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Automata learning without counterexamples
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In Angluin's automata learning framework, a student aims to learn a regular language $Lsubseteq Sigma^*$ by asking two types of questions to his teacher:
Word queries: given $win Sigma^*$, is $win L$?
Equivalence queries: given a language $Ksubseteq Sigma^*$, is $K=L$? If not, the teacher gives a counterexample, i.e. a word $win Ksetminus L cup Lsetminus K$.
Using Angluin's algorithm, the student learns $L$ with polynomially many queries in the number of states of the minimal DFA of $L$ and the size of the counterexamples.
Now, consider a restricted scenario where the teacher no longer gives counterexamples. Is it still possible to learn L with a polynomial number of queries? I conjecture that this is not the case because for every polynomial-length sequence of queries and answers, one can find several regular languages consistent with the answers.
Does anyone see how to prove this?
automata-theory regular-language
asked 4 hours ago
user49692
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up vote
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In Angluin's automata learning framework, a student aims to learn a regular language $Lsubseteq Sigma^*$ by asking two types of questions to his teacher:
Word queries: given $win Sigma^*$, is $win L$?
Equivalence queries: given a language $Ksubseteq Sigma^*$, is $K=L$? If not, the teacher gives a counterexample, i.e. a word $win Ksetminus L cup Lsetminus K$.
Using Angluin's algorithm, the student learns $L$ with polynomially many queries in the number of states of the minimal DFA of $L$ and the size of the counterexamples.
Now, consider a restricted scenario where the teacher no longer gives counterexamples. Is it still possible to learn L with a polynomial number of queries? I conjecture that this is not the case because for every polynomial-length sequence of queries and answers, one can find several regular languages consistent with the answers.
Does anyone see how to prove this?
automata-theory regular-language
asked 4 hours ago
user49692
391
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
In Angluin's automata learning framework, a student aims to learn a regular language $Lsubseteq Sigma^*$ by asking two types of questions to his teacher:
Word queries: given $win Sigma^*$, is $win L$?
Equivalence queries: given a language $Ksubseteq Sigma^*$, is $K=L$? If not, the teacher gives a counterexample, i.e. a word $win Ksetminus L cup Lsetminus K$.
Using Angluin's algorithm, the student learns $L$ with polynomially many queries in the number of states of the minimal DFA of $L$ and the size of the counterexamples.
Now, consider a restricted scenario where the teacher no longer gives counterexamples. Is it still possible to learn L with a polynomial number of queries? I conjecture that this is not the case because for every polynomial-length sequence of queries and answers, one can find several regular languages consistent with the answers.
Does anyone see how to prove this?
automata-theory regular-language
asked 4 hours ago
user49692
391
In Angluin's automata learning framework, a student aims to learn a regular language $Lsubseteq Sigma^*$ by asking two types of questions to his teacher:
Word queries: given $win Sigma^*$, is $win L$?
Equivalence queries: given a language $Ksubseteq Sigma^*$, is $K=L$? If not, the teacher gives a counterexample, i.e. a word $win Ksetminus L cup Lsetminus K$.
Using Angluin's algorithm, the student learns $L$ with polynomially many queries in the number of states of the minimal DFA of $L$ and the size of the counterexamples.
Now, consider a restricted scenario where the teacher no longer gives counterexamples. Is it still possible to learn L with a polynomial number of queries? I conjecture that this is not the case because for every polynomial-length sequence of queries and answers, one can find several regular languages consistent with the answers.
Does anyone see how to prove this?
automata-theory regular-language
automata-theory regular-language
asked 4 hours ago
user49692
391
asked 4 hours ago
user49692
391
asked 4 hours ago
user49692
391
asked 4 hours ago
user49692
391
asked 4 hours ago
user49692
391
391
add a comment |Â
add a comment |Â
1 Answer
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Consider password automata: for each $win0,1^n$, the DFA $M_w$ accepts the language $w$. In this case, a membership query is the same as an equivalence query --- and clearly, you'll need exponentially many of these to find the "needle in the haystack". (This is even if the learner knows in advance that the target automaton is of this form.) For a formal proof, a teacher could simply respond to the first $2^n-1$ membership/equivalence queries with NO, adaptively choosing the target automaton as the unique remaining possibility.
Update. I neglected to mention that the password automaton has $n+1$ states and is learnable in $n+1$ usual Angluin queries.
answered 3 hours ago
Aryeh
4,9861835
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
Consider password automata: for each $win0,1^n$, the DFA $M_w$ accepts the language $w$. In this case, a membership query is the same as an equivalence query --- and clearly, you'll need exponentially many of these to find the "needle in the haystack". (This is even if the learner knows in advance that the target automaton is of this form.) For a formal proof, a teacher could simply respond to the first $2^n-1$ membership/equivalence queries with NO, adaptively choosing the target automaton as the unique remaining possibility.
Update. I neglected to mention that the password automaton has $n+1$ states and is learnable in $n+1$ usual Angluin queries.
answered 3 hours ago
Aryeh
4,9861835
add a comment |Â
up vote
3
down vote
Consider password automata: for each $win0,1^n$, the DFA $M_w$ accepts the language $w$. In this case, a membership query is the same as an equivalence query --- and clearly, you'll need exponentially many of these to find the "needle in the haystack". (This is even if the learner knows in advance that the target automaton is of this form.) For a formal proof, a teacher could simply respond to the first $2^n-1$ membership/equivalence queries with NO, adaptively choosing the target automaton as the unique remaining possibility.
Update. I neglected to mention that the password automaton has $n+1$ states and is learnable in $n+1$ usual Angluin queries.
answered 3 hours ago
Aryeh
4,9861835
add a comment |Â
up vote
3
down vote
up vote
3
down vote
Consider password automata: for each $win0,1^n$, the DFA $M_w$ accepts the language $w$. In this case, a membership query is the same as an equivalence query --- and clearly, you'll need exponentially many of these to find the "needle in the haystack". (This is even if the learner knows in advance that the target automaton is of this form.) For a formal proof, a teacher could simply respond to the first $2^n-1$ membership/equivalence queries with NO, adaptively choosing the target automaton as the unique remaining possibility.
Update. I neglected to mention that the password automaton has $n+1$ states and is learnable in $n+1$ usual Angluin queries.
answered 3 hours ago
Aryeh
4,9861835
Consider password automata: for each $win0,1^n$, the DFA $M_w$ accepts the language $w$. In this case, a membership query is the same as an equivalence query --- and clearly, you'll need exponentially many of these to find the "needle in the haystack". (This is even if the learner knows in advance that the target automaton is of this form.) For a formal proof, a teacher could simply respond to the first $2^n-1$ membership/equivalence queries with NO, adaptively choosing the target automaton as the unique remaining possibility.
Update. I neglected to mention that the password automaton has $n+1$ states and is learnable in $n+1$ usual Angluin queries.
answered 3 hours ago
Aryeh
4,9861835
edited 40 mins ago
answered 3 hours ago
Aryeh
4,9861835
answered 3 hours ago
Aryeh
4,9861835
answered 3 hours ago
Aryeh
4,9861835
4,9861835
add a comment |Â
add a comment |Â
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