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Every row of the channel matrix is a permutation of $(1-2\alpha ,\alpha ,\alpha )$, and every column sum is one. It is therefore a weakly symmetric channel. The uniform input makes the output uniform, with entropy ${\log }_{2}3$, while the conditional output entropy is the entropy of one row:The weakly symmetric channel capacity formula now gives the ternary symmetric channel capacitybits per channel use, for $0\le \alpha \le 1/2$, with the convention $0{\log }_{2}0=0$.

For a discrete memoryless channel with $m$ output symbols, suppose every row of the transition matrix is a permutation of every other row and all column sums agree. Such a communication channel hasThe row Shannon entropy is the same for every input, so $H(Y\mid X)$ has that fixed value for every input probability distribution. The remaining term $H(Y)$ is at most ${\log }_{2}m$ by maximum entropy on a finite alphabet.

A uniform input attains the bound. If there are $\ell$ input symbols, the total of all transition probabilities in the matrix is $\ell$; equal column sums must therefore each be $\ell /m$. Multiplication by the uniform input mass $1/\ell$ gives output mass $1/m$ in each column. Hence $H(Y)={\log }_{2}m$, which proves the displayed formula. The cyclic ternary channel capacity, binary symmetric channel capacity, and ternary symmetric channel capacity are concrete applications of this symmetry calculation.