21
\$\begingroup\$

... Ah sorry, no popcorn here, just POPCNT.

Write the shortest program or function which takes in a number n and output all integers from 0 to 2n - 1, in ascending order of number of 1 bits in the binary representation of the numbers (popcount). No duplicates allowed.

The order of numbers with the same popcount is implementation-defined.

For example, for n = 3, all these output are valid:

0, 1, 2, 4, 3, 5, 6, 7
[0, 4, 1, 2, 5, 3, 6, 7]
0 4 2 1 6 5 3 7 

The input and output format are implementation-defined to allow the use of language features to further golf the code. There are a few restrictions on the output:

  • The numbers must be output in decimal format.
  • The output must contain a reasonable separator between the numbers (trailing separator allowed, but not leading).

    Line feed (\n), tab (\t), space, ,, ., ;, |, -, _, / are quite reasonable separator. I don't mind additional spaces for pretty printing, but don't use letter or digits as separators.

  • The numbers and separators can be surrounded by [ ], { } or any array or list notation.
  • Don't print anything else not stated above.

Bonus

Multiply your score by 0.5 if your solution can generate the number on the fly. The spirit of this bonus is that if you were to directly convert your printing solution to a generator, the generator only uses at most O(n) memory where n is the number of bits as defined above. (You don't have to actually convert your solution to generator). Note that while I impose n <= 28, the memory needed to store all numbers still grows exponentially, and a naive sorting solution would hog up at least 4 GB of memory at n = 28.

Please add some simple explanation of how your solution works before claiming this bonus.

\$\endgroup\$
8
  • 4
    \$\begingroup\$ It seems that the challenge as it is quite boring and would result in a bunch of sorting answers. I would like to add some bonus to make the challenge more interesting. Something along the line of "generating the numbers on the fly". If you are OK with it, please upvote this comment, then I will add it to the question. \$\endgroup\$ Feb 27 '15 at 5:29
  • \$\begingroup\$ If you disagree, please upvote this comment. \$\endgroup\$ Feb 27 '15 at 5:30
  • \$\begingroup\$ Please use the sandbox to ask for further suggestions on a question before posting it live. \$\endgroup\$ Feb 27 '15 at 7:30
  • 24
    \$\begingroup\$ @JanDvorak: It was on sandbox for one month. \$\endgroup\$ Feb 27 '15 at 7:59
  • 1
    \$\begingroup\$ I think it's too late for this question. Generally, questions where you have to figure out a non-trivial algorithm aren't well suited for code golf in my opinion. Make them a code challenge instead and pose all the constraints you need. \$\endgroup\$
    – FUZxxl
    Feb 28 '15 at 2:48

21 Answers 21

10
\$\begingroup\$

Pyth, 9 bytes

osjN2U^2Q

order by the sum of the base 2 representation (jN2) over the range (U) of 2 ^ Q.

(Q = eval(input())).

Try it here.

\$\endgroup\$
7
\$\begingroup\$

Python 2, 72 bytes * 0.5 = 36

N=1<<input()
for k in range(N*N):
 if bin(k%N).count('1')==k/N:print k%N

Try it online!

A new method for this now-ancient challenge. The less-golfed below below might be easier to understand:

87 bytes

n=input()
for i in range(n+1):
 for x in range(2**n):
  if bin(x).count('1')==i:print x

Try it online!

We loop over target popcounts i in increasing order, and for each one iterate over the n-bit numbers and prints those with exactly i set bits.

Even though this loops over the n-bit numbers many times, it still satisfies the efficiency bonus criteria of only using O(n) memory if the loop were converted to a generator. If fact, the golfed code allocates n bits to the target popcount as well as n bits to the number being checked. They are stored as a 2*n-bit number, which allows counting up this single number and extracting the first and last n bits as needed.


Python 2, 59 bytes

lambda n:sorted(range(1<<n),key=lambda x:bin(x).count('1'))

Try it online!

A short sorting-based approach that does not qualify for the bonus.


Python 2, 75 bytes * 0.5 = 37.5

N=2**input()-1
v=N-~N
while v:t=1+(v|~-v);v=N&t|~-(t&-t)/(v&-v)/2;print v^N

Try it online!

Repeatedly generates the next highest v with the same POPCOUNT by this bit-twiddling algorithm.

Actually, it turned out easier to generate them in decreasing pop-count, then print the complement to make it increasing. That way, then v overflows 2**n, we simply remove all but n bits with &N where N=2**n-1, and that gives the smallest number one popcount lower. That way, we can just do one loop. There's probably a better solution that directly finds the next lower number with the same POPCOUNT.

Due to a fencepost issue, we need to start with v=2**(n+1)-1 so that the operation produces v=N-1 on the first loop.

Output for 4:

0
8
4
2
1
12
10
9
6
5
3
14
13
11
7
15
\$\endgroup\$
4
  • \$\begingroup\$ There is no need for the number with the same popcount to be increasing. The order is implementation-defined. \$\endgroup\$ Feb 27 '15 at 9:47
  • 1
    \$\begingroup\$ @n̴̖̋h̷͉̃a̷̭̿h̸̡̅ẗ̵̨́d̷̰̀ĥ̷̳ I'm aware, but I don't see how to save characters doing it differently. \$\endgroup\$
    – xnor
    Feb 27 '15 at 9:48
  • \$\begingroup\$ With a naive 3 loops method, I score almost the same in JS (having console.log() vs print). Maybe the bit trick is too heavy. \$\endgroup\$
    – edc65
    Feb 27 '15 at 11:12
  • \$\begingroup\$ One byte saving: v=N-~N \$\endgroup\$
    – Sp3000
    Mar 2 '15 at 10:36
5
\$\begingroup\$

J, 19 characters, no bonus.

[:(/:+/"1@#:)@i.2^]
  • 2 ^ y – two to the power of y.
  • i. 2 ^ y – the integers from 0 to (2 ^ y) - 1.
  • #: i. 2 ^ y – each of these integers represented in base two.
  • +/"1 #: i. 2 ^ y – the sums of each representation
  • (i. 2 ^ y) /: +/"1 #: i. 2 ^ y – the vector i. 2 ^ y sorted by the order of the items of the previous vector, our answer.
\$\endgroup\$
3
\$\begingroup\$

Python, 63 chars

F=lambda n:`sorted(range(1<<n),key=lambda x:bin(x).count('1'))`

>>> F(3)
'[0, 1, 2, 4, 3, 5, 6, 7]'
\$\endgroup\$
2
  • \$\begingroup\$ @Alex: The list of restrictions implied he wanted a string result. \$\endgroup\$ Mar 3 '15 at 0:30
  • \$\begingroup\$ Sorry, missed that. \$\endgroup\$
    – Alex A.
    Mar 3 '15 at 1:43
3
\$\begingroup\$

Mathematica, 50 46

SortBy[Range[0,2^#-1],Tr@IntegerDigits[#,2]&]&

.

SortBy[Range[0,2^#-1],Tr@IntegerDigits[#,2]&]&

{0, 1, 2, 4, 8, 16, 3, 5, 6, 9, 10, 12, 17, 18, 20, 
24, 7, 11, 13, 14, 19, 21, 22, 25, 26, 28, 15, 
23, 27, 29, 30, 31}
\$\endgroup\$
1
  • \$\begingroup\$ @MartinBüttner, Fixed! Thanks!!! \$\endgroup\$ Mar 1 '15 at 21:09
3
\$\begingroup\$

C 179 * 0.5 = 89.5

main(){int n,i=0,m,o;scanf("%d",&n);m=~((~0)<<n);for(;n--;++i){for(o=0;o<m;++o){int bc=0,cb=28;for(;cb--;)bc+=o&(1<<cb)?1:0;if(bc==i)printf("%d ",o);}}printf("%d\n",m);return 0;}

EDIT: 157 * 0.5 = 78.5

main(){int n,i=0,m,o;scanf("%d",&n);m=~((~0)<<n);for(++n;n--;++i){for(o=0;o<=m;++o){int bc=0,cb=28;for(;cb--;)bc+=o&(1<<cb)?1:0;if(bc==i)printf("%d ",o);}}}

EDIT: 132 * 0.5 = 66

main(){int n,i=0,m,o;scanf("%d",&n);m=~((~0)<<n);for(++n;n--;++i){for(o=0;o<=m;++o){if(__builtin_popcount(o)==i)printf("%d ",o);}}}

or a bit nicer formatted:

main()
{
    int n, i = 0, m, o;
    scanf("%d", &n);
    m = ~((~0) << n);
    for(++n; n--; ++i)
    {
        for(o = 0; o <= m; ++o)
        {
            if (__builtin_popcount(o) == i)
                printf("%d ", o);
        }
    }
}

What it does?

m = ~((~0) << n);

calculates the last number to show (pow(2, n) - 1)

    for(++n; n--; ++i)
    {
        for(o = 0; o <= m; ++o)
        {

the outer loop iterates over the bit count (so 0 to n-1) while the inner loop just counts from 0 to m

            if (__builtin_popcount(o) == i)
                printf("%d ", o);

On x86 there is the POPCNT instruction that can be used to count the set bits. GCC and compatible compilers may support the __builtin_popcount function that basically compiles to that instruction.

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2
\$\begingroup\$

CJam, 13 bytes

2ri#,{2b1b}$p

Pretty straight forward implementation.

How it works:

2ri#,             "Get an array of 0 to 2^n - 1 integers, where n is the input";
     {    }$      "Sort by";
      2b1b        "Convert the number to binary, sum the digits";
            p     "Print the array";

Try it online here

\$\endgroup\$
2
\$\begingroup\$

K (ngn/k), 18 9 8 bytes

-9 bytes from @ngn's improvements

-1 byte from using where instead of take

<+/!2+&:

Try it online!

A tacit function, equivalent to {<+/!2+&x}.

  • 2+&: generate x 2's
  • !... generate "odometer" of the input (in this case this provides the binary representation of the numbers from 0 to pow(2,x))
  • +/ sum column-wise
  • < "grade up" the sums (i.e. sort them by the number of 1's in the binary-representation of each value)
\$\endgroup\$
3
  • \$\begingroup\$ here t@<.. is the same as <.. \$\endgroup\$
    – ngn
    Dec 27 '20 at 13:47
  • 1
    \$\begingroup\$ odometer can be useful \$\endgroup\$
    – ngn
    Dec 27 '20 at 13:48
  • \$\begingroup\$ @ngn very true! Somehow I missed that y\!1*/x#y is equivalent to !x#y. \$\endgroup\$
    – coltim
    Dec 27 '20 at 16:34
2
\$\begingroup\$

Husk, 8 bytes

ÖoΣḋŀ`^2

Try it online!

     `^2  flip ^, partially applied, gets fully applied with implicit input, 2 ^ n
    ŀ     range from [0, n)
Ö         sort by
 o        composed function
  Σ       sum
   ḋ      binary digits
\$\endgroup\$
1
\$\begingroup\$

JavaScript (ES6) 41 (82*0.5)

The simplest way, golfed

F=b=>{
  for(l=0;l<=b;l++)
    for(i=1<<b;i;t||console.log(i))
      for(t=l,u=--i;u;--t)
        u&=u-1;
}

Ungolfed

F=b=>
{
  for (l = 0; l <= b; l++)
  {
    for (i = 1 << b; i > 0; )
    {
      --i;
      for (t = 0, u = i; u; ++t) // Counting bits set, Brian Kernighan's way
        u &= u - 1;
      if (t == l) console.log(i);
    }
  }
}

Test In Firefox/FireBug console

F(4)

0
8
4
2
1
12
10
9
6
5
3
14
13
11
7
15

\$\endgroup\$
1
\$\begingroup\$

Bash + coreutils, 66

One to get you started:

jot -w2o%dpc $[2**$1] 0|dc|tr -d 0|nl -ba -v0 -w9|sort -k2|cut -f1
\$\endgroup\$
4
  • \$\begingroup\$ Nothing really exciting here. Given your comment I will happily delete/revise this answer if you want to change the question. \$\endgroup\$ Feb 27 '15 at 5:46
  • \$\begingroup\$ Not sure if I should highlight your program must work for all values of n from 0 to 28 inclusive. I don't know how many answers here pass that requirement. \$\endgroup\$ Feb 27 '15 at 6:36
  • \$\begingroup\$ I have removed the clause, since people don't seem to notice it anyway. \$\endgroup\$ Feb 27 '15 at 10:37
  • \$\begingroup\$ @n̴̖̋h̷͉̃a̷̭̿h̸̡̅ẗ̵̨́d̷̰̀ĥ̷̳ Now, in theory at least it should work up to 28. I've tested up to 22 so far, but of course this makes sort take a long time. With n=28, sort will need to sort 2^28 lines / ~13GB of data. \$\endgroup\$ Feb 27 '15 at 20:32
1
\$\begingroup\$

Haskell, (87 * 0.5) = 43,5

f n=[0..n]>>=(\x->x#(n-x))
a#0=[2^a-1]
0#_=[0]
a#b=[1+2*x|x<-(a-1)#b]++[2*x|x<-a#(b-1)]

Usage example: f 4, which outputs [0,1,2,4,8,3,5,9,6,10,12,7,11,13,14,15]

How it works: neither sorting nor repeatedly iterating over [0..2^n-1] and looking for numbers containing i 1s.

The # helper functions takes two parameters a and b and constructs a list of every number made up of a 1s and b 0s. The main function f calls # for every combination of a and b where a+b equals n, starting with no 1s and n 0s to have the numbers in order. Thanks to Haskell's laziness all those lists don't have to be constructed completely in memory.

\$\endgroup\$
2
  • \$\begingroup\$ Doesn't the ++ in a#b mean that the left hand side (which could be large) needs to be produced entirely and then copied before the first item in the result is produced, thus violating the requirements for the bonus? \$\endgroup\$
    – Jules
    Mar 31 '16 at 19:08
  • \$\begingroup\$ Ah, no, thinking about it can still lazily generate them while they're being produced, it just needs to make a copy of each item, which as both can be garbage collected during processing means that space usage is constant. Ignore me. \$\endgroup\$
    – Jules
    Mar 31 '16 at 19:14
1
\$\begingroup\$

Ruby 47 chars

Much like the Python one from @KeithRandall :

f=->n{(0..1<<n).sort_by{|x|x.to_s(2).count ?1}}
\$\endgroup\$
1
\$\begingroup\$

Mathematica, 26

Tr/@(2^Subsets@Range@#/2)&

Example:

Tr/@(2^Subsets@Range@#/2)&[4]

{0, 1, 2, 4, 8, 3, 5, 9, 6, 10, 12, 7, 11, 13, 14, 15}

\$\endgroup\$
1
\$\begingroup\$

Jelly, 7 bytes

2*’BS$Þ

Try it online!

Explanation

2*’BS$Þ
2*      2^n
  ’     decremented
     $Þ sort the (implicit) range using the following two functions:
   B    binary
    S   summed
\$\endgroup\$
1
\$\begingroup\$

05AB1E, 7 bytes

oL<Σb1¢

Try it online!

oL<Σb1¢  # full program
   Σ     # sort...
 L       # [1, 2, 3...
o        # ...2 **...
         # implicit input
 L       # ]...
  <      # with one subtracted from...
         # (implicit) each element...
   Σ     # by...
      ¢  # number of...
     1   # ones...
      ¢  # in...
         # (implicit) current element in list...
    b    # in binary
         # implicit output

Alternative, 7 bytes

oL<Σ2вO

Try it online!

2вO  # full modified section
  O  # sum of...
 в   # list of (base-10) values of digits of...
     # (implicit) current element in list...
 в   # in base...
2    # literal

Another alternative, 7 bytes

oL<ΣbSO

Try it online!

bSO  # full modified section
  O  # sum of...
 S   # list of characters of...
     # (implicit) current element in list...
b    # in binary
\$\endgroup\$
2
  • 1
    \$\begingroup\$ Since [1,1,1]>[1,1], you can do oL<Σbʒ for 6 bytes. \$\endgroup\$
    – ovs
    Dec 27 '20 at 18:18
  • \$\begingroup\$ @ovs Oh, that trailing ʒ to only keep all 1s is pretty smart. :) \$\endgroup\$ Nov 25 at 10:00
1
\$\begingroup\$

Vyxal, 6 bytes

‹Eʁµb∑

Try it Online!

  ʁ    # 0...
‹E     # 2 ** (n-1)
   µ   # Sorted by
    b  # binary
     ∑ # Summed
\$\endgroup\$
0
\$\begingroup\$

Java 8, 205

public class S{public static void main(String[] n){java.util.stream.IntStream.range(0,1<<Integer.parseInt(n[0])).boxed().sorted((a,b)->Integer.bitCount(a)-Integer.bitCount(b)).forEach(System.out::print);}}
\$\endgroup\$
0
\$\begingroup\$

C++11, 117 characters:

using namespace std;int main(){ set<pair<int,int> > s;int b;cin>>b;int i=0;while(++i<pow(2,b))s.insert({bitset<32>(i).count(),i});for (auto it:s) cout <<it.second<<endl;}

Ungolfed:

using namespace std;
int main()
{
    set<pair<int,int> > s;
    int b;
    cin>>b;
    int i=0;
    while (++i<pow(2,b))  {
        s.insert({bitset<32>(i).count(),i});
    }
    for (auto it:s) {
        cout <<it.second<<endl;
    }
}

Explanation:

Create a set of int,int pairs. The first int is bit count, the second one is the number. Pairs compare themselves according their first parameter, so the set is sorted by bit count.

\$\endgroup\$
0
\$\begingroup\$

MathGolf, 6 bytes

óráÅàç

Try it online.

Explanation:

ó       # Push 2 to the power of the (implicit) input-integer
 r      # Pop and push a list in the range [0,2**input)
  á     # Sort this list by,
   Å    # using the following two characters as inner code-block:
    à   #  Convert it to a binary-string
     ç  #  And only leave the truthy digits, removing the 0s
\$\endgroup\$
-1
\$\begingroup\$

Perl, 64/2 = 32

#!perl -ln
for$i(0..$_){$i-(sprintf"%b",$_)=~y/1//||print for 0..2**$_-1}

Simply iterate over the range [0..2^n-1] n + 1 times. In each iteration print only the numbers that have number of 1 bits equal to the iteration variable ($i). Bits are counted by counting 1's (y/1//) in the number converted to binary string with sprintf.

Test me.

Perl, 63

Sorting approach:

#!perl -l
print for sort{eval+(sprintf"%b-%b",$a,$b)=~y/0//dr}0..2**<>-1
\$\endgroup\$
5
  • 1
    \$\begingroup\$ @Optimizer, It uses O(1) memory. What other definition we have? Oops, that is not true as I print it live :) \$\endgroup\$
    – nutki
    Feb 27 '15 at 8:49
  • \$\begingroup\$ @Optimizer, fixed. \$\endgroup\$
    – nutki
    Feb 27 '15 at 9:02
  • \$\begingroup\$ Well, I am aware of this when I set the condition, but I allow it anyway, since I want to see what convoluted answers people can come up with. \$\endgroup\$ Feb 27 '15 at 9:44
  • 2
    \$\begingroup\$ I just asked "how" as I cannot read perl :) Care to add more explanation ? \$\endgroup\$
    – Optimizer
    Feb 27 '15 at 9:48
  • \$\begingroup\$ @Optimizer, some more explanation added. \$\endgroup\$
    – nutki
    Feb 27 '15 at 11:39

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