Files
JavaScript/Project-Euler/Problem014.js
Roland Hummel 86d333ee94 feat: Test running overhaul, switch to Prettier & reformat everything (#1407)
* chore: Switch to Node 20 + Vitest

* chore: migrate to vitest mock functions

* chore: code style (switch to prettier)

* test: re-enable long-running test

Seems the switch to Node 20 and Vitest has vastly improved the code's and / or the test's runtime!

see #1193

* chore: code style

* chore: fix failing tests

* Updated Documentation in README.md

* Update contribution guidelines to state usage of Prettier

* fix: set prettier printWidth back to 80

* chore: apply updated code style automatically

* fix: set prettier line endings to lf again

* chore: apply updated code style automatically

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Co-authored-by: github-actions <${GITHUB_ACTOR}@users.noreply.github.com>
Co-authored-by: Lars Müller <34514239+appgurueu@users.noreply.github.com>
2023-10-04 02:38:19 +05:30

46 lines
1.4 KiB
JavaScript

/*
Longest Collatz sequence
The following iterative sequence is defined for the set of positive integers:
n → n/2 (n is even)
n → 3n + 1 (n is odd)
Using the rule above and starting with 13, we generate the following sequence:
13 → 40 → 20 → 10 → 5 → 16 → 8 → 4 → 2 → 1
It can be seen that this sequence (starting at 13 and finishing at 1) contains 10 terms. Although it has not been proved yet (Collatz Problem), it is thought that all starting numbers finish at 1.
Which starting number, under one million, produces the longest chain?
NOTE: Once the chain starts the terms are allowed to go above one million.
*/
const getCollatzSequenceLength = (num, seqLength) => {
if (num === 1) {
return seqLength
} else {
let newElement
if (num % 2 === 0) {
newElement = num / 2
} else {
newElement = 3 * num + 1
}
seqLength++
return getCollatzSequenceLength(newElement, seqLength)
}
}
export const findLongestCollatzSequence = (limit = 1000000) => {
let startingPointForLargestSequence = 1
let largestSequenceLength = 1
for (let i = 2; i < limit; i++) {
const currentSequenceLength = getCollatzSequenceLength(i, 1)
if (currentSequenceLength > largestSequenceLength) {
startingPointForLargestSequence = i
largestSequenceLength = currentSequenceLength
}
}
return startingPointForLargestSequence
}