How to properly solve fibonacci series in javascript using matrices - javascript

I am trying to solve the Fibonacci algorithm using matrices. My target time complexity is an o(logn) instead of an o(n). The return output of the program is not the number for the series but the sixth significant digits. Its why I am returning the remainder of the solution divided by a million.
I have written the code and it runs well but I noticed that for extremely large inputs, I get a NAN(not a number) instead of an output
const fib = (n) => {
let fibMatrix = [[1,1], [1,0]]
if(n == 0){
return 0;
}
raiseToPower(fibMatrix, n - 1);
return Math.floor(fibMatrix[0][0] % 1000000)
}
const raiseToPower = (matrix, n) => {
if(n == 0 || n == 1){
return;
}
let newMatrix = [[1,1], [1,0]]
raiseToPower(matrix, Math.floor(n / 2))
multiplyMatrices(matrix, matrix)
if(n % 2 !== 0){
multiplyMatrices(matrix, newMatrix)
}
}
const multiplyMatrices = (matrix, newMatrix) => {
let x = matrix[0][0]*newMatrix[0][0] + matrix[0][1]*newMatrix[1][0];
let y = matrix[0][0]*newMatrix[0][1] + matrix[0][1]*newMatrix[1][1];
let z = matrix[1][0]*newMatrix[0][0] + matrix[1][1]*newMatrix[1][0];
let w = matrix[1][0]*newMatrix[0][1] + matrix[1][1]*newMatrix[1][1];
matrix[0][0] = x;
matrix[0][1] = y;
matrix[1][0] = z;
matrix[1][1] = w;
}
console.log(fib(2000))
Thats my code above. Is there anything I could change to actually make this much more performant?

I actually found the error. My numbers were getting larger than the maximum value.
I changed this by instead returning the remainder of my value divided by a million to the matrix and then returning the value from the matrix instead of returning the value and then dividing by a million. The former is efficient and works for any sized inputs.
const fib = (n) => {
let fibMatrix = [[1,1], [1,0]]
if(n == 0){
return 0;
}
raiseToPower(fibMatrix, n - 1);
return (fibMatrix[0][0])
}
const raiseToPower = (matrix, n) => {
if(n == 0 || n == 1){
return;
}
let newMatrix = [[1,1], [1,0]]
raiseToPower(matrix, Math.floor(n / 2))
multiplyMatrices(matrix, matrix)
if(n % 2 !== 0){
multiplyMatrices(matrix, newMatrix)
}
}
const multiplyMatrices = (matrix, newMatrix) => {
let x = (matrix[0][0]*newMatrix[0][0] + matrix[0][1]*newMatrix[1][0]);
let y = (matrix[0][0]*newMatrix[0][1] + matrix[0][1]*newMatrix[1][1]);
let z = (matrix[1][0]*newMatrix[0][0] + matrix[1][1]*newMatrix[1][0]);
let w = (matrix[1][0]*newMatrix[0][1] + matrix[1][1]*newMatrix[1][1]);
matrix[0][0] = x % 1000000;
matrix[0][1] = y % 1000000;
matrix[1][0] = z % 1000000;
matrix[1][1] = w % 1000000;
}
console.log(fib(10000))

Related

How to determine a Perfect Power efficiently?

Challenge: https://www.codewars.com/kata/57c7930dfa9fc5f0e30009eb/train/javascript
Hi I have been trying this problem for many hours but unfortunately my code is taking too long to pass:
function closestPower(num) {
num = Math.floor(num);
if (num < 4) return 4;
// check if input is perfect power
let base = 2;
while (base < 10) {
let exponent = Math.trunc(getBaseLog(base , num));
if ( Math.pow(base, exponent) === num ) {
return num;
}
base++;
}
// check for upper and lower
base = 2;
const verifyObj = {upper:null, lower:null}; // verify
let upperPower = num + 1;
let lowerPower = num - 1;
while (!verifyObj.upper || !verifyObj.lower)
{
// no perfect power
if (lowerPower <= 2 ) verifyObj.lower = "Not found";
if (upperPower === Infinity ) verifyObj.upper = "Not found";
// up til base 9
if (base === 10) {
if (!verifyObj.upper) upperPower++;
if (!verifyObj.lower) lowerPower--;
base = 2;
}
// upper
if (!verifyObj.upper) {
let exponent = Math.trunc(getBaseLog(base , upperPower));
if ( Math.pow(base, exponent) === upperPower ) {
verifyObj.upper = upperPower;
}
}
// lower
if (!verifyObj.lower) {
let exponent = Math.trunc(getBaseLog(base , lowerPower));
if ( Math.pow(base, exponent) === lowerPower ) {
verifyObj.lower = lowerPower;
}
}
base++;
}
console.log(verifyObj) // {upper:64, lower: 49}
// nearest power
if ((upperPower - num) < (num - lowerPower)) {
return upperPower;
}
else return lowerPower;
}
closestPower(56.5); // 49
function getBaseLog(x, y) {
return Math.log(y) / Math.log(x);
}
I realized that my code is redundant as all i need to know if a “base” and “exponent” are more than 1 to determine a perfect power. Any formulas or ideas?
Some issues:
There is no reason why base should not be allowed to be 10 or more
Trying with upperPower at each increment is taking too many iterations. The distance to the next power might be rather big.
I would suggest the following algorithm:
Let the exponent to try with start at 2, and then increment by 1. Calculate which could be the corresponding base. The real base can be found by raising n to the inverse exponent (i.e. 1/exp). Then there are only 2 interesting integer bases to consider: by rounding downwards and upwards.
Here is an implementation:
function closestPower(n) {
if (n <= 6) return 4;
let result = -1;
let closest = n;
for (let factor, exp = 2; (factor = n ** (1 / exp)) > 1.9; ++exp) {
let above = Math.ceil(factor);
for (let intfactor = Math.floor(factor); intfactor <= above; intfactor++) {
let power = intfactor ** exp;
let diff = Math.abs(power - n);
if (diff == 0) return n;
if (diff < closest || diff == closest && power < n) {
closest = diff;
result = power;
}
}
}
return result;
}
// Some tests:
const tests = [
[0, 4], [9, 9], [30, 32], [34, 32], [56.5, 49],
[123321456654, 123321773584]
];
for (let [n, expected] of tests) {
let result = closestPower(n);
if (result === expected) continue;
console.log(`closestPower(${n}) returned ${result}, but expected ${expected}`);
}
console.log("all tests done");
Here's my algorithm
first i will get the exponent from base that less than of the n then I added the current base of the loop with the n then get the base log.
function closestPower(n) {
if(n < 4) return 4
let closest = []
let base = 2
while(base < n) {
const exponent = Math.floor(Math.log(n + base) / Math.log(base))
const power = Math.pow(base,exponent)
if(exponent === 1) break
if(power === n) return n
closest.push(power)
base++
}
return closest.reduce((prev, curr) => (Math.abs(curr - n) < Math.abs(prev - n) ? curr : prev))
}
console.log(closestPower(0))
console.log(closestPower(9))
console.log(closestPower(30))
console.log(closestPower(34))
console.log(closestPower(56.5))
console.log(closestPower(123321456654))

How to generate a random number based on fixed character set in javascript

I'm generating a number based on a fixed character set.
function generator()
{
var text = "";
var char_list = "LEDGJR", number_list = "0123456789";
for(var i=0; i < 2; i++ )
{
text += char_list.charAt(Math.floor(Math.random() * char_list.length));
}
for(var j=0; j < 2; j++ )
{
text += number_list.charAt(Math.floor(Math.random() *
number_list.length));
}
return text;
}
Result :
RE39, JE12 etc...
Once all the permutation related to the above sequence is done, then the generator should generate string as RE391, JE125 means adding one more number to the complete number.
How can I get the permutation count of sequence?
For simplicity consider the case where:
chars = "AB"
nums = "123";
and we want to generate a 4-digit sequence of two chars and two numbers.
We define these variables
rows = [chars, chars, nums, nums]
rowSizes = rows.map(row => row.length) // [2, 2, 3, 3]
It’s easy to see the set size of all possible permuations equals:
spaceSize = rowSizes.reduce((m, n) => m * n, 1) // 2*2*3*3 = 36
And we define two set of utility functions, usage of which I'll explain in detail later.
decodeIndex() which gives us uniqueness
function euclideanDivision(a, b) {
const remainder = a % b;
const quotient = (a - remainder) / b;
return [quotient, remainder]
}
function decodeIndex(index, rowSizes) {
const rowIndexes = []
let dividend = index
for (let i = 0; i < rowSizes.length; i++) {
const [quotient, remainder] = euclideanDivision(dividend, rowSizes[i])
rowIndexes[i] = remainder
dividend = quotient
}
return rowIndexes
}
getNextIndex() which gives us pseudo-randomness
function isPrime(n) {
if (n <= 1) return false;
if (n <= 3) return true;
if (n % 2 == 0 || n % 3 == 0) return false;
for (let i = 5; i * i <= n; i = i + 6) {
if (n % i == 0 || n % (i + 2) == 0) return false;
}
return true;
}
function findNextPrime(n) {
if (n <= 1) return 2;
let prime = n;
while (true) {
prime++;
if (isPrime(prime)) return prime;
}
}
function getIndexGeneratorParams(spaceSize) {
const N = spaceSize;
const Q = findNextPrime(Math.floor(2 * N / (1 + Math.sqrt(5))))
const firstIndex = Math.floor(Math.random() * spaceSize);
return [firstIndex, N, Q]
}
function getNextIndex(prevIndex, N, Q) {
return (prevIndex + Q) % N
}
Uniqueness
Like mentioned above, spaceSize is the number of all possible permutations, thus each index in range(0, spaceSize) uniquely maps to one permutation. decodeIndex helps with this mapping, you can get the corresponding permutation to an index by:
function getSequenceAtIndex(index) {
const tuple = decodeIndex(index, rowSizes)
return rows.map((row, i) => row[tuple[i]]).join('')
}
Pseudo-Randomness
(Credit to this question. I just port that code into JS.)
We get pseudo-randomness by polling a "full cycle iterator"†. The idea is simple:
have the indexes 0..35 layout in a circle, denote upperbound as N=36
decide a step size, denoted as Q (Q=23 in this case) given by this formula‡
Q = findNextPrime(Math.floor(2 * N / (1 + Math.sqrt(5))))
randomly decide a starting point, e.g. number 5
start generating seemingly random nextIndex from prevIndex, by
nextIndex = (prevIndex + Q) % N
So if we put 5 in we get (5 + 23) % 36 == 28. Put 28 in we get (28 + 23) % 36 == 15.
This process will go through every number in circle (jump back and forth among points on the circle), it will pick each number only once, without repeating. When we get back to our starting point 5, we know we've reach the end.
†: I'm not sure about this term, just quoting from this answer
‡: This formula only gives a nice step size that will make things look more "random", the only requirement for Q is it must be coprime to N
Full Solution
Now let's put all the pieces together. Run the snippet to see result.
I've also includes the a counter before each log. For your case with char_list="LEDGJR", number_list="0123456789", the spaceSize for 4-digit sequence should be 6*6*10*10 = 3600
You'll observe the log bump to 5-digit sequence at 3601 😉
function euclideanDivision(a, b) {
const remainder = a % b;
const quotient = (a - remainder) / b;
return [quotient, remainder];
}
function decodeIndex(index, rowSizes) {
const rowIndexes = [];
let divident = index;
for (let i = 0; i < rowSizes.length; i++) {
const [quotient, remainder] = euclideanDivision(divident, rowSizes[i]);
rowIndexes[i] = remainder;
divident = quotient;
}
return rowIndexes;
}
function isPrime(n) {
if (n <= 1) return false;
if (n <= 3) return true;
if (n % 2 == 0 || n % 3 == 0) return false;
for (let i = 5; i * i <= n; i = i + 6) {
if (n % i == 0 || n % (i + 2) == 0) return false;
}
return true;
}
function findNextPrime(n) {
if (n <= 1) return 2;
let prime = n;
while (true) {
prime++;
if (isPrime(prime)) return prime;
}
}
function getIndexGeneratorParams(spaceSize) {
const N = spaceSize;
const Q = findNextPrime(Math.floor((2 * N) / (1 + Math.sqrt(5))));
const firstIndex = Math.floor(Math.random() * spaceSize);
return [firstIndex, N, Q];
}
function getNextIndex(prevIndex, N, Q) {
return (prevIndex + Q) % N;
}
function generatorFactory(rows) {
const rowSizes = rows.map((row) => row.length);
function getSequenceAtIndex(index) {
const tuple = decodeIndex(index, rowSizes);
return rows.map((row, i) => row[tuple[i]]).join("");
}
const spaceSize = rowSizes.reduce((m, n) => m * n, 1);
const [firstIndex, N, Q] = getIndexGeneratorParams(spaceSize);
let currentIndex = firstIndex;
let exhausted = false;
function generator() {
if (exhausted) return null;
const sequence = getSequenceAtIndex(currentIndex);
currentIndex = getNextIndex(currentIndex, N, Q);
if (currentIndex === firstIndex) exhausted = true;
return sequence;
}
return generator;
}
function getRows(chars, nums, rowsOfChars, rowsOfNums) {
const rows = [];
while (rowsOfChars--) {
rows.push(chars);
}
while (rowsOfNums--) {
rows.push(nums);
}
return rows;
}
function autoRenewGeneratorFactory(chars, nums, initRowsOfChars, initRowsOfNums) {
let realGenerator;
let currentRowOfNums = initRowsOfNums;
function createRealGenerator() {
const rows = getRows(chars, nums, initRowsOfChars, currentRowOfNums);
const generator = generatorFactory(rows);
currentRowOfNums++;
return generator;
}
realGenerator = createRealGenerator();
function proxyGenerator() {
const sequence = realGenerator();
if (sequence === null) {
realGenerator = createRealGenerator();
return realGenerator();
} else {
return sequence;
}
}
return proxyGenerator;
}
function main() {
const char_list = "LEDGJR"
const number_list = "0123456789";
const generator = autoRenewGeneratorFactory(char_list, number_list, 2, 2);
let couter = 0
setInterval(() => {
console.log(++couter, generator())
}, 10);
}
main();

how to solve number Pelindrom by using for loop in javascript

How to solve this giving undesired result, getting false always.
function isPelindrom(n) {
if (n < 0) return false;
let orgNum = n
let reversed = 0
for (let x = 0; x <= orgNum; x++) {
let lastDigit = orgNum % 10;
reversed = (reversed * 10) + lastDigit;
orgNum = parseInt(orgNum / 10);
}
return n == reversed
}
isPelindrom(12321)
function Palindrome(num) {
let numToStringArray = num.toString();
const reversed = numToStringArray.toString().split('').reverse().join('');
return numToStringArray === reversed ? "It's a palindrome" : "It's not a palindrome";
}
console.log(Palindrome(989));
console.log(Palindrome(23));
console.log(Palindrome(9));
function Palindrome(number) {
var temp = number, final = 0;
while (number > 0) {
rem = number % 10;
number = parseInt(number / 10);
final = final * 10 + rem;
}
if (temp == final) {
return "It's Palindrome";
}
else {
return "It's not Palindrome";
}
}
console.log(Palindrome(989));
console.log(Palindrome(23));
console.log(Palindrome(9));
You can change it to a for loop and either set it to greater or equal to 0.1 or just greater than 0.
function isPelindrom(n) {
if (n < 0) return false;
let orgNum = n
let reversed = 0
while(orgNum>=0.1) {
let lastDigit = orgNum % 10;
reversed = (reversed * 10) + lastDigit;
orgNum = parseInt(orgNum / 10);
}
return n == reversed
}
console.log(isPelindrom(12321))
console.log(isPelindrom(12322))
Does while loop count?
const isPalindrome = num => {
const origin = num
let reversed = 0
while (num !== 0) {
const digit = num % 10
reversed = reversed * 10 + digit
num = Math.floor(num / 10)
}
return origin === reversed
}
console.log(isPalindrome(12321))
console.log(isPalindrome(12331))
You can try this:
function isPelindrom(n) {
if (n < 0) return false;
let orgNum = n
let reversed = 0
var numberOfDigits = 0;
for(i = orgNum; i > 1; ++i){
++numberOfDigits;
i = Math.floor(i/10);
}
for (let x = 0; x < numberOfDigits; x++) {
let lastDigit = orgNum % 10;
reversed = (reversed * 10) + lastDigit;
orgNum = parseInt(orgNum / 10);
}
return n == reversed
}

How can I find the prime factors of an integer in JavaScript?

I was trying to find the prime factors of a number, recorded below as 'integer' using a for loop in javascript. I can't seem to get it working and I'm not sure whether it's my JavaScript or my calculation logic.
//integer is the value for which we are finding prime factors
var integer = 13195;
var primeArray = [];
//find divisors starting with 2
for (i = 2; i < integer/2; i++) {
if (integer % i == 0) {
//check if divisor is prime
for (var j = 2; j <= i / 2; j++) {
if (i % j == 0) {
isPrime = false;
} else {
isPrime = true;
}
}
//if divisor is prime
if (isPrime == true) {
//divide integer by prime factor & factor store in array primeArray
integer /= i
primeArray.push(i);
}
}
}
for (var k = 0; k < primeArray.length; k++) {
console.log(primeArray[k]);
}
The answer above is inefficient with O(N^2) complexity. Here is a better answer with O(N) complexity.
function primeFactors(n) {
const factors = [];
let divisor = 2;
while (n >= 2) {
if (n % divisor == 0) {
factors.push(divisor);
n = n / divisor;
} else {
divisor++;
}
}
return factors;
}
const randomNumber = Math.floor(Math.random() * 10000);
console.log('Prime factors of', randomNumber + ':', primeFactors(randomNumber).join(' '))
You can filter for duplicates as you please!
Here's a working solution:
function getPrimeFactors(integer) {
const primeArray = [];
let isPrime;
// Find divisors starting with 2
for (let i = 2; i <= integer; i++) {
if (integer % i !== 0) continue;
// Check if the divisor is a prime number
for (let j = 2; j <= i / 2; j++) {
isPrime = i % j !== 0;
}
if (!isPrime) continue;
// if the divisor is prime, divide integer with the number and store it in the array
integer /= i
primeArray.push(i);
}
return primeArray;
}
console.log(getPrimeFactors(13195).join(', '));
You were very much on the right track. There were two minor mistakes. The evaluation of integer - 1 seemed to be incorrect. I believe the more appropriate evaluation is <= integer in your outer for loop. This is because when you divide your integer below integer /= i, this results in the final integer evaluation to be 29. The final prime divisor in this case is also 29 and as such will need to be evaluated as <= as oppose to < integer - 1.
As for why the final log statement isn't working, there was a simple typo of primeArray[i] as oppose to primeArray[k].
I do believe there is a mistake in both code above. If you replace the integer by 100 the prime factorization won't work anymore as the factor 2 cannot be considered with those for loops. As j = 2, i = 2 and j<=i/2 in the condition - meaning the loop will never run for i=2, which is a prime factor.
Tried to make it work this way but couldn't figure out.
Had to rely on a different approach with a while loop here :
function getAllFactorsFor(remainder) {
var factors = [], i;
for (i = 2; i <= remainder; i++) {
while ((remainder % i) === 0) {
factors.push(i);
remainder /= i;
}
}
return factors;
}
https://jsfiddle.net/JamesOR/RC7SY/
You could also go with something like that :
let findPrimeFactors = (num) => {
let arr = [];
for ( var i = 2; i < num; i++) {
let isPrime
if (num % i === 0) {
isPrime = true;
for (var j = 2; j <= i; j++) {
if ( i % j === 0) {
isPrime == false;
}
}
}if (isPrime == true) { arr.push(i)}
}console.log(arr)
}
findPrimeFactors(543)
We can find the prime factor numbers up to n with only one loop. It is a very simple solution without any nested loop.
Time complexity would be less than O(n) because we are dividing "n" by "i".
function primeFactors(n) {
let arr=[];
let i = 2;
while(i<=n){
if(n%i == 0) {
n= n/i;
arr.push(i);
} else {
i++;
}
}
return arr;
}
// primeFactors(10) [2,5]
// primeFactors(10) [2,2,5,5]
// primeFactors(2700) [2, 2, 3, 3, 3, 5, 5]
When factorizing an integer (n) to its prime factors, after finding the first prime factor, the problem in hand is reduced to finding prime factorization of quotient (q).
Suppose n is divisible to prime p1 then we have n = p1 * q1 so after finding p1 the problem is reduced to factorizing q1 (quotient). If the function name is primeFactorizer then we can call it recursively and solution for n would be:
n = p1 * primeFactorizer(q1)
n = p1 * p2 * primeFactorizer(q2)
...
Until qn is prime itself.
Also I'm going to use a helper generator function which generates primes for us:
function * primes () {
let n = 2
while (true) {
let isPrime = true
for (let i = 2; i <= n / 2; i++) {
if (n % i === 0) {
isPrime = false
break
}
}
if (isPrime) {
yield n
}
n++
}
}
And function to factorize n would be:
function primeFactorizer (n, result = []) {
for (const p of primes()) {
if (n === p) {
result.push(p)
return result
}
if (n % p === 0) {
result.push(p)
return primeFactorizer(n / p, result)
}
}
}
I've refined this function over time, trying to get it as fast as possible (racing it against others' functions that I've found online, haven't found one that runs consistently faster than it yet).
function primFact(num) {
var factors = [];
/* since 2 is the only even prime, it's easier to factor it out
* separately from the odd factor loop (for loop doesn't need to
* check whether or not to add 1 or 2 to f).
* The condition is essentially checking if the number is even
* (bitwise "&" operator compares the bits of 2 numbers in binary
* and outputs a binary number with 1's where their digits are the
* same and 0's where they differ. In this case it only checks if
* the final digit for num in binary is 1, which would mean the
* number is odd, in which case the output would be 1, which is
* interpreted as true, otherwise the output will be 0, which is
* interpreted as false. "!" returns the opposite boolean, so this
* means that '!(num & 1)' is true when the num is not odd)
*/
while (!(num & 1)) {
factors.push(2);
num /= 2;
}
// 'f*f <= num' is faster than 'f <= Math.sqrt(num)'
for (var f = 3; f*f <= num; f += 2) {
while (!(num % f)) { // remainder of 'num / f' isn't 0
factors.push(f);
num /= f;
}
}
/* if the number is already prime, then this adds it to factors so
* an empty array isn't returned
*/
if (num != 1) {
factors.push(num);
}
return factors;
}
This performs very well at large numbers compared to functions I've run it against, especially when the number is prime, (rarely runs slower than 10ms when I've run it in an online compiler like OneCompiler) so if you want speed I'd say this is a pretty good way to go about it.
Still working on making it even faster, but only way to include all primes without adding new conditions to check is to iterate through all odd numbers.
I just started JavaScript but i managed to come up with my own solution for this while working on a school project with a similar objective.
Only issue is that it takes a very long time for large numbers, its not v ery efficient. But it works perfectly.
function isPrime(n){
if (n === 1){
return false;
}
else if (n === 2){
return true;
}
else{
for (let x = 2; x < n; x ++){
if (n % x === 0){
return false;
}
}
return true;
}
}
let primeFac = []
let num = 30
for (let x = 0; x <= num; x++){
if (num % x === 0 && isPrime(x) === true){
primeFac.push(x);
}
}
console.log(`${primeFac}`)
If you work up from the bottom there's no need to check if any following factor is prime. This is because any lower primes will have already been divided out.
function getPrimeFactorsFor(num) {
const primes = [];
for (let factor = 2; factor <= num; factor++) {
while ((num % factor) === 0) {
primes.push(factor);
num /= factor;
}
}
return primes;
}
console.log("10 has the primes: ", getPrimeFactorsFor(10));
console.log("8 has the primes: ", getPrimeFactorsFor(8));
console.log("105 has the primes: ", getPrimeFactorsFor(105))
console.log("1000 has the primes: ", getPrimeFactorsFor(1000))
console.log("1155 has the primes: ", getPrimeFactorsFor(1155))
In case somebody is looking for the fastest solution, here's one based on my library prime-lib. It can calculate prime factors for any number between 2 and 2^53 - 1, in under 1ms. The function source code is available here.
import {primeFactors} from 'prime-lib';
const factors = primeFactors(600851475143);
//=> [71, 839, 1471, 6857]
Here an other implementation to find prime factors, in three variations. It's more efficient than the other implementations, worst case sqrt(n), because it stops earlier.
The function* means it's a generator function. So a generator is returned instead of an array and the next prime factor is only calculated as soon as it is requested.
// Example: 24 -> 2, 3
function* singlePrimeFactors (n) {
for (var k = 2; k*k <= n; k++) {
if (n % k == 0) {
yield k
do {n /= k} while (n % k == 0)
}
}
if (n > 1) yield n
}
// Example: 24 -> 2, 2, 2, 3
function* repeatedPrimeFactors (n) {
for (var k = 2; k*k <= n; k++) {
while (n % k == 0) {
yield k
n /= k
}
}
if (n > 1) yield n
}
// Example: 24 -> {p: 2, m: 3}, {p: 3, m: 1}
function* countedPrimeFactors (n) {
for (var k = 2; k*k <= n; k++) {
if (n % k == 0) {
var count = 1
for (n /= k; n % k == 0; n /= k) count++
yield {p: k, m: count}
}
}
if (n > 1) yield {p: n, m: 1}
}
// Test code
for (var i=1; i<=100; i++) {
var single = JSON.stringify(Array.from(singlePrimeFactors(i)))
var repeated = JSON.stringify(Array.from(repeatedPrimeFactors(i)))
var counted = JSON.stringify(Array.from(countedPrimeFactors(i)))
console.log(i, single, repeated, counted)
}
// Iterating over a generator
for (var p of singlePrimeFactors(24)) {
console.log(p)
}
// Iterating over a generator, an other way
var g = singlePrimeFactors(24)
for (var r = g.next(); !r.done; r = g.next()) {
console.log(r.value);
}
My solution avoids returning not prime factors:
let result = [];
let i = 2;
let j = 2;
let number = n;
for (; i <= number; i++) {
let isPrime = number % i === 0;
if (isPrime) {
result.push(i);
number /= i;
}
while (isPrime) {
if (number % i === 0) {
result.push(i);
number /= i;
} else {
isPrime = false;
}
}
}
return result;
With so many good solutions above, wanted to make a little bit of improvement by using this theorem in the Math Forum Finding prime factors by taking the square root
.
function primeFactors(n)
{
// Print the number of 2s that divide n
while (n%2 == 0)
{
console.log(2);
n = n/2;
}
// n must be odd at this point. So we can skip
// one element (Note i = i +2)
for (var i = 3; i <= Math.sqrt(n); i = i+2)
{
// While i divides n, print i and divide n
while (n%i == 0)
{
console.log(i);
n = n/i;
}
}
// This condition is to handle the case when n
// is a prime number greater than 2
if (n > 2)
console.log(n);
}
primeFactors(344);
console.log("--------------");
primeFactors(4);
console.log("--------------");
primeFactors(10);
Hope this answer adds value.
Here is a solution using recursion
function primeFactors(num, primes){
let i = 2;
while(i < num){
if(num % i === 0){
primes.push(i);
return primeFactors(num/i, primes);
}
i++
}
primes.push(num);
return primes;
}
console.log(primeFactors(55, []))
console.log(primeFactors(15, []))
console.log(primeFactors(40, []))
console.log(primeFactors(13, []))
// [ 5, 11 ]
// [ 3, 5 ]
// [ 2, 2, 2, 5 ]
// [ 13 ]
I found this solution by chance when i was trying to simplify several
solutions that i saw here. Although it doesn't check if the divisor
is a prime number somehow it seems to work, i tested it with
miscellaneous numbers but i could not explain how this was possible.
function start() {
var integer = readInt("Enter number: ");
println("The prime factorization is: ");
for(var i = 2; i <= integer; i++) {
if (integer % i == 0) {
println(i);
integer = integer / i;
i = i - 1;
}
}
}
I checked the algorithm with yield, but that is a lot slower than recursive calls.
function rootnth(val, power=2) {
let o = 0n; // old approx value
let x = val;
let limit = 100;
let k = BigInt(power);
while(x**k!==k && x!==o && --limit) {
o=x;
x = ((k-1n)*x + val/x**(k-1n))/k;
}
return x;
}
// Example: 24 -> 2, 2, 2, 3
function repeatedPrimeFactors (n,list) {
if (arguments.length == 1) list = "";
if (n % 2n == 0) return repeatedPrimeFactors(n/2n, list + "*2")
else if (n % 3n == 0) return repeatedPrimeFactors(n/3n, list + "*3")
var sqrt = rootnth(n);
let k = 5n;
while (k <= sqrt) {
if (n % k == 0) return repeatedPrimeFactors(n/k, list + "*" + k)
if (n % (k+2n) == 0) return repeatedPrimeFactors(n/(k+2n), list + "*" + (k+2n))
k += 6n;
}
list = list + "*" + n;
return list;
}
var q = 11111111111111111n; // seventeen ones
var t = (new Date()).getTime();
var count = repeatedPrimeFactors(BigInt(q)).substr(1);
console.log(count);
console.log(("elapsed=" + (((new Date()).getTime())-t)+"ms");
Here I try for the factors 2 and 3, followed by alternatingly adding 2 anf 4 (5,7,11,13,17,...) until the square root of the number.
Seventeen ones (which is not prime) takes about 1 second and nineteen ones (which is prime) eight seconds (Firefox).
Here is the solution with the nested function using the filter method.
function primeFactors(params) {
function prime(number) {
for (let i = 2; i < number + 1; ) {
if (number === 2) {
return true;
}
if (number % i === 0 && number !== i) {
return false;
} else if (i < number) {
i++;
} else {
return true;
}
}
}
let containerPrime = [];
let containerUnPrime = [];
for (let i = 0; i < params; i++) {
if (prime(i)) {
containerPrime.push(i);
} else {
containerUnPrime.push(i);
}
}
return containerPrime.filter((e) => params % e === 0);
}
console.log(primeFactors(13195));
function primeFactorization(n) {
let factors = [];
while (n % 2 === 0) {
factors.push(2);
n = n / 2;
}
for (let i = 3; i <= Math.sqrt(n); i += 2) {
while (n % i === 0) {
factors.push(i);
n = n / i;
}
}
if (n > 2) {
factors.push(n);
}
return factors;
}
console.log(primeFactorization(100));
The answer with O(sqrt(n)) complexity, it's faster than O(n):
const number = 13195;
let divisor = 2;
const result = [];
let n = number;
while (divisor * divisor <= number) {
if (n % divisor === 0) {
result.push(divisor);
n /= divisor;
} else {
divisor++;
}
}
if (n > 1) {
result.push(n);
}
console.log(result);
The above code (the code which has while loop) is correct, but there is one small correction in that code.
var num, i, factorsArray = [];
function primeFactor(num) {
for (i = 2; i <= num; i++) {
while (num % i == 0) {
factorsArray.push(i);
num = num / 2;
}
}
}
primeFactor(18);
var newArray = Array.from(new Set(factorsArray));
document.write(newArray);
This is my solution
function prime(n) {
for (var i = 1; i <= n; i++) {
if (n%i===0) {
myFact.push(i);
var limit = Math.sqrt(i);
for (var j = 2; j < i; j++) {
if (i%j===0) {
var index = myFact.indexOf(i);
if (index > -1) {
myFact.splice(index, 1);
}
}
}
}
}
}
var n = 100, arr =[],primeNo = [],priFac=[];
for(i=0;i<=n;i++){
arr.push(true);
}
//console.log(arr)
let uplt = Math.sqrt(n)
for(j=2;j<=uplt;j++){
if(arr[j]){
for(k=j*j;k<=n;k+=j){
arr[k] = false;
}
}
}
for(l=2;l<=n;l++){
if(arr[l])
primeNo.push(l)
}
for(m=0;m<primeNo.length;m++){
if(n%primeNo[m]==0)
priFac.push(primeNo[m])
}
console.log(...priFac);

finding sum of prime numbers under 250

var sum = 0
for (i = 0; i < 250; i++) {
function checkIfPrime() {
for (factor = 2; factor < i; factor++) {
if (i % factor = 0) {
sum = sum;
}
else {
sum += factor;
}
}
}
}
document.write(sum);
I am trying to check for the sum of all the prime numbers under 250. I am getting an error saying that i is invalid in the statement if (i % factor = 0) I know was creating in the original for statement, but is there any way to reference it in the if statement?
With the prime computation, have you considered using Sieve of Eratosthenes? This is a much more elegant way of determining primes, and, summing the result is simple.
var sieve = new Array();
var maxcount = 250;
var maxsieve = 10000;
// Build the Sieve, marking all numbers as possible prime.
for (var i = 2; i < maxsieve; i++)
sieve[i] = 1;
// Use the Sieve to find primes and count them as they are found.
var primes = [ ];
var sum = 0;
for (var prime = 2; prime < maxsieve && primes.length < maxcount; prime++)
{
if (!sieve[prime]) continue;
primes.push(prime); // found a prime, save it
sum += prime;
for (var i = prime * 2; i < maxsieve; i += prime)
sieve[i] = 0; // mark all multiples as non prime
}
document.getElementById("result").value =
"primes: " + primes.join(" ") + "\n"
+ "count: " + primes.length + "\n"
+ "sum: " + sum + "\n";
#result {
width:100%;
height:180px
}
<textarea id="result">
</textarea>
(EDIT) With the updated algorithm, there are now two max involved:
maxcount is the maximum number of prime numbers you wish to find
maxsieve is a guess of sieve large enough to contain maxcount primes
You will have to validate this by actually checking the real count since there are two terminating conditions (1) we hit the limit of our sieve and cannot find any more primes, or (2) we actually found what we're looking for.
If you were to increase the number to numbers much greater than 250, than the Sieve no longer becomes viable as it would be consume great deals of memory. Anyhow, I think this all makes sense right? You really need to play with the Sieve yourself at this point than rely on my interpretation of it.
You can equally use this
let sum = 0;
let num = 250;
for (let i = 2; i < num; i++) {
let isPrime = true;
for (let j = 2; j < i; j++) {
if (i % j === 0) {
isPrime = false;
}
}
if (isPrime) {
sum += i;
}
}
console.log(sum);
i % factor === 0
Use === for comparison. = is for assignment. Yeah I said triple equals. Type coercion is annoying.
You need a == or ===: if (i % factor == 0)
Here's a pretty decent way to do it. It's not as advanced as the sieve but it's a decent starting point. NOTE: I'm using ES6 syntax.
/*
* Sum the first n prime numbers
*
* #param n (integer)
* #return integer
*
*/
function sumNprimes(n){
const arr = [];
let i = 2
while (arr.length < n) {
if (isPrime(i)) {
arr.push(i)
}
i++
}
return arr.reduce( (x,y) => x+y );
/*
* #param n (integer)
* #return Boolean
*
*/
function isPrime(n) {
if ( n < 2 ) {
return false
}
for ( let i = 2; i <= Math.sqrt(n); i++ ) {
if ( n % i === 0 ) {
return false;
}
}
return true
}
}
So i had to face a similar challenge and here is my solution, i hope you find it helpful:
function sumPrimes(num) {
// determine if a number is prime
function isPrime(n) {
if (n === 2) return true;
if (n === 3) return true;
if (n % 2 === 0) return false;
if (n % 3 === 0) return false;
var i = 5;
var w = 2;
while (i * i <= n) {
if (n % i === 0) {
return false;
}
i += w;
w = 6 - w;
}
return true;
}
// subtract 1 for 'not being prime' in my context
var sum = isPrime(num) ? num - 1 : -1;
for (var x = 0; x < num; x++) {
if (isPrime(x) === true) {
sum += x;
}
}
return sum;
}
As per the "Sieve of Eratosthenes", I have implemented the code using JS:
function isPrime(n){
return ((n/2 === 1 || n/3 === 1 || n/5 === 1 || n/7 === 1)?true:(n%2===0 || n%3 === 0 || n%5 ===0 || n%7 === 0)?false:true);
};
var val = 250;
let outArr = [];
for(let i=2;i<val;i++){
if(isPrime(i)){
outArr.push(i);
}
}
console.log("Prime number between 0 - "+val+" : "+outArr.join(","));
Here is a simple way of looping through array and implementing the sieve of Eratosthenes...
function sumPrimes(num) {
var t, v = [],
w = [],
x = [],
y = [],
z = 0;
//enumerating Vee array starts at 2 as first prime number
for (let a = 2; a <= num; a++) {
v.push(a)
}
//creating a moving loop by splicing its first index
for (let i = 0; i < v.length; i) { //ensure all items spliced
t = v[i]; // t as prime to be removed from Vee array
x.push(t); // x storage of primes
z += t // total of peculiar primes
w.push(v.splice(i, 1)) //tested to move all one by one
// prompt(v) //tested that v loses its v[i] every iteration
//= now trying to remove others using remainder (%) vs prime t
for (let vi in v) {
v[vi] % t === 0 ? y.push(v.splice(vi, 1)) : ""; //recursive removal of composite items by prime t
}
}
return z // returns sum of primes
}
sumPrimes(250);
You generate the array beginning with 2 as first prime,
You sieve the array removing items by the remainder of prime using % === 0.
The you loop through the remaining array by using the next prime until the last remaining prime is pushed to the prime arrays. Add all primes to get the Sum.
If the question is purely academical, earlier answers are better suited.
The example below uses modern libraries, in case you need an efficient and elegant solution.
import {generatePrimes} from 'prime-lib';
import {from, reduce, takeWhile} from 'rxjs';
from(generatePrimes())
.pipe(takeWhile(p => p < 250), reduce((a, c) => a + c))
.subscribe(sum => {
// sum = 5830
});
Performance-wise, it will take significantly less than 1ms.
How would it affect the code if I wanted say the sum of the first 250 prime numbers instead of the prime numbers under 250?
You would just replace takeWhile(p => p < 250) with take(250):
import {generatePrimes} from 'prime-lib';
import {from, reduce, take} from 'rxjs';
from(generatePrimes())
.pipe(take(250), reduce((a, c) => a + c))
.subscribe(sum => {
// sum = 182109
});
P.S. I am the author of prime-lib.

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