Print the amount of ones in a binary number without using bitwise operators

Question

Description

Given a number, print the amount of 1s it has in binary representation.

Input

A number >= 0 in base 10 that won't exceed the highest number your language is able to handle.

Output

The amount of 1s in binary representation.

Winning condition

The shortest code wins.

Disallowed

• Bitwise operators. Other operators, like addition and multiplication, are allowed.
• Built-in base conversion functions.

Examples

Input:     Ouput:

56432      8


Input:     Output:

45781254   11


Input:     Output:

0          0

Answer 1

APL, 9 12 characters

+/2|⌊⎕÷2*⍳32

This assumes that the interpreter uses 32-bit integers, and that ⎕IO is set to 0 (meaning that monadic ⍳ begins with 0, rather than 1). I used the 32-bit version of Dyalog APL.

Explanation, from right to left:

• ⍳32 generates a vector of the first 32 integers (as explained before, because ⎕IO is 0, this vector begins with 0).
• * is the power function. In this case, it generates 2 to the power of each element of the vector supplied as its right argument.
• ÷ is the divided-by function. It gives us ⎕ (evaluated user input) divided by each element of the vector to its right (each power of two).
• ⌊ floors each element of the argument to its right.
• 2| gives us the remainder of each element of to its right divided by 2.
• / reduces (folds) its right argument using the function to its left, +.

Not quite 9 characters anymore. :(

Old, rule-breaking version:

+/⎕⊤⍨32/2

Explanation, from right to left:

• 32/2: Replicate 2, 32 times.
• ⍨ commutes the dyadic function to its left, which in this case is ⊤ (i.e., X⊤⍨Y is equivalent to Y⊤X).
• ⊤ is the encode function. It encodes the integer to its right in the base given to its left. Note that, because of the commute operator, the right and left arguments are switched. The base is repeated for the number of digits required, hence 32/2.
• ⎕ is a niladic function that accepts user input and evaluates it.
• +/ reduces (folds) its right argument using +. (We add up the 0's and 1's.)

Answer 2

Brainbool, 2

,.

The most reasonable interpretation, in my opinion (and what most of the answers use) of "highest number your language is able to handle" is "largest number your language natively supports". Brainbool is a brainfuck derivative that uses bits rather than bytes, and takes input and output in binary (0 and 1 characters) rather than character codes. The largest natively supported number is therefore 1, and the smallest is 0, which have Hamming weights 1 and 0 respectively.

Brainbool was created in 2010, according to Esolang.

Answer 3

Brainfuck, 53 characters

This was missing an obligatory Brainfuck solution, so I made this one:

[[->+<[->->>>+<<]>[->>>>+<<]<<<]>>>>[-<<<<+>>>>]<<<<]

Takes number from cell 1 and puts the result into cell 6.

Unenrolled and commented version:

[  while n != 0
  [  div 2 loop
    -
    >+<  marker for if/else
    [->->>>+<<]  if n != 0 inc n/2
    >
    [->>>>+<<]  else inc m
    <<<
  ]
  >>>>  move n/2 back to n
  [-<<<<+>>>>]
  <<<<
]

Answer 4

J, 13 characters

(+ the number of digits in the number)

+/2|<.n%2^i.32

Usage: replace the n in the program with the number to be tested.

Examples:

+/2|<.56432%2^i.32
8
+/2|<.45781254%2^i.32
11
+/2|<.0%2^i.32
0

There's probably a way of rearranging this so the number can be placed at the beginning or end, but this is my first J entry and my head's hurting slightly now.

Explanation(mainly so that I understand it in the future)

i.32 - creates an array of the numbers 1 to 32

2^ - turns the list into the powers of two 1 to 4294967296

n% - divides the input number by each element in the list

<. - rounds all the divison results down to the next integer

2| - same as %2 in most languages - returns 0 if even and 1 if odd

+/ - totals the items in the list (which are now just 1s or 0s)

Answer 5

Python 2.6, 41 characters

t,n=0,input()
while n:t+=n%2;n/=2
print t

note: My other answer uses lambda and recursion and this one uses a while loop. I think they are different enough to warrant two answers.

Answer 6

Ruby, 38 characters

f=->u{u<1?0:u%2+f[u/2]}
p f[gets.to_i]

Another solution using ruby and the same recursive approach as Steven.

Answer 7

GolfScript, 17 16 characters

~{.2%\2/.}do]0-,

Edit: new version saves 1 character by using list operation instead of fold (original version was ~{.2%\2/.}do]{+}*, direct count version: ~0\{.2%@+\2/.}do;).

Answer 8

C, 45

f(n,c){for(c=0;n;n/=2)c+=n%2;printf("%d",c);}

Nothing really special here for golfing in C: implicit return type, implicit integer type for parameters.

Answer 9

Python 2.6, 45 characters

b=lambda n:n and n%2+b(n/2) 
print b(input())

Answer 10

Perl, 45 43 36 Characters

$n=<>;while($n){$_+=$n%2;$n/=2}print

Thanks to Howard for 45->43, and to User606723 for 43->36.

Answer 11

Perl, 30 chars

$==<>;1while$_+=$=%2,$=/=2;say

Based on PhiNotPi's solution, with some extra golfing. Run with perl -M5.010 to enable the Perl 5.10 say feature.

Answer 12

Common Lisp, 12 chars

(assuming a 1 char variable name - i.e.: 11 + number length)

It's not a base conversion function, so it should work:

(logcount x)

Examples:

[1]> (logcount 0)
0
[2]> (logcount 1)
1
[3]> (logcount 1024)
1
[4]> (logcount 1023)
10
[5]> (logcount 1234567890123456789012345678901234567890)
68

(Using GNU CLISP.)

Answer 13

C, 61 60 57 53 characters

void f(x){int i=0;for(;x;x/=2)i+=x%2;printf("%u",i);}

The function body only is 38 characters. Edit: removed bitwise operator Edit: put printf out of the loop as suggested in the comments Edit: switch to K&R declaration; also, this is no longer C99-specific

Answer 14

dc – 26 chars

This is rather long, mostly due to the lack of loop constructs in dc.

0?[d2%rsi+li2/d0<x]dsxx+p

Keeps adding up the modulo 2 of the number and dividing the number by to until it reaches zero. Can handle arbitrarily long integers.

Example:

$ dc -e '0?[d2%rsi+li2/d0<x]dsxx+p' <<< 127
7
$ dc countones.dc <<< 1273434547453452352342346734573465732856238472384263456458235374653784538469120235
138

Answer 15

C, 66 characters

main(int n,char **a){printf("%u",__builtin_popcount(atoi(a[1])))};

Note: requires gcc or gcc-compatible compiler (e.g. ICC, clang).

For some CPUs __builtin_popcount compiles to a single instruction (e.g. POPCNT on x86).

Answer 16

D (70 chars)

int f(string i){int k=to!int(i),r;while(k){if(k%2)r++;k/=2;}return r;}

Answer 17

JavaScript, 78 72 71 characters

I'll post my initial solution which I came up with before posting the question as well. There is already a much better JavaScript answer though :)

for(n=prompt(a=0),j=1;j<=n;j*=2)for(i=j;i<=n;i+=2*j)n<i+j&&a++;alert(a)

http://jsfiddle.net/Mk8zd/1/

The idea comes from certain "mind reading cards" which enable you to obtain the number someone else has in mind, by showing them cards and let them say on which cards their number is apparent.

It works because each number is a unique combination of 1s / 0s in binary. My solution checks on which "cards" the number is apparent so as to determine how many 1s it has. It's just not very efficient, though...

I found this document which outlines the mind reading technique.

Answer 18

Haskell (60 chars)

f n=sum[1|x<-[0..n],odd$n`div`2^x]
main=interact$show.f.read

Answer 19

PHP, 57

$i=$s=0;for(;$i<log($n,2);){$s+=$n/pow(2,$i++)%2;}echo$s;

This assumes that $n holds the value to be tested.

PHP, 55 (alternative solution)

function b($i){return$i|0?($i%2)+b($i/2):0;}echo b($n);

Again, this assumes that $n holds the value to be tested. This is an alternative because it uses the or-operator to floor the input.

Both solutions work and do not cause notices.

Answer 20

Ocaml, 45 characters

Based on @Leah Xue's solution. Three spaces could be removed and it's sligthly shorter (~3 characters) to use function instead of if-then-else.

let rec o=function 0->0|x->(x mod 2)+(o(x/2))  

Answer 21

Java 7, 36 bytes

int b(Long a){return a.bitCount(a);}

Because of course this, of all things, is something that Java has a builtin for...

Answer 22

Mathematica 26

Count[n~IntegerDigits~2,1]

Answer 23

386 opcode, 10 Bytes

2BC0                       SUB EAX,EAX
01C9                       ADD ECX,ECX
1400                       ADC AL,0
41                         INC ECX
E2F9                       LOOP $-5
C3                         RET

Thank Peter Cordes for 1B save and 1B rule save

Answer 24

Vyxal l, 49 bits¹, 6.125 bytes

ʀEṗ'∑?=;f

Try it Online!

Brute force solution, so the online interpreter times out. May potentially be made shorter by porting others, but I thought this solution was fun

ʀ         # range 0..input+1
 E        # raise 2^n for each value in that range
  ṗ       # powerset
   '   ;  # filtered by
    ∑     # the sum
     ?=   # is equal to the input?
        f # flatten
          # after which the l flag takes the length of the top of the stack and implicitly outputs it

Answer 25

Scala, 86 characters

object O extends App{def f(i:Int):Int=if(i>0)i%2+f(i/2)else 0
print(f(args(0).toInt))}

Usage: scala O 56432

Answer 26

R, 53 characters

o=function(n){h=n%/%2;n%%2+if(h)o(h)else 0};o(scan())

Examples:

> o=function(n){h=n%/%2;n%%2+if(h)o(h)else 0};o(scan())
1: 56432
2: 
Read 1 item
[1] 8
> o=function(n){h=n%/%2;n%%2+if(h)o(h)else 0};o(scan())
1: 45781254
2: 
Read 1 item
[1] 11
> o=function(n){h=n%/%2;n%%2+if(h)o(h)else 0};o(scan())
1: 0
2: 
Read 1 item
[1] 0

If inputting the number is not part of the character count, then it is 43 characters:

o=function(n){h=n%/%2;n%%2+if(h)o(h)else 0}

with test cases

> o(56432)
[1] 8
> o(45781254)
[1] 11
> o(0)
[1] 0

Answer 27

OCaml, 52 characters

let rec o x=if x=0 then 0 else (x mod 2) + (o (x/2))

Answer 28

Scheme

I polished the rules a bit to add to the challenge. The function doesn't care about the base of the number because it uses its own binary scale. I was inspired by the way analog to numeric conversion works. I just use plain recursion for this:

(define (find-ones n)
  (define (nbits n)
    (let nbits ([i 2])
      (if (< i n) (nbits (* i 2)) i)))
  (let f ([half (/ (nbits n) 2)] [i 0] [n n])
    (cond [(< half 2) i]
      [(< n i) (f (/ half 2) i (/ n 2))]
      [else (f (/ half 2) (+ i 1) (/ n 2))])))

Answer 29

Isn't reading a number into binary or printing the number from binary a "builtin base conversion function", thus invalidating every answer above that prints an integer? If you permit reading and printing an integer, like almost all the above answers do, then I'll make claims using a builtin popcount function :

Haskell, 50

There was a popCount routine added to the Data.Bits module for GHC v7.2.1/v7.4.1 this summer (see tickets concerning the primop and binding).

import Data.Bits
main=interact$show.popCount.read

I cannot beat the above Python and Perl scores using their GMPY or GMP::Mpz modules for GMP sadly, although GMP does offer a popcount function too.

Answer 30

JavaScript, 57 55 51 50 bytes

a=prompt(b=0);while(a){b+=a%2;a=(a-a%2)/2}alert(b)

Modulo is not bitwise operator ☺