All ASCII characters with a given bit count

Question

(Title with thanks to @ChasBrown)

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The Background

This challenge is inspired by a question that I recently posted on Puzzling Stack Exchange. Please feel free to follow the link if you are interested in the original question. If not then I won't bore you with the details here.

The Facts

Every printable standard ASCII character has a decimal value between 32 and 126 inclusive. These can be converted to their corresponding binary numbers in the range 100000 to 1111110 inclusive. When you sum the bits of these binary numbers you will always end up with an integer between 1 and 6 inclusive.

The Challenge

Given an integer between 1 and 6 inclusive as input, write a program or function which will output in any acceptable format all of the printable standard ASCII characters where the sum of the bits of their binary value is equal to the input integer.

The Examples/Test Cases

1 -> ' @'
2 -> '!"$(0ABDHP`'
3 -> '#%&)*,1248CEFIJLQRTXabdhp'
4 -> ''+-.3569:<GKMNSUVYZ\cefijlqrtx'
5 -> '/7;=>OW[]^gkmnsuvyz|'
6 -> '?_ow{}~'

An ungolfed Python reference implementation is available here (TIO).

The Rules

1. Assume the input will always be an integer (or string representation of an integer) between 1 and 6 inclusive.
2. You may write a program to display the results or a function to return them.
3. Output may be in any reasonable format but must be consistent for all inputs. If you choose to output a quoted string then the same type of quotes must be used for all inputs.
4. Standard loopholes prohibited as usual.
5. This is code golf so shortest code in each language wins.

Answer 1

x86-16 machine code, IBM PC DOS, 29 25 bytes

Machine code:

00000000: be81 00ad b330 b108 d0c8 12dd e2fa 3adc  .....0........:.
00000010: 7502 cd29 fec0 79ec c3                   u..)..y..

Listing:

BE 0081     MOV  SI, 081H   ; SI = memory address of command line string
AD          LODSW           ; AL = start ASCII value (is 20H from space on cmd line)
                            ; AH = target number of bits (in ASCII)
        CHR_LOOP:
B3 30       MOV  BL, '0'    ; BL = counter of bits, reset to ASCII zero
B1 08       MOV  CL, 8      ; loop through 8 bits of AL
        BIT_LOOP:
D0 C8       ROL  AL, 1      ; rotate LSB of AL into CF
12 DD       ADC  BL, CH     ; add CF to BL (CH is always 0) 
E2 FA       LOOP BIT_LOOP   ; loop to next bit
3A DC       CMP  BL, AH     ; is current char the target number of bits?
75 02       JNE  NO_DISP    ; if not, do not display
CD 10       INT  29H        ; display ASCII char in AL (current char in loop)
        NO_DISP: 
FE C0       INC  AL         ; increment char to next ASCII value
79 EC       JNS  CHR_LOOP   ; if char <= 127, keep looping
C3          RET             ; return to DOS

Standalone PC DOS executable program, input number from command line. Output is displayed to console window.

Answer 2

CP-1610 assembly (Intellivision), 20 DECLEs¹ = 25 bytes

Takes \$N\$ in R0 and a pointer to the output buffer in R4. Writes all matching characters in the buffer and marks the end of the results with NUL.

                ROMW    10              ; use 10-bit ROM width
                ORG     $4800           ; map this program at $4800

                ;; ------------------------------------------------------------- ;;
                ;;  test code                                                    ;;
                ;; ------------------------------------------------------------- ;;
4800            EIS                     ; enable interrupts

4801            MVII    #$103,    R4    ; set the output buffer at $103 (8-bit RAM)
4803            MVII    #2,       R0    ; test with N = 2
4805            CALL    getChars        ; invoke our routine

4808            MVII    #$103,    R4    ; R4 = pointer into the output buffer
480A            MVII    #$215,    R5    ; R5 = backtab pointer

480C  draw      MVI@    R4,       R0    ; read R0 from the buffer
480D            SLL     R0,       2     ; R0 *= 8
480E            SLL     R0
480F            BEQ     done            ; stop if it's zero

4811            ADDI    #7-256,   R0    ; draw it in white
4815            MVO@    R0,       R5

4816            B       draw            ; go on with the next entry

4818  done      DECR    R7              ; loop forever

                ;; ------------------------------------------------------------- ;;
                ;;  routine                                                      ;;
                ;; ------------------------------------------------------------- ;;
      getChars  PROC

4819            MVII    #32,      R1    ; start with R1 = 32

481B  @loop     MOVR    R1,       R3    ; copy R1 to R3
481C            CLRR    R2              ; clear R2
481D            SETC                    ; start with the carry set

481E  @count    ADCR    R2              ; add the carry to R2
481F            SARC    R3              ; shift R3 to the right (the least
                                        ; significant bit is put in the carry)
4820            BNEQ    @count          ; loop if R3 is not zero

4822            CMPR    R2,       R0    ; if R2 is equal to R0 ...
4823            BNEQ    @next

4825            MVO@    R1,       R4    ; ... write R1 to the output buffer

4826  @next     INCR    R1              ; advance to the next character
4827            CMPI    #127,     R1    ; and loop until 127 is reached
4829            BLT     @loop

482B            MVO@    R3,       R4    ; write NUL to mark the end of the output

482C            JR      R5              ; return

                ENDP

Output for N=2

NB: The opening parenthesis looks a lot like an opening square bracket in the Intellivision font. Both characters are distinct, though.

^{screenshot from jzIntv}

---

_{1. A CP-1610 opcode is encoded with a 10-bit value, known as a 'DECLE'. This routine is 20 DECLEs long, starting at $4819 and ending at $482C (included).}

Answer 3

Python 2, 62 bytes

lambda n:[chr(i)for i in range(32,127)if bin(i).count('1')==n]

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Answer 4

Perl 6, 41 34 bytes

{chrs grep *.base(2)%9==$_,^95+32}

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Anonymous code block that takes a number and returns a string of valid characters.

Explanation:

{                                }  # Anonymous code block taking a number
      grep                ,^95+32   # Filter from the range 32 to 126
           *.base(2)                # Where the binary of the digit
                    %9                # When parsed as a decimal modulo 9
                      ==$_            # Is equal to the input
 chrs                               # And convert the list of numbers to a string

It can be proven that for any number \$n\$ in base \$b\$, \$n \equiv \text{digitsum}(n) \pmod{b-1}\$ (clue: remember that \$b \pmod{b-1}=1\$).

We can use this to get the digitsum of our binary number by parsing it as a decimal number and moduloing by 9, which is valid because the range of numbers we are using is guaranteed to have less than 9 bits. This is helped along by Perl 6's automatic casting of the binary string to a decimal number when used in a numeric context.

Answer 5

05AB1E, 8 bytes

žQʒÇbSOQ

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Explanation

žQ        # push the printable ascii characters
  ʒ       # filter, keep elements whose
   Ç      # character code
    b     # converted to binary
     SO   # has a digit sum
       Q  # equal to the input

Answer 6

Jelly, 8 bytes

ØṖOB§=ʋƇ

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ØṖ       printable ascii character list
  OB     to binary
    §    popcount
     =   equal to input?
      ʋƇ filter (implicitly output)

Answer 7

Brachylog, 7 bytes

∈Ṭ&ạhḃ+

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A predicate which functions as a generator, takes input through its output variable, and produces each character through its input variable. Because Brachylog.

           The input variable (which is an element of the output)
∈          is an element of
 Ṭ         the string containing every printable ASCII character
  &        and the input
   ạh      converted to a codepoint
     ḃ     converted to a list of binary digits
      +    sums to
           the output variable (which is the input).

Answer 8

JavaScript (Node.js), 60 bytes

Using Jo King's modulo trick

n=>(g=x=>x>>7?'':Buffer(x.toString(2)%9-n?0:[x])+g(x+1))(32)

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---

JavaScript (Node.js),  70  69 bytes

n=>(g=x=>x>>7?'':Buffer((h=x=>x&&x%2+h(x>>1))(x)-n?0:[x])+g(x+1))(32)

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Commented

n => (              // n = input
  g = x =>          // g = recursive function, taking a byte x
    x >> 7 ?        //   if x = 128:
      ''            //     stop recursion and return an empty string
    :               //   else:
      Buffer(       //     create a Buffer:
        (h = x =>   //       h = recursive function taking a byte x
          x &&      //         stop if x = 0
          x % 2 +   //         otherwise, add the least significant bit
          h(x >> 1) //         and do a recursive call with floor(x / 2)
        )(x)        //       initial call to h
        - n ?       //       if the result is not equal to n:
          0         //         create an empty Buffer (coerced to an empty string)
        :           //       else:
          [x]       //         create a Buffer consisting of the character x
      ) +           //     end of Buffer()
      g(x + 1)      //     append the result of a recursive call to g with x + 1
)(32)               // initial call to g with x = 32

Answer 9

Japt, 9 bytes

;EƶXc¤è1

Try it or test all inputs

;EƶXc¤è1     :Implicit input of integer U
;E            :Printable ASCII
  Æ           :Filter each X
   ¶          :Test U for equality with
    Xc        :  Character code of X
      ¤       :  To binary string
       è1     :  Count the 1s

Answer 10

Excel (2016 or later), 76 bytes

=CONCAT(IF(LEN(SUBSTITUTE(DEC2BIN(ROW(32:126)),0,))=A1,CHAR(ROW(32:126)),""))

Takes input from A1, outputs in whatever cell you put this formula. This is an array formula, so you need to press Ctrl+Shift+Enter to input it. The "2016 or later" is because it needs the CONCAT function (the deprecated CONCATENATE won't take an array as argument).

Answer 11

C (standard library), 74 67 bytes

i;j;k;f(n){for(i=31;i<126;k||puts(&i))for(k=n,j=++i;j;j/=2)k-=j&1;}

Using only standard library functions. Thanks go to @gastropner for improvement from 74 to 67 bytes.

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Answer 12

R, 77 68 bytes

Approach using for loop

-9 bytes thanks to Giuseppe

n=scan();for(i in 32:126)if(sum(intToBits(i)>0)==n)cat(intToUtf8(i))

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Previously:

R, 78 69 66 bytes

-12 bytes thanks to Giuseppe

a=32:126;cat(intToUtf8(a[colSums(sapply(a,intToBits)>0)==scan()]))

Turns the numbers 32 to 126 into a matrix of bits then sums across the rows to find which match the input number.

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Answer 13

Java 10, 98 97 94 70 67 bytes

n->{for(var c='';c-->31;)if(n.bitCount(c)==n)System.out.print(c);}

-24 bytes thanks to NahuelFouilleul.

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Explanation:

Contains an unprintable character with unicode value 127.

n->{                         // Method with Integer parameter and no return-type
  for(var c='';c-->31;)     //  Loop character `c` in the range ['~', ' '] / (127,31):
    if(n.bitCount(c)         //   If the amount of 1-bits in the two's complement binary
                             //   representation of the current characters
                    ==n)     //   equals the input:
      System.out.print(c);}  //    Print the current character

Answer 14

Java 8, 131 71 bytes

-60 bytes thanks to everyone in the comments

Returns a java.util.stream.IntStream of codepoints

n->java.util.stream.IntStream.range(32,127).filter(i->n.bitCount(i)==n)

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Using HashSet, 135 bytes. Returns a Set<Object>:

n->new java.util.HashSet(){{for(int i=31;i++<126;add(Long.toBinaryString(i).chars().map(c->c-48).sum()==n?(char)i+"":""),remove(""));}}

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Answer 15

C# (Visual C# Interactive Compiler), 86 bytes

n=>Enumerable.Range(32,95).Where(x=>"0123456".Sum(g=>x>>g-48&1)==n).Select(x=>(char)x)

Thanks to @ExpiredData for giving me the idea to use Sum()! When I get back to my PC, I will replace the string "0123456" with unprintables, saving three bytes.

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Answer 16

Dyalog APL Extended, 24 22 bytes

⎕ucs a⌿⍨⎕=+⌿2⊤a←32…126

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-2 bytes thanks to ngn

Alternative 22 bytes in regular Dyalog APL by ngn:

⎕ucs 32+⍸⎕=32↓+/↑,⍳7⍴2

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Answer 17

Factor, 52 bytes

[ 32 126 [a,b] [ bit-count = ] with filter >string ]

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• 32 126 [a,b] Create a range of printable ascii values
• [ bit-count = ] with filter Select values with the bit count indicated by the input
• >string Convert the result from a sequence of code points to a printable string

Answer 18

J, ^{31 27} 23 bytes

((=1#.#:)u:@#])32+i.@95

-2 thanks to south

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J, ³¹ 27 bytes

-4 bytes thanks to Galen

[:u:32+[:I.]=1#.32#:@+i.@95

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Original Answer

a.#~&(95{.32}.])]=1#.2#:@i.@^8:

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• 2#:@i.@^8: produces the binary numbers 0 through 255 (2 ^ 8 is 256)
• 1#. sums each one
• ]= produces a binary mask showing where the sum equals the original input
• a.#~ mask uses that binary mask to filter J's full ascii alphabet a.
• &(95{.32}.]) but before doing so take only elements 32...126 from both the alphabet and the mask

Answer 19

Python 2, 69 bytes

lambda n:[chr(i)for i in range(32,127)if sum(map(int,bin(i)[2:]))==n]

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Answer 20

Gaia, 10 bytes

₵R⟪¤cbΣ=⟫⁇

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		| implicit input, n
₵R		| push printable ascii
  ⟪	⟫⁇	| filter the list where:
   ¤cbΣ		| the sum of the code point in binary
       =	| is equal to n

Answer 21

Ruby, 48 bytes

->n{(' '..?~).select{|x|x.ord.digits(2).sum==n}}

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Answer 22

Perl 5 -a, 50 43 bytes

@NahuelFouilleul saves 7 bytes

map{$_=chr;unpack('B*')%9-"@F"||say}32..126

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Answer 23

K (ngn/k), 20 bytes

Solution:

`c$32+&(+/2\32+!95)=

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Explanation:

Evaluated right-to-left:

`c$32+&(+/2\32+!95)= / the solution
                   = / equals?
       (          )  / do this together
               !95   / range 0..94
            32+      / add 32, so range 32..126
          2\         / break into base-2
        +/           / sum up
      &              / indices where true
   32+               / add 32
`c$                  / cast to character

Answer 24

6502 assembly (NES), 22 bytes

Machine code:

a0 1f a6 60 c8 98 30 fb ca 0a b0 fc d0 fb e8 d0 f1 8c 07 20 f0 ec

Assembly:

    ldy #$1f ; Y holds the current character code
NextCharacter:
    ldx $60 ; load parameter into X
    iny
    tya
    bmi (NextCharacter + 1) ; exit at char 128, #$60 is the return opcode

CountBits:
    dex
Continue:
    asl
    bcs CountBits
    bne Continue

CompareBitCount:
    inx ; fixes off-by-one error and sets Z flag if bit count matches
    bne NextCharacter
    sty $2007
    beq NextCharacter ; always branches

Full program. Tested with FCEUX 2.2.3, should work on any standard NES emulator.

Inspired by Ryan Russell's answer. Input given at CPU address $60. Outputs to the console's Picture Processing Unit memory.

Answer 25

VBA, 146 bytes

Sub Main()
n=3
For i=32 To 127
b=CStr(Application.WorksheetFunction.Dec2Bin(i))
If Len(b)-Len(Replace(b,"1",""))=n Then MsgBox Chr(i)
Next
End Sub

I wonder if somehow it could be done using Excel alone, I am not a excel formula pro....

Answer 26

Nibbles, 5 bytes

|_~`^+`@ 2

That's 10 nibbles each encoded as half a byte in the binary form.

Translation:

|          filter
 _         printables (since stdin is empty)
 ~         not
  `^       xor
    +      sum
     `@2   to base 2
        $  filter var (implicit)
    @      first input arg (implicit)

It uses not of the xor to set itself up for implicit args (instead of == which wouldn't be able to use any)

Usage:

nibbles <filename>.nbl <input number>

Nibbles isn't on TIO yet...

Answer 27

Python 3.10, 58 bytes

lambda n:(chr(i)for i in range(32,127)if i.bit_count()==n)

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It's kind of like Lynn's answer.
Just it uses the .bit_count introduced in Python3.10, so its shorter

More info about .bit_count here: https://docs.python.org/3/library/stdtypes.html

Answer 28

Kotlin, 60 bytes

{n->(32..127).filter{n==it.countOneBits()}.map{it.toChar()}}

Note:
For the moment Kotlin's STDOUT in ATO is always empty.
And TIO doesn't support countOneBits().

More information about how to count bytes for a Kotlin answer: https://codegolf.stackexchange.com/a/202985/112099

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Answer 29

MATL, 10 bytes

6Y2tB!si=)

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Explanation

6Y2        % All ASCII Chars
   t       % dup
    B      % convert each to binary
     !     % transpose
      s    % sum
       i=  % equal to the input?
         ) % Keep truthy elements

Answer 30

Wolfram Language (Mathematica), 70 bytes

FromCharacterCode/@Select[32~Range~126,s=#;Tr@IntegerDigits[#,2]==s&]&

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