Collatz Conjecture (OEIS A006577)

Question 1

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

Question 2

Octave, 65 bytes

@(x)eval 'k=0;do++k;if mod(x,2)x=3*x+1;else x/=2;end,until x<2,k'

Try it online!

It's time we have an Octave answer here. It's a straight forward implementation of the algorithm, but there are several small golfs.

Using do ... until instead of while ... end saves some bytes. Instead of having while x>1,k++;...end, we could have do++k;...until x<2, saving two bytes. Using eval in an anonymous function saves a few bytes, compared to having input(''). Also, skipping the parentheses in the eval call saves some bytes.

Question 3

Ruby, 48 bytes

(f=->n{n>1&&1+f[n%2<1?n/2:3*n+1]||0})[gets.to_i]

Same as other Ruby, but using n%2?a:b syntax instead of [a,b][n%2]. Saves one char.

Question 4

You mention that you save one byte off of the other Ruby answer yet your answer is 13 bytes longer because of (...)[gets.to_i]. Even if I remove that (which doesn't break your answer) you still have the same length as the other answer. This is ok, I just thought maybe you might have made a mistake.

Question 5

PHP, (削除) 80 (削除ここまで) 73 Bytes

Tried a recursive function Try it here! (80 Bytes)

Try it online (73 Bytes)

Code (recursive function)

function f($n,$c=0){echo($n!=1)?(($n%2)?f($n*3+1,$c+1):f($n/2,$c+1)):$c;}

Output

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

Explanation

function f($n,$c=0){ //$c counts the iterations, $n the input
echo($n!=1)? 
(($n%2)?
 f($n*3+1,$c+1): //$n is odd
 f($n/2,$c+1)) //$n is even
:$c; //echo $c (counter) when n ==1
}

Question 6

No, the test cases aren't part of the byte count. Your submission looks fine as it is.

Question 7

Python 2, 48 bytes

f=lambda n,s=0:s*(n<2)or f((n/2,3*n+1)[n%2],s+1)

Try it online!

Aaah, recursion.

# s*0 or s*1.
s*(n<2)
# while n>1, this will evaluate to 0 or f(n,s+1).
# Since positive integers are Truthy, this will return f().
# when n<2, this will return s without evaluating f().
s*(n<2)or f(...)

Question 8

JavaScript, 35 bytes

f=(n,c)=>n<2?c:f(n%2?n*3+1:n/2,-~c)

Try it online!

Question 9

Japt, (削除) 18 (削除ここまで) 15 bytes

É©Òß[U*3ÄUz]gUv

Try it

Question 10

Wren, 68 bytes

Fn.new{|n|
var c=0
while(n>1){
n=n%2==0?n/2:n*3+1
c=c+1
}
return c
}

Try it online!

Explanation

Fn.new{|n| // New anonymous function with param n
var c=0 // Declare a variable c as 0
while(n>1){ // While n is larger than 1:
n=n%2==0? // If n is divisible by 2:
 n/2: // Halve n
 n*3+1 // Otherwise, triple n & increment.
c=c+1 // Increment the counter
} // This is here due to Wrens bad brace-handling system
return c // Return the value of the counter
}

Question 11

Keg `-rR`, 23 bytes (SBCS)

^{I really need to remember those new instructions.}

0&{:1>|:2%[3*9|1⁄2](6)

Try it online!

Question 12

-5 bytes with -rR and new register commands

Question 13

Kotlin, 63 bytes

{var n=it
var c=0
while(n>1){n=if(n%2==0)n/2 else n*3+1
c++}
c}

Try it online!

Question 14

MAWP, 36 bytes

@[!!2P2WA{%3W1M}<%2P>1A{1M}/1M\]%1A:

Works as per the basic rules. Increments existing 1 in stack for each step.

Prints out n-1.

Try it!

Question 15

Rust, 62 bytes

fn c(x:u8)->u8{if x==1{0}else{c(if x%2==0{x/2}else{x*3+1})+1}}

This recursively determines the total. For 2 extra bytes u8 can be changed to u64 to support all 64-bit integers instead of just 8-bit ones.

Question 16

Welcome to the site, and nice first answer! Be sure to check out our Tips for golfing in Rust page for ways you can golf your program

Question 17

Lua, 75 bytes

function C(x)z=0 while x>1 do x=({x//2,3*x+1})[x%2+1]z=z+1 end return z end

Try it online!

Question 18

Python - 101 bytes

n=int(input())
m=0
while n>1:
 if n%2==0:
 n=n/2
 m+=1
 else:
 n=3*n+1
 m+=1
if n==1:
 print(m)

This assumes n is inputted to STDIN as an integer. If it is not explicitly that, a type check is most certainly possible, but would cost a few bytes, i.e.

if type(n) != int: 
 print(N/A)

(edit 1: input is so expensive)

Question 19

71 bytes

Question 20

Burlesque, 26 bytes

1{J2dv{2./}{3.*+.}IE}C~1Fi

Try it online!

1 #Needed for C~
{
 J #Duplicate
 2dv #Even
 {2./} #Halve
 {3.*+.} #3n+1
 IE #If even, else
}
C~ #Continue indefinitely
1Fi #Find index of 1

Question 21

Python 3, 64 bytes

def f(n,a=0):
 while n>0:n=[n//2,n*3+1][n%2];a+=1
 yield a

Try it online!

Question 22

INPUT A
Y:
IF A MOD 2=0 THEN
 B=A/2
 PRINT B
 A=B
 ELSE
 B=A*3+1
 PRINT B
 A=B
 END IF
IF A>1 THEN
 GOTO Y
 ELSE
 END IF

Question 23

Welcome to Code Golf! Could you edit in the language used, along with the length (in bytes) of your code, as this is a [code-golf] challenge? I've edited your answer slightly to format the code properly

Question 24

tinylisp, (削除) 68 (削除ここまで) 67 bytes

(load library
(d f(q((n)(i(e n 1)0(inc(f(i(odd? n)(a(* 3 n)1)(/ n 2

Try it online!

This is the same recursive solution as, e.g., Carcigenicate's Clojure answer. Because tinylisp has only addition and subtraction built in, I load the standard library to get odd?, /, *, and inc. Other library functions would make the code longer; for instance, I'm defining the function manually with (q((n)(...))) rather than using (lambda(n)(...)). Here's how it would look ungolfed and indented:

(load library)
(def collatz
 (lambda (n)
 (if (equal? n 1)
 0
 (inc
 (collatz
 (if (odd? n)
 (add2 (* 3 n) 1)
 (/ n 2)))))))

Here's a 101-byte solution that doesn't use the library. The E function returns n/2 if n is even and the empty list (falsey) if n is odd, so it can be used both to test evenness and to divide by 2.*

(d E(q((n _)(i(l n 2)(i n()_)(E(s n 2)(a _ 1
(d f(q((n)(i(e n 1)0(a 1(f(i(E n 0)(E n 0)(a(a(a n n)n)1

_{* Only works for strictly positive integers, but that's exactly what we're dealing with in this challenge.}

Question 25

Desmos, 63 bytes

i->.5si(5k+1)+sk-s+1,o->o+s
i=\ans_0
o=0
k=mod(i,2)
s=sign(i-1)

Output is the value of o after the code finishes running.

Have fun trying to figure out how this works! (It's really not as complicated as it seems)

Try It On Desmos!

Try It On Desmos! - Prettified

Question 26

Ruby, 60 bytes

n,i=gets.to_i,0;while n>1 do n=n%2==0?n/2:n*3+1;i+=1 end;p i

Pretty readable and easy to understand compared to the previous Ruby submission.

Attempt This Online!

Question 27

56 bytes

Question 28

Factor + `project-euler.014`, 22 bytes

[ collatz length 1 - ]

Try it online!

Question 29

Haskell (54 bytes)

c n a|n<2=a|odd n=c(3*n+1)(a+1)|even n=c(div n 2)(a+1)

This function takes two arguments, the value n and an accumulator a. The type signature is: c :: Int -> Int -> Int.

In expanded form:

collatz n acc
 | n < 2 = acc
 | odd n = collatz (3 * n + 1) (acc + 1)
 | even n = collatz (div n 2) (acc + 1)

Question 30

I forget where this is spelled out on Meta, but you can't expect extra input. Fortunately, it can be golfed more than enough to compensate (and without abandoning the accumulator, which would result in nimi's existing solution).

Question 31

brev, 71 bytes

(rec(f x)(cond((= x 1)0)((odd? x)(+(f(+(* 3 x)1))1))(1(+(f(/ x 2))1))))

Question 32

Fig, \15ドル\log_{256}(96)\approx\$ 12.347 bytes

#?{x}oX?Ox}*3xH

Try it online!

This challenge does not lend itself well to Fig's implicit inputs...

#?{x}oX?Ox}*3xH
 {x # Decrement the input
#? # If ^ is false, return ^, else return...
 oX # Call this function with the following argument:
 ?Ox # If odd
 *3x # Multiply by 3
 } # Add 1
 # Else
 H # Halve
 } # After calling this function, increment the result

Question 33

Vyxal, (削除) 17 (削除ここまで) (削除) 12 (削除ここまで) 11 bytes

∷[T›|1⁄2)İ2ḟ⇧

Try it Online!

-5 bytes thanks to lyxal

Removed flag thanks to emanresu A.

Question 34

Try it Online! for 12 bytes

Question 35

Flagless 11 (husk port)

Question 36

@emanresuA Alternative: ‡(½‡T›iİ2ḟ⇧

Question 37

Try it Online! for 10 bytes/7.875 bytes. Uses newer functionality that didn't exist back when this answer was made.

Question 38

Thunno 2, 17 bytes

(×ばつ+:1⁄2;ẋ)x−

Attempt This Online!

Only works in Thunno \$\le 2.1.9\$. In versions \$\ge 2.1.10\$, ẋ can be replaced with ẋ+.

Explanation

(×ばつ+:1⁄2;ẋ)x− # implicit input; x is initialised to 1
(1Q; ) # while TOS != 1:
 D # duplicate
 ɗ? # if TOS is odd:
 ×ばつ+ # triple and increment
 : # else:
 1⁄2 # halve
 ;ẋ # increment x
 x− # push x and decrement
 # implicit output

Question 39

><> (Fish), 31 bytes

:1=?\::2%?2,円
;nl~/i+1*3/
|.!0/

Try it

Question 40

Nekomata, 10 bytes

l{1>e1⁄23*→I

Attempt This Online!

l{1>e1⁄23*→I
l{ Repeat the following function until it fails, and count the number of steps:
 1> Check if greater than 1
 e Parallelly apply the following two functions:
 1⁄2 Check if it is even, and divide by 2
 3*→ Multiply by 3 and add 1
 I Choose the first result that does not fail

Question 41

Punchtape, 405 bytes & Punchcode, 50 bytes

I did this just for fun and showcase

Punchtape is an esoteric language trying to imitate punched tape, a form of data storage widely used by computers in the 1950s and 1960s.

Punchcode is the compiled (UTF-8) form of it.

There is no compiler or interpreter publicly accessible at the moment because I am still working on it.

punchtape

START|
O-OO-|
OOOO-|
----O|
---O-|
----O|
-----|
----O|
---O-|
----O|
---OO|
----O|
----O|
----O|
O--OO|
----O|
-----|
----O|
-----|
OOOO-|
---OO|
-----|
-OO--|
---O-|
O---O|
OOO-O|
--OO-|
--O-O|
OO-OO|
O--O-|
O--OO|
OOO-O|
--OOO|
--O--|
O-O--|
---O-|
-O--O|
OO-OO|
OOOO-|
---OO|
-----|
O----|
--OO-|
O--OO|
---O-|
--OOO|
-OO-O|
O-O--|
---O-|
-O--O|
-O-OO|

punchcode

(This contains characters that a lot of web browsers and fonts dont display or display them in a way that is inconvenient for showing code. To combat this, all characters have been shifted 32 times byte-wise to show as basic latin characters.)

6>!"! !"!#!!!3! ! ># ,"1=&%;23='4ドル");># 0&3"'-4")+

Here is the original form of the code, if you are curious:

Question 42

Uiua, 23 bytes

⧻⍢(×ばつ3⟩◿2.|¬∊2)¤

Explain:

⧻ # length of list
 ⍢( | ) # do while
 ⊂0 # prepend 0 to the list
 ×ばつ3⟩◿2. # apply collatz the the whole list
 ¬∊2 # while the list doesn't contain 2
 ¤ # arg as list

Question 43

TI-84 BASIC, 53 bytes

Input I
DelVar LWhile I≠1
L+1→L
remainder(I,2
(1+3I)Ans+.5Inot(Ans→I
End
Disp L

Stewie Griffin 46.8k16 gold badges137 silver badges304 bronze badges · Answer 1 · 2018-05-12 20:29:06Z

Octave, 65 bytes

@(x)eval 'k=0;do++k;if mod(x,2)x=3*x+1;else x/=2;end,until x<2,k'

Try it online!

It's time we have an Octave answer here. It's a straight forward implementation of the algorithm, but there are several small golfs.

Using do ... until instead of while ... end saves some bytes. Instead of having while x>1,k++;...end, we could have do++k;...until x<2, saving two bytes. Using eval in an anonymous function saves a few bytes, compared to having input(''). Also, skipping the parentheses in the eval call saves some bytes.

Joey 111 bronze badge · Answer 2 · 2018-06-14 17:35:03Z

1

\$\begingroup\$

Ruby, 48 bytes

(f=->n{n>1&&1+f[n%2<1?n/2:3*n+1]||0})[gets.to_i]

Same as other Ruby, but using n%2?a:b syntax instead of [a,b][n%2]. Saves one char.

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edited Jun 17, 2020 at 9:04

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1

answered Jun 14, 2018 at 17:35

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Joey

111 bronze badge

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1

\$\begingroup\$ You mention that you save one byte off of the other Ruby answer yet your answer is 13 bytes longer because of (...)[gets.to_i]. Even if I remove that (which doesn't break your answer) you still have the same length as the other answer. This is ok, I just thought maybe you might have made a mistake. \$\endgroup\$

Wheat Wizard
– Wheat Wizard ♦

2018年06月19日 14:11:14 +00:00
Commented Jun 19, 2018 at 14:11

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Francisco Hahn 6013 silver badges10 bronze badges · Answer 3 · 2018-03-26 13:23:17Z

PHP, (削除) 80 (削除ここまで) 73 Bytes

Tried a recursive function Try it here! (80 Bytes)

Try it online (73 Bytes)

Code (recursive function)

function f($n,$c=0){echo($n!=1)?(($n%2)?f($n*3+1,$c+1):f($n/2,$c+1)):$c;}

Output

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

Explanation

function f($n,$c=0){ //$c counts the iterations, $n the input
echo($n!=1)? 
(($n%2)?
 f($n*3+1,$c+1): //$n is odd
 f($n/2,$c+1)) //$n is even
:$c; //echo $c (counter) when n ==1
}

No, the test cases aren't part of the byte count. Your submission looks fine as it is.

Triggernometry 8314 silver badges11 bronze badges · Answer 4 · 2018-09-13 16:28:42Z

Python 2, 48 bytes

f=lambda n,s=0:s*(n<2)or f((n/2,3*n+1)[n%2],s+1)

Try it online!

Aaah, recursion.

# s*0 or s*1.
s*(n<2)
# while n>1, this will evaluate to 0 or f(n,s+1).
# Since positive integers are Truthy, this will return f().
# when n<2, this will return s without evaluating f().
s*(n<2)or f(...)

Oliver 7,9422 gold badges19 silver badges38 bronze badges · Answer 5 · 2019-01-23 22:05:37Z

1

\$\begingroup\$

JavaScript, 35 bytes

f=(n,c)=>n<2?c:f(n%2?n*3+1:n/2,-~c)

Try it online!

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edited Jun 17, 2020 at 9:04

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1

answered Jan 23, 2019 at 22:05

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Oliver

7,9422 gold badges19 silver badges38 bronze badges

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Shaggy 45k4 gold badges39 silver badges95 bronze badges · Answer 6 · 2018-02-21 15:14:09Z

1

\$\begingroup\$

Japt, (削除) 18 (削除ここまで) 15 bytes

É©Òß[U*3ÄUz]gUv

Try it

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edited Jun 17, 2020 at 9:04

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answered Feb 21, 2018 at 15:14

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user85052 · Answer 7 · 2019-12-12 12:59:23Z

Wren, 68 bytes

Fn.new{|n|
var c=0
while(n>1){
n=n%2==0?n/2:n*3+1
c=c+1
}
return c
}

Try it online!

Explanation

Fn.new{|n| // New anonymous function with param n
var c=0 // Declare a variable c as 0
while(n>1){ // While n is larger than 1:
n=n%2==0? // If n is divisible by 2:
 n/2: // Halve n
 n*3+1 // Otherwise, triple n & increment.
c=c+1 // Increment the counter
} // This is here due to Wrens bad brace-handling system
return c // Return the value of the counter
}

user85052 · Answer 8 · 2019-12-12 12:41:05Z

1

\$\begingroup\$

Keg `-rR`, 23 bytes (SBCS)

^{I really need to remember those new instructions.}

0&{:1>|:2%[3*9|1⁄2](6)

Try it online!

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edited Dec 14, 2019 at 0:56

answered Dec 12, 2019 at 12:41

user85052

\$\endgroup\$

1

\$\begingroup\$ -5 bytes with -rR and new register commands \$\endgroup\$

lyxal
– lyxal ♦

2019年12月13日 20:24:15 +00:00
Commented Dec 13, 2019 at 20:24

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snail_ 2,68013 silver badges16 bronze badges · Answer 9 · 2020-03-22 05:56:18Z

1

\$\begingroup\$

Kotlin, 63 bytes

{var n=it
var c=0
while(n>1){n=if(n%2==0)n/2 else n*3+1
c++}
c}

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answered Mar 22, 2020 at 5:56

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snail_

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Razetime 27.6k3 gold badges31 silver badges77 bronze badges · Answer 10 · 2020-08-26 09:26:51Z

1

\$\begingroup\$

MAWP, 36 bytes

@[!!2P2WA{%3W1M}<%2P>1A{1M}/1M\]%1A:

Works as per the basic rules. Increments existing 1 in stack for each step.

Prints out n-1.

Try it!

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answered Aug 26, 2020 at 9:26

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Razetime

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smitop 1516 bronze badges · Answer 11 · 2021-02-28 00:08:09Z

1

\$\begingroup\$

Rust, 62 bytes

fn c(x:u8)->u8{if x==1{0}else{c(if x%2==0{x/2}else{x*3+1})+1}}

This recursively determines the total. For 2 extra bytes u8 can be changed to u64 to support all 64-bit integers instead of just 8-bit ones.

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answered Feb 28, 2021 at 0:08

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1

1

\$\begingroup\$ Welcome to the site, and nice first answer! Be sure to check out our Tips for golfing in Rust page for ways you can golf your program \$\endgroup\$

caird coinheringaahing
– caird coinheringaahing ♦

2021年02月28日 00:33:16 +00:00
Commented Feb 28, 2021 at 0:33

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J. А. del Rio - слава СССР 3593 silver badges6 bronze badges · Answer 12 · 2021-03-01 14:28:56Z

1

\$\begingroup\$

Lua, 75 bytes

function C(x)z=0 while x>1 do x=({x//2,3*x+1})[x%2+1]z=z+1 end return z end

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edited Mar 1, 2021 at 14:54

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answered Mar 1, 2021 at 14:28

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J. А. del Rio - слава СССР

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user100887 · Answer 13 · 2021-03-15 20:56:07Z

Python - 101 bytes

n=int(input())
m=0
while n>1:
 if n%2==0:
 n=n/2
 m+=1
 else:
 n=3*n+1
 m+=1
if n==1:
 print(m)

This assumes n is inputted to STDIN as an integer. If it is not explicitly that, a type check is most certainly possible, but would cost a few bytes, i.e.

if type(n) != int: 
 print(N/A)

(edit 1: input is so expensive)

1

\$\begingroup\$ 71 bytes \$\endgroup\$

oeuf
– oeuf

2022年04月10日 03:55:47 +00:00
Commented Apr 10, 2022 at 3:55

DeathIncarnate 2,6108 silver badges9 bronze badges · Answer 14 · 2021-11-22 17:56:51Z

Burlesque, 26 bytes

1{J2dv{2./}{3.*+.}IE}C~1Fi

Try it online!

1 #Needed for C~
{
 J #Duplicate
 2dv #Even
 {2./} #Halve
 {3.*+.} #3n+1
 IE #If even, else
}
C~ #Continue indefinitely
1Fi #Find index of 1

Alan Bagel 4,2691 gold badge13 silver badges46 bronze badges · Answer 15 · 2021-12-07 18:24:59Z

1

\$\begingroup\$

Python 3, 64 bytes

def f(n,a=0):
 while n>0:n=[n//2,n*3+1][n%2];a+=1
 yield a

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answered Dec 7, 2021 at 18:24

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serdar evren 111 bronze badge · Answer 16 · 2022-01-23 20:29:46Z

1

\$\begingroup\$

INPUT A
Y:
IF A MOD 2=0 THEN
 B=A/2
 PRINT B
 A=B
 ELSE
 B=A*3+1
 PRINT B
 A=B
 END IF
IF A>1 THEN
 GOTO Y
 ELSE
 END IF

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edited Jan 23, 2022 at 20:31

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caird coinheringaahing ♦

50.9k11 gold badges133 silver badges364 bronze badges

answered Jan 23, 2022 at 20:29

serdar evren's user avatar

serdar evren

111 bronze badge

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1

7

\$\begingroup\$ Welcome to Code Golf! Could you edit in the language used, along with the length (in bytes) of your code, as this is a [code-golf] challenge? I've edited your answer slightly to format the code properly \$\endgroup\$

caird coinheringaahing
– caird coinheringaahing ♦

2022年01月23日 20:31:29 +00:00
Commented Jan 23, 2022 at 20:31

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DLosc 40.7k6 gold badges87 silver badges142 bronze badges · Answer 17 · 2017-10-26 21:43:57Z

tinylisp, (削除) 68 (削除ここまで) 67 bytes

(load library
(d f(q((n)(i(e n 1)0(inc(f(i(odd? n)(a(* 3 n)1)(/ n 2

Try it online!

This is the same recursive solution as, e.g., Carcigenicate's Clojure answer. Because tinylisp has only addition and subtraction built in, I load the standard library to get odd?, /, *, and inc. Other library functions would make the code longer; for instance, I'm defining the function manually with (q((n)(...))) rather than using (lambda(n)(...)). Here's how it would look ungolfed and indented:

(load library)
(def collatz
 (lambda (n)
 (if (equal? n 1)
 0
 (inc
 (collatz
 (if (odd? n)
 (add2 (* 3 n) 1)
 (/ n 2)))))))

Here's a 101-byte solution that doesn't use the library. The E function returns n/2 if n is even and the empty list (falsey) if n is odd, so it can be used both to test evenness and to divide by 2.*

(d E(q((n _)(i(l n 2)(i n()_)(E(s n 2)(a _ 1
(d f(q((n)(i(e n 1)0(a 1(f(i(E n 0)(E n 0)(a(a(a n n)n)1

_{* Only works for strictly positive integers, but that's exactly what we're dealing with in this challenge.}

Aiden Chow 14.5k1 gold badge22 silver badges61 bronze badges · Answer 18 · 2022-04-10 03:08:40Z

Desmos, 63 bytes

i->.5si(5k+1)+sk-s+1,o->o+s
i=\ans_0
o=0
k=mod(i,2)
s=sign(i-1)

Output is the value of o after the code finishes running.

Have fun trying to figure out how this works! (It's really not as complicated as it seems)

Try It On Desmos!

Try It On Desmos! - Prettified

oeuf 1,9556 silver badges19 bronze badges · Answer 19 · 2022-04-10 03:51:05Z

1

\$\begingroup\$

Ruby, 60 bytes

n,i=gets.to_i,0;while n>1 do n=n%2==0?n/2:n*3+1;i+=1 end;p i

Pretty readable and easy to understand compared to the previous Ruby submission.

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1

\$\begingroup\$ 56 bytes \$\endgroup\$

naffetS
– naffetS

2022年04月10日 20:17:11 +00:00
Commented Apr 10, 2022 at 20:17

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chunes 27.8k3 gold badges32 silver badges55 bronze badges · Answer 20 · 2022-05-08 20:19:03Z

1

\$\begingroup\$

Factor + `project-euler.014`, 22 bytes

[ collatz length 1 - ]

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answered May 8, 2022 at 20:19

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Ed The ''Pro'' 5433 silver badges10 bronze badges · Answer 21 · 2022-05-10 13:04:42Z

1

\$\begingroup\$

Haskell (54 bytes)

c n a|n<2=a|odd n=c(3*n+1)(a+1)|even n=c(div n 2)(a+1)

This function takes two arguments, the value n and an accumulator a. The type signature is: c :: Int -> Int -> Int.

In expanded form:

collatz n acc
 | n < 2 = acc
 | odd n = collatz (3 * n + 1) (acc + 1)
 | even n = collatz (div n 2) (acc + 1)

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answered May 10, 2022 at 13:04

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1

1

\$\begingroup\$ I forget where this is spelled out on Meta, but you can't expect extra input. Fortunately, it can be golfed more than enough to compensate (and without abandoning the accumulator, which would result in nimi's existing solution). \$\endgroup\$

Unrelated String
– Unrelated String

2022年08月21日 03:04:46 +00:00
Commented Aug 21, 2022 at 3:04

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Sandra 4622 silver badges5 bronze badges · Answer 22 · 2022-08-21 19:50:46Z

1

\$\begingroup\$

brev, 71 bytes

(rec(f x)(cond((= x 1)0)((odd? x)(+(f(+(* 3 x)1))1))(1(+(f(/ x 2))1))))

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Seggan 7,12228 silver badges53 bronze badges · Answer 23 · 2022-10-03 20:28:18Z

Fig, \15ドル\log_{256}(96)\approx\$ 12.347 bytes

#?{x}oX?Ox}*3xH

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This challenge does not lend itself well to Fig's implicit inputs...

#?{x}oX?Ox}*3xH
 {x # Decrement the input
#? # If ^ is false, return ^, else return...
 oX # Call this function with the following argument:
 ?Ox # If odd
 *3x # Multiply by 3
 } # Add 1
 # Else
 H # Halve
 } # After calling this function, increment the result

naffetS 13.1k2 gold badges22 silver badges54 bronze badges · Answer 24 · 2022-04-10 01:10:07Z

1

\$\begingroup\$

Vyxal, (削除) 17 (削除ここまで) (削除) 12 (削除ここまで) 11 bytes

∷[T›|1⁄2)İ2ḟ⇧

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-5 bytes thanks to lyxal

Removed flag thanks to emanresu A.

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answered Apr 10, 2022 at 1:10

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\$\begingroup\$ Try it Online! for 12 bytes \$\endgroup\$

lyxal
– lyxal ♦

2022年04月10日 03:04:27 +00:00
Commented Apr 10, 2022 at 3:04
\$\begingroup\$ Flagless 11 (husk port) \$\endgroup\$

emanresu A
– emanresu A

2023年01月30日 07:00:06 +00:00
Commented Jan 30, 2023 at 7:00
\$\begingroup\$ @emanresuA Alternative: ‡(½‡T›iİ2ḟ⇧ \$\endgroup\$

naffetS
– naffetS

2023年01月30日 18:20:09 +00:00
Commented Jan 30, 2023 at 18:20
\$\begingroup\$ Try it Online! for 10 bytes/7.875 bytes. Uses newer functionality that didn't exist back when this answer was made. \$\endgroup\$

lyxal
– lyxal ♦

2023年07月18日 13:28:24 +00:00
Commented Jul 18, 2023 at 13:28

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The Thonnu 18.7k3 gold badges19 silver badges76 bronze badges · Answer 25 · 2023-06-03 18:15:16Z

Thunno 2, 17 bytes

(×ばつ+:1⁄2;ẋ)x−

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Only works in Thunno \$\le 2.1.9\$. In versions \$\ge 2.1.10\$, ẋ can be replaced with ẋ+.

Explanation

(×ばつ+:1⁄2;ẋ)x− # implicit input; x is initialised to 1
(1Q; ) # while TOS != 1:
 D # duplicate
 ɗ? # if TOS is odd:
 ×ばつ+ # triple and increment
 : # else:
 1⁄2 # halve
 ;ẋ # increment x
 x− # push x and decrement
 # implicit output

chunes 27.8k3 gold badges32 silver badges55 bronze badges · Answer 26 · 2023-06-04 01:33:13Z

1

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><> (Fish), 31 bytes

:1=?\::2%?2,円
;nl~/i+1*3/
|.!0/

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alephalpha 51.9k7 gold badges75 silver badges196 bronze badges · Answer 27 · 2023-06-04 02:38:59Z

Nekomata, 10 bytes

l{1>e1⁄23*→I

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l{1>e1⁄23*→I
l{ Repeat the following function until it fails, and count the number of steps:
 1> Check if greater than 1
 e Parallelly apply the following two functions:
 1⁄2 Check if it is even, and divide by 2
 3*→ Multiply by 3 and add 1
 I Choose the first result that does not fail

madeforlosers 2,0181 silver badge11 bronze badges · Answer 28 · 2024-03-11 19:24:51Z

Punchtape, 405 bytes & Punchcode, 50 bytes

I did this just for fun and showcase

Punchtape is an esoteric language trying to imitate punched tape, a form of data storage widely used by computers in the 1950s and 1960s.

Punchcode is the compiled (UTF-8) form of it.

There is no compiler or interpreter publicly accessible at the moment because I am still working on it.

punchtape

START|
O-OO-|
OOOO-|
----O|
---O-|
----O|
-----|
----O|
---O-|
----O|
---OO|
----O|
----O|
----O|
O--OO|
----O|
-----|
----O|
-----|
OOOO-|
---OO|
-----|
-OO--|
---O-|
O---O|
OOO-O|
--OO-|
--O-O|
OO-OO|
O--O-|
O--OO|
OOO-O|
--OOO|
--O--|
O-O--|
---O-|
-O--O|
OO-OO|
OOOO-|
---OO|
-----|
O----|
--OO-|
O--OO|
---O-|
--OOO|
-OO-O|
O-O--|
---O-|
-O--O|
-O-OO|

punchcode

(This contains characters that a lot of web browsers and fonts dont display or display them in a way that is inconvenient for showing code. To combat this, all characters have been shifted 32 times byte-wise to show as basic latin characters.)

6>!"! !"!#!!!3! ! ># ,"1=&%;23='4ドル");># 0&3"'-4")+

Here is the original form of the code, if you are curious:

Zylviij 4702 silver badges5 bronze badges · Answer 29 · 2024-03-15 22:14:03Z

Uiua, 23 bytes

⧻⍢(×ばつ3⟩◿2.|¬∊2)¤

Explain:

⧻ # length of list
 ⍢( | ) # do while
 ⊂0 # prepend 0 to the list
 ×ばつ3⟩◿2. # apply collatz the the whole list
 ¬∊2 # while the list doesn't contain 2
 ¤ # arg as list

madeforlosers 2,0181 silver badge11 bronze badges · Answer 30 · 2024-05-09 15:03:53Z

1

\$\begingroup\$

TI-84 BASIC, 53 bytes

Input I
DelVar LWhile I≠1
L+1→L
remainder(I,2
(1+3I)Ans+.5Inot(Ans→I
End
Disp L

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answered May 9, 2024 at 15:03

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Collatz Conjecture (OEIS A006577)

149 Answers 149

Octave, 65 bytes

Ruby, 48 bytes

PHP, (削除) 80 (削除ここまで) 73 Bytes

Python 2, 48 bytes

JavaScript, 35 bytes

Japt, (削除) 18 (削除ここまで) 15 bytes

Wren, 68 bytes

Explanation

Keg -rR, 23 bytes (SBCS)

Kotlin, 63 bytes

MAWP, 36 bytes

Rust, 62 bytes

Lua, 75 bytes

Python - 101 bytes

Burlesque, 26 bytes

Python 3, 64 bytes

tinylisp, (削除) 68 (削除ここまで) 67 bytes

Desmos, 63 bytes

Ruby, 60 bytes

Factor + project-euler.014, 22 bytes

Haskell (54 bytes)

brev, 71 bytes

Fig, \15ドル\log_{256}(96)\approx\$ 12.347 bytes

Vyxal, (削除) 17 (削除ここまで) (削除) 12 (削除ここまで) 11 bytes

Thunno 2, 17 bytes

Explanation

><> (Fish), 31 bytes

Nekomata, 10 bytes

Punchtape, 405 bytes & Punchcode, 50 bytes

punchtape

punchcode

Uiua, 23 bytes

TI-84 BASIC, 53 bytes

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Keg `-rR`, 23 bytes (SBCS)

Factor + `project-euler.014`, 22 bytes