I don't know what the differences are between a transition diagram and finite automata. When I google for 'transition diagram', I get state diagrams as a result.
Is there a difference between transition diagrams and finite automata? Or is finite automata a form of transition diagrams?
Thanks.
A transition diagram is a way of visually representing finite state machines. It's kind of on the borderline between flowcharts and source code; it contains enough information to completely describe the finite state machine, but when implementing FSMs on a computer, we generally use other representations that are easier for the computer to process.
A transition diagram for DFA, is a graph shows moment or transition
between states For each state in Q there is a node represented by the
circle.3 main components are initial state,final state and inputs
. Finite machine . It is an abstract machine shows finite number
of states it is the simplest machine to recognize patterns.
Hope this will be helpfull for you.
Finite Automata is a machine where you feed the machine with some input and the machine produces a respective output(Mealy Machine, Moore machine) or no output at all(Deterministic Finite Automata, Non Deterministic Finite Automata) depending on the machine.
Whereas, a transition diagram is used to show the transition from one state to another which is used by all of the above machines. For Example transition from Q1 (initial state) to QF (Final state).
A finite automaton (FA) as name implies Finite number of states
is a simple idealized machine used to recognize patterns within input taken from some character set (or alphabet) .
The job of an FA is to accept or reject an input string depending on whether that string being accepted by FA or not.
whereas ;
Transition daigram can be interpreted as a flowchart for an algorithm recognizing a language ; show the transition form one state to other after recieving input strings consists of three things:
A finite set of states, at least one of which is designated the start state and some of which are designated as final states
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I have a burning question about finite state machines, that how we can know that this language needs 2 states or 3 states? I mean is there any formula for that?
Although I believe, we would always work to minimize the number of states but still How can we determine the number of states to be created according to any language or string (without actually constructing the DFA)?
You are in effect asking about DFA minimization. It is a well-studied problem for which a number of algorithms have been developed. The Wikipedia article on it is a good starting point.
The theoretical result which governs the number of states is the Myhill-Nerode theorem, but this theorem doesn't give any quick formula. You have to determine the number of equivalence classes in an equivalence relation defined in terms of the language. Hopcroft's algorithm for DFA minimization is essentially an algorithm for determining the equivalence classes in the Myhill-Nerode equivalence relation. I suspect that any attempt to use Myhill-Nerode more directly is going to lead to something similar to Hopcroft's algorithm, though I am not an expert in the field.
Aho-corasick multiple pattern matching algorithm is a finite state machine with only 1 state.
as I understand the latest ANTLR4 went away from building static DFA tables for the lexer and for parser and now they do it at runtime. Is this correct? Could someone please explain in general how ANTLR4 works?
Our recent paper describes the mechanism in excruciating, academic detail. Section 3 provides a high level overview:
Instead of relying on static grammar analysis, an ALL(*) parser adapts to the input sentences presented to it at parse- time. The parser analyzes the current decision point (nonterminal with multiple productions) using a GLR-like mechanism to explore all possible decision paths with respect to the current “call” stack of in-process nonterminals and the remaining input on-demand. The parser incrementally and dynamically builds a lookahead DFA per decision that records a mapping from lookahead sequence to predicted production number. If the DFA constructed to date matches the current lookahead, the parser can skip analysis and immediately expand the predicted alternative.
We use an augmented transition network (ATN) to represent the grammar but build DFA using an algorithm very similar to the subset-construction algorithm of NFA-to-DFA conversion fame.
Hope this helps.
I want to know how to represent a coffee machine using a Deterministic finite automata?
I've tried a lot to do this job.
I represented each and every processes as a set,by putting one to one correspondence with Natural numbers.
But I still don't know how to represent it using DFA.
First, try to imagine the states your automaton can be in. Something like:
Off, Ready, Working
Afterwards imagine the buttons or inputs you have to perform to switch between these states. Do not forget to define every input on every state. If you leave out several transitions, the automaton is not deterministic therefore is an NFA. Transitions could be:
0 for power off/on
1 for start/stop working
Off -0-> Ready
Ready -1-> Working
Ready -0-> Off
Working -1-> Ready (4 for the actual working process)
Off -1-> Off
Working -0-> Working (nothing happens in this cases)
Just connect the states with the given transitions, and voilá!
How could I show that there exist infinitely many DFA's each of which recognises the language {ϵ,a,b}.
That depends on how you are counting DFAs. Clearly there is one DFA for the language, and you can always add an unreachable state to the automaton. Ordinarily, though, such trivial differences are discounted, and with a finite language, there are only a finite number of different DFAs. With an infinite language there would be cycles in the transition graph which can be expanded again and again, which makes for a more significant difference. Another way to put it, you cannot show that there exist an infinite number of different DFAs for the given finite language.
Im am working on a VHDL project that includes an fsm.
Some states change according to a counter. It dit not work until i put 'clk' in the sensitivity list, besides the current state and the input.
I know that during synthesis, the sensitivity not used, or discarded. But how can that have such an impact on the result in the simulation? if a leave this 'clk', would the fsm perform as i want op an FPGA?
thanks,
David
This is the simple explanation:
The simulator uses the sensitivity list to figure out when it needs to run the process. The reason why the simulator needs hints to figure out when to run the process is because computer processors can only do one (or only a few in multicore systems) thing at a time and the processor will have to take turns running each part of your design. The sensitivity list allows simulation to run in a reasonable time frame.
When you synthesize code into an ASIC or FPGA, the process is always "running" since it has dedicated hardware.
When you simulate a state machine without the clock in the sensitivity list, the process will never run on the clock edges, but only on changes to your input. If you have the state transition implemented as a flip flop (if clk'event and clk = '1') then your state transition will never be simulated unless you happen to change your input at the same time as the clock's rising edge.
You should probably leave the clock in the sensitivity list, assuming the FSM changes on clock edges.
Also, try to proofread your questions.
Synthesis tools focus on logic design (FPGA, ASIC) and ignore sensitivity lists because there are only three basic types of logic: Combinational logic, edge sensitive storage (flip-flops and some RAM), and level sensitive storage (latches and some RAM).
Combinational logic requires all input signals to be on the sensitivity list. From a synthesis tool perspective, if one is missing, they can either ignore the sensitivity list and treat it as if all inputs were on the sensitivity list, or produce some complicated combination of flip-flops and combinational logic that probably will not do what the user wanted anyway. Both of these have an implementation cost to the vendor, hence, why invest money (development time) to create something that is not useful. As a result, the only good investment is to simplify and ignore the sensitivity list.
Simulators on the other hand, have a bigger perspective than just logic design. The language defines sensitivity lists as to indicate when the code should run. So simulators implement that semantic with a high fidelity.
Long term it may make you happy to know that VHDL-2008 allows the keyword "all" to be used in a sensitivity list to replace the signal inputs there. This is intended to simplify the modeling of combinational logic. Syntax is as follows:
MyStateMachine : process(all)
begin
-- my statemachine logic
end process MyStateMachine ;
Turn on the VHDL-2008 switch and it out in your synthesis tool. If it does not work, be sure to submit a bug against the tool.