Collatz conjecture; Count the tries to reach $1$
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!
JavaScript, 28 bytes …
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[Sclipting], (UTF-16) 44 34 32 …
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[Python 3], 48 42 39 bytes …
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[Haskell], 43 39 bytes …
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Lua 5.4, 67 60 bytes ``` lu …
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[BQN], 31 28 bytes {1+( …
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Japt, 15 bytes É©Òß[U3Ä …
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J, 35 char ``` :@:&3)@.(2& …
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[Wolfram Language (Mathematica …
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C (gcc), 43 38 bytes ``` C …
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[AWK], 46 bytes {c=0;fo …
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[shortC], 47 bytes a,b; …
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[Ruby], 33 bytes Recursive la …
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Scala, 50 bytes ```scala Str …
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[Java (JDK)], 150 bytes …
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15 answers
Sclipting, (UTF-16) 44 34 32 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
貶 Decrement n
要 While n (is non-zero)
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)
終 End while
并長貶 Join stack into a list, take the length and decrement by one
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Python 3, 48 42 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])
Japt, 15 bytes
É©Òß[U*3ÄUz]gUv
É©Òß[U*3ÄUz]gUv :Implicit input of integer U
É :Subtract 1
© :Logical AND with
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?0 comment threads
BQN, 31 28 bytes
{1+(1≠𝕩)◶¯1‿𝕊2(|⊑÷˜∾1+3×⊢)𝕩}
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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J, 35 char
<:#-:`(>:@:*&3)@.(2&|)^:(1&<)^:(<_)
How it works:
NB. <: subtract one from number result on right
NB. # count number of items from list result on right
NB. -: if intermediate result is even, half the value
NB. ` make a gerund from verb to left and verb to right
NB. (>:@:*&3) if intermediate result is odd, multiply by 3
NB. @. choose from gerund using index on right
NB. (2&|) mod 2 used as index into gerund on left
NB. ^:(1&<) repeat verb on left while intermediate value is >1
NB. ^:(<_) repeat verb on left until intermediate value stops changing and keep intermediate values in a list
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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
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Scala, 50 bytes
Stream.iterate(_)(x=>Seq(x/2,3*x+1)(x%2))indexOf 1
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Wolfram Language (Mathematica), 38 bytes
Tr[1^ResourceFunction["Collatz"]@#]-1&
Don't Try it online!
Doesn't work on TIO due to using the Collatz builtin which needs to be downloaded from the Function Repository
Relatively badly-golfed version without ResourceFunction (52 bytes):
i=0;If[#!=1,i++;#0[If[EvenQ@#,Floor[#/2],3 #+1]],i]&
Had to add a slightly janky print statement into the main body for the value of i to get reset on each evaluation.
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);}}
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0 comment threads