Check whether an integer is a power of 2 without using +,- operations [closed]

Question

Write a program that checks if the integer is a power of 2.

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Sample input:

8

Sample output:

Yes

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Sample input:

10

Sample output:

No

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Rules:

• Don't use +,- operations.

• Use some sort of input stream to get the number. Input is not supposed to be initially stored in a variable.

• The shortest code (in bytes) wins.

You can use any truthy/falsy response (for instance, true/false). You may assume that input number is greater than 0.

Answer 1

GolfScript, 6 chars, no decrements

~.3/&!

Here's a solution that doesn't use the x & (x-1) method in any form. It uses x & (x/3) instead. ;-) Outputs 0 if false, 1 if true.

Explanation:

• ~ evals the input string to turn it into a number,
• . duplicates it (for the subsequent &),
• 3/ divides it by three (truncating down),
• & computes the bitwise AND of the divided value with the original, which will be zero if and only if the input is zero or a power of two (i.e. has at most one bit set), and
• ! logically negates this, mapping zero to one and all other values to zero.

Notes:

• Per the clarified rules, zero is not a valid input, so this code is OK, even though it outputs 1 if the input is zero.

• If the GolfScript decrement operator ( is allowed, then the 5-character solution ~.(&! posted by aditsu is enough. However, it seems to go against the spirit of the rules, if not the letter.

• I came up with the x & (x/3) trick years ago on the Fun With Perl mailing list. (I'm sure I'm not the first to discover it, but I did (re)invent it independently.) Here's a link to the original post, including a proof that it actually works.

Answer 2

APL (7)

Yes, that's 7 bytes. Assume for the moment that I'm using IBM codepage 907 instead of Unicode and then each character is a byte :)

0=1|2⍟⎕

i.e. 0 = mod(log(input(),2),1)

Answer 3

GolfScript, 11 (for 1 (true) and 0 (false))

.,{2\?}%?0>

Put the number on the stack and then run.

GolfScript, 22 (for Yes/No)

.,{2\?}%?0>'Yes''No'if

I love how converting 1/0 to Yes/No takes as much code as the challenge itself :D

Warning: EXTREMELY inefficient ;) It does work fine for numbers up to 10000, but once you get that high you start to notice slight lag.

Explanation:

• .,: turns n into n 0..n (. duplicate, , 0..n range)
• {2\?}: to the power of 2
• %: map "power of 2" over "0..n" so it becomes n [1 2 4 8 16 ...]
• ?0>: checks to see if the array contains the number (0 is greater than index)

Answer 4

Mathematica 28

Numerator[Input[]~Log~2]==1

For integer powers of 2, the numerator of the base 2 log will be 1 (meaning that the log is a unit fraction).

Here we modify the function slightly to display the presumed input. We use # in place of Input[] and add & to define a pure function. It returns the same answer that would be returned if the user input the number in the above function.

    Numerator[#~Log~2] == 1 &[1024]
    Numerator[#~Log~2] == 1 &[17]

True
False

Testing several numbers at a time.

    Numerator[#~Log~2] == 1 &&/@{64,8,7,0}

{True, True, False, False}

Answer 5

Perl 6 (17 characters)

say get.log(2)%%1

This program gets a line from STDIN get function, calculates logarithm with base 2 on it (log(2)), and checks if the result divides by 1 (%%1, where %% is divides by operator). Not as short as GolfScript solution, but I find this acceptable (GolfScript wins everything anyway), but way faster (even considering that Perl 6 is slow right now).

~ $ perl6 -e 'say get.log(2)%%1'
256
True
~ $ perl6 -e 'say get.log(2)%%1'
255
False

Answer 6

Octave (15 23)

EDIT: Updated due to user input requirement;

~mod(log2(input('')),1)

Lets the user input a value and outputs 1 for true, 0 for false.

Tested in Octave, should work in Matlab also.

Answer 7

R, 13 11

Based on the Perl solution. Returns FALSE or TRUE.

!log2(i)%%1

The parameter i represents the input variable.

An alternative version with user input:

!log2(scan())%%1

Answer 8

GolfScript, 5

Outputs 1 for true, 0 for false. Based on user3142747's idea:

~.(&!

Note: ( is decrement, hopefully it doesn't count as - :)
If it does (and the OP's comments suggest that it might), then please refer to Ilmari Karonen's solution instead.
For Y/N output, append 'NY'1/= at the end (7 more bytes).

Answer 9

Python, 31

print 3>bin(input()).rfind('1')

Answer 10

C, 48

main(x){scanf("%i",&x);puts(x&~(x*~0)?"F":"T");}

Answer 11

I decided to to use another approach, based on the population count or sideways sum of the number (the number of 1-bits). The idea is that all powers of two have exactly one 1 bit, and no other number does. I added a JavaScript version because I found it amusing, though it certainly won't win any golfing competition.

J, 14 15 chars (outputs 0 or 1)

1=##~#:".1!:1]1

JavaScript, 76 chars (outputs true or false)

alert((~~prompt()).toString(2).split("").map(Number).filter(Boolean).length)

Answer 12

Clip, 9 8 7

!%lnxWO

Reads a number from stdin.

Explanation:

To start with, Z = 0, W = 2 and O = 1. This allows placing of W and O next to each other, whereas using 2 and 1 would be interpreted as the number 21 without a separating space (an unwanted extra character). In Clip, the modulo function (%) works on non-integers, so, to work out if some value v is an integer, you check if v mod 1 = 0. Using Clip syntax, this is written as =0%v1. However, as booleans are stored as 1 (or anything else) and 0, checking if something is equal to 0 is just 'not'ing it. For this, Clip has the ! operator. In my code, v is lnx2. x is an input from stdin, n converts a string to a number and lab is log base b of a. The program therefore translates (more readably) to 0 = ((log base 2 of parseInt(readLine)) mod 1).

Examples:

8

outputs

1

and

10

outputs

0

Edit 1: replaced 0, 1 and 2 with Z, O and W.

Edit 2: replaced =Z with !.

Also:

Pyth, 5

Compresses the Clip version even further, as Pyth has Q for already evaluated input and a log2(a) function instead of just general log(a, b).

!%lQ1

Answer 13

Javascript (37)

a=prompt();while(a>1)a/=2;alert(a==1)

Simple script that just divides by 2 repeatedly and checks the remainder.

Answer 14

Mathematica (21)

IntegerQ@Log2@Input[]

Without input it is a bit shorter

IntegerQ@Log2[8]

True

IntegerQ@Log2[7]

False

Answer 15

Ruby — 17 characters (fourth try)

p /.1/!~'%b'%gets

My current best is a fusion of @steenslag's answer with my own. Below are my previous attempts.

Ruby — 19 characters (third try)

p /10*1/!~'%b'%gets

Ruby — 22 characters (second try)

p !('%b'%gets)[/10*1/]

Ruby — 24 characters (first try)

p !!('%b'%gets=~/^10*$/)

Answer 16

JavaScript, 41 40 characters

l=Math.log;alert(l(prompt())/l(2)%1==0);

How this works: you take the logarithm at base 2 using l(prompt()) / l(2), and if that result modulo 1 is equal to zero, then it is a power of 2.

For example: after taking the logarithm of 8 on base 2, you get 3. 3 modulo 1 is equal to 0, so this returns true.

After taking the logarithm of 7 on base 2, you get 2.807354922057604. 2.807354922057604 modulo 1 is equal to 0.807354922057604, so this returns false.

Answer 17

JavaScript, 35

Works for bytes.

alert((+prompt()).toString(2)%9==1)

46 character version, Works for 16 bit numbers.

x=(+prompt()).toString(2)%99;alert(x==1|x==10)

The trick works in most dynamic languages.

Explanation: Convert the number to base 2, interpret that string as base 10, do modulo 9 to get the digit sum, which must be 1.

Answer 18

Perl 5.10+, 13 + 1 = 14 chars

say!($_&$_/3)

Uses the same method from an old FWP thread as my GolfScript entry. Prints 1 if the input is a power of two, and an empty line otherwise.

Needs to be run with perl -nE; the n costs one extra char, for a total of 14 chars. Alternatively, here's an 18-character version that doesn't need the n:

say!(($_=<>)&$_/3)

Answer 19

python 3, 38

print(1==bin(int(input())).count('1'))

python, 32

However, the code doesn't work in every version.

print 1==bin(input()).count('1')

Notice that the solution works also for 0 (print False).

Answer 20

K/Kona (24 17)

d:{(+/(2_vs x))~1

Returns 1 if true and 0 if false. Any power of 2 has a single bit equal to 1:

2_vs'_(2^'(!10))
(,1
1 0
1 0 0
1 0 0 0
1 0 0 0 0
1 0 0 0 0 0
1 0 0 0 0 0 0
1 0 0 0 0 0 0 0
1 0 0 0 0 0 0 0 0
1 0 0 0 0 0 0 0 0 0)

(this prints out all the powers of 2 (from 0 to 9) in binary)

So I sum up all the components of the binary expression of x and see if it's equal to 1; if yes then x=2^n, otherwise nope.

...knew I could make it smaller

Answer 21

C# (54 characters)

 Math.Log(int.Parse(Console.ReadLine()),2)%1==0?"Y":"N"

Answer 22

Rebmu (9 chars)

z?MOl2A 1

Test

>> rebmu/args [z?MOl2A 1] 7
== false

>> rebmu/args [z?MOl2A 1] 8 
== true

>> rebmu/args [z?MOl2A 1] 9 
== false

Rebmu is a constricted dialect of Rebol. The code is essentially:

z? mo l2 a 1  ; zero? mod log-2 input 1

Alternative

14 chars—Rebmu does not have a 'mushed' bitwise AND~

z?AND~aTIddA 3

In Rebol:

zero? a & to-integer a / 3

Answer 23

OCaml - 42 or 45

Since no-one has submitted an OCaml solution, I submit one inspired by nightcrackers C solution.

There is some debate as to whether the operator - includes unary negation. Other entries have managed to sidestep the issue by finding a solution that is equally short without unary negation. I instead give two answers:

let x=read_int()in 0=x land lnot(x* -1);;

and

let x=read_int()in 0=x land lnot(x*lnot 0);;

the ;; ends a group of statements in the OCaml interpreter and causes the result to be printed as - : bool = false or - : bool = true, which are clear negative/positive answers as required by the question.

Answer 24

GTB, 46 bytes

2→P`N[@N=Pg;1P^2→P@P>1E90g;2]l;2~"NO"&l;1"YES"

Answer 25

Python, 35

print bin(input()).strip('0')=='b1'

Doesn't use not only +/- operations, but any math operations aside from converting to binary form.

Other stuff (interesting, but not for competition):

I have also a regexp version (61):

import re;print re.match(r'^0b10+$',bin(input())) is not None

(Love the idea, but import and match function make it too long)

And nice, but boring bitwise operations version (31):

x=input();print x and not x&~-x

(yes, it's shorter, but it uses ~-x for decrement which comtains - operation)

Answer 26

Python 2.7 (30 29 39 37)

EDIT: Updated due to user input requirement;

a=input()
while a>1:a/=2.
print a==1

Brute force, try to divide until =1 (success) or <1 (fail)

Answer 27

Python (33)

print int(bin(input())[3:]or 0)<1

Answer 28

Ruby, 33, 28, 25

p /.1/!~gets.to_i.to_s(2)

Answer 29

APL (12 for 0/1, 27 for yes/no)

≠/A=2*0,ιA←⍞ 

or, if we must output text:

3↑(3x≠/A=2*0,ιA←⍞)↓'YESNO '

Read in A. Form a vector 0..A, then a vector 2⁰..2^A (yes, that's way more than necessary), then a vector comparing A with each of those (resulting in a vector of 0's and at most one 1), then xor that (there's no xor operator in APL, but ≠ applied to booleans will act as one.) We now have 0 or 1.

To get YES or NO: multiply the 0 or 1 by 3, drop this number of characters from 'YESNO ', then take the first 3 characters of this.

Answer 30

C, 65 bytes

main(k){scanf("%i",&k);while(k&&!(k%2))k/=2;puts(k==1?"T":"F");}