Alternate Start Infinite Answer
15-211: Finite State Machine practice problems 1 (answers)Finite State Machines - Exercises 1AnswersFollowing are some exercises on finite state machines. You shouldattempt to work through these before checking the answers.For the problems in this section, draw a deterministic finite statemachine which implements the specification. Some machines may beimpossible to construct; explain why if you think so. Whereappropriate, the alphabet (allowable input characters) for the machineis listed in brackets. It's not important to be precise about everytransition; it is sufficient to draw an unadorned arrow to mean 'allother characters'.
Remember to indicate the starting state.1. Consider machines which accept a stream of 1 or more bits(the alphabet is limited to 1's and 0's), and whose output is 1(accepting) or 0 (rejecting). Construct FSMs which implement thefollowing operations:(a) orThat is, the output should equal the C expressionc 1 c 2.
c i(b) andc 1 &&c 2 &&. && c i(c) xorc 1 ^c 2 ^.
^ c i2.(a) Accepts just the stringMURMUR by itself. MRU(b) Accepts CYBOT followed by any characters.BCOTY(c) Accepts FROG with any prefix. FGORBe careful. Does yourmachine accept FROFROG?(d) Accepts any string containing FROG. FGOR(e) Accepts any string containing MURMURS. MRSUDoes yours get MURMURMURS?(f) Accepts strings consisting of only 0 or more repetitions of 15211.
125(g) Accepts strings starting with 0 or more repetitions of 15211. 125(all strings start with 0 or more repetitions of 15211.)3.(a) Accepts CAT or DOG alone.
ACDGOT(b) Accepts strings containing CAT or DOG anywhere. ACDGOT(c) Accepts strings containing ART or ARC anywhere. ACRT(d) Accepts strings which are any of ART or ARTS or ARTIST or ABLE. ABEILRSTWhat does this remind you of?It's a trie!4. # c means the number of character c occuringin the string.(a) Accepts strings of even length.
AB(b) Accepts strings with exactly 3 A's. AB(c) Accepts strings with at least 3 A's. AB(d) Accepts strings where #a #b. Since #a is unbounded, this machine would needan infinite amount of states, or some infinite auxilliary storage.A counting argument shows why: Since at any point inthe processing of the string we can encounter any number of B's, we'dneed to keep track of the number of A's encountered so far.
This is anunbounded number. In order for a machine to keep track of exactly howmany A's we've had, we'd need at least one state for each possibility.But there are an infinite number of possibilities for #a, so we can'tdo this with a finite state machine.(e) Accepts strings where #a = #b. Since the A's and B's can come inany order, this can't be done for the same reason as 4d.(f) Accepts strings where (#a mod 3) = (#b mod 3). AB In contrast to 4d and 4e, this machine isable to 'count' since #a mod 3 and #b mod 3 are always bounded. In what otherways can and can't finite state machines count?( ).
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