Collatz Conjecture (OEIS A006577)

Question

This is the Collatz Conjecture (OEIS A006577):

• Start with an integer n > 1.
• Repeat the following steps:
  • If n is even, divide it by 2.
  • If n is odd, multiply it by 3 and add 1.

It is proven that for all positive integers up to 5 * 2⁶⁰, or about 5764000000000000000, n will eventually become 1.

Your task is to find out how many iterations it takes (of halving or tripling-plus-one) to reach 1.

Relevant xkcd :)

Rules:

• Shortest code wins.
• If a number < 2 is input, or a non-integer, or a non-number, output does not matter.

Test cases

2  -> 1
16 -> 4
5  -> 5
7  -> 16

Answer 1

C, 70 69 chars

Quite simple, no tricks.
Reads input from stdin.

a;
main(b){
    for(scanf("%d",&b);b-1;b=b%2?b*3+1:b/2)a++;
    printf("%d",a);
}

Answer 2

Q,46

{i::0;{x>1}{i+:1;$[x mod 2;1+3*x;(_)x%2]}\x;i}

Answer 3

C++ (51 48)

This is a recursive function that does this; input reading comes separately.

int c(n){return n==1?0:1+(n%2?c(n*3+1):c(n/2));}

I'm sure I can do some sort of "and/or" trick with the == 0 stuff, but I have no idea how.

Answer 4

~-~! (No Comment) - 71 53

This language is obviously not the best for golfing since it lacks a large amount of native functionality, but that's the beauty of it.

'=|*;~~[*,~~~-~]*/~~|:''=|'''==~[*]'''='&''':''&*+~|:

First, set ''' to your input. The function '' can then be called with % as it's input and will return the answer, like so:

'''=~~~~~:''&%:

This will return ~~~~~. It actually works for n==1 (it loops forever with n==0).

As always with this language, untested.

Answer 5

JavaScript (ES6) - 29 Characters

f=x=>x>1?f(x%2?x*3+1:x/2)+1:0

Creates a function f which accepts a single argument and returns the number of iterations.

JavaScript - 31 Characters

for(c=0;n>1;n=n%2?n*3+1:n/2)++c

Assumes that the input is in the variable n and creates a variable c which contains the number of iterations (and will also output c to the console as its the last command).

Answer 6

Perl 6, 40 bytes

Recursive function method, as per Valentin CLEMENT and daniero: 40 characters

sub f(\n){n>1&&1+f n%2??3*n+1!!n/2}(get)

Lazy list method: 32 characters

+(get,{$_%2??$_*3+1!!$_/2}...^1)

Answer 7

Ruby, 35 bytes

f=->n{n<2?0:1+f[n*3/(6-5*w=n%2)+w]}

Try it online!

How it works

Instead of getting the 2 values and choosing one, multiply by 3, divide by 1 if odd, or 6 if even, and then add n modulo 2.

Answer 8

MathGolf, 7 bytes

kÅ■┐▲î 

Don't get fooled, there's a non-breaking space at the end of the program.

Try it online!

Explanation

k        Read input as integer
 Å       Start a block of length 2
  ■      Map TOS to the next item in the collatz sequence
   ┐     Push TOS-1 without popping
    ▲    Do block while TOS is true
     î   Push the length of the last loop
         Discard everything but top of stack

MathGolf, 14 bytes (no built-ins, provided by JoKing)

{_¥¿É3*)½┐}▲;î

Explanation

{               Start block of arbitrary length
 _              Duplicate TOS
  ¥             Modulo 2
   ¿            If-else (works with TOS which is 0 or 1 based on evenness)
    É3*)        If true, multiply TOS by 3 and increment
        ½       Otherwise halve TOS
         ┐      Push TOS-1 (making the loop end when TOS == 1)
          }▲    End block, making it a do-while-true with pop
            ;   Discard TOS
             î  Print the loop counter of the previous loop (1-based)

Ideally, this solution could become 13 bytes, since it's not neccessary to have the ending of the block be explicit when the loop type instruction comes right after. I'll see when I get around to coding implicit block ending when loop type is present.

Try it online!

Answer 9

C (gcc), 43 33 bytes

f(x){x=~-x?f(x&1?3*x+1:x/2)+1:0;}

Try it online!

Answer 10

BitCycle -u, 90 bytes

 ~  ~!
?v C/v
v<   <
A\\ B^
>/\/C =v
  Cvv  <
  v~v/
  >   ^
  v =
>> >>^
\~~~
 ~v~^
^ + ~

Try it online! Or, watch it in action here.

Algorithm

The main loop starts with the current number \$n\$ in unary in the A collector. We divide the number by 2, splitting off two bits at a time; one of the halves, \$\lfloor \frac n 2 \rfloor\$, goes into the uppermost C collector; the other half goes into the middle C collector; and the remainder, \$n\text{ mod }2\$, goes into the bottom C collector.

Once the number is completely divided up in this way, the C collectors open.

• The top C collector sends a 1 to the sink at the top right, adding 1 to the output, and sends all of its bits back into A.
• If the bottom C collector is empty (i.e. \$n\text{ mod }2 = 0\$, i.e. \$n\$ was even), the first bit from the middle collector hits the bottommost switch = and activates it pointing right, which discards the bits. This leaves A with just the \$\lfloor\frac n 2\rfloor\$ bits it got from the top C collector: \$n\text{ even}\to n/2\$.
• If the bottom C collector contains a 1 bit (i.e. \$n\text{ mod }2 = 1\$, i.e. \$n\$ was odd), a negated copy of it hits the bottommost switch and activates it pointing left. This sends the bits from the bottom and middle C collectors into the big collection of dupnegs ~ at the bottom, which makes five copies of its input and discards one bit. All the copies are then sent back into A: \$n\text{ odd}\to \lfloor\frac n 2\rfloor + 5\left( \lfloor\frac n 2\rfloor + 1\right) - 1 = 6\lfloor\frac n 2\rfloor + 3 + 1 = 3n + 1\$.

This whole process repeats until \$n=1\$, at which point the two halves are 0; this means the only C collector with data is the bottommost one that holds the remainder. The remainder bit is directed up to the uppermost switch =. Normally, this switch would have been activated by the bits from the middle C collector already, and the remainder bit would follow them into the 5-times circuitry. But since the middle C collector is empty, the remainder bit passes through the switch and continues northward off the playfield. Since there are no bits remaining on the playfield, the program halts and displays the number of steps taken.

Answer 11

><>, 28 bytes

:1=?v::2%?v2,
+c0.\l1-n;\3*1

This takes input from the stack, computes the different steps on the stack, then returns its size when 1 is reached.

---

Improved version by JoKing, 24 bytes :

:1=?\::2%b$.2,
3*1+\~ln;

Answer 12

x86 machine code, 15 bytes

xxd -g1:

00000000: 99 42 8d 4c 40 01 d1 e8 0f 42 c1 75 f4 4a c3     [email protected].

Commented assembly (NASM syntax):

    [bits 32]
    global collatz
    ; input: eax, assumed positive and > 1
    ; output: edx
    ; clobbers: eax, ecx, edx
collatz:
    cdq                       ; count = 0 (abuses eax > 1)
.Lloop:
    inc    edx                ; increment count
    lea    ecx, [eax+2*eax+1] ; tmp = 3*n + 1
    shr    eax, 1             ; n = n / 2, sets flags
    cmovc  eax, ecx           ; swap with 3n+1 if it was originally odd (does not set flags)
    jnz    .Lloop             ; shr also sets ZF if the shr result was zero, end condition
    dec    edx                ; Correct the off by one
    ret                       ; return

Try it online! (converted to GAS Intel syntax and wrapped in C++)

Notes

The way this works is by using shr magic, allowing me to calculate n/2 and also test if n was originally odd (CF=1) or originally 1 (ZF=1).

Unfortunately, this results in an off by one since it will run when n == 1, but it is correctable via a simple dec.

Note that while this is larger than the other x86 solution, the other solution is a snippet, not a complete function, and it doesn't even count the steps, only calculating the sequence.

If that version were to count the steps, while it would be more efficient, it would be larger because the bsr complicates the bookkeeping, unless I can be proven otherwise.

Answer 13

Julia, 54 chars

f(n,i)=n==1 ? i : n%2==0 ? f(n/2,i+=1) : f(n*3+1,i+=1)

Try it online!

Julia, 45 chars

>(n,i=0)=n<2 ? i : n%2<1 ? n/2>i+1 : 3n+1>i+1

Try it online!

Thanks to the brilliant suggestion by MarcMush!

Answer 14

Pyth 23byte

J0WtQ=Q@,/Q2h*3QQ=JhJ;J

Answer 15

Acc!!, 127 75 70 bytes

-5 bytes thanks to Mukundan314

Count u while N {
u
}
Count i while _-1 {
_/2+_%2*(5*_/2+2)
Write 49
}

The program takes input and produces output in unary. Try it online!

(Here's a decimal I/O version in 209 bytes.)

With comments

The input method technically takes advantage of undefined behavior: the official implementation of Acc!! adds a newline character to the end of each line of input.

# Read input in unary
Count u while N {    # Increment u from 0 while not EOF
  u                  # Set accumulator to u
}
# For a unary number X, the loop will run X+1 times (including the
# trailing newline); because the loop variable starts at 0, this sets
# the accumulator to X

# Main loop
Count i while _-1 {  # Increment i from 0 while accumulator is not equal to 1
  _/2+_%2*(5*_/2+2)  # Apply one step of Collatz function to accumulator
  Write 49           # Write "1" to output
}

The expression _/2+_%2*(5*_/2+2) boils down to

_/2,               if _%2 is 0
_/2 + (5*_)/2 + 2, if _%2 is 1

This is integer division, so the latter case comes out to

_/2 + 2*_ + _/2 + 2
 = 2*_ + (_/2)*2 + 2
 = 2*_ + _ + 1
 = 3*_ + 1

Answer 16

newLISP - 94 chars

Strangely similar to Valentin's Scheme answer... :) I'm let down here by verbosity of the language but there's a bitshift division which appears to work...

(let(f(fn(x)(cond((= x 1)0)((odd? x)(++(f(++(* 3 x)))))(1(++(f(>> x)))))))(f(int(read-line))))

Answer 17

Python (73):

Can probably be golfed a heck of a lot more.

i=0
while 1:
 i+=1;j=i;k=0
 while j!=1:j=(j/2,j*3+1)[j%2];k+=1
 print i,k

Answer 18

Haskell 73 Bytes 73 Chars

r n |even n=n`quot`2
    |otherwise=3*n+1
c=length.takeWhile(/=1).iterate r

Answer 19

Fish (33 chars including whitespace, 26 without)

:2%?v:2,  >:1=?v
    >:3*1+^;nl~<

The whitespace is necessary for it to function, as ><> is a 2D language. Example run:

$ python3 fish.py collatz.fish -v 176
18

Answer 20

K, 24 bytes

#1_(1<){(x%2;1+3*x)x!2}\

With test cases:

  (#1_(1<){(x%2;1+3*x)x!2}\)'2 16 5 7
1 4 5 16

This uses a bit of a cute trick to avoid conditionals- (x%2;1+3*x) builds a list of the potential next term and then the parity calculated by x!2 indexes into that list. Otherwise it's a straightforward application of the "do while" form of \, given the tacit predicate (1<) (while greater than 1) as a stopping condition:

  (1<){(x%2;1+3*x)x!2}\5
5 16 8 4 2 1

The example output indicates that we need to drop the first (1_) of this sequence before taking the count (#). This is slightly shorter than taking the count and then subtracting one.

Answer 21

Befunge, 42 40 bytes

Surprisingly short to be an esolang! I thank @Sok for showing how to avoid one extra branching in his answer. Saved 2 bytes after a complete rewriting of the code.

0&>\1+\:2/\:3v
.$<v_v#%2\+1*<@
`!|>\>$:1

Original answer:

1&>:2%v>2v
^\+1*3_^ /
>+v  v`1:<
^1\#\_$.@

Shold be compatible with both Befunge 93 and Befunge 98. Interpretor available here.

There is no need for a trailing white space after @, so I count it as 42. However, 2D languages are often counted by their bounding box.

Answer 22

Haskell, 43 Bytes

c 1=0
c x|odd x=1+c(3*x+1)|1<2=1+c(x`div`2)

Usage: c 7-> 16

Answer 23

Python 2, 59 57 55 54 bytes

i=0;n=input()
while~-n:n=[n/2,n*3+1][n%2];i+=1
print i

Answer 24

Clojure, 60 bytes

(fn c[n](if(= n 1)0(inc(c(if(even? n)(/ n 2)(+(* n 3)1))))))

Pretty standard. Recursive function that recurses when n isn't equal to one. Each iteration, one is added to the accumulator via inc.

While this uses unoptimized recursion, I'm currently testing to see when it fails. It's at 1711000000, and is still going. The highest number of steps I've seen so far is 1008, so I don't expect it to fail anytime soon.

Pregolfed:

(defn collatz-conj [n]
  (if (= n 1)
    0 ; Base case
    (inc ; Add one to step
      (collatz-conj ; Recurse
        (if (even? n) ; The rest should be be self-explanatory
          (/ n 2)
          (+ (* n 3) 1))))))

Answer 25

TCL 8.5 (71 70 68) (67)

TCL has no real chance of ever winning, but it is a fun way to oil the machine:

proc c x {while \$x>1 {set x [expr $x%2?3*$x+1:$x/2];incr k};set k}

formatted for readability:

proc c x {
    while {$x>1} {
    set x [expr $x%2 ? 3*$x+1 : $x/2]
    incr k
    }
    set k
}

Edits: many suggestions (inspired) by sergiol. I guess the answer is more theirs than mine, by now :-)

Answer 26

Game Maker Language, 63 61 60 bytes

Make script/function c with this code and compile with uninitialized variables as 0:

a=argument0while(a>1){i++if i mod 2a=a*3+1else a/=2}return i

Call it with c(any number) and it will return how many times it took to become 1.

Answer 27

Emacs/Common Lisp, 61 bytes

(defun f(n)(if(= 1 n)0(1+(f(if(oddp n)(1+(* 3 n))(/ n 2))))))

alternatively:

(defun f(n)(if(= 1 n)0(1+(f(if(oddp n)(+ n n n 1)(/ n 2))))))

Answer 28

Python 2, 38 37 bytes

f=lambda n:n<3or-~f([n/2,n*3+1][n%2])

Thanks to @user84207 for a suggestion that saved 1 byte!

Note that this returns True instead of 1.

Try it online!

Answer 29

Befunge-93, 29 bytes

&<\+1\/2+*%2:+2*5:_$#-.#1@#:$

Try it online!

A nice and concise one-liner. This uses the formula (n+(n*5+2)*(n*5%2))/2 to calculate the next number in the series.

Answer 30

Emojicode, 157 bytes

🐖🎅🏿➡️🔡🍇🍮a🐕🍮c 0🔁▶️a 1🍇🍊😛🚮a 2 0🍇🍮a➗a 2🍉🍓🍇🍮a➕✖️a 3 1🍉🍮c➕c 1🍉🍎🔡c 10🍉

Try it online!

Explanation:

🐋🚂🍇    
🐖🎅🏿➡️🔡🍇
🍮a🐕      👴 input integer variable 'a'
🍮c 0         👴 counter variable
🔁▶️a 1🍇      👴 loop while number isn’t 1
🍊😛🚮a 2 0🍇     👴 if number is even
🍮a➗a 2       👴 divide number by 2
🍉
🍓🍇      👴 else
🍮a➕✖️a 3 1   👴 multiply by 3 and add 1
🍉
🍮c➕c 1     👴 increment counter
🍉
🍎🔡c 10   👴 return final count as string
🍉
🍉
🏁🍇
 😀🎅🏿 16
🍉