Develop Reference

Sum All Primes Below a Given Number | Intermediate Javascript Algorithm | Recursion

Question
A prime number is a whole number greater than 1 with exactly two divisors: 1 and itself. For example, 2 is a prime number because it is only divisible by 1 and 2. In contrast, 4 is not prime since it is divisible by 1, 2 and 4.
Rewrite sumPrimes so it returns the sum of all prime numbers that are less than or equal to num.
My Attempt
const isPrime = a => {
for(let i = 2; i < a; i++)
if(num % i === 0) return false;
return a > 1;
}
function sumPrimes(num, total = []) {
let numVar = num;
let n = total.reduce((aggregate, item)=>{
return aggregate + item;
}, 0);
if(n > numVar){
return n;
}
for(let i = 1; i <= numVar; i++){
if(isPrime(i)== true){
total.push(i);
}
}
return sumPrimes(num, total);
}
sumPrimes(10);
The Problem
It says: 'Num is not defined'
I am not sure if there are other errors.
My Question
Please could you help me find the error, and fix the code to solve the algorithm?

This was a simple syntax error identified by #Jonas Wilms (upvote him in the comment above :))!
By replacing the 'a' with 'num' the function was fixed.
const isPrime = a => {
for(let i = 2; i < a; i++)
if(a % i === 0) return false;
return a > 1;
}
function sumPrimes(num, total = []) {
let numVar = num;
let n = total.reduce((aggregate, item)=>{
return aggregate + item;
}, 0);
if(n > numVar){
return n;
}
for(let i = 1; i <= numVar; i++){
if(isPrime(i)== true){
total.push(i);
}
}
return sumPrimes(num, total);
}
console.log(sumPrimes(10));

this simple function may help you,
function isPrime(num) {
for (var i = 2; i < num; i++)
if (num % i === 0) return false;
return num > 1;
}
function sumPrimes(num) {
let tot = 0;
for (let i = 0; i < num; i++)
if (isPrime(i))
tot += i;
return tot;
}
console.log(sumPrimes(10));

Finding sequence of prime numbers with JS

I have to get a sequence of prime numbers. But my code does not work. How is it possible to fix it?
var num1 = parseInt(prompt('Enter a number'));
var num2 = parseInt(prompt('Enter a number'));
var num3 = 0;
function primeSeq(num1, num2) {
var b = 1;
var c = '';
if (num1 > num2) {
num3 = num2;
num2 = num1;
num1 = num3;
}
for (var i = num1; i < num2; i++) {
for (var j = 2; j < i; j++) {
if (i % j == 0) {
b++;
}
if (b <= 1) {
c += i + ' ';
}
}
}
return c;
}
alert(primeSeq(num1, num2));

I guess you wanted something like this
var num1 = parseInt(prompt('Enter a number'));
var num2 = parseInt(prompt('Enter a number'));
var num3 = 0;
if (num1 > num2) {
num3 = num2;
num2 = num1;
num1 = num3;
}
function primeSeq(num1, num2) {
var b;
var c = '';
for (var i = num1; i < num2; i++) {
b = 1;
for (var j = 2; j < i; j++) {
if (i % j === 0) {
b++;
}
}
if (b === 1) {
c += i + ' ';
}
}
return c;
}
alert(primeSeq(num1, num2));
So in short, b should reset to 1 on every new prime candidate (i loop) and check of b should be outside of inner (j) loop.
Please note that there are more optimal algorithms.

The lot easier way is to use a sieve system, if a number is divisible by another prime number it is not a prime. You can write a function like this:
function primes(max) {
let primes = [2];
for (let i = 3; i <= max; i++) {
let found = true;
for (let j = 0; j < primes.length; j++) {
if (i % primes[j] == 0) found = false;
}
if (found) primes.push(i);
}
return primes;
}
Explanation
What you know by default is that 2 is a prime, so you start at 3. You don't want to exeed the max, that is the i <= max statement. Assume it is a prime, then search in the array if it is divisible by primes you found before, if that is the case, set found to false.
Now check if is was found, push is to the array and return the primes.

Here is bit more optimized algorithm:
function primeSeq(num1, num2) {
var primes = [];
var isPrime;
var j;
var results = [];
for (var i = 2; i < num2; i++) {
isPrime = true;
j = 0;
while (j < primes.length) {
if (i % primes[j] === 0) {
isPrime = false;
break;
}
j++;
}
if (isPrime) {
primes.push(i);
if (i >= num1) {
results.push(i);
}
}
}
return results.join(' ');
}
In order for number to be prime it must not be dividable with all the smaller primes, so we are generating an array of primes to check upon.
There is a theory also that every prime bigger than 3 has a following form:
6k+1 or 6k-1
So this would simplify it bit more.

Try this one
<input ng-model="range" type="number" placeholder="Enter the range">
<button ng-click="findPrimeNumber(range)">
$scope.findPrimeNumber=function(range){
var tempArray=[];
for(var i=0;i<=range;i++){
tempArray.push(i)
}
var primenumbers= tempArray.filter((number) => {
for (var i = 2; i <= Math.sqrt($scope.range); i++) {
if (number % i === 0) return false;
}
return true;
});
console.log(primenumbers);
}

Factorialize a Number

I'm taking the freecodecamp course one of the exercises it's to create a Factorialize function, I know there is several ways to do it just not sure what this one keeps returning 5
function factorialize(num) {
var myMax = num;
var myCounter = 1;
var myTotal = 0;
for (i = 0; i>= myMax; i++) {
num = myCounter * (myCounter + 1);
myCounter++;
}
return num;
}
factorialize(5);

This is a recursive solution of your problem:
function factorialize(num) {
if(num <= 1) {
return num
} else {
return num * factorialize(num-1)
}
}
factorialize(5)
This is the iterative solution:
function factorialize(num) {
var cnt = 1;
for (var i = 1; i <= num ; i++) {
cnt *= i;
}
return cnt;
}
factorialize(5)
with argument 5, it will return the 5! or 120.

To answer your question, why your function is returning 5:
Your function never reaches the inner part of the for-loop because your testing if i is greater than myMax instead of less than.
So you are just returning your input parameter which is five.
But the loop does not calculate the factorial of num, it only multiplies (num+1) with (num+2);

My solution in compliance with convention for empty product
function factorializer(int) {
if (int <= 1) {
return 1;
} else {
return int * factorializer(int - 1);
}
}

Here is another way to solve this challenge and I know it is neither the shortest nor the easiest but it is still a valid way.
function factorialiaze(num){
var myArr = []; //declaring an array.
if(num === 0 || num === 1){
return 1;
}
if (num < 0){ //for negative numbers.
return "N/A";
}
for (var i = 1; i <= num; i++){ // creating an array.
myArr.push(i);
}
// Reducing myArr to a single value via .reduce:
num = myArr.reduce(function(a,b){
return a * b;
});
return num;
}
factorialiaze(5);

Maybe you consider another approach.
This solution features a very short - cut to show what is possible to get with an recursive style and a implicit type conversion:
function f(n) { return +!~-n || n * f(n - 1); }
+ convert to number
! not
~ not bitwise
- negative
function f(n) { return +!~-n || n * f(n - 1); }
var i;
for (i = 1; i < 20; i++) {
console.log(f(i));
}
.as-console-wrapper { max-height: 100% !important; top: 0; }

Try this function
const factorialize = (num) => num === 0 ? 1 : num * factorialize(num-1)
Use it like this:
factorialize(5) // returns 120

Try this :
function factorialize(num) {
var value = 1;
if(num === 1 || num ===0) {
return value;
} else {
for(var i = 1; i<num; i++) {
value *= i;
}
return num * value;
}
}
factorialize(5);

// My solution
const factorialize = num => {
let newNum = 1;
for (let i = 1; i <= num; i++) {
newNum *= i
}
return newNum;
}

I love syntactic sugar, so
let factorialize = num => num <= 1 ? num : num * factorialize(num -1)
factorialize(5)

Create range of letters and numbers

I'm creating a form where users can input a range. They are allowed to input letters and numbers. Some sample input:
From: AA01
To: AZ02
Which should result in:
AA01
AA02
AB01
AB02
And so on, till AZ02
And:
From: BC01
To: DE01
Should result in:
BC01
BD01
BE01
CC01
CD01
CE01
Etc
I managed to get it working for the input A01 to D10 (for example)
jsFiddle
However, i can't get it to work with multiple letters.
JS code:
var $from = $('input[name="from"]');
var $to = $('input[name="to"]');
var $quantity = $('input[name="quantity"]');
var $rangeList = $('.rangeList');
var $leadingzeros = $('input[name="leadingzeros"]');
$from.on('keyup blur', function () {
$(this).val($(this).val().replace(/[^a-zA-Z0-9]/g, ''));
updateQuantity();
});
$to.on('keyup blur', function () {
$(this).val($(this).val().replace(/[^a-zA-Z0-9]/g, ''));
updateQuantity();
});
$leadingzeros.on('click', function () {
updateQuantity();
});
function updateQuantity() {
var x = parseInt($from.val().match(/\d+/));
var y = parseInt($to.val().match(/\d+/));
var xl = $from.val().match(/[a-zA-Z]+/);
var yl = $to.val().match(/[a-zA-Z]+/);
var result = new Array();
if (xl != null && yl != null && xl[0].length > 0 && yl[0].length > 0) {
xl = xl[0].toUpperCase();
yl = yl[0].toUpperCase();
$rangeList.html('');
var a = yl.charCodeAt(0) - xl.charCodeAt(0);
for (var i = 0; i <= a; i++) {
if (!isNaN(x) && !isNaN(y)) {
if (x <= y) {
var z = (y - x) + 1;
$quantity.val(z * (a + 1));
$rangeList.html('');
for (var b = z; b > 0; b--) {
var c = ((y - b) + 1);
if ($leadingzeros.prop('checked')) {
c = leadingZeroes(c, y.toString().length);
}
result.push(String.fromCharCode(65 + i) + c);
}
} else {
$rangeList.html('');
$quantity.val(0);
}
} else {
$rangeList.html('');
$quantity.val(0);
}
}
} else if (!isNaN(x) && !isNaN(y)) {
if (x < y) {
var z = (y - x) + 1;
$quantity.val(z);
$rangeList.html('');
for (var i = z; i > 0; i--) {
var c = (y - i) + 1;
if ($leadingzeros.prop('checked')) {
c = leadingZeroes(c, y.toString().length);
}
result.push(c);
}
} else {
$rangeList.html('');
$quantity.val(0);
}
} else {
$rangeList.html('');
$quantity.val(0);
}
$rangeList.html('');
for (var i = 0; i < result.length; i++) {
$rangeList.append(result[i] + '<br />');
}
}
function leadingZeroes(number, size) {
number = number.toString();
while (number.length < size) number = "0" + number;
return number;
}

This is perfect for a recursive algorithm:
function createRange(from, to) {
if (from.length === 0) {
return [ "" ];
}
var result = [];
var innerRange = createRange(from.substring(1), to.substring(1));
for (var i = from.charCodeAt(0); i <= to.charCodeAt(0); i++) {
for (var j = 0; j < innerRange.length; j++) {
result.push(String.fromCharCode(i) + innerRange[j]);
}
}
return result;
}
Called as follows:
createRange('BC01', 'DE02'); // Generates an array containing all values expected
EDIT: Amended function below to match new test case (much more messy, however, involving lots of type coercion between strings and integers).
function prefixZeroes(value, digits) {
var result = '';
value = value.toString();
for (var i = 0; i < digits - value.length; i++) {
result += '0';
}
return result + value;
}
function createRange(from, to) {
if (from.length === 0) {
return [ "" ];
}
var result = [];
if (from.charCodeAt(0) < 65) {
fromInt = parseInt(from);
toInt = parseInt(to);
length = toInt.toString().length;
var innerRange = createRange(from.substring(length), to.substring(length));
for (var i = fromInt; i <= toInt; i++) {
for (var j = 0; j < innerRange.length; j++) {
result.push(prefixZeroes(i, length) + innerRange[j]);
}
}
} else {
var innerRange = createRange(from.substring(1), to.substring(1));
for (var i = from.charCodeAt(0); i <= to.charCodeAt(0); i++) {
for (var j = 0; j < innerRange.length; j++) {
result.push(String.fromCharCode(i) + innerRange[j]);
}
}
}
return result;
}

Please note that because of your strict logic in how the value increments this method requires exactly 4 characters (2 letters followed by 2 numbers) to work. Also, this might not be as efficient/tidy as it can be but it took some tinkering to meet your logic requirements.
function generate(start, end) {
var results = [];
//break out the start/end letters/numbers so that we can increment them seperately
var startLetters = start[0] + start[1];
var endLetters = end[0] + end[1];
var startNumber = Number(start[2] + start[3]);
var endNumber = Number(end[2] + end[3]);
//store the start letter/number so we no which value to reset the counter to when a maximum boundry in reached
var resetLetter = startLetters[1];
var resetNumber = startNumber;
//add first result as we will always have at least one
results.push(startLetters + (startNumber < 10 ? "0" + startNumber : "" + startNumber));
//maximum while loops for saefty, increase if needed
var whileSafety = 10000;
while (true) {
//safety check to ensure while loop doesn't go infinite
whileSafety--;
if (whileSafety == 0) break;
//check if we have reached the maximum value, if so stop the loop (break)
if (startNumber == endNumber && startLetters == endLetters) break;
//check if we have reached the maximum number. If so, and the letters limit is not reached
//then reset the number and increment the letters by 1
if (startNumber == endNumber && startLetters != endLetters) {
//reset the number counter
startNumber = resetNumber;
//if the second letter is at the limit then reset it and increment the first letter,
//otherwise increment the second letter and continue
if (startLetters[1] == endLetters[1]) {
startLetters = '' + String.fromCharCode(startLetters.charCodeAt(0) + 1) + resetLetter;
} else {
startLetters = startLetters[0] + String.fromCharCode(startLetters.charCodeAt(1) + 1);
}
} else {
//number limit not reached so just increment the number counter
startNumber++;
}
//add the next sequential value to the array
results.push(startLetters + (startNumber < 10 ? "0" + startNumber : "" + startNumber));
}
return results;
}
var results = generate("BC01", "DE01");
console.log(results);
Here is a working example, which uses your second test case

Using #Phylogenesis' code, i managed to achieve my goal.
jsFiddle demo
function updateQuantity() {
var x = parseInt($from.val().match(/\d+/));
var y = parseInt($to.val().match(/\d+/));
var xl = $from.val().match(/[a-zA-Z]+/);
var yl = $to.val().match(/[a-zA-Z]+/);
var result = new Array();
var r = createRange(xl[0], yl[0]);
var z = (y - x) + 1;
if (x <= y) {
for (var j = 0; j < r.length; j++) {
var letters = r[j];
for (var i = z; i > 0; i--) {
var c = (y - i) + 1;
if ($leadingzeros.prop('checked')) {
c = leadingZeroes(c, y.toString().length);
}
if (i == z) {
r[j] = letters + c + '<br />';
} else {
j++;
r.splice(j, 0, letters + c + '<br />');
}
}
}
} else {
for (var i = 0; i < r.length; i++) {
r[i] += '<br />';
}
}
$quantity.val(r.length);
$rangeList.html('');
for (var i = 0; i < r.length; i++) {
$rangeList.append(r[i]);
}
}
This works for unlimited letters and numbers, as long as the letters are first.
Thanks for your help!

Javascript Loto Game

How can I check for matching numbers in this script, stuck here, I need to compare the array of user numbers with the array of lotto numbers and display how many numbers they got correct if any along with their prize value.
function numbers() {
var numbercount = 6;
var maxnumbers = 40;
var ok = 1;
r = new Array(numbercount);
for (var i = 1; i <= numbercount; i++) {
r[i] = Math.round(Math.random() * (maxnumbers - 1)) + 1;
}
for (var i = numbercount; i >= 1; i--) {
for (var j = numbercount; j >= 1; j--) {
if ((i != j) && (r[i] == r[j])) ok = 0;
}
}
if (ok) {
var output = "";
for (var k = 1; k <= numbercount; k++) {
output += r[k] + ", ";
}
document.lotto.results.value = output;
} else numbers();
}
function userNumbers() {
var usersNumbers = new Array(5);
for (var count = 0; count <= 5; count++) {
usersNumbers[count] = window.prompt("Enter your number " + (count + 1) + ": ");
}
document.lotto.usersNumbers.value = usersNumbers;
}

Here is a lotto numbers generator and a scoring system. I'm going to leave it to you to validate the user input.
function lottoGen(){
var lottoNumbers = [];
for(var k = 0; k<6; k++){
var num = Math.floor(Math.random()*41);
if(lottoNumbers.indexOf(num) != -1){
lottoNumbers.push(num);
}
}
return lottoNumbers;
}
function scoreIt(){
var usersNumbers = document.getElementsByName('usersNumbers').item(0);
usersNumbers = String(usersNumbers)
usersNumbers = usersNumbers.split(' ');
var matches = 0;
for(var i = 0; i<6; i++){
if(lottoNumbers.indexOf(usersNumbers[i]) != -1){matches++;}
}
return matches;
}

Hi I'm new to this and trying to learn off my own back so obviously I'm no expert but the code above makes a lot of sense to me, apart from the fact I can't get it to work.. I tried to console.log where it says RETURN so I could see the numbers but it just shows an empty array still. I assumed this was to do with it being outside the loop..
I've tried various ways but the best I get is an array that loops the same number or an array with 6 numbers but some of which are repeated..
function lottoGen(){
var lottoNumbers = [];
for(var k = 0; k<6; k++){
var num = Math.floor(Math.random()*41);
if(lottoNumbers.indexOf(num) != -1){
lottoNumbers.push(num);
}
}
return lottoNumbers;
}

Lotto JS: CODEPEN DEMO >> HERE <<
(function(){
var btn = document.querySelector("button");
var output = document.querySelector("#result");
function getRandom(min, max){
return Math.round(Math.random() * (max - min) + min);
}
function showRandomNUmbers(){
var numbers = [],
random;
for(var i = 0; i < 6; i++){
random = getRandom(1, 49);
while(numbers.indexOf(random) !== -1){
console.log("upps (" + random + ") it is in already.");
random = getRandom(1, 49);
console.log("replaced with: (" + random + ").");
}
numbers.push(random);
}
output.value = numbers.join(", ");
}
btn.onclick = showRandomNUmbers;
})();