CS计算机代考程序代写 algorithm CSE 565, lecture 18 , 10/27/2021

CSE 565, lecture 18 , 10/27/2021
Def 1 PSPACE) : the set of decision problems that are
solvable in polynomial space .

=aH P C- PSPACE .

Q : NP vs. PSPACE ?

Fait : It ✗ =p Y and Y c- PSPACE , then ✗ c- PSPACE .

Fanti Let ✗ c- NPC , It ✗ c- PSPACE, then NP C-PSPACE .

Preet : Let X’ be an arbitrary problem in NP
X

Ep X ( b. e. ✗ c-NPC) .

✗ c- PSPACE ⇒ X
‘ C- PSPACE ⇒ NPE PSPACE ☒.

Claire , 3. SAT C- PSPACE .

proot : Algo- for – 3SAT I CNT-E-tx.VE#Al 7×1 ) ) .
with n variables , m clauses .

idea : Enumeration . zn
binary
array of size n

|
“””” “””

Xi ← o . Kien

while I counter < zn) B← evaluate (E) :QH-sp / it D= true : return true end increase - count and return false Fonti above algorithm runs in 0cal space . Claim : NP C- PSPACE . claim co- NP C- PSPACE proof: 3¥ c- PSPACE Hierarchy : ☒ Np co -NP ' ' PSPACE Note: P F- PSPACE, ( open . consensus 😛 =/ PSPACE) Nod any two are not known yet to be equal / not equal . Prob6m_ ( QSAT : Quantified SAT) lnisodd Input : CNF I 1×1 , " - , Xn) =L ) ✗ I 7N ) outputi-H-7-XIV-xz7-XSV-X47-xs-i-V-xn-17-XNTO-lxr.mn) is true . I existential & universal quantifiers alternates . F-1×1,112,1131--1×1 VXZVXJINXNXZVX} ) ✗ ( XTVIKVXJ ) T IIX , ,XzX3) = IXIVXJVXJINXNXJVX} ) ✗ ( XTVIKVXJ ) /YXTVXZVX}) F Note:1isky that QSAT is in NP or co -NP . Clavin ' QSAT C- PSPACE . proof: Idea : to evaluate quantified formula 7-XiVvI↳V-xaven

P2 has a winning strategy .

Geograph problem is in PSPACE . I homeworry

Det_ ( PSPACE – complete ) : ✗ c- PSPACE – complete if
I ‘ ✗ c- PSPACE

2. f X’ C- PSPACE , X’ Epx .

Claim : QSAT is PSPACE – complete .

Claim : Geo graph problem is PSPACE – complete .
proofs : to show

QSA-t-p-egraphyp-6m.FM#T-lXnX2iXD=lXiVX2VX3)AlXTVXaVXJ
) .

if 7- Xi V-X.7-X3-0-lxi.xz.XD.is true .

o•

× , Pi
: OFF

a

pz ☐

☒←④-
Pl

×, pz*É⑤←
÷

P2


I

claim : I is

true

⇐ Pi has a winning strategy .