CS计算机代考程序代写 algorithm CSE 565 , lecture 13 . 10/11/2021

CSE 565 , lecture 13 . 10/11/2021

Runningtimeof pretlow – push i OIIVP
– IED .

Best algorithm so far : -0114M (

T.ODin.no/Prob6m-lmax-matching)
2npu undirected graph A- = IKE )

outputs a matching M , sit . 1Mt is maximized .

Detlmatehiñg) : MEE sit . M don’t share common
vertices .


1Mt =3 .

HI =3 .

Problems / minimum vertex-cover)

Inputs undirected graph a-= 1h E)

output a VC Vi EV , sit .lv , / is minimized

Det Nc) : Vi EV sit. V1 covers all edges .

Fa Let M be any matching , let V1 be any
lrertex cover , of the same G- = ME), then

1Mt E Wil .

max-matching
µ ,1Mt Iv

;pE→ R
min – Vc .

Q : Do d-hey↳ meet ?
A : No .

IM-4–1counterexample : OA
IVF / = 2

⑤codd-lengthcy#
Q : what if G-does-mt-n-aiod-d-ek.rs
A : Yes , they away meet .

Fait : An undirected glraph a-= IV. E) does not contain

any odd
– length cycle it and only if G- is a bipartitegraph .

③÷¥i÷
.

G- = I XUY, E)

Problems : find it of a bipartite graph G-=/XUY.FI .

Algorithm : cpolynomiattiinerdutiou)
Steph for any instance G-

= IXUY. E), create an

instance of the max- flow problem .

GEW.EY.de#t-V.stepItindf*otG’stepI
find M* from f-

*
,

G- IXVY, E)
G’ = ( XUYU {sit} , EY

c /e) =p

:
:::
t*le

let f-
* be an maximum integration, 1-+19 c-{ on}

step return ME { e EE 11*14–1 } .

Fanti There is a one – to – one correspondence between any

matching Mot BG and any integral flow f- of
the network sit . IMI = Ifl .

⇒ IMA = 11-+4 .

pnbtemi find Vi of a bipartite graph 4-=/XVY, E) .

{ a-}

*•÷☒D¥÷:÷÷¥¥÷¥¥¥E.i.i://t.is} Hint} St
residual graph wry µ ,


original network

Algorithm :
1. construct Gf*
2. construct s – t cut (541-+1)| 3. return VF:-# AT’t) U ( Yns*) .

end

Claim : In the original network, there is no edge from
✗ ns* to YAT’t

UES* VE Tt .

proof: lby contradiction) . Assume: •u→•u
case 1 : f-

*
let = 0

:-O. ÷
U U V

original network residual graph
⇒ contradiction , as v will be reachable from s

case 2 : f-
*
14=1www..IT/WsY.#..iEt:–;c-;It

original network
residual graph

⇒ contradiction . as u won’t be balanced .

Fonti VF is a Vc .

privet : directly proved by above claim .

East : V7 is a minimum VC .

proot : 159T¥ is a min s -t cut .
⇒ c (5×11-+1) =/ f-* 1 . I by max-flow min – cut theorem
is
‘t
, -1*1 . = IVFY lby above claims

* I = 1Mt ) cby the one-to-one correspondence) .
⇒ WEI = 1Mt
⇒ Vi is a minimum VC . Ias they meet.)

Summaryi bipartite
general graph

m÷:÷i•÷::TÉÉÉ÷” .
/