Smallest Bit Rotation

Question

For a given positive integer, try to find out the smallest possible rotation resulted by rotating it 0 or more bits.

For example, when the given number is 177, whose binary representation is \$10110001_{(2)}\$:

• \$ 10110001_{(2)}=177 \$
• \$ 01100011_{(2)}=99 \$
• \$ 11000110_{(2)}=198 \$
• \$ 10001101_{(2)}=141 \$
• \$ 00011011_{(2)}=27 \$
• \$ 00110110_{(2)}=54 \$
• \$ 01101100_{(2)}=108 \$
• \$ 11011000_{(2)}=216 \$

27 is the smallest rotating result. So we output 27 for 177.

Input / Output

You may choose one of the following behaviors:

• Input a positive integer \$n\$. Output its smallest bit rotation as defined above.
• Input a positive integer \$n\$. Output smallest bit rotation for numbers \$1\dots n\$.
• Input nothing, output this infinity sequence.

Due to definition of this sequence. You are not allowed to consider it as 0-indexed, and output smallest bit rotate for \$n-1\$, \$n+1\$ if you choose the first option. However, if you choose the second or the third option, you may optionally include 0 to this sequence, and smallest bit rotation for \$0\$ is defined as \$0\$. In all other cases, handling \$0\$ as an input is not a required behavior.

Test cases

So, here are smallest bit rotate for numbers \$1\dots 100\$:

 1  1  3  1  3  3  7  1  3  5
 7  3  7  7 15  1  3  5  7  5
11 11 15  3  7 11 15  7 15 15
31  1  3  5  7  9 11 13 15  5
13 21 23 11 27 23 31  3  7 11
15 13 23 27 31  7 15 23 31 15
31 31 63  1  3  5  7  9 11 13
15  9 19 21 23 19 27 29 31  5
13 21 29 21 43 43 47 11 27 43
55 23 55 47 63  3  7 11 15 19

Notes

• This is code-golf as usual.
• This is A163381.
• The largest bit rotation is A163380. A233569 is similar but different. (The first different item is the 37^th).

Answer 1

x86-64 machine code, 18 16 bytes

0F BD D7 0F 43 C7 0F B3 D7 D1 D7 39 F8 75 F4 C3

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Following the standard calling convention for Unix-like systems (from the System V AMD64 ABI), this takes \$n\$ in EDI and returns its smallest bit rotation in EAX.

In assembly:

f:  bsr edx, edi    # Set EDX to the highest position of a 1 bit in n.
r:  cmovnc eax, edi # (EAX will hold the lowest rotation found so far.)
                    #  Set EAX to EDI if CF=0. The first iteration, CF=0 always:
                    #  the value of CF after BSR is officially undefined, but it is
                    #  consistently 0 (as this code needs) on at least some CPUs.
                    #  (To not rely on this, add CLC before this for +1 byte.)
                    #  On later iterations, CF=1 if EDI ≤ EAX (from the cmp).
    btr edi, edx    # Set the bit at position EDX in EDI to 0.
                    #  Set CF to the previous value of that bit.
    rcl edi, 1      # Rotate EDI and CF together left by 1,
                    #  putting the removed bit back in at the end.
    cmp eax, edi    # Compare EAX and EDI.
    jne r           # Jump back if they are not equal.
                    #  (Once a value repeats, all possible values have been seen.)
    ret             # Return.

Answer 2

05AB1E, 6 bytes

bā._Cß

Input \$n\$; outputs the smallest bit rotation.

Try it online or verify all test cases.

Both the \$[1,n]\$ list with an input \$n\$ and infinite list would be 2 bytes longer with a leading Lε or ∞ε respectively.
Try the infinite version online.

Explanation:

b       # Convert the (implicit) input-integer to a binary-string
 ā      # Push a list in the range [1,length] (without popping the string)
  ._    # Map each integer v in the list to a v-times (left-)rotated version of the
        # binary-string
    C   # Convert each binary-string in the list back to an integer
     ß  # Pop and leave the minimum
        # (which is output implicitly as result)

Answer 3

Excel, 81 80 71 68 bytes

=LET(a,DEC2BIN(n),b,LEN(a),c,SEQUENCE(b),MIN(BIN2DEC(MID(a&a,c,b))))

Input n

With many thanks to @KevinCruijssen for the great suggestions and 11-byte improvement!

Answer 4

Octave, 73 bytes

It's not pretty, and it doesn't work on TIO. But it works in the program itself.

x=de2bi(k=input(''));for i=1:k,[ans,bi2de(circshift(x',i)')];end,min(ans)

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x=de2bi(k=input(''));for i=1:k,[ans,bi2de(circshift(x',i)')];end,min(ans)

        k=input('')                   % Take input, k = 177
x=de2bi(k=input(''));                 % Convert input to binary, k = [1 0 0 0 1 1 0 1], 
                                      % x = 177
for i=1:k,[...];end                   % Do k times (177 times)
                                      % Shortest way to make something happen enough times
               circshift(x',i)')      % Transpose x to make a vertical vector
                                      % Shift it i times, and transpose it back
                                      % Must be transposed for circshift to work
         bi2de(circshift(x',i)')      % Convert to decimal
    [ans,bi2de(...)]                  % Concatenate into a long vector 'ans'
min(ans)                              % The minimum of this vector       

Answer 5

Jelly, 5 bytes

BṙRḄṂ

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Explanation

BṙRḄṂ  # Implicit input (n)
B      # Convert n into binary
 ṙ     # Rotate this string this many times:
  R    #  Each of the numbers in [1..n]
       # (this will generate some duplicates, but that's fine)
   Ḅ   # Convert each one back from binary
    Ṃ  # Get the minimum of this list
       # Implicit output

Answer 6

Python, 61 bytes

-5 bytes thanks to Shaggy and -5 bytes thanks to M Virts.

Takes a positive integer n as input. Outputs its smallest bit rotation in integer form.

lambda n:min(int((a:=f'{n:b}')[i:]+a[:i],2)for i in range(n))

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Answer 7

J, 16 bytes

[:<./2#.i.|."{#:

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• #:: Convert \$n\$ to binary.
• i. |. " {: i. makes a list of the integers from 0 (inclusive) to \$n\$ (exclusive), then |. uses that to rotate the binary expansion of \$n\$, with its ranks being set (") to the ranks of { so that it takes each integer as a separate rotation.
• 2 #.: Convert back from binary expansions to numbers.
• [: <. /: Take the minimum, using a 'cap' to do so monadically.

Answer 8

><> (Fish), 91 bytes

i0$1>:@$:@)?!v\
 ;n\^@+1$*2@ <r
$@:\?=@:$@:r-1/&~$?)@:&:
2+*2~v?:}:{%2:/.2b-%1:,
1$*2$/.38-

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Instead of counting how many iterations happened, which would be expensive, this exits if the current value equals the minimum value.

Answer 9

R, 62 59 bytes

Edit: -3 bytes thanks to pajonk

\(n,j=2^(0:(i=log2(n))))min(j%*%array(n%/%j%%2,2:3+i)[-1,])

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Instead of iteratively rotating bits, we use them to fill a 2D array with one-too-many rows, with recycling, so that each column ends-up with the bits rotated (plus an extra useless row).
We can then use matrix multiplication (%*% in R) with a vector of powers-of-2 to easily calculate the values encoded by each column, and output the minimum one.

Answer 10

JavaScript (Node.js), 63 bytes

x=>'0b'+[...y=x.toString(2)].map(t=>y=y.slice(1)+t).sort()[0]-0

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t keeps being the needed digit

Answer 11

APL (Dyalog Extended), 9 bytes

Anonymous tacit prefix function.

⌊/⍳⊥⍤⌽¨⊤¨

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⌊/ the smallest (lit. minimum-reduction) of

⍳…¨⊤¨ for each integer in the range 1 through \$n\$, combined with the binary representation of \$n\$:

 …⍤⌽ rotate that representation that integer steps left, and then:

  ⊥ convert back to a regular number

Answer 12

R, 80 78 75 71 bytes

f=\(n,b=2^(0:log2(n)),x=n%/%b%%2)if(n)min(x%*%b,f(n-1,b,c(x[-1],x[1])))

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Recursive function that in every iteration rotates the binary representation by one place and keeps track of the minimum value.

Answer 13

Fig, \$18\log_{256}(96)\approx\$ 14.816 bytes

[KeBtLbxGWbxO'J]xq

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Fig is missing a few important builtins, so this is longer than expected.

[KeBtLbxGWbxO'J]xq
          bx       # Get the binary form of this number
        GW         # Generate an infinite list from the binary
            O'     # Using the function...
              J]xq # Rotate by prepending the last element to the tailless list
     Lbx           # Length of the binary representation
    t              # Take that many items from the infinite list
  eB               # Convert each from binary
 K                 # Sort ascending
[                  # Take the first element (i.e. the minimum one)

Answer 14

C (gcc), ⁷³ 61 bytes

s;l;r;f(n){for(l=r=log2(s=n);r--;s=s<n?s:n)n=n%2<<l|n/2;r=s;}

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Saved a whopping 12 bytes thanks to c--!!!

Answer 15

Python, 57 bytes

lambda n:int(min(h:=f"{n:b}",*[h:=h[1:]+k for k in h]),2)

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Old Python, 58 bytes

lambda n:int(min([h:=f"{n:b}"]+[h:=h[1:]+k for k in h]),2)

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Answer 16

C (gcc) with -lgmp, 233 231 227 bytes

• -6 thanks to ceilingcat

As it uses the GMP library, the maximum integer that this function handles is quite large. Takes a number as a string to avoid overflowing native types.

#import<gmp.h>
f(s,v,w,x,y)char*s,*v;{mpz_t c,l;mpz_init_set_str(c,s,10);mpz_init(l);for(y=x=strlen(v=mpz_get_str(0,2,c));y--;mpz_cmp(c,l)>0?mpz_set(c,l):0)w=v[x-1],bcopy(v,v+1,x-1),*v=w,mpz_set_str(l,v,2);gmp_printf("%Zd",c);}

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Ungolfed:

#import<gmp.h>
f(s, // input value
  v, // bit representation of value
  w, // rotated bit
  x,y // bit string length and loop counter
  )char*s,*v; {
  mpz_t c,l; // current lowest value, test value
  mpz_init_set_str(c,s,10); // initialize variables
  mpz_init(l);
  for(
    y=x=strlen(v=mpz_get_str(0,2,c)); // initialize bit string
    y--;
    mpz_cmp(c,l)>0?mpz_set(c,l):0) { // adjust lowest value
    w=v[x-1]; // get rotated bit
    bcopy(v,v+1,x-1); // rotate
    *v=w; // put rotated bit at top
    mpz_set_str(l,v,2); // get test value
  }
  gmp_printf("%Zd",c);
}

Answer 17

Python 2, 76 bytes

i,o=bin(input())[2:],[]
for x in i:i=i[-1]+i[:-1];o+=[int(i,2)]
print min(o)

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Python2 because it accepts integers directly as input (no int() required) and it saves a space on the print statement by not needing brackets.

Answer 18

Retina 0.8.2, 55 bytes

.+
$*
+r`\1(1+)
0$1
10
1
.
$&$'$`¶
1
10
+`01
110
O`
\G1

Try it online! Explanation:

.+
$*

Convert to unary.

+r`\1(1+)
0$1
10
1

Convert to mirrored binary i.e. LSB first, MSB last.

.
$&$'$`¶

Generate all of the rotations.

1
10
+`01
110

Convert each rotation back to unary.

O`

Sort.

\G1

Convert the shortest to decimal.

Answer 19

Factor, 43 bytes

[ >bin all-rotations [ bin> ] map infimum ]

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>bin              ! convert input from decimal to a binary string
all-rotations     ! get every rotation as a sequence of strings
[ bin> ] map      ! convert each to decimal from binary
infimum           ! get the smallest one

Answer 20

K (ngn/k), 16 bytes

&/2/'{1_x,*x}\2\

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• 2\ convert (implicit) input to its binary representation
• {1_x,*x}\ build a list of all rotations of the bits
• 2/' convert each back to its decimal representation
• &/ calculate, and (implicitly) return the minimum

Answer 21

Python 2.7, 73 bytes

x=bin(input())[2:]
print min([int(x[i:]+x[:i],2)for i in range(len(x))])

Answer 22

Vyxal g, 7 6 bytes

ƛ?b$ǓB

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• -1 thanks to a suggestion from Shaggy

Explanation

ƛ?b$ǓB  # Implicit input
ƛ       # Map over the range:
 ?b     #  Get the input in binary
   $Ǔ   #  Rotate ^ ^^ times
     B  #  Convert it back from binary
        # g flag gets the minimum of this list
        # Implicit output

Previous 7-byter:

b:ż¨VǓB  # Implicit input
b        # Convert the input to binary
 :ż      # Duplicate and push range(len(^))
   ¨V    # Map over each element in ^:
     Ǔ   #  Rotate ^^^ ^ places to the left
      B  # Convert each back from binary
         # g flag gets the minimum of this list
         # Implicit output

Answer 23

JavaScript (ES6), 57 bytes

-2 bytes thanks to @l4m2

n=>(m=g=q=>x--?g(q%2<<Math.log2(n,m=m<q?m:q)|q/2):m)(x=n)

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Commented

n => (              // n = input
  m =               // m = minimum value, initially non-numeric
  g =               // g is a recursive function taking
  q =>              // the current rotation q
  x-- ?             // if x is not equal to 0 (decrement it afterwards):
    g(              //   do a recursive call:
      q % 2 <<      //     using the number of bits in the original
      Math.log2(    //     input, left-shift the LSB so that it takes
        n,          //     the place of the MSB
        m = m < q ? //     update m to min(m, q)
          m         //     (this always updates m to q if m is
        :           //     still non-numeric)
          q         //
      ) | q / 2     //     right-shift all other bits by 1 position
    )               //   end of recursive call
  :                 // else:
    m               //   stop and return m
)(x = n)            // initial call to g with q = x = n

Answer 24

MATL, 13 12 bytes

:"GB@YSXBvX<

1 byte saved thanks to @lmendo

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Explanation

       % Implicitly grab the input (N)
:      % Create an array from [1...N]
"      % For each value in the array
  G    % Grab the input as a number
  B    % Convert to an array of booleans representing the binary equivalent
  @YS  % Circularly shift by the loop index
  XB   % Convert the result from binary to decimal
  v    % Vertically concatenate the entire stack
  X<   % Compute the minimum value so far
       % Implicitly display the result

Answer 25

Julia 1.0, 84 78 bytes

!x=(c=string(x,base=2);r=keys(c);min((r.|>i->parse(Int,"0b"*(c^2)[i.+r]))...))

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• c is the bitstring of input x. Another method would be lstrip(bitstring(x),'0').
• r contains the indices in c and is iterated over the repeated bitstring c^2 to get each rotation.
• min treats a vector V as a single value, so the splat operator is used: min(V...). Another method would be minimum(V).

-6 bytes thanks to MarcMush:

• Replace range 1:length(c) with iterator keys(c)
• Use broadcasting

Answer 26

Java 10, 118 bytes

n->{var b=n.toString(n,2);for(int l=b.length(),i=l,t;i-->0;n=t<n?t:n)t=n.parseInt((b+b).substring(i,i+l),2);return n;}

Input \$n\$; outputs the smallest bit rotation.

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Explanation:

n->{                       // Method with Integer as both parameter and return-type
  var b=n.toString(n,2);   //  Convert the input-integer to a binary-String
  for(int l=b.length(),    //  Set `l` to the length of this binary-String
      i=l,t;i-->0          //  Loop `i` in the range (l,0]:
      ;                    //    After every iteration:
       n=t<n?              //     If `t` is smaller than `n`:
             t:n)          //      Set `n` to this new minimum `t`
    t=                     //   Set `t` to:
      n.parseInt(          //    The following binary converted to a base-10 integer:
         (b+b)             //     Append the binary-String to itself
         .substring(i,i+l) //     And then get its substring in the range [i,i+l)
       ,2);
  return n;}               //  Return the modified `n` holding the smallest result

Answer 27

Charcoal, 18 bytes

≔⍘Ｎ²θＩ⍘⌊Ｅθ⭆θ§θ⁺κμ²

Try it online! Link is to verbose version of code. Explanation:

≔⍘Ｎ²θ

Convert the input to binary.

Ｉ⍘⌊Ｅθ⭆θ§θ⁺κμ²

Generate all of the rotations, take the minimum, and convert back to decimal.

Answer 28

Zsh, 70 bytes

b=$[[##2]$1];for i ({1..$#b})m+=($[2#${b:$i}${b:0:$i}]);printf ${(n)m}

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Answer 29

Husk, 7 bytes

ḋ▼U¡ṙ1ḋ

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      ḋ  # convert to bits,
   ¡     # make infinite list by repeatedly
    ṙ1   #   rotating by one position,
  U      # get the longest unique prefix,
 ▼       # get the minimum of this,
ḋ        # and get the value from the bits.

Answer 30

Desmos, 95 bytes

L=[floor(log_2N)...0]
B=mod(floor(N/2^L),2)
f(N)=[total(join(B[i+1...],B[1...i])2^L)fori=L].min

Even worse than the fish answer... at least it beats Java :P

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