A Pythagorean triplet is a set of three natural numbers, {a, b, c}, for which,
a**2 + b**2 = c**2
For example,
3**2 + 4**2 = 9 + 16 = 25 = 5**2.
There exists exactly one Pythagorean triplet for which a + b + c = 1000.
Find the product a * b * c.
To run the tests, run the command busted
from within the exercise directory.
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.
Problem 9 at Project Euler http://projecteuler.net/problem=9
It's possible to submit an incomplete solution so you can see how others have completed the exercise.
local pythagorean = require('pythagorean-triplet')
describe('pythagorean-triplet', function()
local function sort(triplets)
table.sort(triplets, function(a, b) return a[1] < b[1] end)
return triplets
end
describe('is_triplet', function()
it('should identify triplets', function()
assert.is_true(pythagorean.is_triplet(3, 4, 5))
assert.is_true(pythagorean.is_triplet(5, 12, 13))
end)
it('should identify non-triplets', function()
assert.is_false(pythagorean.is_triplet(3, 4, 6))
assert.is_false(pythagorean.is_triplet(5, 6, 17))
end)
end)
describe('triplets_with', function()
it('should generate all triplets with a specified maximum factor', function()
assert.same(
{ { 3, 4, 5 }, { 5, 12, 13 }, { 6, 8, 10 }, { 8, 15, 17}, { 9, 12, 15 } },
sort(pythagorean.triplets_with{ max_factor = 17 })
)
end)
it('should generate all triplets with a specified minimum and maximum factor', function()
assert.same(
{ { 6, 8, 10 }, { 9, 12, 15 } },
sort(pythagorean.triplets_with{ min_factor = 6, max_factor = 15 })
)
end)
it('should generate all triplets with a specified maximum factor and a specified sum', function()
assert.same(
{ { 18, 80, 82 }, { 30, 72, 78 }, { 45, 60, 75 } },
sort(pythagorean.triplets_with{ sum = 180, max_factor = 100 })
)
end)
end)
end)
local function is_triplet(a, b, c)
return a^2 + b^2 == c^2
end
local function is_natural_number(n)
return n==math.floor(n) and n > 0
end
local function triplets_with(constraints)
local min = constraints.min_factor or 3
local max = constraints.max_factor or assert(fail,"max_factor is mandatory")
local result = {}
for a = min, max do
for b = a + 1, max do
local c = math.sqrt(a^2 + b^2)
if (is_natural_number(c) and c <= max and (a+b+c) == (constraints.sum or a+b+c)) then table.insert(result, {a, b, c}) end
end
end
return result
end
return {
is_triplet = is_triplet,
triplets_with = triplets_with
}
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Community comments
Cool, I like how you avoid iterating through c. I wonder how the performance of this compares to mine. I think for large max it will be significantly faster.