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