I learned how to convert context-free grammar to pushdown automata but how can I do the opposite? to convert PDA to CFG?
For example: to write CFG for the automata
My attempt:
$S=A_{03}$ because $q_{\color{blue}0}$ is the initial state and $q_{\color{blue}3}$ is the final state.
There are $4$ states so we will have $4^2$ variables:
$$A_{00},A_{01},A_{02},A_{03},\\ A_{10},A_{11},A_{12},A_{13},\\ A_{20},A_{21},A_{22},A_{23},\\ A_{30},A_{31},A_{32},A_{33}$$
for each state $q_{i}$ of the automata $A_{ii}\to\varepsilon$
$$A_{00}\to \varepsilon\\ A_{11}\to \varepsilon\\ A_{22}\to \varepsilon\\ A_{33}\to \varepsilon\\$$
For each triplet of states $q_i,q_j,q_k$, we add the rule $A_{ij}\to A_{ik}A_{jk}$, this gives us $4^3=64$ rules:
$$A_{00}\to A_{00}A_{00}|A_{01}A_{10}|A_{02}A_{20}|A_{03}A_{30}\\ A_{01}\to A_{00}A_{01}|A_{01}A_{11}|A_{02}A_{21}|A_{03}A_{31}\\ A_{02}\to A_{00}A_{02}|A_{01}A_{12}|A_{02}A_{22}|A_{03}A_{32}\\ \vdots\\ A_{33}\to A_{30}A_{03}|A_{31}A_{13}|A_{32}A_{23}|A_{33}A_{33}\\$$
I am stuck here.