§6 PSPACE and Polynomial Hierarchy

1. §6 PSPACE and Polynomial Hierarchy • PSPACE-complete problemsQBF, 3QBF, GRAPH • (Savitch's Theorem: PSPACE = NPSPACE) • Polynomial Hierarchy: syntactically / semantically • (Baker, Gill & Solovay)

2. QBF and PSPACE coNP ? NP ? 3QBF: Given Boolean term Φ(Y1,…Ym), does it hold y1y2 y3 …: Φ(y1,…ym)=1 ? PSPACE Recursively evaluate quantifiers: s(m) = s(m-1) + poly(n) Theorem:QBF is PSPACE-complete. Proof:Let A decide LPSPACE in space s:=poly(n). Encode configurations ofAinsbitsu=(u1,…us). edge fromutovifvencodesA's unique config afteru. G=GA,sdigraph with u=start config oninputx, wunique accepting config. Aacceptsx  PG(u,w,s), : "path of length ≤2mfromutow"  v: PG(u,v,s-1)  PG(v,w,s-1)  v s,t: (s=u  t=v)  (s=v  t=w) → PG(s,t,s-1) Draw w u v of length ≤2s SAT: Given Boolean term Φ(Y1,…Yn), does it hold y1,…, yn{0,1}: Φ(y1,… yn)=1 ? NPc

3. Two-Player Game on Graphs 3QBF: Given Boolean term Φ(X1,…Xm)in 3CNF, does it hold x1x2 x3 …: Φ(x1,…xm)=1 ? PSPACEc • Fix digraph G with start vertex s. Rules: • Red (start) and blue player alternatingly • mark current vertex, and follow any outgoing edge • to a yet unmarked vertex. • Who cannot move, loses. Red has a winning strategy if, however bluereacts,red can follow such that, however ……… blue looses. Decision problem: Given (G,s), does red have a winning strategy?

4. 7 n n Two-Player Game on Graphs 3QBF: Given Boolean term Φ(X1,…Xm)in 3CNF, does it hold x1x2 x3 …: Φ(x1,…xm)=1 ? PSPACEc Proof (reduction from 3QBF): Let Φ = C1 C2 …  Ck See illustration for = (x1  x2  x3)  (x2  x3  x7)  …  Ck Theorem: The followingis PSPACE-complete: Decision problem: Given (G,s), does red have a winning strategy?

5. Oracle Complexity Theory Fix ON. An OWHILE+program AOhas test instructions "xjO?" Definition: Fix some class C of subsets LN. PC := { LN decided by polytime OWHILE+AO, OC } NPC := { LN accep. nondet.poly. OWHILE+AO, OC } Examples: a) MinBF NPSAT = NPNP  PNPNP b) PP=P, NPP=NP, PSPACEPSPACE=PSPACE c) NP  coNP  PNP; „≠“ unless NP=coNP

6. : Semantic Polynomial Hierarchy Δ3P Def:Δ0P=Σ0P=Π0P:= P • Δk+1P := PΣkP • Σk+1P := NPΣkP • Πk+1P:= coNPΣkP • PH := ΣkP Lemma: a) ΔkP=co-ΔkP b) ΔkP ΣkP ΠkP c) ΣkP ΠkPΔk+1P d) PH  PSPACE Σ2PΠ2P = PΠkP coNPNP NPNP = NPΠkP = Π2P =Σ2P = coNPΠkP PNP =Δ2P NPcoNP coNP NP =Π1P =Σ1P P

7. Syntactic Polynomial Hierarchy Abbreviate Nn := {yN : ℓ(y)≤n} Theorem: a) LN belongs to coNP iff L = {x | yNpoly(n) : x,yV}for someVP b) L belongs to k+1 iff L = {x | yNpoly(n) : x,yW}for someWk ork c) L belongs to k+1 iff L = {x | yNpoly(n) : x,yZ}for someZk ork d) L belongs to k iff L = {x | y1Npoly(n) y2Npoly(n) y3… QkykNpoly(n) : x,y1,y2,…ykA }for someAP n:=ℓ(x) "" if k odd, "" else k+1P = NPkP =NPkPk+1P=coNPΣkP =coNPkP

8. Walter Savitch's Theorem Def: For nondecreasingf:N→N, let TIME(f) := {L {0,1}* decidable by WHILE+ program in timef(n) } Similarly NTIME(f), SPACE(f), NSPACE(f): nondet. WHILE+ Theorem: For any polynom. p, NSPACE(p)SPACE(p²) Proof:Let nondet.A accept Lin space s:=p(n) Encode configurations ofAinsbitsu=(u1,…us). edge fromutovifvencodesA's possible config afteru. G=GA,sdigraph with u=start config oninputx, wunique accepting config. Aacceptsx  PG(u,w,s), : "path of length ≤2mfromutow"  v: PG(u,v,s-1)  PG(v,w,s-1) Recursive algorithm of depth s stores v in s bits: O(s²) Draw u w of length ≤2s v SAT: Given Boolean term Φ(Y1,…Yn), does it hold y1,…, yn{0,1}: Φ(y1,… yn)=1 ?