[PPT] - Compressed Membership for NFA (DFA) with Compressed Labels is in NP PowerPoint Presentation

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Compressed Membership for NFA (DFA) with Compressed Labels is in NP (P)

Artur Jeż

Wrocław, Poland

7 September 2011

Artur Jeż Compressed membership for NFA 7 September 2011 1 / 25

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Straight Line Programms SLPs

Definition (Straight Line Programms (SLP))

Context free grammar defining a single word. (Chomsky normal form).

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Straight Line Programms SLPs

Definition (Straight Line Programms (SLP))

Context free grammar defining a single word. (Chomsky normal form). Much smaller, than the word. Word an: grammar size O(log n) A1 → a A2 → A1A1 . . . Aℓ+1 → AℓAℓ A → Aℓ0Aℓ1 . . .

Artur Jeż Compressed membership for NFA 7 September 2011 2 / 25

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Straight Line Programms SLPs

Definition (Straight Line Programms (SLP))

Context free grammar defining a single word. (Chomsky normal form). Much smaller, than the word. Word an: grammar size O(log n) A1 → a A2 → A1A1 . . . Aℓ+1 → AℓAℓ A → Aℓ0Aℓ1 . . .

SLPs as a compression model

application (LZ, logarithmic transformation) theory (formal languages) up to exponential compression preserves/captures word properties

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Usage and work on SLP

Theory

word equations (Plandowski: satisfiability in PSPACE)

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Usage and work on SLP

Theory

word equations (Plandowski: satisfiability in PSPACE)

String algorithms

equality pattern matching

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Usage and work on SLP

Theory

word equations (Plandowski: satisfiability in PSPACE)

String algorithms

equality pattern matching

LZW/LZ dealing algorithms

O(n log n) pattern matching for LZ compressed text O(n) pattern matching for fully LZW compressed text

Artur Jeż Compressed membership for NFA 7 September 2011 3 / 25

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Usage and work on SLP

Theory

word equations (Plandowski: satisfiability in PSPACE)

String algorithms

equality pattern matching

LZW/LZ dealing algorithms

O(n log n) pattern matching for LZ compressed text O(n) pattern matching for fully LZW compressed text

Independent interest

indexing structure for SLP

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Compressed membership

SLPs are used membership problem develop tools/gain understanding

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Compressed membership

SLPs are used membership problem develop tools/gain understanding

Compressed membership [Plandowski & Rytter; Jewels are forever 1999]

In membership problems, words are given as SLPs.

Artur Jeż Compressed membership for NFA 7 September 2011 4 / 25

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Compressed membership

SLPs are used membership problem develop tools/gain understanding

Compressed membership [Plandowski & Rytter; Jewels are forever 1999]

In membership problems, words are given as SLPs.

Known results

RE, CFG, Conjunctive grammars . . .

Artur Jeż Compressed membership for NFA 7 September 2011 4 / 25

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Compressed membership

SLPs are used membership problem develop tools/gain understanding

Compressed membership [Plandowski & Rytter; Jewels are forever 1999]

In membership problems, words are given as SLPs.

Known results

RE, CFG, Conjunctive grammars . . .

Open questions

Posted in Jewels are forever some solved Compressed membership for NFA

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Compressed membership for NFA

Input: SLP, NFA N Output: Yes/No

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Compressed membership for NFA

Input: SLP, NFA N Output: Yes/No Simple dynamic algorithm: for Xi calculate {(p, q) | δ(p, word(Xi), q)}

Artur Jeż Compressed membership for NFA 7 September 2011 5 / 25

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Compressed membership for NFA

Input: SLP, NFA N Output: Yes/No Simple dynamic algorithm: for Xi calculate {(p, q) | δ(p, word(Xi), q)} Where is the hardness?

Artur Jeż Compressed membership for NFA 7 September 2011 5 / 25

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Compressed membership for NFA

Input: SLP, NFA N Output: Yes/No Simple dynamic algorithm: for Xi calculate {(p, q) | δ(p, word(Xi), q)} Where is the hardness? Compress N as well: allow transition by words.

Artur Jeż Compressed membership for NFA 7 September 2011 5 / 25

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Compressed membership for NFA

Input: SLP, NFA N Output: Yes/No Simple dynamic algorithm: for Xi calculate {(p, q) | δ(p, word(Xi), q)} Where is the hardness? Compress N as well: allow transition by words. Fully compressed NFA membership SLP for w NFA N, compressed transitions

Artur Jeż Compressed membership for NFA 7 September 2011 5 / 25

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Compressed membership for NFA

Input: SLP, NFA N Output: Yes/No Simple dynamic algorithm: for Xi calculate {(p, q) | δ(p, word(Xi), q)} Where is the hardness? Compress N as well: allow transition by words. Fully compressed NFA membership SLP for w NFA N, compressed transitions

a X Y b A X

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Compressed membership for NFA: complexity

Complexity

NP-hardness (subsum), already for

◮ acyclic NFA ◮ unary alphabet

in PSPACE: enough to store positions inside decompressed words

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Compressed membership for NFA: complexity

Complexity

NP-hardness (subsum), already for

◮ acyclic NFA ◮ unary alphabet

in PSPACE: enough to store positions inside decompressed words

Conjecture

In NP.

Partial results

Plandowski & Rytter (unary in NP) Lohrey & Mathissen (highly periodic in NP, highly aperiodic in P)

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New results

Theorem

Fully compressed membership for NFA is in NP.

Theorem

Fully compressed membership for DFA is in P.

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New results

Theorem

Fully compressed membership for NFA is in NP.

Theorem

Fully compressed membership for DFA is in P.

New technique

New, interesting technique.

Artur Jeż Compressed membership for NFA 7 September 2011 7 / 25

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Convention

SLPs given as a single grammar Xi → XjXk, implies i > j, i > k input word: Xn word(Xi)

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Idea: Recompression

Difficulty: the words are long. Shorten them.

Artur Jeż Compressed membership for NFA 7 September 2011 9 / 25

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Idea: Recompression

Difficulty: the words are long. Shorten them. a b c a a b

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Idea: Recompression

Difficulty: the words are long. Shorten them. d c a d

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Idea: Recompression

Difficulty: the words are long. Shorten them. d c a d

Deeper understanding

New production: d → ab. Building new SLP (recompression). SLP problems: hard, as SLP are different. Building canonical SLP for the instance.

Artur Jeż Compressed membership for NFA 7 September 2011 9 / 25

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Idea: Recompression

Difficulty: the words are long. Shorten them. d c a d

Deeper understanding

New production: d → ab. Building new SLP (recompression). SLP problems: hard, as SLP are different. Building canonical SLP for the instance. What to do with an? a a c a a a

Artur Jeż Compressed membership for NFA 7 September 2011 9 / 25

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Idea: Recompression

Difficulty: the words are long. Shorten them. d c a d

Deeper understanding

New production: d → ab. Building new SLP (recompression). SLP problems: hard, as SLP are different. Building canonical SLP for the instance. What to do with an? Replace each non-extendible an by a single symbol. a2 c a3

Artur Jeż Compressed membership for NFA 7 September 2011 9 / 25

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Idea: Recompression

Difficulty: the words are long. Shorten them. d c a d

Deeper understanding

New production: d → ab. Building new SLP (recompression). SLP problems: hard, as SLP are different. Building canonical SLP for the instance. What to do with an? Replace each non-extendible an by a single symbol. a2 c a3

Problems

easy for text, what about grammar? what to do with the NFA?

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Local decompression

Re-compression

decompressed text: easy; size: large, compressed text: hard; size: small.

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Local decompression

Re-compression

decompressed text: easy; size: large, compressed text: hard; size: small.

Local decompression

Decompress locally the SLP: X → uYvZ u, v: blocks of letters, linear size Y , Z: nonterminals recompression inside u, v

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Outline

Outline of the algorithm

while | word(Xn) > n| do LΣ ← list of letters, LP ← list of pairs for ab ∈ LP do compress pair ab, modify N accordingly for a ∈ LΣ do compress a non-extendible appearances, modify N accordingly Decompress the word and solve the problem naively.

Artur Jeż Compressed membership for NFA 7 September 2011 11 / 25

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Outline

Outline of the algorithm

while | word(Xn) > n| do LΣ ← list of letters, LP ← list of pairs for ab ∈ LP do compress pair ab, modify N accordingly for a ∈ LΣ do compress a non-extendible appearances, modify N accordingly Decompress the word and solve the problem naively.

Theorem

There are O(log word(Xn)) iterations.

Proof.

Consider two consecutive letters ab. One of them is compressed. So word shortens by a constant factor.

Artur Jeż Compressed membership for NFA 7 September 2011 11 / 25

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Formalisation

New symbols: letters.

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Formalisation

New symbols: letters.

Grammar invariants

the set of nonterminals is {Xn, . . . , X1} (the input ones) the productions are of the form Xi → uXjvXk

r

Xi → uXjv

r

Xi → u, for Xi → uXjvXk, the input had a production Xi → XjXk

Artur Jeż Compressed membership for NFA 7 September 2011 12 / 25

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Formalisation

New symbols: letters.

Grammar invariants

the set of nonterminals is {Xn, . . . , X1} (the input ones) the productions are of the form Xi → uXjvXk

r

Xi → uXjv

r

Xi → u, for Xi → uXjvXk, the input had a production Xi → XjXk In this way the grammar does not blow up (the skeleton is the same).

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NFA invariants

NFA cannot blow up either.

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NFA invariants

NFA cannot blow up either.

NFA invariants

transitions of N: letter transitions labelled by a single letter nonterminal transition labelled by nonterminal nonterminal transition have counterparts in the input NFA

◮ the same nonterminal ◮ corresponding start and end ◮ the same multiplicity Artur Jeż Compressed membership for NFA 7 September 2011 13 / 25

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NFA invariants

NFA cannot blow up either.

NFA invariants

transitions of N: letter transitions labelled by a single letter nonterminal transition labelled by nonterminal nonterminal transition have counterparts in the input NFA

◮ the same nonterminal ◮ corresponding start and end ◮ the same multiplicity

The nonterminal part is ‘the same’ (of size n).

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What is hard, what is easy

What is hard to compress, what easy?

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What is hard, what is easy

What is hard to compress, what easy?

Hard

a letter a is outer, if it is the first or last letter of some word(Xi) a pair ab is crossing if

◮ Xi → uaXjvXk, where word(Xj) = b . . . ◮ ab spreads over transitions

at least one nonterminal a b

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What is hard, what is easy

What is hard to compress, what easy?

Hard

a letter a is outer, if it is the first or last letter of some word(Xi) a pair ab is crossing if

◮ Xi → uaXjvXk, where word(Xj) = b . . . ◮ ab spreads over transitions

at least one nonterminal a b

Easy

a letter a is inner otherwise a pair ab is non-crossing otherwise

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A little detailed outline

Detailed outline

while | word(Xn) > n| do while possible do for a: inner letter do compress appearances of a for non-crossing pair ab in word(Xn) do compress ab

Artur Jeż Compressed membership for NFA 7 September 2011 15 / 25

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A little detailed outline

Detailed outline

while | word(Xn) > n| do while possible do for a: inner letter do compress appearances of a for non-crossing pair ab in word(Xn) do compress ab L ← list of all outer letters for a ∈ L do compress appearances of a for each akb in word(Xn) do compress akb

Artur Jeż Compressed membership for NFA 7 September 2011 15 / 25

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A little detailed outline

Detailed outline

while | word(Xn) > n| do while possible do for a: inner letter do compress appearances of a for non-crossing pair ab in word(Xn) do compress ab L ← list of all outer letters for a ∈ L do compress appearances of a for each akb in word(Xn) do compress akb Decompress Xn and solve the problem naively.

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Non-crossing pair compression

for each production Xi → uXjvXk do replace each ab in u, v by c

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Non-crossing pair compression

for each production Xi → uXjvXk do replace each ab in u, v by c for states p, q do if δN(p, ab, q) then put a transition δN(p, c, q)

Artur Jeż Compressed membership for NFA 7 September 2011 16 / 25

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Non-crossing pair compression

for each production Xi → uXjvXk do replace each ab in u, v by c for states p, q do if δN(p, ab, q) then put a transition δN(p, c, q)

a b c

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Non-crossing pair compression

for each production Xi → uXjvXk do replace each ab in u, v by c for states p, q do if δN(p, ab, q) then put a transition δN(p, c, q)

a b c

Lemma

It works.

Proof.

The pair is non-crossing: it always appears inside production.

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Inner letter compression

Appearance compression for and inner letter a

compute the lengths ℓ1, . . . , ℓk of a’s non-extendible appearance for each aℓm do for each production Xi → uXjvXk do replace non-extendible aℓm in in u, v by aℓm

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Inner letter compression

Appearance compression for and inner letter a

compute the lengths ℓ1, . . . , ℓk of a’s non-extendible appearance for each aℓm do for each production Xi → uXjvXk do replace non-extendible aℓm in in u, v by aℓm for states p, q in N do guess, whether δN(p, aℓm, q) if guess is positive then verify the guess put a transition δN′(p, aℓm, q)

Artur Jeż Compressed membership for NFA 7 September 2011 17 / 25

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Inner letter compression

Appearance compression for and inner letter a

compute the lengths ℓ1, . . . , ℓk of a’s non-extendible appearance for each aℓm do for each production Xi → uXjvXk do replace non-extendible aℓm in in u, v by aℓm for states p, q in N do guess, whether δN(p, aℓm, q) if guess is positive then verify the guess put a transition δN′(p, aℓm, q)

a a a a a a2 a

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Inner letter compression

Appearance compression for and inner letter a

compute the lengths ℓ1, . . . , ℓk of a’s non-extendible appearance for each aℓm do for each production Xi → uXjvXk do replace non-extendible aℓm in in u, v by aℓm for states p, q in N do guess, whether δN(p, aℓm, q) if guess is positive then verify the guess put a transition δN′(p, aℓm, q)

a a a a a a2 a a4

Artur Jeż Compressed membership for NFA 7 September 2011 17 / 25

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Inner letter compression

Appearance compression for and inner letter a

compute the lengths ℓ1, . . . , ℓk of a’s non-extendible appearance for each aℓm do for each production Xi → uXjvXk do replace non-extendible aℓm in in u, v by aℓm for states p, q in N do guess, whether δN(p, aℓm, q) if guess is positive then verify the guess put a transition δN′(p, aℓm, q)

a a a a a a2 a a4

Lemma

It works. how many lengths: inner letter verification: Plandowski & Rytter result

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Crossing pairs and outer letters

Lemma

O(n) outer letters and O(n2) crossing pairs appearing in word(Xn)

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Crossing pairs and outer letters

Lemma

O(n) outer letters and O(n2) crossing pairs appearing in word(Xn)

Proof.

charge outer letter to the non-terminal

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Crossing pairs and outer letters

Lemma

O(n) outer letters and O(n2) crossing pairs appearing in word(Xn)

Proof.

charge outer letter to the non-terminal pairs in word(Xn):

◮ appears is some explicit word uXjvXk, total size O(n2). ◮ spreads over u word(Xj),

charged to a production Xi → uXjvXk, total size O(n).

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Convert hard to easy

Convert outer letters to inner, crossing pairs to non-crossing (Sequentially).

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Convert hard to easy

Convert outer letters to inner, crossing pairs to non-crossing (Sequentially). fix an outer letter a convert it to inner compress a appearances make each pair of the form aℓb non-crossing compress each such pair

Artur Jeż Compressed membership for NFA 7 September 2011 19 / 25

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Turning a into an inner letter

Cut a-prefix and a-suffix from each nonterminal. Represent word(Xi) as aℓiwari, turn it into w.

◮ Xi → uXjvXk ◮ word(Xi) = u word(Xj)v word(Xk) Artur Jeż Compressed membership for NFA 7 September 2011 20 / 25

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Turning a into an inner letter

Cut a-prefix and a-suffix from each nonterminal. Represent word(Xi) as aℓiwari, turn it into w.

◮ Xi → uXjvXk ◮ word(Xi) = u word(Xj)v word(Xk)

Changing an outer letter a to an inner one

for i = 1 . . n do let Xi → uXjvXk replace Xi → uaℓjX ′

j arjvaℓkX ′ kark

calculate the a-prefix aℓi and a-suffix ari, remove them

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Turning a into an inner letter

Cut a-prefix and a-suffix from each nonterminal. Represent word(Xi) as aℓiwari, turn it into w.

◮ Xi → uXjvXk ◮ word(Xi) = u word(Xj)v word(Xk)

Changing an outer letter a to an inner one

for i = 1 . . n do let Xi → uXjvXk replace Xi → uaℓjX ′

j arjvaℓkX ′ kark

calculate the a-prefix aℓi and a-suffix ari, remove them

Lemma

The algorithm makes a an inner letter.

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Adjusting the NFA

Adjust the transitions δ(p, Xi, q)

Xi Xi aℓi ari

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Adjusting the NFA

Adjust the transitions δ(p, Xi, q)

Xi Xi aℓi ari

Lemma

This adjusts the NFA properly.

Artur Jeż Compressed membership for NFA 7 September 2011 21 / 25

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Adjusting the NFA

Adjust the transitions δ(p, Xi, q)

Xi Xi aℓi ari

Lemma

This adjusts the NFA properly.

aℓ

aℓ in NFA and grammar.

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Adjusting the NFA

Adjust the transitions δ(p, Xi, q)

Xi Xi aℓi ari

Lemma

This adjusts the NFA properly.

aℓ

aℓ in NFA and grammar.

Appearance compression

guess and verify, whether δN(p, aℓ, q) by Plandowski & Rytter result (unary case in NP)

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Crossing to non-crossing

Turn pairs aℓb into non-crossing.

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Crossing to non-crossing

Turn pairs aℓb into non-crossing. aℓ is inner ‘pop’ first letter of each non-terminal replace: word(Xi) = bu → word(Xi) = u

a b Xi

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Crossing to non-crossing

Turn pairs aℓb into non-crossing. aℓ is inner ‘pop’ first letter of each non-terminal replace: word(Xi) = bu → word(Xi) = u

a b Xi

Lemma

After popping letters, no pair aℓb is crossing.

Proof.

Easy, but goes into details.

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Sizes and running time

Running time

All algorithms run in time npoly(n, |G|, |Σ|, |N|).

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Sizes and running time

Running time

All algorithms run in time npoly(n, |G|, |Σ|, |N|).

Size of G

abbbcceaXjaddfeaaf Xk In each iteration

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Sizes and running time

Running time

All algorithms run in time npoly(n, |G|, |Σ|, |N|).

Size of G

abbbcceabhaXjabaddfeaaf cdaXk In each iteration O(n) new letters

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Sizes and running time

Running time

All algorithms run in time npoly(n, |G|, |Σ|, |N|).

Size of G

abbbcceabhaXjabaddfeaaf cdaXk In each iteration O(n) new letters shrinking by a constant factor

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Sizes and running time

Running time

All algorithms run in time npoly(n, |G|, |Σ|, |N|).

Size of G

uvbhaXjabxyzcdaXk In each iteration O(n) new letters shrinking by a constant factor

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Sizes and running time

Running time

All algorithms run in time npoly(n, |G|, |Σ|, |N|).

Size of G

uvbhaXjabxyzcdaXk In each iteration O(n) new letters shrinking by a constant factor

New letters (|Σ|)

noncrossing pairs, inner letters appearance compression (shrinks |G|)

uter letters and crossing pairs:

there are O(n) outer letters and O(n2) pairs in word(Xn)

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SLIDE 77

Proof end

NFA size

size O(|Q| · |Σ| + n) new states: replacing nonterminal transitions by chains, (for outer letters, poly(n))

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SLIDE 78

Proof end

NFA size

size O(|Q| · |Σ| + n) new states: replacing nonterminal transitions by chains, (for outer letters, poly(n)) And this is it.

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DFA

DFA vs NFA

Non-determinism: guessing and verifying transitions δ(p, aℓ, q). for DFA: easy

perations preserve determinism of DFA

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