Challenges

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# Background

Check out this video on the Collatz conjecture, also known as A006577.

If you don't know what this is, we're given an equation of $3x + 1$, and it is applied this way:

• If $x$ is odd, then $3x + 1$.
• If $x$ is even, then $\frac{x}{2}$.

This will send us in a loop of 4 → 2 → 1 → 4 → 2 → 1..., which got me into making this challenge.

# Challenge

Write a program that establishes the Collatz conjecture:

• Take input of a positive integer. This will be the $x$ of the problem.
• Read the background for how it works, or watch the video for further explanation.
• The result should be how many turns it would take before reaching $1$. There, the sequence stops.
• This is code-golf, so the shortest program in each language wins!

# Test Cases

From 1 to 10:

1  → 0  (1)
2  → 1  (2 → 1)
3  → 7  (3 → 10 → 5 → 16 → 8 → 4 → 2 → 1)
4  → 2  (4 → 2 → 1)
5  → 5  (5 → 16 → 8 → 4 → 2 → 1)
6  → 8  (6 → 3 → 10 → 5 → 16 → 8 → 4 → 2 → 1)
7  → 16 (7 → 22 → 11 → 34 → 17 → 52 → 26 → 13 → 40 → 20 → 10 → 5 → 16 → 8 → 4 → 2 → 1)
8  → 3  (8 → 4 → 2 → 1)
9  → 19 (9 → 28 → 14 → 7 → 22 → 11 → 34 → 17 → 52 → 26 → 13 → 40 → 20 → 10 → 5 → 16 → 8 → 4 → 2 → 1)
10 → 6  (10 → 5 → 16 → 8 → 4 → 2 → 1)


More of these can be found on OEIS. Thanks to @Shaggy for the link!

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# JavaScript, 28 bytes

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


Try it online!

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# Python 3, 4842 39 bytes

Saved 6 bytes thanks to Hakerh400‭ in the comments

Saved another 3 bytes thanks to user in the comments

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


Try it online!

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42 bytes (1 comment)
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f 1=0
f n=1+f([div n 2,n*3+1]!!mod n 2)


Try it online!

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# Sclipting, (UTF-16) 443432 bytes

貶要❶갠剩❷隔❸增갰乘嗎終并長貶


Because comparing with 1 is expensive (requires copying and decrementing), we instead use a modified version of the Collatz sequence - namely, we use the sequence where every number is one lower. This allows us to compare with 0 instead.

Input of n

❶갠剩    Take n mod 2
❷隔      Compute n integer divided by 2 (1)
❸增갰乘  Compute n plus 1 and multiplied by 3 (2)
嗎       Condition on n mod 2; if odd, take (1) else take (2)


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# Japt, 15 bytes

É©Òß[U*3ÄUz]gUv


Try it

É©Òß[U*3ÄUz]gUv     :Implicit input of integer U
É                   :Subtract 1
Ò                 :Negate the bitwise NOT of (i.e., increment)
ß                :Recursive call with input
[               :  Array containing
U*3Ä           :    U*3+1
Uz         :    U floor divided by 2
]        :  End array
g       :  Get element at 0-based index
Uv     :    Is U divisible by 2?
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# BQN, 31 28 bytes

{1+(1≠𝕩)◶¯1‿𝕊2(|⊑÷˜∾1+3×⊢)𝕩}


Try it online!

An anonymous function which takes a number.

the ¯1 branch is a bit tacky but saves a byte over (1+𝕊).

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# Ruby, 33 bytes

Recursive lambda solution.

c=->n{n<2?0:1+c[n%2<1?n/2:n*3+1]}

c=->n{                          }  # c = lambda taking n
n<2? :                       # if n < 2...
0                        # return 0...
1+c[               ]   # else return 1 + collatz count for...
n%2<1?   :         # if n is even...
n/2          # n / 2...
n*3+1    # else 3n + 1


Try it online!

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# shortC, 47 bytes

a,b;AOK"%d",&b);b-1;b=b%2?b*3+1:b/2)a++;R"%d",a


Try it online!

From @ugoren's C answer from CGCC.

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# Scala, 50 bytes

Stream.iterate(_)(x=>Seq(x/2,3*x+1)(x%2))indexOf 1


Try it in Scastie!

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# AWK, 46 bytes

{c=0;for(n=\$0;n-1;c++)n=n%2?n*3+1:n/2;print c}


Try it online!

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# Java (JDK), 150 bytes

interface M{static void main(String[]a){int i=new java.util.Scanner(System.in).nextInt(),j=0;for(;i!=1;j++){i=i%2==1?i*3+1:i/2;}System.out.print(j);}}


Try it online!

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