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Daily Challenge #88 - Recursive Ninjas

dev.to staff on October 11, 2019

Ninjas frequently need to signal each other in code. They often employ natural sounds as cover. Our ninja will chirp like a bird, with different am...

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Dimitri Marion • Oct 11 '19

Javascript

const chirp = (n) => n == 1 ? 'chirp.' : 'chirp-' + chirp(n-1);

Davmi Jose Valdez Ogando • Oct 11 '19

Hot

Pierre Bouillon • Oct 11 '19 • Edited

Small solution in Python 3.X 😄

def recursive_ninjas(nb: int) -> str:
  return 'chirp.' if nb == 1 else f'chirp-{recursive_ninjas(nb - 1)}'

Amin • Oct 11 '19

Nice one! But you forgot an edge case. I can still call it with a negative value and get a recursion error.

print(recursive_ninjas(-10));
# RecursionError: maximum recursion depth exceeded in comparison

I think you should add another if else branch for when the value is less or equal than zero.

Pierre Bouillon • Oct 11 '19 • Edited

Given some context, which is ninjas talking, I doubt that they can talk negativly or that not talking at all is considered as talking

However, without those information, I fully agree ! 😄

Amin • Oct 11 '19

I was reading the old ninja scroll all wrong... Thanks sensei 😁

YuJin Kim • Oct 11 '19

#include <sstream>
#include <string>

std::string chirp(size_t n_reps) {
    std::stringstream s;
    for (size_t i = 0; i < n_reps; ++i) {
        s << "chirp-";
    }
    std::string release = s.str();
    release.erase(--release.end());
    return release;
}

My first submission at Dev.To! (Excluding the welcome post). I love computer science.

edh_developer • Oct 11 '19

Java

        public static void chirp(int i)
        {
                if (i == 1)
                        System.out.println("chirp");
                else if (i > 1) {
                        System.out.print("chirp-");
                        chirp(i - 1);
                }
        }

Note that this assumes a few things:

We want a carriage return at the end of the string.
The correct response to chirp(i) for i < 1 is silence.
The value of i is not going to be large enough for us to be worried about blowing the stack.

Blessed T Mahuni • Oct 11 '19

None recursive javascript solution

function chirp(numberOfSignals){
  let signal = ''
  for(let i = 0; i < numberOfSignals; i ++) {
    if(signal === '') {
      signal += 'chirp'
    } else {
      signal += '-chirp'
    }
  }
  return signal
}

console.log(chirp(4)) //chirp-chirp-chirp-chirp
console.log(chirp(5)) //chirp-chirp-chirp-chirp-chirp

Josh • Oct 11 '19 • Edited

As always, there is no need to use any language but Elixir. We'll establish a signature with a guard clause to make sure the user can only pass a natural number, and a private chirp_ signature to do the work of building the list of chirps, with a default empty list as its second argument. We could use List.duplicate, if the point of this exercise wasn't to make a recursive call 😜

defmodule Ninja do
  def chirp(n) when is_integer(n) and n > 0, do: chirp_(n)

  defp chirp_(n, chirps \\ [])
  defp chirp_(0, chirps), do: Enum.join(chirps, "-")
  defp chirp_(n, chirps), do: chirp_(n-1, ["chirp" | chirps])
end

Amin • Oct 11 '19 • Edited

JavaScript

Decided to throw an error if the count is not greater than zero because a ninja won't bother spending energy on calling this secret technique for no purposes.

"use strict";

/**
 * Ninja secret call sound function
 * 
 * @param {number} count The number of times the word "chirp" should be repeated
 * @throws If the argument count is not valid
 * @throws If the first argument is not a number
 * @throws If the first argument is not greater than 0
 * @return {string} The ninja secret call sound
 * @example chirp(15);
 */
function chirp(count) {
    if (arguments.length !== 1) {
        throw new Error("Expected exactly one parameter.");
    }

    if (typeof count !== "number") {
        throw new TypeError("First argument expected to be a number.");
    }

    if (count <= 0) {
        throw new Error("First argument expected to be greater or equal to 1.");
    }

    if (count === 1) {
        return "chirp.";
    }

    const decreasedCount = count - 1;

    return `chirp-${chirp(decreasedCount)}`;
}

console.log(chirp(4));
console.log(chirp(3));
console.log(chirp(2));
console.log(chirp(1));
console.log(chirp(0));

Training dojo available here (say chirp to enter).

Afief S • Oct 13 '19

Reason

let rec chirp = (times: int, ~result: string="", ()) =>
  switch (times) {
  | 0 => result ++ "."
  | _ =>
    switch (result) {
    | "" => chirp(times - 1, ~result=result ++ "chirp", ())
    | __ => chirp(times - 1, ~result=result ++ "-chirp", ())
    }
  };

Elm


chirp : Int -> String
chirp times =
    case times of
        1 ->
            "chirp."
        _ ->
            "chirp-" ++ (chirp <| times - 1)

Craig McIlwrath • Oct 12 '19

Haskell:

chirp 1 = "chirp."
chirp n = "chirp-" ++ chirp (n-1)

Also, you never said anything about Haskell standard functions :)

import Data.List (intercalate) 

chirp = (++".") . intercalate '-' . (flip replicate) "chirp"

K.V.Harish • Oct 12 '19

My solution in js

const chirp = (count = 1) => Array(count).fill('chirp').join('-');

Gaurav Kamath • Oct 15 '19 • Edited

Python

chirp = lambda n : '-'.join(['chirp']*n) + '.'

how do i syntax highlight? :(