Consider a nondeterministic parity automaton \(A = (Q, \Sigma, \Delta, q_0, \Omega),\) where: \(Q\) is a finite set of states, \(\Sigma\) is the input alphabet, \(\Delta \subseteq Q \times \Sigma \times Q\) is the transition relation, \( q_0 \in Q\) is the initial state, and \(\Omega : Q \to \mathbb{N}\) is the parity acceptance condition. We say that a probabilistic automaton \(P\) is obtained from a memoryless resolver for \(A\) if, for each state \(s \in Q\) and input symbol \(a \in \Sigma\), the nondeterministic set of successors \[\{ t \in Q \mid (s, a, t) \in \Delta \} \] is replaced by a probability distribution \(D_s^a\) over that set. The resulting probabilistic automaton \(P\) retains the same state set \(Q\), alphabet \(\Sigma\), and initial state \(q_0\), but transitions are decided by the probability distributions \( \{ D_s^a \}_{s \in Q, a \in \Sigma}, \) which define the probability of moving from state \(s\) to state \(t\) upon reading symbol \(a\).\(\\\)
We say that \(A\) is memoryless stochastically resolvable if there exists a probabilistic automaton \(P\), obtained from a memoryless resolver as described above, such that for every infinite word \(w \in \Sigma^\omega\) accepted by \(A\), the probabilistic automaton \(P\) also accepts \(w\) with probability 1.
Moreover, we say that \(P\) is obtained from uniform memoryless resolver if, for all \(s \in Q\) and \(a \in \Sigma\), the distribution \(D_s^a\) is uniform over the set \( \{ t \in Q \mid (s, a, t) \in \Delta \}. \) This leads to the question: If a nondeterministic parity automaton \(A\) is stochastically resolvable using a memoryless resolver, is it always possible to construct such a resolver using uniform distributions?