finish the above FasterLuc code with python and report results. Please answer full question and make sure code runs!

3. (20 pts) Ron and Hermione are having a competition to see who can compute the nth Lucas number Ln more quickly, without resorting to magic. Recall that the nth Lucas number is defined as Ln-- for n >1 with base cases Lo 2 and L1-1 Ron opens with the classic recursive algorithm: Luc(n) if n0 return 2 else if n-1 { return 1 } else return Luc(n-1) + Luc(n-2) which he claims takes R(n) = R(n-1) + R(n-2) + c 0(0) time. 3. (20 pts) Ron and Hermione are having a competition to see who can compute the nth Lucas number Ln more quickly, without resorting to magic. Recall that the nth Lucas number is defined as Ln-- for n >1 with base cases Lo 2 and L1-1 Ron opens with the classic recursive algorithm: Luc(n) if n0 return 2 else if n-1 { return 1 } else return Luc(n-1) + Luc(n-2) which he claims takes R(n) = R(n-1) + R(n-2) + c 0(0) time