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-x
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 .