# Bit-Reversal Permutations

Your goal is to create a function or a program to reverse the bits in a range of integers given an integer $$\n\$$. In other words, you want to find the bit-reversal permutation of a range of $$\2^n\$$ items, zero-indexed. This is also the OEIS sequence A030109. This process is often used in computing Fast Fourier Transforms, such as the in-place Cooley-Tukey algorithm for FFT. There is also a challenge for computing the FFT for sequences where the length is a power of 2.

This process requires you to iterate over the range $$\[0, 2^n-1]\$$, convert each value to binary and reverse the bits in that value. You will be treating each value as a $$\n\$$-digit number in base 2 which means reversal will only occur among the last $$\n\$$ bits.

For example, if $$\n = 3\$$, the range of integers is [0, 1, 2, 3, 4, 5, 6, 7]. These are

i  Regular  Bit-Reversed  j
0    000        000       0
1    001        100       4
2    010        010       2
3    011        110       6
4    100        001       1
5    101        101       5
6    110        011       3
7    111        111       7


where each index $$\i\$$ is converted to an index $$\j\$$ using bit-reversal. This means that the output is [0, 4, 2, 6, 1, 5, 3, 7].

The output for $$\n\$$ from 0 thru 4 are

n    Bit-Reversed Permutation
0    
1    [0, 1]
2    [0, 2, 1, 3]
3    [0, 4, 2, 6, 1, 5, 3, 7]


You may have noticed a pattern forming. Given $$\n\$$, you can take the previous sequence for $$\n-1\$$ and double it. Then concatenate that doubled list to the same double list but incremented by one. To show,

[0, 2, 1, 3] * 2 = [0, 4, 2, 6]
[0, 4, 2, 6] + 1 = [1, 5, 3, 7]
[0, 4, 2, 6] ⊕ [1, 5, 3, 7] = [0, 4, 2, 6, 1, 5, 3, 7]


where $$\⊕\$$ represents concatenation.

You can use either of the two methods above in order to form your solution. If you know a better way, you are free to use that too. Any method is fine as long as it outputs the correct results.

## Rules

• This is so the shortest solution wins.
• Builtins that solve this challenge as a whole and builtins that compute the bit-reversal of a value are not allowed. This does not include builtins which perform binary conversion or other bitwise operations.
• Your solution must be, at the least, valid for $$\n\$$ from 0 to 31.
• "Builtins that solve this challenge as a whole and builtins that compute the bit-reversal of a value are not allowed." Awww, IntegerReverse[Range[2^#]-1,2,#]&. (I don't know why Mathematica needs that built-in but I guess it's not a lot weirder than Sunset...) Jun 20, 2016 at 20:47
• @MartinEnder Nice find. Someday, it might be that there will be a builtin for everything in Mathematica, including generating random code-golf challenges. Jun 20, 2016 at 20:57
• Can we print 0 instead of  or does it have to be a list? Jun 20, 2016 at 21:03
• @Dennis Good point. I'll allow it, since it's only important that the output represents a valid permutation regardless of format. Jun 20, 2016 at 21:10
• Would returning false instead of 0 be acceptable? Jun 20, 2016 at 21:21

f 0=
f a=[(*2),(+1).(*2)]<*>f(a-1)


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Thanks to @Laikoni for three bytes!

# 05AB1E, 8 bytes

Code:

¾)IF·D>«


Explanation:

¾         # Constant for 0.
)        # Wrap it up into an array.
IF      # Do the following input times.
·     # Double every element.
D    # Duplicate it.
>   # Increment by 1.
«  # Concatenate the first array.


Uses the CP-1252 encoding. Try it online!.

• Nice one! Beats the one I had :) Jun 20, 2016 at 19:44
• @Emigna Thanks! What was your version then? Jun 20, 2016 at 19:45
• 0)ïsF·D>« was close though. Had some issues with the '0'. Jun 20, 2016 at 19:48
• Nice usage of ¾. Gonna have to remember that trick. Jun 20, 2016 at 19:53
• @KevinCruijssen For input 0, it looks prettier to have the int representation of 0 and not the string representation :). Other than that, there are no differences. May 25, 2018 at 12:40

# MATL, 1312109 8 bytes

0i:"EtQh


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Explanation

0       % Push number literal 0 to the stack
i:"     % Loop n times
E   % Multiply by two
t   % Duplicate
h   % Horizontally concatenate the result
% Implicit end of loop, and implicitly display the result



For the sake of completeness, here was my old answer using the non-recursive approach (9 bytes).

W:qB2&PXB


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Explanation

W       % Compute 2 to the power% ofImplicitly thegrab input (n) and compute 2^n
:       % Create an array from [1...2^n]
q       % Subtract 1 to get [0...(2^n - 1)]
B       % Convert to binary where each row is the binary representation of a number
2&P     % Flip this 2D array of binary numbers along the second dimension
XB      % Convert binary back to decimal
% Implicitly display the result


# J, 15 11 bytes

2&(*,1+*)0:


There is an alternative for 15 bytes that uses straight-forward binary conversion and reversal.

2|."1&.#:@i.@^]


## Usage

   f =: 2&(*,1+*)0:
f 0
0
f 1
0 1
f 2
0 2 1 3
f 3
0 4 2 6 1 5 3 7
f 4
0 8 4 12 2 10 6 14 1 9 5 13 3 11 7 15


## Explanation

2&(*,1+*)0:  Input: n
0:  The constant 0
2&(     )    Repeat n times starting with x = 
2      *       Multiply each in x by 2
,          Append that to
2  *           The list formed by multiplying each in x by 2
Return that as the next value of x
Return the final value of x


# Jelly, 5 bytes

2ṗ’UḄ


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-1 thanks to Dennis.

# Octave, 37 bytes

@(n)bin2dec(fliplr(dec2bin(0:2^n-1)))


Creates an anonymous function named ans that can simply be called with ans(n).

Online Demo

# Python 2, 5655 54 bytes

f=lambda n:[n:]or[i+j*2for i in 0,1for j in f(n-1)]


Test it on Ideone.

Thanks to @xnor for golfing off 1 byte!

• You can do [n:]or.
– xnor
Jun 21, 2016 at 1:47

# Java, 422 419 bytes:

import java.util.*;class A{static int[]P(int n){int[]U=new int[(int)Math.pow(2,n)];for(int i=0;i<U.length;i++){String Q=new String(Integer.toBinaryString(i));if(Q.length()<n){Q=new String(new char[n-Q.length()]).replace("\0","0")+Q;}U[i]=Integer.parseInt(new StringBuilder(Q).reverse().toString(),2);}return U;}public static void main(String[]a){System.out.print(Arrays.toString(P(new Scanner(System.in).nextInt())));}}


Well, I finally learned Java for my second programming language, so I wanted to use my new skills to complete a simple challenge, and although it turned out very long, I am not disappointed. I am just glad I was able to complete a simple challenge in Java.

Try It Online! (Ideone)

• You can save a few bytes parsing an int from args rather than reading from StdIn Jan 5, 2017 at 0:06

## Mathematica, 56 33 bytes

Byte count assumes an ISO 8859-1 encoded source.

±0={0};±x_:=Join[y=±(x-1)2,y+1]


This uses the recursive definition to define a unary operator ±.

# Perl, 46 45 bytes

Includes +1 for -p

Give input number on STDIN

#!/usr/bin/perl -p
map$F[@F]=($_*=2)+1,@F for(@F=0)..$_;$_="@F"


# Jelly, 7 6 bytes

Ḥ;‘$$¡  Thanks to @EriktheOutgolfer for golfing off 1 byte! Try it online! ### How it works Ḥ;‘$$¡  Main link. No arguments.
Implicit argument / initial return value: 0

¡  Read an integer n from STDIN and call the link to the left n times.
$Combine the two links to the left into a monadic chain, to be called with argument A (initially 0, later an array). Ḥ Unhalve; yield 2A.$      Combine the two links to the left into a monadic chain, to be called
with argument 2A.
‘         Increment; yield 2A + 1
;          Concatenate 2A and 2A + 1.


# Pyth - 11 bytes

ushMByMGQ]0


# Javascript ES6, 6553 51 bytes

f=(n,m=1)=>n?[...n=f(n-1,m+m),...n.map(i=>i+m)]:


Uses the recursive double-increment-concat algorithm.

Example runs:

f(0) => 
f(1) => [0, 1]
f(2) => [0, 2, 1, 3]
f(4) => [0, 8, 4, 12, 2, 10, 6, 14, 1, 9, 5, 13, 3, 11, 7, 15]

• How about f=n=>n>0?(r=f(n-1).map(i=>i*2)).concat(r.map(i=>i+1)):? Jun 20, 2016 at 20:37
• @miles Whoops, didn't realize I don't need a base case for n==1, thanks. Jun 20, 2016 at 20:43
• I think I managed to shave off 2 bytes by moving the multiplication by two: f=(n,m=1)=>n?[...n=f(n-1,m+m),...n.map(i=>i+m)]:
– Neil
Jun 20, 2016 at 23:30

# Python 3, 67 59 bytes

Thanks to @Dennis for -8 bytes

lambda n:[int(bin(i+2**n)[:1:-1],2)//2for i in range(2**n)]


We may as well have a (modified) straightforward implementation in Python, even if this is quite long.

An anonymous function that takes input by argument and returns the bit-reversed permutation as a list.

How it works

lambda n                 Anonymous function with input n
...for i in range(2**n)  Range from 0 to 2**n-1
bin(i+2**n)[:1:-1]       Convert i+2**n to binary string, giving 1 more digit than needed,
remove '0b' from start, and reverse
int(...,2)               Convert back to decimal
...//2                   The binary representation of the decimal value has one trailing
bit that is not required. This is removed by integer division by 2
:[...]                   Return as list


Try it on Ideone

• This ties with my approach, but it golfiness won't survive a port to Python 3. Jun 21, 2016 at 0:07

# Dyalog APL, 12 bytes

Requires ⎕IO←0 which is default on many systems.

2⊥⊖2⊥⍣¯1⍳2*⎕


2⊥ from-base-2 of

⊖ the flipped

2⊥⍣¯1 inverse of from-base-2 of

⍳ the first n integers, where n is

2* 2 to the power of

⎕ numeric input

TryAPL online!

For comparison, here is the other method:

(2∘×,1+2∘×)⍣⎕⊢0


( the function train...

2∘× two times (the argument)

, concatenated to

1+ one plus

2∘× two times (the argument)

)⍣ applied as many times as specified by

⎕ numeric input

⊢ on

0 zero

• (⍋,⍨)⍣⎕⊢0 (⎕io←0)
– ngn
Apr 25, 2019 at 16:49
• @ngn That has nothing to do with my algorithm.
Apr 27, 2019 at 21:25
• i thought it was similar to your "other method" but ok, i'll write a separate answer
– ngn
Apr 27, 2019 at 21:56

# Ruby, 51 bytes

f=->n{n<1?:(a=f[n-1].map{|i|i*2})+a.map(&:next)}


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# K (ngn/k), 11 8 bytes

2/|!2|&:


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 x:3  / just for testing
&x   / that many zeroes
0 0 0
2|&x / max with 2
2 2 2
!x#2 / binary words of length x, as a transposed matrix
(0 0 0 0 1 1 1 1
0 0 1 1 0 0 1 1
0 1 0 1 0 1 0 1)
|!x#2 / reverse
(0 1 0 1 0 1 0 1
0 0 1 1 0 0 1 1
0 0 0 0 1 1 1 1)
2/|!x#2 / base-2 decode the columns
0 4 2 6 1 5 3 7


& is the last verb in the composition so we need a : to force it to be monadic

# Ruby, 57 47 39 bytes

->n{[*0...2**n].sort_by{|a|a.digits 2}}


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3 years later, 18 bytes golfed.

# Julia, 23 22 bytes

!n=n>0&&[t=2*!~-n;t+1]


Rather straightforward implementation of the process described in the challenge spec.

Try it online!

# Pyth, 8 bytes

iR2_M^U2


Try it online: Demonstration or Test Suite

### Explanation:

iR2_M^U2Q   implicit Q (=input number) at the end
^U2Q   generate all lists of zeros and ones of length Q in order
_M       reverse each list
iR2         convert each list to a number


## Clojure, 78 bytes

Just following the spec...

(defn f[n](if(= n 0)(let[F(map #(* 2 %)(f(dec n)))](concat F(map inc F)))))


# PHP, 57 bytes

while($i<1<<$argv)echo bindec(strrev(decbin($i++))),_;  takes input from command line parameter, prints underscore-delimited values. Run with -nr. recursive solution, 72 bytes function p($n){$r=[$n];if($n)foreach($r=p($n-1)as$q)$r[]=$q+1;return$r;}  function takes integer, returns array ## JavaScript (Firefox 30-57), 48 bytes f=n=>n?[for(x of[0,1])for(y of f(n-1))x+y+y]:  Port of @Dennis♦'s Python 2 solution. • ReferenceError: f is not defined – l4m2 May 25, 2018 at 10:39 # Japt, 14 13 bytes 2pU Ç¤w ú0U Í  Try it online! ### Unpacked & How it works 2pU o_s2 w ú0U n2 2pU 2**n o_ range(2**n).map(...) s2 convert to binary string w reverse ú0U right-pad to length n, filling with '0' n2 convert binary string to number  Straightforward implementation. • There's actually an undocumented shortcut for n2: Í May 25, 2018 at 15:35 # APL (Dyalog Classic), 9 bytes (⍋,⍨)⍣⎕⊢0  Try it online! ⎕ evaluated input ( )⍣⎕⊢0 apply the thing in ( ) that many times, starting with 0 ,⍨ concatenate the current result with itself ⍋ indices of a sort-ascending permutation # Perl 6, 25 bytes {[X+] 0,|(0 X(2 X**^$_))}


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{0,{$^p+^($_-$_/2+>lsb ++$)}...$_}o 1+<*-1  Try it online! Incrementing an integer simply flips a sequence of least-significant bits, for example from xxxx0111 to xxxx1000. So the next bit-reversed index can be obtained from the previous one by flipping a sequence of most-significant bits. The XOR mask can be computed with m - (m >> (ctz(i) + 1)) for m = 2**n or m = 2**n-1. # TI-BASIC, 26 bytes Input A:{0:For(I,1,A:2Ans:augment(Ans,1+Ans:End:Ans  Prompts the user to input n. Outputs the bit-reversal permutation as is specified in the challenge. Explanation: Input A ;prompt the user for "n". store the input into A {0 ;leave {0} in Ans For(I,1,A ;loop A times 2Ans ;double Ans and leave the result in Ans augment(Ans,1+Ans ;add 1 to Ans and append it to the previous result ; leave the new result in Ans End ;loop end Ans ;leave Ans in Ans ;implicit print of Ans  Examples: prgmCDGF25 ?1 {0 1} prgmCDGF25 ?3 {0 4 2 6 1 5 3 7}  Note: TI-BASIC is a tokenized language. Character count does not equal byte count. # Desmos, 71 bytes f(n)=[total(mod(floor(2i/2^{[1...n]}),2)2^{[n-1...0]})fori=[0...2^n-1]]  Straightforward implementation of the challenge (convert to binary, then reverse), but probably optimizable with actions... Try It On Desmos! Try It On Desmos! - Prettified # Factor, 43 bytes [ { 0 } [ 2 v*n dup 1 v+n append ] repeat ]  Try it online! # x86, 31 bytes Takes a sufficiently large int[] buffer in eax and n in ecx, and returns the buffer in eax. Implements the concatenating algorithm given in the challenge statement. It may be possible to save bytes by incrementing the pointers by 4 instead of using array accesses directly, but lea/mov is already pretty short (3 bytes for 3 regs and a multiplier). .section .text .globl main main: mov$buf, %eax          # buf addr
mov     \$3, %ecx            # n

start:
xor     %ebx, %ebx
mov     %ebx, (%eax)        # init buf = 0
inc     %ebx                # x = 1

l1:
mov     %ebx, %edi
dec     %edi                # i = x-1
lea     (%eax,%ebx,4), %edx # buf+x

l2:
mov     (%eax,%edi,4), %esi # z = buf[i]
sal     %esi                # z *= 2
mov     %esi, (%eax,%edi,4) # buf[i] = z
inc     %esi                # z += 1
mov     %esi, (%edx,%edi,4) # buf[x+i] = z

dec     %edi                # --i
jns     l2                  # do while (i >= 0)

sal     %ebx                # x *= 2
loop    l1                  # do while (--n)

ret

.data
buf:    .space 256, -1


Hexdump:

00000507  31 db 89 18 43 89 df 4f  8d 14 98 8b 34 b8 d1 e6  |1...C..O....4...|
00000517  89 34 b8 46 89 34 ba 4f  79 f1 d1 e3 e2 e7 c3     |.4.F.4.Oy......|