marcmartino

function genSeqUntil(valFunc, limitPredicate, seq = []) {
  var nextVal = valFunc(seq.length);
  if (limitPredicate(seq.length, nextVal)) {
    seq[seq.length] = nextVal;
    return genSeqUntil(valFunc, limitPredicate, seq);
  }
  return seq;
}

function sumOfAllMultiplesUnder(multsArr, max) {

marcmartino

function genSeqUntil(valFunc, limitPredicate, seq = []) {
    var nextVal = valFunc(seq.length);
    if (limitPredicate(seq.length, nextVal)) {
      seq[seq.length] = nextVal;
      return genSeqUntil(valFunc, limitPredicate, seq);
    }
    return seq;
}
const memoize = (fn) => (function () {
      var cache = {};

marcmartino

function findPrimeFact(num, accum = []) {
  return 1;
}
function isDivisible(num) {
  return (divisor) => num % divisor === 0;
}
function isPrime(num) {
  return Array.from({length: Math.floor(num/4 - 1)},
      (x, i) => i * 2 + 3)
    .filter(isDivisible(num)).length === 0;

marcmartino

myAudio = new Audio(audioUri + phrase + audioFormat);
myAudio.play();

//anti-pattern: does not work
return Rx.Observable.fromEvent(myAudio, 'onended');


myAudio.onended = () {
  return Rx.Observable.just("success"); 
}

marcmartino

var funcNames = {
  addition: {
    func: function (paramsArr = []) {
      return paramsArr.reduce((tot, num) => tot + num, 0)
    },
    textRep: '+'
  },
  subtraction: {
    func: function ([head, ...tail] = []) {
      return tail.reduce((tot, num) => tot - num, head)

marcmartino

//jshint esnext:true
// curveColors should have a length longer than or equal to 1
// the colors are split evenly- with a curveColors length of 3
// with said curveColors, a score of 75 would respond to halfway between index 1 and 2
const curveColors = [{r: 49, g: 198, b: 84},
                     {r: 46, g: 48, b: 193},
                     {r: 193, g: 138, b: 44},
                     {r: 193, g: 44, b: 44}]
// score should be between 0-100
const score = 50

marcmartino

// 1..100
// 3 5

const seq = min => max => {
    let list = [];
    for (let index = min; index <= max; index++) {
        list.push(index);
    }
    return list;
}

marcmartino

//jshint esnext:true
/*
Multiples of 3 and 5
Problem 1

If we list all the natural numbers below 10 that are multiples of 3 or 5, 
we get 3, 5, 6 and 9. The sum of these multiples is 23.

Find the sum of all the multiples of 3 or 5 below 1000.
*/

marcmartino

//jshint esnext:true
/*
Multiples of 3 and 5 - Problem 1

If we list all the natural numbers below 10 that are multiples of 3 or 5, 
we get 3, 5, 6 and 9. The sum of these multiples is 23.

Find the sum of all the multiples of 3 or 5 below 1000.
*/

marcmartino

maximum [x*y | x <- [100..999] , y <- [100..999], x*y == reverseInt (x*y)]