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Added the optimized solution for promblem 44 #1324

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Added the optimized solution for promblem 44 #1324

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62 changes: 41 additions & 21 deletions Project-Euler/Problem044.js
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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?
*
* @author ddaniel27
* @author utkarsh-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 }