Theory Of Computation

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Concept:- General_idea: click_me L={a^n b^n c^n | n>=1} w={abc,aabbcc,aaabbbccc,....} a a a b b b c c c  X a a Y b b Z c c X X a Y Y b Z Z c Step_1: Replace a by X and go right side. TM tape can move any side. It has infinite memory in both side. And it can remember previous states details. Step_2: Replace b by Y and go right side of tape. Step_3: Replace c by Z and turn left till X .  Step_4: After seeing X(capital X) replace next a by X and do same loop. Step_5: X portions finished. Step_6: Move to Y states. Denote it for one move first and after that notate self loops. Goto final state Z in self loop. Finally stop turing machine. Success. Next step: How to add self loop states? Ok listen me... For that we are taking two steps X a a Y b b Z c c X X a Y Y b Z Z c  Just understand in between symbols. And make it as self loops. If any doubt please post comments..I can explain it. Small editing...This one is right answ

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pf : No need to consider any trees on fewer than 3 vertices (WHY?). If the radius of the tree on n ≥ 3 vertices is one, then there is a vertex with eccentricity 1, which means it is connected to all other points. Claim: this center is unique. Assume otherwise, then another point is connected to all other points as well, including the known center. Taking an edge from this point to the known center, another edge out to a third distinct point, and finally an edge back to this point gives a cycle. That contradicts that this graph is a tree. Now let T be a tree with radius r > 1. Remove all vertices of degree 1 and their associated edges. This creates a graph of radius r − 1, which we can assume by induction has either 1 or 2 centers, which would obviously be the centers T. Reference: [1] 9.1 Introduction to Trees ,Available: click_me 

Arun Anoop M  BTech PG Diploma MTech MPhil (PhD) ASSISTANT PROFESSOR, Department of Computer Science & Engineering MES College Of Engineering,Kuttipuram Kerala, INDIA  E-mail:             arunanoopm@gmail.com     Phone: +91 9497394076 Whatsapp_number:  +91 9497394076 =================================================== Arun Anoop M  BTech PG Diploma MTech MPhil (PhD) PhD Research Scholar(January 2016 Batch). Under Anna University,Chennai Research Center: VCET,Madurai. Guide: Dr.S.Poonkuntran,             Professor,Dept. Of CSE,VCET,Madurai. If anyone want Online class please mail me or whatsapp me. First mail me your university syllabus. GATE Oriented class: 1. Datastructure 2. Theory of Computation 3. Functional dependancy in DBMS 4. Master's theorem in Algorithms

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QUESTION is L=(AB)+  where A=11|10 and B=000|110 Download JFLAP from www. jflap .org/ jflap tmp/ Then Windows_Button+R ie; RUN Type cmd or command Then type java –jar <filename.jar> L=(AB)+      where A=11|10 and B=000|110 From this ….. Before design …what we need….Our question is L=(AB)+  where A=10|11 and B=110|000. W={10110,10000,11110,11000,1011010110,1000010000,……..} First I drawn the diagram based on first four inputs….Then I drawn an epsilon state to Initial state. How to write CFG(Grammar):- q0 as S,q1 as A and so on. S -> 1A A -> 1B | 1A B -> 1C | 0C C -> 1D | 0D           D -> 0E               E -> S | epsilon

Design View: By JFLAP Software Concept :  aaabbb Case1: XaaYbb Case2:XXaYYb Case3:XXXYYY Case 1: Read first 'a' and replace it by 'X' Then read first 'b' and replace it as 'Y' What i am doing is .....Meaning of a^n b^n is " Equal no: of a's must followed by equal no: of b's " In Case 1 : I read first 'a' and First 'b'. Before reading Second 'b'....Should read Second 'a' .... Then continue the steps.....Finally 'B' reads....Machine will stop... Transitions concept will continue ......

-Alan turing. -Language: Recursively Enumerable Language -Unrestricted Grammar (Why because there is no restriction for data reading .... Read data from Either Left or right side) - Input Tape:        ----------- B B  <string divided and keep in each cells> B B --------------- Finite Control Use "Tape head" to read  at a time one letter from input-tape cell. - Formal Definition M=(Q,Sigma,Toe,Delta,q0,B,F)                // 7 tuples or 7 elements...that we need to develop a TM Delta maps from Delta: Q X Toe -> Q X Toe X {L,R,S} Movements are L(Left),R(Right),S(Stationary)