irst calculate the output-source entropy H(Y) and the equivocation entropy H(Y|X). For this, we need the output probability distribution p(y1), p(y2), which is obtained from the transition-matrix elements in Eq. (K2) as follows: p(y1) = p(y1x1)p(x1)+p(y1x2)p(x2) =aq+(1-b)(1-q) p(y2) = p(y2)x1)p(x1) + p(y2)x2)p(x2) = (1-a)q+b(1-q) =1-p(y1). - In Eq. (K4), we have defined the input probability distribution according to p(x1) = q and p(x2) = 1 - q. From Eq. (K4), we can calculate the output-source entropy: H(Y) = p(y1) log p(y1) - - p(y2) log p(y2) = p(y1)logp(y1)-[1-p(y1)]log[1-p(y1) - - = f[p(y1)] = flag+(1-b)(1-q)]. Appendix K where the function f is defined by f(u) =f(1-u) = = -ulogu (1 - u) log(1 u), noting that, as usual, the logarithms are in base two. The next step consists of calculating the equivocation H(Y|X). For this, we need the joint distribution p(x, y). Using Bayes's theorem and the conditional probabilities shown in Eq. (K2) we obtain: p(y1,x1) = p(y1|x1)p(x1) = aq p(y1,x2) = p(y1|x2)p(x2) = (1 b)(1-q) - p(y2,X1) = p(y2)x1)p(x1) = (1 a)q p(y2,x2) = p(y2)x2)p(x2) = b(1 - q). The results in Eq. (K5) now make it possible to calculate the equivocation H(Y|X): = - -[aq log a + (1 log(1 - a) + (1 - -b)(1-q)log(1 -b)+b(1-q) log b] qf(a) (1 - q)f(b). Substituting Eqs. (K7) and (K8) in Eq. (K3), we obtain the mutual information H(X; Y): H(X;Y) = f[aq + (1 - b)(1 - q)] - qf(a) - (1 q)f(b). Setting q = 0 or q = 1 in the result in Eq. (K9) yields, in both cases, H(X; Y) = 0. The case a = 1 - b (or a + b = 1) corresponds to H(X; Y) = 0, as can easily be verified from Eq. (K9). This means that, regardless of the input probability distribution, the mutual information is zero. As discussed in the main text, such a channel is useless. The condition a = 1 - b in the transition matrix (Eq. (K2)) represents the most general definition of useless channels, i.e., including but not limited to the case a = b = E = 1/2. In the general case, we have H(X; Y) > 0, and the function is concave (meaning that a cord between any t