Let L be the language generated by regular expression 0*10* and accepted by…

Answer: There are 3 equivalence classes.

Reasoning: The language described by 0*1 0* is the set of all strings over {0,1} that contain exactly one '1'. Prefixes of input strings fall into three distinct behaviour classes with respect to which continuations will lead to acceptance.

• Prefixes with no '1' seen yet (e.g., ε, 0, 00). From here, appending a single '1' followed by zeros can produce an accepting string.

• Prefixes with exactly one '1' seen (e.g., 1, 01, 0010). These are already in the accepting condition provided only zeros follow.

• Prefixes with two or more '1's (e.g., 11, 101). Once two '1's appear, no continuation can make the string belong to the language (dead class).

Distinguishing the classes:

• No '1' vs exactly one '1': use the empty suffix ε. A prefix with no '1' plus ε is not accepted, whereas a prefix that already is '1' plus ε is accepted.

• Exactly one '1' vs two or more '1's: use the empty suffix ε. A prefix with exactly one '1' is accepted, while a prefix with two '1's is not.

• No '1' vs two or more '1's: use the suffix '1'. Appending '1' to a prefix with no '1' can create an accepted string, while appending '1' to a prefix that already has two '1's cannot.

Conclusion: These three mutually distinguishable classes correspond to three reachable states in the minimal DFA (start/no '1', accepting/exactly one '1', dead/two or more '1's). Therefore R_M has 3 equivalence classes.