Consider the state machine whose states are triples of non-negative integers (r, s, a). The initialstate is (x, y, 0). The transitions are given by the rule d that for s > 0:(3т, s/3, а)(3г, (s — 1)/3, а +r)(3r, (s — 2)/3, а + 2г) otherwise.if 3 sif 3 | (s – 1)6(т, 5, а) —-The predicate axy is true for every state of this state machine.TrueFalse

Consider the state machine whose states are triples of non-negative integers (r, s, a). The initial state is (x, y, 0). The transitions are given by the rule & that for s > 0: (3r, s/3, a) (3r, (s – 1)/3, a +r) (3r, (s – 2)/3, a + 2r) otherwise. if 3 s if 3 | (s – 1) 8(r, s, a) - The predicate a = xy is true for every state of this state machine. O True False

Consider the state machine whose states are triples of non-negative integers (r, s, a). The initial state is (x, y, 0). The transitions are given by the rule & that for s > 0: (3r, s/3, a) (3r, (s – 1)/3, a +r) (3r, (s – 2)/3, a + 2r) otherwise. if 3 s if 3 | (s – 1) 8(r, s, a) - The predicate a = xy is true for every state of this state machine. O True False

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

See similar textbooks

Similar questions

For the finite state automaton given by the transition diagram, find the states, the input symbols, the initial state, the accepting states and write the annotated next state table a σο 01 W b 03 -0 b 02 b b
The correct statements are: Group of answer choices A DPDA must halt. A NDPDA must halt. A NDPDA without ε-transitions must halt. A DPDA or NDPDA without ε-transitions must halt in at most |w| steps.
Convert the NFA defined by 6(q0, a) = {qo, q1} 8(q1,b) = {q1, q2} 8(q2, a) = {q2} With initial state qo and final state q2 into an equivalent DFA
The simplest expression (in sum-of-products form) for the function F(A,B,C,D,E) = m (1,3,5,7,8,12,15,18,19,22,23,24,27,28,31) is O ABE + CD'E + AB'D+ADE+BD'E' O A'B'E + CD'E + AB'D+ADE+BD'E' O ABE + CD'E + AB'D+A'DE+BD'E' O A'B'E + CDE + AB'D + ADE + BD'E' Use the K-map to find the minimum sum of products of F(A,B,C,D) = M (0,2,4,6,8,10,12,14) OF=D O none of these OF = D' O F = A'D Simplify the Expression (wx + yz)(w'y'+xz') using Boolean Algebra: * O wxz' O wx'z Ow'xz Ow'x'y'
5. Consider the following set of Boolean functions: W (p, q, r) = E(1,3, 5,7) X (p, q, r) = E(0,1,5,7) Y (p, q, r) = E(1,3,6) (a) Deduce the optimal ROM truth table for the above functions. (b) Draw the ROM logic diagram to implement the circuit.
Given a digraph D = (V; A; ') with all but one arc (u; v) having non-negative lengths, explain how to determine if D contains a negative circuit in O (m + n log n) of time.
Consider the following NFA (N1) over the alphabet Σ = {a, b}. Compute the ε-closure (ec) for each state in N1.
Given a boolean function F(A.B.C.D)= A@C+B.c@D implement the boolean function F(A,B.C.D) using only NOR gates with only 2 inputs, in other words write the equivalent expression of F using only NAND(variable1 variable2)?
Find a Hamiltonian circuit inCay({(R90, 0), (H, 0), (R0, 1)}:D4 ⊕ Z3).Does this circuit generalize to the case Dn+1 ⊕Zn for all n ≥ 3?
6. Please draw the circuit for Grover's algorithm with N=4 and given that winner is |w)=10), which can be realized with 2 qubits.
Question (4): Build a finite state machine (FSM) with an input X and an output Z. If the sequence '0110' shows up on X, then a value '1' should show up on Z, otherwise '0'. Allow overlap of Sequences (i.e., end of one sequence could be the beginning of another one). Unused states should go to the initial valid state. You need to give two different designs. One for Moore and the other for Mealy. In each design case:- a) Draw the state diagram. b) Write the state stable.
Convert the given finite state machine to a regular expression using the state elimination technique, starting with eliminating (a) state 1 first and (b) state 2 first. The green-outlined state is the accepted state. Show all steps.