CSCE 5400 ASSIGNMENT #1 SOLUTION

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Carmella Horn
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1 CSCE 5400 ASSIGNMENT #1 SOLUTION 1. Design deterministic finite automata to recognize the languages below. Test your DFA s using Prolog on the indicated strings in the language, as well as some strings which are not in the language. (a) L = {w w is a floating point constant in a typical programming language (e.g. C++)}. Example strings include 10.87, 2.0E7, 6.625E 27, etc. Script started on Tue 11 Sep :42:50 AM CDT brb0164@faculty% cat as2f18-1a.pro delta(q0, [- X]) :- delta(q1, X). delta(q0, [D X]) :- D >= 0, D =< 9, delta(q2, X). delta(q1, [D X]) :- D >= 0, D =< 9, delta(q2, X). delta(q2, [. X]) :- delta(q3, X). delta(q2, [D X]) :- D >= 0, D =< 9, delta(q2, X). delta(q3, [D X]) :- D >= 0, D =< 9, delta(q4, X). delta(q4, [e X]) :- delta(q5, X). delta(q4, [D X]) :- D >= 0, D =< 9, delta(q4, X). delta(q4, []). /* q4 is a final state. */ delta(q5, [+ X]) :- delta(q6, X). delta(q5, [- X]) :- delta(q6, X). delta(q5, [D X]) :- D >= 0, D =< 9, delta(q7, X). delta(q6, [D X]) :- D >= 0, D =< 9, delta(q7, X). delta(q7, [D X]) :- D >= 0, D =< 9, delta(q7, X). delta(q7, []). /* q7 is a final state. */ brb0164@faculty% gprolog?- consult( as2f18-1a ). compiling /home/brb0164/csce5400/prolog/as2f18-1a.pro for byte code... /home/brb0164/csce5400/prolog/as2f18-1a.pro compiled, 15 lines read bytes writ?- delta(q0, [-, 1, 0,., 8, 7]).?- delta(q0, [2,., 0, e, 7]).?- delta(q0, [6,., 6, 2, 5, e, -, 2, 7]).

2 Script done on Tue 11 Sep :43:00 AM CDT (b) L = {w w {0, 1}, w contains either three consecutive 0 s or three consecutive 1 s}. Example strings include 0111, 0000 and Script started on Tue 11 Sep :02:20 AM CDT brb0164@faculty% cat as2f18-1b.pro delta(q0, [0 X]) :- delta(q1, X). delta(q0, [1 X]) :- delta(q3, X). delta(q1, [0 X]) :- delta(q2, X). delta(q1, [1 X]) :- delta(q3, X). delta(q2, [0 X]) :- delta(q5, X). delta(q2, [1 X]) :- delta(q3, X). delta(q3, [0 X]) :- delta(q1, X). delta(q3, [1 X]) :- delta(q4, X). delta(q4, [0 X]) :- delta(q1, X). delta(q4, [1 X]) :- delta(q5, X). delta(q5, [0 X]) :- delta(q5, X). delta(q5, [1 X]) :- delta(q5, X). delta(q5, []). brb0164@faculty% gprolog?- consult( as2f18-1b ). compiling /home/brb0164/csce5400/prolog/as2f18-1b.pro for byte code... /home/brb0164/csce5400/prolog/as2f18-1b.pro compiled, 13 lines read bytes writ?- delta(q0, [0, 1, 1, 1]).?- delta(q0, [0, 0, 0, 0]).?- delta(q0, [0, 0, 0, 1, 1, 1, 0]).

3 Script done on Tue 11 Sep :03:47 AM CDT 2. Design nondeterministic finite automata to recognize the languages below. Test your NFA s using Prolog on the indicated strings in the language, as well as some strings which are not in the language. (a) L = {w w {0, 1}, some two 0 s are separated by a string whose length is 4i, for some i 0}. Example strings include (i = 0), (i = 1), (i = 2), etc. Script started on Tue 11 Sep :19:20 AM CDT brb0164@faculty% cat as2f18-2a.pro delta(q0, [0 X]) :- delta(q0, X). delta(q0, [0 X]) :- delta(q1, X). delta(q0, [1 X]) :- delta(q0, X). delta(q1, [0 X]) :- delta(q2, X). /* previous 0 is first in pair */ delta(q1, [0 X]) :- delta(q5, X). /* previous 0 and current 0 4i apart */ delta(q1, [1 X]) :- delta(q2, X). delta(q2, [0 X]) :- delta(q3, X). /* 1st symbol after first 0 */ delta(q2, [1 X]) :- delta(q3, X). delta(q3, [0 X]) :- delta(q4, X). /* 2nd symbol after first 0 */ delta(q3, [1 X]) :- delta(q4, X). delta(q4, [0 X]) :- delta(q1, X). /* 3rd symbol after first 0 */ delta(q4, [1 X]) :- delta(q1, X). delta(q5, [0 X]) :- delta(q5, X). delta(q5, [1 X]) :- delta(q5, X). delta(q5, []). brb0164@faculty:~/csce5400/prolog$ gprolog?- consult( as2f18-2a ). compiling /home/brb0164/csce5400/prolog/as2f18-2a.pro for byte code... /home/brb0164/csce5400/prolog/as2f18-2a.pro compiled, 15 lines read bytes writ?- delta(q0, [1, 0, 1, 0, 0, 1]).

4 ?- delta(q0, [0, 1, 1, 1, 1, 0]).?- delta(q0, [0, 1, 0, 1, 0, 1, 1, 0, 1, 0]). (4 ms) Script done on Tue 11 Sep :19:30 AM CDT (b) L = {w w a b c, the number of a s plus the number of b s is odd or the number of b s plus the number of c s is even}. Example strings include b, cc, aaabbcc, etc. Script started on Wed 11 Sep :05:16 PM CDT brb0164@faculty% cat as2f18-2b.pro delta(q0, [a X]) :- delta(q1, X). delta(q0, [b X]) :- delta(q6, X). delta(q0, [c X]) :- delta(q11, X). delta(q0, []). delta(q1, [a X]) :- delta(q2, X). /* # a s is odd */ delta(q1, [b X]) :- delta(q3, X). delta(q1, [c X]) :- delta(q5, X). delta(q1, []). delta(q2, [a X]) :- delta(q1, X). /* # a s is even */ delta(q2, [b X]) :- delta(q6, X). delta(q2, [c X]) :- delta(q11, X). delta(q2, []). delta(q3, [b X]) :- delta(q4, X). /* # a s is odd, # b s is odd */ delta(q3, [c X]) :- delta(q6, X). delta(q4, [b X]) :- delta(q3, X). /* # a s is odd, # b s is even */ delta(q4, [c X]) :- delta(q5, X). delta(q4, []). delta(q5, [c X]) :- delta(q5, X). delta(q5, []). delta(q6, [b X]) :- delta(q7, X). /* # a s is even, # b s is odd */ delta(q6, [c X]) :- delta(q8, X). delta(q6, []). delta(q7, [b X]) :- delta(q6, X). /* # a s is even, # b s is even */ delta(q7, [c X]) :- delta(q10, X). delta(q7, []). delta(q8, [c X]) :- delta(q9, X). /* # b s is odd, # c s is odd */ delta(q8, []). delta(q9, [c X]) :- delta(q8, X). /* # b s is odd, # c s is even */

5 delta(q10, [c X]) :- delta(q11, X). /* # b s is even, # c s is even */ delta(q10, []). delta(q11, [c X]) :- delta(q10, X). /* # b s is even, # c s is odd */ brb0164@faculty% gprolog?- consult( as2f18-2b ). compiling /home/brb0164/csce5400/prolog/as2f18-2b.pro for byte code... /home/brb0164/csce5400/prolog/as2f18-2b.pro compiled, 31 lines read bytes writ?- delta(q0, [b]).?- delta(q0, [c, c]).?- delta(q0, [a, a, a, b, b, c, c]).?- delta(q0, [a, a, c]). no Script done on Wed 11 Sep :28:24 PM CDT 3. Convert the NFA M=({q 1, q 2, q 3 }), {a, b}, δ,q 1, {q 2 }), where δ is: δ (q 1, ɛ) = {q 2 } δ (q 1, a) = {q 3 } δ (q 2, a) = {q 1 } δ (q 3, a) = {q 2 } δ (q 3, b) = {q 2, q 3 } into an equivalent DFA. Taking the e-closure, we get: d(q1, a) = {q1, q3} Let q4 = {q1, q3}, q5 = {q2, q3}, and q6 = {q1, q2}. Then:

6 d(q1, a) = q4 d(q2, a) = q1 d(q3, a) = q2 d(q3, b) = q5 d(q4, a) = q5 d(q4, b) = q5 d(q5, a) = q6 d(q5, b) = q5 d(q6, a) = q4 Final states are q1, q5 and q6. 4. A deterministic finite transducer is a 6-tuple M=(Q,Σ,,δ,q 0,F ), where 1) Q is a finite set of states. 2) Σ is a finite input alphabet. 3) is a finite output alphabet. 4) δ is a mapping from Q (Σ {ε}) to Q. Determinism implies that 1) either δ(q, a) contains at most one element for each a Σ, and δ(q, ε) =, or 2) δ(q, ε) contains one element, and for all a Σ, δ(q, a) =. That is, there can be no moves involving a choice between making an ε-move or consuming an input symbol. 5) q 0 Q is the initial state. 6) F Q is the set of final states. Note that a finite transducer accepts its input when it reaches a final state and outputs a string over for each input string accepted. (a) Construct a deterministic finite transducer whose input is of the form ( ) + $, the set of all ternary numbers terminated with a $, and whose output is the residue modulo 5 of the input. For example, on input 210$, the output is 1, on input $, the output is 4, and on input 11110$, the output is 0. (b) Test your DFT using Prolog on the above example strings in the language. Note that a DFT transition of the form δ (q 0, a) = (q 1, bc) is written in Prolog as: delta(q0,[a RestOfInput], InitialOutput, FinalOutput) :- append(initialoutput, [b, c], NewOutput), To run on the first string above, use the goal: delta(q0, [2, 1, 0, $ ], [], Output). The others are similar. Script started on Wed 11 Sep :32:28 PM CDT brb0164@faculty% cat dftf18-4.pro delta(p0, [0 RestOfInput], InitialOutput, FinalOutput) :-

7 delta(q0, RestOfInput, NewOutput, FinalOutput). delta(p0, [1 RestOfInput], InitialOutput, FinalOutput) :- delta(p0, [2 RestOfInput], InitialOutput, FinalOutput) :- delta(q2, RestOfInput, NewOutput, FinalOutput). delta(q0, [0 RestOfInput], InitialOutput, FinalOutput) :- delta(q0, RestOfInput, NewOutput, FinalOutput). delta(q0, [1 RestOfInput], InitialOutput, FinalOutput) :- delta(q0, [2 RestOfInput], InitialOutput, FinalOutput) :- delta(q2, RestOfInput, NewOutput, FinalOutput). delta(q0, [ $ RestOfInput], InitialOutput, FinalOutput) :- append(initialoutput, [0], NewOutput), delta(q1, [0 RestOfInput], InitialOutput, FinalOutput) :- delta(q3, RestOfInput, NewOutput, FinalOutput). delta(q1, [1 RestOfInput], InitialOutput, FinalOutput) :- delta(q4, RestOfInput, NewOutput, FinalOutput). delta(q1, [2 RestOfInput], InitialOutput, FinalOutput) :- delta(q0, RestOfInput, NewOutput, FinalOutput). delta(q1, [ $ RestOfInput], InitialOutput, FinalOutput) :- append(initialoutput, [1], NewOutput), delta(q2, [0 RestOfInput], InitialOutput, FinalOutput) :- delta(q2, [1 RestOfInput], InitialOutput, FinalOutput) :- delta(q2, RestOfInput, NewOutput, FinalOutput). delta(q2, [2 RestOfInput], InitialOutput, FinalOutput) :- delta(q3, RestOfInput, NewOutput, FinalOutput). delta(q2, [ $ RestOfInput], InitialOutput, FinalOutput) :- append(initialoutput, [2], NewOutput), delta(q3, [0 RestOfInput], InitialOutput, FinalOutput) :- delta(q4, RestOfInput, NewOutput, FinalOutput). delta(q3, [1 RestOfInput], InitialOutput, FinalOutput) :-

8 delta(q0, RestOfInput, NewOutput, FinalOutput). delta(q3, [2 RestOfInput], InitialOutput, FinalOutput) :- delta(q3, [ $ RestOfInput], InitialOutput, FinalOutput) :- append(initialoutput, [3], NewOutput), delta(q4, [0 RestOfInput], InitialOutput, FinalOutput) :- delta(q2, RestOfInput, NewOutput, FinalOutput). delta(q4, [1 RestOfInput], InitialOutput, FinalOutput) :- delta(q3, RestOfInput, NewOutput, FinalOutput). delta(q4, [2 RestOfInput], InitialOutput, FinalOutput) :- delta(q4, RestOfInput, NewOutput, FinalOutput). delta(q4, [ $ RestOfInput], InitialOutput, FinalOutput) :- append(initialoutput, [4], NewOutput), delta(pf, [], Output, Output). /* pf is the final state */ brb0164@faculty% gprolog?- consult( dftf18-4 ). compiling /home/brb0164/csce5400/prolog/dftf18-4.pro for byte code... /home/brb0164/csce5400/prolog/dftf18-4.pro compiled, 76 lines read bytes written, (4 ms)?- delta(p0, [2, 1, 0, $ ], [], Output). Output = [1]??- delta(p0, [0, 0, 1, 1, 2, 2, $ ], [], Output). Output = [4]??- delta(p0, [1, 1, 1, 1, 0, $ ], [], Output). Output = [0]?

9 Script done on Wed 11 Sep :32:55 PM CDT

1. (10 points) Draw the state diagram of the DFA that recognizes the language over Σ = {0, 1}

1. (10 points) Draw the state diagram of the DFA that recognizes the language over Σ = {0, 1} CSE 5 Homework 2 Due: Monday October 6, 27 Instructions Upload a single file to Gradescope for each group. should be on each page of the submission. All group members names and PIDs Your assignments in