CS计算机代考程序代写 algorithm CSE 565

CSE 565
,

lecture 11
. 0912912021

Fortson algorithm

Det-lrewikag-aphwrto.net flow f) :
G-
g-
= LV
, Ef ) , 4-19 . He

c- Ef .

Exampte
:

¥9:#⑤¥§FÉ⑤ ¥¥04K
(G. c. sit) and f- Gf

¥ : forward edge)
↳ del -fiel : tie)< cle) •Ude¥I A ñu tie > 0G- ( backward edgy Gf

Algorithm Ford – Fulton ( G- =lvitl.de/V-ec-E.s.t.c-V)initf:fie)–oV-eEE. OIIED
while ( true

| construct Gf wrt f. OHEHND| find an s -t path in . 1B¥ 0-(14-1121) .if such path p can be found : augment /f. B) .
otherwise : return f- 0lk .

d- . end

Augment

⑤*¥¥o ⑤¥:¥É
☒0b¥ ×*_11-1=4

( G, c. sit) and f- Gf

⇒ ⑤÷£ÉÉ¥*⇒
function augment If , If

p : °g→→→o→o
in If 4-11

t
II> = meienp 9-19

fl *emains the same . if e is not affected by p .

fk) C- fie -1×117 if e corresponds to a

forward- edge”inP .

☒e) ← fte) – ✗ Ip ) it
e – – – –

“backward ” inp .

Fonti after augmenting, f- is still a valid flow .
µ still satisfies birth conservation and capacity constraints)

Fae ater augmenting . 1ft is increased by ✗ cp) .

G- Gt

⑤¥¥¥⑤ # ⇒ ⑤#¥¥¥×a⇒⇒¥0,4T
so;÷E¥¥+¥ “” *⇒
so;¥É§¥¥ ” ‘⇒

⑤¥÷¥¥¥ “”
*⇒

1-*=/ art}

⑤¥¥¥÷¥”→s¥¥¥a±I

Pet : when FF terminates . denote the returned How as -1$ .

Fact : In Gf* , there is no path from s to t .

Define : s
*
= { a c- V1 s can reach v in Gfx }

1-
* = V1 .

Fait : 151T¥ is an-s-tut.to/- the original network)

claim : If* I = 4511, -1*1.
proof :

② 1-* = fait } .

)¥¥¥É¥ 1+1–9I
s*

tf e e Elst -1*1 , f-*Ie) = Cle) .
proofi because otherwise , new vertex in 1-

*
can be reached from S.

F e c- ELTE 5×1 , 1-
*Ie) = 0 .

proofi because otherwise , new vertex in 1-
*
can be reached from s.

11-4=-2 = -2
£ “” no
fate) – I #e) .

e. c- 01s) ec-EISF-ijec-T-t.FI

= [eels-41-+4
” -0 =ds*

,
-1*1 ☒ .

theorem Imax- flow , wain – out) : For every network ,
its value of maximum flow equals to the capacity of its

minimum cut .

Ruminating : The FF uses at most 11¥ iterations .
0111–4.114-11 EID

-ffffgg@

( assuming the capacities are integers . ) .

Q : Does this algorithm run in polynomial – time ?
A : No . ɥ
Why? 1%4*1 11-+4=211091*1 ) . #¥

c←ÉÉ algo) . |scahñg] 173kt) .

preflow-relabelalgorithm.FI
PR

maintains a flow maintain a pretlow

global_ update Local update

pefeftow) : f : E-→Rt satisfying
1. ( capacity constraint) 0=1-1 a- de ) . He EE
2. Define exceed of a vertex o .

Xflu) = Ifk)
– I tie

eEIw et Qu)

4- IN 30 , tf u c- V1 { sit} .

Exampte :

⑤¥}¥g⑤ 1-
is a pretlow .

⇒ 144
Ietsi wrt a pretlow f)

¥ ±

Detllabeting ) : h : V→Z⇒ .

Pef_ I compatible) : let f- be a pretlow , and h be a

labeling , of the same network . We define f- and h
are compatible . if :

i. his)=n n=#
2. hit) = 0

3. For every edge e= Luv) in Gf we must have
Iffy

that him shiv) -11 . I steepness condition

if :-.:-. \>• •w)-11him = her) huh -111h ‘D hlu)= has -11 ↳•
✓ ✓ ✓ ✗