a symmetric karnaugh map is dual of a finite state machine. true or false?
is transitive closure is suitable to a finite automata?
Dear friend Srinivasa Rao Tadepalli
True, a symmetric Karnaugh Map and a Finite State Machine are in duality.
The Karnaugh Map is a graphical method used to simplify boolean algebra expressions. It is commonly used in digital logic design to minimize the number of logic gates required to implement a boolean function.
On the other hand, Finite State Machines (FSMs) are a mathematical model used to represent and analyze the behavior of digital circuits, such as sequential logic circuits. FSMs are often used to design and verify the operation of digital systems.
A symmetric Karnaugh Map and a Finite State Machine are said to be duals of each other because they can be used to represent the same digital circuit or function. The duality property states that if we take the Karnaugh Map of the complement of a boolean function, we obtain the dual of the Karnaugh Map of the original function.
Regarding your second question, transitive closure is a suitable operation for Finite Automata. The transitive closure of a Finite Automaton can be used to find all the possible paths between two states in the automaton. This can be useful for verifying properties of the automaton, such as whether it satisfies certain safety or liveness conditions.
take the start state So as the 1st vertex of the graph shown in the Karnaugh map. continue the sequence of states as the successor vertices and the reverse arrows of either states or the graph will show that the entries aij and aji have the same 1 or 0 numerals that represent the presence of an edge in the graph. if this pattern is continued, then the dual space is the space of vertices which can also be considered as the space of states of the finite automata.
perhaps, you are carried away.
please visit linear programming problems.
constraints become the coefficients of variables and vice versa.
this is the duality.
similarly, regions of a graph is treated as vertex in another sense and vice versa which allows a state refered to a vertex and vice versa.
Dear Peter sir!
When a researcher posts a question, it is a question to his or her own selves first. Then to the world. So, i am already having answer/s to my question. So, i wish to wait some more time so that I can get a brighter answer than mine. If i am not happy at nth stage, then obviously I disclose my answer. Also answer to every related question there of.
I humbly request you to wait.
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