Kleene's Theorem in TOC | Part-1 Last Updated : 22 Oct, 2025 Comments Improve Suggest changes 31 Likes Like Report A language is said to be regular if it can be represented by using Finite Automata or if a Regular Expression can be generated for it. This definition leads us to the general definition that; For every Regular Expression corresponding to the language, a Finite Automata can be generated.For certain expressions like:- (a+b), ab, (a+b)*; It's fairly easier to make the Finite Automata by just intuition as shown below. The problem arises when we are provided with a longer Regular Expression. This brings about the need for a systematic approach towards Finite Automata generation, which Kleene has put forward in Kleene's Theorem.For any Regular Expression r that represents Language L(r), there is a Finite Automata that accepts same language.Regular ExpressionTo understand Kleene's Theorem-I, Let's take into account the basic definition of Regular Expression where we observe that \phi, \epsilon and a single input symbol "a" can be included in a Regular Language and the corresponding operations that can be performed by the combination of these are: Say, r1 and r2be two regular expressions. Then,r1 + r2 is a regular expression too, whose corresponding language is L(r1) U L(r2)r1 . r2 is a regular expression too, whose corresponding language is L(r1).L(r2)r1* is a regular expression too, whose corresponding language is L(r1)*Observations From Kleene TheoremWe can further use this definition in association with Null Transitions to give rise to a FA by the combination of two or smaller Finite Automata (each corresponding to a Regular Expression). Let S accept L = {a} and T accept L = {b}, then R can be represented as a combination of S and T using the provided operations:Union OperationR = S + TUnion OperationObservations:In the case of union operation, we can have a new start state, from which, the null transition proceeds to the starting state of both the Finite State Machines.The final states of both the Finite Automata are converted to intermediate states. The final state is unified into one which can be traversed by null transitions.Concatenation OperationR = S.T Concatenation OperationObservations:In the case of a concatenation operation we can have the same starting state as that of S, the only change occurs in the end state of S, which is converted to an intermediate state followed by a Null Transition.The Null transition is followed by the starting state of T, and the final state of T is used as the end state of R.Exponential Operation R = S*Exponential Operation Observations:A new starting state is added, and S has been put as an intermediate state so that the self-looping condition could be incorporated.Starting and Ending states have been defined separately so that the self-looping condition is not disturbed.Now that we are aware of the general operations. Let's see how Kleene's Theorem-I can be used to generate a FA for the given Regular Expression.Example: Make a Finite Automata for the expression (ab+a)* Kleene AlgorithmWe see that using Kleene's Theorem - Gives a systematic approach towards the generation of Finite Automata for the provided Regular Expression. Create Quiz Comment A AdityaPati Follow 31 Improve A AdityaPati Follow 31 Improve Article Tags : Theory of Computation Explore Automata _ IntroductionIntroduction to Theory of Computation5 min readChomsky Hierarchy in Theory of Computation2 min readApplications of various Automata4 min readRegular Expression and Finite AutomataIntroduction of Finite Automata3 min readArden's Theorem in Theory of Computation6 min readSolving Automata Using Arden's Theorem6 min readL-graphs and what they represent in TOC4 min readHypothesis (language regularity) and algorithm (L-graph to NFA) in TOC7 min readRegular Expressions, Regular Grammar and Regular Languages7 min readHow to identify if a language is regular or not8 min readDesigning Finite Automata from Regular Expression (Set 1)4 min readStar Height of Regular Expression and Regular Language3 min readGenerating regular expression from Finite Automata3 min readCode Implementation of Deterministic Finite Automata (Set 1)8 min readProgram for Deterministic Finite Automata7 min readDFA for Strings not ending with "THE"12 min readDFA of a string with at least two 0âs and at least two 1âs3 min readDFA for accepting the language L = { anbm | n+m =even }14 min readDFA machines accepting odd number of 0âs or/and even number of 1âs3 min readDFA of a string in which 2nd symbol from RHS is 'a'10 min readUnion Process in DFA4 min readConcatenation Process in DFA3 min readDFA in LEX code which accepts even number of zeros and even number of ones6 min readConversion from NFA to DFA5 min readMinimization of DFA7 min readReversing Deterministic Finite Automata4 min readComplementation process in DFA2 min readKleene's Theorem in TOC | Part-13 min readMealy and Moore Machines in TOC3 min readDifference Between Mealy Machine and Moore Machine4 min readCFGRelationship between grammar and language in Theory of Computation4 min readSimplifying Context Free Grammars6 min readClosure Properties of Context Free Languages11 min readUnion and Intersection of Regular languages with CFL3 min readConverting Context Free Grammar to Chomsky Normal Form5 min readConverting Context Free Grammar to Greibach Normal Form6 min readPumping Lemma in Theory of Computation4 min readCheck if the language is Context Free or Not4 min readAmbiguity in Context free Grammar and Languages3 min readOperator grammar and precedence parser in TOC6 min readContext-sensitive Grammar (CSG) and Language (CSL)2 min readPDA (Pushdown Automata)Introduction of Pushdown Automata5 min readPushdown Automata Acceptance by Final State4 min readConstruct Pushdown Automata for given languages4 min readConstruct Pushdown Automata for all length palindrome6 min readDetailed Study of PushDown Automata3 min readNPDA for accepting the language L = {anbm cn | m,n>=1}2 min readNPDA for accepting the language L = {an bn cm | m,n>=1}2 min readNPDA for accepting the language L = {anbn | n>=1}2 min readNPDA for accepting the language L = {amb2m| m>=1}2 min readNPDA for accepting the language L = {am bn cp dq | m+n=p+q ; m,n,p,q>=1}2 min readConstruct Pushdown automata for L = {0n1m2m3n | m,n ⥠0}3 min readConstruct Pushdown automata for L = {0n1m2n+m | m, n ⥠0}2 min readNPDA for accepting the language L = {ambncm+n | m,n ⥠1}2 min readNPDA for accepting the language L = {amb(m+n)cn| m,n ⥠1}3 min readNPDA for accepting the language L = {a2mb3m|m>=1}2 min readNPDA for accepting the language L = {amb2m+1 | m ⥠1}2 min readNPDA for accepting the language L = {aibjckdl | i==k or j==l,i>=1,j>=1}3 min readConstruct Pushdown automata for L = {a2mc4ndnbm | m,n ⥠0}3 min readNPDA for L = {0i1j2k | i==j or j==k ; i , j , k >= 1}2 min readNPDA for accepting the language L = {anb2n| n>=1} U {anbn| n>=1}2 min readNPDA for the language L ={wÐ{a,b}* | w contains equal no. of a's and b's}3 min readTuring MachineTuring Machine in TOC7 min readTuring Machine for addition3 min readTuring machine for subtraction | Set 12 min readTuring machine for multiplication2 min readTuring machine for copying data2 min readConstruct a Turing Machine for language L = {0n1n2n | nâ¥1}3 min readConstruct a Turing Machine for language L = {wwr | w ∈ {0, 1}}5 min readConstruct a Turing Machine for language L = {ww | w ∈ {0,1}}7 min readConstruct Turing machine for L = {an bm a(n+m) | n,mâ¥1}3 min readConstruct a Turing machine for L = {aibjck | i*j = k; i, j, k ⥠1}2 min readTuring machine for 1's and 2âs complement3 min readRecursive and Recursive Enumerable Languages in TOC6 min readTuring Machine for subtraction | Set 22 min readHalting Problem in Theory of Computation4 min readTuring Machine as Comparator3 min readDecidabilityDecidable and Undecidable Problems in Theory of Computation6 min readUndecidability and Reducibility in TOC5 min readComputable and non-computable problems in TOC6 min readTOC Interview preparationLast Minute Notes - Theory of Computation13 min readTOC Quiz and PYQ's in TOCTheory of Computation - GATE CSE Previous Year Questions2 min read Like