CODEWITHSUNDEEP
ruby-on-railsrubyjruby
Harish Shetty
Harish Shetty

Reputation: 64363

How do I find the value of exponent in Ruby?

Consider the equation below:

2 ** n = A

Let us assume A=64.

What is the easiest way to find the value of n?

I am currently using following two approaches

A= 64; n = 1; n+=1 while (A >> n) > 0; n-1

A= 64; n = 0; n+=1 until (A == ( 2 ** n));n

Is there a better approach?

Other way of stating the same problem:

2 = nth root A If I know the value of A, how do I determine the value of n?

Upvotes: 1

Views: 3324

Answers (5)

Lykos
Lykos

Reputation: 1029

The proposed answer is good, but you have to count up to your logarithm and there is a slightly more efficient way, if you know that the logarithm has an upper bound, i.e. if you know that you never deal with numbers larger than 1 << bound.

def faster_log2(n)
  l = 0; bound = 64
  while bound > 0
    if 1 << bound > n
      n >>= bound; l += bound
    end
    bound >>= 1
  end
  l
end

If you don't have an upper bound, but you still want to use this algorithm, you can also calculate a bound before executing the algorithm, but if performance is not an issue, then I would stick with the shorter solution.

def bound(n)
  bound = 1;
  bound >>= 1 while 1 << bound < n
end

Upvotes: 0

pierrotlefou
pierrotlefou

Reputation: 40751

If you care about the problems mentioned by mobrule. How about this? It is the same as your own method but use built-in function to communicate the idea explictly.

    def exp_value a 
       (1..a).to_a.detect {|n| 2**n >=a}
    end

    exp_value 64
    => 6

Upvotes: -1

mob
mob

Reputation: 118635

Neither of the above answers is better than your first approach: 

A= 64; n = 1; n+=1 while (A >> n) > 0; n-1

Evaluating Math.log(x) takes a lot longer than doing a dozen bit-shifts, and will also give you an answer like 5.99999999999980235 to make sense of.

See this SO question for some better ideas.

Upvotes: 5

JRL
JRL

Reputation: 78013

Try this:

def exp_value(n)
  Math.log(n) / Math.log(2)
end

Upvotes: 9

user187291
user187291

Reputation: 53940

log2n = ln n / ln 2

hence

log_2_64 = Math.log(64) / Math.log(2)

Upvotes: 3

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