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
③
:
:::
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 ,
t¥
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÷” .
/