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Answer:

Stringcacfadefababebbbbfbbcffabcc
Accept or
Reject?
Accept Accept Reject Accept Reject Accept Reject

Epsilon-transitions

automaton

(The two characters that label the transition into state q3 are capital L and lower-case l).

The above automaton accepts strings that are legal Java integer literals, such as the strings:

0 0L +32 -96l 0X235 -0xA2E5L

There are several new features to this automaton. The epsilon (e-like symbol) labeling the transition out of the start state means that the transition can be made without consuming a character. This allows an optional "+" or "-" character at the beginning of the string.

There are two transitions out of state q1 labeled with 0. This means that if the current character is 0 then either of these transitions can be used.


QUESTION 23:

Is the string "+0Xf" accepted?