Relationships between classes

• So far: CS151 L ⊆ P ⊆ PSPACE ⊆ EXP • believe all containments strict Complexity Theory • know L ⊆ PSPACE, P ⊆ EXP Lecture 3 • even before any mention of NP, two major unsolved problems: April 6, 2021 ? ? L = P P = PSPACE

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A P-complete problem A P-complete problem

• We don’t know how to prove L ≠ P • logspace reduction: f computable by TM • But, can identify problems in P least likely that uses O(log n) space to be in L using P- completeness. – denoted “L1 ≤L L2” • need stronger notion of reduction (why?) • If L is P-complete, then L in L implies L = f 2 2 yes yes P (homework problem)

f no no L1 L2

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A P-complete problem A P-complete problem

• Circuit Value (CVAL): given a variable-free • Tableau (configurations written in an Boolean circuit (gates ∧, ∨, ¬, 0, 1), does array) for machine M on input w: it output 1? … w1/qs w2 … wn _ • height = … w1 w2/q1 … wn _ time taken Theorem: CVAL is P-complete. … c w1/q1 a … wn _ = |w| • Proof: . . . . – already argued in P . . • width = … – L arbitrary language in P, TM M decides L in _/qa _ … _ _ space used nc steps ≤ |w|c

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1 A P-complete problem A P-complete problem

• Important observation: contents of cell in • Can build Boolean circuit STEP tableau determined by 3 others above it: – input (binary encoding of) 3 cells – output (binary encoding of) 1 cell

a/q1 b a a b/q1 a a b/q1 a • each output bit is some b/q1 a function of inputs

a b a STEP • can build circuit for each b • size is independent of a size of tableau

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A P-complete problem A P-complete problem w w w … 1 2 n Tableau for w1/qs w2 … wn _ This circuit M on input … w q w … w … _ C has w1 w2/q1 … wn _ 1/ s 2 n M, w w . . inputs . . . . STEP STEP STEP STEP STEP w1w2…wn and outputs 1 iff M • |w|c copies of STEP compute row i from i-1 STEP STEP STEP STEP STEP . . . . accepts input … . . w. STEP STEP STEP STEP STEP logspace STEP STEP STEP STEP STEP ignore these reduction … 2c 1 iff cell contains qaccept Size = O(|w| )

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Answer to question Padding and succinctness

• Can we evaluate an n node Boolean Two consequences of measuring running circuit using O(log n) space? time as function of input length: • “padding” ∧ • NO! (probably) – suppose L ∈ EXP, and define N |x|k ∧ ∨ PADL = { x# : x ∈ L, N = 2 } • CVAL in L if and – TM that decides PADL: ensure suffix of N #s, ∨ ∧ ¬ ignore #s, then simulate TM that decides L only if L = P – running time now polynomial ! 1 0 1 0 1

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2 Padding and succinctness Succinct encodings

• converse (intuition only): “succinctness” • succinct encoding for a directed – suppose L is P-complete graph G= (V = {1,2,3,…}, E): 1 iff (i, j) ∈ E – intuitively, some inputs are “hard” -- require • a succinct encoding for a full power of P variable-free Boolean circuit: – SUCCINCT has inputs encoded in different L i j form than L, some exponentially shorter 1 iff wire type of – if “hard” inputs are exponentially shorter, then from gate i to gate j gate i candidate to be EXP-complete type of gate j i j

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An EXP-complete problem An EXP-complete problem

• Succinct Circuit Value: given a succinctly – tableau for input x = x1x2x3…xn: encoded variable-free Boolean circuit x _ _ _ _ _ (gates ∧, ∨, ¬, 0, 1), does it output 1? height, k Theorem: Succinct Circuit Value is EXP- width 2n complete. • Proof: – Circuit C from CVAL reduction has size – in EXP (why?) k O(22n ). – L arbitrary language in EXP, TM M decides L k – TM M accepts input x iff circuit outputs 1 in 2n steps

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An EXP-complete problem Summary • Remaining TM details: big-oh necessary. – Can encode C succinctly: • First complexity classes: L, P, PSPACE, EXP 1 iff wire type of type of • First separations (via simulation and from gate diagonalization): gate i gate j i to gate j i j P ≠ EXP, L ≠ PSPACE • if i, j within single STEP circuit, easy to compute • First major open questions: output ? ? • if i, j between two STEP circuits, easy to compute L = P P = PSPACE output • First complete problems: • if one of i, j refers to input gates, consult x to – CVAL is P-complete compute output – Succinct CVAL is EXP-complete

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3 Summary Nondeterminism: introduction

A motivating question:

EXP • Can computers replace mathematicians? PSPACE P L L = { (x, 1k) : statement x has a proof of length at most k }

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Nondeterminism: introduction Nondeterminism

• Outline: • Recall deterministic TM – nondeterminism – nondeterministic time classes – Q finite set of states – NP, NP-completeness, P vs. NP – ∑ alphabet including blank: “_” – coNP – qstart, qaccept, qreject in Q – NTIME Hierarchy – transition function: – Ladner’s Theorem δ : Q x ∑ → Q x ∑ x {L, R, -}

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Nondeterminism Nondeterminism

• deterministic TM: given current configuration, • nondeterministic Turing Machine: unique next configuration – Q finite set of states qstartx1x2x3…xn qstartx1x2x3…xn – ∑ alphabet including blank: “_” – q , q , q in Q start accept reject x ∈ L x ∉ L – transition relation ∆ ⊆ (Q x ∑) x (Q x ∑ x {L, R, -}) qaccept qreject • given current state and symbol scanned, several choices of what to do next. • nondeterministic TM: given current configuration, several possible next configurations

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4 Nondeterminism Nondeterminism

qstartx1x2x3…xn qstartx1x2x3…xn • all paths terminate “computation x ∈ L ∉ x L path” “guess” • time used: maximum length of paths from root • space used: maximum # of work tape qaccept squares touched on any path from root qreject • asymmetric accept/reject criterion

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Nondeterminism Nondeterminism

• NTIME(f(n)) = languages decidable by a • Focus on time classes first: multi-tape NTM that runs for at most f(n) steps on any computation path, where n is k the input length, and f :N → N NP = ∪k NTIME(n ) • NSPACE(f(n)) = languages decidable by a multi-tape NTM that touches at most f(n) nk squares of its work tapes along any NEXP = ∪k NTIME(2 ) computation path, where n is the input length, and f :N → N April 6, 2021 CS151 Lecture 3 27 April 6, 2021 CS151 Lecture 3 28

Poly-time verifiers Poly-time verifiers

Very useful alternate“ witnessdefinition” or of NP: • Example: 3SAT expressible as “certificate” 3SAT = {φ : φ is a 3-CNF formula for which Theorem: language L is in NP if andefficiently only if ∃ assignment A for which (φ, A) ∈ R} it is expressible as: verifiable R = {(φ, A) : A is a sat. assign. for φ} L = { x| ∃ y, |y| ≤ |x|k, (x, y) ∈ R } where R is a language in P. – satisfying assignmdnt A is a “witness” of the satisfiability of φ (“certifies” satisfiability of φ) – R is decidable in poly-time • poly-time TM MR deciding R is a “verifier” April 6, 2021 CS151 Lecture 3 29 April 6, 2021 CS151 Lecture 3 30

5 Poly-time verifiers Poly-time verifiers

L = { x | ∃ y, |y| ≤ |x|k, (x, y) ∈ R } Proof: (⇐) given L ∈ NP, describe L as: Proof: (⇒) give poly-time NTM deciding L L = { x | ∃ y, |y| ≤ |x|k, (x, y) ∈ R } – L is decided by NTM M running in time nk – define the language phase 1: “guess” y R = { (x, y) : y is an accepting computation with |x|k history of M on input x} nondeterministic steps – check: accepting history has length ≤ |x|k phase 2: – check: R is decidable in polynomial time decide if – check: M accepts x iff ∃y, |y| ≤ |x|k, (x, y) ∈ R (x, y) ∈ R

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Why NP? Why NP? problem object we •requirementsnot a realistic model of computationare seeking • but, captures important computational • Why not EXP? feature of many problems:

exhaustive search worksefficient test: – too strong! • contains huge number of natural,does practical y meet – important problems not complete. problems requirements? • many problems have form: L = { x | ∃ y s.t. (x,y) ∈ R}

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Relationships between classes NP-completeness

• Easy: P ⊆ NP, EXP ⊆ NEXP • Circuit SAT: given a Boolean circuit (gates – TM special case of NTM ∧, ∨, ¬), with variables y1, y2, …, ym is • Recall: L ∈ NP iff expressible as there some assignment that makes it k L = { x | ∃ y, |y| ≤ |x| s.t. (x,y) ∈ R} output 1? • NP ⊆ PSPACE (try all possible y) • The central question: Theorem: Circuit SAT is NP-complete. P =? NP • Proof: finding a solution vs. recognizing a solution – clearly in NP

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6 NP-completeness NEXP-completeness

– Given L ∈ NP of form • Succinct Circuit SAT: given a succinctly L = { x | ∃ y s.t. (x,y) ∈ R} encoded Boolean circuit (gates ∧, ∨, ¬), with variables y1, y2, …, ym is there some assignment that makes it output 1? x1 x2 … xn y1 y2 … ym Theorem: Succinct Circuit SAT is NEXP- complete. 1 iff (x,y) ∈ R CVAL reduction • Proof: for R – same trick as for Succinct CVAL EXP- – hardwire input x; leave y as variables complete.

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Complement classes coNP

• In general, if C is a complexity class • Is NP closed under complement?

• co-C is the complement class, containing Can we transform x ∈ L x ∉ L all complements of languages in C this machine: – L ∈ C implies (Σ* - L) ∈ co-C – (Σ* - L) ∈ C implies L ∈ co-C x ∈ L x ∉ L qaccept qreject • Some classes closed under complement: into a machine – e.g. co-P = P qaccept with this behavior? qreject April 6, 2021 CS151 Lecture 3 39 April 6, 2021 CS151 Lecture 3 40

coNP coNP

• “proof system” interpretation: • P = NP implies NP = coNP • Recall: L ∈ NP iff expressible as L = { x | ∃ y, |y| ≤ |x|k, (x, y) ∈ R } • Belief: “proof “proof” verifier” NP ≠ coNP • languages in NP have “short proofs” • coNP captures (in its complete problems) – another major open problem problems least likely to have “short proofs”. – e.g., UNSAT is coNP-complete

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7 NTIME Hierarchy Theorem NTIME Hierarchy Theorem

Theorem (Nondeterministic Time Hierarchy inputs Theorem): Proof Y For every proper complexity function f(n) ≥ attempt : Turing n Machines (M, x): n, and g(n) = ω(f(n+1)), (what’s Y does wrong?) n NTM M n accept x in f(n) NTIME(f(n)) ⊆ NTIME(g(n)). Y steps? n

D : n Y n Y Y n Y

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