Solution to InterviewStreet Manipulative Numbers

permutation.rb

print "Please enter the limits"
number_limit=gets
limit=number_limit.to_i
numbers=[]
xor=[]
def compute_k(foo)
      count=0
      foo=foo+1 if foo%2==0
      until foo ==1
       if foo%2==0
        foo=foo/2
        count=count+1
       else
         count=0
      end
      end
    return count
end

unless limit > 100
until limit==0
 numbers.push(gets)
 limit=limit-1
end
end

for i in 0..numbers.length-1
  for j in i+1..numbers.length
    xor.push(numbers[i].to_i^numbers[j].to_i)
  end
end
#  value=min.sort[0]
 #unless temp == 0

  #else
  #   value+=1
  #end
if(numbers.length<3)
  k=compute_k(xor[0].to_i)
else if(numbers.length==3)
  k=compute_k(xor[1].to_i)
else
  k=compute_k(xor[numbers.length-3].to_i)
end
if k >=0
  if k ==0
    puts k+1
   else
     puts k
   end
else
  puts -1
end
end