I am working on an excercise to sum all prime numbers from 2 to the parameter. I have worked this far in the code, but am stuck. I believe by using the splice function, I am actually skipping an element because of a changed indices.
function sumPrimes(num) {
var primearray = [];
var sum = 0;
for(var i =2; i <= num; i++){
primearray.push(i);
}
for(var j = 0; j < primearray.length; j++) {
console.log(primearray[j]);
if ((primearray[j]%2===0) && (primearray[j] >2)) {
primearray.splice(j,1);
} else if ((primearray[j]%3===0) && (primearray[j] > 3)) {
primearray.splice(j,1);
console.log(primearray);
} else if ((primearray[j]%5===0) && (primearray[j] > 5)) {
primearray.splice(j,1);
} else if ((primearray[j]%7===0) && (primearray[j] > 7)) {
primearray.splice(j,1);
}
}
sum = primearray.reduce();
return sum;
}
sumPrimes(30);
I haven't utilized the reduce function yet because I am still working on the if else statements.
I found a pretty good solution to the same problem. afmeva was spot on. This is how it works.
function isPrime(val){
//test if number is prime
for(var i=2; i < val; i++){
if(val % i === 0){
return false;
}
}
return true;
}
In the above code we accept a number to determine whether or not it is prime. We then loop from two all the way up until our number minus one because we know that our number will be divisible by itself and one. If the remainder of our value with the current loop value is zero then we know it is not prime so break out and say so.
This article explains very well
function sumPrimes(num) {
var answer = 0;
//loop through all numbers from 2 to input value
for(var i=2; i <= num; i++){
//sum only prime numbers, skip all others
if(isPrime(i)){
answer += i;
}
}
return answer;
}
sumPrimes(977); // 73156
Here's another good resource
function sumPrimes(num) {
let arr = Array.from({length: num+1}, (v, k) => k).slice(2);
let onlyPrimes = arr.filter( (n) => {
let m = n-1;
while (m > 1 && m >= Math.sqrt(n)) {
if ((n % m) === 0)
return false;
m--;
}
return true;
});
return onlyPrimes.reduce((a,b) => a+b);
}
sumPrimes(977);
I have seen lots of people putting all prime numbers into arrays and in order to check if a number is prime, they check from 2 to the number to see if there's a remainder.
You only need to check odd numbers, and only need to count to half the number because a number can't be divisible by any number greater than it has.
Here's my solution:
function sumPrimes(num){
var sum = num>=2?2:0;
for(var i=3;i<=num;i+=2){
var isPrime=true;
for(var j=3;j<(i/2);j++){
if (i%j==0)
{
isPrime=false;
break;
}
}
sum+=isPrime?i:0;
}
return sum;
}
Note: I started from j=2 because we are only checking odd numbers, so they'd never be divisible by 2.
function sumPrimes(num) {
var sumArr= [];
for(var i=0;i<=num;i++){
if(isPrime(i))
sumArr.push(i);
}
sumArr = sumArr.reduce(function(a,b){
return a+b;
})
return sumArr;
}
function isPrime(num) {
if(num < 2) return false;
for (var i = 2; i < num; i++) {
if(num%i === 0)
return false;
}
return true;
}
sumPrimes(10);
something like this?
function isPrime(_num) {
for(var i = 2; i < _num; i++) {
if(!(_num % i)) {
return false
}
}
return true;
}
function sumPrimes(_num) {
var sum = 0;
for(var i = 2; i <= _num; i++) {
if(isPrime(i)) {
sum += i;
}
}
return sum;
}
sumPrimes(20) // 77
sumPrimes(5) // 10
You could do this as well.
function sumPrimes(num) {
var sum = 0;
for (var i = 0; i <= num; i++) {
if (isPrime(i)) {
sum += i;
}
}
return sum;
}
function isPrime(n) {
if (n < 2) { return false; }
if (n !== Math.round(n)) { return false; }
var result = true;
for (var i = 2; i <= Math.sqrt(n); i++) {
if (n % i === 0) {
result = false;
}
}
return result;
}
Here's my solution. I hope you find it easy to interpret:
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;
}
here is my solution to sum of n prime number
function sumOfNPrimeNumber(num){
var sum = 0;
const isPrime = function(n){
if (isNaN(n) || !isFinite(n) || n%1 || n<2) {
return false;
}
if (n%2==0){
return (n==2);
}
var sqrt = Math.sqrt(n);
for (var i = 3; i < sqrt; i+=2) {
if(n%i == 0){
return false;
}
}
return true;
}
const getNextPrime = function* (){
let nextNumber = 2;
while(true){
if(isPrime(nextNumber)){
yield nextNumber;
}
++nextNumber;
}
}
const nextPrime = getNextPrime();
for (var i = 0; i < num; i++) {
sum = sum + nextPrime.next().value;
}
return sum;
}
console.log(sumOfNPrimeNumber(3));
All the above answers make use of helper functions or aren't time efficients.
This is a quick, recursive solution in O(n) time:
// # signature int -> int
// # interpr: returns sum of all prime integers <= num
// assume: num is positive integer
function sumPrimes(num) {
if (num <= 2) {
return 2;
}
let i = 2;
while (i < num) {
if (num % i === 0) {
return sumPrimes(num - 1)
}
i++;
}
return num + sumPrimes(num - 1)
}
// test
sumPrimes(10); // -> 17
function prime_sum(num){
let count=0; *//tracks the no of times number is divided perfectly*
for(let i=1;i<=num;i++){ *//from 1 upto the number*
if(num%i==0){count++};
}
if(count===2){return "Prime"};
return{"Not prime"};
}
console.log(prime_sum(10));//expected output is 17**
//the code receives a number,checks through the range and returns prime if it meets the condition
The following solution uses the Eratosthenes Sieve to sum all prime numbers lower than or equal to num. The first for loop fills an array with size equal to num with true. The second for loop sets to false all non-prime numbers in the array. Then, the last for loop simply iterates through the array to sum all the array indexes i for which the value in the array, i.e., array[i], is equal to true.
/**
* Sum all primes lower or equal than n.
* Uses the Eratosthenes Sieve to find all primes under n.
*/
function sumPrimes(num) {
let array = [];
let output = 0;
// Fill an array of boolean with 'true' from 2 to n.
for (let i = 0; i <= num; i++) {
array.push(true);
}
// Set all multiples of primes to 'false' in the array.
for (let i = 2; i <= Math.sqrt(num); i++) {
if (array[i]) {
for (let j = i * i; j <= num; j += i) {
array[j] = false;
}
}
}
// All array[i] set to 'true' are primes, so we just need to add them all.
for (var i = 2; i <= num; i++) {
if (array[i]) {
output += i;
}
}
return output;
}
console.log(sumPrimes(10)); // 17
console.log(sumPrimes(977)); // 73156
console.log(sumPrimes(250_000_000)); // 197558914577
function sumPrimes(num) {
let output = 0;
// check if num is a prime number
function isPrime(num) {
for(let i = 2; i < num; i++) {
if(num % i === 0) {
return false;
}
}
return true;
}
for (let i = 2; i <= num; i++) {
if (isPrime(i)) {
output += i;
}
}
return output;
}
console.log(sumPrimes(10)); // 17
This is what I've done to get primes. I don't know if it's the most efficient, but it works. This is in Java, but can be easily converted to JavaScript. Hopefully this will help point you in the right direction.
final int TOTAL = 10;
int primes[] = new int[TOTAL];
int arrPos = 2;
boolean prime = false;
primes[0] = 2;
for (int i = 2; i < TOTAL; i++) {
prime = false;
int sqrt = (int) Math.sqrt(i);
for (int j = 1; j < arrPos && primes[j] < sqrt; j++) {
if (i % primes[j] != 0) {
prime = true;
} else {
prime = false;
break;
}
}
if (prime == true) {
primes[arrPos] = i;
arrPos++;
}
}
Related
I need to write a prime factorisation function that returns an object whose keys are a prime factor and values are the exponents of the corresponding prime factors. E.g.
console.log(primeFactorisation(4)) // { 2: 2 }
console.log(primeFactorisation(6)) // { 2: 1, 3: 1 }
My code so far is below but I'm stuck. Any help much appreciated.
const primeFactorisation = (num) => {
let result = {};
for (let i = 2; i < num; i++) {
if (num % i === 0) {
result[i] = i;
num /= i;
}
}
return result
}
When a number is found, increment it on the result object instead of assigning i to the result object.
You also need a nested loop to keep testing for the number until it no longer divides evenly, and you also need i <= num, not i < num, so that the last factor is caught:
const primeFactorisation = (num) => {
let result = {};
for (let i = 2; i <= num; i++) {
while (num % i === 0) {
result[i] = (result[i] || 0) + 1;
num /= i;
}
}
return result
}
console.log(
primeFactorisation(6),
primeFactorisation(4),
);
I wanted to create a function to check if the number is prime or not. I found a solution for this challenge and I passed it. But I can't quite understand why num 2 returns as true.
If anyone could share your knowledge, it would be great!
function checkNum(i) {
for(let j = 2 ; j < i; j++) {
if(i % j === 0) {
return false;
}
the below is the whole solution.
So.... When i is 2 in this code, why does it return true? Is it because i is not greater than j? so for loop doesn't evaluate it....? and returns true?
function sumPrimes(num) {
let sum = 0;
function checkNum(i) {
for (let j = 2; j < i; j++) {
if (i % j === 0) {
return false;
}
}
return true;
}
for (let i = 2; i <= num; i++) {
if (checkNum(i)) {
sum += i;
}
}
return sum;
}
console.log(sumPrimes(10));
I'm new to javascript and I'm working on an assignment to create a function that calculates the average using the numbers in an array. If the array is [], it should return null.
I have the following code so far but it keeps returning NaN instead of null, can someone please explain this to me? Thanks in advance.
function mean(arr) {
if (arr == []) {
return null;
} else {
var sum = 0;
var average = 0;
for (var i = 0; i < arr.length; i++) {
sum += arr[i];
}
return round(sum / arr.length);
}
}
console.log(mean([6,2,3,3,110,6,1,0,5])); //returns 4
console.log(mean([])); //currently returning NaN, instead of null
Here you go. Just change arr == [] to arr.length == 0.
function mean(arr) {
if (arr.length == 0) {
return null;
} else {
var sum = 0;
var average = 0;
for (var i = 0; i < arr.length; i++) {
sum += arr[i];
}
return Math.round(sum / arr.length);
}
}
console.log(mean([6,2,3,3,110,6,1,0,5])); //returns 4
console.log(mean([])); //currently returning NaN, instead of null
Check the length of the array using arr.length ===0
function mean(arr) {
if (arr.length === 0) {
return null;
} else {
var sum = 0;
var average = 0;
for (var i = 0; i < arr.length; i++) {
sum += arr[i];
}
return round(sum / arr.length);
}
}
console.log(mean([]));
Added additional condition to check if the variable is not an array to avoid runtime exception.
function mean(arr) {
If (arr.constructor != Array) return null;
if (arr.length == 0) {
return null;
} else {
var sum = 0;
var average = 0;
for (var i = 0; i < arr.length; i++) {
sum += arr[i];
}
return Math.round(sum / arr.length);
}
}
This should return the nth prime (n being the number given by the user). It works fine for the first few numbers (1 returns 2, 2 returns 3, 3 returns 5) but when 5 is given, it returns 9 which isn't prime (it should be 11). This happens with other numbers as well above this (7 returns 15 when it should be 17).
The 'document' stuff is to do with HTML where I'm getting the userValue and to display the prime number.
function isPrime(value) {
for(var i = 2; i < value; i++) {
if(value % i === 0) {
return false;
}
}
return value > 1;
}
function generatePrime() {
var userValue = document.getElementById("inputValue").value;
var iter = 1;
var returnValue = 2;
//checks for an integer
if (parseInt(userValue) === parseFloat(userValue)) {
//checks if the user inputted a value above 0
if (userValue > 0) {
//loops to find the correct prime
while (iter < userValue) {
if (isPrime(returnValue)) {
returnValue += 1;
iter += 1;
}
if (!isPrime(returnValue)) {
returnValue += 1;
}
}
}
else {
returnValue = "That is not a number above 0!";
}
}
else {
returnValue = "That is not a number!";
}
document.getElementById("returnValue").innerHTML = returnValue;
}
I need help with making this return the correct number.
Try this one.
function nextPrime(value) {
if (value > 2) {
var i, q;
do {
i = 3;
value += 2;
q = Math.floor(Math.sqrt(value));
while (i <= q && value % i) {
i += 2;
}
} while (i <= q);
return value;
}
return value === 2 ? 3 : 2;
}
function generatePrime() {
var userValue = document.getElementById("inputValue").value;
var value = 0, result = [];
for (var i = 0; i < userValue; i++) {
value = nextPrime(value);
result.push(value);
}
document.getElementById("returnValue").innerHTML = result[userValue-1];
}
<!DOCTYPE html>
<html>
<head>
<script>
</script>
</head>
<body>
Input value: <input type="text" name="inputValue" id="inputValue"/>
<button onclick="generatePrime()">Prime number</button>
<div id="returnValue">Test: </div>
</body>
</html>
You can try something like this:
function getNthPrimeNumber(n){
var count = 0;
var num = 2;
while(count++ != n){
num = getNextPrimeNumber(num);
}
return num;
}
function getNextPrimeNumber(n){
for(var i = ++n; i< n*n; i++){
if(isPrime(i)) return i
}
return 0;
}
function isPrime(n){
for(var i = 2; i< n; i++)
if (n%i===0)
return false;
return true;
}
console.log(getNthPrimeNumber(0))
console.log(getNthPrimeNumber(2))
console.log(getNthPrimeNumber(5))
You could reduce the range of iteration for function isPrime() :
function isPrime(value) {
for(var i = 2; i < Math.sqrt(value) + 1; i++) {
if(value % i === 0) {
return false;
}
}
return value > 1;
}
The following is based on Bhaskara Arani's submission, but I added caching.
Note: I also threw in an addition condition to check for palindrome.
This check can be removed via:
while (i <= root && value % i /* && isPalindrome(value) */)
const main = () => {
for (let i = 15; i >= 0; i--) {
console.log(`Palindrome Prime ${i + 1} = ${generatePrime(i)}`);
}
};
const cachedPrimes = [];
const generatePrime = (index) => {
let value = cachedPrimes[cachedPrimes.length - 1] ?? 0;
for (let i = cachedPrimes.length; i <= index; i++) {
value = nextPrime(value);
cachedPrimes.push(value);
}
return cachedPrimes[index];
};
const nextPrime = (value) => {
if (value > 2) {
let i, root;
do {
i = 3; value += 2; root = ~~Math.sqrt(value);
while (i <= root && value % i && isPalindrome(value)) {
i += 2;
}
} while (i <= root);
return value;
}
return value === 2 ? 3 : 2;
};
const isPalindrome = (n) => n === reverse(n);
const reverse = n => +`${n}`.split('').reverse().join('');
main();
.as-console-wrapper { top: 0; max-height: 100% !important; }
A math problem describes a list of numbers from 1-200, you must skip the number 1, and then for each number that follows, remove all multiples of that number from the list. Do this until you have reached the end of the list.
Here's what I have so far.
var x = []; // creating an array called x with zero numbers
for ( var i = 1; i <= 200; i++ ){
x.push(i);
};
// x now should have an array that contains the intergers from 1-200.
//looping through the array.
for ( var i = 0; i <= x.length; i++ ){ //going from 1-200
if (x[i] == 1){
continue; // skipping 1
} else {
for ( var n = i+1; n <= i; n++){ // take the number 1 index bigger than x[i]
if ( n % i == 0){ //find if the modulus of x[n] and x[i] is zeor, meaning it is divisible by x[i]
x.shift(); //remove that number
console.log(x[n]);
} else {
continue;
}
}
}
};
Instead of adding number 1 to 200 and then removing non prime numbers, try only putting prime numbers into that list. Since this is a school problem (I'm guessing) I don't want to give you the answer, but if you have more questions I can answer.
Also your nested loop will never run, go over that logic again.
Another version (a minute too late, as always ;-), one with comments
// lil' helper
function nextSet(a,n){
while(a[n] == 0) n++;
return n;
}
function setPrimes(a,n){
var k, j, r;
n = n + 1;
k = n;
while(k--)a[k] = 1;
a[0] = 0; // number 0
a[1] = 0; // number 1
// get rid of all even numbers
for(k = 4; k < n; k += 2) {
a[k] = 0;
}
// we don't need to check all of the numbers
// because sqrt(x)^2 = x
r = Math.floor(Math.sqrt(n));
k = 0;
while(k < n){
k = nextSet(a,k+1);
// a test if we had them all
if(k > r){
break;
}
// unmark all composites
for(j = k * k; j < n; j += 2*k){
a[j] = 0;
}
}
return a;
}
function getPrimes(n){
// we know the size of the input
var primearray = new Array(n);
// we don't know the size of the output
// prime-counting is still an unsolved problem
var output = [];
setPrimes(primearray, n);
for(var i = 0; i < n; i++){
if(primearray[i] == 1){
output.push(i);
}
}
return output;
}
getPrimes(200);
You can go through a full implementation of that algorithm at another primesieve.
Here is, I believe, a working example of what you want:
function isPrime(num){
if(num < 2){
return false;
}
for(var i=2,l=Math.sqrt(num); i<=l; i++){
if(num % i === 0){
return false;
}
}
return true;
}
function range(f, t){
for(var i=f,r=[]; i<=t; i++){
r.push(i);
}
return r;
}
function primeRange(from, to){
var a = range(from, to), r = [];
for(var i=0,l=a.length; i<l; i++){
var v = a[i];
if(isPrime(v)){
r.push(v);
}
}
return r;
}
var prmRng = primeRange(1, 200);
console.log(prmRng);
I solved it this way:
let numbers = new Array();
for (let i = 1; i <= 200; i++) {
numbers.push(i);
}
let primeNumbers = (num) => {
let prime = new Array();
for(let i = 0; i < num.length; i++) {
let count = 0;
for (let p = 2; p <= num[i]; p++) {
if(num[i] % p === 0 && num[i] !== 2) {
count++;
} else {
if(num[i] === 2 || count === 0 && num[i]-1 === p) {
prime[i] = num[i];
}
}
}
}
return prime.filter(Boolean);
}
console.log(primeNumbers(numbers));