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

Prolog (SWI), 43 bytes

A*B*O:-A>B,O=0;- \A*B*S,O is S+popcount(A).

Try it online!

Answer 2

Aya, 15 bytes

[2|,10 2.&].FW

Apparently there's no to binary built-in, so it needs to be done manually. Takes inputs on the stack.

Example Session

Given that there's no try it online, I figured I'd provide an example console session, placing the program in a function for convenience. It would work outside the function and without the variable names, as taking input from the stack is an allowed default for stack languages.

aya> {a b, a b[2|,10 2.&].FW} :f
{a b, a b [2|, 10 2 .&] .F W} 
aya> 4 7 f
8 
aya> 10 20 f
27 
aya> 100 200 f
419 
aya> 1 3 f
4 
aya> 1 2 f
2 
aya> 1000 2000 f
5938 

Explained

[2|,10 2.&].FW­⁡​‎‎⁡⁠⁡‏⁠‎⁡⁠⁢‏⁠‎⁡⁠⁣‏⁠‎⁡⁠⁤‏⁠‎⁡⁠⁣⁣‏‏​⁡⁠⁡‌⁢​‎‎⁡⁠⁣⁡‏⁠‎⁡⁠⁣⁢‏‏​⁡⁠⁡‌⁣​‎‎⁡⁠⁢⁡‏⁠‎⁡⁠⁢⁢‏‏​⁡⁠⁡‌⁤​‎‎⁡⁠⁢⁤‏‏​⁡⁠⁡‌⁢⁡​‎‎⁡⁠⁣⁤‏⁠‎⁡⁠⁤⁡‏‏​⁡⁠⁡‌⁢⁢​‎‎⁡⁠⁤⁢‏‏​⁡⁠⁡‌­
[2|,      ]     # ‎⁡Over each number in a range from the first input to the second
        .&      # ‎⁢  Convert the number
    10          # ‎⁣  From base 10
       2        # ‎⁤  To base 2
           .F   # ‎⁢⁡Flatten the resulting list
             W  # ‎⁢⁢And summate
💎

Created with the help of Luminespire.

Answer 3

MathGolf, 7 bytes

╒k(≥mâΣ

Takes two loose inputs in reversed order.

Try it online.

Explanation:

Unfortunately, MathGolf lacks a ranged-list builtin, so 4 bytes are used for that..

╒k(≥     # Push a list in the range [secondInput,firstInput]:
╒        #  Push a list in the range [1, (first) implicit input-integer]
 k       #  Push the second input-integer
  (      #  Decrease it by 1
   ≥     #  Remove that many leading items from the ranged-list
    m    # Map over each integer in this list:
     â   #  Convert it to a binary-list
      Σ  # Take the flattened sum of this list of lists of bits
         # (after which the entire stack is output implicitly as result)

Answer 4

Swift 5.9, 54 bytes

let f={($0+0...$1+0).reduce(0){$0+$1.nonzeroBitCount}}

f(_:_:) takes a lower bound and an upper bound.

There aren't really any fancy tricks here, just a simple call to reduce(_:_:) and some hints for the type checker in the form of +0s.