Consider (a) L = {w elementof {0, 1}*| the number of 0s in w is exactly twice the number of 1s}, (i) Construct a Turing machine M with input alphabet {0, 1} that decides L. Make sure you specify the tape alphabet as well as the transition function. (Give both a state diagram and an implementation-level description of M that clearly shows how your machine decides L.) (ii) Confirm that your machine yields the correct results on the following strings by constructing an appropriate trace: the empty string, 01, and 010010. (b) Let B = {w elementof {0, 1}*| |w| is odd & the number of 0s in w is either less than twice the number of 1s or more than twice the number of 1s}. Using the closure results for Turing decidability, show that B is Turing-decidable