Avoiding Determinization

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Avoiding Determinization
Orna Kupferman Hebrew University
Joint work with Moshe Vardi
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Deterministic (automaton, Turing machine, person)
Success happily ever after in the unique future.
Nondeterministic (automaton, Turing machine,
person)
Success happily ever after in at least one
future.
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A very convenient definition of success Risk is
for free. Goals are achieved in a more succinct
way.
perhaps
Success happily ever after in at least one
future.
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NFW are exponentially more succinct than DFW
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NFW are exponentially more succinct than DFW
Ln (01).0.(01)n
NFW O(n) states
DFW O(2n) states
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Sometimes, nondeterminism causes no problems.
Nonemptiness check L(A) ? Ø ?
Membership check, projection,
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Sometimes, nondeterminism is problematic.
1. Complementation L(A) comp(L(A))
DFW dualize the acceptance condition.
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Sometimes, nondeterminism is problematic.
1. Complementation L(A) comp(L(A))
DFW dualize the acceptance condition.
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0
1
0
L(A) (01).1
comp(L(A)) e (01).0
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Sometimes, nondeterminism is problematic.
1. Complementation L(A) comp(L(A))
NFW dualize the acceptance condition?
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Sometimes, nondeterminism is problematic.
1. Complementation L(A) comp(L(A))
L(A) (01)
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Sometimes, nondeterminism is problematic.
2. Running A on a tree.
Tree automata
Word automata M(q0,a)q1,q2
Tree automata M(q0,a)?q1,q3?, ?q2,q1?
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Sometimes, nondeterminism is problematic.
2. Running A on a tree.
Given an NFW A.
Wanted an NFT At that accepts all trees all of
whose paths are accepted by A.
Mt(q,a) M(q,a) x M(q,a)
M(q0,a)q1,q2
Mt(q0,a)?q1,q1?, ?q1,q2?, ?q2,q1?, ?q2,q2?
Run A on each of the paths of the tree
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Sometimes, nondeterminism is problematic.
2. Running A on a tree.
0, 1
0, 1
0, 1
0, 1
0
n
L(A) (01).0.(01)n
. . .
A whenever you read 0, guess whether the input
ends after exactly n letters.
At whenever you read 0, guess whether all the
paths in the subtree end after exactly n letters.
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Sometimes, nondeterminism is problematic.
2. Running A on a tree.
0, 1
0, 1
0, 1
0, 1
0
?
?
n
L(A) (01).0.(01)n
. . .
A whenever you read 0, guess whether the input
ends after exactly n letters.
At whenever you read 0, guess whether all the
paths in the subtree end after exactly n letters.
Mt(?,0)??,??, ??,??, ??,??, ??,??
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Sometimes, nondeterminism is problematic.
2. Running A on a tree.
0, 1
0, 1
0, 1
0, 1
0
?
?
n
L(A) (01).0.(01)n
. . .
A whenever you read 0, guess whether the input
ends after exactly n letters.
Mt(?,0)??,??, ??,??, ??,??, ??,??
?
n2
?
?
?
?
?
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Sometimes, nondeterminism is problematic.
2. Running A on a tree.

• Applications in
• Decidability of CTL, µ-calculus, SE84, EJ91,
• Solving games with ?-regular goals. Tho95
• LTL realizability and synthesis. RP89

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How to solve complementation, decidability,
games, synthesis,?
DETERMINIZE!
1 problem exponential blow-up.
legitimate (matching lower bounds).
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Automata on infinite words
Büchi acceptance visit a infinitely often
L(A) (01).1?
There is no DBW for L(A) Lan69.
Safras determinization construction 1988 NBW(n)
? DRW(2O(n log n), O(n))
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Safras construction
- First optimal construction!
Each state of A is an ordered tree in which each
node is labeled by a subset of the states of A
such that the label of a node is
- Beautiful!
Very Complicated!!!
MONA implementation of a nonelementary algorithm
Model checking tools! A success
story!! Synthesis no tools, no story.
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Avoid determinization!!!
Weare
Universality is the answer!!!
We need Synthesis
afraid of Safra
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Universal (automaton, Turing machine, person)
Success happily ever after in all futures.
When viewed as an NFW, L(A) (01)
When viewed as a UFW, L(A)
(010)
every 1 is followed by 0
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Universal automata are sufficiently strong to
serve as intermediate automata in many
applications in which deterministic automata are
traditionally used as intermediate automata.

• Complementation constructions
• Decision procedures
• Synthesis

• Talk outline
• Indeed sufficiently strong.
• Much simpler! Promising practical applications.

• You name it (please do)

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Complementation
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Complementation
Given an NBW A, construct a complementary NBW.
Current procedure

• Construct a DRW equivalent to A.
• Dualize the DRW.
• Translate the result to an NBW.

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Complementation

Given an NBW A, construct a complementary NBW.
Our procedure
GSKV03

• NBW ? complementary UCW.
• UCW ? NBW.

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Example
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1
1
0
s1
s2
s3
UCW
L(A) (10)? infinitely many 0s
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s1
0 1 0
s1
L(A) (10)?
s1
s2
s1
s3
a ranking function fV ? 0,,2n
1 1 1
s3
s1
s2
s2
s3
s1
s2
s3
s1
0
s1
s3
0
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s1
3
0 1 0
s1
3
L(A) (10)?
s1
s2
2
3
s1
s3
1
3
a ranking function fV ? 0,,2n
1 1 1
s3
s1
s2
3
1
2
s2
s3
s1
1
2
3
s2
The state space of the NBW subset construction
ranks for the states in the subset
s3
s1
1
2
3
0
s1
s3
3
1
0
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z z z
Complementation
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µ-calculus satisfiability
Is there a tree satisfying ??
Current procedure

• Construct an APT A? that accepts all trees
  satisfying ? EJ91,KVW00.
• APT ? NPT.
• Check emptiness of the NPT.

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µ-calculus satisfiability
Is there a tree satisfying ??
Our procedure

• Construct an APT A? that accepts all trees
  satisfying ? EJ91,KVW00.
• APT ? UCT.
• Check emptiness of the UCT.

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Complementation
µ-calculus satisfiability
Synthesis
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Synthesis
Given an LTL formula ? over I and O, construct a
finite-state strategy f (2I) ? 2O that
generates only computations that satisfy ?.
Open system interacts with an environment!
o0
o1f(i0)
i0
o2f(i0,i1)
i1
o3f(i0,i1,i2)
i2
(f(?)) ? (i0,f(i0)) ? (i1,f(i0,i1)) ?
(i2,f(i0,i1,i2)) ?
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Synthesis
Is ? realizable?
Current procedure PR88

• Construct a DRW A? that accepts all computations
  satisfying ?.
• Run A? on the I-exhaustive tree.
• Check emptiness of the NRT.

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Synthesis
Is ? realizable?
Our procedure

• Construct an NBW A? that accepts all
  computations satisfying ? VW94.
• Run the dual UCW on
  the I-exhaustive tree.
• Check emptiness of the UCT.

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The magic
UCW ? NBW UCT ? NBT
Based on an analysis of accepting runs of
co-Büchi automata
A run is accepting iff the vertices of its run
DAG can get ranks in 0,,k so that ranks along
paths decrease and odd ranks appear only finitely
often.
The NBW/NBT guesses a ranking, checks decrease,
checks infinitely many visits to even ranks.
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A run is accepting iff the vertices of its run
DAG can get ranks in 0,,k so that ranks along
paths decrease and odd ranks appear only finitely
often.
k

• Width of the run DAG
• For UCW bounded by n.
• For UCT ???
• If the UCT accepts some tree, it also accepts a
  tree generated by a transducer with
  k(2n!)n2n3n(n1)/n! states.

We still need Safra!
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• In practice GSKV03
• Incremental search for k.
• Symbolic implementation.

The magic
UCW ? NBW UCT ? NBT
Based on an analysis of accepting runs of
co-Büchi automata
A run is accepting iff the vertices of its run
DAG can get ranks in 0,,k so that ranks along
paths decrease and odd ranks appear only finitely
often.
The NBW/NBT guesses a ranking, checks decrease,
checks infinitely many visits to even ranks.
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• To sum up
• Many applications use determinization.
• The complexity of Safras determinization
  prevents implementations.
• Universality can replace determinization and
  results in much simpler and friendlier procedures.

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The end
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1
0
s1
s2
s3
0,1
0,1
s1
0 1 1 . . .
s1
s1
s2
s2
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1
0
s1
s2
s3
0,1
0,1
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Complementation
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NBW UCW NBW
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NBW UCW NBW
Easy dualize both branching mode and acceptance
condition. O(1)
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NBW UCW NBW
KV97
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(No Transcript)
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1
1
0
s1
s2
s3
UCW
L(A) (10)?
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1
1
0
s1
s2
s3
s1
0 1 0
L(A) (10)?
s1
s1
s2
a ranking function fV ? 0,,2n
s1
s3
1 1 1
s3
s1
s2
s1
s2
s3
The state space of the NBW subset construction
ranks for the states in the subset
s1
s2
s3
0
s1
s3
0
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z z z
Complementation
GSKV03