Refer to image and answer correctly for upvote! (Automata and Computation) The following is a Deterministic Finite Automaton (DFA) for the input alphabet {a,b,c}:WсStrings in the form ofbZa, b, cWhat kinds of strings are accepted in this DFA?a, bambnckYсwhere m,n,k>0O Strings in the form of abn+¹+2where n20O Strings in the form of an+2pn+1 cn where noO None of the answers are accepted

The Deterministic Finite Automaton (DFA) in the image accepts strings that follow these rules:Starts…

Answered: The following is a Deterministic Finite…

Database System Concepts

7th Edition

ISBN:9780078022159

Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Chapter1: Introduction

Section: Chapter Questions

Problem 1PE

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We are given two automatons for the following problem: Left automaton = {w E {a, b}" : w is empty or ends with a b} Right automaton = {w E {a, b}" : w either consists of an even number of a's and no b's, or has at least one b and an even number of a's after the last b} a a,b Construct a DFA/Cartesian product to design a finite automaton that recognizes the union of the two languages given.
applied disreet maths if |A| = n and f: A--->B is injective, what is |f(A)|?
False For each language L over a given alphabet E, let 1(L, E) denote the language over E obtained from L by generating all permutations of its strings - that is, r(L, E) -{we E*: there is a string v in L such that w is a permutation of v ). Specify which one of the following statements is true. Choose None of the other statements is true. The language r((0b1)*, (0, b,1}) is decidable. The language n((Ob1)*, (0, b,1}) is regular. There is no regular language L over an alphabet E such that the language n(L, E) fails to be context-free. Consider alnhaber E-Lo and context-free grammar G over E with precisely one
Draw a deterministic FSM that recognizes strings of 1’s and 0’s specified by each of the following criteria. Each FSM should reject any characters that are not 0 or 1. *(a) The string of three characters, 101. (b) All strings of arbitrary length that end in 101. For example, the FSM should accept 1101 but reject 1011. (c) All strings of arbitrary length that begin with 101. For example, the FSM should accept 1010 but reject 0101. (d) All strings of arbitrary length that contain a 101 at least once anywhere. For example, the FSM should accept all the strings mentioned in parts (a), (b), and (c), as well as strings such as 11100001011111100111.
Given the following CFG S → OB A → 1AA|A B → OAA The following strings can be generated except i. 001 ii. 000 iii.0011 iv. 01 O A. i and iii O B. i and ii O C. iv only O D. ii only
Must be new solution and run on GNU Common Lisp! Using Lisp, write a program that solves the Missionaries and Cannibals problem that uses a DFS( depth first search). It should use (mac start end). Start is the current state (which can be (3 3 l) and End is the goal state (which can be (0 0 r). This should output the sequences of moves needed to reach the end state from the start state. This should print nil if there is no solution. For example, the call should be something like this! Call: (mac '(3 3 l) '(0 0 r)) Output: ((3 3 l) (2 2 r) (3 2 l) (3 0 r) (3 1 l) (1 1 r) (2 2 l) (0 2 r) (0 3 l) (0 1 r) (1 1 l) (0 0 r))
NFA 0’s near end or 1 at end: Write a nondeterministic finite automaton with at most 10 states that decides whether its input is contained in the following set: {x ∈ {0, 1}∗ | x[|x| − 5] = x[|x| − 2] = 0 or x[|x|] = 1}. We index strings starting at 1, so x[1] is the first symbol, and x[|x|] is the last symbol.
Finite language is a language with finite number of strings in it, i.e., there exist exactly k strings in this language such that k eNand k #00. For a finite language L, let |L| denote the number of elements of L. For example, |{A, a, ababb}| = 3. (Do not mix up with the length |x| of a string x.) The statement |L,L2| = |L1||L2| says that the number of strings in the concatenation LL2 is the same as the product of the two numbers |L1| and |L2|. Is this always true? If so, prove, and if not, find two finite languages L1, L2 S {a, b}* such that |L1L2| # |Li||L2l.
Create a Deterministic finite automaton that takes in binary strings and accepts them which has both an odd number of zeros and a sum that is divisible by 3 (but no others). For example, 0111 should be accepted, but not 011 or 111.
Give English language translations of the following wffs if L(x, y): x loves y H(x): x is handsome M(x): x is a man P(x): x is pretty W(x): x is a woman j: John k. Kathy L(x. y): x loves y а. Н() Л L(#) b. (Vx)[M(x) → H(x)] c. (Vx)(W(x) → (Vy)[L(x, y) → M(y) ^H(y)]) d. (3x)[M(x) ^ H(x) ^ L(x, k)] е. Вx) (W(x) Л Р() Л ()Le, y) — НО) Л М(У)) f. (Vx)[W(x) ^ P(x) →L(j, x)]
Convert the following DFA to an equivalent regular expression: Deterministic finite automaton a 93 a b b b a q2 q1 b a 44 Grafstate® M 1. Create an initial GNFA GO that is equivalent to M. Here are suggested steps: a. Choose a state in Q. Modify GO to create an equivalent GNFA called G1 that contains all states in GO except for the state you chose. b. Choose another state in Q. Modify G1 to create an equivalent GNFA called G2 that contains all states in G1 except for the state you chose. c. Choose another state in Q. Modify G2 to create an equivalent GNFA called G3 that contains all states in G2 except for the state you chose. d. Choose another state in Q. Modify G3 to create an equivalent GNFA called G4 that contains all states in G3 except for the state you chose.
Code in C++ only. Correct answer will upvoted else downvoted. framework of size n×m, with the end goal that every cell of it contains either 0 or 1, is considered lovely if the total in each adjoining submatrix of size 2×2 is actually 2, i. e. each "square" of size 2×2 contains precisely two 1's and precisely two 0's. You are given a network of size n×m. At first every cell of this network is unfilled. How about we indicate the cell on the crossing point of the x-th line and the y-th segment as (x,y). You need to handle the inquiries of three sorts: x y −1 — clear the cell (x,y), in case there was a number in it; x y 0 — compose the number 0 in the cell (x,y), overwriting the number that was there already (assuming any); x y 1 — compose the number 1 in the cell (x,y), overwriting the number that was there beforehand (assuming any). After each question, print the number of ways of filling the unfilled cells of the grid so the subsequent network is delightful. Since the appropriate…

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