CS计算机代考程序代写 algorithm CSE 565 , lecture 14, 10/13/2021

CSE 565 , lecture 14, 10/13/2021

projection problem

Description ④ -8

-30B€ b
G=N
PW )

: profit of v.

t.tt
④→⑤
-7 10

DI Hea side 🙂 Vi EV is feasible , if V1 satisfies :

if u C- V , and Lui V1 C- E , then UE V1 .

Input : G-=/VIE) . plusV-vc-V-uut.IOa feasible V1 c- V sit . ¥yPM is maximized .

Algorithm : reduce into the max- flow problem .

VIVI

step create G-= IVIE) . Cle ) He c- E , sit,

1-*

¥÷¥÷É¥¥÷:*¥ .
step find f-

*
of this network .

an s-t cut 15×1-1*1

step return 1-
* 1ft}

Fa -1*1 It} is feasible .

proof :

1-*

⇒ 45×1-1*1 _→ , a contradiction .

FAI ¥g*yµPM
is maximized .

*

s*

clS¥T*)=¥g¥P
‘” t ¥
,#

1- PWD

=/I PM -[ PH – ¥g±PMv c- 540T¥ VE -1¥
=¥µ,>0PM

– ( ¥¥PMt¥g±Pm)
= c- I

VE-1*1 /ty
P”” ☒ .

Cawnstant)

Baseb.at/Fliminationproblemourrentswringboard- schedule of remaining games .
I w ‘ (Ti,Tj)✗XijgamÉ
T2 WZ

: :

Tn Wa

In teams I # wins)

Question :
is IT -☒ill possible to win the 1st place ?

I #wins 1-* gets is among thehighest.IT

Is T2 -13 -14 TsE×aM
#wing * a 41 43 39 –

M= 43 42 42 43

scheduling text I :O
(1–74)×-2 2 :O

¥9
(Tz , -13) ✗ 3

Tc={Tz , -13, -14} . ( Tz, Ta) ✗ 1 I :O

(-13,1-4) ✗ 1 I :O

(-13,1-5) ✗ 2 .

Preprocessing : assume Ti wins all remaining games
⇒ use m to denote the # total wins) of -1 ,

Question : Is there awntigurationofallremaininggamess.fi#Wittt-fi=m.V-i >2. ?
Algorithm : reduce into Max – flow problem .

step create G- = IV.E) Cle) Yet E. sit .

m-Wi

i.
2
s-% 4

(games ) (teams

skp find f-
*

.
15%-1*1

.

step 3 : It 11*1 = ¥gzXij ,
then answer ” yes

otherwise
, answer


no


.

Proof of correctness :

11-+1 = ,§⇒xij ⇐ such a Intention exists ☒

certificate-1,75¥ :-b .

: Tn } that shows -11 is eliminated :

Hi , -1J ) c- To
✗ it > I Lm

-Wi .
TIETC

Q : Does a Tc always exist when 11–4 < §g⇒Xij ? A : Yes . constructively f ' 4551-7-6 m-Wi :* ¥É÷F ⇒ Claim : If ITi. Tj) Est , then Ti c- stand Tj c- s*, ±t¥i÷÷* ⇒ 45×1-1*1=0 , a contradiction ☒ Clavin : If Ti , -j c- s*, then (Ti-Tj) c- s* . proofs , 1- * By moving IT i.Tjlto 51 that edge is excluded from cut-edges - ⇒ CIST-1*1 won't be minimum , a contradiction . ☒ claim: :TofTiES*} isauria . it HHi§⇒×ij . ie' (¥g%×ij > ¥ M
– Wi

proof : 11*1-+1511-+4

I

=¥Ém
-Witt i¥z×ij – (Ti , ETC

sine 11*1 < ¥>.< Xij ⇒ ¥TEKXY " > ¥ M

– Wi ☒