Sum All Odd Fibonacci Numbers

JavaScript
1 
function sumFibs(num) {

let sum = 0;  
for (let i = 0; i < num.length; i++){
  if (num % 2 == 0) {

  } else if (num % 2 == 1){
    sum = num + num
  }
}
  return sum;
}

//Fibonacci sequence is to add first two numbers, answer takes up next spot, then repeat
console.log(sumFibs(4));

Need help. Need to find a way to tell them to add all the odd numbers in the sequence. Do I need to write a line of code to tell them what the sequence is? Or is the Fibonacci sequence already known for this problem?

2

share that step url as well, thanks :slight_smile:

3

Sorry about that, thanks for reminding me

4

Yes, you will need to find an approach to generate the Fibonacci numbers you need.

5 
function sumFibs(num) {

// firstTerm =  0
// secondTerm = 1

// nextTerm = firstTerm + secondTerm; (0 + 1)
// firstTerm = secondTerm; (1)
// secondTerm = nextTerm; (1)

// nextTerm = firstTerm + secondTerm; (1 + 1)


sumFibs(4);

I understand what the pattern is in the Fibonacci sequence but I am stuck when it comes to writing the code for it

6

It looks like youā€™ve got the correct logical flow. But you need to keep running that logic, until a condition is met that tells your code to stop. Do you know the syntax for something like that? (Thereā€™s a few options here)

7 
function sumFibs(num) {

// firstTerm =  0
// secondTerm = 1

// nextTerm = firstTerm + secondTerm; (0 + 1)
// firstTerm = secondTerm; (1)
// secondTerm = nextTerm; (1)

// nextTerm = firstTerm + secondTerm; (1 + 1)
// when secondTerm <= num...STOP

for (let i = 0; i <= num; i++){
  
}


sumFibs(4);

Do you mean something like this?

8

That looks like a good starting point, yes!

1 Like
9 
function sumFibs(num) {

// firstTerm =  0
// secondTerm = 1

// nextTerm = firstTerm + secondTerm; (0 + 1)
// firstTerm = secondTerm; (1)
// secondTerm = nextTerm; (1)

// nextTerm = firstTerm + secondTerm; (1 + 1)
// when secondTerm <= num...STOP


for (let i = 0; i <= num; i++){
  if (i + (i + 1) == (i + 2)) {
    (i + 1) + (i + 2) == (i + 3)
  }
}

console.log(sumFibs(4));

Iā€™m trying to get it to understand the pattern but Iā€™m almost positive I am taking the wrong approach. Been stuck on the next steps

10

how about you try to put them in more ā€œobviousā€ way, and see if you can spot that sequence going more easily, here is an example of first ā€œsevenā€ numbers from ā€œfibonacci seriesā€

0 + 0 = 0
1 + 0 = 1
1 + 1 = 2
2 + 1 = 3
3 + 2 = 5
5 + 3 = 8
8 + 3 = 11

hope this helps, happy learning :slight_smile:

2 Likes
11 
function sumFibs(num) {
  let prevNumber = 0;
  let currNumber = 1;
  let result = 0;

  // before number reaches num 
  while (currNumber <= num) {
    // if that number is odd
    if (currNumber % 2 == 1) {
      // then those odd numbers to the result [1+1+3] = 5
      result += currNumber;
    } // let them know how the Fibonacci sequence works
    currNumber += prevNumber;// 1 = 0 + 1 aka 0 + 1 = 1
    prevNumber = currNumber - prevNumber; // 0 = 1 - 0; letting them know what the previous number is creating the cycle/pattern
  }

  return result;
}

sumFibs(4);

What do you think?

12

I always get a little confused when I see a big jump in the complexity of your solution relative to your described plan, but this is a clever trick here.

13

This is great! The other common solution pattern is a functional approach using array methods. Something like this:

getAllFibsLte(num)
  .filter(isOdd)
  .reduce(sum)

(youā€™d need to implement the getAllFibsLte, isOdd and sum functions).

The trade-off compared to your imperative solution is reduced performance due to higher memory consumption and more loops, but is arguably more legible (it reads like the English instructions).

1 Like
14

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