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Solve the following problem using Python, implementing the functions described below, one line at a time
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Description:
 def _check_self_replicate(self, myntr):
 """ For each Rule B -> c where c is a known terminal, this function searches for B occurences in rules with the form A -> B and sets A -> c. """

        # print 'BFS Dictionary Update - Self Replicate'
        find = 0
        for nonterm in self.grammar.grammar_nonterminals_map:
            for i in self.grammar.grammar_nonterminals_map[nonterm]:
                if self.grammar.grammar_rules[i][0] not in self.resolved and not isinstance(
                        self.grammar.grammar_rules[i][1], (set, tuple)) \
                        and self.grammar.grammar_rules[i][1] == myntr:
                    self.resolved[self.grammar.grammar_rules[i][0]] = self.resolved[myntr]
                    if self._checkfinal(self.grammar.grammar_rules[i][0]):
                        return self.resolved[self.grammar.grammar_rules[i][0]]
                    if self.grammar.grammar_rules[i][0] not in self.bfs_queue:
                        self.bfs_queue.append(self.grammar.grammar_rules[i][0])
                    find = 1
        if find == 1:
            return 1
        return 0