Regular Languages are closed under intersection, i e , if L1 and L2 are regular then L1 ∩ L2 is also regular Cross-Product Construction Let M1 = (Q1,Σ,δ1,q1,F1) and M2 = (Q2,Σ,δ2,q2,F2) be DFAs recognizing L1 and L2, respec- tively
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Intuitively, a string is in L1 ∪ L2 ∪ Lk iff it is in An operator is idempotent if the result of applying it to two of the same values as arguments with the same set of variables, is: 1 If L is a regular language over alphabet Σ then L = Σ∗ \ L is also regular Proof: Let L closed under complement and intersection Automata
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(15 points) Prove or disprove the following statement: If M = (Q,Σ, δ, q0,F) is a minimal DFA for a regular language L, then M = (Q,Σ, δ, q0,Q − F) is a minimal DFA for L (10 points) The symmetric difference of 2 sets S1 and S2 is defined as (a) Construct regular set L1 − L2 = L1 ∩ L2 = L This can be done since there
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DFA by having a set of start states (rather than exactly one), and having as a co- domain of the two parties to compute the regular language L = L1 ∩ L2 in in the presence of semi-honest adversaries if for each party i ∈ {1 n} there exists
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(a) Union of two non-regular languages cannot be regular Ans: False Let L1 Since, L1 is regular, hence its intersection with L i e L1 ∩ L = L2 is regular (since regular (c) The set of finite length strings over a countably infinite alphabet is countably infinite Ans: True To detect this, the machine cannot have less than
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that operation to any members of the set always yields a member of That is, if L1 and L2 are regular languages, then each of L1 ∪ L2 , L1 L2 and L1 ∗ Closure under Intersection Fact The for these languages both have two states: A
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A language is called regular if it is accepted by a finite state automaton Let A and B be languages (remember they are sets) We define the following operations
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Regular expressions can be seen as a system of notations for denoting ϵ-NFA They form an there is no intersection operation 1 union L1 ∪ L2 of two languages L1 and L2 2 concatenation L1L2 this is the set of all words x1x2 with xi ∈ Li If L1 or L2 ϵ /∈ L(R) then this is the only solution of the equation x = Rx + S
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States of the new DFA correspond to sets of states of the NFA ○ all possible transitions on a for each state in S, then taking the set of If L1 and L2 are regular languages, is L1 ∪ L2? start The Intersection of Two Languages ○ If L 1
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A second set of questions about language families deals with our ability to decide on certain explicitly how to construct a finite accepter for the intersection of two reg- other words, we want to show that if L1 and L2 are regular, then L1 - L2
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Closure under ?. 1. Page 2. Proposition 4. Regular Languages are closed under intersection i.e.
If L1 and L2 are regular sets then intersection of these two will be : A. Regular. B. Non Regular. C. Recursive. D. Non Recursive.
Typically if A is an automaton recognizing both L1 and L2. (thanks to two sets of accepting states)
which is bounded whether L1 e L2 and whether L2 e L1. (The same problems (For if it does
If L1 and L2 are not regular languages then L1 ? L2 is not regular. These are two independent uses of the variable name w. It just happens that the ...
02?/10?/2020 pushdown automaton M and a regular set of configurations C ... store languages of these models could be accepted with only the counters
19?/09?/2017 A set of infinite words is called ?-regular if it is equal to L(A) for some ... These maps will help us to define some of the basic.
F of sets is said to have the finite intersection property if every ?? c0
03?/10?/2018 We give an overview of techniques that will be heavily used ... case then the literal intersection of these two sets is L1 ? L2 = {C}.
28?/09?/2020 Let ? be an alphabet and let L1 and L2 be two sets of reduced pointed. ?-labelled finite rooted trees. If L1 and L2 are regular then L1 · L2 is ...
If L1 and L2 are two regular languages then L1[L2 L1L2 L1 are regular 2 How about L1L2 L1 L2 ? also regular 3 Regular languages are closed under union concata-nation Kleene star set intersection set di erence etc 4 Given two FAs M1 and M2 can we construct new FAs to accept the the languages L(M1) [ L(M2) L(M1)L(M2) L(M1) L1
Thm 4 4: If L1 and L2 are regular languages then L1=L2 is regu-lar: The family of regular languages is closed under right quotient with a regular language Proof: 1 Assume that L1 and L2 are regular and let DFA M = (Q;?;–;q0;F) accept L1 2 We construct DFA Md = (Q;?;q0;Fc) as follows (a) For each qi 2 Q determine if there is a y 2
Since regular languages are sets we cancombine them with the usual set operations UnionIntersectionDifference THEOREM If L1and L2are regular languages so are L1[L2L1L2and L1 L2 PROOFIDEA Construct cross-product DFAs CROSS-PRODUCTDFAS single DFA which simulates operation of twoDFAs in parallel!Let the two DFAs beM1andM2languagesL1andL2
Regular Languages are closed under intersection i e if L 1 and L 2 are regular then L 1 L 2 is also regular Proof Observe that L 1 L 2 = L 1 [L 2 Since regular languages are closed under union and complementation we have L 1 and L 2 are regular L 1 [L 2 is regular Hence L 1 L 2 = L 1 [L 2 is regular Is there a direct proof for
L1? L2is regular since regular languages are closed under complement ? L1? L2 is regular Proof via finite automata construction Let M? M?? and M be the formal definition of the finite automata that recognizes L1 L2 and L1? L2respectively M?=(Q? ? ?? q0? F?) M??=(Q???????? q0?? F??)
=Two views of L?L?: The set of all strings that can be made by concatenating a string in L? with a string in L? The set of strings that can be split into two pieces: a piece from L? and a piece from L? Conceptually similar to the Cartesian product of two sets only with strings
What is the intersection of L1 and L2?
The intersection of these two languages is L1? L2= {ajbjcj?j? 0}, and in the example 41it is proven by the Bar-Hillel lemma that this language is not context-free. QED. Theorem 20. The context-free language class is not closed under the complement operation. Proof. Now we use proof by contradiction.
How many perpendiculars are there to L1 and L2?
A simpler set is 2, — 3, 5. In the general case, since L is perpendicular to both L1 and L2, it follows that (~~3) 1 + mim + nln = 0, 121+ m2m+ n2n = 0. * From now on we drop the qualifying adjective "undirected," and speak merely of lines and directed lines, as usual. t There are infinitely many common perpendiculars to L1 and L2.
Are L1 and L2 intersecting lines for exfig 7 ample?
Suppose, for exFIG. 7 ample, that L1 and L2 are, respectively, x+2=0 and x -y=. * The figure shows L1 and L2 as intersecting lines, but formula (1) and the deduction of it are valid also in case L1 and L2 are parallel.
When are L1 and L2 the same line?
For, L1 and L2 are the same line when and only when they have the same slope and the same intercept on the axis of y, that is, when and only when -A1 A2 and C C2 B1 B2 B1 B2' * Or, in a single case, identical. Cf. Th. 5. Page 47 THE STRAIGHT LINE 47 or A1: A2=B1: B2 and B1: B2= C1: C2, or, finally, Al: A2 = B: B2 = C0: C2.
1 Closure Properties