CS计算机代考程序代写 CSE 565 , lecture 17 , 10/25/2021

CSE 565 , lecture 17 , 10/25/2021

tact : P C- Np .

Question Pt NP I open , consensus opinion : PHP)
.

Fact : Let ✗ c- NPC . It ✗ EP , then p=Np .

Proof : Let X

be an arbitrary problem in NP

✗ c- Npc ⇒ ✗

Ep X . ⇒ YEP 1 because ✗ c- P)

⇒ NPEP ⇒ p=NP ☒ .

Hi⇒ : p ⇒ NP

⑤ ↳NpnNpc=0_
p=NP=NPc NP

Def_lw_¥ problem) : Let ✗ be a decision problem .
It’s complementary problem , denoted as I , is defined as
the same set instances with X but with opposite ground truth .

I 😐

¥÷¥:÷
a
– a


r a

✗ ☒

Exampte:

3.SAT ( CNF) 3SAT
true : ← 7- assignment sit . CNF is true . → false
false :-c V- assignment sit CNF is false . → true

=a¥ I =X

Fonti XEP ⇐ I c- P
Question it ✗ c- NP ⇒ I c- NP

Det ko-µ :p) : Let ✗ be a decision problem . We say XC-co-NP.it
there exists a polynomial – time procedure A aerified , sit .

i. input of A is an instance ✗ of ✗ plus %%ÉÑiticate c.^
Isolation

2. For any instance × of ✗

if ✗ is a

false

– instance

,
then 7- c. sit . Alx . =

no

if ✗ is a “true

– instance
,
then V-c.s.t-AH.cl =

yes

.

Faet ✗ c- NP it and only if I c- co-NP
.

proofs by the defs of co-NP , NP , and complementary problem .

Fanti PE co- NP

Inti P E NP A co
-Np

Question : NP ?= co – Np . Copen . consensus : Np + co -Np)

=ae# If p=NP , then NP = co -Np .

proof ; let ✗ c- Np .⇒ ✗ c-P ⇒ Iep ⇒ IENP ⇒ ☒ c- co
-Np⇒ ✗ c- warp.

let ✗ c- co – Np ⇒ IENP ⇒ IEP ⇒ XEP ⇒ ✗ c-NP ☒ .

Fa It NP # co – NP . then P # Np .

Hierarchy : p # NP
← ↳

NP ?= co -NP
0
P=Np=w-µp * ↳

P=? NPN co- Np
.

yes/ ↳⑤
④ ④NP=co-NP
Np co -NP NP a-NP

Question : P ?= NPA co -NP Copen : mixed)

problem X: Does a G-ME ) contains a flow with 1ft > N ?

✗ c- NP .arhfiaatef-XC-co-NP.mn/-f1owf*–.s-twtlS.T) as certificate .
false instance ⇒ 7-1 f- sit . Ifl > N

⇐ V-fsit.lt/sN-1 .
⇐ * ISN -1

⇐ 454T¥ a- N – I
function verifier 16–14 El . N, )

verify Is , -11 is an s -t cut .

I verity 451) IN -1
end .

Probtem ( integer factorization) .

Input an integer X, an integer n .
output’s it there exists a factor of ✗ ( ✗ mod ✗ ⇒ sit . ✗en .

Algo – for
– IF IX. n)

for K= 2 to n

1 it ✗
mod K=o return true1 end

end .
return false

Running time
i Qln)_= -012109¥ exponential- time

siu-tip-uillog.lt/ogzn)
Question : IF & P I open) .

FAI IF c- NP

proot : certificate ✗ lfaetor)

verifier IX. n , x)

verity ✗ enI verity ✗ mod ✗ =0
end ☒ .

Claim : IF c- co – NP

Fae false instance ( Xin) of IF
⇐ V- factor ✗ of ✗ . sit . × > n .
⇐ it prime factor X of X , sit . X > n .

Exampte : 11=7×11×133 W= 6 , only need to check {7.11.13}
.

✗ =p ,M . pztkpgn} . . . prank ( facto’t:-. ration)
K

toga = nilogp , 1- nztogzpz
+ – – – nk.bg?k–E,ni.logzP;

2

Pi > 2 , ni> 1 .

109N = ,É
,
ni.log.Pi-3-i.kz

,

ni
. ⇒ K= 0110g X) .

¥iat : ( Pnh ,) , 1Pa, nd . ‘
‘ ‘

, CPK, NK) .

function verifier IX. n . { cpi.mil/1–i–k}) .

verity ✗ = II
,

Pini .

| verity Pi > n . KKK
end
veritypiis-apr.me. kick

T-ati-PRIMES-ltodediide.it a number is a prime. c- P –
l AKS

, 20021 .

Fonti Above verifier runs in poly -time .

Eatin : ✗ d- IPP ?