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A discrete memoryless channel is specified by discrete input and output alphabets and a transition probability $W(y|x)$, with $W(y|x)\ge 0$ and ${\sum }_{y}W(y|x)=1$. Its successive outputs are conditionally independent given the inputs, with the same transition rule at every use:Operationally, its channel capacity is the supremum of communication rates in bits per use attainable by increasingly long block codes with decoding error probability tending to zero. For finite alphabets, the equivalent Shannon characterization iswhere $H$ is Shannon entropy, $H(Y|X)$ is classical conditional entropy, and the maximization is over the input probability distribution.

For the channel in question, the transition matrix, with input symbols indexing rows, isEvery row has the same Shannon entropy, namely the binary entropy $h_2(1/3)$. Consequently $H(Y|X)=h_2(1/3)$ for every input probability distribution. Since the output has three symbols, the maximum entropy on a finite alphabet givesChoose a uniform input. Each column of $W$ sums to one, so $P_Y(y)=1/3$ for every output symbol and the upper bound is attained. FinallyThus the cyclic ternary channel capacity isachieved by a uniform input probability distribution. This is a weakly symmetric channel capacity calculation: the matrix rows are permutations and the column sums agree. The channel is not the usual ternary symmetric-error channel, since an error has only one possible changed output for each input.

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.