Avatar of JaeHyoLee

JaeHyoLee's solution

to Pascal's Triangle in the Lua Track

Published at Jul 13 2018 · 0 comments
Instructions
Test suite
Solution

Compute Pascal's triangle up to a given number of rows.

In Pascal's Triangle each number is computed by adding the numbers to the right and left of the current position in the previous row.

    1
   1 1
  1 2 1
 1 3 3 1
1 4 6 4 1
# ... etc

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

Pascal's Triangle at Wolfram Math World http://mathworld.wolfram.com/PascalsTriangle.html

Submitting Incomplete Solutions

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

pascals-triangle_spec.lua

local Triangle = require('pascals-triangle')

describe('pascals-triangle', function()
  it('should generate a triangle with one row', function()
    assert.same({ { 1 } }, Triangle(1).rows)
  end)

  it('should generate a triangle with two rows', function()
    assert.same({ { 1 }, { 1, 1 } }, Triangle(2).rows)
  end)

  it('should generate a triangle with three rows', function()
    assert.same({ { 1 }, { 1, 1 }, { 1, 2, 1 } }, Triangle(3).rows)
  end)

  it('should allow the last row to be accessed directly', function()
    assert.same({ 1, 2, 1 }, Triangle(3).last_row)
  end)

  it('should generate the fourth row correctly', function()
    assert.same({ 1, 3, 3, 1 }, Triangle(4).last_row)
  end)

  it('should generate the fifth row correctly', function()
    assert.same({ 1, 4, 6, 4, 1 }, Triangle(5).last_row)
  end)

  it('should generate the twentieth row correctly', function()
    local twentieth = { 1, 19, 171, 969, 3876, 11628, 27132, 50388, 75582, 92378, 92378, 75582, 50388, 27132, 11628, 3876, 969, 171, 19, 1 }
    assert.same(twentieth, Triangle(20).last_row)
  end)
end)
return function (nth)
  local rows = {}
  for i = 1, nth do
    local elements = {1}
    for j = 2, i do
      elements[j] = rows[i-1][j-1] + (rows[i-1][j] or 0)
    end
    table.insert(rows,elements)
  end

  return {
    rows = rows,
    last_row = rows[nth]
  }
end

Community comments

Find this solution interesting? Ask the author a question to learn more.

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?
  • Are there new concepts here that you could read more about to improve your understanding?