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Finite Automata
DFA Definition: A DFA is 5-tuple or quintuple M = (Q, ∑, δ, q0, A) where Q is non-empty, finite set of states. ∑ is non-empty, finite set of input alphabets. δ is transition function, which is a mapping from Q x ∑ to Q. q0 ∈ Q is the start state. A ⊆ Q is set of accepting or final states. Note: For each input symbol a, from a given state there is exactly one transition (there can be no transitions from a state also) and we are sure (or can determine) to which state the machine enters. So, the machine is called deterministic machine. Since it has finite number of states the machine is called Deterministic finite machine or Deterministic Finite Automaton or Finite State Machine (FSM). The language accepted by DFA is L(M) = { w | w ∈ ∑* and δ*(q0, w) ∈ A } The non-acceptance of the string w by an FA or DFA can be defined in formal notation as: L(M) = { w | w ∈ ∑* and δ*(q0, w) ∉ A }
DFA to accepting strings of a's and b's starting with the string ab
a,b q0 b q3 a,b Fig.2.7 Transition diagram to accept string ab(a+b)* So, the DFA which accepts strings of a's and b's starting with the string ab is given by M = (Q, ∑ , δ, q0, A) where Q = {q0, q1, q2, q3} ←Σ→ ∑ = {a, b} a b δ q0 is the start state q1 q3 →q0 A = {q2}. δ is shown the transition table 2.4. q1 q3 q2 ← States → q2 q3 q2 q3 q2 q3 a a q1 b q2
Draw a DFA to accept string of 0's and 1's ending with the string 011.
1 q0
0 q1
1 0
q2 0
Obtain a DFA to accept strings of a's and b's having a sub string aa
b q0 a b q1 a q2
Obtain a DFA to accept strings of a's and b's except those containing the substring aab.
b q0 a b q1 a
a q2 b q3
Obtain DFAs to accept strings of a's and b's having exactly one a, b q0 a b q0 b q0 a b q1 a b q2 a a b q1 a a,b q2
a, q1b b q3 a q4 a, b
Obtain a DFA to accept strings of a's and b's having even number of a's and b's
The machine to accept even number of a's and b's is shown in fig.2.22.
a q0 a b q2 a Fig.2.22 DFA to accept even no. of a's and b's b a q3 b b q1
Regular language
Definition: Let M = (Q, ∑, δ, q0, A) be a DFA. The language L is regular if there exists a machine M such that L = L(M).
Applications of Finite Automata
String matching/processing Compiler Construction The various compilers such as C/C++, Pascal, Fortran or any other compiler are designed using the finite automata. The DFAs are extensively used in the building the various phases of compiler such as Lexical analysis (To identify the tokens, identifiers, to strip of the comments etc.) Syntax analysis (To check the syntax of each statement or control statement used in the program) Code optimization (To remove the un wanted code) Code generation (To generate the machine code)
Other applications
The concept of finite automata is used in wide applications. It is not possible to list all the applications, as there are infinite numbers of applications. This section lists some applications: 1. Large natural vocabularies can be described using finite automaton which includes the applications such as spelling checkers and advisers, multilanguage dictionaries, to indent the documents, in calculators to evaluate complex expressions based on the priority of an operator etc. to name a few. Any editor that we use uses finite automaton for implementation. 2. Finite automaton is very useful in recognizing difficult problems i.e., sometimes it is very essential to solve an un-decidable problem. Even though there is no general solution exists for the specified problem, using theory of computation, we can find the approximate solutions.
3. Finite automaton is very useful in hardware design such as circuit verification, in design of the hardware board (mother board or any other hardware unit), automatic traffic signals, radio controlled toys, elevators, automatic sensors, remote sensing or controller etc. 4. In game theory and games wherein we use some control characters to fight against a monster, economics, computer graphics, linguistics etc., finite automaton plays a very important role.
Non-deterministic finite automata (NFA)
Definition: An NFA is a 5-tuple or quintuple M = (Q, ∑, δ, q0, A) where Q = finite non empty set of states. ∑ = Non-empty, finite set of input alphabets. δ = Transition function which is a mapping from Q x {∑ U ε} 2Q. This function shows the change of state from one state to a set of states based on the input symbol. q0 ∈ Q is the start state. A ⊆ Q is set of final states.
Acceptance of language
Definition: Let M = (Q, ∑, δ, q0, A) be a DFA where Q is set of finite states, ∑ is set of input alphabets (from which a string can be formed), δ is transition function from Q x {∑Uε} to 2Q, q0 is the start state and A is the final or accepting state. The string (also called language) w accepted by an NFA can be defined in formal notation as: L(M) = { w | w ∈ ∑* and δ*(q0, w) = Q with atleast one Component of Q in A}
Obtain an NFA to accept the following language L = {w | w ∈ ababn or aban where n ≥ 0} The machine to accept either ababn or aban where n ≥ 0 is shown below:
b q1 a ε q0 ε a q5 a q6 b q7 q2 b q3 a q4
Conversion from NFA to DFA
Let MN = (QN, ∑N, δN, q0, AN) be an NFA and accepts the language L(MN). There should be an equivalent DFA MD = (QD, ∑D, δD, q0, AD) such that L(MD) = L(MN). The procedure to convert an NFA to its equivalent DFA is shown below: Step1: The start state of NFA MN is the start state of DFA MD. So, add q0(which is the start state of NFA) to QD and find the transitions from this state. The way to obtain different transitions is shown in step2. Step2: For each state [qi, qj,….qk] in QD, the transitions for each input symbol in ∑ can be obtained as shown below: 1. δD([qi, qj,….qk], a) = δN(qi, a) U δN(qj, a) U ……δN(qk, a) = [ql, qm,….qn] say. 2. Add the state [ql, qm,….qn] to QD, if it is not already in QD. 3. Add the transition from [qi, qj,….qk] to [ql, qm,….qn] on the input symbol a iff the state [ql, qm,….qn] is added to QD in the previous step. Step3: The state [qa, qb,….qc] ∈ QD is the final state, if at least one of the state in qa, qb, ….. qc ∈ AN i.e., at least one of the component in [qa, qb,….qc] should be the final state of NFA. Step4: If epsilon (∈) is accepted by NFA, then start state q0 of DFA is made the final state.
Convert the following NFA into an equivalent DFA.
0 q0 0, q 0, 1 q 1 2 1
Step1: q0 is the start of DFA (see step1 in the conversion procedure). So, QD = {[q0]} (2.7)
Step2: Find the new states from each state in QD and obtain the corresponding transitions. Consider the state [q0]: When a = 0 δD([q0], 0) = δN([q0], 0) = [q0, q1] (2.8) = δN([q0], 1) = [q1] (2.9)
When a = 1 δD([q0], 1)
Since the states obtained in (2.8) and (2.9) are not in QD(2.7), add these two states to QD so that QD = {[q0], [q0, q1], [q1] } (2.10)
The corresponding transitions on a = 0 and a = 1 are shown below. ∑ 0 δ [q0] [q0, q1] Q [q0, q1] [q1] Consider the state [q0, q1]: When a = 0 δD([q0, q1], 0) = = = = δN([q0, q1], 0) δN(q0, 0) U δN(q1, 0) {q0, q1} U {q2} [q0, q1, q2] (2.11) δN([q0, q1], 1) δN(q0, 1) U δN(q1, 1) {q1} U {q2} [q1, q2] (2.12) 1 [q1]
When a = 1 δD([q0, q1], 1)
= = = =
Since the states obtained in (2.11) and (2.12) are the not defined in QD(see 2.10), add these two states to QD so that QD = {[q0], [q0, q1], [q1], [q0, q1, q2], [q1, q2] } (2.13) 6
[q1. q1] [q0. q2] [q2] 0 [q0.16) 7 . q2] When a = 1 δD([q1]. q2]. q1]. q1. q1.and add the transitions on a = 0 and a = 1 as shown below: ∑ δ [q0] Q [q0. q2] [q1. 1) = [q2] (2. 0) = δN([q1]. q2] 1 [q1] [q1. 1) = δN([q1]. q1] [q0. q2]. [q2]} and add the transitions on a = 0 and a = 1 as shown below: ∑ δ [q0] Q [q0. q2] [q2] 1 [q1] [q1. q2] [q1.15) Since the states obtained in (2. q1] [q1] [q0. q2] Consider the state [q1]: When a = 0 δD([q1]. [q0.13). q1. [q0.14) and (2. q2] [q2] (2. add the state q2 to QD so that QD = {[q0]. q1] [q1] [q0.15) are same and the state q2 is not in QD(see 2. q1. [q1]. 0) = [q2] (2.14) 0 [q0. q1.
q1.q1. 1) δN(q0. 1) U δN(q1. 0) δN(q0. q2] [q2] [q1. q1] [q0.q1.19) When a = 1 8 .16).q2].18) are not new states (are already in QD.q2] (2. q2] [q2] 0 [q0. But. see 2. 1) = = = = δN([q0. q2] (2.q2]. 0) U δN(q2.q2]: When a = 0 δD([q0.q1} U {q2} U {φ} [q0.q2]: When a = 0 δD([q1. q2] Q Consider the state [q1.Consider the state [q0.q1.q1. 1) {q1} U {q2} U {q2} [q1.q1.q2]. 0) = = = = δN([q0. 0) U δN(q2. 0) U δN(q1. do not add these two states to QD. 0) {q2} U {φ} [q2] (2. q2] [q1.q2] 1 [q1] [q1. q1. 0) = = = = δN([q1.18) Since the states obtained in (2.q1.17) and (2. the transitions on a = 0 and a = 1 should be added to the transitional table as shown below: ∑ δ [q0] [q0.q2].q2].17) When a = 1 δD([q0.q2]. 0) δN(q1. q2] [q2] [q0. 1) U δN(q2.q1. 0) {q0. q1] [q1] [q0.
The final transitional table is shown in table 2. q1] [q1] [q0. q2] [q2] Q Consider the state [q2]: When a = 0 δD([q2].q2 ] [q1.35. q2] [q1. the transitions on a = 0 and a = 1 should be added to the transitional table as shown below: ∑ δ [q0] [q0. q1. q2] [q2] 1 [q1] [q1. But. 1) = = = = δN([q1. q1. q1.q2] [q0. q2] [q2] [q1. q2] [q2] 0 [q0. see 2.22) Since the states obtained in (2. 1) {q2} U {q2} [q2] (2.21) and (2.16). q2] [q2] 1 [q1] [q1.22) are not new states (are already in QD. q1] [q0.20) are not new states (are already in QD see 2.14. and final DFA is shown in figure 2. do not add these two states to QD. q2] [q2] [q1. 0) When a = 1 δD([q2].q2].q2]. q1. 1) = [q2] (2. But.q1] [q1] 0 [q0.q1.δD([q1. 1) = δN([q2]. 1) U δN(q2.q2] [q2] [q2] φ 9 . δ [q0] [q0. q2] [q2] [q2] [q0. the transitions on a = 0 and a = 1 should be added to the transitional table.16).21) = δN([q2]. 0) = {φ} (2. q1] [q0.q1.19) and (2. do not add these two states to QD. 1) δN(q1.20) Since the states obtained in (2. q2] [q2] [q0.
a) = {1} (B) (A) When input is b: = δN(0. a) = δN(0.35 The DFA Convert the following NFA to its equivalent DFA.[q 0 ] 0 1 [q 0 . q 2 ] 1 [q 1 . 1 [q 0 . 1 [q 2 ] Fig. q 1 ] [q 1] 0 1 0. 0 1 ε ε 0 a 1 b 2 ε 3 ε 6 b ε 7 ε 4 a 5 ∈ 8 ∈ 9 Let QD = {0} Consider the state [A]: When input is a: δ(A. q 2 ] 0. b) δ( A. q 1 . b) = {φ} Consider the state [B]: 10 .2.
b) = {2} = {2.When input is a: = δN(1. 6}.9 also. 9}. 7.9}. a) = {φ} When input is b: = δN(1. 7. a) {5} {5. 6} {3. 9. 8. 8. a) = = = δN({2.e. So. 9} (ascending order) (D) 11 . 8. due to ε-transitions (or without giving any input) there can be transition to states 3.4. δ(C. 4. 4. Therefore. b) δ( C. 5. 9. a) = {3.4. 8. 9} = D When input is b: = δN({2. 4.6.3. 4.6. 6}.3. 3.4. 5. δ(C. Therefore. 4. b) = {7} = {7. b) δ( B. 8. 6} {3. 8. 5. 4.5.4. 6. b) = {3. in state 5 due to ε-transitions.8.3. 8. from state 7 the states that are reachable without any input (i. 6. 4. the states reachable are {8.6. 3. Therefore. b) = {2. 6. 6. 4. 6. 4. all these states are reachable from state 2.6. 9. a) = = = = δN({3. 4. 4.9} (C) This is because. a) δ(B. a) {5} {5. 3. 6} = {3. 9} (ascending order) (D) This is because. 3. 9} = E Consider the state [D]: When input is a: δ(D.6. 9}(ascending order) (E) This is because.9}. δ(B. 3. 9.9} = C Consider the state [C]: When input is a: = δ(C. ε-transition) are {8.. 3.4. in state 2. 9. 6. 8.
36 a A a B b C b E b Fig.4. 8. 5. b) = = = δN({3. D and E. since 9 (final state of NFA) is present in C.7. 3.When input is b: δ(D. a) = = = When input is b: = δ(E. 6. 4. 8. 8. 8.15. 6} {3. 9}(ascending order) (D) δN({3. 4. 4. b) {7} {7.9}. 6. 3. 4. The final transition diagram of DFA is shown in figure 2.D and E are final states. we can stop at this point and the transition table for the DFA is shown in table 2.4. ∑ δ A B C D E a B D D D b C E E E Q Table 2. 8.6. 4.8. 7. 9.7. 6} {3. 9}(ascending order) (E) Since there are no new states.9}. 9. a) {5} {5. 3.6.15 Transitional table The states C. 9. 6} {3. 9} (ascending order) (E) Consider the state [E]: When input is a: = δ(E. 2.9}.5. 184.108.40.206.8. 4. 7. b) = = = = δN({3. b) {7} {7.36 The DFA a a b D 12 . 8.
LS. Set of strings of a's and b's ending with the string abb Set of strings of a's and b's starting with the string ab. The expressions obtained by applying any of the rules from 1-4 are regular expressions. b. 2. Set of strings of a's and b's having a sub string aa. a is a regular expression which indicates the language containing only {a} 4. then a. Regular expressions (a+b)* (a+b)*abb ab(a+b)* (a+b)*aa(a+b)* a*b*c* Meaning Set of strings of a's and b's of any length including the NULL string. Set of string consisting of at least one 'a' followed by string consisting of at least one 'b' followed by string consisting of at least one 'c'. φ is a regular expression denoting an empty language.1 shows some examples of regular expressions and the language corresponding to these regular expressions.Regular Languages Regular expression Definition: A regular expression is recursively defined as follows. R.. R+S is a regular expression corresponding to the language LRULS.S is a regular expression corresponding to the language LR. The table 3. If R is a regular expression denoting the language LR and S is a regular expression denoting the language LS. ε-(epsilon) is a regular expression indicates the language containing an empty string. Set of string consisting of at least one 'a' followed by string consisting of at least one 'b' followed by string consisting of at least 13 a+b+c+ aa*bb*cc* . R* is a regular expression corresponding to the language LR*. 1. c. 3. Set of string consisting of any number of a's(may be empty string also) followed by any number of b's(may include empty string) followed by any number of c's(may include empty string). 5.
a variety of regular expressions can be obtained for a language and all are equivalent. ab. Note: This regular expression can also be represented using set notation as L(R) = {(aa + ab + ba + bb)n | n ≥ 0} Obtain a regular expression to accept a language consisting of strings of a's and b's of odd length. The language may even consist of an empty string denoted by ε. String of a's and b's of odd length can be obtained by the combination of the strings aa. So. ab. String of a's and b's of even length can be obtained by the combination of the strings aa. ba and bb. So. the regular expression can be of the form (aa + ab + ba + bb)* The * closure includes the empty string. Set of strings of a's and b's ending with either a or bb Set of strings consisting of even number of a's followed by odd number of b's Set of strings of 0's and 1's ending with three consecutive zeros(or ending with 000) Set consisting of even number of 1's Table 3. the regular expression can also be represented as (a+b) (aa + ab + ba + bb)* Note: Even though these two expression are seems to be different. ab. the language corresponding to those two expression is same.1 Meaning of regular expressions Obtain a regular expression to accept a language consisting of strings of a's and b's of even length. the regular expression can be of the form (aa + ab + ba + bb)* (a+b) String of a's and b's of odd length can also be obtained by the combination of the strings aa. 14 .(a+b)* (a + bb) (aa)*(bb)*b (0+1)*000 (11)* 'b' followed by string consisting of at least one 'c'. So. ba and bb followed by either a or b. ba and bb preceded by either a or b. So.
q1. We can construct an NFA which accepts either L(R1) or L(R2) which can be represented as L(R1 + R2) as shown in figure 3. φ. ∑2. Let M2 = (Q2.3 To accept the language L(R1 + R2) 15 . ε and a are regular expressions. δ.3. δ2. L(R1) ε q0 ε q1 f q2 f M1 M2 L(R2) ε qf ε Fig. 3. the corresponding machines to recognize these expressions are shown in figure 3. q0.1. Proof: By definition. Case 1: R = R1 + R2. where q is the start state and f is the final state of machine M. So.1. f2) be a machine which accepts the language L(R2) corresponding to the regular expression R2.S and R* are regular expressions which clearly uses three operators '+'.Obtain NFA from the regular expression Theorem: Let R be a regular expression. 3. δ1. ∑1. L(R) q M f Fig 3.1. A) which accepts L(R). ∑.2.2 Schematic representation of FA accepting L(R) In the definition of a regular expression it is clear that if R and S are regular expression.c respectively.'.b and 3.1 NFAs to accept φ. Let M1 = (Q1.a. '-' and '. f1) be a machine which accepts the language L(R1) corresponding to the regular expression R1. q0 qf q0 ε (b) qf q0 a (c) qf φ (a) Fig 3. q2. ε and a The schematic representation of a regular expression R to accept the language L(R) is shown in figure 3. Let us take each case separately and construct equivalent machine. Then there exists a finite automaton M = (Q. then R+S and R.
Thus. Step 1: The machine to accept 'a' is shown below. without any input there will be a transition from state f1 to state q2. 3. We can construct an NFA which accepts L(R1) followed by L(R2) which can be represented as L(R1 . 16 . ε (a) q0 ε q1 M1 f L(R1) ε ε (b) q0 ε q1 f M1 ε ε qf ε qf Fig.5 To accept the language L(R1)* It is clear from figure 3. becomes the final state of machine M and accepts the language L(R1.5. 3. R2) It is clear from figure 3.4To accept the language L(R1 . q0 is the start state of the combined machine and qf is the final state of combined machine M.4 that the machine after accepting L(R1) moves from state q1 to f1.5. L(R1) L(R2) ε q1 M1 q2 M2 f f Fig.R2). We can construct an NFA which accepts either L(R1)*) as shown in figure 3. Here. q1 which is the start state of machine M1 becomes the start state of the combined machine M and f2 which is the final state of machine M2. It can also be represented as shown in figure 3. Case 3: R = (R1)*. the machine moves to f2 which is the final state. Since there is a ε-transition. q0 is the start state qf is the final state. Obtain an NFA which accepts strings of a's and b's starting with the string ab.It is clear from figure 3. R2) as shown in figure 3. The regular expression corresponding to this language is ab(a+b)*. upon accepting L(R2).5 that the machine can either accept ε or any number of L(R1)s thus accepting the language L(R1)*.a.3 that the machine can either accept L(R1) or L(R2).4.b. Case 2: R = R1 . In state q2. Here. R2.
ε a ε 0 a 1 b 2 ε 3 ε 4 5 ε 8 ε 9 6 b ε 7 ε Fig. 0 a 1 b 2 Step 6: The machine to accept ab(a+b)* is shown below. 3.6 To accept the language L(ab(a+b)*) 17 .4 a 5 Step 2: The machine to accept 'b' is shown below. ε a ε 2 ε 3 ε 4 5 ε 8 ε 9 6 b ε 7 ε Step 5: The machine to accept ab is shown below. ε 3 ε 6 b 7 ε 4 a 5 ε 8 Step 4: The machine to accept (a+b)* is shown below. 6 b 7 Step 3: The machine to accept (a + b) is shown below.
Then substitute the regular expressions appropriately in the equation 3.3) Obtain a regular expression for the FA shown below: 0 q0 0 q2 1 1 1 0 q3 0. Any graph can be reduced to the graph shown in figure 3.9. The general procedure to obtain a regular expression from FA is shown below. 2.2) 3.1) Note: 1.Obtain the regular expression from FA Theorem: Let M = (Q.9 Generalized transition graph r q1 r where r1. The regular expression for this can take the form: r = r1*r2 (r4 + r3r1*r2)* (3. Consider the generalized graph r1 q0 r Fig. δ. the regular expression can be of the form r = r1*r2 r4* (3. q0.9. Then there exists an equivalent regular expression R for the regular language L such that L = L(R). r2. If q0 and q1 are the final states then the regular expression can be of the form r = r1* + r1*r2 r4* (3. A) be an FA recognizing the language L. r3 and r4 are the regular expressions and correspond to the labels for the edges. If r3 is not there in figure 3. 3.1 and obtain the final regular expression. ∑.1 q1 The figure can be reduced as shown below: 01 q0 10 18 .
0 q0 1 q1 1 The state q0 is the final state and at this point it can accept any number of 0's which can be represented using notation as 0* q1 is also the final state. We can match an identifier or a decimal number or we can search for a string in the text. So. q2 1 Since. state q2 is the dead state. Applications of Regular Expressions Pattern Matching refers to a set of objects with some common properties. So. An application of regular expression in UNIX editor ed. it can be removed and the following FA is obtained. the final regular expression is obtained by adding 0* and 0*11*. the regular expression is R. to reach q1 one can input any number of 0's followed by 1 and followed by any number of 1's and can be represented as 0*11* So.It is clear from this figure that the machine accepts strings of 01's and 10's of any length and the regular expression can be of the form (01 + 10)* What is the language accepted by the following FA 0 q0 1 q1 1 0 0.E = 0* + 0*11* = 0* ( ∈ + 11*) = 0* ( ∈ + 1+) = 0* (1*) = 0*1* It is clear from the regular expression that language consists of any number of 0's (possibly ε) followed by any number of 1's(possibly ε). 19 .
Note that the editor ed accepts the regular expression and searches for that particular pattern in the text.In UNIX operating system. For example. As the input can vary dynamically. if the command specified is /acb*c/ then the editor searches for a string which starts with ac followed by zero or more b's and followed by the symbol c. 20 . it is challenging to write programs for string patters of these kinds. we can use the editor ed to search for a specific pattern in the text.
21 .Properties of regular languages Pumping Lemma Used to prove certain languages like L = {0n1n | n ≥ 1} are not regular. Closure properties of regular languages Used to build recognizers for languages that are constructed from other languages by certain operations. 1 0 1 4 0 1 5 0. Automata for intersection of two regular languages Decision properties of regular languages Used to find whether two automata define the same language Used to minimize the states of DFA eg. which varies from 1 to ∞. This implies DFA has no memory to remember arbitrary 'n'. 00*11* is not the regular expression defining L. state 3 remembers that two 0's have come and from there state 5 remembers that two 1's are accepted. which is not possible with a finite state machine. we have to have infinite states. Design of switching circuits. In other words if we have to remember n. which has finite number of states. Let L= {0212} 0 1 2 0 3 1 1 6 0. 1 State 6 is a trap state. Pumping Lemma for regular languages Let L = {0n1n | n ≥ 1} There is no regular expression to define L. Ex.
y=an and z = ε.a3 ----an-1 .Pumping Lemma for Regular Languages Theorem: Let L be a regular language. |xy| ≤ n 3.a2 . Let w= a1. |y| > 0 2. Then there exists a constant 'n' (which depends on L) such that for every string w in L such that |w| ≥ n.an+9 is accepted and so on. For all k ≥ 0.a3----an and is in L. the string xykz is also in L. Therefore. Let the start state be P1. we can break w into three strings. xykz = a1 -----. Let w = xyz where x= a1. w=xyz. such that: 1.an is accepted a1 -----. |w| = n ≥ n.an-1 is accepted a1 -----.a2 .an-1 (an)k ε k=0 k=1 k=2 a1 -----. PROOF: Let L be regular defined by an FA having 'n' states. 22 .an+1 is accepted k=10 a1 -----.
-----. It should never be used to show that some language is regular.2. (Because |xy| ≤ n).∞ Put k=0. abbbba. To prove that L={w|w is a palindrome on {a. the break up of x. we get an-l bn ∉ L. xy contain only a's. aba. That is an-l (al)k bn ∈L..…} 23 . Hence the Language is not regular. resulting string is not in L. To prove that L={w|w ε anbn.b}*} is not regular. Example 2.------∞. w=xykz. should belong to L.This is to be used to show that. where l > 0 (Because |y| > 0). i.Uses of Pumping Lemma: . PL. Since 2n > n and L is regular it must satisfy Example 1. General Method of proof: (i) Select w such that |w| ≥ n (ii) Select y such that |y| ≥ 1 (iii) Select x such that |xy| ≤ n (iv) Assign remaining string to z (v) Select k suitably to show that. Let n is the constant (PL Definition). Then.e. where k=0.1. If you want to show that language is regular.1. such that |w|=2n. DFA or NFA. certain languages are not regular. Let L be regular. write separate expression. Contradiction. where n ≥ 1} is not regular Proof: L. Consider a word w in Let w = anbn. y and z can be as follows from the definition of PL . Let |y|=l. L={aabaa.2. for all k=0.
2. y and z can be as follows from the definition of PL w=xykz. where k=0. Let n is the constant (PL Definition). such that |w|=2n+1. (1+m) ≥ 2 because m ≥ 1 Limiting case p=n+2 (p-m) ≥ 2 since m ≤n Example 4. To prove that L={ all strings of 1's whose length is prime} is not regular.e..17 . Let w = anban. Example 3. where l > 0 (Because |y| > 0). it is not a palindrome.----} Proof: Let L be regular.------∞. That is an-l (al)k ban ∈L.1. 04 .1. Consider a word w in L.111 .----} 24 .025 . That is.Proof: Let L be regular.∞. xy contain only a's. for all k=0.016 . hence the language is not regular . L={12. the break up of x. we get an-l b an∉ L. should belong to L. To prove that L={ 0i2 | i is integer and i >0} is not regular. i.09 .. Contradiction.-----. Put k=0.15 . because. (Because |xy| ≤ n). L={02.2. i. by PL xykz ∈L | xykz | = | xz | + | yk | Let k = p-m be prime = (p-m) + m (p-m) = (p-m) (1+m) -----this can not if p-m ≥ 2 or 1+m ≥ 2 1.2. Let w = 1p where p is prime and | p| = n +2 Let y = m. Let |y|=l. Since 2n+1 > n and L is regular it must satisfy PL. 13 .e.
Let w = 0n2 where |w| = n2 ≥ n by PL xykz ∈L. m ≥0 } 25 . Exercises for Readers: a) Show that following languages are not regular L={anbm | n.Proof: Let L be regular. m ≥1 } L={an | n is a perfect square } L={an | n is a perfect cube } b) Apply pumping lemma to following languages and understand why we cannot complete proof L={anaba | n ≥0 } L={anbm | n. m ≥0 and n<m } L={anbm | n.1. m ≥0 and n>m } L={anbmcmdn | n. for all k = 0.--Select k = 2 | xy2z | = | xyz | + | y | = n2 + Min 1 and Max n Therefore n2 < | xy2z | ≤ n2 + n n2 < | xy2z | < n2 + n + 1+n adding 1 + n ( Note that less than or equal to is n2 < | xy2z | < (n + 1)2 replaced by less than sign) Say n = 5 this implies that string can have length > 25 and < 36 which is not of the form 0i2.
a3 b3. a5b5----} RE=ab(ab)* Closure Under Complementation Theorem : If L is a regular language over alphabet S.L is also a regular language.-----} L1∪L2 = {a. The closure (star) of a regular language is regular. The complement of a regular language is regular.a2b2. The difference of two regular languages is regular. The concatenation of regular languages is regular.a4.-----} L2={ab.a5. 6. a2 b2.a6. The inverse homomorphism of a regular language is regular Closure under Union Theorem: If L and M are regular languages. then so is L ∪M.-----} L1∪L2 = {ab. 5. The reversal of a regular language is regular. The intersection of two regular languages is regular. then L = Σ* . a3b3. L1={ab. 26 . 4.-----} L2={a2.a3. 8. 2. a4b4. The union of two regular languages is regular. 9.a3. 3.a2. L1={a. a3b3. a4b4. 7.Closure Properties of Regular Languages 1. A homomorphism (substitution of strings for symbols) of a regular language is regular. Example 1.----} RE=a(a)* Example 2.a5b5.a4.
That is L(A) = (0+1)*01.-----} Σ* -L1 = {e.a3.a4.a6.a5.a2. The complement of L(A) is therefore all string of 0's and 1's that do not end in 01 27 .-----} RE=(aa)* Example 2. A that accepts all and only the strings of 0's and 1's that end in 01. Consider a DFA.Example 1. L1={a.
a6.----} RE=aa(aa)* Example 2. Σ.a7b7-----} L2={a2 b2. Example 1.-----} L1L2 = {a2. where B is the DFA (Q. L1={a. Consider a DFA that accepts all those strings that have a 0. Then L = L(B). F). a6b6. δ. Σ. Proof: .L is also a regular language.a4. q0.a4. A=(Q.Theorem: .a3. Consider a DFA that accepts all those strings that have a 1.-----} L1∩L2 = φ RE= φ Example 3. then. q0. but the accepting states of A have become non-accepting states of B.a5. 28 . Q-F). L = Σ* . and vice versa.a4. w) is in Q-F. then w is in L(B) if and only if δ^ ( q0.a2. which occurs if and only if w is not in L(A).a3b3. then so is L ∩ M.-----} L2={a2. Closure Under Intersection Theorem : If L and M are regular languages.a5b5. δ. a4b4.Let L =L(A) for some DFA. B is exactly like A.a6. That is. L1={ab.If L is a regular language over alphabet Σ.a6.
The product of above two automata is given below. Then state qr means that we have seen only once 0's. DFA for L1 ∩ L2 = φ (as no string has reached to final state (2.4)) 29 . Then pr represents only the initial condition. while state ps represents the condition that we have seen only 1's. The accepting state qs represents the condition where we have seen both 0's and 1's. in which we have seen neither 0 nor 1. Example 4 Write a DFA to accept the intersection of L1=(a+b)*a and L2=(a+b)*b that is for L1 ∩ L2. This automaton accepts the intersection of the first two languages: Those languages that have both a 0 and a 1.
30 . then so is L – M.Example 5 Find the DFA to accept the intersection of L1=(a+b)*ab (a+b)* L2=(a+b)*ba (a+b)* that is for L1 ∩ L2 and DFA for L1 ∩ L2 Closure Under Difference Theorem: If L and M are regular languages. Example.
01} LR={100.a4.111. we can use ∈. Proof.a) -> p Since MR is derivable from M.q0R.a6. 2. -----}.FR) defines the language LR.111. bab. Let FA. F). That is LR = {100. The FA is as given below 31 . MR = (QR.a7. 111}. Let L = {001. ΣR .-----} L1-L2 = {a. q0. Swap initial and final states.10} To prove that regular languages are closed under reversal.-----} L2={a2. That is L = {ab.a3.1}.a5. having only one final state. Reverse all the transitions.transitions from the final states going to a common final state. ΣR = Σ. Σ . 3. If there are more than one final states. iff δ (q. M = (Q. If L is regular we can show that LR is also regular. δ. L={001.01. and δR (p. aab. so is LR Ex. q0R=F.a3.10. R L is a language consisting of the reversals of the strings of L. δ R. As L is regular it can be defined by an FA.aaab.a)-> q. Where QR = Q. The proof implies the following method 1.a5.111}. be a language over Σ={0. LR is also regular. Create a new start state p0 with transition on ∈ to all the accepting states of original DFA Example Let r=(a+b)* ab define a language L.a7----} RE=a(a)* Reversal Theorem : If L is a regular language.FR=q0.01. 10.L1={a.
32 . It is shown in the following figure.The FA for LR can be derived from FA for L by swapping initial and final states and changing the direction of each edge.
Homomorphism A string homomorphism is a function on strings that works by substituting a particular string for each symbol. The function h defined by h(0)=ab h(1)=c is a homomorphism. Theorem: If L is a regular language over alphabet Σ. h applied to the string 00110 is ababccab L1= (a+b)* a (a+b)* h : {a. b} {0. 1}* Resulting : h1(L) = (01 + 11)* 01 (01 + 11)* h2(L) = (101 + 010)* 101 (101 + 010)* h3(L) = (01 + 101)* 01 (01 + 101)* Inverse Homomorphism 33 . then h (L) is also regular. Example. and h is a homomorphism on Σ.
Is the language described empty? 2. Then h-1(L) is the language of regular expression (ba)*. then 1. Let L be the language of regular expression (00+1)*. R=R1 + R2. If we convert an NFA to DFA and then convert the DFA to a regular expression it takes the time O(n34n 2n) Regular Expression to Automaton Conversion Regular expression to ε-NFA takes linear time – O(n) on a regular expression of length n. R=R1* Then L(R) is not empty. R=(R1) Then L(R) is empty if and only if L(R1) is empty since they are the same language. Converting Among Representations Converting NFA's to DFA's Time taken for either an NFA or -NFA to DFA can be exponential in the number of states of the NFA. Example. The running time of NFA to DFA conversion including ε transition is O(n3 2n). 2. 3. Let h be the homomorphism defined by h(a)=01 and h(b)=10. Do two descriptions of a language actually describe the same language? This question is often called "equivalence" of languages.Theorem: If h is a homomorphism from alphabet S to alphabet T. DFA to NFA Conversion Conversion takes O(n) time for an n state DFA. Therefore the bound on the running time is O(n3s) where s is the number of states the DFA actually has. then h-1 (L) is also a regular language. Testing Emptiness of Regular Languages Suppose R is regular expression. 3 34 . It always includes at least ε 4. and L is a regular language over T. Decision Properties of Regular Languages 1. conversion takes O(n34n) by substitution method and by state elimination method conversion takes O(n3) time. Computation of DFA takes O(n3) time where number of states of DFA can be 2n. Is a particular string w in the described language? 3. Computing ε-Closure of n states takes O(n3) time. Conversion from ε-NFA to NFA takes O(n3) time. Then L(R) is empty if and only if either L(R1) or L(R2) is empty. R= R1R2. Automaton to Regular Expression Conversion For DFA where n is the number of states. Then L(R) is empty if and only if both L(R1) and L(R2) are empty.
we can convert to an ε -NFA with almost 2s states. each of which requires O(s2) time.a) = p. in O(s) time. …. If L is represented by a DFA. R=(R1)* Then L(R) is not empty. has 2 stages. 3.closure has to be computed. stop. mark any pair (r. The FA is shown in Fig 1. After a pass in which no new pairs are marked. On each subsequent pass. a. Algorithm 1: List all unordered pair of states (p. This test takes O(n) time If the representation is NFA. Let L = {∈. if there is at least one string. such that one of δ^ (p. w. is w in L. 2. if w is of length n. Simulation of the above takes O(ns2) time on an input w of length n Minimization of Automata (Method 1) Let p and q are two states in DFA. Then L(R) is empty if and only if either L(R1) or L(R2) is empty.b}.w) is accepting and the other is not accepting. If the representation of L is a Regular Expression of size s. NFA has s states. Examples: 1. R= R1R2. a6. a4. R = R1 + R2. Two states p and q are said to be distinguishable.q) is already marked. else it is 'no'. simulate the DFA processing the string of input symbol w.q) are distinguishable.NFA. a2.} be a regular language over ∑ = {a.q) for which p ≠ q. If DFA ends in accepting state the answer is 'Yes'. Then L(R) is empty if and only if both L(R1) and L(R2) are empty. 35 . R=(R1) Then L(R) is empty if and only if L(R1) is empty since they are the same language. On first pass.s) if there is an a∈∑ for which δ (r. The marked pair (p. Our goal is to understand when p and q (p ≠ q) can be replaced by a single state. It always includes at least ε 4. ε . Testing Membership in a Regular Language Given a string w and a Regular Language L. and (p.a) = q. then processing of each input symbol. running time of this algorithm is O(ns2) If the representation is ε . Make a sequence of passes through these pairs. beginning in start state. then 1. mark each pair of which exactly one element is in F. δ (s.Testing Emptiness of Regular Languages Suppose R is regular expression.w) and δ^ (q.
2).φ). (Method1): Let r= (0+1)*10.q) with p ≠ q. The FA is given below Following fig shows all unordered pairs (p. This implies state 1 and 3 are equivalent and cannot be divided further. In pass 2 no boxes are marked because. The resultant FA is shown is Fig 3.q) with p ≠ q 36 .b) φ and δ (3.3) are marked in the first pass according to the algorithm 1.00010. where φ is a non-final state. Fig 3.3) δ(1.3) (φ. That is (1. Minimal Automata corresponding to FA in Fig 1 Minimization of Automata (Method 2) Consider set {1. then L(r) = {10. δ(1. (1.b) φ.φ). gives the list of all unordered pairs of states (p. φ and δ (3. Example 2.a) (φ. That is (1.3}.3) are equivalent and can replaced by a single state A.3) (φ.3) (2. This gives us two states 2.2) and (2. where φ and 3 are non final states. ---}.a) 2. This implies that (1.2) and (1.Fig 2.110. The boxes (1.A.010.
and 4 can be replaced by a single state 124 and states 3. For example (5.2) (6.6) this implies that 2 and 3 belongs to different group hence they are split in level 2.4) was marked on pass 1.2) is one of these. (Method1): (2. They are marked on pass 1.4). From this we can make out that 1. and the pair (6.3) (4. Similarly it can be easily shown for the pairs (4. 5. 2. 6 The transitions of fig 4 are mapped to fig 6 as shown below Example 2. The pairs marked 2 are those marked on the second pass.5) and so on. The resultant minimal FA is shown in Fig. since (5.7) and (2. and 7 can be replaced by the single state 357.5) (1. 37 .The pairs marked 1 are those of which exactly one element is in F.
Tree 38 . T. T the terminals.Context Free Grammar Context Free grammar or CGF. G is represented by four components that is G= (V. 3. 5. S). {0. P P P P P ∈ 0 1 0P0 1P1 Derivation using Grammar Example 1: Leftmost Derivation The inference that a * (a+b00) is in the language of variable E can be reflected in a derivation of that string. A. Example: The grammar Gpal for palindromes is represented by Gpal = ({P}. 1}. P the set of productions and S the start symbol. where V is the set of variables. P) Where A represents the set of five productions 1. Here is one such derivation: E E*E I*E a*E a * (E) a * (E + E) a * (I + E) a * (a + E) a * (a + I) a * (a + I0) a * (a + I00) a * (a + b00) Leftmost Derivation . P. 4. 2. starting with the string E.
Thus. Rightmost Derivation . we can describe the same derivation by: E E*E E *(E) E * (E + E) E * (E + I00) E * (E + b00) a * (a + b00) I * (a + b00) E * (E + I) E * (I + b00) E * (E +I0) E * (a +b00) We can also summarize the leftmost derivation by saying E a * (a + b00). or express several steps of the derivation by expressions such as E*E a * (E).Tree There is a rightmost derivation that uses the same replacements for each variable. although it makes the replacements in different order. This rightmost derivation is: 39 .Example 2: Rightmost Derivations The derivation of Example 1 was actually a leftmost derivation.
For example.S) is a CFG. That is if G = (V. P.a) Rightmost Derivation S (L) (L.E E*E E * (E) E * (E + E) E * (E + I00) E * (E + b00) a * (a + b00) a * (a + b00) I * (a + b00) E * (E + I) E * (I + b00) E * (E + I0) E * (a + b00) This derivation allows us to conclude E Consider the Grammar for string(a+b)*c E E+T|T T F T*F|F (E)|a|b|c (E+T)*F (T+T)*F Leftmost Derivation E T T*F F*F (E)*F (a+b)*F (a+b)*c Rightmost derivation E T T*F T*c F*c (E)*c (E+T)*c (E+b)*c (T+b)*c (F+b)*c (a+b)*c Example 2: Consider the Grammar for string (a.T. T. L(G) = {w in T | S w} Sentential Forms Derivations from the start symbol produce strings that have a special role called "sentential forms". denoted by L(G). then is a right – sentential form.a) S->(L)|a L->L. that is they consist solely of terminals.a) (a. is the set of terminal strings that have derivations from the start symbol.S) (L.S) (a. then any string in (V ∪ T)* such that S α is a sentential form.S) (S. E * (I + E) is a sentential form. and if S α.a) The Language of a Grammar If G = (V. since there is a derivation E E*E E * (E) E * (E + E) E * (I + E) (E+F)*c (F+T)*F (a+T)*F (a+F)*F 40 .a) (S. S) is a CFG.S) (a.P. then is a left – sentential form. Note that the language L(G) is those sentential forms that are in T*.S|S Leftmost derivation S (L) (L. If S α. the language of G.
I a|b|c Consider two derivation trees for a + b * c. +. As an example of a left – sentential form. T={a. consider a * E. ambiguity implies the existence of two or more left most or rightmost derivations. )}. E E+E. *. c. P) with V={E. E E*E. E (E). with the leftmost derivation. T. the derivation Ambiguity A context – free grammar G is said to be ambiguous if there exists some w ∈L(G) which has at least two distinct derivation trees. (. 41 .However this derivation is neither leftmost nor rightmost. E I. since at the last step. Alternatively. I}. the middle E is replaced. Example Consider the grammar G=(V. I*E E*E a*E E * (E) E * (E + E) Additionally. b. and productions. E E*E E Shows that E * (E + E) is a right – sentential form. E.
Example: Condider the Grammar for string aabbccdd S AB | C A C aAb | ab aCd | aDd B cBd | cd D->bDc | bc Parse tree for string aabbccdd 42 . I a|b|c Inherent Ambiguity A CFL L is said to be inherently ambiguous if all its grammars are ambiguous. F (E). T F. E E+T.Now unambiguous grammar for the above Example: E T. T T*F. F I.
} %{ 43 .Applications of Context – Free Grammars Parsers The YACC Parser Generator Markup Languages XML and Document typr definitions The YACC Parser Generator E E+E | E*E | (E)|id %{ #include <stdio. } void yyerror (char *s) { fprintf (stderr.} | Exp '*' Exp {$$ = $1 * $3. "%s\n".h> %} %token ID id %% Exp : id { $$ = $1 . $1). %% int main (void) { return yyparse ( ). } | '(' Exp ')' {$$ = $2.} | Exp '+' Exp {$$ = $1 + $3. s). printf ("result is %d\n". } .
' {printf ("result is %d\n". {ECHO.} | exp '+' term {$$ = $1 + $3.} %% Example 2: %{ #include <stdio. "%s\n".} .} %% 44 .} .} . {ECHO.tab. $1). yyerror ("unexpected character").tab.} [ \t\n] . [-+*/().$3.h> %} %start line %token <a_number> number %type <a_number> exp term factor %% line : exp '.} term : factor {$$ = $1. exp : term {$$ = $1.} | term '*' factor {$$ = $1 * $3.] {return yytext[0].} | term '/' factor {$$ = $1 / $3. } %{ #include "y.} . factor : number {$$ = $1.} | '(' exp ')' {$$ = $2.#include "y.h" %} %% [0-9]+ {yylval.a_number = atoi(yytext). yyerror ("unexpected character").} [ \t \n] . return number.} . return id.ID = atoi(yytext). } void yyerror (char *s) { fprintf (stderr. %% int main (void) { return yyparse ( ). s). [+ * ( )] {return yytext[0].} | exp '-' term {$$ = $1 .h" %} %% [0-9]+ {yylval.
A (E1)* A BA 45 . Moldy bread 2. A BC B E1 C E2 2. A E1 A E2 3.E2. A E1. A E1 | E2.Markup Languages: Functions •Creating links between documents •Describing the format of the document Example The Things I hate 1. People who drive too slow In the fast lane HTML Source <P> The things I <EM>hate</EM>: <OL> <LI> Moldy bread <LI>People who drive too slow In the fast lane </OL> HTML Grammar Char Text Doc Element List-Item List a|A|… e | Char Text e | Element Doc Text | <EM> Doc </EM>| <p> Doc |<OL> List </OL>| … <LI> Doc e | List-Item List Start symbol XML and Document type definitions. 1.
A CFG can be simplified by eliminating 1.1. A (E1)? A ε A E1 EXERCISE QUESTIONS 1) Design context-free grammar for the following cases a) L={ 0n1n | n≥l } b) L={aibjck| i≠j or j≠k} 2) The following grammar generates the language of RE 0*1(0+1)* S A|B A 0A|ε B 0B|1B|ε Give leftmost and rightmost derivations of the following strings a) 00101 b) 1001 c) 00011 3) Consider the grammar S aS|aSbS|ε Show that deviation for the string aab is ambiguous 3) Suppose h is the homomorphism from the alphabet {0. Useless symbols 46 . P.b} defined by h(0) = a. what is h-1(L) ? Simplification of CFG The goal is to take an arbitrary Context Free Grammar G = (V. what is h(L) ? e) If L is the language L(a(ba)*) .A ε B E1 4. what is h(L) ? d) If L is the language L(0+12). h(1) = ab & h(2) = ba a) What is h(0120) ? b) What is h(21120) ? c) If L is the language L(01*2). S) and perform transformations on the grammar that preserve the language generated by the grammar but reach a specific format for the productions. A (E1)+ A BA A B B E1 5. T.2} to the alphabet { a.
P. If the order of eliminations is 2 47 . Eliminate useless symbols: Definition: Symbol X is useful for a grammar G = (V. P. then G1=(V1. ie if S *⇒ αXβ Theorem: Let G = (V. T. A → aA. w∈Γ*. Example: Consider a grammar with productions: S → AB|a. 2. S. 1. A→aA | ε. X is generating variable. Unit production A → B. A → a. S) if there is S *⇒ αXβ *⇒w. Eliminating symbols that are non reachable Elimination has to be performed only in the order of 1 followed by 2. ie if X *⇒ w. X is reachable. T1. Eliminating non generating symbols 2. Here B is a useless symbol Symbol X is useful if both the conditions given below are satisfied in that order itself. where w∈Γ* 2. If the order of eliminations is 1 followed by 2 gives S → a. where A is a variable 3. S) be a CFG and assume that L(G) ≠ φ.Those variables or terminals that do not appear in any derivation of a terminal string starting from Start variable. Here A is a useless symbol S→A. then it is useless. Omitting useless symbols from a grammar does not change the language generated Example: S → aSb | ε | A. ε . S) be a grammar without useless symbols by 1. If X is not useful.Productions A → ε. T. Otherwise the grammar produced will not be completely useless symbols eliminated. A and B are variables 1. B → bB. P1.
Every symbol of T is generating 2. B → aa Eliminate symbols that are non reachable: Dependency Graph: For the production C → xDy. {S → a. A →a. C → aCb Solution: C is a non-generating symbol. A →a}. Example: G= ({S. Here A is still useless symbol. the dependency graph is given below. {a}. Example: Eliminate non-reachable symbols from G= ({S. {S → a. A →a}. then A is generating Non-generating symbols = V. A →a. After eliminating B. After eliminating C gets. B → aa. Eliminate non-generating symbols: Generating symbols follow to one of the categories below: 1. C C D D Draw dependency graph for all productions.followed by 1 gives S → a. {a}. S → aS | A.generating symbols. B}. A. then X is non reachable. S) 48 . A}. A →a}. S → AB | a. If there is no edge reaching a variable X from Start symbol S. S) Example: Eliminate the non-generating symbols from S → aS | A | C. S) Solution: B is a non-generating symbol. {a}. A →a. A}. so completely useless symbols are not eliminated. we get G1= ({S. If A → α and α is already generating.
A → a. {S → a}. C. S) Example: Eliminate non-reachable symbols from S → aS | A.Solution: S S A A Draw the dependency graph as given above. C → AB | b Step 1: Eliminate non-generating symbols V1 = {A. A →a. G1= ({S}. C. So the final variables and productions are same V1 and P1. V2 = {A. A is non-reachable from S. All Variables are reachable. {a}. C → b} Exercises: Eliminate useless symbols from the grammar 49 . B is non-reachable from S. B → aa S S A A B B Draw the dependency graph as given above. we get the grammar with productions S → aS | A. A →a. S} P1 = {S → CA. A → a Example: Eliminate useless symbols from the grammar with productions S → AB | CA. A →a. After eliminating A. C → b} Step 2: Eliminate symbols that are non reachable S S A A C C Draw the dependency graph as given above. B →BC | AB. S} P2 = {S → CA. After eliminating B.
3. A → aA | ε. Find nullable variables 2. then we have to find out an equivalent G without εproductions.Productions: Most theorems and methods about grammars G assume L(G) does not contain ε. A is nullable because of the production A → ε. B → bcc. E →d} aAa. A → bBB. If A *⇒ ε then A is a nullable variable. C →abb. A → aA | ε. E → aC} aBa | BC. A → bA. III. D → E. B → AA} Eliminate ε . In the grammar with productions S → ABA. C → aB} aS | AB. B is nullable because of the production B → ε.1. Remove ε-productions and duplicates Step 1: Find set of nullable variables Nullable variables: Variables that can be replaced by null (ε). B → bB | ε. Example for a grammar G with ε . A →Sb | bCC. 3. C →a. A → aC | BCC.productions is S → ABA. S is nullable because both A and B are nullable. B → bB | ε The procedure to find out an equivalent G with out ε-productions 1. So if ε is not there in L(G). P= {S P= {S P= {S P= {S → → → → aAa. 2. Add productions with nullable variables removed. 50 . B → ab. 4.
{a}. where w' is obtained from w by removing one or more occurrences of nullable variables. B→ ε. α in Ni-1} until Ni = Ni-1 Step 2: Add productions with nullable variables removed For each production of the form A → w. {S → aS | AB. B → bB | ε. production A → ε} Repeat Ni ← Ni-1U {A| A in V. minus any productions of the form A → ε. A.Productions G = ({S. create all possible productions of the form A → w'. Example: For the above example we get the final grammar with productions S → ABA | BA | AA | AB | A | B A → aA | a B → bB | b Example: Find out the grammar without ε . B. So after removing nullable variables we get the productions S → ABA | BA | AA | AB | A | B | ε A → aA | ε | a B → bB | ε | b Step 3: Remove ε-productions and duplicates The desired grammar consists of the original productions together with the productions constructed in step 2. A → α. B and S are nullable variables. Example: In the grammar with productions S → ABA. A → aA | ε. D →b}.Algorithm to find nullable variables is given below: V: set of variables N0 ←{A | A in V. A. D}. S) 51 . A → ε.
productions and obtain G1 2. X → Y| ε. S) Exercise: Eliminate ε . 3. X → aX | bX | ε XY. D →b}. Y → b | X Xa. C → aCD. {S → aS | a | AB | A | B. B →aA. Z →AB. B} New Set of productions: S → aS | a S → AB | A | B D→b G1= ({S. D}. A. 4. A → aAS | a. S→ S→ S→ S→ a |Xb | aYa.productions from the grammar 1. Eliminate useless symbols from G1 and obtain G2 Example: Eliminate ε .productions and useless symbols from the grammar S →a |aA|B|C. Y → bW. 2. B. A → aA | bB | ε. Eliminate ε . B → SbS | A| bb But if you have to get a grammar without ε . D → ddd Step 2: Eliminating useless symbols Step 2a: Eliminate non-generating symbols 52 . A →aB| ε. W →Z. B → Ba | Bb| ε ASB | ε. B →aA|a. {a}.productions Nullable ={A} P1={S →a |aA | B | C. D →ddd Step 1: Eliminate ε .productions and useless symbols. A →aB. X →Zb. C →aCD.Solution: Nullable variables = {S. follow the sequence given below: 1.
A → aB. where A and B are variables. The algorithm for eliminating unit productions from the set of production P is given below: 1. A → aA | a. A. Add all non unit productions of P to P1 2. B. Eliminate unit productions Definition: Unit production is of form A → B. Delete all the unit productions Example: Eliminate unit productions from S → ABA | BA | AA | AB | A | B. So A and B are derivable. For each unit production A → B. Unit productions could complicate certain proofs and they also introduce extra steps into derivations that technically need not be there. add to P1 all productions A → α. Now add productions from derivable. A → aB.Generating ={D. B → aA|a. eliminating C gets P3= S →a |aA|B. D → ddd} Step 2b: Eliminate non -reachable symbols S A B D C is non-reachable. where B → α is a non-unit production in P. S} P2={S →a | aA| B. 53 . B → bB|b Solution: The unit productions are S → A. 3. S →B. B →aA|a} I.
S→ ABA | BA | AA | AB | A | B | aA | a | bB | b A → aA | a B → bB | b Remove unit productions from above productions to get S→ ABA | BA | AA | AB | aA | a | bB | b A → aA | a B → bB | b Example: Eliminate unit productions from S → Aa | B. Add productions from derivable and eliminate unit productions to get. useless symbols and unit productions. follow the sequence given below: 1. D → ddd Step 1: Eliminate ε -productions 54 . A → aB | ε.productions from G and obtain G1 2. C → cCD. Eliminate useless symbols from G2and obtain G3 Example: Eliminate useless symbols. A → B and B → A. B →A | bb Solution: Unit productions are S → B. ε -productions and unit productions from the grammar with productions: S → a | aA | B| C. Eliminate unit productions from G1 and obtain G2 3. Eliminate ε . A → a | bc | B.productions. So A and B are derivable. S → Aa | bb | a | bc A → a | bc | bb B → bb | a | bc Reduced Grammar If you have to get a grammar without ε . B → aA.
B. Eliminate useless symbols. C ∈ V 2. ε -productions and unit productions from the grammar 2. Chomsky Normal Form (CNF) Every nonempty CFL without ε. then add a new variable Cai to V1 and a new production Cai → ai to P1. Replace Xi in A production of P by Cai 55 . A. A → aB. D → ddd} After eliminate symbols that are non reachable S A B D P4 = {S → a | aA. B → aA | a}. A → aB. If Xi is a terminal say ai. S) II. where A ∈ V and a ∈ T Algorithm to produce a grammar in CNF: 1. where A. B → aA | a. D → ddd} Step 2: Eliminate unit productions Unit productions are S → B and S→ C. A → a. D →ddd} Step 3: Eliminate useless symbols After eliminate non-generating symbol C we get P3 = {S → a | aA.Nullable = {A} P1 = {S → a | aA | B | C. C → cCD. B}. A → aB. A → aB. A → aB. P2 = {S → a | aA | cCD. {S → a | aA. B → aA | a. Elimination of terminals on RHS of a production a) Add all productions of the form A → BC or A → a to P1 b) Consider a production A → X1X2…Xn with some terminals of RHS. B → aA | a} So the equivalent grammar G1 = ({S. C → cCD. So Derivable variables are B and C. B → aA | a. has a grammar in CNF with productions of the form: 1. {a}. A → BC.
C2}. Cb. D →d Solution: Step1: Simplify the grammar Grammar is already simplified. A → CaB| CbC2. C2→AB}. C1 → AD A → CbAB to A → CbC2. C2→ AB Grammar converted to CNF is given below: G1=({S. A → aAS | a. Ca → a A →aB to A → CaB A → bAB to A → CbAB. C2. … . Ca. B. D.c) Consider A → X1X2…Xn. b}. where n ≥3 and all Xi's are variables. B →b. Introduce new productions A → X1C1. Cb→ b Step2b: Reduce RHS with 2 variables Change S → CaAD to S → CaC1. C1. A. A →aB | bAB. Cn-2 →Xn-1Xn to P1 and C1. B →SbS | A | bb} Solution: Step1: Simplify the grammar Step 1a: Eliminate ε -productions: Consists of S → ε Eliminating ε -productions from P to get: 56 . C1 → X2C2. Step2a: Elimination of terminals on RHS Change S → aAD to S → CaAD. {a. Cn-2 to V1 Example: Convert to CNF: S → aAD. C1 →AD. S → CaC1. … . Ca→a. S) Example: Convert the grammar with following productions to CNF: P={S → ASB | ε. Cb→b.
A →aAS|aA|a. Conversion to GNF is a complex process. Cb → b. A → 1 57 . B → AC S → 0A0 |1B1 | BB. C → CDE |ε. A → aAS | aA | a. Converting CFG in GNF to PDA gets a PDA without ε-rules. A → CaC2 | CaA | a. D → A | B | ab III. A → bAA | aS | a. 3. B → S|A. Ca→ a. C2 → AS. 4. Greibach Normal Form (GNF) Every nonempty language without ε is L(G) for some grammar G with productions are of the form A → aα. A → aab. B → aBB | bS | b S→ Aba.P1={S →ASB|AB. where a ∈ T and α is a string of zero or more variables. B → SC3 | SCb | CbS | b | CbCb | CaC2 | CaA | a. B→ SbS | Sb | bS | b | A | bb} Step 1b: Eliminate unit productions: B→ A P2= {S → ASB|AB. Example: Transform into Greibach normal form the grammar with productions S → 0S1 | 01 Solution: S → 0SA | 0A. A → C|a. B → C | b. 5. B →SbS | Sb | bS | b | bb | aAS | aA | a} Step 1c: Eliminate useless symbols: no useless symbols Step2: Convert to CNF P3={S → AC1 | AB. S → aSa | bSb | a | b | aa | bb S → bA | aB. C → S| ε S → aAa | bBb| ε. C1 → SB. C3 → CbS} Exercises: Convert the following grammar with productions to CNF: 1. A → C. 2.
Figure 1: Parse tree for a4b4 Figure 2: Extended Parse tree for a4b4 58 . pushes the variables A1A2…An on the stack and enters q1. The remainder of the computation uses the input symbol and the stack top to determine the appropriate transition. consider a CFL L={anbn | n ≥ 1}. inserting as many copies of the sub-string as we like always yields a string in the regular language. The pumping lemma for CFL's states that there are always two short sub-strings close together that can be repeated. as often as we like. Both leftmost derivation and rightmost derivation have same parse tree because the grammar is unambiguous. Assume that the PDA has two states: start state q0 and accepting state q1. The parse tree for the string a4b4 is given in figure 1. C → SB. Pumping Lemma for CFL The pumping lemma for regular languages states that every sufficiently long string in a regular language contains a short sub-string that can be pumped. An S rule of the form S → aA1A2…An generates a transition that processes the terminal a. B → b. Equivalent CNF grammar is having productions S → AC | AB. both the same number of times. A → a. For example.Uses of GNF Construction of PDA from a GNF grammar can be made more meaningful with GNF. That is.
w = ab is the string generated by lower occurrence of S and vwx = aabb is the string generated by upper occurrence of S. Then there exists a constant k≥ 0 such that if z is any string in L such that |z| ≥ k. the middle portion is not too long). The longest path in the parse tree is at least n+1. Number of symbols at each level is at most twice of previous level. 2 symbols at 1. By pigeonhole principle. 1 symbols at level 0. 59 . consider the first pair of same variable along the path. 2. so this path must contain at least n+1 occurrences of the variables. tree must be having at least depth of n and level of at least n+1. 4 symbols at 2 …2i symbols at level i. vx ≠ ε (since v and x are the pieces to be "pumped". Figure 4: sub.trees Figure 3: Parse tree for a4b4 with repeated occurrences of S circled. Example parse tree: For the above example S has repeated occurrences. To have 2n symbols at bottom level. then we can write z = uvwxy such that 1. y=bb. x=b. For all i ≥ 0. and the parse tree is shown in figure 3. Any parse tree for z must be of depth at least n+1. at least one of the strings we pump must not be empty).Extend the tree by duplicating the terminals generated at each level on all lower levels. Suppose z∈ L(G) and |z| ≥ k. Reading from bottom to top. So here u=aa. w=ab. Let k = 2n+1. uviwxiy is in L. Pumping Lemma Theorem: Let L be a CFL. v=a. Say X has 2 occurrences. 3. Proof: The parse tree for a grammar G in CNF will be a binary tree. where n is the number of variables of G. |vwx| ≤ k (that is. some variables occur more than once along the path. The extended parse tree for the string a4b4 is given in figure 2. Break z into uvwxy such that w is the string of terminals generated at the lower occurrence of X and vwx is the string generated by upper occurrence of X.
Figure 5: Parse tree for uv2wx2y ∈L Figure 6: Parse tree for uwy ∈ L To get uwy ∈ L. cut T out of the original tree and replace it with t to get a parse tree of uv0wx0y = uwy as shown in figure 6. 4. 5. 6. Cutting out t and replacing it with copy of T as many times to get a valid parse tree for uviwxiy for i ≥ 1. 3. assume L is context free. Pumping Lemma game: 1. cut out t and replace it with copy of T. 2. for some k The above contradicts the Pumping Lemma Our assumption that L is context free is wrong Example: Show that L = {aibici | i ≥1} is not CFL Solution: 60 . Choose an "appropriate" string z in L Express z = uvwxy following rules of pumping lemma Show that uvkwxky is not in L. The parse tree for uv2wx2y ∈L is given in figure 5. To get uv2wx2y ∈L. These parse trees are shown in figure 4. To show that a language L is not a CFL.Let T be the subtree rooted at upper occurrence of S and t be subtree rooted at lower occurrence of S.
Then uwy will be some string in the form of tt. uv2wx2y will not be having an equal number of a's. 61 . x = b. then |t| =2n-k/2. n |t| ≥ 3n/2. x = ε. If uwy is some repeating string tt. satisfying the conditions |vwx| ≤ n and vx ≠ε. uwy ∈ L. x = a and y = b2c2 then. According to pumping lemma. w = c. in the pumping lemma satisfying the condition uviwxiy for i ≥0. So L is not contextfree. where n is pumping lemma constant. w = ε. v = b. Some a's and some b's 3. b's and c's. Some b's and some c's Case 1: vwx consists of all a's If z = a2b2c2 and u = ε.n ≥ 0}. Can't contradict the pumping lemma! Our original assumption must be wrong. If so. Choose an appropriate z = anbncn = uvwxy. It is sufficient to show that L1= {0m1n0m1n | m. Since |vwx| ≤ n then vwx can either consists of 1. Pick any z = 0n1n0n1n = uvwxy. y = c. This language we prove by taking the case of i = 0. then uv2wx2y will be a3b3c2 ≠L Case 3: vwx consists of some b's and some c's If z = a2b2c2 and u = a2b. t does end in 0 but tt ends with 1. where t is repeating. Example: Show that L = {ww |w ∈{0. w = ε. Then uwy begins with 0n-k1n |uwy| = 4n-k. Suppose vwx is within first n 0's: let vx consists of k 0's. y = bc2.Assume L is CFL. v = a. 1}*} is not CFL Solution: Assume L is CFL. All a's or all b's or all c's 2. then uv2wx2y will be a2b3c2 ≠L If you consider any of the above 3 cases. v = a. So second t is not a repetition of first t. uv2wx2y will be a4b2c2≠L Case 2: vwx consists of some a's and some b's If z = a2b2c2 and u = a. is a CFL. z is having a length of 4n. So if |vwx| ≤ n. But Pumping Lemma says uv2wx2y ∈L. then |uwy| ≥ 3n.
so first t = 0130 and second 0213. Case 2: vwx consists of 2 adjacent symbols say 1 & 2 Then uwy is missing some 1's or 2's and uwy is not in L. {0p | p is a prime} 2. we get uwy. Pick z = uvwxy = 0n1n2n3n where |vwx| ≤ n and vx ≠ ε. Then uwy is not in L. Example: Show that L={0i1j2i3j | i ≥ 1. In all cases uwy is expected to be in the form of tt. So uwy is not in L and L is not context free. t= Suppose vwx consists of 1st block of 0's and first block of 1's: vx consists of only 0's if x= ε. vx = 02 then uwy = tt = 0130313. So L is not a CFL. If we consider any combinations of above cases. Exercises: Using pumping lemma for CFL prove that below languages are not context free 1. If vx has at least one 1. Very similar explanations could be given for the cases of vwx consists of first block of 1's and vwx consists of 1st block of 1's and 2nd block of 0's. This contradicts the assumption. then uwy is not in the form tt. then |t| is at least 3n/2 and first t ends with a 0. Case 1: vwx consists of a substring of one of the symbols Then uwy has n of 3 different symbols and fewer than n of 4th symbol. which is not CFL. j ≥ 1} is not CFL Solution: Assume L is CFL.Example: z = 03130313. Both t's are not same. vwx can consist of a substring of one of the symbols or straddles of two adjacent symbols. {anbnci | i ≤ n} 62 . not a 1. But first t and second t are not the same string.
Rename second and third S's to S0 and S1. homomorphism and inverse homomorphism. reversal. S) is a grammar for s(L) where V′ = V ∪ Va T′= union of Ta's all for a ∈ Σ P′ consists of All productions in any Pa for a ∈ Σ The productions of P. Theorem: If a substitution s assigns a CFL to every symbol in the alphabet of a CFL L. Then G′= (V′. respectively. Σ. closure (star). The CFL's are closed under substitution. Substitution: By substitution operation. then s(L) is a CFL. s(0) = {anbm | m ≤ n}. union. A few do not produce CFL. s(1) = {ab. Proof: Let G = (V. P′. Let Ga = (Va. generated by the grammar S → abA. Pa. CFL's are not closed under intersection (but the intersection of a CFL and a regular language is always a CFL). 1}. Example: L = {0n1n| n ≥ 1}. A → aA | ab. Rename second A to B. T′. Sa) be the grammar corresponding to each terminal a ∈ Σ and V ∩ Va = φ. Resulting grammars are: 63 . complementation. with each terminal a is replaced by Sa everywhere a occurs. and setdifference. concatenation. Example: S(0) = {anbn| n ≥1}. Closure properties consider operations on CFL that are guaranteed to produce a CFL. A → c |ε. abc}. I. S(1)={aa. Ta. generated by the grammar S → aSb | A.bb} is a substitution on alphabet Σ ={0.Closure Properties of CFL Many operations on Context Free Languages (CFL) guarantee to produce CFL. each symbol in the strings of one language is replaced by an entire CFL language. generated by the grammar S → 0S1 | 01. S) be grammar for the CFL L. P.
Closure under union of CFL's L1 and L2: Use L={a. S) where V = V1 ∪ V2 ∪ {S}. s(a)=L1 and s(b)=L2. Their corresponding grammars are G1: S1 → aS1b | ε. S ∉ V1 ∪ V2 and P = P1 ∪ P2 ∪ {S → S1S2} 64 . B → c | ε In the first grammar replace 0 by S0 and 1 by S1. L2 = {bnan | n ≥ 0}. P. S∉ (V1 ∪ V2) and P = {P1 ∪ P2 ∪ {S → S1 | S2 }} Example: L1 = {anbn | n ≥ 0}. Closure under concatenation of CFL's L1 and L2: Let L={ab}. s(a)=L1 and s(b)=L2. b}. b. b}. S). G2 : S2 → bS2a | ε The grammar for L1 ∪ L2 is G = ({S. {S → S1 | S2. B→ c | ε II. T. The resulted grammar after substitution is: S → S0SS1 | S0S1 S0→ aS0b | A. S2}. S1. S1 → aS1b | ε. {a. A → aA | ab S1 → abB. T. P. Then s(L)=L1L2 How to get grammar for L1L2? Add new start symbol and rule S → S1S2 The grammar for L1L2 is G = (V. A →aA | ab S1→abB.S → 0S1 | 01 S0 → aS0b | A. Then s(L)= L1 ∪ L2. How to get grammar for L1 ∪ L2 ? Add new start symbol S and rules S → S1 | S2 The grammar for L1 ∪ L2 is G = (V. S2 → bS2a}. Application of substitution: a. S) where V = {V1 ∪ V2 ∪ S}.
Then s(L) = L1* (or s(L) = L1+). a{nk}b{nk} | k ≥ 0 and ni ≥ 0 for all i} L2 = {a{n2} | n ≥ 1}. Closure under Reversal: L is a CFL.. It is enough to reverse each production of a CFL for L. Closure under homomorphism of CFL Li for every ai∈∑: Suppose L is a CFL over alphabet ∑ and h is a homomorphism on ∑. S). i. G2 : S2 → bS2a | ε The grammar for L1L2 is G = ({S.. m ≥ 0} Their corresponding grammars are G1: S1 → aS1b | ε. S). by h(a). (L2)* = a* How to get grammar for (L1)*: Add new start symbol S and rules S → SS1 | ε. ie h(L) ={h(a1)…h(ak) | k ≥ 0} where h(ai) is a homomorphism for every ai ∈ ∑. {a.Example: L1 = {anbn | n ≥ 0}. c. S2}. III. where V = V1 ∪{S}. {S → S1S2. Example: L1 = {anbn | n ≥ 0} (L1)* = {a{n1}b{n1} . P= P1 ∪{S → SS1 | ε} d. ie s(a) = {h(a)}. Then h(L) = s(L).e. Closure under Kleene's star (closure * and positive closure +) of CFL's L1: Let L = {a}* (or L = {a}+) and s(a) = L1. S2 → bS2a}. to substitute each production A→α by A→αR. S1 → aS1b | ε. Intersection: The CFL's are not closed under intersection 65 . Let s be a substitution that replaces every a ∈ ∑. IV. so LR is a CFL. The grammar for (L1)* is G = (V. S ∉ V1. b}. L2 = {bnan | n ≥ 0} then L1L2 = {anb{n+m}am | n. S1.. T. P.
∑. A→0A | 0. ε.Example: The language L = {0n1n2n | n ≥ 1} is not context-free. ∑. ∑. FP) be PDA to accept L by final state. Z) |-*P (q. w. But L1 = {0n1n2i | n ≥ 1. F) where Q = (Qp X QA) qo = (qp. B→2B | 2 and corresponding grammars for L2: S →AB. we make the same move in PDA P′ and also we carry along the state of DFA A in a second component of P′. a) 2. FA AND PDA Accept/ Reject Stack Proof: Figure 1: PDA for L ∩ R P = (QP. qA). X) = ((r.γ) moves and p = δ*(qA. A→0A1 | 01. a. i ≥ 1} is a CFL and L2 = {0i1n2n | n ≥ 1. γ) are possible if and only if (qp. then L ∩ R is a CFL. Corresponding grammars for L1: S→AB. To get L ∩ R. 66 . g) is in δP(q. s). (r. δP. s = δA(p. P′ accepts a string w if and only if both P and A accept w. δA. B→1B2 | 12. However. Γ. Γ. qA. qo. thus intersection of CFL's is not CFL a. p). a. w) transitions are possible. FA) for DFA to accept the Regular Language R. Let A = (QA. g) such that 1. X) That is for each move of PDA P. we have to run a Finite Automata in parallel with a push down automata as shown in figure 1. Z0. i ≥ 1} is also a CFL. qP. δ. The moves ((qp. p). ε. Construct PDA P′ = (Q. ie w is in L ∩ R. qA) F = (FPX FA) δ is in the form δ ((q. Z) |*P′ ((q. Intersection of CFL and Regular Language: Theorem: If L is CFL and R is a regular language. w. Z0. But L = L1∩ L2. L = L1 ∩ L2 .
L = LC would always be a CFL. So our assumption is false. 1.L = LC. called an inverse homomorphism. Contradiction! .R = L ∩ RC. So CFLs are not closed under setdifference. and L is any language. But (L1C ∪ L2C)C = L1 ∩ L2. CFL and RL properties: Theorem: The following are true about CFL's L. ie if L is a CFL then LC is a CFL.R is a CFL. ie L . Closure of CFL's under set-difference with a regular language. and L2. CFL is not closed under complementation LC is not necessarily a CFL Proof: Assume that CFLs were closed under complement. But ∑* . We have already proved the closure of intersection of a CFL and a regular language. L1. 3. then ∑* . So CFL is closed under set difference with a Regular language. which we just showed isn't necessarily a CFL. We know that L . ∑* is regular and is also CFL. L1C ∪ L2C is a CFL.b. then h-1(L). is the set of all strings w such that h(w) ∈ L. V. The CFL's are closed under inverse homomorphism. Proof: Let L1 = ∑* . Since CFLs are closed under union. Theorem: 67 . and a regular language R. Inverse Homomorphism: Recall that if h is a homomorphism. Proof: R is regular and regular language is closed under complement. CFL is not closed under complementation. If CFLs were closed under set difference. CFLs are not closed under set-difference. ie L1 .L2 is not necessarily a CFL.L. So RC is also regular. But CFL's are not closed under complementation. By our assumption (L1C ∪ L2C)C is a CFL. 2.
X)} If δ(q. where q is an accepting state of P. After input a is read.If L is a CFL and h is a homomorphism. X) = {((p. (F x ε)) to accept h-1(L). We construct a new PDA P′ = (Q′. X) = {(p. q0. h(a) is placed in a buffer. x) such that q is a state in Q x is a suffix of some string h(a) for some input string a in Σ δ′ is defined by δ′ ((q. Only when the buffer is empty does the PDA read another of its input symbol and apply homomorphism to it. δ. h(a)). T. Let PDA P = (Q. because of the way the accepting states of P′ are defined. a. Γ. b. x). where Q′ is the set of pairs (q. Once we accept the relationship between P and P′. Σ. (q0. ε). Thus L(P′)=h-1(L(P)) 68 . Z0. Suppose h applies to symbols of alphabet Σ and produces strings in T*. Γ. ε). γ)} where b∈ T or b = ε then δ′ ((q. X) = {((q. bx). F) that accept CFL L by final state. γ)} The start state of P′ is (q0. Z0. ε. ε). P accepts h(w) if and only if P′ accepts w. ε) The accepting state of P′ is (q. then h-1(L) is a CFL Buffer a Input h h(a) PDA Accept/ Reject Stack Figure 2: PDA to simulate inverse We can prove closurehomomorphism of CFL under inverse homomorphism by designing a new PDA as shown in figure 2. Symbols of h(a) are used one at a time and fed to PDA being simulated. δ′.
Figure 1 shows a diagrammatic representation of PDA. current state and the top symbol on the stack. where Q = a finite set of states. z ∈ Γ is the stack start symbol. Σ = a the input alphabet. Stack functions as the required memory. Γ = the stack alphabet. FSC does some state transitions and does some operations to the stack content.Introduction to Push Down Automata Just as finite-state automata correspond to regular languages. has the form δ: Q X (Σ ∪{∈}) X Γ → finite subsets of Q X Γ* q0 ∈Q is the initial state. and F ⊆ Q is a set of final states. one symbol at a time. 69 . Stack could be kept unchanged. The Finite State Control (FSC) reads inputs. n≥ 0. because FA does not have any memory to remember the number of a's it has already seen. the context-free languages (CFL) have corresponding machines called pushdown automata (PDA). q0. So. Regular expressions are generators for regular languages and Finite Automata's are recognizers for them. Γ. z. or some thing could be pushed into the stack and could be popped out of the stack. a PDA is an NFA with a stack. δ. Based on the input symbol. PDA is NFA with an added memory. for equating with number of b's found. F). An FA cannot recognize the language anbn. Context Free Grammars (CFG) are generators and Pushdown Automata (PDA) are recognizers. Σ. δ = transition function. PDA is more powerful than FA. Input Tape Finite State Control stack Figure 1: PushDown Automaton Formal Definition: A nondeterministic pushdown automaton or npda is a 7-tuple defined as M = (Q. Similarly for Context-free Languages.
it has the form: (P. then move to q2 the final state. 2. ε. then t is replaced with x and w is pushed into the stack. {q2}). Z0) = {(q2. b}. and replace the symbol 't' on top of the stack with the symbol 'u' ". b. stack top is a. If u = t. b. move from state P to state Q. If u = ∈. and symbol on input tape is at ε. If current state is q0. t) → (Q. Z0) = {(q0. 3. Z0. and stack top is Z0. where δ is defined by following rules: δ(q0. t ∈ Γ and P and Q are states. 5. Means "read the symbol 'a' from the input. then stay in q0 and push 'a' to the stack. ε)} δ(q. and stack top is a. {a.Transition function: for any given state. If current state is q1. u) Basically. ε)} δ(q0. stay in q1 and pop the top symbol of the stack 6. if. q0. stack top is a. a) = {(q1. input tape symbol is b. input tape symbol is ε and stack top is Z0. The following transitions are possible: 1. If current state is q0. {a. Z0) = {(q2. a. q2}. and input tape symbol is a. then stack is popped If u= wx.Y) = φ for all other possibilities 70 . b ∈ Σ n ≥ 0}. If current state is q1. b. gives a new state and stack symbol. then stack is unchanged.e. Z0}. a) = {(q0. a ∈ Σ. So we can define PDA as M = ({q0. input tape symbol is 'a'. Start at state q0 and keep Z0 in the stack. Example 1: Construct PDA for the language L= {anbn | a. move to q2 the final state. aa)} δ(q0. δ. If current state is q0. move to state q1 and pop the top symbol of the stack. input tape symbol is b. i. If current state is q0. a. and stack top Z0. input symbol and stack symbol. aZ0)} δ(q0. ε)} δ(q1. q1.x. ε. a) = {(q1. stay in q0 and push 'a' to the stack. 4. a. ε)} δ(q1.
X/α) from state q to state p means that an input tape head positioned at symbol a and stack top with X. with the leftmost symbol at the top of the stack. So u ∈ Γ* Moves of A PDA: Let the symbol "|-" indicates a move of the nPDA.a/aa b. moves automaton to state q and replaces the stack top with α. a. Instantaneous Description is a represented by a triplet (q. where 1. The transition diagram for the above example PDA is given in Figure 2. q is the current state of the automaton. There are two types of moves possible for a PDA. X) = {(p.Graphical Notation of PDA: To understand the behavior or PDA clearer. u). Move by consuming input symbol 71 .a/ ε b. w. Arc corresponds to transitions of PDA. Doubly circled states are final states 4.a/ ε Start q0 q1 ε. Arrow labeled Start indicates start state 3. u is the stack contents. α)…} is an arc labeled (a. the transition diagram of PDA can be used. Z0/aZ0 a. If δ(q. 2. Z0 / ε q2 ε. a. written as a string. w ∈ Σ* 3. 1. Transition diagram of PDA is generalization of transition diagram of FA. w is the unread part of the input string. Z0 / ε Figure 2: Transition diagram Instantaneous Description: Instantaneous Description or configuration of a PDA describes its execution status at any time. Node corresponds to states of PDA 2. 1.
b) Moves for the input string aaabb: (q0. a. ε. ε)} PDA reached a configuration of (q2. aabb. ε)} |. aaZ0) as per transition rule δ(q0. aZ0) as per transition rule δ(q0. b. a. c) Moves for the input string aabbb: (q0. and Z indicates the rest of the stack contents underneath the x. Z0) = {(q0.ε). stack is empty and PDA has reached a final state. an input symbol 'a' is consumed from the input string aW.(q1. a) = {(q0. aZ0)} |. a) = {(q1. a) = {(q1. b. aaZ0) as per transition rule δ(q0. aZ0)} |. abb. aW. aZ0) as per transition rule δ(q0.ε) as per transition rule δ(q1. ε. aaaZ0) as per transition rule δ(q0.(q0.(q0. aa)} |. The input tape is empty. Z0) |. a. So the automaton stops and the string is not accepted. bb.There is no defined move. W.(q1. leaving yZ on the stack. ε.(q1.(q2. The above example PDA with a few example input strings. aaabb. abb. b. a) = {(q0. ε)} 72 . Z0) as per transition rule δ(q0. abbb. aa)} |.(q0. This notation says that in moving from state q1 to state q2. a. aaZ0) as per transition rule δ(q0. aZ0)} |. b..(q0. Z0) = {(q0. b. xZ) |.(q0. and the symbol 'x' at the top (left) of the stack xZ is replaced with symbol 'y'. .(q0. Then the following move by consuming an input symbol is possible: (q1. aabb.(q1. a) = {(q1. Z0) = {(q2. aZ0) as per transition rule δ(q0. a) = {(q1. aa)} |. b. ε)} |.Suppose that δ(q1. aZ0) as per transition rule δ(q0. bb.(q2. a. bbb. a. Z0) |. ε. a) = {(q0. the moves are given below: a) Moves for the input string aabb: (q0. aaZ0) as per transition rule δ(q0. Z0) = {(q0. Z0) |. aZ0) as per transition rule δ(q0. yZ). ε)} |. x) = {(q2. aZ0) as per transition rule δ(q0. aabbb. ε)} |. aa)} |.(q1.. a. bb. a) = {(q1. So the string is accepted. a.(q0. y). where W indicates the rest of the input string following the a.}. b. ε. a) = {(q0.
00)} δ (c. 0)} 73 . aW. δ.|. 0) ={(b. xZ) |. an input symbol 'a' is not consumed from the input string aW. x) = {(q2. where δ is given by: δ (a. ε. the tape head position will not move forward.(q2. aW. 1. ε. 0Z)} δ (b. . yZ). {0. 0) ={(d.used to represent zero or more moves of PDA. 0. leaving yZ on the stack. ε)} |.}. In this move. Figure 3: Transition diagram M = ({a. Z. 0. For the above example. 00)} δ (b. a. Z) ={(b. ε)} δ (c. c. Z0) |-* (q2. Example 2: Design a PDA to accept the set of all strings of 0's and 1's such that no prefix has more 1's than 0's. 0) ={(b.There is no defined move. 0) ={(c. 2. 1.Z}. b.1}. 0. a) = {(q1.1. So the automaton stops and the string is not accepted.. Z0) as per transition rule δ(q0.(q1. y).ε).move Suppose that δ(q1. ε. aabb. {0. d}. {d}).. The relation |-* is the reflexive-transitive closure of |. 0) ={(c. ε)} δ (b. Then the following move without consuming an input symbol is possible: (q1. (q0. b. and the symbol 'x' at the top (left) of the stack xZ is replaced with symbol 'y'. Solution: The transition diagram could be given as figure 3. This notation says that in moving from state q1 to state q2. b. ε. This move is usually used to represent non-determinism.
0Z)} δ (c.(b. Accepting the set of all strings over {a. Z) ={(d. 0Z). For the language L = {wcwR | w ∈ {a. n ≥ 0} 4. ε. For the language L = {anb2n | a. 0Z) |. (consider any type of parentheses) 74 . c ∈ Σ } 2. 0Z) |. 0)} δ (c. Z) |. 0110.Z) |.(b. 0.0Z) |.11. b} with equal number of a's and b's.1.(c. Moves for the input string 0010110 is given by: (a. 00Z) |(c.011. 10.(b. 0) ={(d.Z) |-no move. 00Z) |. So the input string is accepted by the PDA. ε. so PDA stops Exercises: Construct PDA: 1. 110. ε. PDA stops. 010110.(d. Z) |.10110.(c. 0Z) |.δ (b. Z) ={(b. 0. Z)} δ (c. b}*. Moves for 011 (a. Z)} For all other moves. b ∈ Σ.(b. 0Z) |. ε.(c. ε. Accepting the language of balanced parentheses.(b. 0010110. Z) ={(d. Show the moves for abbaba 3.
{0. 1. δ(q0. Z0) = {(q0. Z0)} ε. PDA enters a final state. the appropriate PDA could be by final state. δ(q0.Z0}. Z0) = {(q0. 10)} ε. Then L(P). the difficulty is how to decide the middle of the input string? The 3rd 1 can be part of w or can be part of wR. Both methods are equivalent. 1. The current state could be final or non-final state. stack of the PDA will be empty. Accept by Final state: Let P = (Q. 101101. Z0. For example the language {L = anbm | n ≥ m}. The PDA could be constructed as below. Γ. the language accepted by P by the final state is {w | (q0. It is possible to covert a PDA accept by final state to another PDA accept by empty stack and also the vice versa. 1Z0)} δ(q0. Languages of PDA There are 2 ways of accepting an input string PDA a. Z0) |-* (q. 11)} δ(q0. α)}. 0) = {(q0. δ(q0. Z0) = {(q1. 1111. M = ({q0. 01)} 1. δ(q0. b. Acceptable input strings are like 00. 0. Z0. q1. and 110011. Usually the languages that a PDA accept by final state and PDA by empty stack are different. 0Z0)} δ(q0. 1) = {(q0. 0) = {(q0. F) be a PDA. the language for even length palindrome. Accept by Final state After consuming the input. 1. After consuming the input. 1) = {(q1. q0. Σ. 0110. w. q0. 1)} 75 . where δ is defined by: δ(q0. δ. 0. In the string 0110. where q ∈ F and α ∈ Γ* Example: L = {wwR | w is in (0 + 1)*}. The content of the stack is irrelevant. Accept by empty stack After consuming the input. 1}. the stack may not be empty. {0. 00)} δ(q0. 0) = {(q1. 1) = {(q0. 0. 0)} ε. ε. δ. q2}.Languages of PDA 1. q2).
Z0/ Z0 q0 q1 ε. Z0) = {(q2.0/ ε 1.δ(q1. Z0/1Z0 0. ε.1/ ε ε.1/11 0. 1. wRZ0) |. 0. Z0/0Z0 1. 1) = {(q1. ε)} δ(q1. wwR.0/00 1.0/0 ε.1/ 01 1.(q1. ε)} δ(q1. Z0) The transition diagram for the PDA is given in figure 1. Z0) |-* (q0. 0) = {(q1.(q2. Z0)} (q0. Z0) |. ε. 0.0/ 10 0. Figure 2: moves of PDA for string 101101 76 .1/1 ε. ε. wRZ0) |-* (q1. wR. wR. Z0 / Z0 q2 Figure 1: Transition Diagram for L = {wwR} The moves of the PDA for the input string 101101 are given figure 2.
Z0/Z0 s p ε. then there is a PDA PF such that L = L(PF) Proof: p0 ε. δ. where q ∈ Q Example: Construct PDA to accept by empty stack for the language L = {wwR | w is in (0 + 1)*} Instead of the transition δ(q1. Z0/Z0 q ε. q0. From Empty Stack to Final State: Theorem: If L = N(PN) for some PDA PN= (Q. X0/ε (add this transition from all states of PN to new state Pf) Pf Figure 4: PF simulates PN The method of conversion is given in figure 4. ε)}. Z0)} give δ(q1. ε. δN. Z0) = {(q2. 0/ε 1. ε. 0/ε 1. Z0/0Z0 1. Z0). w. Z0) |-* (q.Accept by empty stack: Let PDA P = (Q. Z0) = {(q2. Γ . q0. 77 . 0. Z0/Z0 1. Z0). The set of accepting states are irrelevant.Z0/Z0 1. ε)} to get accept by empty stack. X0/Z0X0 PN q0 ε. Γ. ε. Conversion between the two forms: a. Σ. Z0/Z0 ε. This example also shows that L(P) = N(P) Example: j Construct PDA to accept by empty stack for the language L={0i1 | 0 ≤ i ≤ j} The transition diagram for the PDA is given in Figure 3. We define the language accepted by empty stack by N(P) = {w | (q0. Σ.Z0/ε r j Figure 3: transition diagram of 0i1 | 0 ≤ i ≤ j 2. 0/00 0.
Z0. The moves of PN to accept a string w can be written like: (p0. δF. w. δF. ε. Γ∪{X0}. ε. Z0 X0)} to push X0 to the bottom of the stack δF(q. X0/Z0X0 PF q0 ε. w. δN.(p. Γ. Γ/ε ε. X0. Z0X0) |-*PF (q. Z0X0) |-*PN (q. y ∈ Γ or y = X0. then there is a PDA PN such that L = N(PN) Proof: p0 ε. a. To avoid PF accidentally empting its stack. same for both PN and PF. X0) = {(q0. y ∈ Γ or y = X0 .We use a new symbol X0. q ∈ F. X0) |. a. where δN is defined by: δN(p0. X0) = {(Pf. p0. y). a. Also add a new start state p0 and final state pf for PF. y) = {(p. ε)} to accept the string by moving to final state. ε. w. ε)}. y) = δF(q . X0) |-PF (p0. δF(q. Also add a new start state p0 and final state p for PN. where δF is defined by δF(p0. ε. initially change the stack start content from Z0 to Z0X0. ε. {Pf}). Σ. a ∈ Σ or a = ε and y ∈ Γ. Σ. which must be not symbol of Γ to denote the stack start symbol for PF. The moves of PF to accept a string w can be written like: (p0. Z0X0)} to change the stack content initially δN(q . y) = δN(q. w. p}. Let PF = (Q∪{p0. ε. Γ/ε p ε. p0. q0.(Pf . Γ/ε Figure 5: PN simulates PF The method of conversion is given in figure 5. ε ) b. X0) |. ε. From Final State to Empty Stack: Theorem: If L = L(PF) for some PDA PF= (Q. same for both δN(p . X0) = {(q0. Let PN = (Q∪{p0. same for both δN(q . X0) |-PN (q0. y) a ∈ Σ or a = ε and y ∈ Γ. to pop the remaining stack contents. F). Γ∪{X0}. X0). pf}. y) = {(p . a. Σ. ε ) 78 . ε. ε. ε)}.
ε. {q1}). 0)} For all other moves. 1}. 0Z)} δ(q0. Also convert the PDA to accept by empty stack. j ≥ 0} 3. 00)} δ(q0. ε)} δ(q0.Z) = {(q1. Z. Solution: PDA by final state is given by M = ({q0. where δ' is the union of δ and the transitions given below: δ(q1. ε. q1}.Z}.1) = {(q0. Z) = {(q0.0) = {(q1. j. 1) = {(q0.Z) = {(q2. Z)} δ(q0. q2}. Z) = {(q0.0) = {(q0. 1. q0. ε.Example: Construct PDA to accept by final state the language of all strings of 0's and 1's such that number of 1's is less than number of 0's.0) = {(q2. q1. {0. ε)} δ(q2. 0) = {(q0. ε)} δ(q0. 1. δ'. 1. {0. where δ is given by: δ(q0. Z). q0. Z}. 1. PDA by empty stack is given by M = ({q0. 11)} δ(q0. ε. {aibjci+j | i. j ≥ 1} 79 . 1Z)} δ(q0. {aibjck | i. PDA stops. {aibi+jcj | i ≥ 0. 1. ε. 0. k ≥ 0 and i = j or i = k} 2. ε. ε)} δ(q2. 0. δ.0) = {(q2. {0. ε)} Exercises: Design nPDA to accept the language: 1. 0. ε)} δ(q1. 1}. {0.Z) = {(q2.
all the rest of the information is encoded in the stack. New transitions are added. include a transition δ(q. Context Free Language defined by CFG 2. For each variable A include a transition. Language accepted by PDA by final state 3. a) = {(q. For each terminal a. For every intermediate sentential form uAα in the leftmost derivation of w (initially w = uv for some v). The stack symbols of the new PDA contain all the terminal and non-terminals of the CFG. one for each production. b) | A → b is a production of Q} 2. A) = {(q. δ(q. Equivalence of PDA and CFG The aim is to prove that the following three classes of languages are same: 1. Most transitions are on ε. There is only one state in the new PDA. each one corresponding to terminals of G. a. S) be a CFG. At the end (case u = w) the stack will be empty. T. δ. Language accepted by PDA by empty stack It is possible to convert between any 3 classes. we construct a PDA P that simulates the leftmost derivations of G. V ∪ T. The PDA which accepts L(G) by empty stack is given by: P = ({q}. q.Equivalence of PDA and CFG I. The representation is shown in figure 1. Let G = (V. T. ε)} 80 . S) where δ is defined by: 1. Q. M will have Aα on its stack after reading u. ε. CFG PDA by empty stack PDA by Final state Figure 1: Equivalence of PDA and CFG From CFG to PDA: Given a CFG G.
ε)} and δ(q. {i. i. q. A)} ε. S). 0S1). 1) = {(q. ε)} From PDA to CFG: Let P = (Q. G has productions S → [q0Z0 p] 2. A.CFG to PDA conversion is another way of constructing PDA. where δ is given by: δ(q. where δ is given by: δ(q. δ. For all states p. {0. ε)} 0. where p. δ(q. δ(q. Z). R. (q. (This does not accept if –if –else-else statements). Let δ(q. e. Y1Y2…Yk)} where a ∈ Σ or a = ε. ε)} 1. S). {X.Z}. 1. ε. δ. productions of R consists of 1. Z) = {(q. e}. and then convert CFG to PDA. A) = {(q.a. ε)} Solution: Equivalent productions are: 81 . First construct CFG. 1}. G has productions [qXrk] → a[rY1r1] [r1Y2r2] … [rk-1Ykrk] If k = 0 then [qXr] →a Example: Construct PDA to accept if-else of a C program and convert it to CFG. q. Σ. δ(q. 0) = {(q. (q. S) = {(q. q0. S). [pXq]}. where V = {S. q ∈ Q and X ∈ Γ. {0. Γ. (q. S}. Z0) be a PDA. δ. XZ)}. Let the PDA P = ({q}. 1A0). Σ. Z) = {(q. k can be 0 or any number and r1r2 …rk are list of states.X) = {(r. δ(q. An equivalent CFG is G = (V. ε. Example: Convert the grammar with following production to PDA accepted by empty stack: S → 0S1 | A A → 1A0 | S | ε Solution: P = ({q}. X) = {(q.
S → [qZq] [qZq] → i[qXq][qZq] [qXq] → e [qZq] → ε If [qZq] is renamed to A and [qXq] is renamed to B, then the CFG can be defined by: G = ({S, A, B}, {i, e}, {S→A, A→iBA | ε, B→ e}, S) Example: Convert PDA to CFG. PDA is given by P = ({p,q}, {0,1}, {X,Z}, δ, q, Z)), Transition function δ is defined by: δ(q, δ(q, δ(q, δ(q, δ(p, δ(p, Solution: Add productions for start variable S → [qZq] | [qZp] For δ(q, 1, Z)= {(q, XZ)} [qZq] → 1[qXq][qZq] [qZq] → 1[qXp][pZq] [qZp] → 1[qXq][qZp] [qZp] → 1[qXp][pZp] For δ(q, 1, X)= {(q, XX)} [qXq] → 1[qXq][qXq] [qXq] → 1[qXp][pXq] [qXp] → 1[qXq][qXp] [qXp] → 1[qXp][pXp] For δ(q, ε, X) = {(q, ε)} [qXq] → ε For δ(q, 0, X) = {(p, X)} 82 1, Z) = {(q, XZ)} 1, X) = {(q, XX)} ε, X) = {(q, ε)} 0, X) = {(p, X)} 1, X) = {(p, ε)} 0, Z) = {(q, Z)}
[qXq] → 0[pXq] [qXp] → 0[pXp] For δ(p, 1, X) = {(p, ε)} [pXp] → 1 For δ(p, 0, Z) = {(q, Z)} [pZq] → 0[qZq] [pZp] → 0[qZp] Renaming the variables [qZq] to A, [qZp] to B, [pZq] to C, [pZp] to D, [qXq] to E [qXp] to F, [pXp] to G and [pXq] to H, the equivalent CFG can be defined by: G = ({S, A, B, C, D, E, F, G, H}, {0,1}, R, S). The productions of R also are to be renamed accordingly. Exercises: a. Convert to PDA, CFG with productions: 1. A → aAA, A → aS | bS | a 2. S → SS | (S) | ε 3. S → aAS | bAB | aB, A → bBB | aS | a, B → bA | a b. Convert to CFG, PDA with transition function: δ(q, δ(q, δ(q, δ(q, δ(p, δ(p, δ(p, 0, Z) = {(q, XZ)} 0, X) = {(q, XX)} 1, X) = {(q, X)} ε, X) = {(p, ε)} 1, X) = {(p, XX)} ε, X) = {(p, ε)} 1, Z) = {(p, ε)}
Deterministic PDA:
NPDA provides non-determinism to PDA. Deterministic PDA's (DPDA) are very useful for use in programming languages. For example Parsers used in YACC are DPDA's. Definition: A PDA P= (Q, Σ, Γ, δ, q0, Z0, F) is deterministic if and only if, 1.δ(q,a,X) has at most one move for q∈Q, a ∈ Σ or a= ε and X∈Γ
2.If δ(q,a,X) is not empty for some a∈Σ, then δ(q, ε,X) must be empty DPDA is less powerful than nPDA. The Context Free Languages could be recognized by nPDA. The class of language DPDA accept is in between than of Regular language and CFL. NPDA can be constructed for accepting language of palindromes, but not by DPDA. Example: Construct DPDA which accepts the language L = {wcwR | w ∈ {a, b}*, c ∈ Σ}. The transition diagram for the DPDA is given in figure 2.
0,0/ ε 1,1/ ε c,0/0 c,1/1 c, Z0/ Z0
ε, Z0 / Z0
Figure 2: DPDA L = {wcwR} Exercises: Construct DPDA for the following: 1. Accepting the language of balanced parentheses. (Consider any type of parentheses) 2. Accepting strings with number of a's is more than number of b's 3. Accepting {0n1m| n ≥ m} DPDA and Regular Languages: The class of languages DPDA accepts is in between regular languages and CFLs. The DPDA languages include all regular languages. The two modes of acceptance are not same for DPDA. Acceptance with final state: If L is a regular language, L=L(P) for some DPDA P. PDA surely includes a stack, but the DPDA used to simulate a regular language does not use the stack. The stack is
L={0*} could be accepted by DPDA with final state. ie. q ∈ Q such that δA(q.a. If L = L(P) for DPDA P. but not with empty stack. But the language. or vice versa.Z)} for all p. Acceptance with empty stack: Every regular language is not N(P) for some DPDA P. x and y satisfy the prefix property.inactive always.So add a new production $→ε as a variable of G. y ∈ L. This is language is accepted by only nPDA. because strings of this language do not satisfy the prefix property. y∈ L(WcWR). Definition of prefix property of L states that if x. then L has unambiguous CFG. then surely L has unambiguous CFG. To convert L(P) to N(P) to have prefix property by adding an end marker $ to strings of L. N(P) ⊆ L DPDA and Ambiguous grammar: DPDA is very important to design of programming languages because languages DPDA accept are unambiguous grammars. Then convert N(P) to CFG G'. Non-Regular language L=WcWR could be accepted by DPDA with empty stack. So N(P) are properly included in CFL L. From G' we have to construct G to accept L by getting rid of $ . But all unambiguous grammars are not accepted by DPDA. For example S → 0S0|1S1| ε is an unambiguous grammar corresponds to the language of palindromes. If L = N(P) for DPDA P. then x should not be a prefix of y. If A is the FA for accepting the language L. A language L = N(P) for some DPDA P if and only if L has prefix property. because if you take any x. 85 . then δP(q.Z)={(p.a)=p.
For example to recognize string that is a multiple of 4 1 0 0 0 S2 0 S1 S0 1 But can regular language recognize strings of the form 0n1n ? No Context Free Language A language is called Context free language if and only if some pushdown automaton recognizes it. 86 . language Context free language DFA Regular language Regular language A language is called a regular language if some finite automaton recognizes it.e.Language Hierarchy and Turing Machine (TM) A Hierarchy of Formal Languages Turing machines PDA r.
Ex: To recognize a string of the form 0n1n S → 0 S1 / λ Limitation again Can context free language recognize strings of the form 0n1n2n ? No Recursively Enumerable Language (r.X.2) →(S 3 .1.0 .Y.Y.Y)→(SA.0 .1.R ) ( S 0 .1 ) →(S 2 .1 ) → (S 2 .0 ) →(S 1.1 ) →(S 3.e language if some Turing machine recognizes it.X) → (S 0 .R ) ( S 1.e.0 ) →(S 1.R) ( S 2 .R ) ( S 1.R) ( S 2 .Y.L) ( S 3 .Z.R) SA : Accepting state 87 .L) ( S 3 .0 ) →(S 3. language) A language is called r. Ex: To recognize strings of the form 0n1n2n ( S 0 .Y)→ (S 3 .L) ( S 3.L) ( S 3.X.
(0 | 1)* Regular Expression NFA with λ-moves NFA 2-way DFA DFA 88 .Formal Machines Overview Regex operators: * (Kleene *) | (choice) DFA Regex example: . (Concatenation) Regular Lang.
His theorem states that in any mathematical system. there exist certain obviously true assertions that cannot be proved to be true by the system. 89 . For example: anbncn Multi-stack PDA PDA with queues PDA Some Historical Notes At the turn of the 20th century the German mathematician David Hilbert proposed the Entscheidungsproblem. is it possible to formulate a general procedureto decide whether a given element of U is a member of X? Hilbert's grand ideas were watered down by the incompleteness theorem proposed by the Austrian Kurt Gödel.PDA CFL example: 0n1n CFL CFG But there are strings that cannot be recognized by PDAs. and a criteria for membership. and a universe of elements U. Given a set X.
Notation for the Turing Machine: TM = <Q.In 1936 the British cryptologist Alan Turing. and showed that they represent algorithmic computation. Church and Turing proved the equivalence of Turing Machines and lcalculus. q0. addressed Hilbert's Entscheidungsproblem using a different approach. But Turing's a-machine became famous for something else other than its original intentions. The amachines came to be more popularly known as Turing Machines The Princeton mathematician Alonzo Church recognized the power of amachines. on which input is provided as a finite sequence of symbols. A Turing Machine consists of a tape of infinite length. and showed the existence of some problems where membership cannot be determined. This is called the Church-Turing thesis.R} is the transition function {L. mathematical model that describes what can and cannot be computed. T. He invited Turing to Princeton to compare Turing Machines with his own l-calculus. Turing Machines Definition: A Turing Machine (TM) is an abstract. δ.R} is direction in which the head moves L : Left R: Right 90 . H> where. Q = a set of TM states T = a set of tape symbols q0 = the start state H⊂S is a set of halting states δ : Q x T Q x T x {L. changes its internal state and moves one cell to the right or left. The a-machine was found to be more expressive than CFGs. He proposed two kinds of mathematical machines called the a-machine and the c-machine respectively. On reading an input symbol it optionally replaces it with another symbol. The Turing Machine starts at "start state" S0. It can recognize strings that cannot be modeled by CFLs like anbncn. A head reads the input tape.
2. Some of the characteristics of a Turing machine are: 1.) The input symbols occupy some of the tape's cells. 1. 1. Transitions: (S0. 1) (S0.input symbols on infinite length tape 1 0 1 0 1 1 1 1 1 1 0 Read/Write head The Turing machine model uses an infinite tape as its unlimited memory. Solved examples: TM Example 1: Turing Machine U+1: Given a string of 1s on a tape (followed by an infinite number of 0s). Initially the TM is in Start state S0. 3. Move right as long as the input symbol is 1. (This is important because it helps to show that there are tasks that these machines cannot perform. Input :#111100000000……. The TM head can move in either direction – Left or Right. The tape is of infinite length The special states. and other cells contain blank symbols. replace it with 1 and halt. The symbols can be both read from the tape and written on it. add one more 1 at the end of the string. take immediate effect. STOP) 91 . 4. even though unlimited memory and unlimited time is given. R) (S0. 0) ( h . Halting states and Accepting states. When a 0 is encountered. Output : #1111100000000……….
which we should replace. R) b) Once the first 1 of y is reached. _) (S4. So. _. L) (S6. L) c) State s3 is when corresponding 1s from both x and y have been stamped out. STOP) (S1. This means we have stamped out one extra 1 in x. L) (S0._) (S2. 0) (S3.b) (S4. STOP) 92 . #) (S3. (S3. _) (S5. R) (h. _. 1) (S6. While searching for the next 1 from x. #___11b___000… a) Stamp out the first 1 of x and seek the first 1 of y. For purposes of simplicity we shall be using multiple tape symbols. _. (S0. L) (S4. 1. compute |x-y| using a TM. (S1.. _. (S1. 1. R) (S2. If instead the input ends. we have stamped out one extra 1. then y has finished.TM Example 2 : TM: X-Y Given two unary numbers x and y. 1) (S2. Example: 5 (11111) – 3 (111) = 2 (11) #11111b1110000…. But in x. stamp it out. 0.b) (S6. go to some state s5 which can handle this. 1) b) 1) b) (S1. then stop. L) (S4. if we reach the head of tape. Now go back to x to find the next 1 to stamp. 1) (S4. _) (S5. L) (S2. STOP) d) State s5 is when y ended while we were looking for a 1 to stamp. b. #. 1. 1. _. and replace the blank character with 1 and stop the process. b. So. R) (S5. R) (h. L) (h. L) (S6. _) (S3. _. b. go back to x. b. (S2. (S5. (S0.
R) ( S 2 .Y)→(S 3.R) ( S 1.R) S3 = Seeking X.0 ) →(S 1.1.0 ) →(S 1. then move one step right. stamp it with X S1 = Seeking 1.Z.Y.Y)→ (S 4 .R ) ( S 4 .2) →(S 3 .Y. Step 3: Move right until the end of the input denoted by blank( _ ) is reached passing through X Y Z s only. then seek the first 1 and stamp it with Y.1.Y.R) ( S 4 .Y.0. ( S 0 .R) ( S 1.0 . ( S 0 .R) ( S 4 .0 )→(S3._ ) → (SA.L) (S 3.Y)→ (S 4 .R) ( S 2 .Z) → (S 4 . (S 3. Step 1: Stamp the first 0 with X. and then seek the first 2 and stamp it with Z and then move left. to repeat the process.STOP) S4 = Seeking blank 93 ._ .1 ) →(S 2 .L) S0 = Start State. stamp it with Y S2 = Seeking 2.Design of Turing Machines and Universal Turing Machine Solved examples: TM Example 1: Design a Turing Machine to recognize 0n1n2n ex: #000111222_ _ _ _ _…….1)→(S3.Z. seeking 0.X)→(S 0.L) (S 3. then the final state SA is reached. stamp it with Z Step 2: Move left until an X is reached.X.X.1 ) →(S 2 .L) (S 3.
_ .L) ( S 4 .L) ( S 4 .0 ._ .STOP) ( S 2 ._ .L) ( S 4 .2 ) →(h.0 .0 .R) Step 3: If a blank ( _ ) is reached when seeking next pair of characters to match or when seeking a matching character. then accepting state is reached.1) → (S 4 .2 .L) ( S 6 . ( S 0 .2 ) →(h. then move left until a blank is reached to start the process again._)→(SA.STOP) ( S 4 .STOP) (S 5._ .R ) ( S 1.1.1.STOP) ( S 0 .L) ( S 5 .R ) ( S 0 .2 ) →(h._) →(SA._ .1) →(S 6 .STOP) ( S 0 .2._ ) → (S 5 ._ .L) ( S 6 .L) ( S 6 .1 ) → (h. ( S 4 .1.0 ) → (S 1. ( S 3 . then seek the last character by moving till a _ is reached.STOP) ( S 1.0) → (S 4 .1.1) → (S 6 .2.STOP) TM Example 2 : Design a Turing machine to accept a Palindrome ex: #1011101_ _ _ _ _……._ .STOP) 94 .1 ) →(h.L) ( S 3 .0 ) → (h.STOP) (S 0 ._ ) →(h._) → (S 0 ._) →(S 0 .STOP) Step 2: If the last character is 0/1 accordingly.0) → (S 4 . If the last character is not 0/1 (as required) then halt the process immediately._ .R) ( S 5 .STOP) ( S 4 ._ .1._ ._ ) →(SA.1 ) →(S 2 .These are the transitions that result in halting states._)→(SA.0) → (S 6 . Step 1: Stamp the first character (0/1) with _. (S 3._ .STOP ) ( S 2 .1 ) →(h._ ._ ._ ) → (S 3.
. Design a TM to recognize a string of 0s and 1s such that the number of 0s is not twice as that of 1s.. 2-seeker and 2-stamper. Modularization of TMs Designing complex TM s can be done using modular approach. For example. Inside each module. #_ _ _ _ s5 _ _ _ _ _ _ #_ _ _ _ sA _ _ _ _ _ _ Exercises: n 2n 1. 1seeker... #_0101s5_ _ _ #_010s6_ _ _ _ #_s60101_ _ _ #_s00101_ _ _ . Design a TM to recognize a string of the form a b . there could be several state transitions.. 1-stamper. the problem of designing Turing machine to recognize the language 0n1n2n can be divided into modules such as 0-stamper. 2. 0-seeker. The main problem can be divided into sequence of modules.The sequence of events for the above given input are as follows: #s010101_ _ _ #_s20101_ _ _ #_0s2101_ _ _ .. The associations between the modules are shown in the following figure: 95 .
0) (d. (c. (a.1) (c. S0=a.H.0. The encoding can be as follows: 96 .d}.1.R) .d) Suppose. H={b.d} T={0.TM: 0 n1 n2 n 0Stamper 1Seeker 1Stamper 2Seeker 2Stamper 0Seeker Load → Decode → Execute → Store Universal Turing Machine (UTM) A Universal Turing Machine UTM takes an encoding of a TM and the input data as its input in its tape and behaves as that TM on the input data. where $ is delimiter This spec along with the actual input data would be the input to the UTM.R) and so on then TM spec: $abcd$a$bd$01$a0b1Ra1c1Rc0d0R…….b.T.S0.c.R) . S={a. A TM spec could be as follows: TM = (S.1.1} d : (a. This can be encoded in binary by assigning numbers to each of the characters appearing in the TM spec.0) (b.
97 .0100 …… Hence TM spec can be regarded just as a number. Go back and find which transition to apply.0011.0000. the sequence goes through the cycle: Load → Decode → Execute → Store Extensions to Turing Machines Proving Equivalence For any two machines M1 from class C1 and M2 from class C2: M2 is said to be at least as expressive as M1 if L(M2) = L(M1) or if M2 can simulate M1.0000.0010. Hence.0001. Sequence of actions in UTM: Initially UTM is in the start state S0. where necessary. Then store the changes. Then repeat the steps with next input. Load the input that is TM spec. Make changes. Composite Tape TMs Track 0 0 1 1 0 1 0 1 0 0 … Track 1 A composite tape consists of many tracks which can be read or written simultaneously.0001.$ : 0000 0 : 0101 a : 0001 1 : 0110 b : 0010 L : 0111 c : 0011 R : 1000 d : 0100 So the TM spec given in previous slide can be encoded as: 0000. M1 is said to be at least as expressive as M2 if L(M1) = L(M2) or if M1 can simulate M2.0010.0100.
Equivalence of 2TMs and TMs 2TM = TM: Just don't use the left part of the tape… TM = 2TM: Simulate a 2-way infinite tape on a one-way infinite tape… 98 . T = {a. There is no # that delimits the left end of the tape. 01. R) δ(s''.(s'. the tape is infinite on both sides. b.a) |-. c. STM = <S. Equivalence of CTMs and TMs A CTM is simply a TM with a complex alphabet.S) TM: δ(s.A composite tape TM (CTM) contains more than one tracks in its tape. H> δ: S x T à S x T x {L. T..(s'. Move right.*.a) |-.b.L) 2-way Infinite Turing Machine In a 2-way infinite TM (2TM). S} Equivalence of STMs and TMs STM = TM: Just don't use the S option… TM = STM: For L and R moves of a given STM build a TM that moves correspondingly L or R… TM = STM: For S moves of the STM.Move back left without changing the tape 3.(s''. d} T' = {00. 2. 10. 11} Turing Machines with Stay Option Turing Machines with stay option has a third option for movement of the TM head: left. right or stay.*) |-.STM: δ(s. δ. b. do the following: 1. R. s0.
99 . Equivalence of MTMs and TMs MTM = TM: Use just the first tape… TM = MTM: Reduction of multiple tapes to a single tape.… -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 … 0 –1 1 –2 2 –3 3 –4 4 –5 5 … Multi-tape Turing Machines A multi-tape TM (MTM) utilizes many tapes. A single tape TM that is equivalent can be constructed by reducing m tapes to a single tape. Consider an MTM having m tapes.
Multi-dimensional TMs Multi-dimensional TMs (MDTMs) use a multi-dimensional space instead of a single dimensional tape.A B C 0 1 2 3 4 5 6 7 … 0 1 2 3 4 5 6 7 … 0 1 2 3 4 5 6 7 … TM A0 B0 C0 A1 B1 C1 A2 B2 C2 A3 B3 . TM 100 ..
{(s3. 0 1 2 3 4 5 6 7 8 9 10 0 1 2 3 4 Non-deterministic TM A non-deterministic TM (NTM) is defined as: NTM = <S. T. s0.a). a.a) |-.b.a. b.(s3. H> where δ: S x T x 2SxTx{L.(s4. δ.R)} |-at (s2.Equivalence of MDTMs to TMs Reducing a multi-dimensional space to a single dimensional tape. L). R)} Equivalence of NTMs and TMs A "concurrent" view of an NTM: (s2.a. (s4.L) (s2.R} Ex: (s2. a) |-.b.{(s3.a) |-.R) 101 . two TMs are spawned: (s2.L) (s4.a) |-.
b.{(s3.a) |-. R)} bccaaabccacb s2 bccbaabccacb s3 bccaaabccacb s4 102 . a.Simulating an NTM with an MDTM Consider an MDTM to simulate an NTM: (s2. L) (s4.
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