Computational Complexity Lecture 5 in which we relate space and - - PowerPoint PPT Presentation

Computational Complexity

Lecture 5 in which we relate space and time, and see the essence of PSPACE (TQBF)

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SPACE and TIME

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SPACE and TIME

DTIME(F) NTIME(F)

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SPACE and TIME

DTIME(F) NTIME(F)

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SPACE and TIME

DTIME(F) DTIME(2O(F)) NTIME(F) NTIME(2O(F))

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SPACE and TIME

DTIME(F) DTIME(2O(F)) NTIME(F) NTIME(2O(F))

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SPACE and TIME

DTIME(F) DTIME(2O(F)) NTIME(F) NTIME(2O(F))

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SPACE and TIME

DTIME(F) DSPACE(F) NSPACE(F) DTIME(2O(F)) NTIME(F) NTIME(2O(F))

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SPACE and TIME

DTIME(F) DSPACE(F) NSPACE(F) DTIME(2O(F)) NTIME(F) NTIME(2O(F))

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SPACE and TIME

DTIME(F) DSPACE(F) NSPACE(F) DTIME(2O(F)) NTIME(F) NTIME(2O(F))

F=!(n) F=!(log n)

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SPACE and TIME

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SPACE and TIME

In time T(n), can use at most T(n) space

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SPACE and TIME

In time T(n), can use at most T(n) space DTIME(T) ⊆ DSPACE(T)

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SPACE and TIME

In time T(n), can use at most T(n) space DTIME(T) ⊆ DSPACE(T) In fact, NTIME(T) ⊆ DSPACE(O(T)) (try all certificates of length at most T, one after the other)

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SPACE and TIME

In time T(n), can use at most T(n) space DTIME(T) ⊆ DSPACE(T) In fact, NTIME(T) ⊆ DSPACE(O(T)) (try all certificates of length at most T, one after the other) With space S(n), only 2O(S(n)) configurations (for S(n) = !(log n)). So can take at most 2O(S(n)) time (else gets into an infinite loop)

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SPACE and TIME

In time T(n), can use at most T(n) space DTIME(T) ⊆ DSPACE(T) In fact, NTIME(T) ⊆ DSPACE(O(T)) (try all certificates of length at most T, one after the other) With space S(n), only 2O(S(n)) configurations (for S(n) = !(log n)). So can take at most 2O(S(n)) time (else gets into an infinite loop) DSPACE(S) ⊆ DTIME(2O(S))

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SPACE and TIME

In time T(n), can use at most T(n) space DTIME(T) ⊆ DSPACE(T) In fact, NTIME(T) ⊆ DSPACE(O(T)) (try all certificates of length at most T, one after the other) With space S(n), only 2O(S(n)) configurations (for S(n) = !(log n)). So can take at most 2O(S(n)) time (else gets into an infinite loop) DSPACE(S) ⊆ DTIME(2O(S)) In fact, NSPACE(S) ⊆ DTIME(2O(S))

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NSPACE(S) ⊆ DTIME(2O(S))

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NSPACE(S) ⊆ DTIME(2O(S))

h=2(O(S))

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NSPACE(S) ⊆ DTIME(2O(S))

Configuration graph as a DAG is of size 2O(S)

h=2(O(S))

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NSPACE(S) ⊆ DTIME(2O(S))

Configuration graph as a DAG is of size 2O(S) Write down all configurations and edges

h=2(O(S))

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NSPACE(S) ⊆ DTIME(2O(S))

Configuration graph as a DAG is of size 2O(S) Write down all configurations and edges Can do it less explicitly if space were a concern (but it’ s not, here)

h=2(O(S))

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NSPACE(S) ⊆ DTIME(2O(S))

Configuration graph as a DAG is of size 2O(S) Write down all configurations and edges Can do it less explicitly if space were a concern (but it’ s not, here) Run (in poly time) any reachability algorithm (say, breadth-first search) to see if there is a (directed) path from start config. to an accept config.

h=2(O(S))

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NSPACE(S) ⊆ DTIME(2O(S))

Configuration graph as a DAG is of size 2O(S) Write down all configurations and edges Can do it less explicitly if space were a concern (but it’ s not, here) Run (in poly time) any reachability algorithm (say, breadth-first search) to see if there is a (directed) path from start config. to an accept config. poly(2O(S)) = 2O(S)

h=2(O(S))

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SPACE and TIME

DTIME(F) DSPACE(F) NSPACE(F) DTIME(2O(F)) NTIME(F) NTIME(2O(F))

F=!(n) F=!(log n)

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SPACE and TIME

DTIME(F) DSPACE(F) NSPACE(F) DTIME(2O(F)) NTIME(F) NTIME(2O(F))

F=!(n) F=!(log n)

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SPACE and TIME

DTIME(F) DSPACE(F) NSPACE(F) DTIME(2O(F)) NTIME(F) NTIME(2O(F))

F=!(n) F=!(log n)

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NSPACE and DSPACE: Savitch’ s Theorem

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NSPACE and DSPACE: Savitch’ s Theorem

NSPACE(S) ⊆ DSPACE(S2)

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NSPACE and DSPACE: Savitch’ s Theorem

NSPACE(S) ⊆ DSPACE(S2)

h=2(O(S))

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NSPACE and DSPACE: Savitch’ s Theorem

NSPACE(S) ⊆ DSPACE(S2) Naive DFS (or BFS) has stack depth h=2(O(S))

h=2(O(S))

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NSPACE and DSPACE: Savitch’ s Theorem

NSPACE(S) ⊆ DSPACE(S2) Naive DFS (or BFS) has stack depth h=2(O(S)) Look for C s.t. Start!C in h/2 steps and C!Accept in h/2 steps

h=2(O(S))

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NSPACE and DSPACE: Savitch’ s Theorem

NSPACE(S) ⊆ DSPACE(S2) Naive DFS (or BFS) has stack depth h=2(O(S)) Look for C s.t. Start!C in h/2 steps and C!Accept in h/2 steps Recursively! Depth of recursion only log h; at each level remember one configuration

h=2(O(S))

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NSPACE and DSPACE: Savitch’ s Theorem

NSPACE(S) ⊆ DSPACE(S2) Naive DFS (or BFS) has stack depth h=2(O(S)) Look for C s.t. Start!C in h/2 steps and C!Accept in h/2 steps Recursively! Depth of recursion only log h; at each level remember one configuration Space needed = O(log h)*O(S) = O(S2)

h=2(O(S))

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SPACE and TIME

DTIME(F) DSPACE(F) NSPACE(F) DTIME(2O(F)) NTIME(F) NTIME(2O(F))

F=!(n) F=!(log n)

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SPACE and TIME

DTIME(F) DSPACE(F) NSPACE(F) DTIME(2O(F)) NTIME(F) NTIME(2O(F)) DSPACE(F2)

F=!(n) F=!(log n)

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SPACE and TIME

DTIME(F) DSPACE(F) NSPACE(F) DTIME(2O(F)) NTIME(F) NTIME(2O(F)) DSPACE(F2)

F=!(n) F=!(log n)

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Zoo, so far

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Zoo, so far

Major classes of interest (so far):

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Zoo, so far

Major classes of interest (so far): P, EXP; NP, NEXP; L, NL; PSPACE, NPSPACE

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Zoo, so far

P

PSPACE

EXP NP NEXP L NL

NPSPACE

Major classes of interest (so far): P, EXP; NP, NEXP; L, NL; PSPACE, NPSPACE

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Zoo, so far

P

PSPACE

EXP NP NEXP L NL

NPSPACE

Major classes of interest (so far): P, EXP; NP, NEXP; L, NL; PSPACE, NPSPACE PSPACE = NPSPACE (by Savitch)

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Zoo, so far

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PSPACE

EXP NP NEXP L NL

NPSPACE

Major classes of interest (so far): P, EXP; NP, NEXP; L, NL; PSPACE, NPSPACE PSPACE = NPSPACE (by Savitch) Coming up: PSPACE-completeness

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PSPACE completeness

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PSPACE completeness

A language L is PSPACE-Complete if for all L ’ in PSPACE, L ’ !p L and L in PSPACE

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PSPACE completeness

A language L is PSPACE-Complete if for all L ’ in PSPACE, L ’ !p L and L in PSPACE Trivial PSPACE-complete problem: SPACETM = { (M,z,1n) | TM M accepts z within space n }

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PSPACE completeness

A language L is PSPACE-Complete if for all L ’ in PSPACE, L ’ !p L and L in PSPACE Trivial PSPACE-complete problem: SPACETM = { (M,z,1n) | TM M accepts z within space n } (An) essence of PSPACE: Understanding 2-player games

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PSPACE completeness

A language L is PSPACE-Complete if for all L ’ in PSPACE, L ’ !p L and L in PSPACE Trivial PSPACE-complete problem: SPACETM = { (M,z,1n) | TM M accepts z within space n } (An) essence of PSPACE: Understanding 2-player games Can the first/second player always win?

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QBF Game

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QBF Game

Two players: Alice and Adversary, each given n (mutually disjoint) sets of variables (sets numbered [1,n])

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QBF Game

Two players: Alice and Adversary, each given n (mutually disjoint) sets of variables (sets numbered [1,n]) Given a boolean formula over these variables

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QBF Game

Two players: Alice and Adversary, each given n (mutually disjoint) sets of variables (sets numbered [1,n]) Given a boolean formula over these variables In ith round players set the values of the variables in their ith sets. Say Alice moves first.

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QBF Game

Two players: Alice and Adversary, each given n (mutually disjoint) sets of variables (sets numbered [1,n]) Given a boolean formula over these variables In ith round players set the values of the variables in their ith sets. Say Alice moves first. When all variables set, formula evaluated. If true Alice wins, else adversary wins

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QBF Game

Two players: Alice and Adversary, each given n (mutually disjoint) sets of variables (sets numbered [1,n]) Given a boolean formula over these variables In ith round players set the values of the variables in their ith sets. Say Alice moves first. When all variables set, formula evaluated. If true Alice wins, else adversary wins Given a QBF game does Alice have a sure-to-win strategy

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QBF game: examples

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QBF game: examples

Vars: x1, y1, x2, y2, x3, y3. Formula: φ(x1,y1,x2,y1,x3,y3)

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QBF game: examples

Vars: x1, y1, x2, y2, x3, y3. Formula: φ(x1,y1,x2,y1,x3,y3) Say, no variables for Adversary. Only x1

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QBF game: examples

Vars: x1, y1, x2, y2, x3, y3. Formula: φ(x1,y1,x2,y1,x3,y3) Say, no variables for Adversary. Only x1 Strategy for Alice? Is “∃x1 φ(x1 )” true?

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QBF game: examples

Vars: x1, y1, x2, y2, x3, y3. Formula: φ(x1,y1,x2,y1,x3,y3) Say, no variables for Adversary. Only x1 Strategy for Alice? Is “∃x1 φ(x1 )” true? Say, no variables for Alice. Only y1

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QBF game: examples

Vars: x1, y1, x2, y2, x3, y3. Formula: φ(x1,y1,x2,y1,x3,y3) Say, no variables for Adversary. Only x1 Strategy for Alice? Is “∃x1 φ(x1 )” true? Say, no variables for Alice. Only y1 “Strategy” for Alice? Is “∀y1 φ(y1)” true?

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QBF game: examples

Vars: x1, y1, x2, y2, x3, y3. Formula: φ(x1,y1,x2,y1,x3,y3) Say, no variables for Adversary. Only x1 Strategy for Alice? Is “∃x1 φ(x1 )” true? Say, no variables for Alice. Only y1 “Strategy” for Alice? Is “∀y1 φ(y1)” true? Say only x1, y1 (now, that’ s more like a game):

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QBF game: examples

Vars: x1, y1, x2, y2, x3, y3. Formula: φ(x1,y1,x2,y1,x3,y3) Say, no variables for Adversary. Only x1 Strategy for Alice? Is “∃x1 φ(x1 )” true? Say, no variables for Alice. Only y1 “Strategy” for Alice? Is “∀y1 φ(y1)” true? Say only x1, y1 (now, that’ s more like a game): Strategy for Alice? Is “∃x1 ∀y1 φ(x1,y1 )” true?

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QBF game: examples

Vars: x1, y1, x2, y2, x3, y3. Formula: φ(x1,y1,x2,y1,x3,y3) Say, no variables for Adversary. Only x1 Strategy for Alice? Is “∃x1 φ(x1 )” true? Say, no variables for Alice. Only y1 “Strategy” for Alice? Is “∀y1 φ(y1)” true? Say only x1, y1 (now, that’ s more like a game): Strategy for Alice? Is “∃x1 ∀y1 φ(x1,y1 )” true? In general, winning strategy for Alice exists iff

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QBF game: examples

Vars: x1, y1, x2, y2, x3, y3. Formula: φ(x1,y1,x2,y1,x3,y3) Say, no variables for Adversary. Only x1 Strategy for Alice? Is “∃x1 φ(x1 )” true? Say, no variables for Alice. Only y1 “Strategy” for Alice? Is “∀y1 φ(y1)” true? Say only x1, y1 (now, that’ s more like a game): Strategy for Alice? Is “∃x1 ∀y1 φ(x1,y1 )” true? In general, winning strategy for Alice exists iff ∃x1 ∀y1 ... ∃xn ∀yn φ(x1,y1,...,xn,yn ) is true

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QBF game: examples

Vars: x1, y1, x2, y2, x3, y3. Formula: φ(x1,y1,x2,y1,x3,y3) Say, no variables for Adversary. Only x1 Strategy for Alice? Is “∃x1 φ(x1 )” true? Say, no variables for Alice. Only y1 “Strategy” for Alice? Is “∀y1 φ(y1)” true? Say only x1, y1 (now, that’ s more like a game): Strategy for Alice? Is “∃x1 ∀y1 φ(x1,y1 )” true? In general, winning strategy for Alice exists iff ∃x1 ∀y1 ... ∃xn ∀yn φ(x1,y1,...,xn,yn ) is true Else adversary has a winning strategy

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TQBF , the language

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TQBF , the language

True Quantified Boolean Formula:

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TQBF , the language

True Quantified Boolean Formula: ψ := ∃x1 ∀y1 ... ∃xn ∀yn φ(x1,y1,...,xn,yn )

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TQBF , the language

True Quantified Boolean Formula: ψ := ∃x1 ∀y1 ... ∃xn ∀yn φ(x1,y1,...,xn,yn ) TQBF = {ψ | ψ is true}

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TQBF , the language

True Quantified Boolean Formula: ψ := ∃x1 ∀y1 ... ∃xn ∀yn φ(x1,y1,...,xn,yn ) TQBF = {ψ | ψ is true} e.g. ψ1: ∃x∀y (x=y), ψ2: ∀y∃x (x=y)

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TQBF is in PSPACE

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TQBF is in PSPACE

When is a QBF true?

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TQBF is in PSPACE

When is a QBF true? e.g. ∃a,b ∀c φ(a,b,c)

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TQBF is in PSPACE

When is a QBF true? e.g. ∃a,b ∀c φ(a,b,c) Game-Tree

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TQBF is in PSPACE

When is a QBF true? e.g. ∃a,b ∀c φ(a,b,c) ∃a ∃b ∀c ∃b

φ(0,0,0) φ(0,0,1)

Game-Tree

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TQBF is in PSPACE

When is a QBF true? e.g. ∃a,b ∀c φ(a,b,c) ∃a ∃b ∀c ∃b

φ(0,0,0) φ(0,0,1)

Game-Tree

Winning strategy for Alice if game gets here?

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TQBF is in PSPACE

When is a QBF true? e.g. ∃a,b ∀c φ(a,b,c) Ask if winning strategy from each node ∃a ∃b ∀c ∃b

φ(0,0,0) φ(0,0,1)

Game-Tree

Winning strategy for Alice if game gets here?

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TQBF is in PSPACE

When is a QBF true? e.g. ∃a,b ∀c φ(a,b,c) Ask if winning strategy from each node Yes from ∃ node if yes from either child. Yes from ∀ node if yes from both. ∃a ∃b ∀c ∃b

φ(0,0,0) φ(0,0,1)

Game-Tree

Winning strategy for Alice if game gets here?

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TQBF is in PSPACE

When is a QBF true? e.g. ∃a,b ∀c φ(a,b,c) Ask if winning strategy from each node Yes from ∃ node if yes from either child. Yes from ∀ node if yes from both. Naive evaluation takes exponential space (and time) ∃a ∃b ∀c ∃b

φ(0,0,0) φ(0,0,1)

Game-Tree

Winning strategy for Alice if game gets here?

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TQBF is in PSPACE

When is a QBF true? e.g. ∃a,b ∀c φ(a,b,c) Ask if winning strategy from each node Yes from ∃ node if yes from either child. Yes from ∀ node if yes from both. Naive evaluation takes exponential space (and time) Can reuse left child computation space for the right child ∃a ∃b ∀c ∃b

φ(0,0,0) φ(0,0,1)

Game-Tree

Winning strategy for Alice if game gets here?

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TQBF is in PSPACE

When is a QBF true? e.g. ∃a,b ∀c φ(a,b,c) Ask if winning strategy from each node Yes from ∃ node if yes from either child. Yes from ∀ node if yes from both. Naive evaluation takes exponential space (and time) Can reuse left child computation space for the right child ∃a ∃b ∀c ∃b

φ(0,0,0) φ(0,0,1)

Space needed = O(depth) + for evaluation = poly(|QBF|) Game-Tree

Winning strategy for Alice if game gets here?

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TQBF is PSPACE-hard

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TQBF is PSPACE-hard

For L in PSPACE (i.e., TM ML decides L in space poly(n), or with configs of size S(n)=poly(n) ), show L !p TQBF

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TQBF is PSPACE-hard

For L in PSPACE (i.e., TM ML decides L in space poly(n), or with configs of size S(n)=poly(n) ), show L !p TQBF Given x, output f(x) = ψ, s.t. ψ is true iff ML accepts x

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TQBF is PSPACE-hard

For L in PSPACE (i.e., TM ML decides L in space poly(n), or with configs of size S(n)=poly(n) ), show L !p TQBF Given x, output f(x) = ψ, s.t. ψ is true iff ML accepts x x→ψ in poly time. In particular size of ψ is poly(n)

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TQBF is PSPACE-hard

For L in PSPACE (i.e., TM ML decides L in space poly(n), or with configs of size S(n)=poly(n) ), show L !p TQBF Given x, output f(x) = ψ, s.t. ψ is true iff ML accepts x x→ψ in poly time. In particular size of ψ is poly(n) Note: As in Cook’ s theorem, can build an unquantified formula φ (even 3CNF) s.t. φ is true iff ML accepts x

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TQBF is PSPACE-hard

For L in PSPACE (i.e., TM ML decides L in space poly(n), or with configs of size S(n)=poly(n) ), show L !p TQBF Given x, output f(x) = ψ, s.t. ψ is true iff ML accepts x x→ψ in poly time. In particular size of ψ is poly(n) Note: As in Cook’ s theorem, can build an unquantified formula φ (even 3CNF) s.t. φ is true iff ML accepts x But size is poly(time bound on ML) = exp(n)

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TQBF is PSPACE-hard

For L in PSPACE (i.e., TM ML decides L in space poly(n), or with configs of size S(n)=poly(n) ), show L !p TQBF Given x, output f(x) = ψ, s.t. ψ is true iff ML accepts x x→ψ in poly time. In particular size of ψ is poly(n) Note: As in Cook’ s theorem, can build an unquantified formula φ (even 3CNF) s.t. φ is true iff ML accepts x But size is poly(time bound on ML) = exp(n) Use power of quantification to write it succinctly

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TQBF is PSPACE-hard

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TQBF is PSPACE-hard

An exponential QBF:

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TQBF is PSPACE-hard

An exponential QBF: ∃ C1 C2 ... CT ψ0(Cstart,C1) ∧ ψ0(C1,C2) ∧ ... ψ0(CT,Caccept)

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TQBF is PSPACE-hard

An exponential QBF: ∃ C1 C2 ... CT ψ0(Cstart,C1) ∧ ψ0(C1,C2) ∧ ... ψ0(CT,Caccept) Here Ci are variables whose value assignments correspond to configurations. |Ci| = O(S(n)), |ψ0(C,C’)| = O(S(n)), T=2O(S(n))

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TQBF is PSPACE-hard

An exponential QBF: ∃ C1 C2 ... CT ψ0(Cstart,C1) ∧ ψ0(C1,C2) ∧ ... ψ0(CT,Caccept) Here Ci are variables whose value assignments correspond to configurations. |Ci| = O(S(n)), |ψ0(C,C’)| = O(S(n)), T=2O(S(n)) ψ0(C,C’) is an unquantified formula (only variables being C,C’), s.t. it is true iff C evolves into C’ in one step

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TQBF is PSPACE-hard

An exponential QBF: ∃ C1 C2 ... CT ψ0(Cstart,C1) ∧ ψ0(C1,C2) ∧ ... ψ0(CT,Caccept) Here Ci are variables whose value assignments correspond to configurations. |Ci| = O(S(n)), |ψ0(C,C’)| = O(S(n)), T=2O(S(n)) ψ0(C,C’) is an unquantified formula (only variables being C,C’), s.t. it is true iff C evolves into C’ in one step F be the (const. sized) formula to derive each bit of new config from a few bits in the previous config

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TQBF is PSPACE-hard

An exponential QBF: ∃ C1 C2 ... CT ψ0(Cstart,C1) ∧ ψ0(C1,C2) ∧ ... ψ0(CT,Caccept) Here Ci are variables whose value assignments correspond to configurations. |Ci| = O(S(n)), |ψ0(C,C’)| = O(S(n)), T=2O(S(n)) ψ0(C,C’) is an unquantified formula (only variables being C,C’), s.t. it is true iff C evolves into C’ in one step F be the (const. sized) formula to derive each bit of new config from a few bits in the previous config Then, ψ0(C,C’) is ∧j(C’(j) = F(C(j-c),...,C(j+c)))

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TQBF is PSPACE-hard

An exponential QBF: ∃ C1 C2 ... CT ψ0(Cstart,C1) ∧ ψ0(C1,C2) ∧ ... ψ0(CT,Caccept) Here Ci are variables whose value assignments correspond to configurations. |Ci| = O(S(n)), |ψ0(C,C’)| = O(S(n)), T=2O(S(n)) ψ0(C,C’) is an unquantified formula (only variables being C,C’), s.t. it is true iff C evolves into C’ in one step F be the (const. sized) formula to derive each bit of new config from a few bits in the previous config Then, ψ0(C,C’) is ∧j(C’(j) = F(C(j-c),...,C(j+c))) |ψ0(C,C’)| = O(|C|)

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TQBF is PSPACE-hard

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TQBF is PSPACE-hard

Plan for a more succinct ψ: A partially quantified boolean formula ψi s.t. ψi(C,C’) (fully quantified) is true iff C’ reachable from C in the configuration graph G(ML,x) within 2i steps. Output ψ=ψS(n)(start,accept)

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TQBF is PSPACE-hard

Plan for a more succinct ψ: A partially quantified boolean formula ψi s.t. ψi(C,C’) (fully quantified) is true iff C’ reachable from C in the configuration graph G(ML,x) within 2i steps. Output ψ=ψS(n)(start,accept) Base case (i=0): an unquantified formula, ψ0

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TQBF is PSPACE-hard

Plan for a more succinct ψ: A partially quantified boolean formula ψi s.t. ψi(C,C’) (fully quantified) is true iff C’ reachable from C in the configuration graph G(ML,x) within 2i steps. Output ψ=ψS(n)(start,accept) Base case (i=0): an unquantified formula, ψ0 ψi+1(C,C’) := ∃C’’ ψi(C,C’’) ∧ ψi(C’’,C’)

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TQBF is PSPACE-hard

Plan for a more succinct ψ: A partially quantified boolean formula ψi s.t. ψi(C,C’) (fully quantified) is true iff C’ reachable from C in the configuration graph G(ML,x) within 2i steps. Output ψ=ψS(n)(start,accept) Base case (i=0): an unquantified formula, ψ0 ψi+1(C,C’) := ∃C’’ ψi(C,C’’) ∧ ψi(C’’,C’)

c.f. Savitch’ s theorem

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TQBF is PSPACE-hard

Plan for a more succinct ψ: A partially quantified boolean formula ψi s.t. ψi(C,C’) (fully quantified) is true iff C’ reachable from C in the configuration graph G(ML,x) within 2i steps. Output ψ=ψS(n)(start,accept) Base case (i=0): an unquantified formula, ψ0 ψi+1(C,C’) := ∃C’’ ψi(C,C’’) ∧ ψi(C’’,C’) Can be rewritten as a QBF in “Prenex Normal form”

c.f. Savitch’ s theorem

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TQBF is PSPACE-hard

Plan for a more succinct ψ: A partially quantified boolean formula ψi s.t. ψi(C,C’) (fully quantified) is true iff C’ reachable from C in the configuration graph G(ML,x) within 2i steps. Output ψ=ψS(n)(start,accept) Base case (i=0): an unquantified formula, ψ0 ψi+1(C,C’) := ∃C’’ ψi(C,C’’) ∧ ψi(C’’,C’) Can be rewritten as a QBF in “Prenex Normal form” Problem: |ψS(n)| still exponential in S(n)

c.f. Savitch’ s theorem

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TQBF is PSPACE-hard

Plan for a more succinct ψ: A partially quantified boolean formula ψi s.t. ψi(C,C’) (fully quantified) is true iff C’ reachable from C in the configuration graph G(ML,x) within 2i steps. Output ψ=ψS(n)(start,accept) Base case (i=0): an unquantified formula, ψ0 ψi+1(C,C’) := ∃C’’ ψi(C,C’’) ∧ ψi(C’’,C’) Can be rewritten as a QBF in “Prenex Normal form” Problem: |ψS(n)| still exponential in S(n) In fact, same as naive formula!

c.f. Savitch’ s theorem

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TQBF is PSPACE-hard

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TQBF is PSPACE-hard

ψi+1(C,C’) := ∃C’’ ψi(C,C’’) ∧ ψi(C’’,C’)

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TQBF is PSPACE-hard

ψi+1(C,C’) := ∃C’’ ψi(C,C’’) ∧ ψi(C’’,C’) Problem: |ψS(n)| exponential in S(n)

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TQBF is PSPACE-hard

ψi+1(C,C’) := ∃C’’ ψi(C,C’’) ∧ ψi(C’’,C’) Problem: |ψS(n)| exponential in S(n) More variables/ quantification to “reuse” formula

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TQBF is PSPACE-hard

ψi+1(C,C’) := ∃C’’ ψi(C,C’’) ∧ ψi(C’’,C’) Problem: |ψS(n)| exponential in S(n) More variables/ quantification to “reuse” formula ψi+1(C,C’) := ∃C’’ ∀(D,D’) ((D,D’)=(C,C’’) ∨ (D,D’)=(C’’,C’)) ⇒ ψi(D,D’)

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TQBF is PSPACE-hard

ψi+1(C,C’) := ∃C’’ ψi(C,C’’) ∧ ψi(C’’,C’) Problem: |ψS(n)| exponential in S(n) More variables/ quantification to “reuse” formula ψi+1(C,C’) := ∃C’’ ∀(D,D’) ((D,D’)=(C,C’’) ∨ (D,D’)=(C’’,C’)) ⇒ ψi(D,D’) = and ⇒ shorthands for slightly longer formulas

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TQBF is PSPACE-hard

ψi+1(C,C’) := ∃C’’ ψi(C,C’’) ∧ ψi(C’’,C’) Problem: |ψS(n)| exponential in S(n) More variables/ quantification to “reuse” formula ψi+1(C,C’) := ∃C’’ ∀(D,D’) ((D,D’)=(C,C’’) ∨ (D,D’)=(C’’,C’)) ⇒ ψi(D,D’) = and ⇒ shorthands for slightly longer formulas

c.f. Savitch’ s theorem

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TQBF is PSPACE-hard

ψi+1(C,C’) := ∃C’’ ψi(C,C’’) ∧ ψi(C’’,C’) Problem: |ψS(n)| exponential in S(n) More variables/ quantification to “reuse” formula ψi+1(C,C’) := ∃C’’ ∀(D,D’) ((D,D’)=(C,C’’) ∨ (D,D’)=(C’’,C’)) ⇒ ψi(D,D’) = and ⇒ shorthands for slightly longer formulas |ψS(n)| = O(S(n)) + |ψS(n)-1| = O(S(n)2) + |ψ0| = O(S(n)2)

c.f. Savitch’ s theorem

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TQBF is PSPACE-hard

ψi+1(C,C’) := ∃C’’ ψi(C,C’’) ∧ ψi(C’’,C’) Problem: |ψS(n)| exponential in S(n) More variables/ quantification to “reuse” formula ψi+1(C,C’) := ∃C’’ ∀(D,D’) ((D,D’)=(C,C’’) ∨ (D,D’)=(C’’,C’)) ⇒ ψi(D,D’) = and ⇒ shorthands for slightly longer formulas |ψS(n)| = O(S(n)) + |ψS(n)-1| = O(S(n)2) + |ψ0| = O(S(n)2) “Quantification is a powerful programming language”

c.f. Savitch’ s theorem

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TQBF

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TQBF

PSPACE-complete

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TQBF

PSPACE-complete Generalizes SAT and SATc (which have only one quantifier)

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TQBF

PSPACE-complete Generalizes SAT and SATc (which have only one quantifier) How about 2, 3, 4, ... quantifier alternations?

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TQBF

PSPACE-complete Generalizes SAT and SATc (which have only one quantifier) How about 2, 3, 4, ... quantifier alternations? Coming soon!

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Today

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Today

Zoo (more later)

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Today

Zoo (more later) TQBF

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Today

Zoo (more later) TQBF PSPACE complete

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Today

Zoo (more later) TQBF PSPACE complete Will see more of it soon

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Today

Zoo (more later) TQBF PSPACE complete Will see more of it soon Next Lecture: NL

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Today

Zoo (more later) TQBF PSPACE complete Will see more of it soon Next Lecture: NL NL-completeness

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Today

Zoo (more later) TQBF PSPACE complete Will see more of it soon Next Lecture: NL NL-completeness NL = co-NL

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