Lecture 23 Scribble Topics Reductions Chat Moderators Junyeob Cliques Independent so Emerson SAT Two Problems XEpY Yes 4 R y AY µ Ax A finding shortest path from 5 tot Djikostoa's Ay finding all shortest paths Bettman Ford Ojikstra's E Ballmautford easy has a Yes polynomial time 4 R s Ay deterministic solution No A hard X Ep Id X is no harder than't If X is hard it should also be hard Y is atleast as hard as X
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Lecture 23 Scribble Topics Reductions
Chat Moderators Junyeob Cliques Independent so
Emerson SAT
Two Problems XEpY
Yes4 R yAYµ
Ax
A finding shortest path from 5 tot Djikostoa's
Ay finding all shortest paths Bettman Ford
Ojikstra's E Ballmautford
easy has a
Yes polynomial time4 R s Ay deterministicsolution
No
A hard
X Ep Id X is no harder than't
If X is hard it should also be hardY is atleast as hard as X
Example CliqueyIndependent SetGCU E
eeIndependent set I C V St
Tryno Yu C I are connected
6 eby an edge aod
Clique C E U sat u u EE offor all u v C C e f
Given GCU E and an integer k i
Does G have an II SI 3 K MAX INO SETDoes G have a clique let 3k MAX CLIQUE
Ay Solves MAX CLIQUE
V EGCU E
a a
dd
t.LIe f
yesMIS Ine T R finding the edgeG R Max Clique
9 No complement of theG graphMax IS 0Gt
Maxis Ep MarxClique Max IS p Max Clique
yesX Ep Y
4 R yAYµ
A
If Y is hard is also hard FALSE
If X is known to be hard then Y does not havea efficient algorithm TRUEWWant to prove HW P 4G is hardReduce a hard problem to X
GIS or Clique
If Y has a polynomial time algorithm then it impliesX has a polynomial time algorithm D pendsIf R is polynomial then true
If R is hard the false
X Ep Y implies Yep X FALSE
f X Ep Y and Y Ep Z then X Ep 2 TRUE
no
SAT Problems CNF SAT 3CNF 3 SAT
X x xz Xn V o G A and o
x U zV Iii N xz V x 5 Axs Is there a assignment to
fr the boolean variables etclause the expression is true
conjunctive normal form Iff.ITmaIIIi3CNFx VxzVx a N xioVxzVxzDN
Every boolean expression can be expressed as 3 CNF
aba Ab EUWE r a Ub N auto
i
1
Problem SAT satisfiability NRGiven a CDF formula is there an assignment which results
in the formula as evaluating true
Every CNF expression caru be reduced to 3 CNFif clause 3 literals Ca V love Lavrov e Acc VEX Y are boolean expressions if X V Y are SATVybve U d Caubuz A evolve then X z h Y E are SAT
if clause has 3k literal
aub aub Vu A fav 6 VE
if clause has 1 literala La Vu Vu a Cavivu a Cavuv Native