Different number, same weight

Question

Background

The Hamming weight of an integer is the number of ones in its binary representation. For this challenge, integers are represented with 32 bits, and they are unsigned.

Challenge

Given an integer between 0 and 2^32-1 (non-inclusive), output a different integer within the same range, and also with the same Hamming weight.

Examples

Input (Decimal) | Input (Binary) | Hamming weight | Possible output (Decimal)
       46       |   0b0010 1110  |       4        |      15
       12       |   0b0000 1100  |       2        |      3
        1       |   0b0000 0001  |       1        |      2
        3       |   0b0000 0011  |       2        |      6
      2^31      |   0b1000....0  |       1        |      1
      2^31+2    |   0b1000...10  |       2        |      3
      2^32-5    |   0b1111..011  |       31       |      2^31-1
      2^32-2    |   0b1111....0  |       31       |      2^31-1
        0       |   0b0000 0000  |       0        | None (This case need not be handled)
      2^32-1    |   0b1111....1  |       32       | None (This case need not be handled)

Scoring

This is code-golf, so the solution in the fewest bytes in each language wins.

Answer 1

x86-64 assembly, 5 4 bytes

   0:   97                      xchg   %eax,%edi
   1:   d1 c0                   rol    %eax
   3:   c3                      retq   

A function using the C calling convention that bitwise rotates its argument left by 1 bit.

Answer 2

Python, 20 bytes

lambda x:x*2%~-2**32

Bitwise rotation left by 1 bit.

Answer 3

Jelly, 10 8 bytes

‘&~^^N&$

Swaps the least significant set and unset bit.

Try it online!

How it works

‘&~^^N&$  Main link. Argument: n

‘         Increment; yield n+1, toggling all trailing set bits and the rightmost
          unset bit.
  ~       Bitwise NOT; yield ~n, toggling ALL bits of n.
 &        Bitwise AND; yield (n+1)&~n, keeping the only bit that differs in n+1 and
          ~n, i.e., the rightmost unset bit.
   ^      Perform bitwise XOR with n, toggling the rightmost unset bit.
       $  Combine the two links to the left into a monadic chain.
     N        Negate; yield -n. Since ~n = -(n+1) in 2's complement, -n = ~n+1.
      &       Take the bitwise AND of n and -n. Since -n = ~n + 1 and n = ~~n, the
              same reasoning that applied for (n+1)&~n applies to -n&n; it yields
              the rightmost unset bit of ~n, i.e., the rightmost set bit of n.
    ^      XOR the result to the left with the result to the right, toggling the
           rightmost set bit of the left one.

Answer 4

MATL, 9 bytes

32&B1YSXB

Circularly shifts the 32-digit binary representation one step to the right.

Try it online!

Answer 5

JavaScript (ES6), 35 31 bytes

Looks for the first bit transition (0 → 1 or 1 → 0) and inverts it.

f=(n,k=3)=>(n&k)%k?n^k:f(n,k*2)

Demo

f=(n,k=3)=>(n&k)%k?n^k:f(n,k*2)

;[ 46, 12, 255, 2**32-2 ]
.map(n => {
  r = f(n);
  console.log((n >>> 0), '=', (n >>> 0).toString(2), '-->', (r >>> 0), '=', (r >>> 0).toString(2));
})

Bit rotation, 14 bytes

Much shorter but less fun.

n=>n>>>31|n<<1

Demo

let f =

n=>n>>>31|n<<1

;[ 46, 12, 255, 2**32-2 ]
.map(n => {
  r = f(n);
  console.log((n >>> 0), '=', (n >>> 0).toString(2), '-->', (r >>> 0), '=', (r >>> 0).toString(2));
})

Answer 6

Brain-Flak, 78 bytes

(([()()])[[]()]){((){}<({({})({}())}{})>)}{}([(({}(({}){})())<>)]){({}())<>}{}

Try it online!

Returns 2n if n < 2^31, and 2n+1-2^32 otherwise. Unfortunately, because Brain-Flak doesn't have any fast way to determine the sign of a number, the program times out on TIO if the input differs from 2^31 by more than about 500000.

Explanation

First, push -2^32 onto the stack:

(([()()])[[]()])                               push (initial value) -2 and (iterator) -5
                {((){}<                >)}     do 5 times:
                       ({({})({}())}{})        replace the current (negative) value with the negation of its square
                                            {}   pop the (now zero) iterator

Then, compute the desired output:

      (({}){})                        replace n by 2n on left stack
   ({}        ())                     push 2n+1-2^32 on left stack
  (              <>)                  push again on right stack
([                  ])                push its negation on right stack
                      {({}())<>}      add 1 to the top value of each stack until one of them reaches zero
                                {}    pop this zero, and implicitly print the number below it on the stack

Answer 7

dc, 10

?2~z31^*+p

Try it online.

This is an arithmetic implementation of a 32bit right-rotate:

?           # input
 2~         # divmod by 2 - quotient pushed first, then the remainder
   z        # z pushes the size of the stack which will be 2 (quotient and remainder) ...
    31^     #  ... and take that 2 to the 31st power
       *    # multiply the remainder by 2^31
        +   # add
         p  # output

Answer 8

Java 8, 117 17 29 bytes

n->n*2%~-(long)Math.pow(2,32)

+12 bytes by changing int to long, because int's max size is 2³¹-1

100 89 bytes saved by creating a port of @AndersKaseorg's amazing Python answer.

Try it here.

Outputs:

46 (101110):                                     92 (1011100)
12 (1100):                                       24 (11000)
1 (1):                                           2 (10)
3 (11):                                          6 (110)
10000 (10011100010000):                          20000 (100111000100000)
987654 (11110001001000000110):                   1975308 (111100010010000001100)
2147483648 (10000000000000000000000000000000):   1 (1)
4294967294 (11111111111111111111111111111110):   4294967293 (11111111111111111111111111111101)

Old answer (117 118 bytes):

n->{long r=0;for(;!n.toBinaryString(++r).replace("0","").equals(n.toBinaryString(n).replace("0",""))|r==n;);return r;}

+1 byte by changing int to long, because int's max size is 2³¹-1

Try it here.

Outputs:

46 (101110):                                     15 (1111)
12 (1100):                                       3 (11)
1 (1):                                           2 (10)
3 (11):                                          5 (101)
10000 (10011100010000):                          31 (11111)
987654 (11110001001000000110):                   255 (11111111)
2147483648 (10000000000000000000000000000000):   1 (1)

Answer 9

APL, 12 bytes

(2⊥32⍴1)|2×⊢

How?

           ⊢  ⍝ monadic argument
         2×   ⍝ shift left (×2)
(2⊥32⍴1)|     ⍝ modulo 2^32 - 1

Answer 10

Mathematica, 29 bytes

Mod@##+Quotient@##&[2#,2^32]&

Try it at the Wolfram sandbox

Rotates left arithmetically: first multiply by 2, which possibly shifts the number out of range, then cut off the out-of-range digit with Mod[...,2^32] and add it back on the right with +Quotient[...,2^32].

(Mathematica does have a single builtin that gives the modulus and the quotient in one go, but it's QuotientRemainder, which is a bit of a golfing handicap…)

Answer 11

05AB1E, 5 bytes

·žJ<%

Try it online!

Explanation

Uses the trick to rotate the binary representation left by 1 bit from Anders Kaseorg's python answer.

·       # input * 2
    %   # modulus
 žJ<    # 2^32-1

Answer 12

R, 42 63 bytes

function(x){s=x;while(s==x){sample(binaryLogic::as.binary(x))}}

Shuffles the bits around randomly, but checks to make sure it didn't return the same number by chance.

Answer 13

Whitespace, 81 80 bytes

(1 byte saved thanks to @Ørjan Johansen reminding me dup is shorter than push 0)

Try it online!

Basically implements a cyclic right bitshift using integer arithmetic. Pushing a large constant is expensive in Whitespace so we save some bytes by pushing 2^8 and squaring it twice. (Saves 1 byte over (2^16)^2 and 10 bytes over pushing 2^32 directly.)

Explanation

sssn  ; push 0
sns   ; dup
tntt  ; getnum from stdio
ttt   ; retrieve n from heap and put it on the stack
sns   ; dup
ssstsn ; push 2
tstt  ; mod - check if divisible by 2 (i.e. even)
ntsn  ; jez "even"
ssstssssssssn ; push 2^8
sns   ; dup
tssn  ; mul - square it to get 2^16
sns   ; dup
tssn  ; mul - square it to get 2^32
tsss  ; add 2^32 so MSB ends up set after the divide
nssn  ; even:
ssstsn ; push 2
tsts  ; divide by 2, aka shift right
tnst  ; putnum - display result

Answer 14

Python 2.7, 89 bytes

Full Program:

from random import*;a=list(bin(input())[2:].zfill(32));shuffle(a);print int(''.join(a),2)

Try it online!

Suggestions welcome! :)

Answer 15

Pari/GP, 15 bytes

n->2*n%(2^32-1)

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Answer 16

Japt, 5 bytes

Ñ%3pH

Bitwise rotation, like most answers here.

Try it

Answer 17

Perl 5 -p, 39 bytes

$_=sprintf'0b%b',$_;$_=oct$_.int s/10//

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Answer 18

C++ (gcc), 45 39 bytes

-6 bytes thanx to ceilingcat

auto h(unsigned x){return x%2<<31|x/2;}

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Answer 19

Python 3, 45 bytes

lambda i:int(f'{i:32b}'[1:]+f'{i:32b}'[:1],2)

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