= Weakly symmetric channel capacity
{title2=$C=\log_2m-H(\text{one row})$}
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> has
$$
C=\log_2m-H(\text{one row}).
$$
The 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 $\log_2m$ 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 $H(Y)=\log_2m$, 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.
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