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My solution for Chop guess exercise (Programming Elixir 1.3)
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############################## | |
##Exercise: ModulesAndFunctions-6 | |
defmodule Chop do | |
def guess(actual, lo..hi) when (lo <= hi) and (actual in lo..hi) do | |
current = guessing(lo, hi) | |
IO.puts "Is it #{current}" | |
guess(current, actual, lo..hi) | |
end | |
defp guess(value, actual, _) when value == actual, do: IO.puts value | |
defp guess(value, actual, lo.._) when actual < value, do: guess actual, lo..(value - 1) | |
defp guess(value, actual, _..hi) when actual > value, do: guess actual, (value + 1)..hi | |
defp guessing(lo, hi), do: lo + div(hi - lo, 2) | |
end | |
Chop.guess 273, 1..1000 | |
# Is it 500 | |
# Is it 250 | |
# Is it 375 | |
# Is it 312 | |
# Is it 281 | |
# Is it 265 | |
# Is it 273 | |
# 273 |
I am a newbie in Elixir coming from Ruby. I'd written my solution for this but I felt like it could be better and then I came across your solution. Brilliant! I'd never thought about matching that range as lo..hi
in the function head, I was still doing the matching in the function body. 😄 I was also doing the guessing in each of the functions separately, your code helped me learn a lot!
Hello, @veelenga ! Although your solution works, I would like to make a couple of notes:
- An extra work for the given exercise;
- Both your solution and the book author's solution use private functions,
which seems unfair to me, since this is the topic of the next section.
Thus, the solution can be rewritten as follows:
defmodule Chop do
def guess(actual, low..high) do
guessed = div(low + high, 2)
IO.puts "Is it #{guessed}"
processing_guessed =
fn
n when n == actual -> IO.puts n
n when n > actual -> guess actual, low..(n - 1)
n when n < actual -> guess actual, (n + 1)..high
end
processing_guessed.(guessed)
end
end
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@pdbradley yes, the signature of
def
anddefp
functions are different, so they are valid.Keep in mind, that solution was written 3 years ago, when the language was slightly different :)