Design DFA to check even number of 1s using 2 states.
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The alphabet is {0,1}. Zero 1s in the string should be rejected. This can be done easily using 3 states. Can this be done usinng 2 states?
computer-science automata
asked Jan 13 at 6:57
user415612user415612
1
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add a comment |
$begingroup$
The alphabet is {0,1}. Zero 1s in the string should be rejected. This can be done easily using 3 states. Can this be done usinng 2 states?
computer-science automata
asked Jan 13 at 6:57
user415612user415612
1
$endgroup$
$begingroup$
You need a start state and a state that says there are an odd number of $1$s. Since you are rejecting zero $1$s, the start state cannot correspond to the end state so you need at least $3$ states.
$endgroup$
– John Douma
Jan 13 at 7:06
add a comment |
$begingroup$
The alphabet is {0,1}. Zero 1s in the string should be rejected. This can be done easily using 3 states. Can this be done usinng 2 states?
computer-science automata
asked Jan 13 at 6:57
user415612user415612
1
$endgroup$
The alphabet is {0,1}. Zero 1s in the string should be rejected. This can be done easily using 3 states. Can this be done usinng 2 states?
computer-science automata
computer-science automata
asked Jan 13 at 6:57
user415612user415612
1
asked Jan 13 at 6:57
user415612user415612
1
asked Jan 13 at 6:57
user415612user415612
1
asked Jan 13 at 6:57
user415612user415612
1
asked Jan 13 at 6:57
user415612user415612
1
1
$begingroup$
You need a start state and a state that says there are an odd number of $1$s. Since you are rejecting zero $1$s, the start state cannot correspond to the end state so you need at least $3$ states.
$endgroup$
– John Douma
Jan 13 at 7:06
add a comment |
$begingroup$
You need a start state and a state that says there are an odd number of $1$s. Since you are rejecting zero $1$s, the start state cannot correspond to the end state so you need at least $3$ states.
$endgroup$
– John Douma
Jan 13 at 7:06
$begingroup$
You need a start state and a state that says there are an odd number of $1$s. Since you are rejecting zero $1$s, the start state cannot correspond to the end state so you need at least $3$ states.
$endgroup$
– John Douma
Jan 13 at 7:06
$begingroup$
You need a start state and a state that says there are an odd number of $1$s. Since you are rejecting zero $1$s, the start state cannot correspond to the end state so you need at least $3$ states.
$endgroup$
– John Douma
Jan 13 at 7:06
add a comment |
1 Answer
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$begingroup$
How about $s_0rightarrow_0 s_0$, $s_0rightarrow_1 s_1$, and $s_1rightarrow_{0,1} s_1$ with start state $s_0$ and end state $s_1$.
Ahh, okay. Sorry for misunderstanding. An even number of 1's. Then the automaton looks as follows:
$s_0rightarrow_0 s_0$, $s_0rightarrow_1 s_1$, $s_1rightarrow_0 s_1$, and $s_1rightarrow_{1} s_0$, where $s_0$ is the starting and ending state.
answered Jan 13 at 8:00
WuestenfuxWuestenfux
5,1271513
$endgroup$
$begingroup$
In the title the OP requires also an even number of 1s, not just a non-zero number.
$endgroup$
– Peter Leupold
Jan 14 at 19:37
$begingroup$
"Zero 1s in the string should be rejected". This will accept 0.
$endgroup$
– user415612
Jan 15 at 18:25
add a comment |
Your Answer
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1 Answer
1
active
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1 Answer
1
active
oldest
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$begingroup$
How about $s_0rightarrow_0 s_0$, $s_0rightarrow_1 s_1$, and $s_1rightarrow_{0,1} s_1$ with start state $s_0$ and end state $s_1$.
Ahh, okay. Sorry for misunderstanding. An even number of 1's. Then the automaton looks as follows:
$s_0rightarrow_0 s_0$, $s_0rightarrow_1 s_1$, $s_1rightarrow_0 s_1$, and $s_1rightarrow_{1} s_0$, where $s_0$ is the starting and ending state.
answered Jan 13 at 8:00
WuestenfuxWuestenfux
5,1271513
$endgroup$
$begingroup$
In the title the OP requires also an even number of 1s, not just a non-zero number.
$endgroup$
– Peter Leupold
Jan 14 at 19:37
$begingroup$
"Zero 1s in the string should be rejected". This will accept 0.
$endgroup$
– user415612
Jan 15 at 18:25
add a comment |
$begingroup$
How about $s_0rightarrow_0 s_0$, $s_0rightarrow_1 s_1$, and $s_1rightarrow_{0,1} s_1$ with start state $s_0$ and end state $s_1$.
Ahh, okay. Sorry for misunderstanding. An even number of 1's. Then the automaton looks as follows:
$s_0rightarrow_0 s_0$, $s_0rightarrow_1 s_1$, $s_1rightarrow_0 s_1$, and $s_1rightarrow_{1} s_0$, where $s_0$ is the starting and ending state.
answered Jan 13 at 8:00
WuestenfuxWuestenfux
5,1271513
$endgroup$
$begingroup$
In the title the OP requires also an even number of 1s, not just a non-zero number.
$endgroup$
– Peter Leupold
Jan 14 at 19:37
$begingroup$
"Zero 1s in the string should be rejected". This will accept 0.
$endgroup$
– user415612
Jan 15 at 18:25
add a comment |
$begingroup$
How about $s_0rightarrow_0 s_0$, $s_0rightarrow_1 s_1$, and $s_1rightarrow_{0,1} s_1$ with start state $s_0$ and end state $s_1$.
Ahh, okay. Sorry for misunderstanding. An even number of 1's. Then the automaton looks as follows:
$s_0rightarrow_0 s_0$, $s_0rightarrow_1 s_1$, $s_1rightarrow_0 s_1$, and $s_1rightarrow_{1} s_0$, where $s_0$ is the starting and ending state.
answered Jan 13 at 8:00
WuestenfuxWuestenfux
5,1271513
$endgroup$
How about $s_0rightarrow_0 s_0$, $s_0rightarrow_1 s_1$, and $s_1rightarrow_{0,1} s_1$ with start state $s_0$ and end state $s_1$.
Ahh, okay. Sorry for misunderstanding. An even number of 1's. Then the automaton looks as follows:
$s_0rightarrow_0 s_0$, $s_0rightarrow_1 s_1$, $s_1rightarrow_0 s_1$, and $s_1rightarrow_{1} s_0$, where $s_0$ is the starting and ending state.
answered Jan 13 at 8:00
WuestenfuxWuestenfux
5,1271513
edited Jan 15 at 8:47
answered Jan 13 at 8:00
WuestenfuxWuestenfux
5,1271513
answered Jan 13 at 8:00
WuestenfuxWuestenfux
5,1271513
answered Jan 13 at 8:00
WuestenfuxWuestenfux
5,1271513
5,1271513
$begingroup$
In the title the OP requires also an even number of 1s, not just a non-zero number.
$endgroup$
– Peter Leupold
Jan 14 at 19:37
$begingroup$
"Zero 1s in the string should be rejected". This will accept 0.
$endgroup$
– user415612
Jan 15 at 18:25
add a comment |
$begingroup$
In the title the OP requires also an even number of 1s, not just a non-zero number.
$endgroup$
– Peter Leupold
Jan 14 at 19:37
$begingroup$
"Zero 1s in the string should be rejected". This will accept 0.
$endgroup$
– user415612
Jan 15 at 18:25
$begingroup$
In the title the OP requires also an even number of 1s, not just a non-zero number.
$endgroup$
– Peter Leupold
Jan 14 at 19:37
$begingroup$
In the title the OP requires also an even number of 1s, not just a non-zero number.
$endgroup$
– Peter Leupold
Jan 14 at 19:37
$begingroup$
"Zero 1s in the string should be rejected". This will accept 0.
$endgroup$
– user415612
Jan 15 at 18:25
$begingroup$
"Zero 1s in the string should be rejected". This will accept 0.
$endgroup$
– user415612
Jan 15 at 18:25
add a comment |
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$begingroup$
You need a start state and a state that says there are an odd number of $1$s. Since you are rejecting zero $1$s, the start state cannot correspond to the end state so you need at least $3$ states.
$endgroup$
– John Douma
Jan 13 at 7:06