Saya pikir dugaan Collatz sudah terkenal. Tetapi bagaimana jika kita membalikkan aturan?

Mulai dengan bilangan bulat n> = 1.

Ulangi langkah-langkah berikut:

Jika n adalah genap , kalikan dengan 3 dan tambahkan 1.

Jika n ganjil , kurangi 1 dan bagi dengan 2.

Berhenti ketika mencapai 0

Cetak nomor yang diulang.

Kasus uji:

 1        => 1, 0
 2        => 2, 7, 3, 1, 0
 3        => 3, 1, 0
10        => 10, 31, 15, 7, 3...
14        => 14, 43, 21, 10, ...

Aturan:

• Urutan ini tidak berfungsi untuk banyak angka karena masuk dalam loop tak terbatas. Anda tidak perlu menangani kasus-kasus itu. Hanya mencetak test case di atas sudah cukup.

• Saya menyarankan untuk mengurangi 1 dan membaginya dengan dua untuk memberikan integer yang valid untuk melanjutkan, tetapi tidak harus dihitung seperti itu. Anda dapat membaginya dengan 2 dan dilemparkan ke integer atau apa pun metode lain yang akan memberikan hasil yang diharapkan.

• Anda perlu mencetak input awal juga.

• Output tidak perlu diformat sebagai kasus uji. Itu hanya saran. Namun, perintah yang diulang harus dihormati.

• Kode terkecil menang.

Perl 6 , 30 byte

{$_,{$_%2??$_+>1!!$_*3+1}...0}

Cobalah online!

Blok kode anonim yang mengembalikan urutan.

Penjelasan:

{$_,{$_%2??$_+>1!!$_*3+1}...0}
{                            }   # Anonymous code block
   ,                     ...     # Define a sequence
 $_                              # That starts with the given value
    {                   }        # With each element being
     $_%2??     !!               # Is the previous element odd?
           $_+>1                 # Return the previous element bitshifted right by 1
                  $_*3+1         # Else the previous element multiplied by 3 plus 1
                            0    # Until the element is 0

Haskell , 40 39 byte

f 0=[]
f n=n:f(cycle[3*n+1,div n 2]!!n)

Cobalah online!

Sekarang tanpa 0 akhir.

Bersihkan , 53 byte

import StdEnv
$0=[0]
$n=[n: $if(isOdd n)(n/2)(n*3+1)]

Cobalah online!

Jelly , 9 byte

:++‘ƊḂ?Ƭ2

Cobalah online!

Python 2, 54 52 44 byte

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

-2 byte terima kasih kepada Tn. Xcoder

Pasti ada cara yang lebih cepat. Anehnya, ketika saya mencoba lambda itu adalah bytecount yang sama. Saya mungkin berhalusinasi.

Cobalah online!

Haskell , 76 69 61 56 byte

Saya merasa ini terlalu lama. Di sini lmenghasilkan daftar tak terbatas dari urutan inverse-collatz, dan fungsi anonim di baris pertama hanya memotongnya di tempat yang tepat.

Terima kasih untuk -5 byte @ ØrjanJohansen!

fst.span(>0).l
l r=r:[last$3*k+1:[div k 2|odd k]|k<-l r]

Cobalah online!

Rust , 65 64 byte

|mut n|{while n>0{print!("{} ",n);n=if n&1>0{n>>1}else{n*3+1};}}

Cobalah online!

05AB1E , 15 14 byte

[Ð=_#Èi3*>ë<2÷

-1 byte terima kasih @MagicOctopusUrn .

Cobalah online.

Penjelasan:

[             # Start an infinite loop
 Ð            #  Duplicate the top value on the stack three times
              #  (Which will be the (implicit) input in the first iteration)
  =           #  Output it with trailing newline (without popping the value)
   _#         #  If it's exactly 0: stop the infinite loop
     Èi       #  If it's even:
       3*     #   Multiply by 3
         >    #   And add 1
      ë       #  Else:
       <      #   Subtract 1
        2÷    #   And integer-divide by 2

JAEL , 18 byte

![ؼw>î?èÛ|õÀ

Cobalah online!

JavaScript (ES6), 31 bytes

f=n=>n&&n+' '+f(n&1?n>>1:n*3+1)

Try it online!

Or 30 bytes in reverse order.

Pyth, 12 bytes

.u?%N2/N2h*3

Try it here as a test suite!

Wolfram Language (Mathematica), 35 bytes

0<Echo@#&&#0[3#+1-(5#+3)/2#~Mod~2]&

Try it online!

0<Echo@# && ...& is short-circuit evaluation: it prints the input #, checks if it's positive, and if so, evaluates .... In this case, ... is #0[3#+1-(5#+3)/2#~Mod~2]; since #0 (the zeroth slot) is the function itself, this is a recursive call on 3#+1-(5#+3)/2#~Mod~2, which simplifies to 3#+1 when # is even, and (#-1)/2 when # is odd.

Common Lisp, 79 bytes

(defun x(n)(cons n(if(= n 0)nil(if(=(mod n 2)0)(x(+(* n 3)1))(x(/(- n 1)2))))))

Try it online!

PowerShell, 53 52 bytes

param($i)for(;$i){$i;$i=(($i*3+1),($i-shr1))[$i%2]}0

Try it Online!

Edit:
-1 byte thanks to @mazzy

Emojicode 0.5, 141 bytes

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

Try it online!

🐖🎅🏿🍇
🍮a🐕      👴 input integer variable 'a'
😀🔡a 10      👴 print input int
🔁▶️a 0🍇      👴 loop while number isn’t 0
🍊😛🚮a 2 1🍇     👴 if number is odd
🍮a➗a 2       👴 divide number by 2
🍉
🍓🍇      👴 else
🍮a➕✖️a 3 1   👴 multiply by 3 and add 1
🍉
😀🔡a 10     👴 print iteration
🍉🍉

Stax, 11 bytes

₧↑╔¶┘tÇ╣;↑è

Run and debug it

MathGolf, 12 bytes

{o_¥¿½É3*)}∟

Try it online!

Explanation

{             Start block of arbitrary length
 o            Output the number
  _           Duplicate
   ¥          Modulo 2
    ¿         If-else with the next two blocks. Implicit blocks consist of 1 operator
     ½        Halve the number to integer (effectively subtracting 1 before)
      É       Start block of 3 bytes
       3*)    Multiply by 3 and add 1
          }∟  End block and make it do-while-true

x86 machine code, 39 bytes

00000000: 9150 6800 0000 00e8 fcff ffff 5958 a901  .Ph.........YX..
00000010: 0000 0074 04d1 e8eb 066a 035a f7e2 4009  ...t.....j.Z..@.
00000020: c075 dec3 2564 20                        .u..%d 

Assembly (NASM syntax):

section .text
	global func
	extern printf
func:					;the function uses fastcall conventions
	xchg eax, ecx			;load function arg into eax
	loop:
		push eax
		push fmt
		call printf	;print eax
		pop ecx
		pop eax
		test eax, 1	;if even zf=1
		jz even		;if eax is even jmp to even
		odd:		;eax=eax/2
			shr eax, 1
			jmp skip
		even:		;eax=eax*3+1
			push 3
			pop edx
			mul edx
			inc eax
		skip:
		or eax, eax
		jne loop	;if eax!=0, keep looping
	ret			;return eax
section .data
	fmt db '%d '

Try it online!

R, 66 61 bytes

-5 bytes thanks to Robert S. in consolidating ifelse into if and removing brackets, and x!=0 to x>0

print(x<-scan());while(x>0)print(x<-`if`(x%%2,(x-1)/2,x*3+1))

instead of

print(x<-scan());while(x!=0){print(x<-ifelse(x%%2,(x-1)/2,x*3+1))}

Try it online!

perl -Minteger -nlE, 39 bytes

{say;$_=$_%2?$_/2:3*$_+1 and redo}say 0

Add++, 38 35 33 bytes

D,f,@:,d3*1+$2/iA2%D
+?
O
Wx,$f>x

Try it online!

How it works

First, we begin by defining a function $f(x)$, that takes a single argument, performs the inverted Collatz operation on $x$ then outputs the result. That is,

When in function mode, Add++ uses a stack memory model, otherwise variables are used. When calculating $f(x)$, the stack initially looks like $S = [x]$.

We then duplicate this value (d), to yield $S = [x, x]$. We then yield the first possible option, $3x + 1$ (3*1+), swap the top two values, then calculate $\lfloor\frac{x}{2}\rfloor$, leaving $S = [3x+1, \lfloor\frac{x}{2}\rfloor]$.

Next, we push $x$ to $S$, and calculate the bit of $x$ i.e. $x \: \% \: 2$, where $a \: \% \: b$ denotes the remainder when dividing $a$ by $b$. This leaves us with $S = [3x+1, \lfloor\frac{x}{2}\rfloor, (x \: \% \: 2)]$. Finally, we use D to select the element at the index specified by $(x \: \% \: 2)$. If that's $0$, we return the first element i.e. $3x+1$, otherwise we return the second element, $\lfloor\frac{x}{2}\rfloor$.

That completes the definition of $f(x)$, however, we haven't yet put it into practice. The next three lines have switched from function mode into vanilla mode, where we operate on variables. To be more precise, in this program, we only operate on one variable, the active variable, represented by the letter x. However, x can be omitted from commands where it is obviously the other argument.

For example, +? is identical to x+?, and assigns the input to x, but as x is the active variable, it can be omitted. Next, we output x, then entire the while loop, which loops for as long as $x \neq 0$. The loop is very simple, consisting of a single statement: $f>x. All this does is run $f(x)$, then assign that to x, updating x on each iteration of the loop.

Retina 0.8.2, 46 bytes

.+
$*
{*M`1
^(..)+$
$&$&$&$&$&$&111
1(.*)\1
$1

Try it online! Explanation:

.+
$*

Convert to unary.

{

Repeat until the value stops changing.

*M`1

Print the value in decimal.

^(..)+$
$&$&$&$&$&$&111

If it is even, multiply by 6 and add 3.

1(.*)\1
$1

Subtract 1 and divide by 2.

The trailing newline can be suppressed by adding a ; before the {.

Red, 70 bytes

func[n][print n if n = 0[exit]either odd? n[f n - 1 / 2][f n * 3 + 1]]

Try it online!

Racket, 75 bytes

(define(f n)(cons n(if(= n 0)'()(if(odd? n)(f(/(- n 1)2))(f(+(* 3 n)1))))))

Try it online!

Equivalent to JRowan's Common Lisp solution.

C# (.NET Core), 62 bytes

a=>{for(;a>0;a=a%2<1?a*3+1:a/2)Console.Write(a+" ");return a;}

Try it online!

Ungolfed:

a => {
    for(; a > 0;                // until a equals 0
        a = a % 2 < 1 ?             // set a depending on if a is odd or even
                a * 3 + 1 :             // even
                a / 2                   // odd (minus one unnecessary because of int casting)
    )
        Console.Write(a + " "); // writes the current a to the console
    return a;                   // writes a to the console (always 0)
}

Dart, 49 bytes

f(n){for(;n>0;n=n%2>0?(n-1)>>1:(n*3)+1)print(n);}

Try it online!

Gambit Scheme (gsi), 74 bytes

(define(f n)(if(= 0 n)'()(cons n(f(if(even? n)(+ (* 3 n)1)(/(- n 1)2))))))

Try it online!