[AB 2]. 2 3 3 Def: A Non-deterministic Turing machine (NDTM) accepts L iff there exists a path from...
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Transcript of [AB 2]. 2 3 3 Def: A Non-deterministic Turing machine (NDTM) accepts L iff there exists a path from...
![Page 1: [AB 2]. 2 3 3 Def: A Non-deterministic Turing machine (NDTM) accepts L iff there exists a path from an initial configuration to an accepting configuration.](https://reader037.fdocuments.us/reader037/viewer/2022110400/56649d9e5503460f94a8921d/html5/thumbnails/1.jpg)
Non-deterministic time
[AB 2]
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Next:
•Let us now consider a non realistic computational model: NONDETERMIONISTIC
Which:
•Mathematically natural
•can be simulated by DTMs
•However, with an exponential blowup in time.
2
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• finite set of statesQ• input alphabet: a finite set • tape alphabet • :QP(Q{L,R}) - transition function• start stateq0
• accept stateQacc
• reject stateQrej3
Non-deterministic Turing Machines
3
Excluding “_”
and _
qrejectqaccept
power set
P(A)={B | BA}
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The accepted language
Def: A Non-deterministic Turing machine (NDTM) accepts L iff there exists a path from an initial configuration to an accepting configuration
In particular if it rejects x, there is no path from an initial configuration to an accepting configuration in the configuration graph of x
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Time complexity
Def: The running time of a NDTM T on input x
is the length of the longest path from an initial configuration to a terminating configuration.
Def: A NDTM T runs in time f(n) if for every input x, the running time of T on x is at most f(|x|).
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• Let t:NN be a complexity function
Definition, Non-deterministic time:
Nondet. Polynomial time:
6
Time-Complexity
k
knNTIMENP
TM non time- by decided | ticdeterminisntOLLntNTIME
EXPNP
P
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Non-deterministic computation
tree
Deterministic computation
7
Deterministic vs. Nondeterministic
Time
accepts if some computation accepts
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Perfect Matching – Input G=(V,E). Yes instances: G has a perfect matching.No instances: G does not have a perfect matching.
MaxClique-atleast-kInput G=(V,E),k. Yes instances: G has a k-clique.No instances: G does not have a k-clique.
Examples
8
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No Perfect Matching – Input G=(V,E). No instances: G has a perfect matching.Yes instances: G does not have a perfect matching.
MaxClique-atmost-kInput G=(V,E),k. No instances: G has a k-clique.Yes instances: G does not have a k-clique.
What about
9
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Co-NP
NP
P
10
P, NP and co-NPcoL= *-L
coNP== {coL | LNP}
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A language L belongs to NP if there exists A polynomial p(n)A polynomial time TM M
Such that
xL iff u {0,1}p(n) s.t. M(x,u)=1
Any u s.t. M(x,u)=1 is called a witness for xL
NP – second definition
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Nondet. TM
A verifier
magically guess which
transitions to take to eventually accept if possible
Verifies a witness,
certificate to the fact that xL
Witness Verification Program
12
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13
Nodeterministic
GuessTraverse from s to t
A prime factorization
Isomorphism
VerifyIs it a path from s to
t?
Primes, product =N
Does transform G into G’?
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Proof:
Suppose L belongs to NP, solvable by M(x,u).
Algorithm in PSPACE:
Given x {0,1}n , try sequentially all u {0,1}p(n)
and accept iff for some u, M(x,u)=1.
NP PSPACE
14
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Defined complexity classes via bounds on TMs:
• Polynomial timeP
• Nondeterministic Poly time
NP
• Complement of NPcoNP
• Exponential timeEXPTIME
• Logarithmic spaceL
• Polynomial SpacePSPACE15
EXP
PSPACE
NPcoNP
P
L
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anbncn
STCON
Perfect
matching
Hamiltonian
Path
Factoring
Name the Class
16
EXPPSPACENP
P
NL
L
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• A Boolean formula.
SAT Instance:
• Is the formula satisfiable?
Decision Problem:
17
SAT
SA
T o
r u
nS
AT
?
)x(x) x) xx((x 231321 11 xx
EXP
PSPACENPcoNP
P
L
Theorem:
• SAT is in NP
Proof:
• Can verify an ass. efficiently
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• SAT is NP-Complete
Theorem:
• Given an NP machine M and an input w, construct a Boolean formula M,w M,w satisfiable M accepts w.
Proof Outline:
18
Cook/LevinSIP 254-259, AB 2.3
LM M,w SAT
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# q0 X2 X3 Xn _ _ _ #
# ? q? X3 Xn _ _ _ #
# #
# #
# #
# #
# #
# #
# #
# #
# qacc _ _ _ _ _ _ _ _ #
State +cell
content
Tapeends
Input
19
TableauxRepresent a computation as a configurations` table
cne
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20
Example• = {q0,q1,qaccept,qreject}
Q
• = {0, 1}
• = {0,1,q*,_}
• (q0,1)={(q1,_,R)}• (q1,1)={(q0,_,R)}• (q0,0)={(q0,_,R)}• (q1,0)={(q1,_,R)}• (q0,_)={(qacc,_,L)}• (q1,_)={(qrej,_,L)}
# q00 0 1 1 1 #
# _ q00 1 1 1 #
# _ _ q01 1 1 #
# _ _ _ q11 1 #
# _ _ _ _ q01 #
# _ _ _ _ _ q1_
# _ _ _ _ qrej _
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21
M(Q{#})4• = {q0,q1,qaccept,qreject}
Q • = {0, 1}
• = {0, 1, q*, _}
• (q0,1)={(q1,_,R)}• (q1,1)={(q0,_,R)}• (q0,0)={(q0,_,R)}• (q1,0)={(q1,_,R)}• (q0,_)={(qacc,_,L)}• (q1,_)={(qrej,_,L)}
q00 0 0
.. q30 ..
# q00 0
.. q01 ..
_ 0 0
.. q00 ..
1 0 0
.. 1 ..
1 0 0
.. 0 ..
q00 0 0
.. q31 ..
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# q0 W1 W2 Wn _ _ _ #
# ? q? W2 Wn _ _ _ #
# #
# #
# S #
# #
# #
# #
# #
# #
# qacc _ _ _ _ _ _ _ _ #22
M,w variablesBoolean Xi,j,S standing for:does cell i,j has value S? cn
e
i
j
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1 value for each entry
input consistent
transitions legal
machine accepts
0,0,# 1,0, 2,0, ,0, 1,0,_ ,0,_ 1,0,#21e eX X X X X X Xq w n w n cn cniw n
wM ,
23
M,w
acce qjicnjiX ,,
,0
, , 1, 1, 1 , 1, 2 1, 1, 3( , 1, 2, 3)1 , e i j s i j s i j s i j ss s s s Validi j cn
X X X X
input blanks
accepting configuration
start
locally legal transition
)xx(x tj,i,sj,i,#ts
sj,i,#sji,1 QQcne
≥1 value
<2 values
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wM ,
24
Q.E.D.
acce qjicnjiX ,,
,0
, , 1, 1, 1 , 1, 2 1, 1, 3( , 1, 2, 3)1 , e i j s i j s i j s i j ss s s s Validi j cnX X X X
)xx(x tj,i,sj,i,#ts
sj,i,#sji,1 QQcne
• tableau legal i,j transition is locally legal
Claim:
• M,w Satisfiable WLM
Corollary:
• Size of M,w polynomial in |W|
Claim:
0,0,# 1,0, 2,0, ,0, 1,0,_ ,0,_ 1,0,#21e eX X X X X X Xq w n w n cn cniw n
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ANY NP LANGUAG
E
•JUST SHOWN
SAT
25
SAT is NPCWe have just shown SAT is NP-hard, as any NP language can be reduced to SAT
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Co-NPNP
PNPC
26
P, NP, co-NP and NPC
SAT
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ANY NP LANGUAG
E
• JUST SHOWN
SAT • SHOW
A
27
NPCClaim: Tshow a problem A is NP-hard, it suffices to reduce SAT to
A
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ANY NP LANGUAG
E
• SHOWN
A • SHOW
B
28
NPCFurthermore, once we’ve shown A is NP-hard, we can reduce from it to show other problems NP-hard
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proved SAT is NP-Complete
Consider SAT the Genesis problem, show other problems are NP-hard
29
Summary so far
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30
Goal:
•introduce some additional NP-Complete problems.
Plan:
•3SAT•CLIQUE & INDEPENDENT-SET
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So far we only showed one such problem: SAT
• 3SAT
Next we show a special case of SAT is NPC:
• 3CNF formula
3SAT Instance:
• Is it satisfiable?
Decision Problem:
31
3SAT• L In NP• L NP-hard – via Karp-
reduction
Recall: L is NPC if
Conjunctive Normal
Form – 3 literals in each
clause
3C
NF:
(xyz)(xyz)
(xxx)(xxx)
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• CNF NPClaim:
• CNF NP-hardClaim:
• amend our SAT formula, so it becomes CNF• To make it a CNF: use DNFCNF on 3rd line
Proof:
32
CNF is NPCSIP 259-260
CNF is a
special case of
SAT.
acce qjicnjiX ,,
,0
)xx(x tj,i,sj,i,#ts
sj,i,#sji,1 QQcne
What is the size of new
formula?
wM ,
, , 1, 1, 1 , 1, 2 1, 1, 3( , 1, 2, 3)1 , e i j s i j s i j s i j ss s s s Validi j cnX X X X
0,0,# 1,0, 2,0, ,0, 1,0,_ ,0,_ 1,0,#21e eX X X X X X Xq w n w n cn cniw n
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(xy)(x1x2... xt)...
33
CNF3CNF
Complexi
ty©D.Moshkovitz
clauses with 1 or 2 literals
(xyx)
replication
clauses with more than 3 literals
split
(x1 x2 c11)(c11 x3 c12)... (c1t-3 xt-1xt)
QED
3SAT is NP-
Complete
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• CLIQUE NP
Observation:
• Given C, verify all inner edges are in G
Proof:
• A graph G=(V,E) and a threshold k
CLIQUE instance:
• Is there a set of nodes C={v1,...,vk}V, s.t. u,vC: (u,v)E
Decision problem:
34
CLIQUE is NPC
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3S
AT
C
LIQ
UE
G,k
35
3SAT L CLIQUE
001 010 111
101
7 vertex for each clauseedge consistency
k = number of clauses
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SAT p CLIQUE: proof
• Let A be a satisfying assignment to , C(A) is the set of vertices that correspond to this assignment.
Completeness:
• A clique C in G of size k corresponds to a satisfying assignment to
Soundness:
36
SIP 251-253
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• A graph G=(V,E) and a threshold k
IS instance:
• Is there a set of nodes I={v1,...,vk}V, s.t. u,vI: (u,v)E
Decision problem:
37
INDEPENDENT-SET is NPC
• IS NPObservation:
• Given I, verify all inner edges not in GProof:
• IS is NP-hard
Observation:
CliqueIS on complement graph