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BooleanCat's solution

to Pythagorean Triplet in the Lua Track

Published at Jan 05 2019 · 0 comments
Instructions
Test suite
Solution

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.

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

Problem 9 at Project Euler http://projecteuler.net/problem=9

Submitting Incomplete Solutions

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

pythagorean-triplet_spec.lua

local triplets_with_sum = 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('triplets_with_sum', function()
    it('finds triplets whose sum is 12', function()
      assert.same(
        { { 3, 4, 5 } },
        sort(triplets_with_sum(12))
      )
    end)

    it('finds triplets whose sum is 108', function()
      assert.same(
        { { 27, 36, 45 } },
        sort(triplets_with_sum(108))
      )
    end)

    it('finds triplets whose sum is 1000', function()
      assert.same(
        { { 200, 375, 425 } },
        sort(triplets_with_sum(1000))
      )
    end)

    it('finds no triplets whose sum is 1001', function()
      assert.same({}, triplets_with_sum(1001))
    end)

    it('finds triplets whose sum is 90', function()
      assert.same(
        {
          { 9, 40, 41 },
          { 15, 36, 39 },
        },
        sort(triplets_with_sum(90))
      )
    end)

    it('finds triplets whose sum is 840', function()
      assert.same(
        {
          { 40, 399, 401 },
          { 56, 390, 394 },
          { 105, 360, 375 },
          { 120, 350, 370 },
          { 140, 336, 364 },
          { 168, 315, 357 },
          { 210, 280, 350 },
          { 240, 252, 348 },
        },
        sort(triplets_with_sum(840))
      )
    end)

    it('finds triplets whose sum is a large number (30000)', function()
      assert.same(
        {
          { 1200, 14375, 14425 },
          { 1875, 14000, 14125 },
          { 5000, 12000, 13000 },
          { 6000, 11250, 12750 },
          { 7500, 10000, 12500 },
        },
        sort(triplets_with_sum(30000))
      )
    end)
  end)
end)
return function(sum)
  local function from_a(x)
    -- Given that:
    -- (1) a + b + c = p
    -- (2) a ^ 2 + b ^ 2 = c ^ 2
    -- (2) c = sqrt(a ^ 2 + b ^ 2)
    --
    -- (2) -> (1) a + b + sqrt(a ^ 2 + b ^ 2) = p
    --
    -- Rearranging we get:
    -- b = 0.5 * (p - (pa / (p - a)))

    return (sum - sum * x / (sum - x)) / 2
  end

  local triplets = {}

  for a=1, sum do
    local b = from_a(a)

    if a > b then
      break
    end

    local frac
    b, frac = math.modf(b)
    if frac == 0.0 then
      -- Mathematically from_a might yield non-natural numbers which are then
      -- truncated by integer division, so here we filter only for integer
      -- triplets.
      table.insert(triplets, { a, b, sum - (a + b) })
    end
  end

  return triplets
end

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