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## to Nth Prime in the Lua Track

Published at Jul 13 2018 · 3 comments
Instructions
Test suite
Solution

Given a number n, determine what the nth prime is.

By listing the first six prime numbers: 2, 3, 5, 7, 11, and 13, we can see that the 6th prime is 13.

If your language provides methods in the standard library to deal with prime numbers, pretend they don't exist and implement them yourself.

## 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

A variation on Problem 7 at Project Euler http://projecteuler.net/problem=7

## Submitting Incomplete Solutions

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

### nth-prime_spec.lua

``````local nth = require('nth-prime')

describe('nth-prime', function()
local function benchmark(f)
local start = os.clock()
f()
return os.clock() - start
end

it('should give 2 as the first prime', function()
assert.equal(2, nth(1))
end)

it('should give 3 as the second prime', function()
assert.equal(3, nth(2))
end)

it('should be able to calculate the nth prime for small n', function()
assert.equal(13, nth(6))
end)

it('should be able to calculate the nth prime for large n', function()
assert.equal(104743, nth(10001))
end)

it('should be efficient for large n', function()
local execution_time = benchmark(function()
nth(10001)
end)

assert(execution_time < 1, 'should take less than a second to execute')
end)

it('should raise an error for n <= 0', function()
assert.has_error(function()
nth(0)
end)

assert.has_error(function()
nth(-1)
end)
end)
end)``````
``````local function is_prime(number)
local to = math.sqrt(number)
for from = 2, to do
if number % from == 0 then return false end
end
return true
end

return function(nth)
local candidate = 1
local count = 0
assert(nth > 0, 'nth should be nature number')

repeat
candidate = candidate + 1
if is_prime(candidate) then count = count + 1 end
until count == nth

return candidate
end`````` An improvement: instead of iterate until to := math.sqrt(number), you could iterate until from² > number. A square is faster than a square root. @carlos-pavanetti yeap you are right. square is faster than a square root :) @carlos-pavanetti great catch!

### What can you learn from this solution?

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?