Consider the language L = {w |w ∈ {0, 1} ∗ , 2 × #0 (w) = #1 (w)} , here #0(w) and #1(w) means the numbers of 0s and 1s in w respectively. That is, this is the language consisting of all strings with exactly twice as many 1s as 0s: for example, 101 ∈ L, 010 ∈/ L. In this problem we will show how to construct a context-free grammar for it. (a) (2 points) For a length n word x, we define a function f (i) = (2 × number of 0s in locations [1 . . . i]) − (number of 1s in locations [1 . . . i]). Show that we have f(i) = f(j) for some i < j if and only if the substring in indices [i + 1, . . . j] has exactly twice as many 1s as 0s (i. e. xi+1,i+2,...,j ∈ L). (b) (2 points) Show that if x ∈ L starts with 0, then it can be written as x = 0w1y1z where w, y, and z are each strings with twice as many 1s as 0s (aka. w, y, z ∈ L).