Question

I am making a program where one of the problems is that I need to do some analysis of the bit pattern in some integers.

Because of this I would like to be able to do something like this:

#Does **NOT** work:
num.each_bit do |i|
   #do something with i
end

I was able to make something that works, by doing:

num.to_s(2).each_char do |c|
   #do something with c as a char
end

This however does not have the performance I would like.

I have found that you can do this:

0.upto(num/2) do |i|
   #do something with n[i]
end

This have even worse performance than the each_char method

This loop is going to be executed millions of times, or more, so I would like it to be as fast as possible.

For reference, here is the entirety of the function

@@aHashMap = Hash.new(-1)

#The method finds the length of the longes continuous chain of ones, minus one 
#(101110 = 2, 11 = 1, 101010101 = 0, 10111110 = 4)

def afunc(n) 
if @@aHashMap[n] != -1
    return @@aHashMap[n]
end

num = 0
tempnum = 0
prev = false

(n.to_s(2)).each_char do |i|
    if i
        if prev
            tempnum += 1
            if tempnum > num
                num = tempnum
            end
        else
            prev = true
        end
    else
        prev = false
        tempnum = 0
    end
end

@@aHashMap[n] = num
return num
end

Answer 1

To determine the length of the longest sequence of consecutive 1’s, this is more efficient:

def longest_one_chain(n)
  c = 0
  while n != 0
    n &= n >> 1
    c += 1
  end
  c
end

The method simply counts how many times you can “bitwise AND” the number with itself shifted 1 bit to the right until it is zero.

Example:

                 ______ <-- longest chain
    01011011100001111110011110101010 c=0
AND  0101101110000111111001111010101
        1001100000111110001110000000 c=1, 1’s deleted
AND      100110000011111000111000000
            100000011110000110000000 c=2, 11’s deleted
AND          10000001111000011000000
                    1110000010000000 c=3, 111’s deleted
AND                  111000001000000
                     110000000000000 c=4, 1111’s deleted
AND                   11000000000000
                      10000000000000 c=5, 11111’s deleted
AND                    1000000000000
                                   0 c=6, 111111’s deleted