Is this number evil?

Question

Introduction

In number theory, a number is considered evil if there are an even number of 1's in its binary representation. In today's challenge, you will be identifying whether or not a given number is evil.

Challenge

Your job is to write a full program or function which accepts a single, non-negative integer as input and outputs (or returns) whether or not that number is evil.

You may output any truthy value if the number is evil, and any falsy value if the number is not evil.
You may input and output in any acceptable format.
Standard loopholes are disallowed.
OEIS sequence A001969 is the sequence containing all evil numbers.
Here is a list of the first 10000 evil numbers, for reference (and more test cases!)
This question is code-golf, so the shorter, the better.
Don't be put off by extremely short answers in golfing languages. I encourage you to submit in any language you like.

Here are some test cases:

3 => True
11 => False
777 => True
43 => True
55 => False
666 => False

The Leaderboard

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# Language Name, N bytes

where N is the size of your submission. If you improve your score, you can keep old scores in the headline, by striking them through. For instance:

# Ruby, <s>104</s> <s>101</s> 96 bytes

If there you want to include multiple numbers in your header (e.g. because your score is the sum of two files or you want to list interpreter flag penalties separately), make sure that the actual score is the last number in the header:

# Perl, 43 + 2 (-p flag) = 45 bytes

You can also make the language name a link which will then show up in the leaderboard snippet:

# [><>](http://esolangs.org/wiki/Fish), 121 bytes

/* Configuration */

var QUESTION_ID = 169724; // Obtain this from the url
// It will be like https://XYZ.stackexchange.com/questions/QUESTION_ID/... on any question page
var ANSWER_FILTER = "!t)IWYnsLAZle2tQ3KqrVveCRJfxcRLe";
var COMMENT_FILTER = "!)Q2B_A2kjfAiU78X(md6BoYk";
var OVERRIDE_USER = 81420; // This should be the user ID of the challenge author.

/* App */

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getAnswers();

var SCORE_REG = /<h\d>\s*([^\n,<]*(?:<(?:[^\n>]*>[^\n<]*<\/[^\n>]*>)[^\n,<]*)*),.*?(\d+)(?=[^\n\d<>]*(?:<(?:s>[^\n<>]*<\/s>|[^\n<>]+>)[^\n\d<>]*)*<\/h\d>)/;

var OVERRIDE_REG = /^Override\s*header:\s*/i;

function getAuthorName(a) {
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}

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  answers.forEach(function(a) {
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        body = '<h1>' + c.body.replace(OVERRIDE_REG, '') + '</h1>';
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        user: getAuthorName(a),
        size: +match[2],
        language: match[1],
        link: a.share_link,
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    //else console.log(body);
  });

  valid.sort(function(a, b) {
    var aB = a.size,
      bB = b.size;
    return aB - bB
  });

  var languages = {};
  var place = 1;
  var lastSize = null;
  var lastPlace = 1;
  valid.forEach(function(a) {
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    lastSize = a.size;
    ++place;

    var answer = jQuery("#answer-template").html();
    answer = answer.replace("{{PLACE}}", lastPlace + ".")
      .replace("{{NAME}}", a.user)
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      .replace("{{LINK}}", a.link);
    answer = jQuery(answer);
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  langs.sort(function(a, b) {
    if (a.lang_raw.toLowerCase() > b.lang_raw.toLowerCase()) return 1;
    if (a.lang_raw.toLowerCase() < b.lang_raw.toLowerCase()) return -1;
    return 0;
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body {
  text-align: left !important;
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}

#answer-list {
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  float: left;
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#language-list {
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table thead {
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table td {
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<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
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EDIT: I believe this question is not a duplicate of this, because whereas that question is asking to count the number of ones, this question is asking whether the number of ones is even. Although you can accomplish this question by simply counting the bits, there are other approaches too.

Related (XOR-ing every binary digit is the same as taking the sum modulo-2). — Kevin Cruijssen, Aug 1, 2018 at 15:25
Possible duplicate of Count the number of ones in an unsigned 16-bit integer — Beta Decay, Aug 1, 2018 at 17:11
@BetaDecay but that doesn't work in reverse: i.e. you cannot take all of these answers and remove the mod 2. Therefore, this challenge invites some new methods. — Amphibological, Aug 1, 2018 at 20:29
Yes, 666 is not evil, but 616 is. More evidence corroborating Papyrus 115! — aschepler, Aug 7, 2018 at 11:41

Aiden4 · Accepted Answer · 2021-03-02 17:31:28Z

1

Rust, 21 bytes

|n|n.count_ones()%2<1

Try it online!

Nothing fancy going on here, just counting the ones of the input then checking if it is even.

answered Mar 2, 2021 at 17:31

Aiden4

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Aiden Chow · Accepted Answer · 2021-06-22 03:24:38Z

1

Desmos, 251 201 197 178 79 bytes

Wow... Coming back to this after months, and finding the most obvious golfs... lol

a=[0,...floor(\log_2N)]
f(N)=\{\mod(\total(\floor(\mod(N,2^a2)/2^a)),2)=0:1,0\}

Try It On Desmos!

Try It On Desmos! - Prettified

edited Jun 22, 2021 at 3:24

answered Oct 16, 2020 at 23:18

Aiden Chow

11.3k1 gold badge11 silver badges49 bronze badges

\$\begingroup\$ u dont need the comma before ... \$\endgroup\$
– naffetS
May 6, 2022 at 3:37
\$\begingroup\$ and u can call \total with dot notation \$\endgroup\$
– naffetS
May 6, 2022 at 3:39

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NoLongerBreathedIn · Accepted Answer · 2021-06-22 03:27:02Z

1

MMIX, 12 bytes (3 instrs)

00000000: db000000 7f000001 f8010000           ṃ¡¡¡¶¡¡¢ẏ¢¡¡

isevil  SADD $0,$0,0
        ZSEV $0,$0,1
        POP  1,0

answered Jun 22, 2021 at 3:27

NoLongerBreathedIn

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chunes · Accepted Answer · 2021-06-22 15:05:35Z

1

Factor, 19 bytes

[ bit-count even? ]

Try it online!

Should be self-explanatory, I hope.

answered Jun 22, 2021 at 15:05

chunes

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1

\$\begingroup\$ tfw not a native english speaker \$\endgroup\$
– Razetime
Jun 22, 2021 at 15:25

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Michael · Accepted Answer · 2021-07-12 14:00:57Z

Pyth, 10 bytes

!%/.BQ"1"2

Try it online!

Trying to learn Pyth, so this is my go at it. How it works:

Convert the input to binary, count the number of "1"s, check whether they're even, invert the output (so it checks whether they're odd. In more detail:

!                 # negate output (0 -> 1, 1 -> 0)
    %             # check whether the following block returns an even number
        /         # count string occurences in...
            .B    # ...the binary string version of...
                Q # ...the input
            "1"   # the string 1; the string to be count the occurences of
    2             # the 2 used by the modulo parity check

Seggan · Accepted Answer · 2022-10-03 16:07:44Z

1

Fig, \$3\log_{256}(96)\approx\$ 2.469 bytes

OSb

Try it online!

Outputs using inverted truthiness (0 -> true, 1 -> false) because of a bug with E.

OSb
  b # To binary
 S  # Sum
O   # Is odd?

answered Oct 3, 2022 at 16:07

Seggan

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Conor O'Brien · Accepted Answer · 2018-08-01 21:19:26Z

0

Stacked, 15 bytes

[bits sum 2%0=]

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Checks whether or not the sum of the bits of the input mod 2 (2%) is 0 (0=).

answered Aug 1, 2018 at 21:19

Conor O'Brien

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Charlie · Accepted Answer · 2018-08-02 07:54:26Z

0

C# (.NET Core), 58 bytes

n=>System.Convert.ToString(n,2).Replace("0","").Length%2<1

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I could do something similar to the Java answer with

n=>System.Convert.ToString(n,2).Sum(c=>c)%2<1

That's 45 bytes but then I would need to add another 18 for using System.Linq;. So I needed to find another approach.

answered Aug 2, 2018 at 7:54

Charlie

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Galen Ivanov · Accepted Answer · 2018-08-02 08:18:07Z

0

Red, 57 bytes

func[n][s: on until[if n % 2 = 1[s: not s]1 > n: n / 2]s]

Try it online!

answered Aug 2, 2018 at 8:18

Galen Ivanov

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ბიმო · Accepted Answer · 2018-08-02 16:51:11Z

dc, 27 bytes

0sb[2~lb+sbd1<a]dsaxlb1++2%

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Instead of accumulating the bits in register b, we can also leave them on the stack (replace lb+sb by r) which essentially converts the input to binary, push 1, sum the stack and do the mod 2. But it's the same byte-count:

[2~rd1<a]dsax1[+z1<a]dsax2%

Try it online!

Explanation

Register b = 0, divmod on ToS, add remainder to b and loop until ToS is less or equal to 1, then compute ToS + b + 1 mod 2:

0sb[2~lb+sbd1<a]dsaxlb1++2%  # example input 5                     |  stack
0sb                          # initialize register b with 0        |  5
   [2~lb+sbd1<a]             # push the string [..]                |  [2~lb+sbd1<a]
                dsa          # duplicate & store it to register a  |  [2~lb+sbd1<a]
                   x         # execute the string*                 |  1
                    lb       # push register b                     |  1 1
                      1+     # increment by 1                      |  2 1
                        +    # add                                 |  3
                         2%  # modulo 2                            |  1

# Recursively called string:
2~lb+sbd1<a  # recursively called string   |  5
2~           # divmod 2                    |  1 2
  lb         # push register b             |  0 1 2
    +        # add                         |  1 2
     sb      # store register b            |  2
       d     # duplicate                   |  2 2
        1<   # if top (popped) is > 1:
          a  # | execute register a        |  2

user3490 · Accepted Answer · 2018-08-03 13:22:58Z

0

SAS, 35 bytes

e=1-mod(count(put(i,binary.),1),2);

This expression takes a variable i as input and populates the result in a new variable, e. It relies on implicit conversion from character to numeric when specifying which digit to count.

edited Aug 3, 2018 at 13:22

answered Aug 2, 2018 at 11:00

user3490

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Community · Accepted Answer · 2020-06-17 09:04:33Z

0

Python 3, 69 53 39 bytes

print(bin(int(input())).count('1')%2<1)

Works by using the built in python binary to converter, bin(), on the input, then counts the number of ones in there, checks if it is equal to 0 mod 2, and prints the result (outputs True/False).

Probably needs some more golfing, but this was my first whirl.

Try it online!

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Aug 2, 2018 at 19:19

Auden Young

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\$\begingroup\$ You can't assume inputs preassigned to a variable, though you can use a lambda: lambda x:1if len([i for i in bin(x)[2:]if i=='0'])%2==0else 0 which is shorter anyways (I included some basic golfing). \$\endgroup\$
– ბიმო
Aug 2, 2018 at 19:25
\$\begingroup\$ Some more golfing on your approach (using that b has a higher ascii-codepoint than '0') leaves you with: lambda x:len([1for i in bin(x)if'0'<i])%2 \$\endgroup\$
– ბიმო
Aug 2, 2018 at 19:31
\$\begingroup\$ @OMᗺ you should probably post that as your own answer, as it's such a better approach than mine. I'd feel bad taking credit for it =) \$\endgroup\$
– Auden Young
Aug 2, 2018 at 19:56
\$\begingroup\$ I won't - there are already shorter ones, but your submission still uses a preassigned variable as input which is not allowed. \$\endgroup\$
– ბიმო
Aug 2, 2018 at 20:17
\$\begingroup\$ @OMᗺ fixed by switching to input() \$\endgroup\$
– Auden Young
Aug 2, 2018 at 20:24

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Community · Accepted Answer · 2020-06-17 09:04:33Z

0

Math++, 40 bytes

?>a
3+3*!a>$
b+a%2>b
_(a/2)>a
2>$
!(b%2)

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Aug 3, 2018 at 19:10

SuperJedi224

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Joseph · Accepted Answer · 2018-08-05 03:50:55Z

0

Python 2, 46 bytes

lambda n:sum(int(d)for d in str(bin(n))[2:])%2

Outputs 1 for True or 0 for false

answered Aug 5, 2018 at 3:50

Joseph

9356 silver badges18 bronze badges

\$\begingroup\$ You don't need to stringify the binary \$\endgroup\$
– Jo King ♦
Aug 7, 2018 at 23:03

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Zoltán · Accepted Answer · 2018-08-06 08:10:15Z

0

Scala, 39 bytes

readInt.toBinaryString.count('1'==)%2<1

Try it online

This can be run in a Scala console, after which the user has to input the integer and press enter.

edited Aug 6, 2018 at 8:10

answered Aug 3, 2018 at 13:51

Zoltán

1011 bronze badge

1

\$\begingroup\$ Welcome to PPCG! \$\endgroup\$
– Luis felipe De jesus Munoz
Aug 3, 2018 at 13:56
1

\$\begingroup\$ This submission seems to be a snippet which is disallowed. Please change your submission to either a full program or a function definition. \$\endgroup\$
– Jonathan Frech
Aug 3, 2018 at 14:04
\$\begingroup\$ Also, could you please make the title of your answer into a header (i.e. # Scala, 34 bytes) to fix the scoreboard? \$\endgroup\$
– Amphibological
Aug 4, 2018 at 15:43
\$\begingroup\$ @JonathanFrech I updated my answer, it now reads the input from user input, and is executable from the Scala console, or as a Scala script \$\endgroup\$
– Zoltán
Aug 6, 2018 at 7:56

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aaaaa says reinstate Monica · Accepted Answer · 2018-08-08 17:52:20Z

0

MATLAB 31 bytes

Anonymous function, use as f=@(n)...;f(3)

@(n)1-mod(sum(dec2bin(n)-48),2)

answered Aug 8, 2018 at 17:52

aaaaa says reinstate Monica

3811 silver badge8 bronze badges

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Alion · Accepted Answer · 2018-08-11 13:05:23Z

0

AsciiDots, 86 64 bytes

/{&}\
|(*)?#-. 
*{>}:@1({+}*@:-{%}$#&
*{+}\#1/.^-/.-#2/
\-*.<#1)

Returns 1 if the number is evil, otherwise 0. Try it online!

edited Aug 11, 2018 at 13:05

answered Aug 10, 2018 at 19:38

Alion

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AlexRacer · Accepted Answer · 2018-08-11 15:21:04Z

0

Pascal (FPC), 84 bytes

var n,s:word;begin read(n);repeat s:=n mod 2xor s;n:=n div 2until n=0;write(s=0)end.

Try it online!

No conversion to binary :(

answered Aug 11, 2018 at 15:21

AlexRacer

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Community · Accepted Answer · 2020-06-17 09:04:33Z

0

Scala, 43 41 bytes

def e(n:Int):Int=if(n>0)n%2^e(n/2)else 1

Test: https://scalafiddle.io/sf/VlU7nun/2

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Aug 4, 2018 at 11:39

Kjetil S

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\$\begingroup\$ Could you please add a comma after Scala in the title of your answer so it doesn't break the scoreboard? \$\endgroup\$
– Amphibological
Aug 4, 2018 at 15:38

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u-ndefined · Accepted Answer · 2018-10-21 03:14:27Z

0

CJam, 15 10 bytes

qi2b1e=2%!

Try it online!

Explanation
qi2b1e=2%! %whole code
qi         %Read input as number | Example stack: 10
  2b       %Convert to binary    | Example stack: 1010
    1e=    %Count ones           | Example stack: 22
       2%  %Mod 2                | Example stack: 0
         ! %Invert               | Example stack: 1

edited Oct 21, 2018 at 3:14

u-ndefined

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answered Aug 9, 2018 at 10:02

lolad

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Gegell · Accepted Answer · 2018-10-21 13:00:48Z

Gol><>, 14 bytes

IW2SD@+$|~2%zh

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Explanation:

IW2SD@+$|~2%zh  //Program for reference

I               //Input a integer
 W              //Loop until the number given is 0 (due to the int div 2 inside the loop)
  2SD           //  Push the div and mod of the remaining number by 2
     @          //  Rotate the top three elements on the stack (stack now is: curNumber/2, lastBitSet, bitCount)
      +         //  Add the bit count and the last bit flag
       $|       //  Swap the bitCount with the curNumer/2 and repeat until curNumber/2 == 0
         ~      //Remove the loop exit flag (0) 
          2%zh  //Check wether the bitCount is even, output & exit: 1 == truthy, 0 == falsy

Superloach · Accepted Answer · 2020-02-21 00:48:24Z

0

chevron, 45 bytes

method borrowed from Lynn's python answer.

>:>^n
^n,10,2~x>>^n
b^n,13~x>>^n
^n%2>>^n
>^n

notes:

outputs :0 for falsy and :1 for truthy.
online interpreter defaults to printing the input at prompts, uncheck "print inp" to disable.
replace : with ^__ for no prompt.

answered Feb 21, 2020 at 0:48

Superloach

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Pseudo Nym · Accepted Answer · 2020-02-21 15:04:32Z

0

GolfScript, 14 bytes

2base{^}*!

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Explanation:

2base         convert input to binary
     {^}*     xor all bits
         !    flip answer

edited Feb 21, 2020 at 15:04

answered Feb 20, 2020 at 21:18

Pseudo Nym

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\$\begingroup\$ This appears to be only 10 bytes \$\endgroup\$
– Jo King ♦
Jul 31, 2022 at 1:40

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2014MELO03 · Accepted Answer · 2020-10-17 02:39:34Z

0

GolfScript, 11 bytes

~2base{^}*!

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~2base        # Convert to binary
      {^}*    # Xor all the numbers in the array
          !   # Negate the output

answered Oct 17, 2020 at 2:39

2014MELO03

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Zsolt Szilagy · Accepted Answer · 2020-10-17 22:47:20Z

0

PHP 7.4, 65 bytes

Input is given in $n

echo array_reduce(str_split(decbin($n)),fn($c,$i) => $i xor$c,1);

or prior to PHP 7.4: 76 bytes

echo array_reduce(str_split(decbin($n)),function($c,$i){return$i xor$c;},1);

Try it online

Ungolfed

echo array_reduce(
    str_split(decbin($n)),
    function ($c,$i) {
        return $i xor $c;
    },
1);

edited Oct 17, 2020 at 22:47

answered Oct 17, 2020 at 22:41

Zsolt Szilagy

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Wasif · Accepted Answer · 2021-03-02 09:51:33Z

0

PowerShell, 70 bytes

if(([convert]::tostring($args[0],2)-replace"0","").length%2){0}else{1}

Try it online!

Sorry don't know if any powershell solution is posted, if so, I am ready to delete my submission.

answered Mar 2, 2021 at 9:51

Wasif

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emanresu A · Accepted Answer · 2021-07-12 21:24:28Z

0

Vyxal, 10 3 bytes

b∑₂

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b∑  # Sum of binary digits
  ₂ # Is even?

edited Jul 12, 2021 at 21:24

answered Jun 4, 2021 at 7:38

emanresu A

32.8k5 gold badges77 silver badges186 bronze badges

\$\begingroup\$ 3 bytes \$\endgroup\$
– Aaroneous Miller
Jul 12, 2021 at 19:38

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Alan Bagel · Accepted Answer · 2022-01-13 20:34:33Z

0

Python 3, 34 bytes

lambda n:not f"{n:b}".count("1")&1

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answered Jan 13, 2022 at 20:34

Alan Bagel

3,1991 gold badge6 silver badges36 bronze badges

\$\begingroup\$ You can do lambda x:~x.bit_count()%2 \$\endgroup\$
– TKirishima
May 12, 2022 at 22:20

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Bbrk24 · Accepted Answer · 2022-03-03 00:41:03Z

Swift, 24 bytes

{$0.nonzeroBitCount%2<1}

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Type information not included, since the argument can be any FixedWidthInteger. For example, all of the following are valid:

let f: (Int) -> Bool = {$0.nonzeroBitCount%2<1}
let g: (UInt64) -> Bool = {$0.nonzeroBitCount%2<1}
let h: (CChar) -> Bool = {$0.nonzeroBitCount%2<1}

Unlike many other C-family languages, integers cannot implicitly be converted into booleans, and characters cannot be implicitly converted to numbers, so the approaches taken by the top C++ and C# answers are too verbose.

Oliver · Accepted Answer · 2022-03-10 18:43:51Z

0

Python, 34 bytes

lambda x:not bin(x).count('1')%2

Converts input to bin, then counts 1s, mod 2 (invert because of even=1)

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Try it online (OLD) (Python 3.8.0b4)

edited Mar 10, 2022 at 18:43

answered Feb 5, 2022 at 17:34

Oliver

1556 bronze badges

\$\begingroup\$ This looks like a snippet which just evaluates to the answer, which isn't one of our allowed output methods. You'd need to include print() around the expression in your answer, or convert your answer to a function, like lambda x:not bin(x).count('1')%2 \$\endgroup\$
– pxeger
Mar 8, 2022 at 7:50
\$\begingroup\$ You can do lambda x:~x.bit_count()%2 \$\endgroup\$
– TKirishima
May 12, 2022 at 22:20

Add a comment |

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