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Javascript Function Issues for Prime Numbers

So what I'm trying to do is use a function that prompts to enter a number and then prints all prime numbers up till that number. I have the code as far as I know but I keep getting an error that the bracket following prime(num) is wrong. Not sure what I'm doing wrong here, would appreciate any help.
function p5Bela() {
var num = prompt("Please enter a number: ", "");
for (i = 2; i <= num; i++) {
if (prime(i)==true) {
alert(i);
}
}
prime(num) {
var flag = true;
var d = 2;
while (flag==true && d <= num/2) {
if (num%d == 0) {
flag = false;
}
d++;
}
return flag;
}
}

Simplest Way
function isPrime(num) {
for(var i = 2; i < num; i++)
if(num % i === 0) return false;
return num > 1;
}
With the ES6 syntax:
const isPrime = num => {
for(let i = 2; i < num; i++)
if(num % i === 0) return false;
return num > 1;
}
You can also decrease the complexity of the algorithm from O(n) to O(sqrt(n)) if you run the loop until square root of a number:
const isPrime = num => {
for(let i = 2, s = Math.sqrt(num); i <= s; i++)
if(num % i === 0) return false;
return num > 1;
}

count 9's from 1 to n - 5kyu Kata

I am working on this kata https://www.codewars.com/kata/count-9-s-from-1-to-n/train/javascript
and i have written this code for it, but its not working. This question is similar to this one Count the number of occurrences of 0's in integers from 1 to N
but it is different because searching for 9's is practically very different to searching for 0's.
think part of the problem with this code is that it takes too long to run...
any advice appreciated!
function has9(n) {
var nine = [];
var ninearr = n.toString().split('');
for (var j = 0; j < ninearr.length; j++) {
if (ninearr[j] == '9') {
nine.push(ninearr[j]);
}
}
return nine.length;
}
function number9(n) {
var arr = [];
var arrnew = [];
for (var i = 0; i <= n; i++) {
arr.push(i);
}
for (var l = 0; l < arr.length; l++) {
arrnew.push(has9(l));
}
var sum = arrnew.reduce((a, b) => a + b, 0);
return sum;
}

Why not a regex based solution? (Too slow as well?)
const count9s = num => num.toString().match(/9/g).length
console.log(count9s(19716541879)) // 2
console.log(count9s(919191919191919)) // 8
console.log(count9s(999)) // 3
console.log(count9s(999999)) // 6

I have taken the above hint and completely re written the code, which I now feel should work, and it does for most inputs, but codewars is saying it fails on some of them. any ideas why?
function nines(n){
if(n>=100){
var q= Math.floor(n/100);
var nq= q * 20;
var r = (n%100);
var s = Math.floor(r/9);
if (r<=90){
return s + nq;
}
if (r == 99){
return 20 + nq;
}
if (90 < r < 100 && r!= 99){
var t = (r-90);
return nq + s + t;
}
}
if (n<100){
if (n<=90){
var a = Math.floor(n/9);
return a ;
}
if (n == 99){
return 20
}
if (90 < n < 100 && n!= 99){
var c = (n-90);
return 10 + c;
}
}
}

=== UPDATE ===
I just solved your kata using
function number9Helper(num) {
var pow = Math.floor(Math.log10(num));
var round = Math.pow(10, pow);
var times = Math.floor(num / round);
var rest = Math.abs(num - (round * times));
var res = pow * (round==10 ? 1 : round / 10) * times;
if (num.toString()[0] == '9') res += rest;
if (rest < 9) return res;
else return res + number9Helper(rest);
}
function number9(num) {
var res = number9Helper(num);
res = res + (num.toString().split('9').length-1);
return res;
}
== Function below works but is slow ===
So, could something like this work for you:
for (var nines=0, i=1; i<=n; i++) nines += i.toString().split('9').length-1;
Basically, there are many way to achieve what you need, in the end it all depends how do you want to approach it.
You can test it with
function nines(n) {
for (var nines=0, i=1; i<=n; i++) nines += i.toString().split('9').length-1;
return nines;
}

function number9(n) {
if (n < 8) {
return 0
};
if (n === 9) {
return 1
};
if (n > 10) {
let str = ''
for (let i = 9; i <= n; i++) {
str += String(i)
}
return str.match(/[9]/g).length
}
}

finding the nth prime javascript

the functions below are supposed to spit out the nth prime number. However, it keeps on spitting out 3. Can somebody please help? Cheers, Anthony
function Prime(num) {
output = true
for (i=2 ; i<num ; i++) {
if (num%i === 0) {
output = false ; break
}
}
return output
}
function PrimeMover(num) {
var count = 0
for (i=2 ; i<10000 ; i++) {
if (Prime(i) === true) {
count = count + 1
}
if (count === num) {
return i
break
}
}
}

You have created loop counter i in global scope.so both PrimeMover and Prime mutates same global i.In every iteration ,PrimeMover assigns i=2.After that Prime assigns i=2.your i variable's value will be changed between 2 and 3.use local loop counter variable var i=0;
function Prime(num) {
output = true
for (var i=2 ; i<num ; i++) { //var i=2
if (num%i === 0) {
output = false ; break
}
}
return output
}
function PrimeMover(num) {
var count = 0
for (var i=2 ; i<10000 ; i++) { //var i=2
if (Prime(i) === true) {
count = count + 1
}
if (count === num) {
return i
break
}
}
}

For minimal code lovers,
function nthprime(n)
{
var prime=[], i=1
while (i++ && prime.length<n) prime.reduce((a,c)=>(i%c)*a,2) && prime.push(i)
return prime.length?prime.pop():-1
}
[-1,0,1,2,3,5,10,100].forEach(n=>console.log(`nthprime(${n})=${nthprime(n)}`))

function main(inp) {
var count = 0;
for (var i = 2; i <= 100000; i++) {
if (isPrime(i)) count = count + 1;
if (count == inp) return i;
}
}
function isPrime(i) {
for (var j = 2; j < i; j++) {
//instead of `j < i` it can be reduced using other conditions
if (i % j == 0) {
return false
}
}
return true
}
main(5) // any number

This might be a bit more optimal
function nthPrime(n) {
var P = 0;
function isPrime(x) {
var isPrime= true;
for (var d = 2; d <= Math.sqrt(x); d++) {
if((x/d) % 1 == 0) {
isPrime = false;
break;
}
}
return isPrime;
}
for (var i = 1; 0 < n; i++) {
if(isPrime(i)) {
P = i; n--;
}
// we can skip the even numbers
if(3 <= i){
i++;
}
}
return P;
}

Try this
var pos=10001;
console.log(primeNumforPos(pos));
function primeNumforPos(pos){
var num=2,curPos=0;
while(curPos<=pos){
if(isPrime(num)){
curPos++;
}
if(curPos==pos){
return num;
}else{
num++;
}
}
}
function isPrime(num){
for(var i=2;i<=Math.sqrt(num);i++){
if(num%i==0){
return false;
}
}
return true;
}

So, I decided to optimise the hell out of the code (cuz why not). It is almost 6 times as fast as that of ppseprus (297ms vs 1773ms in nth_prime(100000)).
let primes = [2, 3];
const nth_prime = (n) => {
if (n <= primes.length) return primes[n - 1]; // handle values which have been cached
let i = 1;
while (1){
const a = 6 * i - 1, b = 6 * i + 1, a_limit = Math.sqrt(a), b_limit = Math.sqrt(b); // the 6n - 1 and 6n + 1 rule for primes
let a_prime = true, b_prime = true;
i++;
// prime check
for (const prime of primes){
if (prime > a_limit) break;
if (a % prime == 0){
a_prime = false;
break;
}
}
if (a_prime){
if (primes.length + 1 == n) return a;
primes.push(a); // cache
}
for (const prime of primes){
if (prime > b_limit) break;
if (b % prime == 0){
b_prime = false;
break;
}
}
if (b_prime){
if (primes.length + 1 == n) return b;
primes.push(b); // cache
}
}
}

const findPrime = num => {
let i, primes = [2, 3], n = 5
const isPrime = n => {
let i = 1, p = primes[i],
limit = Math.ceil(Math.sqrt(n))
while (p <= limit) {
if (n % p === 0) {
return false
}
i += 1
p = primes[i]
}
return true
}
for (i = 2; i <= num; i += 1) {
while (!isPrime(n)) {
n += 2
}
primes.push(n)
n += 2
};
return primes[num - 1]
}
console.time('Time')
let x = findPrime(9999)
console.timeEnd('Time')
console.log(x)

for loop number sequence of (1,1,2,2,3,3, etc.)

I looked it up and this pattern is Hofstadter Female sequence. The equations are:
M(n) = n-F(M(n-1))
F(n) = n-M(F(n-1))
but I'm not sure how to put that into code.
So far I have:
while () {
_p++
_r++
if (_p % 2 === 0) {
_r = _p - 1;
}
}
Any help?

Without memoization:
function F(n)
{
return 0 < n ? n - M(F(n-1)) : 1
}
function M(n)
{
return 0 < n ? n - F(M(n-1)) : 0
}
var N = 10;
var f = [];
var m = [];
for (var i = 0; i <= N; ++i) {
f.push(F(i));
m.push(M(i));
}
console.log('F: ' + f.join(','))
console.log('M: ' + m.join(','))
Output:
F: 1,1,2,2,3,3,4,5,5,6,6
M: 0,0,1,2,2,3,4,4,5,6,6
http://jsfiddle.net/KtGBg/1/

Recursion should be avoided, if possible, so you can cache the already-calculated values for F(n) and M(n) :
var f = new Array();
var m = new Array();
function F(n){
if(f[n] != undefined) {
return f[n];
}
if (n==0) {
value = 1;
} else {
value = n - M(F(n-1));
}
f[n] = value;
return value;
}
function M(n){
if(m[n] != undefined) {
return m[n];
}
if (n==0) {
value = 0;
} else {
value = n - F(M(n-1));
}
m[n] = value;
return value;
}
This yields a much faster result for greater numbers (try it with 10000)

how about:
function F(n){
if (n==0) return 1
else return n - M(F(n-1))
}
function M(n){
if (n==0) return 0
else return n - F(M(n-1))
}
var str = ""
for(var i=0; i<=10; i++) str += F(i) + ", "
console.log(str.substr(0,str.length-2))

Similar to GaborSch's answer, you could use Doug Crockford's memoizer function, which can be found in Chapter 4 of Javascript: The Good Parts. Using memoization took the calculation time for the first 150 terms of the male and female Hofstadter sequences down to 256 ms as compared to almost 8 seconds without memoization.
var memoizer = function (memo, formula) {
var recur = function (n) {
var result = memo[n];
if (typeof result !== 'number') {
result = formula(recur, n);
memo[n] = result;
}
return result;
};
return recur;
};
var maleHofstadter = memoizer([0], function (recur, n) {
return n - femaleHofstadter(recur(n-1));
});
var femaleHofstadter = memoizer([1], function (recur, n) {
return n - maleHofstadter(recur(n-1));
});
var N = 150;
var f = [];
var m = [];
for (var i = 0; i <= N; ++i) {
f.push(femaleHofstadter(i));
m.push(maleHofstadter(i));
}
console.log('F: ' + f.join(','));
console.log('M: ' + m.join(','));

Prime Number Simple Test

the following code always seems to return true whatever value of n I plug in and I can't see why. If n = 8, then arr2 should contain the value of i for 2, 4 and 8?
Can someone please explain? Thank you very much.
var primetest = function(n){
var divisor = [2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20];
for (var i = 0; i < divisor.length; i++) {
var arr2 = [];
if(n%divisor[i] == 0) {arr2.push(i);}
if(arr2.length > 1) {prime = false;}
else {prime = true;}
return prime;
};
};

Couple of problems in your original code:
You are doing only one iteration
You declare the array holding divisions inside the loop, causing it to always have maximum 1 item.
Quick fix of the above would be:
var primetest = function(n){
var divisor = [2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20];
var arr2 = [];
for (var i = 0; i < divisor.length; i++) {
if(n%divisor[i] == 0) {
arr2.push(i);
}
}
return arr2.length <=1;
}
Live test case.
Optimized code that does not iterate through the whole list of divisors if not reuired (guess that's what you were trying to achieve) is:
var primetest = function(n){
var divisor = [2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20];
var arr2 = [];
for (var i = 0; i < divisor.length; i++) {
if(n % divisor[i] === 0) {
arr2.push(i);
if (arr2.length > 1)
return false;
}
}
return true;
};
Updated fiddle.

Two problems:
Don't declare you array inside the loop, because it will reset in every loop.
And move your return outside of the loop because your function will return at first loop, wich always true for odd numbers(n%2 = 0 for even numbers).
JSFiddle
function primetest(n){
var divisor = [2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20];
var arr2 = [];
for (var i = 0; i < divisor.length; i++) {
if((n%divisor[i]) === 0)
arr2.push(i);
if(arr2.length > 1)
prime = false;
else prime = true;
};
return prime;
}
$(function (){
var pprime = 7;
if(primetest(pprime))
alert("ok");
});

Return the first 1001 positions into an array of prime numbers.
Usefull code to modify, enjoy!
//cum calculam daca un numar este prim
//chiar mai mult - care este al 1001-lea nr prim
function isPrime(num) {
if(num < 2) return false;
for (var i = 2; i < num; i++) {
if(num%i===0)
return false;
}
return true;
}
var shir=[];
var j=0;
var i=1;
while(j<1001)
{
if(isPrime(i))
{shir[j]=i;
j++;i++}
else{i++}
}
console.log(shir);
console.log (shir[0]);
console.log (shir[1000]);
alert("Al 1001-lea numar prim este " + shir[1000]);

Your condition
if(arr2.length > 1) {prime = false;}
is checking for >1 instead of >0 or >=1 which is ignoring the prime condition that you specified. Hence your corrected code would be something like this:
var primetest = function(n){
var divisor = [2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20];
for (var i = 0; i < divisor.length; i++) {
var arr2 = [];
if(n%divisor[i] == 0) {arr2.push(i);}
if(arr2.length > 0) {prime = false;}
else {prime = true;}
return prime;
};
};
if(arr2.length > 1) {prime = false;}
There are better ways to check for prime number. Here's the fastest one I could lay my hands on:
var primetest = function(n) {
if (isNaN(n) || !isFinite(n) || n%1 || n<2) return false;
if (n==leastFactor(n)) return true;
return false;
}
// leastFactor(n)
// returns the smallest prime that divides n
// NaN if n is NaN or Infinity
// 0 if n=0
// 1 if n=1, n=-1, or n is not an integer
leastFactor = function(n){
if (isNaN(n) || !isFinite(n)) return NaN;
if (n==0) return 0;
if (n%1 || n*n<2) return 1;
if (n%2==0) return 2;
if (n%3==0) return 3;
if (n%5==0) return 5;
var m = Math.sqrt(n);
for (var i=7;i<=m;i+=30) {
if (n%i==0) return i;
if (n%(i+4)==0) return i+4;
if (n%(i+6)==0) return i+6;
if (n%(i+10)==0) return i+10;
if (n%(i+12)==0) return i+12;
if (n%(i+16)==0) return i+16;
if (n%(i+22)==0) return i+22;
if (n%(i+24)==0) return i+24;
}
return n;
}
Source: http://www.javascripter.net/faq/numberisprime.htm