41
\$\begingroup\$

There are clever ways of determining whether a number is a power of 2. That's no longer an interesting problem, so let's determine whether a given integer is an integer power of -2. For example:

-2 => yes: (-2)¹
-1 => no
0 => no
1 => yes: (-2)⁰
2 => no
3 => no
4 => yes: (-2)²

Rules

  • You may write a program or a function and use any of the standard methods of receiving input and providing output.

  • Your input is a single integer, and output must be a truthy value if the integer is an integer power of -2, and a falsy value otherwise. No other output (e.g. warning messages) is permitted.

  • The usual integer overflow rules apply: your solution must be able to work for arbitrarily large integers in a hypothetical (or perhaps real) version of your language in which all integers are unbounded by default, but if your program fails in practice due to the implementation not supporting integers that large, that doesn't invalidate the solution.

  • You may use any programming language, but note that these loopholes are forbidden by default.

Winning condition

This is a contest: the answer which has the fewest bytes (in your chosen encoding) is the winner.

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  • 17
    \$\begingroup\$ @KritixiLithos I don't see why it should. There is no integer i such that (-2)^i = 2 \$\endgroup\$ – Fatalize Apr 6 '17 at 12:10
  • 2
    \$\begingroup\$ Are the exponents positive or -0.5 should be valid since it's 2^(-1). \$\endgroup\$ – Mr. Xcoder Apr 6 '17 at 12:13
  • 1
    \$\begingroup\$ @Mr.Xcoder, Since inputs are always integer values, a negative exponent won't be required (or possible). \$\endgroup\$ – Toby Speight Apr 6 '17 at 12:16
  • 1
    \$\begingroup\$ @SIGSEGV maybe whereas i is not natural \$\endgroup\$ – Mr. Xcoder Apr 6 '17 at 12:55
  • 2
    \$\begingroup\$ @Jason, as many as supported/natural in your language - see the third rule. And it's code-golf because it needs an objective winning criterion to be on-topic here - "a pleasing solution" doesn't cut it (though I do like the Mathematica answer - that surprised me). \$\endgroup\$ – Toby Speight Apr 6 '17 at 14:55

40 Answers 40

28
\$\begingroup\$

Mathematica, 22 bytes

EvenQ@Log2@Max[#,-2#]&

Try it online! (Using Mathics instead, where this solution also works.)

I tried to find a solution with bitwise operators for a while, and while one definitely exists, I ended up finding something which is probably simpler:

  • Max[#,-2#] multiplies the input by -2 if it's negative. Multiplying by another factor of -2 doesn't change whether the value is a power of -2 or not. But now all odd powers of -2 have been turned into even powers of -2.
  • But even powers of -2 are also even powers of 2, so we can use a simple Log2@... and check if the result is an integer (to check whether it's a power of 2). This already saves two bytes over Log[4,...] (another way to look at even powers of -2).
  • As an added bonus, checking whether a value is an even integer is shorter than just checking whether it's an integer: we can save three more bytes by using EvenQ instead of IntegerQ.
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  • \$\begingroup\$ Does it help to consider that even powers of -2 are integer powers of 4? I like the idea of multiplying by -2 to get everything positive - though disappointed to see no bit-twiddling so far. \$\endgroup\$ – Toby Speight Apr 6 '17 at 13:04
  • 5
    \$\begingroup\$ @TobySpeight Treating them as powers of 2 actually saves 5 bytes. I used powers of 4 at first, but Log[4,...] is longer than Log2@... and IntegerQ is longer than EvenQ. \$\endgroup\$ – Martin Ender Apr 6 '17 at 13:05
16
\$\begingroup\$

Jelly, 5 bytes

æḟ-2=

Try it online!

How it works

æḟ-2=  Main link. Argument: n

æḟ-2   Round n towards 0 to the nearest power of -2.
    =  Test if the result is equal to n.
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12
\$\begingroup\$

Python, 46 bytes

-2 bytes thanks to @ovs.

def g(x):
 while x%-2==0!=x:x/=-2
 return x==1

Function with usage:

g(4) # put your number between the brackets

Try it online!

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  • \$\begingroup\$ print g(8) prints False \$\endgroup\$ – Felipe Nardi Batista Apr 6 '17 at 13:02
  • 2
    \$\begingroup\$ @FelipeNardiBatista shouldn't it? \$\endgroup\$ – Mr. Xcoder Apr 6 '17 at 13:04
  • 2
    \$\begingroup\$ sorry, my example was a bad one, print g(4) does the same \$\endgroup\$ – Felipe Nardi Batista Apr 6 '17 at 13:05
  • \$\begingroup\$ Wait, there is a small error, fixing it shortly \$\endgroup\$ – Mr. Xcoder Apr 6 '17 at 13:06
  • 1
    \$\begingroup\$ I've put a ; instead of a newline... sorry for that. Fixed @FelipeNardiBatista \$\endgroup\$ – Mr. Xcoder Apr 6 '17 at 13:08
11
\$\begingroup\$

Jelly, 6 bytes

b-2S⁼1

Try it online!

This is based on how Jelly converts an integer N to any arbitrary base B, doing so by converting N to an array, in which each integer is a digit d of (N)B, which can have a value 0≤Vd<B. Here, we will 0-index digits from the right, so every digit adds VdBd to form N. Vd<BVdBd<BBd=Bd+1, therefore every possible N has only one unique representation, if we ignore leading 0s in (N)B.

Here, d=input, B=-2. N=Bd=1Bd=VdBd⇔1=VdVd=1, and, since we're not adding any other multiples of powers of B, every other V would be 0. Right now, the array should be a 1 concatenated with d 0s. Since Jelly 1-indexes from the left, we should check whether the array's 1st element is 1, and all other elements are 0.

Hmm... all good, right? No? What's going on? Oh yeah, I have a better idea! First, let's take the sum of all of the integers in the array, treating it as if it was an integer array and not a number in base -2. If it is 1, it means that there is only one 1, and all other integers are 0. Since there can't be leading zeroes, except in the case of 0-2 (where the sum would be 0≠1 anyways), the 1st integer must be non-zero. The only non-zero integer in the array is the 1, so it must be the first one. Therefore, this is the only case that the sum of all of the integers in the array would be 1, because the smallest possible sum of a pair of positive integers is Σ{1,1}=2, since the smallest positive integer is 1. Every integer in a base representation is non-negative, so the only way the sum is 1 is to only have one 1, and all other integers are 0. Therefore, we can just check if the sum of all of the integers in the array is 1.

Here is what the code does:

b-2S⁼1 Main link. Arguments: d
b-2    Convert d to base -2.
   S   Take the sum.
    ⁼1 Check if the sum is equal to 1.
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  • 1
    \$\begingroup\$ Phew, that explanation took time to write... \$\endgroup\$ – Erik the Outgolfer Apr 6 '17 at 13:34
  • \$\begingroup\$ I'd hate to see what an explanation for a long program would look like then... \$\endgroup\$ – boboquack Apr 8 '17 at 8:32
  • 1
    \$\begingroup\$ @boboquack Here I am explaining why I use the base conversion stuff. I don't think explanation for long programs would be this long. A post can contain up to 30000 markdown characters, and explanations for longer programs would be more terse anyways. Also, I have read much longer explanations, and they are not that boring. \$\endgroup\$ – Erik the Outgolfer Apr 8 '17 at 8:38
11
\$\begingroup\$

Python 2, 35 34 32 bytes

f=lambda n:n==1or n!=n%2<f(n/-2)

Try it online!

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11
\$\begingroup\$

Python 2, 98 50 bytes

lambda x:x*(x&-x==abs(x))*((x<0)^x.bit_length()&1)

Try it online!

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10
\$\begingroup\$

Excel, 40 36 bytes

Saved 4 bytes by CallumDA

Excel can certainly do it but correcting errors adds 11 bytes

=IFERROR(-2^IMREAL(IMLOG2(A1)),1)=A1

Input is in cell A1. Output is TRUE or FALSE

If it was allowed to return either FALSE or #NUM! error for false values, it would be only 25 bytes:

=-2^IMREAL(IMLOG2(A1))=A1
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  • \$\begingroup\$ Heres a small improvement: =IFERROR(-2^IMREAL(IMLOG2(A1)),1)=A1 \$\endgroup\$ – CallumDA Apr 6 '17 at 23:49
  • 1
    \$\begingroup\$ @CallumDA Thanks! I tried to figure a way to use the complex number functions but everything I came up with was longer. \$\endgroup\$ – Engineer Toast Apr 7 '17 at 3:43
9
\$\begingroup\$

05AB1E, 8 bytes

Y(IÄÝm¹å

Try it online! or as a Test suite

Explanation

Y(         # push -2
  IÄÝ      # push range [0 ... abs(input)]
     m     # element-wise power
      ¹å   # check if input is in the resulting list
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  • \$\begingroup\$ Why the downvote? \$\endgroup\$ – Cows quack Apr 6 '17 at 15:28
  • \$\begingroup\$ @KritixiLithos: Seems like someone has downvoted all golfing languages. \$\endgroup\$ – Emigna Apr 6 '17 at 15:29
  • 6
    \$\begingroup\$ Noticed that too. Although I haven't been around PPCG for long, I've learned that creative and interesting solutions in standard languages are far more appreciated than 3-byte solutions in golfing languages. However, there are some people who (unfortunately) downvote very creative solutions in golfing languages, simply because they think everything is built-in, and don't understand how good the algorithms (although written in golfing languges) are. +1 for the incredible solution @Emigna \$\endgroup\$ – Mr. Xcoder Apr 6 '17 at 15:32
  • \$\begingroup\$ ÄLY(småO for 8. Y(sÄLm¢Z for 8... Nevermind, all 8. \$\endgroup\$ – Magic Octopus Urn Jul 13 '17 at 14:04
9
\$\begingroup\$

JavaScript (ES6), 37 28 24 bytes

f=x=>!x|x%2?x==1:f(x/-2)

Saved 4 bytes thanks to Arnauld.

f=x=>!x|x%2?x==1:f(x/-2)

console.log(f(-2));
console.log(f(-1));
console.log(f(0));
console.log(f(1));
console.log(f(2));
console.log(f(3));
console.log(f(4));

\$\endgroup\$
  • \$\begingroup\$ Why do I see some errors (before the true/false values) when I click on "Run code snippet"? \$\endgroup\$ – numbermaniac Apr 6 '17 at 13:49
  • \$\begingroup\$ @numbermaniac I'm not sure, maybe you're using a browser that doesn't fully support ES6? \$\endgroup\$ – Tom Apr 6 '17 at 13:51
  • \$\begingroup\$ Welp, refreshed and tried again, no errors. Not sure what happened the first time. \$\endgroup\$ – numbermaniac Apr 6 '17 at 13:54
9
\$\begingroup\$

C (gcc), 34 29 bytes

f(n){n=n%2?n==1:f(n?n/-2:2);}

Try it online!

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8
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MATL, 9 8 bytes

2_y|:q^m

Try it online! Or verify all test cases.

How it works

Consider input -8 as an example

2_    % Push -2
      % STACK: -2
y     % Implicit input. Duplicate from below
      % STACK: -8, -2, -8
|     % Absolute value
      % STACK: -8, -2, 8
:     % Range
      % STACK: -8, -2, [1 2 3 4 5 6 7 8]
q     % Subtract 1, element-wise
      % STACK: -8, -2, [0 1 2 3 4 5 6 7]
^     % Power, element-wise
      % STACK: -8, [1 -2 4 -8 16 -32 64 -128]
m     % Ismember. Implicit display
      % STACK: 1
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  • \$\begingroup\$ If I understand your explanation correctly, then given input n, this creates an array of size n as an intermediate step. Good job that efficiency is not a criterion here! \$\endgroup\$ – Toby Speight Apr 6 '17 at 15:03
  • 2
    \$\begingroup\$ @Toby Of course! This is code golf, who cares about efficiency? :-D \$\endgroup\$ – Luis Mendo Apr 6 '17 at 15:05
6
\$\begingroup\$

Octave, 28 bytes

@(n)any((-2).^(0:abs(n))==n)

This defines an anonymous function. The approach is similar to that in my MATL answer.

Try it online!

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6
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PHP, 41 Bytes

for(;$argn%-2==0;)$argn/=-2;echo$argn==1;

PHP, 52 Bytes

echo($l=log(abs($argn),2))==($i=$l^0)&&$argn>0^$i%2;

PHP, 64 Bytes

Working with a Regex

echo preg_match("#^".($argn>0?1:"1+0")."(00)*$#",decbin($argn));
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5
\$\begingroup\$

Python 3, 34 bytes

lambda n:n==(-2)**~-n.bit_length()
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5
\$\begingroup\$

JavaScript (ES6), 21 bytes

A recursive function that returns 0 or true.

f=n=>n==1||n&&f(n/-2)

How it works

This doesn't include any explicit test -- like n being odd or abs(n) being less than one -- to stop the recursion early when the input is not an exact power of -2.

We exit only when n is exactly equal to either 1 or 0.

This does work however because any IEEE-754 float will eventually be rounded to 0 when divided by 2 (or -2) enough times, because of arithmetic underflow.

Test cases

f=n=>n==1||n&&f(n/-2)

console.log(f(-2));
console.log(f(-1));
console.log(f(0));
console.log(f(1));
console.log(f(2));
console.log(f(3));
console.log(f(4));

\$\endgroup\$
5
\$\begingroup\$

C (gcc), 33 bytes

f(n){return n%2?n==1:n&&f(n/-2);}

Try it online!

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4
\$\begingroup\$

Java 7, 55 bytes

boolean c(int n){return n==0?0>1:n%-2==0?c(n/-2):n==1;}

Explanation:

boolean c(int n){  // Method with integer parameter and boolean return-type
  return n==0 ?    //  If n is zero:
    0>1//false     //   Return false
   : n%-2==0 ?     //  Else-if n mod -2 is zero:
    c(n/-2)        //   Recursive call for the input divided by -2
   :               //  Else:
    n==1;          //   Return if n is one
}                  // End of method

Test code:

Try it here.

class M{
  static boolean c(int n){return n==0?0>1:n%-2==0?c(n/-2):n==1;}

  public static void main(String[] a){
    for(int i = -2; i <= 4; i++){
      System.out.println(i + ": " + c(i));
    }
  }
}

Output:

-2: true
-1: false
0: false
1: true
2: false
3: false
4: true
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  • \$\begingroup\$ The non-recursive way is shorter by 5 bytes: boolean c(int n){while(0==n%-2)n/=-2;return 1==n;}. \$\endgroup\$ – Olivier Grégoire Apr 7 '17 at 14:14
  • \$\begingroup\$ @OlivierGrégoire Unfortunately that one doesn't work for n=0 in Java, because 0%-2==0 will be true and 0/-2 is still 0, causing an infinite loop, which is why I added the n==0?0>1 part to my recursive method. \$\endgroup\$ – Kevin Cruijssen Apr 7 '17 at 14:40
  • \$\begingroup\$ Nicely spotted! \$\endgroup\$ – Olivier Grégoire Apr 7 '17 at 14:51
4
\$\begingroup\$

Haskell, 24 23 bytes

f 0=0
f 1=1
f n=f(-n/2)

Defines a function f which returns 1 for powers of -2 and 0 otherwise.

A golfed version of my first submission to the other challenge.

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3
\$\begingroup\$

Javascript(ES7), 45 bytes

x=>-1**Math.log2(Math.abs(x))*Math.abs(x)==x
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  • \$\begingroup\$ Math.abs(x) is longer than x>0?x:-x, 11 bytes to 8 bytes. You should also be able to do -2**... instead of -1... to remove the second Math.abs(x) \$\endgroup\$ – fəˈnɛtɪk Apr 7 '17 at 1:38
  • \$\begingroup\$ What's ES7 specific in this? \$\endgroup\$ – Arjun Apr 8 '17 at 3:36
  • \$\begingroup\$ @DobbyTheFree-Elf, ** is. \$\endgroup\$ – Qwertiy Apr 10 '17 at 20:11
3
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Perl 6, 21 bytes

{$_==(-2)**(.lsb//0)}

Try it

Expanded:

{  # bare block lambda with implicit parameter 「$_」

  $_                  # is the input
  ==                  # equal to
  (-2)**( .lsb // 0 ) # -2 to the power of the least significant bit of the input
}

Note that 0.lsb returns Nil which produces a warning when used as a number, so the defined or operator // is used.
(Think of // as || with a different slant)

A method call with no invocant where a term is expected is implicitly called on $_. (.lsb)

Also works with .msb.

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  • \$\begingroup\$ I like this one! \$\endgroup\$ – tale852150 Apr 7 '17 at 14:03
3
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Prolog (SWI), 44 bytes

p(X):-X=1;X\=0,X mod 2=:=0,Z is X/(-2),p(Z).

Online interpreter

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3
\$\begingroup\$

Python, 24 bytes

lambda n:n*n&n*n-1<n%3%2

Try it online!

The bit trick k&k-1==0 checks whether k is a power of 2 (or k==0). Checking this for k=n*n as n*n&n*n-1==0 tells us whether abs(n) is a power of 2.

To further see if n is a power of -2, we need only check that n%3==1. This works because mod 3, the value -2 is equal to 1, so its powers are 1. In contrast, their negations are 2 mod 3, and of course 0 gives 0 mod 3.

We combine the checks n*n&n*n-1==0 and n%3==1 into a single expression. The first can be written with <1 for ==0, since it's never negative. The n%3==1 is equivalent to n%3%2, giving 0 or 1. So, we can combine them as n*n&n*n-1<n%3%2.

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2
\$\begingroup\$

R, 22 bytes

Takes input from stdin, returns TRUE or FALSE accordingly.

scan()%in%(-2)^(0:1e4)

I'm not 100% sure that this is a valid answer, as it only works for integers up to R's size limit, and if the integers were unbounded it wouldn't work. However, the rules state:

The usual integer overflow rules apply: your solution must be able to work for arbitrarily large integers in a hypothetical (or perhaps real) version of your language in which all integers are unbounded by default, but if your program fails in practice due to the implementation not supporting integers that large, that doesn't invalidate the solution.

In a hypothetical version of R which does allow unbounded integers, then we could use the following code, for the same byte count:

scan()%in%(-2)^(0:Inf)

Of course, in real R, the above code just gives Error in 0:Inf : result would be too long a vector.

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2
\$\begingroup\$

bc 88 bytes

bc -l <<< "n=$1;q=l(sqrt(n*n));p=4*a(1);((n<1)*c(q/l(2)*p/2)+(n>1)*(s(q/l(4)*p)))^2==0"

I have this in a file neg2.sh and it prints 1 for powers of -2 and 0 otherwise

I know it's really long, but it was fun

Test

$ for i in {-129..257}; do echo -n "$i: "; ./neg2.sh $i; done | grep ': 1'
-128: 1
-32: 1
-8: 1
-2: 1
1: 1
4: 1
16: 1
64: 1
256: 1

Explanation

The main body has two halves, both are trying to equal zero for powers of -2.

q=l(sqrt(n*n))               % ln of the absolute value of the input
p=4*a(1)                     % pi: arctan(1) == pi/4
q/l(2) -> l(sqrt(n*n))/l(2)  % change of base formula -- this gives
                             % the power to which 2 is raised to equal
                             % sqrt(n*n). It will be an integer for 
                             % numbers of interest
n<1                          % 1 if true, 0 if false. for negative
                             % numbers check for powers of 2
n>1                          % for positive numbers, check for powers
                             % of 4
c(q/l(2)*p/2)                % cos(n*pi/2) == 0 for integer n (2^n)
s(q/l(4)*p)                  % sin(n*pi) == 0 for integer n (4^n)
(....)^2==0                  % square the result because numbers are
                             % not exactly zero and compare to 0
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  • \$\begingroup\$ I never expected trigonometry! Good answer! \$\endgroup\$ – Toby Speight Apr 11 '17 at 8:24
2
\$\begingroup\$

Julia 0.5, 20 bytes

!n=n∈(-2).^(0:n^2)

Try it online!

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2
\$\begingroup\$

Fourier, 53 bytes

I~X1~N~G0(0-2*G~GX*X~PG*G>P{1}{0~O~N}G{X}{1~O0~N}N)Oo

I'll work on golfing this later, but the outline of this is:

X = User input
G = N = 1
Loop until N = 0
    G = -2 * G
    P = X*X 
    If G*G > P then
        N = O = 0
    End if
    If G = X then
        O = 1
        N = 0
    End if
End loop
Print O

Where the output is 0 for falsey and 1 for truthy.

Try it online!

\$\endgroup\$
  • \$\begingroup\$ In the algo description not would be better not use P variable and write If G*G > X*X then...? \$\endgroup\$ – RosLuP Apr 12 '17 at 15:13
  • \$\begingroup\$ @RosLuP That would be better, but Fourier would simply treat that as (G*G > X)*X \$\endgroup\$ – Beta Decay Apr 12 '17 at 19:44
2
\$\begingroup\$

Casio BASIC, 76 bytes

Note that 76 bytes is what it says on my calculator.

?→X
0→O
While Abs(X)≥1
X÷-2→X
If X=1
Then 1→O
IfEnd
WhileEnd
O

This is my first venture into Casio BASIC... I never realised I could write such decent programs on a calculator :D

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1
\$\begingroup\$

Python 2.7, 40 bytes

a=input()
while a%-2==0:a/=-2
print a==1

Credits to Mr. Xcoder for the original code of length 43 bytes. Had to post as a separate answer since I don't have enough reputation to comment.

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  • \$\begingroup\$ It's kind of the same thing, since I've made my answer version-universal, so it works in both Python 2 and 3. If you were to do this in Python 3, you should have added int(input()) which would have gone over the limit of the def-like function. Additionally, In python 3, you must use print() which would of wasted 1 byte. That's why I chose that way, because in Python 3 it gets longer... \$\endgroup\$ – Mr. Xcoder Apr 6 '17 at 15:27
1
\$\begingroup\$

Retina, 27 bytes

+`(1+)\1
$1_
^(1|-1_)(__)*$

Try it online!

Takes input in unary, which is fairly standard for Retina. The first two lines do partial unary to binary conversion based on the first two lines of code from the Tutorial entry (any extraneous 1s will cause the match to fail anyway), while the last line checks for a power of four or a negative odd power of two.

+`(1+)\1\1\1
$1_
^(-1)?1_*$

Try it online!

This time I do partial unary to base four conversion. Powers of four end up as ^1_*$ while negative odd powers of two end up as ^-11_*$.

+`\b(1111)*$
$#1$*
^(-1)?1$

Try it online!

This time I just keep dividing by four as much as I can and check for 1 or -11 at the end.

+`\b(1+)\1\1\1$
$1
^(-1)?1$

Try it online!

Another way of dividing by four. And still annoyingly 27 bytes...

\$\endgroup\$
1
\$\begingroup\$

Scheme, 60 bytes

(define(f n)(cond((= 1 n)#t)((<(abs n)1)#f)(#t(f(/ n -2)))))

Recursive solution.

\$\endgroup\$

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