const primes = num => {
  let arr = Array.from({ length: num - 1 }).map((x, i) => i + 2),
    sqroot = Math.floor(Math.sqrt(num)),
    numsTillSqroot = Array.from({ length: sqroot - 1 }).map((x, i) => i + 2);
  numsTillSqroot.forEach(x => (arr = arr.filter(y => y % x !== 0 || y === x)));
  return arr;
};

primes(10); // [2, 3, 5, 7]

class SequentialQueue{	//if you want it to go backwards, too bad
	next(){
		return this.i++;
	}
	constructor(start = 0){
		this.i = start;
	}
}
const que = new SequentialQueue(0);
for(let i = 0; i < 10; i++){
	console.log(que.next());
}

# Acronyms

## Computer Numerical Control (CNC)

- Typicalldfdfsdffdafdasgfdgfgfdfdafdasfdafda

// There are n rings and each ring is either red, green, or blue. The rings are distributed across ten rods labeled from 0 to 9.

// You are given a string rings of length 2n that describes the n rings that are placed onto the rods. Every two characters in rings forms a color-position pair that is used to describe each ring where:

// The first character of the ith pair denotes the ith ring's color ('R', 'G', 'B').
// The second character of the ith pair denotes the rod that the ith ring is placed on ('0' to '9').
// For example, "R3G2B1" describes n == 3 rings: a red ring placed onto the rod labeled 3, a green ring placed onto the rod labeled 2, and a blue ring placed onto the rod labeled 1.

// Return the number of rods that have all three colors of rings on them.


let rings = "B0B6G0R6R0R6G9";
var countPoints = function(rings) {
    
    let sum = 0;
    
	// Always 10 Rods
    for (let i = 0; i < 10; i++) {
        if (rings.includes(`B${i}`) && rings.includes(`G${i}`) && rings.includes(`R${i}`)) {
            sum+=1;
        }
    }
    
    return sum;
    
};

console.log(countPoints(rings));

//use this with canvas file finder
function NewTab(testing) {
  window.open(testing, "_blank");
}

for(test in results) {
  var testing = 'https://unt.instructure.com/files/' + test + '/download';
  NewTab(testing);
}

const getSubsets = arr => arr.reduce((prev, curr) => prev.concat(prev.map(k => k.concat(curr))), [[]]);

// Examples
getSubsets([1, 2]);     // [[], [1], [2], [1, 2]]
getSubsets([1, 2, 3]);  // [[], [1], [2], [1, 2], [3], [1, 3], [2, 3], [1, 2, 3]]