How even is a number?

Question

The ancient Greeks had these things called singly and doubly even numbers. An example of a singly even number is 14. It can be divided by 2 once, and has at that point become an odd number (7), after which it is not divisible by 2 anymore. A doubly even number is 20. It can be divided by 2 twice, and then becomes 5.

Your task is to write a function or program that takes an integer as input, and outputs the number of times it is divisible by 2 as an integer, in as few bytes as possible. The input will be a nonzero integer (any positive or negative value, within the limits of your language).

Test cases:

14 -> 1

20 -> 2

94208 -> 12

7 -> 0

-4 -> 2

The answer with the least bytes wins.

Tip: Try converting the number to base 2. See what that tells you.

Answer 1

Vyxal, 2 bytes

2Ǒ

Try it Online!

This one by @lyxal

2Ǒ # I'm running out of things to say for this comment line
 Ǒ # How many times is the input divisible by...
2  # 2

Vyxal, 4 bytes

Eġ∆l

Try it Online!

Ported from the Desmos answer.

Eġ∆l # 4 bytes!
E    # 2^input
 ġ   # GCD of that and the input
  ∆l # log2

Vyxal, 5 bytes

bṘȧ1ḟ

Try it Online!

Strategy from the osabie answer. Suprised that there has been no Vyxal answer for a such a popular question. Bit twiddling also gives 5 bytes.

bṘȧ1ḟ # This comment line needs some love
b     # Convert to binary
 Ṙ    # Reverse
  ȧ   # Absolute value of each element in the list. This is to handle negative numbers correctly
   1ḟ # First index of 1

Answer 2

CJam, 8 bytes

rizmf2e=

Read integer, absolute value, prime factorize, count twos.

Answer 3

JavaScript ES6, 36 38 bytes

Golfed two bytes thanks to @ETHproductions

Fairly boring answer, but it does the job. May actually be too similar to another answer, if he adds the suggested changes then I will remove mine.

b=>{for(c=0;b%2-1;c++)b/=2;alert(c)}

To run, assign it to a variable (a=>{for...) as it's an anonymous function, then call it with a(100).

Answer 4

PowerShell, 36 bytes

param($a)for(;!($a%2)){$a/=2;$o++}$o

Takes input $a, then enters a for() loop. There is no setup, but the conditional means the loop ends when $a is no longer even. Inside the loop, we just divide $a by 2 and increment a counter, then output the counter.

The above correctly accounts for negative numbers (in PowerShell, the % operator follows the sign of the dividend, but any non-zero number is truthy, the ! of which is falsey).

Answer 5

DUP, 20 bytes

[$2/%0=[2/f;!1+.][0]?]f:

Try it here!

Converted to recursion, output is now the top number on stack. Usage:

94208[2/\0=[f;!1+][0]?]f:f;!

Explanation

[                ]f: {save lambda to f}
 2/\0=               {top of stack /2, check if remainder is 0}
      [     ][ ]?    {conditional}
       f;!1+         {if so, then do f(top of stack)+1}
              0      {otherwise, push 0}

Answer 6

Japt, 9 5 bytes

¢w b1

Test it online!

The previous version should have been five bytes, but this one actually works.

How it works

       // Implicit: U = input integer
¢      // Take the binary representation of U.
w      // Reverse.
b1     // Find the first index of a "1" in this string.
       // Implicit output

Answer 7

C, 44 40 38 36 bytes

2 bytes off thanks @JohnWHSmith. 2 bytes off thanks @luserdroog.

a;f(n){for(;~n&1;n/=2)a++;return a;}

Test live on ideone.

Answer 8

Mathematica, 20 bytes

#~IntegerExponent~2&

Yet another long, un-golfable built-in...

Answer 9

R, 30 bytes

sum(gmp::factorize(scan())==2)

Assumes gmp package installed

Answer 10

Oracle SQL 11.2, 111 bytes

WITH v(i)AS(SELECT 1 FROM DUAL UNION ALL SELECT i+1 FROM v WHERE MOD(:1/POWER(2,i),1)=0)SELECT MAX(i)-1 FROM v;

Un-golfed

WITH v(i) AS 
(
  SELECT 1 FROM DUAL 
  UNION ALL 
  SELECT i+1 FROM v WHERE MOD(:1/POWER(2,i),1)=0
)
SELECT MAX(i)-1 FROM v;

Answer 11

Javascript ES6, 39 chars

n=>n.toString(2).match(/0*$/)[0].length

Test:

[14,20,94208,7,-4].map(n=>n.toString(2).match(/0*$/)[0].length) == "1,2,12,0,2"

Answer 12

PHP, 36 28 bytes

Used a different approach than most others. I'm checking divisibility by 2^N where I'm increasing N until it's no longer divisible by it.

for(;0==$argv[1]%2**++$b;);echo$b-1;

Run like this (-d added for aesthetics only):

php -d error_reporting=32757 -r 'for(;0==$argv[1]%2**++$b;);echo$b-1; echo"\n";' -- -65536

Implementing orlp's log algorithm would be even shorter. I don't like the requirement to create a file for PHP golfs, but this would be the shortest:

<?=log(($x=$argv[1])&-$x,2);

Edit: I found out you can actually run that without creating a file, by piping it like this:

echo '<?=log(($x=$argv[1])&-$x,2);' | php -- -65536

Answer 13

𝔼𝕊𝕄𝕚𝕟, 8 chars / 10 bytes

ïⓑᴙą1

Try it here (Firefox only).

Explanation

Converts input to binary, reverses it, then gets index of first 1.

Answer 14

Python, 48 chars

print len(str(bin(int(input()))).split("1")[-1])

Simply counts the number of 0s at the end of the binary number

Answer 15

R, 56 46 40 bytes

x=scan();a=0;while(!x%%2){x=x/2;a=a+1};a

Another answer than @mnel's one without the gmp package.

Thanks to @user5957401 for saving 10 bytes

Thanks to @Frédéric for saving 6 bytes

Answer 16

Excel, 20 bytes

Works up to 2^53 (9,007,199,254,740,990)

=LOG(GCD(A1,2^53),2)

Using Binarys, a 36 byte solution that only works up to 511:

=10-FIND(2,DEC2BIN(A2)+DEC2BIN(-A2))

Answer 17

Desmos, 22 bytes

f(n)=log_2(gcd(n,2^n))

Try It On Desmos!

Doesn't work for 94208 because it's too large for the program. Below is one that supports much more numbers:

43 bytes

f(n)=log_2(gcd(n,2^{floor(log_2(abs(n)))}))

Try It On Desmos!

Answer 18

K (ngn/k), 9 8 5 7 bytes

*&|~~2\

Try it online!

Answer 19

Prolog (SWI), 39 33 32 bytes

A-B:-A/\1<1,A//2-D,B is D+1;B=0.

Try it online!

-6 bytes thanks to Jo King!

-1 byte thanks to Steffan!

Answer 20

Seriously, 9 bytes

,wii2=*.

Contains an unprintable (0x7F) at the end. Hexdump:

2c77 6969 323d 2a2e 7f

Try it online!

Explanation:

,wii2=*.<0x7F>
,w              get prime factorization of input (list of base, exp pairs)
  ii            flatten first (base, exp) pair so that base, exp is top of stack
    2=*         multiply exponent by 1 if base is 2 else 0
       .<0x7F>  print top item and exit

Answer 21

Javascript, 45 39 38 bytes

1 byte off thanks @manatwork.

i=>/0*$/.exec(i.toString(2))[0].length

f=
i=>/0*$/.exec(i.toString(2))[0].length

F=i=>document.body.innerHTML+='<pre>f('+i+') -> '+f(i)+'\n</pre>'

F(14)
F(20)
F(94208)

Answer 22

jq, 26 characters

[while(.%2==0;./2)]|length

Sample run:

bash-4.3$ jq '[while(.%2==0;./2)]|length' <<< 94208
12

bash-4.3$ jq '[while(.%2==0;./2)]|length' <<< -4
2

On-line test:

• 94208
• -4

Answer 23

Perl 6  28  27 bytes

{($_+&-$_).polymod(2 xx*)-1}
{($_+&-$_).base(2).chars-1}

Usage:

my &code = {($_+&-$_).base(2).chars-1}

say code    14; # 1
say code    20; # 2
say code 94208; # 12
say code     7; # 0
say code    -4; # 2

Answer 24

Java, 44 39 bytes

int f(int n){return n%2==0?1+f(n/2):0;}

Works for odd, zero, and negative numbers.

Golfed 5 bytes because input will not be zero.

Answer 25

PHP, 40 bytes

function e($i){return $i%2?0:e($i/2)+1;}

Answer 26

POSIX shell and GNU/BSD utilities, 43 30 bytes

factor ${1#-}|rs -T|grep -xc 2

We simply count the number of 2s in the output of the factor command.

Answer 27

Groovy, 83 bytes

There was not a groovy answer yet, so here goes. Definitely room for improvement.

int n=args[0].toInteger();def e(int n){x=0;while(n%2==0){n/=2;x++;};print x;};e(n);

You can use it with: groovy filename.groovy "94208"

Answer 28

Pure Bash, 40

If 0 could not be submited as input... Thanks to @TobySpeight for help me to drop a lot.

for((o=0;1<<o&~i;++o));do :;done;echo $o

Proof

pureBashStr='for((o=0;1<<o&~i;o++));do :;done;echo $o'
echo ${#pureBashStr}
40

for i in 14 20 64#w0000 94208 7 -4 ;do
    printf " %8s: %4d\n" $i $(
        eval $pureBashStr)
  done
       14:    1
       20:    2
 64#w0000:   29
    94208:   12
        7:    0
       -4:    2

+10 to support 0 case: 50

pureBashStr='for((o=0;1<<o&~i;o++));do((i))||break;done;echo $o'
i=0
printf " %8s: %4d\n" $i $(eval $pureBashStr)
        0:    0

Answer 29

Python 2, 27 bytes

e=lambda n:~n%2and e(n/2)+1

In Python 3, you'd have to use e(n//2), since ~ operator doesn't work with floats.

Answer 30

SmileBASIC, 45 bytes

INPUT N@L
IF!(N<<31)THEN N=N>>1Q=Q+!GOTO@L
?Q

I'm pretty sure N<<31 is the shortest way to check the lowest bit in SB, since ​ MOD ​ and ​ AND ​ are so long.