I have this code to return true if num is a power of 2.
def is_power_of_two?(num)
result = num.inject(0) {|n1, n2| n2 ** n1}
if result == num
true
else
false
end
end
p is_power_of_two?(16)
I keep getting an error though. How could I fix and simplify this code?
Clearly, n is a non-negative integer.
Code
def po2?(n)
n.to_s(2).count('1') == 1
end
Examples
po2? 0 #=> false
po2? 1 #=> true
po2? 32 #=> true
po2? 33 #=> false
Explanation
Fixnum#to_s provides the string representation of an integer (the receiver) for a given base. The method's argument, which defaults to 10, is the base. For example:
16.to_s #=> "16"
16.to_s(8) #=> "20"
16.to_s(16) #=> "10"
15.to_s(16) #=> "f"
It's base 2 we're interested in. For powers of 2:
1.to_s(2) #=> "1"
2.to_s(2) #=> "10"
4.to_s(2) #=> "100"
8.to_s(2) #=> "1000"
16.to_s(2) #=> "10000"
For a few natural numbers that are are not powers of 2:
3.to_s(2) #=> "11"
5.to_s(2) #=> "101"
11.to_s(2) #=> "1011"
We therefore wish to match binary strings that contain one 1.
Another Way
R = /
\A # match beginning of string ("anchor")
10* # match 1 followed by zero or more zeroes
\z # match end of string ("anchor")
/x # free-spacing regex definition mode
def po2?(n)
(n.to_s(2) =~ R) ? true : false
end
po2?(4) #=> true
po2?(5) #=> false
One more
This uses Fixnum#bit_length and Fixnum#[]:
def po2?(n)
m = n.bit_length-1
n[m] == 1 and m.times.all? { |i| n[i].zero? }
end
po2? 0 #=> false
po2? 1 #=> true
po2? 32 #=> true
po2? 33 #=> false
And a fourth, perhaps the best
def po2?(n)
return false if n <= 0
while n.even?
n /= 2
end
n == 1
end
po2? 0 #=> false
po2? 1 #=> true
po2? 32 #=> true
po2? 33 #=> false
Try:
def is_power_of_two?(num)
num != 0 && (num & (num - 1)) == 0
end
It is well explained here (for C#, but @GregHewgill's explanation applies here as well)
I would do something like this, using Ruby's Math module.
def power_of_two?(n)
Math.log2(n) % 1 == 0
end
Or, if you wanted to be really cool:
def power_of_two?(n)
(Math.log2(n) % 1).zero?
end
Some IRB output:
2.1.0 :004 > power_of_two?(2)
=> true
2.1.0 :005 > power_of_two?(32768)
=> true
2.1.0 :006 > power_of_two?(65536)
=> true
This method assumes that the input is a positive integer.
Another way to solve this is to go the other way around than most of the answers here - we can use the number 1 to start and find out if the number is the power of two. Like this:
def power_of_two?(num)
product = 1
while product < num
product *= 2
end
product == num
end
We start with 1. Then we multiply the 1 by 2, and keep multiplying by 2 until the product is larger than num (product < num). Once we hit that condition, we stop, exit the loop, and check if it's equal to num (product == num). If it is, the num is the power of 2.
As was pointed out in the comments above, you were getting errors because you're trying to use the inject method on a non-iterable (an int). Here's a solution using the suggested log2
def is_power_of_two?(num)
result = Math.log2(num)
result == Integer(result)
end
Note: will fail with very big numbers close to binaries (like 2 ^ 64 - 1). A foolproof version (but slower) would be:
def is_power_of_two?(num)
while (num % 2 == 0 and num != 0)
num /= 2
end
num == 1
end
Please comment any improvements that any of you may find.
Here is another solution that uses recursion:
def power_of_2?(number)
return true if number == 1
return false if number == 0 || number % 2 != 0
power_of_2?(number / 2)
end
In my opinion, the easiest -- but maybe a little long -- way of doing what you need to do is just writing this recursive method like so:
def power_of_two?(number)
continue = true
if number == 1
return true
end
if number % 2 != 0
return false
else
while continue == true do
if number.to_f / 2.0 == 2.0
continue = false
return true
else
if number % 2 != 0
continue = false
return false
else
number /= 2
continue = true
end
end
end
end
end
One is a power of two (2^0), so it first checks if the number given is 1. If not, it checks if it is odd, because 1 is the only odd number that is a power of two.
If it is odd it returns false and moves on to the else statement. It will check if the number divided by 2 is two, because then it would obviously be a power of 2. It does this as a float, because 5/2 in Ruby would return 2.
If that is false, it then again checks if the number is odd -- unnecessary on the first round, necessary after that. If the number is not odd, it will divide the number by two and then do the loop another time.
This will continue until the program resolves itself by getting 2 or any odd number, and returns true or false, respectively.
I ran into this one in a bootcamp application prep. I'm not a math person and don't understand a few of these methods, so I wanted to submit a common sense approach for people like me. this requires little knowledge of math, except to know a number to the second power will be the result of some number multiplied by itself.
def is_power_of_two?(num)
num.times {|n| return true if (n+1) * (n+1) == num}
false
end
this method counts up to the num variable starting at 1 and returns true if (any of those numbers in the sequence multiplied by itself) is equal to num & if num is not 0 (more on that below).
example: num = 9
1 * 1 == 9 #=> false
2 * 2 == 9 #=> false
3 * 3 == 9 #=> true
true is returned and method is finished running.
the #times method requires an integer > 0, so this edge case is "handled" by virtue of the fact that #times does nothing with "0" as the variable and returns false when outside of the #times iteration.
