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

K (ngn/k), 19 13 bytes

{+//2\x_!1+y}

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{ } is a function with arguments x and y

!1+y is the list 0 1 ... y

x_ drops the first x elements

2\ encodes each int as a list of binary digits of the same length (this is specific to ngn/k)

+/ sum

+// sum until convergence; in this case sum of the sum of all binary digit lists

Answer 2

Perl 6, 32 30 bytes

-1 bytes thanks to Brad Gillbert

{[…](@_)>>.base(2).comb.sum}

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Explanation:

[…](@_)    #Range of parameter 1 to parameter 2
       >>    #Map each number to
                      .sum  #The sum of
                 .comb      #The string of
         .base(2)    #The binary form of the number

Answer 3

Julia 0.6, 28 bytes

(a,b)->sum(count_ones.(a:b))

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---

If input can be sent in in the form of a range (i.e. c(4:7) instead of c(4,7)), that saves 6 bytes:

Julia 0.6, 22 bytes

r->sum(count_ones.(r))

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Answer 4

Pari/GP, 31 bytes

Saved one byte thanks to Mr. Xcoder.

a->b->sum(i=a,b,sumdigits(i,2))

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Answer 5

APL (Dyalog Extended), 6 bytes^SBCS

≢⍤⍸⍤⊤…

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 ≢ tally

⍤ of

 ⍸ where true

⍤ in

 ⊤ the binary representation of

… the range

Answer 6

PHP, 68 bytes

The mapping variant already has been posted (though not yet in it´s mostly golfed version), so here is a looping solution:

for($i=$argv[2];$i>=$argv[1];)for($n=$i--;$n;$n>>=1)$s+=$n&1;echo$s;

Run with -nr or try it online.

Answer 7

6510 machine code, 29 28 bytes

sub routine;
takes input from A (lower bound) and X (upper bound) registers;
returns result in A (MSB) and Y (LSB)

machine code:

85 02 A0 00 84 FC E8 CA
E4 02 30 OD 8A F0 F8 46
90 FB E8 90 F8 E6 FC D0
F4 A5 FC 60

source code:

        STA $02     store lower bound in $02
        LDY #0      init result to 0 (Y = LSB, $FC=MSB)
        STY $FC
        INX         increment upper bound
LOOP1:  DEX         decrement upper bound
        CPX $02     compare to lower bound
        BMI :FINISH if smaller, return
        TXA         copy X to A
LOOP2:  BEQ :LOOP1  if 0, next outer loop
        LSR         shift right
        BCC :LOOP2  if carry is clear, next inner loop
        INY         else increment result
        BCC :LOOP2
        INC $FC
        BNE :LOOP2  next inner loop
FINISH: LDA $FC
        RTS

notes

• With only 8 bit input possible, the maximum number of set bits is 1024; so incrementing the MSB (INC $FC) always has a non-zero result; hence BNE :LOOP always branches.
• BEQ following that BNE never branches, even not if the accumulator is zero (so I could actually add two to the BEQ parameter and save one cycle); but that doesn´t matter: LSR will clear the carry and set the zero flag, BCC will hop to LOOP2 and the BEQ to LOOP1.
• I´m not completely sure (it´s been so long I actually coded on the C64), but it may fail if the range is larger than 127: CPX $02 is actually a substraction; if the result is >127, the negative flag may be set, so BMI would end the routine.
• I hope I got the branching parameters correct - I assembled the machine code manually.

Answer 8

Perl 5 -p, 39 bytes

map$\+=(sprintf'%b',$_)=~y/1//,$_..<>}{

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Answer 9

Japt -x, 10 8 7 bytes

òV ®¤è1

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Answer 10

Wolfram Language (Mathematica), 28 bytes

Tr@DigitCount[Range@##,2,1]&

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Answer 11

Scala, 45 44 bytes

_.to(_)flatMap(_.toBinaryString)count(49==)

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Answer 12

Thunno 2 S, 4 bytes

I2Bʂ

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Explanation

I2Bʂ  # Implicit input
I     # Inclusive range
 2B   # Convert each to binary
   ʂ  # Sum each inner list
      # Implicit output of sum

Answer 13

Rust, 54 50 btyes

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|l:u32,h:u32|(l..=h).fold(0,|c,x|c+x.count_ones())

count_ones does the heavy lifting of counting the number of ones in each number here.

Used fold to avoid turbofish from sum.

Ungolfed:

|low: u32, high: u32| (low..=high).fold(0, |total, x| total + x.count_ones())

Answer 14

J-uby, 26 bytes

:!~|:sum+(~:digits&2|:sum)

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Answer 15

Japt -x, 8 7 6 bytes

õV cì2

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õV cì2     :Implicit input of integers U & V
õV         :Range [U,V]
   c       :Flat map
    ì2     :  Convert to binary digit array
           :Implicit output of sum

Answer 16

Retina 0.8.2, 43 bytes

\d+
$*
M!&`(?<=^(1+),.*)\1.*
+`(1+)\1
$1x
1

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Answer 17

RProgN 2, 10 bytes

R²2Br.`0-L

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Answer 18

Javascript ES6, 78 bytes

With currying syntax

x=>y=>[...Array(y-x+1)].map((e,i)=>(i+x).toString`2`).join``.split`1`.length-1

Answer 19

Red, 82 bytes

func[a b][s: 0 until[n: a until[if n % 2 = 1[s: s + 1]1 > n: n / 2]b < a: a + 1]s]

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Answer 20

F#, 80 bytes

let c s e=Seq.sumBy(fun x->seq{for i=0 to 31 do yield x>>>i&&&1}|>Seq.sum){s..e}

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For each number x in the range, shift the number by i bytes, AND it with 1, and add that to the sum for that number. Finally add all the shifting results for each number and return it.

It does the shift 32 times, since the starting and ending numbers are of type int32.

Answer 21

sed 4.2.2, 59

:
s/\b(1+) (11\1)/\1 1\1 \2/
t
:a
s/(1+)\1/\10/
ta
s/0| //g

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Input and output as unary. Input is two space-separated unary integers.

The TIO has a footer line to convert to decimal, as a convenience.

The 1000, 2000 testcase takes too long - TIO times out after 1 minute.

Answer 22

Tcl, 80 bytes

proc P a\ b {while \$a<=$b {incr c [regexp -all 1 [format %b $a]]
incr a}
set c}

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Answer 23

Pyt, 3 bytes

Input is the larger number then the smaller number

ŘĦƩ

Explanation:

      Implicitly get the two numbers x and y (x<y)
Ř     Push [x,x+1,...,y]
Ħ     Get the Hamming weight of each element of the array
Ʃ     Sum the list of Hamming weights
      Implicit output

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Answer 24

APL(NARS), 15 chars, 30 bytes

{+/∊(⍵⍴2)⊤⍺..⍵}

test

  f←{+/∊(⍵⍴2)⊤⍺..⍵}
  1000 f 2000
5938

This seems ok even in the case 0 f 0 because +/⍬ is 0.

Answer 25

C# (.NET Core) with LINQ, 82 bytes

(a,b)=>{int c=0;for(;a<=b;a++)c+=Convert.ToString(a,2).Count(x=>x=='1');return c;}

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Ungolfed:

(a, b) => {                     // takes in two integer inputs, separated by a comma
    int c = 0;                  // initialize the sum variable
    for(; a <= b; a++)          // from a (inclusive) to b (exclusive)
        c +=                            // add to c:
            Convert.ToString(a, 2)          // convert a to binary
                .Count(x => x == '1');      // count the number of ones in the binary form
    return c;                   // print c after the loop
}

Answer 26

Burlesque - 12 bytes

per@b2\['1CN

pe             Parse eval
  r@           range
    b2         convert to base2
      \[       concat
        '1CN   count the `1`.

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Answer 27

C (GCC), 49 bytes

This function takes lower (a) and upper (b) bounds. Output is either through global c or by return value (if your ABI uses the same register from accumulating and function return).

c;f(a,b){for(c=b-a?f(a+1,b):0;a;a/=2)c+=a&1;a=c;}

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Acknowledgments

• -5 bytes thanks to GPS
• -1 byte thanks to ceilingcat

Answer 28

Powershell, 59 58 bytes

param($n,$m)$n..$m|%{for(;$_){$s+=$_-band1;$_=$_-shr1}};$s

Test script:

$f = {
param($n,$m)$n..$m|%{for(;$_){$s+=$_-band1;$_=$_-shr1}};$s
}

@(
,(4,7,8)
,(10,20,27)
,(100,200,419)
,(1,3,4)
,(1,2,2)
,(1000,2000,5938)
) | % {
    $n,$m,$e=$_
    $r = &$f $n $m
    $c = $r -eq $e

    "$c : $n , $m ---> $e = $r"
}

Output:

True : 4 , 7 ---> 8 = 8
True : 10 , 20 ---> 27 = 27
True : 100 , 200 ---> 419 = 419
True : 1 , 3 ---> 4 = 4
True : 1 , 2 ---> 2 = 2
True : 1000 , 2000 ---> 5938 = 5938

Answer 29

x86 machine code, 14 bytes

00000000: 31c0 f30f b8d9 01d8 4139 d17e f5c3       1.......A9.~..

Takes bottom of range in ecx, and top of range in ebx. Returns in eax.

Assembly (NASM syntax):

section .text
	global func
func:
					;1st arg = ecx, 2nd arg = edx
	xor eax, eax			;reset total
	loop:
		popcnt ebx, ecx		;get number of binary 1's in current number and store in ebx
		add eax, ebx		;add ebx to total (eax)
		inc ecx			;increase the current number (ecx) in range
		cmp ecx, edx
		jle loop		;repeat while current number (ecx) >= range max (edx)
	ret				;return the register eax

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C code equivalent:

/*
ecx=range_start
edx=range_end
eax=total
ebx=ones_count
*/

int ones_in_range(int range_start, int range_end){
	int total = 0;
	do {
		int ones_count = __builtin_popcount(range_start);
		total += ones_count;
		range_start++;	
	} while(range_start <= range_end);
	return total;
}

Answer 30

x86-16 machine code, 16 bytes

Binary:

00000000: 33c0 8bd1 d1e2 7301 4075 f93b cbe0 f3c3  3.....s.@u.;....

Listing:

33 C0       XOR  AX, AX         ; clear 1's counter in AX 
        INTLOOP: 
8B D1       MOV  DX, CX         ; current number into DX 
        BITLOOP: 
D1 E2       SHL  DX, 1          ; MSb into CF, ZF = ( DX == 0 )
73 01       JNC  BITZERO        ; if a zero, check ZF
40          INC  AX             ; increment counter 
        BITZERO: 
75 F9       JNZ  BITLOOP        ; if DX > 0 keep looping 
3B CB       CMP  CX, BX         ; is CX < BX? 
E0 F3       LOOPNZ INTLOOP      ; if not, decrement CX and loop 
C3          RET                 ; return to caller

Callable function, low range number in BX, high number in CX. Result in AX.

Note: There is another IA machine code submission which uses the POPCNT instruction introduced with SSE4 requiring an Intel Core or later architecture. This one will run on any 8086 or later CPU.

Test program: