Code Golf: Collatz Conjecture

Answer 1

Haskell, 62 chars 63 76 83, 86, 97, 137

c 1=[1]
c n=n:c(div(n`mod`2*(5*n+2)+n)2)
main=readLn>>=print.c

User input, printed output, uses constant memory and stack, works with arbitrarily big integers.

A sample run of this code, given an 80 digit number of all '1's (!) as input, is pretty fun to look at.

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Original, function only version:

Haskell 51 chars

f n=n:[[],f([n`div`2,3*n+1]!!(n`mod`2))]!!(1`mod`n)

Who the @&^# needs conditionals, anyway?

(edit: I was being "clever" and used fix. Without it, the code dropped to 54 chars. edit2: dropped to 51 by factoring out f())

Answer 2

ruby, 43, possibly meeting the I/O requirement

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Run with ruby -n hail

n=$_.to_i
(n=n%2>0?n*3+1: n/2
p n)while n>1

Answer 3

Scala + Scalaz

import scalaz._
import Scalaz._
val collatz = 
   (_:Int).iterate[Stream](a=>Seq(a/2,3*a+1)(a%2)).takeWhile(1<) // This line: 61 chars

And in action:

scala> collatz(7).toList
res15: List[Int] = List(7, 22, 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2)

Scala 2.8

val collatz = 
   Stream.iterate(_:Int)(a=>Seq(a/2,3*a+1)(a%2)).takeWhile(1<) :+ 1

This also includes the trailing 1.

scala> collatz(7)
res12: scala.collection.immutable.Stream[Int] = Stream(7, 22, 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1)

With the following implicit

implicit def intToEven(i:Int) = new {
  def ~(even: Int=>Int, odd: Int=>Int) = { 
    if (i%2==0) { even(i) } else { odd(i) }
  }
}

this can be shortened to

val collatz = Stream.iterate(_:Int)(_~(_/2,3*_+1)).takeWhile(1<) :+ 1

Edit - 58 characters (including input and output, but not including initial number)

var n=readInt;while(n>1){n=Seq(n/2,n*3+1)(n%2);println(n)}

Could be reduced by 2 if you don't need newlines...

Answer 4

Golfscript : 20 chars

  ~{(}{3*).1&5*)/}/1+`
# 
# Usage: echo 21 | ruby golfscript.rb collatz.gs

This is equivalent to

stack<int> s;
s.push(21);
while (s.top() - 1) {
  int x = s.top();
  int numerator = x*3+1;
  int denominator = (numerator&1) * 5 + 1;
  s.push(numerator/denominator);
}
s.push(1);
return s;

Answer 5

Ruby, 50 chars, no stack overflow

Basically a direct rip of makapuf's Python solution:

def c(n)while n>1;n=n.odd?? n*3+1: n/2;p n end end

Ruby, 45 chars, will overflow

Basically a direct rip of the code provided in the question:

def c(n)p n;n.odd?? c(3*n+1):c(n/2)if n>1 end

Answer 6

not the shortest, but an elegant clojure solution

(defn collatz [n]
 (print n "")
 (if (> n 1)
  (recur
   (if (odd? n)
    (inc (* 3 n))
    (/ n 2)))))