Algorithms Scripting Notes and Examples: Part 4

• 7:45 AM Trying to do my studies early in the morning and some in the afternoon. In order for me to become who I wanna be I need to continue not giving up no matter how hard.
• Anyways moving on. Now Today we're figuring out how to return the sum of all odd Fibonacci Numbers that are less than or equal to num.

• Basically its just every additional number in the sequence is the sum of the previous numbers. Ex: the first two numbers in the Fibonacci sequence starts with 1 and 1. After it should be followed by 2, 3, 5, 8, and so forth.

• For example, sum(10) should return 10 because all odd Fibonacci numbers less than or equal to 10 are 1, 1, 3, and 5.
  

function sum(num) {
  return num;
}

sum(4); // this should return 5 because all odd Fibonacci numbers less than or equal to `4` are 1, 1, 3

• Answer:

function sum(num) {
  let sequence = [0, 1]
  let count = sequence[sequence.length - 2] + sequence[sequence.length - 1];


    while (count <= num) {
     sequence.push(count);
     count = sequence[sequence.length - 2] + sequence[sequence.length - 1];
    }

     let sumOfAllOdds = 0

    sequence.forEach(function(num) {
    if (num % 2 != 0) {
      sumOfAllOdds += num; 
  }
    });

  return sumOfAllOdds;
}

console.log(sum(4)); // want to return 5 because that's the sum of all odd Fibonacci numbers [ 1, 1, 3];

Alright onto the next one! This time they want us to check return the sum of all prime numbers that are less than or equal to num.

• If you don't know what a prime number is, basically it's 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. While something like 4 is not because it is divisible by 1, 2 and 4.
• Now lets rewrite SumOfAllPrimes so it returns the sum of all prime numbers that are less than or equal to num.

function sumOfAllPrimes(num) {
  return num;
}

sumOfAll(10);

• Answer:

function sumOfAllPrimes(num) {
  function isPrime(num) {
    for (let x = 2; x < num; x++) {
      if (num % x === 0) {
        return false;
      }
    }
    return true;  
  }

  let range = [] 

  for (let i = 2; i <= num; i++) {
    if (isPrime(i)) {
      range.push(i)
    }
  }

  return range.reduce((a, b) =>  a + b)









}
console.log(sumOfAllPrimes(10)); // will display 17 because 2 + 5 + 10 = 17



// Recommended (reduce with default value)
// Array.prototype.reduce can be used to iterate through the array, adding the current element value to the sum of the previous element values.

// console.log(
//   [1, 2, 3, 4].reduce((a, b) => a + b, 0)
// )
// console.log(
//   [].reduce((a, b) => a + b, 0)
// )

//  a prime number can only return false - your code --> "if num % x ===0 { return false}" if it % 1 OR ITSELF. But you put x start from 2, so it won't check 1. then u put "x < num"  so it won't check itself