How to connect DFA state which had no transitions after minimizing it?
I have the following DFA, let it be $A$:
The problem asks to find the minimal connected DFA for $A$, it is as follows according to the solution (the state ${d,e}$ is called $e$ for simplicity's sake):
This is my solution:
In order to find the minimal DFA we need to find the equivalence classes for $E_k$:
$$
pi_0={{a,b,c,d}, {d,e}}\
pi_1={{a,b,d},{c}, {d,e}}\
pi_2={{a,b},{d},{c}, {d,e}}\
pi_3=pi_2
$$
because we're arrived to the stage that $pi_3=pi_2$, we're done so we can present the minimal DFA:
I don't understand why in the solution there's a transition from $e$ to $d$ via $a,b$?
proof-explanation formal-languages automata
asked Nov 24 at 13:10
Yos
1,137723
add a comment |
I have the following DFA, let it be $A$:
The problem asks to find the minimal connected DFA for $A$, it is as follows according to the solution (the state ${d,e}$ is called $e$ for simplicity's sake):
This is my solution:
In order to find the minimal DFA we need to find the equivalence classes for $E_k$:
$$
pi_0={{a,b,c,d}, {d,e}}\
pi_1={{a,b,d},{c}, {d,e}}\
pi_2={{a,b},{d},{c}, {d,e}}\
pi_3=pi_2
$$
because we're arrived to the stage that $pi_3=pi_2$, we're done so we can present the minimal DFA:
I don't understand why in the solution there's a transition from $e$ to $d$ via $a,b$?
proof-explanation formal-languages automata
asked Nov 24 at 13:10
Yos
1,137723
add a comment |
I have the following DFA, let it be $A$:
The problem asks to find the minimal connected DFA for $A$, it is as follows according to the solution (the state ${d,e}$ is called $e$ for simplicity's sake):
This is my solution:
In order to find the minimal DFA we need to find the equivalence classes for $E_k$:
$$
pi_0={{a,b,c,d}, {d,e}}\
pi_1={{a,b,d},{c}, {d,e}}\
pi_2={{a,b},{d},{c}, {d,e}}\
pi_3=pi_2
$$
because we're arrived to the stage that $pi_3=pi_2$, we're done so we can present the minimal DFA:
I don't understand why in the solution there's a transition from $e$ to $d$ via $a,b$?
proof-explanation formal-languages automata
asked Nov 24 at 13:10
Yos
1,137723
I have the following DFA, let it be $A$:
The problem asks to find the minimal connected DFA for $A$, it is as follows according to the solution (the state ${d,e}$ is called $e$ for simplicity's sake):
This is my solution:
In order to find the minimal DFA we need to find the equivalence classes for $E_k$:
$$
pi_0={{a,b,c,d}, {d,e}}\
pi_1={{a,b,d},{c}, {d,e}}\
pi_2={{a,b},{d},{c}, {d,e}}\
pi_3=pi_2
$$
because we're arrived to the stage that $pi_3=pi_2$, we're done so we can present the minimal DFA:
I don't understand why in the solution there's a transition from $e$ to $d$ via $a,b$?
proof-explanation formal-languages automata
proof-explanation formal-languages automata
asked Nov 24 at 13:10
Yos
1,137723
asked Nov 24 at 13:10
Yos
1,137723
asked Nov 24 at 13:10
Yos
1,137723
asked Nov 24 at 13:10
Yos
1,137723
asked Nov 24 at 13:10
Yos
1,137723
1,137723
add a comment |
add a comment |
1 Answer
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Normally, state minimisation is performed on a complete DFA, in which there is a transition from every state on every symbol.
To make a DFA complete, a "sink" state is added and made the target of all missing transitions. The sink state is not final and all of its transitions are to itself.
In this case, during the minimisation the sink state will be merged with state $d$, which is identical.
answered Nov 24 at 14:58
rici
36829
add a comment |
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1 Answer
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1 Answer
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active
oldest
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active
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votes
Normally, state minimisation is performed on a complete DFA, in which there is a transition from every state on every symbol.
To make a DFA complete, a "sink" state is added and made the target of all missing transitions. The sink state is not final and all of its transitions are to itself.
In this case, during the minimisation the sink state will be merged with state $d$, which is identical.
answered Nov 24 at 14:58
rici
36829
add a comment |
Normally, state minimisation is performed on a complete DFA, in which there is a transition from every state on every symbol.
To make a DFA complete, a "sink" state is added and made the target of all missing transitions. The sink state is not final and all of its transitions are to itself.
In this case, during the minimisation the sink state will be merged with state $d$, which is identical.
answered Nov 24 at 14:58
rici
36829
add a comment |
Normally, state minimisation is performed on a complete DFA, in which there is a transition from every state on every symbol.
To make a DFA complete, a "sink" state is added and made the target of all missing transitions. The sink state is not final and all of its transitions are to itself.
In this case, during the minimisation the sink state will be merged with state $d$, which is identical.
answered Nov 24 at 14:58
rici
36829
Normally, state minimisation is performed on a complete DFA, in which there is a transition from every state on every symbol.
To make a DFA complete, a "sink" state is added and made the target of all missing transitions. The sink state is not final and all of its transitions are to itself.
In this case, during the minimisation the sink state will be merged with state $d$, which is identical.
answered Nov 24 at 14:58
rici
36829
edited Nov 24 at 20:43
answered Nov 24 at 14:58
rici
36829
answered Nov 24 at 14:58
rici
36829
answered Nov 24 at 14:58
rici
36829
36829
add a comment |
add a comment |
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