Published at Aug 24 2018
·
0 comments

Instructions

Test suite

Solution

The Collatz Conjecture or 3x+1 problem can be summarized as follows:

Take any positive integer n. If n is even, divide n by 2 to get n / 2. If n is odd, multiply n by 3 and add 1 to get 3n + 1. Repeat the process indefinitely. The conjecture states that no matter which number you start with, you will always reach 1 eventually.

Given a number n, return the number of steps required to reach 1.

Starting with n = 12, the steps would be as follows:

- 12
- 6
- 3
- 10
- 5
- 16
- 8
- 4
- 2
- 1

Resulting in 9 steps. So for input n = 12, the return value would be 9.

Go through the setup instructions for ECMAScript to install the necessary dependencies:

http://exercism.io/languages/ecmascript

Install assignment dependencies:

```
$ npm install
```

Execute the tests with:

```
$ npm test
```

In the test suites all tests but the first have been skipped.

Once you get a test passing, you can enable the next one by
changing `xtest`

to `test`

.

An unsolved problem in mathematics named after mathematician Lothar Collatz https://en.wikipedia.org/wiki/3x_%2B_1_problem

It's possible to submit an incomplete solution so you can see how others have completed the exercise.

```
import { steps } from './collatz-conjecture';
describe('steps()', () => {
test('zero steps for one', () => {
expect(steps(1)).toEqual(0);
});
xtest('divide if even', () => {
expect(steps(16)).toEqual(4);
});
xtest('even and odd steps', () => {
expect(steps(12)).toEqual(9);
});
xtest('Large number of even and odd steps', () => {
expect(steps(1000000)).toEqual(152);
});
xtest('zero is an error', () => {
expect(() => {
steps(0);
}).toThrow(new Error('Only positive numbers are allowed'));
});
xtest('negative value is an error', () => {
expect(() => {
steps(-15);
}).toThrow(new Error('Only positive numbers are allowed'));
});
});
```

```
function steps(n,nSteps=0) {
if (n<1) {
throw('Only positive numbers are allowed');
} else if (n===1) {
return(nSteps);
} else if (n%2===0) {
n/=2;
} else {
n = n*3 +1;
}
return steps(n, ++nSteps);
}
module.exports={steps}
```

A huge amount can be learned from reading other people’s code. This is why we wanted to give exercism users the option of making their solutions public.

Here are some questions to help you reflect on this solution and learn the most from it.

- What compromises have been made?
- Are there new concepts here that you could read more about to improve your understanding?

Level up your programming skills with 3,373 exercises across 50 languages, and insightful discussion with our volunteer team of welcoming mentors.
Exercism is
**100% free forever**.

## Community comments