Created
July 1, 2015 19:49
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Pascal's triangle in Elixir (using recursion where possible).
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| # Author: MrSaints | |
| # License: MIT | |
| defmodule Factorial do | |
| def recursive(n, acc) when n === 1 or n === 0 do | |
| acc | |
| end | |
| def recursive(n, acc) do | |
| temp = n - 1 | |
| recursive(temp, acc * temp) | |
| end | |
| def recursive(n) do | |
| acc = n | |
| recursive(n, acc) | |
| end | |
| def enum(n) do | |
| Enum.reduce n..1, fn(x, acc) -> | |
| x * acc | |
| end | |
| end | |
| end | |
| defmodule PascalsTriangle do | |
| def element(row, column) when column === 0 or row === column do | |
| 1 | |
| end | |
| def element(row, column) when column === 1 do | |
| row + 1 | |
| end | |
| def element(row, column) do | |
| result = Factorial.recursive(row) / | |
| (Factorial.recursive(column) * Factorial.recursive(row - column)) | |
| trunc result | |
| end | |
| def row(n, current, acc) when current > n do | |
| acc | |
| end | |
| def row(n, current, acc) do | |
| newTuple = Tuple.insert_at(acc, current, element(n, current)) | |
| row n, current + 1, newTuple | |
| end | |
| def row(n) do | |
| row n, 1, {element(n, 0)} | |
| end | |
| def rows(n, current, acc) when n === current do | |
| acc | |
| end | |
| def rows(n, current, acc) do | |
| rows n, current + 1, acc ++ [row(current)] | |
| end | |
| def rows(n) do | |
| rows n, 1, [row(0)] | |
| end | |
| end |
Here is another way:
defmodule PascalTriangle do
def draw(1), do: (IO.puts("1"); [1])
def draw(current_level) do
list = draw(current_level - 1)
new_list = [1] ++ for(x <- 0..length(list)-1, do: Enum.at(list, x) + Enum.at(list, x+1, 0))
Enum.join(new_list, " ") |> IO.puts
new_list
end
end
PascalTriangle.draw(5)
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You could have written:
def pascal(n), do: pascal(n, [1, 1])
defp pascal(0, _), do: [1] # Not really coherent -- Should be 1 but it is easier for later use
defp pascal(1, acc), do: acc
defp pascal(n, acc) do
r = Enum.map(Enum.zip(acc ++ [0], [0 | acc]), fn {x, y} -> x+y end)
pascal(n-1, r)
end