so related
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* that the sum of the proper divisors of each is equal to the other number.
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* (A proper divisor of a number is a positive factor of that number other than the number itself.
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* For example, the proper divisors of 6 are 1, 2, and 3.)
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* A pair of amicable numbers constitutes an aliquot sequence of period 2.
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* It is unknown if there are infinitely many pairs of amicable numbers.
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* *
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* <p>
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* * link: https://en.wikipedia.org/wiki/Amicable_numbers
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* * </p>
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* <p>
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* Simple Example : (220,284) 220 is divisible by {1,2,4,5,10,11,20,22,44,55,110 } <- Sum = 284
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* 284 is divisible by -> 1,2,4,71,142 and the Sum of that is. Yes right you probably expected it 220
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*/
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public class AmicableNumber {
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public static void main(String[] args) {
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AmicableNumber.findAllInRange(1,3000);
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/* Res -> Int Range of 1 till 3000there are 3Amicable_numbers These are 1: = ( 220,284) 2: = ( 1184,1210)
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3: = ( 2620,2924) So it worked */
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}
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/**
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* @param startValue
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* @param stopValue
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* @return
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*/
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static void findAllInRange(int startValue, int stopValue) {
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/* the 2 for loops are to avoid to double check tuple. For example (200,100) and (100,200) is the same calculation
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* also to avoid is to check the number with it self. a number with itself is always a AmicableNumber
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* */
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StringBuilder res = new StringBuilder();
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int countofRes = 0;
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for (int i = startValue; i < stopValue; i++) {
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for (int j = i + 1; j <= stopValue; j++) {
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if (isAmicableNumber(i, j)) {
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countofRes++;
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res.append("" + countofRes + ": = ( " + i + "," + j + ")" + "\t");
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}
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}
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}
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res.insert(0, "Int Range of " + startValue + " till " + stopValue + " there are " + countofRes + " Amicable_numbers.These are \n ");
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System.out.println(res.toString());
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}
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/**
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* Check if {@code numberOne and numberTwo } are AmicableNumbers or not
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*
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* @param numberOne numberTwo
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* @return {@code true} if {@code numberOne numberTwo} isAmicableNumbers otherwise false
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*/
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static boolean isAmicableNumber(int numberOne, int numberTwo) {
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return ((recursiveCalcOfDividerSum(numberOne, numberOne) == numberTwo && numberOne == recursiveCalcOfDividerSum(numberTwo, numberTwo)));
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}
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/**
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* calculated in recursive calls the Sum of all the Dividers beside it self
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*
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* @param number div = the next to test dividely by using the modulo operator
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* @return sum of all the dividers
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*/
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static int recursiveCalcOfDividerSum(int number, int div) {
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if (div == 1) {
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return 0;
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} else if (number % --div == 0) {
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return recursiveCalcOfDividerSum(number, div) + div;
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} else {
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return recursiveCalcOfDividerSum(number, div);
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}
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}
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}
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