Construct an NFA over alphabet {, } that accepts all words with more a’s modulo 4 than b’s modulo 3. For example, “” is accepted as it has two a’s and one b. The word “” is accepted as it has six a’s and four b’s.
I must be missing something obvious, as any solution I come up with gets increasingly messy.
I would love to hear anyone else's solutions.
So I build an automata which end states accepted are colored in red and to avoid confusion, I pointed the initial state; here is the link for the quiver diagram
I forgot to color $(2,2)$ in red! and the above $b$ is in fact an $a$
Here is a screenshot of the diagram: