TUGAS 4

TEORI BAHASA DAN OTOMATA

1.
      

  Dari diagram state di atas tentukan :
  a. ABAAAAB
  b. BBBBAAA
  c. BABABAB

                                                                              JAWABAN

a. ABAAAAB

 b.BBBBAAA

c.BABABAB

TUGAS 3

TEORI BAHASA & OTOMATA

DFSA/DFA

 

 

T = ({Q},{∑},δ,S,f)

Dimana

Q = (q0,q1,q2)

∑= a,b

S = q0

F = q1

Ditanya :      1. Buatlah tabel transisinya

                         2.bacalah input    a = abbabbaaa

                                                      b = bbbabbaa

                                                      c = ab

JAWABAN

1.      Tabel Transisi

	δ	a	b
→	q0	q0,q2	q1
*	q1	q1,q2	q1
	q2	-	q0,q1

2.      a. Jika T diberi input abbabbaaa dengan State awal (q0, abbabbaaa)

            maka :

               q0, abbabbaaa ┣ T (q0, bbabbaaa)

                                          ┣ T (q1, babbaaa)

                                          ┣ T (q1, abbaaa)

                                          ┣ T (q2, bbaaa)

                                          ┣ T (q1,baaa)

                                          ┣ T (q1,aaa)

                                          ┣ T (q1,aa)

                                                ┣ T (q1,a)

                              ┣ T (q1,e)

                                  Karena (q0, abbabbaaa) ┣ * T jadi abbabbaaa diterima T

         b. Jika T diberi input bbbabbaa dengan State awal(q0, bbbabbaa)

              maka :

             q0, bbbabbaa ┣ T (q1,bbabbaa)

                                      ┣ T (q1,babbaa)

                                      ┣ T (q1,abbaa)

                                      ┣ T (q2,bbaa)

                                      ┣ T (q0,baa)

                                      ┣ T(q1,aa)

                                      ┣ T(q1,a)

                                                                            ┣ T (q1,e)

         Karena (q0,bbbabbaa) ┣ * T jadi bbbabbaa diterima T

          c. Jika T diberi input ab dengan State awal (q0,ab)

              Maka :

             q0, ab ┣ T (q0,b)

                         ┣ T (q1,e)

Karena (q0,ab) ┣ * T jadi ab diterima T

TUGAS 2

TEORI BAHASA & AUTOMATA

 TUGAS 2 

1. UNION 
L = {0.1, 100, 110}
M = {100, 101, 0.00} UNION = {0.1, 100, 110, 101, 0.00}


 2. KONKATENASE 
L = {0.1, 100, 110}
M = {100, 101, 0.00}
KONKATENASE : {0.1, 100, 110, 0.1100, 0.1101, 0.10.00, 100100, 100101, 1000.00, 110100, 110101, 1100.00}


3. KLOSUR 
P = { A, BBB }
Q = { XX, Y }
 KLOSUR : P = {0,111}, {0,1111}, {111,0} Q ={11,0}, {111,0}

Teori Bahasa & Automata

nama : Novita Sabuna
nim    : 13110239

KONSEP SENTRAL – PANGKAT ALFABET

∑ = {A,B,C}

∑³=

{AAA,AAB,AAC,ABA,ACA,ACB,ABC,ACC,ABB,BBB,BBA,BBC,BAB,BAC,BCA,

BCB,BAA,BCC,CCC,CCA,CCB,CAC,CAB,CBC,CBA,CAA,CBB}

∑⁴=

{AAAA,AAAB,AAAC,AABA,AACA,AACB,AABC,ABBB,ACCC,ABAA,ACAA,

ACBA,ABCA,ABAC,ACAB,ABCB,ACBC,ABBC,ACCB,ABAB,ACAC,ABBA,ACCA,

AABB,AACC,ABCC,ACBB,BBBB,BBBA,BBBC,BBAB,BBAC,BBCA,BBAC,BAAA,

BCCC,BABB,BCBB,BACB,BCAB,BABC,BCBA,BACA,BCAC,BAAC,BCCA,BABA,

BCBC,BAAB,BCCB,BBAA,BBCC,BACC,BCAA,CCCC,CCCA,CCCB,CCAC,CCBC,

CCAB,CCBA,CAAA,CBBB,CACC,CBCC,CABC,CBAC,CACB,CBCA,CBAB,CABA,

CBBA,CAAB,CBCB,CACA,CBBC,CAAC,CCBB,CCAA,CBAA,CABB}