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# JBaack's solution

## to Collatz Conjecture in the Lua Track

Published at Sep 25 2019 · 1 comment
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.

## Examples

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

1. 12
2. 6
3. 3
4. 10
5. 5
6. 16
7. 8
8. 4
9. 2
10. 1

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

## Running the tests

To run the tests, run the command `busted` from within the exercise directory.

## Further information

For more detailed information about the Lua track, including how to get help if you're having trouble, please visit the exercism.io Lua language page.

## Source

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

## Submitting Incomplete Solutions

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

### collatz-conjecture_spec.lua

``````local conjecture = require('collatz-conjecture')

describe('collatz-conjecture', function()
it('zero steps for one', function()
assert.are.equal(0, conjecture(1))
end)

it('divide if even', function()
assert.are.equal(4, conjecture(16))
end)

it('even and odd steps', function()
assert.are.equal(9, conjecture(12))
end)

it('large number of even and odd steps', function()
assert.are.equal(152, conjecture(1000000))
end)

it('zero is an error', function()
assert.has_error(
function() conjecture(0) end,
'Only positive numbers are allowed'
)
end)

it('negative value is an error', function()
assert.has_error(
function() conjecture(-15) end,
'Only positive numbers are allowed'
)
end)
end)``````
``````local calc_collatz
local steps = 0

calc_collatz = function(number)

if(number > 1) then
if((number % 2) == 0) then
number = (number / 2)
else
number = (number * 3) + 1
end
steps = steps + 1
calc_collatz(number)
end

end

return function(number)

if(number <= 0) then
error("Only positive numbers are allowed")
else
steps = 0
calc_collatz(number)
end
return steps

end``````

JBaack
Solution Author
commented over 1 year ago

Did a while (number > 1) do if (number % 2 == 0) then ...

solution first.

Thought recursion would be fun.

Then tried a solution for neg numbers(works with 3n - 1 for odd numbers and reduces to -1)

### What can you learn from this solution?

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

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