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# Haikyo's solution

## to Nth Prime in the Lua Track

Published at Jul 20 2018 · 0 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 isPrime(n)
if n <= 3 then
return n > 1
end
if n % 2 == 0 or n % 3 == 0 then
return false
end
for i = 5, math.sqrt(n), 6 do
if n % i == 0 or n % (i + 2) == 0 then
return false
end
end
return true
end

local function nth(n)
assert(n > 0)
if n < 3 then
return n + 1
end

local prime
local i, j, k = 2, 5, 1
repeat
local p = k%2 == 0 and j+2 or j
if isPrime(p) then
i = i + 1
prime = p
end
j = j + (k%2 == 0 and 6 or 0)
k = k + 1
until i == n
return prime
end

return nth