May 12, 2023
diff --git a/Project-Euler/Problem044.js b/Project-Euler/Problem044.js
 * It can be seen that P4 + P7 = 22 + 70 = 92 = P8. However, their difference, 70 − 22 = 48, is not pentagonal.
 * Find the pair of pentagonal numbers, Pj and Pk, for which their sum and difference are pentagonal and D = |Pk − Pj| is minimised; what is the value of D?
 *
 * @authorddaniel27
 * @authorutkarsh-shrivastav77
 */

 function problem44 (k) {
  if (k < 1) {
 function getPentagonalNumber(n) {
  return n * (3 * n - 1) / 2;
 }

 // The function takes a limit parameter that determines the maximum pentagonal number to consider. It first defines a getPentagonalNumber function to calculate the nth pentagonal number using the given formula.

 function findMinPentagonalDifference(limit) {

  if (limit < 1) {
    throw new Error('Invalid Input')
  }

  while (true) {
    k++
    const n = k * (3 * k - 1) / 2 // calculate Pk
  // Then, it initializes two arrays: pentagonalNumbers to store the pentagonal numbers and pentagonalIndices to store their indices. We use a Set for pentagonalIndices to ensure that checking if a number is pentagonal is O(1) on average.

  const pentagonalNumbers = [];
  const pentagonalIndices = new Set();

  // The function then iterates over all pairs of indices in pentagonalNumbers and checks if their sum and difference are also pentagonal. If they are, it checks if the difference D is smaller than the current minimum difference minD. If it is, it updates minD to the new minimum value and minPair to the corresponding pair of pentagonal numbers.

  for (let i = 1; i <= limit; i++) {
    const pentagonalNumber = getPentagonalNumber(i);
    pentagonalNumbers.push(pentagonalNumber);
    pentagonalIndices.add(pentagonalNumber);
  }

  let minD = Infinity;
  let minPair;

    for (let j = k - 1; j > 0; j--) {
      const m = j * (3 * j - 1) / 2 // calculate all Pj < Pk
      if (isPentagonal(n - m) && isPentagonal(n + m)) { // Check sum and difference
        return n - m // return D
  for (let j = 1; j < pentagonalNumbers.length; j++) {
    for (let k = j + 1; k < pentagonalNumbers.length; k++) {
      const sum = pentagonalNumbers[j] + pentagonalNumbers[k];
      const diff = pentagonalNumbers[k] - pentagonalNumbers[j];
      if (pentagonalIndices.has(sum) && pentagonalIndices.has(diff)) {
        if (diff < minD) {
          minD = diff;
          minPair = [pentagonalNumbers[j], pentagonalNumbers[k]];
        }
      }
    }
  }

  return minD;
 }

 /**
 * Function to check if a number is pentagonal or not
 * This function solves n
 * applying the solution for a quadratic function
 * @see {@link https://en.wikipedia.org/wiki/Quadratic_function}
 */
 // You can call the function like this to find the minimum difference between a pair of pentagonal numbers whose sum and difference are also pentagonal up to the pentagonal number limit of 10000


 export { findMinPentagonalDifference }


 function isPentagonal (n) {
  const pent = (Math.sqrt(24 * n + 1) + 1) / 6
  return pent === Math.floor(pent)
 }

 export { problem44 }
Original file line number	Diff line number	Diff line change
Expand Up		@@ -8,37 +8,57 @@
		* It can be seen that P4 + P7 = 22 + 70 = 92 = P8. However, their difference, 70 − 22 = 48, is not pentagonal.
		* Find the pair of pentagonal numbers, Pj and Pk, for which their sum and difference are pentagonal and D = \|Pk − Pj\| is minimised; what is the value of D?
		*
		* @authorddaniel27
		* @authorutkarsh-shrivastav77
		*/

		function problem44 (k) {
		if (k < 1) {
		function getPentagonalNumber(n) {
		return n * (3 * n - 1) / 2;
		}

		// The function takes a limit parameter that determines the maximum pentagonal number to consider. It first defines a getPentagonalNumber function to calculate the nth pentagonal number using the given formula.

		function findMinPentagonalDifference(limit) {

		if (limit < 1) {
		throw new Error('Invalid Input')
		}

		while (true) {
		k++
		const n = k * (3 * k - 1) / 2 // calculate Pk
		// Then, it initializes two arrays: pentagonalNumbers to store the pentagonal numbers and pentagonalIndices to store their indices. We use a Set for pentagonalIndices to ensure that checking if a number is pentagonal is O(1) on average.

		const pentagonalNumbers = [];
		const pentagonalIndices = new Set();

		// The function then iterates over all pairs of indices in pentagonalNumbers and checks if their sum and difference are also pentagonal. If they are, it checks if the difference D is smaller than the current minimum difference minD. If it is, it updates minD to the new minimum value and minPair to the corresponding pair of pentagonal numbers.

		for (let i = 1; i <= limit; i++) {
		const pentagonalNumber = getPentagonalNumber(i);
		pentagonalNumbers.push(pentagonalNumber);
		pentagonalIndices.add(pentagonalNumber);
		}

		let minD = Infinity;
		let minPair;

		for (let j = k - 1; j > 0; j--) {
		const m = j * (3 * j - 1) / 2 // calculate all Pj < Pk
		if (isPentagonal(n - m) && isPentagonal(n + m)) { // Check sum and difference
		return n - m // return D
		for (let j = 1; j < pentagonalNumbers.length; j++) {
		for (let k = j + 1; k < pentagonalNumbers.length; k++) {
		const sum = pentagonalNumbers[j] + pentagonalNumbers[k];
		const diff = pentagonalNumbers[k] - pentagonalNumbers[j];
		if (pentagonalIndices.has(sum) && pentagonalIndices.has(diff)) {
		if (diff < minD) {
		minD = diff;
		minPair = [pentagonalNumbers[j], pentagonalNumbers[k]];
		}
		}
		}
		}

		return minD;
		}

		/**
		* Function to check if a number is pentagonal or not
		* This function solves n
		* applying the solution for a quadratic function
		* @see {@link https://en.wikipedia.org/wiki/Quadratic_function}
		*/
		// You can call the function like this to find the minimum difference between a pair of pentagonal numbers whose sum and difference are also pentagonal up to the pentagonal number limit of 10000


		export { findMinPentagonalDifference }


		function isPentagonal (n) {
		const pent = (Math.sqrt(24 * n + 1) + 1) / 6
		return pent === Math.floor(pent)
		}

		export { problem44 }