Computational Complexity Lecture 2 in which we talk about - - PowerPoint PPT Presentation

Computational Complexity

Lecture 2 in which we talk about NP-completeness (reductions, reductions)

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Recap

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Recap

Languages in NP are of the form:

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Recap

Languages in NP are of the form: L= { x | ∃w, |w| < poly(|x|) s.t. (x,w) ∈ L ’ }, where L ’ is in P

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Recap

Languages in NP are of the form: L= { x | ∃w, |w| < poly(|x|) s.t. (x,w) ∈ L ’ }, where L ’ is in P Today: Hardest problems in NP

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Reductions

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Reductions

At the heart of today’ s complexity theory

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Reductions

At the heart of today’ s complexity theory L1 ! L2 if problem of deciding L1 “reduces to that of deciding” L2

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Reductions

At the heart of today’ s complexity theory L1 ! L2 if problem of deciding L1 “reduces to that of deciding” L2 if can decide L2, can decide L1

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Turing and Many-One

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Turing reduction:

Turing and Many-One

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Turing reduction: Build a TM (oracle machine) ML1, s.t. using the oracle OL2 which decides L2, ML1^OL2 decides L1

Turing and Many-One

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Turing reduction: Build a TM (oracle machine) ML1, s.t. using the oracle OL2 which decides L2, ML1^OL2 decides L1 ML1 may query OL2 many times (with different inputs)

Turing and Many-One

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Turing reduction: Build a TM (oracle machine) ML1, s.t. using the oracle OL2 which decides L2, ML1^OL2 decides L1 ML1 may query OL2 many times (with different inputs) Many-One:

Turing and Many-One

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Turing reduction: Build a TM (oracle machine) ML1, s.t. using the oracle OL2 which decides L2, ML1^OL2 decides L1 ML1 may query OL2 many times (with different inputs) Many-One: ML1 can query OL2 only once, and must output what OL2 outputs

Turing and Many-One

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Turing reduction: Build a TM (oracle machine) ML1, s.t. using the oracle OL2 which decides L2, ML1^OL2 decides L1 ML1 may query OL2 many times (with different inputs) Many-One: ML1 can query OL2 only once, and must output what OL2 outputs ML1 maps its input x to an input f(x) for OL2

Turing and Many-One

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Turing reduction: Build a TM (oracle machine) ML1, s.t. using the oracle OL2 which decides L2, ML1^OL2 decides L1 ML1 may query OL2 many times (with different inputs) Many-One: ML1 can query OL2 only once, and must output what OL2 outputs ML1 maps its input x to an input f(x) for OL2 x ∈ L1 ⇒ f(x) ∈ L2 and x ! L1 ⇒ f(x) ! L2

Turing and Many-One

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Turing reduction: Build a TM (oracle machine) ML1, s.t. using the oracle OL2 which decides L2, ML1^OL2 decides L1 ML1 may query OL2 many times (with different inputs) Many-One: ML1 can query OL2 only once, and must output what OL2 outputs ML1 maps its input x to an input f(x) for OL2 x ∈ L1 ⇒ f(x) ∈ L2 and x ! L1 ⇒ f(x) ! L2 L1 L2

Turing and Many-One

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Turing reduction: Build a TM (oracle machine) ML1, s.t. using the oracle OL2 which decides L2, ML1^OL2 decides L1 ML1 may query OL2 many times (with different inputs) Many-One: ML1 can query OL2 only once, and must output what OL2 outputs ML1 maps its input x to an input f(x) for OL2 x ∈ L1 ⇒ f(x) ∈ L2 and x ! L1 ⇒ f(x) ! L2 L1 L2

Turing and Many-One

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Turing reduction: Build a TM (oracle machine) ML1, s.t. using the oracle OL2 which decides L2, ML1^OL2 decides L1 ML1 may query OL2 many times (with different inputs) Many-One: ML1 can query OL2 only once, and must output what OL2 outputs ML1 maps its input x to an input f(x) for OL2 x ∈ L1 ⇒ f(x) ∈ L2 and x ! L1 ⇒ f(x) ! L2 L1 L2

Turing and Many-One

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Polynomial-Time Reduction

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Polynomial-Time Reduction

Many-one reduction, where ML1 runs in polynomial time

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Polynomial-Time Reduction

Many-one reduction, where ML1 runs in polynomial time L1 !p L2

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Polynomial-Time Reduction

Many-one reduction, where ML1 runs in polynomial time L1 !p L2 L2 is “computationally (almost) as hard or harder” compared to L1

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Polynomial-Time Reduction

Many-one reduction, where ML1 runs in polynomial time L1 !p L2 L2 is “computationally (almost) as hard or harder” compared to L1 “almost”: reduction overheads (reduction time, size blow-up)

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Polynomial-Time Reduction

Many-one reduction, where ML1 runs in polynomial time L1 !p L2 L2 is “computationally (almost) as hard or harder” compared to L1 “almost”: reduction overheads (reduction time, size blow-up) L2 may be way harder

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Cook, Karp, Levin

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Cook, Karp, Levin

Polynomial-time reduction

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Cook, Karp, Levin

Polynomial-time reduction Cook: Turing reduction

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Cook, Karp, Levin

Polynomial-time reduction Cook: Turing reduction Karp: Many-one reduction

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Cook, Karp, Levin

Polynomial-time reduction Cook: Turing reduction Karp: Many-one reduction We use this for !p

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Cook, Karp, Levin

Polynomial-time reduction Cook: Turing reduction Karp: Many-one reduction We use this for !p Between NP languages

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Cook, Karp, Levin

Polynomial-time reduction Cook: Turing reduction Karp: Many-one reduction We use this for !p Between NP languages Levin: Karp + witnesses easily transformed back and forth

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Cook, Karp, Levin

Polynomial-time reduction Cook: Turing reduction Karp: Many-one reduction We use this for !p Between NP languages Levin: Karp + witnesses easily transformed back and forth Parsimonious: Karp + number

• f witnesses doesn’

t change

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NP-completeness

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NP-completeness

A language L is NP-Hard if for all L ’ in NP, L ’ !p L

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NP-completeness

A language L is NP-Hard if for all L ’ in NP, L ’ !p L A language L is NP-Complete if it is NP-Hard and is in NP

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NP-completeness

A language L is NP-Hard if for all L ’ in NP, L ’ !p L A language L is NP-Complete if it is NP-Hard and is in NP To efficiently solve all problems in NP, you need to efficiently solve L and nothing more

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A simple NPC language

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A simple NPC language

TMSAT = { (M,z,1n,1t) | ∃w, |w|<n, s.t. TM represented by M accepts (z,w) within time t }

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A simple NPC language

TMSAT = { (M,z,1n,1t) | ∃w, |w|<n, s.t. TM represented by M accepts (z,w) within time t } TMSAT is in NP: TMVAL = { (M,z,1n,1t,w) | |w|<n and TM represented by M accepts (z,w) within time t } is in P

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A simple NPC language

TMSAT = { (M,z,1n,1t) | ∃w, |w|<n, s.t. TM represented by M accepts (z,w) within time t } TMSAT is in NP: TMVAL = { (M,z,1n,1t,w) | |w|<n and TM represented by M accepts (z,w) within time t } is in P TMSAT is NP-hard: Given a language L in NP defined as L = { x | ∃w, |w|<n s.t. ML

’ accepts (x,w) } and ML ’ runs

within time t, (where n,t are poly(|x|) ), let the Karp reduction be f(x) = (ML

’,x,1n,1t)

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A simple NPC language

TMSAT = { (M,z,1n,1t) | ∃w, |w|<n, s.t. TM represented by M accepts (z,w) within time t } TMSAT is in NP: TMVAL = { (M,z,1n,1t,w) | |w|<n and TM represented by M accepts (z,w) within time t } is in P TMSAT is NP-hard: Given a language L in NP defined as L = { x | ∃w, |w|<n s.t. ML

’ accepts (x,w) } and ML ’ runs

within time t, (where n,t are poly(|x|) ), let the Karp reduction be f(x) = (ML

’,x,1n,1t)

Any “natural” NPC language?

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Boolean Circuits

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Boolean Circuits

Boolean valued wires, AND, OR, NOT, CONST gates, inputs, output, directed acyclic graph

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Boolean Circuits

Boolean valued wires, AND, OR, NOT, CONST gates, inputs, output, directed acyclic graph Circuit evaluation CKT-VAL: given (ckt,inputs) find ckt’ s boolean

• utput value

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Boolean Circuits

Boolean valued wires, AND, OR, NOT, CONST gates, inputs, output, directed acyclic graph Circuit evaluation CKT-VAL: given (ckt,inputs) find ckt’ s boolean

• utput value

Can be done very efficiently: CKT-VAL is in P

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Boolean Circuits

Boolean valued wires, AND, OR, NOT, CONST gates, inputs, output, directed acyclic graph Circuit evaluation CKT-VAL: given (ckt,inputs) find ckt’ s boolean

• utput value

Can be done very efficiently: CKT-VAL is in P CKT-SAT: given ckt, is there a “satisfying” input (output=1). In NP.

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CKT-SAT is NP-Complete

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CKT-SAT is NP-Complete

Reduce any NP language L to CKT-SAT

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CKT-SAT is NP-Complete

Reduce any NP language L to CKT-SAT Let’ s start from the TM for verifying membership in L, with time bound T

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CKT-SAT is NP-Complete

Reduce any NP language L to CKT-SAT Let’ s start from the TM for verifying membership in L, with time bound T Build a circuit which on input w outputs what the TM outputs on (x,w), within T steps

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CKT-SAT is NP-Complete

Reduce any NP language L to CKT-SAT Let’ s start from the TM for verifying membership in L, with time bound T Build a circuit which on input w outputs what the TM outputs on (x,w), within T steps This circuit is an instance of CKT-SAT

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CKT-SAT is NP-Complete

Reduce any NP language L to CKT-SAT Let’ s start from the TM for verifying membership in L, with time bound T Build a circuit which on input w outputs what the TM outputs on (x,w), within T steps This circuit is an instance of CKT-SAT Ensure reduction is poly-time

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TM to Circuit

(x,w)

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TM to Circuit

Wires for configurations: a bundle for each tape cell, encoding (content,state), where state is encoded in the cell with the head (x,w)

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TM to Circuit

Wires for configurations: a bundle for each tape cell, encoding (content,state), where state is encoded in the cell with the head (x,w)

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TM to Circuit

Wires for configurations: a bundle for each tape cell, encoding (content,state), where state is encoded in the cell with the head Circuitry for evolution: each bundle depends (as specified by the TM) on 3 neighboring bundles at the previous level (x,w)

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TM to Circuit

Wires for configurations: a bundle for each tape cell, encoding (content,state), where state is encoded in the cell with the head Circuitry for evolution: each bundle depends (as specified by the TM) on 3 neighboring bundles at the previous level (Part of) initial configuration, namely w, to be plugged in as input (x,w)

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TM to Circuit

Wires for configurations: a bundle for each tape cell, encoding (content,state), where state is encoded in the cell with the head Circuitry for evolution: each bundle depends (as specified by the TM) on 3 neighboring bundles at the previous level (Part of) initial configuration, namely w, to be plugged in as input x,q0 (x,w)

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TM to Circuit

Wires for configurations: a bundle for each tape cell, encoding (content,state), where state is encoded in the cell with the head Circuitry for evolution: each bundle depends (as specified by the TM) on 3 neighboring bundles at the previous level (Part of) initial configuration, namely w, to be plugged in as input x,q0 (x,w) w

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TM to Circuit

Wires for configurations: a bundle for each tape cell, encoding (content,state), where state is encoded in the cell with the head Circuitry for evolution: each bundle depends (as specified by the TM) on 3 neighboring bundles at the previous level (Part of) initial configuration, namely w, to be plugged in as input T configurations, T bundles each x,q0 (x,w) w

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TM to Circuit

Wires for configurations: a bundle for each tape cell, encoding (content,state), where state is encoded in the cell with the head Circuitry for evolution: each bundle depends (as specified by the TM) on 3 neighboring bundles at the previous level (Part of) initial configuration, namely w, to be plugged in as input T configurations, T bundles each Circuit size = O(T2) x,q0 (x,w) w

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TM to Circuit

x,q0 (x,w) w

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TM to Circuit

Reducing any NP language L to CKT-SAT x,q0 (x,w) w

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TM to Circuit

Reducing any NP language L to CKT-SAT TM for verifying membership in L, time-bound T, and input x " A circuit which on input w

• utputs what the TM outputs
• n (x,w) within T steps

x,q0 (x,w) w

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TM to Circuit

Reducing any NP language L to CKT-SAT TM for verifying membership in L, time-bound T, and input x " A circuit which on input w

• utputs what the TM outputs
• n (x,w) within T steps

Poly-time reduction x,q0 (x,w) w

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TM to Circuit

Reducing any NP language L to CKT-SAT TM for verifying membership in L, time-bound T, and input x " A circuit which on input w

• utputs what the TM outputs
• n (x,w) within T steps

Poly-time reduction CKT-SAT is NP-complete x,q0 (x,w) w

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Other NP-complete problems

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Other NP-complete problems

SAT and 3SAT

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Other NP-complete problems

SAT and 3SAT SAT: Are all given “clauses” simultaneously satisfiable? (Conjunctive Normal Form)

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Other NP-complete problems

SAT and 3SAT SAT: Are all given “clauses” simultaneously satisfiable? (Conjunctive Normal Form) 3SAT: Each clause has at most 3 literals

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Other NP-complete problems

SAT and 3SAT SAT: Are all given “clauses” simultaneously satisfiable? (Conjunctive Normal Form) 3SAT: Each clause has at most 3 literals CLIQUE, INDEP-SET, VERTEX-COVER

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Other NP-complete problems

SAT and 3SAT SAT: Are all given “clauses” simultaneously satisfiable? (Conjunctive Normal Form) 3SAT: Each clause has at most 3 literals CLIQUE, INDEP-SET, VERTEX-COVER Hundreds (thousands?) more known

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Other NP-complete problems

SAT and 3SAT SAT: Are all given “clauses” simultaneously satisfiable? (Conjunctive Normal Form) 3SAT: Each clause has at most 3 literals CLIQUE, INDEP-SET, VERTEX-COVER Hundreds (thousands?) more known Shown using already known ones:

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Other NP-complete problems

SAT and 3SAT SAT: Are all given “clauses” simultaneously satisfiable? (Conjunctive Normal Form) 3SAT: Each clause has at most 3 literals CLIQUE, INDEP-SET, VERTEX-COVER Hundreds (thousands?) more known Shown using already known ones: If L !p L1 and L1 !p L2, then L !p L2

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CKT-SAT #p SAT

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CKT-SAT #p SAT

Converting a circuit to a collection of clauses:

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CKT-SAT #p SAT

Converting a circuit to a collection of clauses: For each wire (connected component), add a variable

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CKT-SAT #p SAT

Converting a circuit to a collection of clauses: For each wire (connected component), add a variable For each gate, add a clause involving variables for wires connected to the gate:

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CKT-SAT #p SAT

Converting a circuit to a collection of clauses: For each wire (connected component), add a variable For each gate, add a clause involving variables for wires connected to the gate: e.g. : (z⇒x), (z⇒y), (¬z ⇒ ¬x ∨ ¬y). i.e., (¬z ∨ x), (¬z ∨ y), (z ∨ ¬x ∨ ¬y).

AND

x y z

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CKT-SAT #p SAT

Converting a circuit to a collection of clauses: For each wire (connected component), add a variable For each gate, add a clause involving variables for wires connected to the gate: e.g. : (z⇒x), (z⇒y), (¬z ⇒ ¬x ∨ ¬y). i.e., (¬z ∨ x), (¬z ∨ y), (z ∨ ¬x ∨ ¬y). and : (z⇒x ∨ y), (¬z⇒¬x), (¬z⇒¬y).

AND

x y z

OR

x y z

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SAT #p 3SAT

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SAT #p 3SAT

Previous reduction was to 3SAT, so 3SAT is NP-complete. And SAT is in NP. So SAT !p 3SAT.

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SAT #p 3SAT

Previous reduction was to 3SAT, so 3SAT is NP-complete. And SAT is in NP. So SAT !p 3SAT. More directly:

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SAT #p 3SAT

Previous reduction was to 3SAT, so 3SAT is NP-complete. And SAT is in NP. So SAT !p 3SAT. More directly: (a ∨ b ∨ c ∨ d ∨ e) " (a ∨ b ∨ x), (¬x ∨ c ∨ d ∨ e) " (a ∨ b ∨ x), (¬x ∨ c ∨ y), (¬y ∨ d ∨ e)

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SAT #p 3SAT

Previous reduction was to 3SAT, so 3SAT is NP-complete. And SAT is in NP. So SAT !p 3SAT. More directly: (a ∨ b ∨ c ∨ d ∨ e) " (a ∨ b ∨ x), (¬x ∨ c ∨ d ∨ e) " (a ∨ b ∨ x), (¬x ∨ c ∨ y), (¬y ∨ d ∨ e) Reduction needs 3SAT

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SAT #p 3SAT

Previous reduction was to 3SAT, so 3SAT is NP-complete. And SAT is in NP. So SAT !p 3SAT. More directly: (a ∨ b ∨ c ∨ d ∨ e) " (a ∨ b ∨ x), (¬x ∨ c ∨ d ∨ e) " (a ∨ b ∨ x), (¬x ∨ c ∨ y), (¬y ∨ d ∨ e) Reduction needs 3SAT 2SAT is in fact in P! [Exercise]

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SAT #p 3SAT

Previous reduction was to 3SAT, so 3SAT is NP-complete. And SAT is in NP. So SAT !p 3SAT. More directly: (a ∨ b ∨ c ∨ d ∨ e) " (a ∨ b ∨ x), (¬x ∨ c ∨ d ∨ e) " (a ∨ b ∨ x), (¬x ∨ c ∨ y), (¬y ∨ d ∨ e) Reduction needs 3SAT 2SAT is in fact in P! [Exercise] Reduction not parsimonious (can you make it? [Exercise])

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3SAT #p CLIQUE

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3SAT #p CLIQUE

Clauses " Graph

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3SAT #p CLIQUE

Clauses " Graph

(x ∨ ¬y ∨ ¬z) (w ∨ x ∨ ¬z) (w ∨ y)

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3SAT #p CLIQUE

Clauses " Graph vertices: each clause’ s satisfying assignments (for its variables)

(x ∨ ¬y ∨ ¬z) (w ∨ x ∨ ¬z) (w ∨ y)

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3SAT #p CLIQUE

Clauses " Graph vertices: each clause’ s satisfying assignments (for its variables)

(x ∨ ¬y ∨ ¬z) (w ∨ x ∨ ¬z)

0*1* 1*1* 1*0*

(w ∨ y)

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3SAT #p CLIQUE

Clauses " Graph vertices: each clause’ s satisfying assignments (for its variables)

(x ∨ ¬y ∨ ¬z) (w ∨ x ∨ ¬z)

0*1* 1*1* 1*0*

(w ∨ y)

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3SAT #p CLIQUE

Clauses " Graph vertices: each clause’ s satisfying assignments (for its variables)

(x ∨ ¬y ∨ ¬z)

*000 *001 *010 *100 *101 *110 *111

(w ∨ x ∨ ¬z)

0*1* 1*1* 1*0*

(w ∨ y)

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3SAT #p CLIQUE

Clauses " Graph vertices: each clause’ s satisfying assignments (for its variables)

(x ∨ ¬y ∨ ¬z)

*000 *001 *010 *100 *101 *110 *111

(w ∨ x ∨ ¬z)

0*1* 1*1* 1*0*

(w ∨ y)

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3SAT #p CLIQUE

Clauses " Graph vertices: each clause’ s satisfying assignments (for its variables)

(x ∨ ¬y ∨ ¬z)

*000 *001 *010 *100 *101 *110 *111

(w ∨ x ∨ ¬z)

00*0 01*1 01*0 10*0 10*1 11*0 11*1 0*1* 1*1* 1*0*

(w ∨ y)

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3SAT #p CLIQUE

Clauses " Graph vertices: each clause’ s satisfying assignments (for its variables) edges between consistent assignments

(x ∨ ¬y ∨ ¬z)

*000 *001 *010 *100 *101 *110 *111

(w ∨ x ∨ ¬z)

00*0 01*1 01*0 10*0 10*1 11*0 11*1 0*1* 1*1* 1*0*

(w ∨ y)

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3SAT #p CLIQUE

Clauses " Graph vertices: each clause’ s satisfying assignments (for its variables) edges between consistent assignments

(x ∨ ¬y ∨ ¬z)

*000 *001 *010 *100 *101 *110 *111

(w ∨ x ∨ ¬z)

00*0 01*1 01*0 10*0 10*1 11*0 11*1 0*1* 1*1* 1*0*

(w ∨ y)

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3SAT #p CLIQUE

Clauses " Graph vertices: each clause’ s satisfying assignments (for its variables) edges between consistent assignments

(x ∨ ¬y ∨ ¬z)

*000 *001 *010 *100 *101 *110 *111

(w ∨ x ∨ ¬z)

00*0 01*1 01*0 10*0 10*1 11*0 11*1 0*1* 1*1* 1*0*

(w ∨ y)

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3SAT #p CLIQUE

Clauses " Graph vertices: each clause’ s satisfying assignments (for its variables) edges between consistent assignments

(x ∨ ¬y ∨ ¬z)

*000 *001 *010 *100 *101 *110 *111

(w ∨ x ∨ ¬z)

00*0 01*1 01*0 10*0 10*1 11*0 11*1 0*1* 1*1* 1*0*

(w ∨ y)

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3SAT #p CLIQUE

Clauses " Graph vertices: each clause’ s satisfying assignments (for its variables) edges between consistent assignments

(x ∨ ¬y ∨ ¬z)

*000 *001 *010 *100 *101 *110 *111

(w ∨ x ∨ ¬z)

00*0 01*1 01*0 10*0 10*1 11*0 11*1 0*1* 1*1* 1*0*

(w ∨ y)

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3SAT #p CLIQUE

Clauses " Graph vertices: each clause’ s satisfying assignments (for its variables) edges between consistent assignments

(x ∨ ¬y ∨ ¬z)

*000 *001 *010 *100 *101 *110 *111

(w ∨ x ∨ ¬z)

00*0 01*1 01*0 10*0 10*1 11*0 11*1 0*1* 1*1* 1*0*

(w ∨ y)

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3SAT #p CLIQUE

Clauses " Graph vertices: each clause’ s satisfying assignments (for its variables) edges between consistent assignments m-clique iff all m clauses satisfiable

(x ∨ ¬y ∨ ¬z)

*000 *001 *010 *100 *101 *110 *111

(w ∨ x ∨ ¬z)

00*0 01*1 01*0 10*0 10*1 11*0 11*1 0*1* 1*1* 1*0*

(w ∨ y)

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3SAT #p CLIQUE

Clauses " Graph vertices: each clause’ s satisfying assignments (for its variables) edges between consistent assignments m-clique iff all m clauses satisfiable

(x ∨ ¬y ∨ ¬z)

*000 *001 *010 *100 *101 *110 *111

(w ∨ x ∨ ¬z)

00*0 01*1 01*0 10*0 10*1 11*0 11*1 0*1* 1*1* 1*0*

(w ∨ y)

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3SAT #p CLIQUE

Clauses " Graph vertices: each clause’ s satisfying assignments (for its variables) edges between consistent assignments m-clique iff all m clauses satisfiable

(x ∨ ¬y ∨ ¬z)

*000 *001 *010 *100 *101 *110 *111

(w ∨ x ∨ ¬z)

00*0 01*1 01*0 10*0 10*1 11*0 11*1 0*1* 1*1* 1*0*

(w ∨ y)

1*1* *110 11*0

• 1110

sat assignment 3-Clique

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INDEP-SET and VERTEX-COVER

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INDEP-SET and VERTEX-COVER

CLIQUE !p INDEP-SET

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INDEP-SET and VERTEX-COVER

CLIQUE !p INDEP-SET G has an m-clique iff G’ has an m-independent-set

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INDEP-SET and VERTEX-COVER

CLIQUE !p INDEP-SET G has an m-clique iff G’ has an m-independent-set INDEP-SET !p VERTEX-COVER

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INDEP-SET and VERTEX-COVER

CLIQUE !p INDEP-SET G has an m-clique iff G’ has an m-independent-set INDEP-SET !p VERTEX-COVER G has an m-indep-set iff G has an (n-m)-vertex-cover

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NP, P, co-NP and NPC

P NP coNP

NPC coNPC

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NP, P, co-NP and NPC

We say class X is “closed under polynomial reductions” if (L1 !p L2 and L2 in class X) implies L1 in X

P NP coNP

NPC coNPC

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NP, P, co-NP and NPC

We say class X is “closed under polynomial reductions” if (L1 !p L2 and L2 in class X) implies L1 in X e.g. P, NP are closed under polynomial reductions

P NP coNP

NPC coNPC

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NP, P, co-NP and NPC

We say class X is “closed under polynomial reductions” if (L1 !p L2 and L2 in class X) implies L1 in X e.g. P, NP are closed under polynomial reductions So is co-NP (If X is closed, so is co-X. Why?)

P NP coNP

NPC coNPC

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NP, P, co-NP and NPC

We say class X is “closed under polynomial reductions” if (L1 !p L2 and L2 in class X) implies L1 in X e.g. P, NP are closed under polynomial reductions So is co-NP (If X is closed, so is co-X. Why?) If any NPC language is in P, then NP = P

P NP coNP

NPC coNPC

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NP, P, co-NP and NPC

We say class X is “closed under polynomial reductions” if (L1 !p L2 and L2 in class X) implies L1 in X e.g. P, NP are closed under polynomial reductions So is co-NP (If X is closed, so is co-X. Why?) If any NPC language is in P, then NP = P If any NPC language is in co-NP, then NP = co-NP

P NP coNP

NPC coNPC

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NP, P, co-NP and NPC

We say class X is “closed under polynomial reductions” if (L1 !p L2 and L2 in class X) implies L1 in X e.g. P, NP are closed under polynomial reductions So is co-NP (If X is closed, so is co-X. Why?) If any NPC language is in P, then NP = P If any NPC language is in co-NP, then NP = co-NP Note: X ⊆ co-X ⇒ X = co-X (Why?)

P NP coNP

NPC coNPC

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NP, P, co-NP and NPC

We say class X is “closed under polynomial reductions” if (L1 !p L2 and L2 in class X) implies L1 in X e.g. P, NP are closed under polynomial reductions So is co-NP (If X is closed, so is co-X. Why?) If any NPC language is in P, then NP = P If any NPC language is in co-NP, then NP = co-NP Note: X ⊆ co-X ⇒ X = co-X (Why?) L is NP-complete iff Lc is co-NP-complete (Why?)

P NP coNP

NPC coNPC

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NP, P, co-NP and NPC

We say class X is “closed under polynomial reductions” if (L1 !p L2 and L2 in class X) implies L1 in X e.g. P, NP are closed under polynomial reductions So is co-NP (If X is closed, so is co-X. Why?) If any NPC language is in P, then NP = P If any NPC language is in co-NP, then NP = co-NP Note: X ⊆ co-X ⇒ X = co-X (Why?) L is NP-complete iff Lc is co-NP-complete (Why?) co-NP complete = co-(NP-complete)

P NP coNP

NPC coNPC

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Today

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Today

Polynomial-time reductions

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Today

Polynomial-time reductions NP-completeness (using Karp reductions)

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Today

Polynomial-time reductions NP-completeness (using Karp reductions) Trivially, TMSAT

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Today

Polynomial-time reductions NP-completeness (using Karp reductions) Trivially, TMSAT Interestingly, CKT-SAT, SAT, 3SAT, CLIQUE, INDEP-SET, VERTEX-COVER

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Today

Polynomial-time reductions NP-completeness (using Karp reductions) Trivially, TMSAT Interestingly, CKT-SAT, SAT, 3SAT, CLIQUE, INDEP-SET, VERTEX-COVER If any NPC language in P, then P=NP

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Next Time

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Next Time

Ladner’ s Theorem: If NP ! P, then non-P, non-NPC languages

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Next Time

Ladner’ s Theorem: If NP ! P, then non-P, non-NPC languages Time hierarchy theorems: More time, more power, strictly!

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