count ones in range

Question

Challenge :

Count the number of ones 1 in the binary representation of all number between a range.

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Input :

Two non-decimal positive integers

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Output :

The sum of all the 1s in the range between the two numbers.

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Example :

4 , 7        ---> 8
4  = 100 (adds one)   = 1
5  = 101 (adds two)   = 3
6  = 110 (adds two)   = 5
7  = 111 (adds three) = 8

10 , 20     ---> 27
100 , 200   ---> 419
1 , 3       ---> 4
1 , 2       ---> 2
1000, 2000  ---> 5938

I have only explained the first example otherwise it would have taken up a huge amount of space if I tried to explain for all of them.

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Note :

• Numbers can be apart by over a 1000
• All input will be valid.
• The minimum output will be one.
• You can accept number as an array of two elements.
• You can choose how the numbers are ordered.

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Winning criteria :

This is code-golf so shortest code in bytes for each language wins.

Answer 1

Python 2, 47 bytes

f=lambda x,y:y/x and bin(x).count('1')+f(x+1,y)

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Answer 2

JavaScript (ES6), 38 bytes

Takes input in currying syntax (a)(b).

a=>b=>(g=c=>a>b?0:1+g(c^c&-c||++a))(a)

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Commented

a => b => (         // given the input values a and b
  g = c =>          // g = recursive function taking c = current value
    a > b ?         // if a is greater than b:
      0             //   stop recursion and return 0
    :               // else:
      1 +           //   add 1 to the final result
      g(            //   and do a recursive call to g() with:
        c ^ c & -c  //     the current value with the least significant bit thrown away
        || ++a      //     or the next value in the range if the above result is 0
      )             //   end of recursive call
)(a)                // initial call to g() with c = a

Answer 3

Jelly, 4 bytes

rBFS

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Explanation

rBFS – Full program. Takes the two inputs from the commands line arguments.
r    – Range.
 B   – For each, convert to binary.
  FS – Flatten and sum.

Answer 4

05AB1E, 4 bytes

ŸbSO

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Answer 5

Java (JDK 10), 55 bytes

a->b->{int c=0;for(;a<=b;)c+=a.bitCount(b--);return c;}

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Answer 6

Python 2, 45 bytes

lambda x,y:`map(bin,range(x,y+1))`.count('1')

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Answer 7

Pyth, 8 7 bytes

1 byte thanks to Mr. Xcoder.

ssjR2}F

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Answer 8

MATL, 5 4 bytes

&:Bz

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Thanks to Luis Mendo for saving a byte!

(implicit input a and b, a<b)
&:                              % two-element input range, construct [a..b]
  B                             % convert to Binary as a logical vector (matrix)
   z                            % number of nonzero entries
(implicit output of the result)

Answer 9

R, 41 34 bytes

function(a,b)sum(intToBits(a:b)>0)

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Heavily inspired by the other R solution by ngm. This uses a different approach after the conversion to bits. Huge thanks to Giuseppe for hinting at a possible 34 bytes solution.

Answer 10

Ruby, 38 bytes

->a,b{("%b"*(b-a+1)%[*a..b]).count ?1}

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Answer 11

APL (Dyalog Unicode), 16 bytes

{≢⍸(⍵⍴2)⊤⍺↓0,⍳⍵}

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-1 thanks to H.PWiz.

Left argument = min
Right argument = max

Answer 12

Octave with Communication toolbox, 21 bytes

@(a,b)nnz(de2bi(a:b))

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The code should be fairly obvious. Number of nonzero elements in the binary representation of each of the numbers in the range.

This would be @(a,b)nnz(dec2bin(a:b)-48) without the communication toolbox.

Answer 13

Python 3, 56 54 52 bytes

This can be golfed more imo. -2 Bytes thanks to Mr.Xcoder -2 More bytes thanks to M. I. Wright

lambda a,b:''.join(map(bin,range(a,b+1))).count('1')

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Answer 14

Vyxal d, 2 bytes

ṡb

Try it online or see a four byte flagless version.

Gets the range between the two inputs, converts each to a list of 1s and 0s, then deep flattens and sums with the d flag.

Answer 15

Stax, 6 bytes

çy╠Ƽ☻

Run and debug it

Answer 16

PowerShell, 72 bytes

param($x,$y)$x..$y|%{$o+=([convert]::ToString($_,2)-replace0).length};$o

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Long because of the conversion to binary [convert]::ToString($_,2) and getting rid of the zeros -replace0. Otherwise we just take the input numbers, make a range $x..$y and for each number in the range convert it to binary, remove the zeros, take the .length thereof (i.e., the number of ones remaining), and add it to our $output.

Answer 17

Bash + common utilities, 50

jot -w%o - $@|tr 247356 1132|fold -1|paste -sd+|bc

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Converting integers to binary strings is always a bit of pain in bash. The approach here is slightly different - convert the integers to octal, then replace each octal digit with the number of binary 1s it contains. Then we can just sum all converted digits

Answer 18

APL+WIN, 33 26 bytes

Prompts for vector of integers:

+/,((↑v)⍴2)⊤(1↓v)+0,⍳-/v←⎕

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Explanation:

v←⎕ prompt for input of a vector of two integers max first

(v←1↓v)+0,⍳-/ create a vector of integers from min to max

(↑v)⍴2 set max power of 2 to max 

⊤ convert integers to a matrix of binaries

+/, convert matrix to a vector and sum

Answer 19

R, 44 40 37 bytes

function(a,b)sum(c(0,intToBits(a:b)))

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Previously:

function(a,b)sum(strtoi(intToBits(a:b)))
function(a,b)sum(as.integer(intToBits(a:b)))

Answer 20

Brachylog, 8 bytes

⟦₂ḃᵐcọht

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Explanation

⟦₂         Ascending range between the two elements in the input
  ḃᵐ       Map to base 2
    c      Concatenate
     ọ     Occurrences of each element
      h    Head: take the list [1, <number of occurrences of 1>]
       t   Tail: the number of occurrences of 1

Answer 21

Forth (gforth), 69 bytes

: f 1+ 0 -rot swap do i begin 2 /mod -rot + swap ?dup 0= until loop ;

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Explanation

The basic algorithm is to loop over every number in range, and sum the binary digits (divide by two, add remainder to sum, repeat until number is 0)

Code Explanation

1+                   \ add one to the higher number to make it inclusive
0 -rot swap          \ create a sum value of 0 and put loop parameters in high low order
do                   \ start a loop over the range provided
  i                  \ place the index on the stack
  begin              \ start an indefinite loop
    2 /mod           \ get the quotient and remainder of dividing by 2
    -rot             \ move the quotient to the back
    + swap           \ add the remainder to the sum and move it down the stack
    ?dup             \ duplicate the quotient unless it equals 0
    0=               \ check if it equals 0
  until              \ if it does equal 0, end the inner loop
loop                 \ end the outer loop

Answer 22

J, 16, 15 14 bytes

1 byte saved thanks to FrownyFrog!

+/@,@#:@}.i.,]

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Explanation:

A dyadic verb, the left argument is the lower bound m of the range, the right one - the upper n.

            ,    append                      
             ]   n to the
          i.     list 0..n-1
         }.      drop m elements from the beginning of that list 
      #:@        and convert each element to binary 
    ,@           and flatten the table
 +/@             and find the sum

Answer 23

QBasic, 95 93 83 82 bytes

@DLosc saved me some a lot of bytes!

Saved another byte using this technique!

INPUT a,b
FOR i=a TO b
k=i
FOR j=i TO 0STEP-1
x=k>=2^j
s=s-x
k=k+x*2^j
NEXT j,i
?s

Language of the Month FTW!

Explanation

INPUT a,b           Ask user for lower and upper bound
FOR i=a TO b        Loop through that range
k=i                 we need a copy of i to not break the FOR loop
FOR j=i TO 0STEP-1  We're gonna loop through exponents of 2 from high to low.
                    Setting the first test up for 4 to 2^4 (etc) we know we're overshooting, but that 's OK
x=k>=2^j            Test if the current power of 2 is equal to or smaller than k 
                    (yields 0 for false and -1 for true)
s=s-x               If k is bigger than 2^j, we found a 1, so add 1 to our running total s
                    (or sub -1 from the total s...)
k=k+x*2^j           Lower k by that factor of 2 if the test is true, else by 0
NEXT                Test the next exponent of 2
NEXT                process the next number in range
?s                  print the total

Last testcase of 1000 to 2000 actually works, in QBasic 4.5 running on Dosbox:

Answer 24

Husk, 4 bytes

Σṁḋ…

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Explanation

Σṁḋ…
   …     Get the (inclusive) range.
 ṁḋ      Convert each to binary and concatenate.
Σ        Get the sum.

Answer 25

PHP, 97 Bytes

(sure this can be shortened, but wanted to use the functions)

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Code

<?=substr_count(implode(array_map(function($v){return decbin($v);},
 range($argv[0],$argv[1]))),1);

Explanation

<?=
 substr_count(   //Implode the array and count every "1"
  implode(
    array_map(function($v){return decbin($v);}, //Transform every decimal to bin
          range($argv[0],$argv[1])   //generate a range between the arguments
     )
),1);   //count "1"'s

Answer 26

Haskell, 42 bytes

import Data.Bits
a%b=sum$popCount<$>[a..b]

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Answer 27

Pip, 10 bytes

$+JTB:a\,b

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Explanation

            a and b are command-line args (implicit)
      a\,b  Inclusive range from a to b
   TB:      Convert to binary (: forces TB's precedence down)
  J         Join into a single string of 1's and 0's
$+          Sum (fold on +)

Answer 28

Proton, 40 37 bytes

x=>y=>str(map(bin,x..y+1)).count("1")

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Answer 29

Charcoal, 10 bytes

ＩΣ⭆…·ＮＮ⍘ι²

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

     ＮＮ     Input numbers
   …·       Inclusive range
  ⭆         Map over range and join
        ι   Current value
         ²  Literal 2
       ⍘    Convert to base as string
 Σ          Sum of digits
Ｉ           Cast to string
            Implicitly print

Answer 30

Bash + coreutils, 38 32 bytes

seq -f2o%.fn $*|dc|tr -d 0|wc -c

Thanks to @Cowsquack for golfing off 6 bytes!

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