Develop Reference

Prime Number Check Javascript

So I mede the prime number generator in JavaScript:
function prime(from, to) {
for (var i = from; i <= to; i++) {
var IsPrime = false;
for (var j = from; j < i; j++) {
if (i % j == 0) {
IsPrime = true;
}
}
if (IsPrime == false) {
console.log([i]);
}
}
}
prime(2, 9999);
But the output looks like this...
[2]
[3]
[5]
[7]
[11]
...
...and I want it to look like this:
[2,3,5,7,11...]
can someone help me?
thanks for your ideas;-)

function prime(from, to) {
var result = []; // creating a variable for accumulating
for (var i = from; i <= to; i++) {
var IsPrime = false;
for (var j = from; j < i; j++) {
if (i % j == 0) {
IsPrime = true;
}
}
if (IsPrime == false) {
// console.log([i]);
result.push(i); // adding value
}
}
console.log(result) // console.log result
return result // returning for further usage
}
var a = prime(2, 9999); // put result into a variable
console.log(a); // console.log again for example

You're logging inside your loop. instead, push the results into an array, and just log the array at the end
function prime(from, to) {
var primearray = [];
for (var i = from; i <= to; i++) {
var IsPrime = false;
for (var j = from; j < i; j++) {
if (i % j == 0) {
IsPrime = true;
}
}
if (IsPrime == false) {
primearray.push(i);
}
}
console.log(primearray)
}
prime(2, 9999);

That's because you are logging each element separately. If you would like to have the output as an array - you can create an empty array and .push() elements to it.
See the example below.
function prime(from, to) {
let primeArray = [];
for (var i = from; i <= to; i++) {
var IsPrime = false;
for (var j = from; j < i; j++) {
if (i % j == 0) {
IsPrime = true;
}
}
if (IsPrime == false) {
primeArray.push([i]);
}
}
console.log(primeArray)
}
prime(2, 9999);
Cheers.

You can just add numbers to array:
function prime(from, to) {
var arr = [];
for (var i = from; i <= to; i++) {
var IsPrime = false;
for (var j = from; j < i; j++) {
if (i % j == 0) {
IsPrime = true;
}
}
if (IsPrime == false) {
arr.push(i);
}
if(i === to) {
return arr;
}
}
}
console.log(prime(2, 10));

Try this below:
function prime(from, to) {
let output = []
for (var i = from; i <= to; i++) {
var IsPrime = false;
for (var j = from; j < i; j++) {
if (i % j == 0) {
IsPrime = true;
}
}
if (IsPrime == false) {
output.push([i]);
}
}
print(output)
}
prime(2, 9999);

finding prime numbers in an array from 1-200

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));

How to generate 100 random numbers and checking each one for primality?

Write a Boolean function named isPrime which takes an integer as an argument and returns true if the argument is a prime number of false otherwise. Generate 100 random numbers and display the results of checking each one for primality.
This is supposed to output the random numbers that are prime (after the true or false check) but I am getting the results of 2 sets of numbers in order.
Here is my code:
var arr = []
while(arr.length < 100){
var randomnumber=Math.ceil(Math.random()*100)
var found=false;
for(var i=0;i<arr.length;i++){
if(arr[i]==randomnumber){found=true;break}
}
if(!found)arr[arr.length]=randomnumber;
}
console.log(arr);
for(i = 0; i < 100; i++){
if(isPrime(i)) console.log(i);
}
function isPrime(num) {
if(num < 2) return false;
for (var i = 2; i < num; i++) {
if(num%i==0)
return false;
}
return true;
}

You need to check for primality of arr[i] instead of i:
for(i = 0; i < 100; i++){
if(isPrime(arr[i])) console.log(arr[i]);
}

function isPrime(value) {
for(var i = 2; i < value; i++) {
if(value % i === 0) {
return false;
}
}
return value > 1;
}
function myFunction() {
for(var i = 0; i < 100; i++) {
$('#demo').append('<div>'+i+'*****'+isPrime(i)+'</div>');
}
}
try it in an html page and dont forget to add jquery

The below code part, you are trying to check the numbers from 0 to 100
for(i = 0; i < 100; i++){
if(isPrime(i)) console.log(i);
}
But you should check the arr array one by one
for(i = 0; i < 100; i++){
if(isPrime(arr[i])) console.log(arr[i]);
}

Using an object to store the numbers let's you avoid the loop to check for duplicates and moving the check for isPrime () let's you skip the second loop. (Your issue was, as #Andrea pointed out, not passing arr [i] to isPrime)
var myRandObj = {};
var randomNumber = 0;
for (var i = 0; i < 100; i++) {
do {
randomNumber = Math.ceil(Math.random() * 100);
} while (typeof myRandObj[randomNumber] !== 'undefined');
myRandObj[randomNumber] = 0;
if (isPrime (randomNumber))
console.log (randomNumber);
}

console.log(arr) is printing the entire array to the console. Remove that line to debug further.
var arr = []
for(x = 0; x < 100; x++){
arr[x] = Math.ceil(Math.random()*100)
}
for(i = 0; i < 100; i++){
if(isPrime(arr[i])) console.log(arr[i]);
}
function isPrime(num) {
if(num < 2) return false;
for (var i = 2; i < num; i++) {
if(num%i==0)
return false;
}
return true;
}

How do I sum all prime numbers?

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++;
}
}

Why isn't this simple javascript code working?

The code is supposed to check whether the first few triangular numbers are prime (they are not), but it does not run.
<!DOCTYPE HTML>
<html>
<head><title>C1E9P1</title>
<script>
for(i = 3; i < 13; i++){
prime = true;
n = (i*(i+1))/2;
for(i = 2; i < Math.sqrt(n)+1; i++){
if(n%i == 0){
prime = false;
}
}
if(prime){
document.write(n+" is prime.");
document.write("<br>");
} else {
document.write(n+" is composite.");
document.write("<br>");
}
}
</script>
</head>
</html>

You used same variable for your two for loops that are nested!
for(i = 3; i < 13; i++){ <-- i
for(i = 2; i < Math.sqrt(n)+1; i++){ <-- i

Try this and look at your JavaScript console :
for(var i = 3; i < 13; i++){
var prime = true;
var n = (i*(i+1))/2;
for(var j = 2; j < Math.sqrt(n)+1; j++){
if(n%j == 0){
prime = false;
}
}
if(prime){
console.log(n+" is prime.");
console.log("<br>");
} else {
console.log(n+" is composite.");
console.log("<br>");
}
}

I've modified script, but don't known if algorithm is correct.
for(var i = 3; i < 13; i++){
var prime = true;
var n = (i*(i+1))/2;
for(var j = 2; j < Math.sqrt(n)+1; j++){
if(n%j == 0){
prime = false;
}
}
if(prime){
document.write(n+" is prime.");
document.write("<br>");
} else {
document.write(n+" is composite.");
document.write("<br>");
}
}

You should change the inner for-loop's variable name to something else other than i.
for(i = 3; i < 13; i++){
prime = true;
n = (i*(i+1))/2;
for(j = 2; j < Math.sqrt(n)+1; j++){
if(n%j == 0){
prime = false;
}
}
You should be all good after that.