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Problems There are 3 mandatory problems and one bonus problem. 1.Give state diagrams of DFAs with as few states as you can recognizing the following languages. (a) L 1 = fw j w begins with 10 and ends with 11g. (b) L 2 = fw j w represents a binary number that is congruent to 2 modulo 3g. In other words, this number minus 2 is divisible by 3.

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Dw Bi 3 DFA idea i see if what we have so is for it i divisible by3 something 12 mod 3 is I o K 5 2 8 T 45,8 had 3 is I QI iasr anything 93,69 27 E a Mafia a go where ...

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a) Construct the Finite Automata for binary umber divisible by 2 b) Design FA for decimal number divisible by 5 c) Give formal definition of Turing Machine d) State and explain closure properties of regular languages e) Construct DFA accepting all the strings corresponding to the Regular expression Q2. a) Construct the following grammar to CNF

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Definition of Divisible explained with real life illustrated examples. Also learn the facts to easily understand math glossary with fun math worksheet online at SplashLearn. 3: If the sum of all of the digits is divisible by three, the number is divisible by 3.

3. L 3 = fx2f0;1g jxis divisible by 3g. ... State diagram of DFA M 3 such that L 3 = L(M 3) Intuition of the various states: - q 0: all strings xsuch that x 0 mod 3. - q This set of Automata Theory Multiple Choice Questions & Answers (MCQs) focuses on “Finite Automata”. 1. Assume the R is a relation on a set A, aRb is partially ordered such that a and b are _____________ a) reflexive b) transitive c) symmetric d) reflexive and transitive 2. Moore Machine is an application of: a) … Design a DFA which accepts all the numbers written in binary and divisible by 3. For example your automaton should accept the words 0, 11, 110, 1001, 1100 … Can you prove that the automaton accepts language ? Can you generalize this to any divisor ‘n’ and any base ‘b’ ? Answers are in the next lecture slides

Problems There are 3 mandatory problems and one bonus problem. 1.Give state diagrams of DFAs with as few states as you can recognizing the following languages. (a) L 1 = fw j w begins with 10 and ends with 11g. (b) L 2 = fw j w represents a binary number that is congruent to 2 modulo 3g. In other words, this number minus 2 is divisible by 3. Why you can add the digits to see if something is divisible by 3...The examples of binary number divisible by 3 are 0, 011, 110, 1001, 1100, 1111, 10010 etc. The DFA corresponding to binary number divisible by 3 can be shown in Figure 4. The above automata will accept all binary numbers divisible by 3. For 1001, the automata will go from q0 to q1, then q1 to q2, then q2 to q1 and finally q2 to q0, hence accepted.

Homework # 3 For each of the following, give a DFA that accepts the speci ed language. Exercises to be handed in from Part I include 1,3 and 28. Exercises to be handed in from Part II include 4 and 6. Part I 1. The set of strings over fa;b;cg in which all a0s preceded the b0s; which in turn precede the c0s: 2. The same as in 1 except for the ... Problem 3. Design a DFA that accepts strings over f0,1gcontaining 101 as a substring. (10 points) Note: Draw the graph. Do not give the transition table. Problem 4. Prove that there exists an integer whose decimal representation consists entirely of 1’s, and which is divisible by 1987. (10 points)

Numbers evenly Divisible by 3. Numbers are divisible by 3 if the sum of all the individual digits is evenly divisible by 3. For example, the sum of the digits for the number 3627 is 18, which is evenly divisible by 3 so the number 3627 is evenly divisible by 3.A DFA for the language of all strings w over Σ = {0,1,2,3,4,5,6,7,8,9} such that the sum of the digits of w is divisible by 7 can be easily built using our knowledge on Number Theory. The idea is to build a DFA in which each state corresponds to a remainder of the division of 3. From each state, go through all of the characters and answer the question “How would reading this character change what I know about my string?” and draw transitions to the appropriate states. Reading a character d should transition to the state representing “the last character of the string is d”. L = { w is divisible by 5 }

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