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On Thursday, October 22, 2009 at 9:57:47 PM UTC-5, Future_News wrote:
> Number Agreement
> A noun and the words that modify that noun must agree in number. Nouns
> that cannot be counted or divided, such as "oil," "happiness," and
> "furniture," require singular modifiers.
> Instead of: I found those five money.
> Consider: I found this money.
> Or consider: I found those five pieces of money.
>
> Instead of: She loved these two pasta.
> Consider: She loved these two types of pasta.
> Or consider: She loved this pasta.
> A Neat Realization
> This is from comp.infosystems and comp.theory on USENET:
> MACHINE PROOFED RESOLUTION TO THE P VERSUS NP PROBLEM PRESENTED BY M.
> MICHAEL MUSATOV WITH A PROMISE TO DONATE ALL $1MM DOLLARS OF THE CLAY
> PRIZE TO CURE CHILDHOOD CANCER WITH HTTP://WWW.ALEXSLEMONADE.ORG/ THE
> P VERSUS NP PROBLEM TWO POSSIBILITIES: 1 OR 2 CHOOSE
> [P =/=NP] AND [P == NP] OR
> [P =/=NP] AND [P == NP] BUT NOT BOTH VERSE THEM EXPLORE THEM ENTWINE
> THEM INT: 2=TPO UNARY 1.=P===NP P=/=NP 2.=P=/=NP P===NP 3.-=P===NP P=/
> =NP 4.=P=/=NP P===NP
>
> ----------------------------------------------------------------------------------------------------------------------------
> +1.=P===NP P=/=NP
> +2.=P=/=NP P===NP
>
> ----------------------------------------------------------------------------------------------------------------------------
> +3.-=P===NP P=/=NP
>
> ----------------------------------------------------------------------------------------------------------------------------
> +4.=P=/=NP P===NP ECHO HELLO WORLD WRITE OUT = 2- : OVER THIS MESSAGE
> AND AFTER THIS MESSAGE ANDE THROUGH THIS MESSEAGE ECHO HOLLA WORLD:
> On Oct 22, 9:03 pm, Future_News <future_n...@brew-master.com wrote:
> On Oct 22, 8:31 pm, Future_News <future_n...@brew-master.com wrote:
> On Oct 22, 4:48 pm, Tegiri Nenashi <tegirinena...@gmail.com wrote:
> On Oct 21, 5:04 pm, cplxphil <cplxp...@gmail.com wrote:
> I'm not sure if you're disagreeing. If you are, my response is that
> you could argue that it preserves the same structure of the decision
> problem, but expressed as a formal language, it's technically
> different. It has to be proven that languages are basically equivalent
> (in terms of decidability, time complexity, and space complexity) and
> independent of encoding scheme, and this proof only applies to
> encoding schemes that use at least two symbols in the alphabet.
> I'm not comfortable with this alphabet symbols count. Do word
> separators count as alphabet symbols or not? Why is the language is
> assumed to be structured into sentences of words? This looks like ad-
> hoc assumption from language theory perspective.
> I guess something I should have mentioned earlier is that Cook's
> theorem would not be true if you use a unary alphabet, meaning that
> SAT would not be NP-complete in that case. If you think you can prove
> Cook's theorem for a unary alphabet, please share this proof.
> http://en.wikipedia.org/wiki/Cook%E2%80%93Levin_theorem
> The only place I noticed depending on the encoding appeared to be the
> last sentence. Changing log into polynomial doesn't change anything
> there. Is there another dependency I'm missing?
> NP 3 334 44 NP NPNP-Co -Co-Com mmp pplet letletene eneenes sss ss
> 34.1 Polynomial time 34.2 Polynomial-time verification 34.3 NP-
> completeness and reducibility 34.4 NP-completeness proofs 34.5 NP-
> complete problems 34 NP-Completeness
> 3 334 44 NP NPNP-Co -Co-Com mmp pplet letletene eneenes sss ss
> ............. polynomial time.... ,.... worst case....... O(nk),.. n .
> input size, k. constant. .. ............ polynomial time..
> Turing.........: Halting problem,.................,.. ... polynomial
> time...... .............,........ O(nk).
> Generally, we think of problems that are solvable by polynomial-time
> algorithms as being tractable, and problems that require
> superpolynomial time as being intractable.
> 34 NP-Completeness
> 3 334 44 NP NPNP-Co -Co-Com mmp pplet letletene eneenes sss ss
> ..... polynomial time.....,... . P......... polynomial
> time... ..........,.. NP....
> . 1971...... P = NP ?...,...
> ............. ....... NP-complete...,... NP .......,...... polynomial
> time ...,............. polynomial time....
> 34 NP-Completeness
> 3 334 44 NP NPNP-Co -Co-Com mmp pplet letletene eneenes sss ss
> .. NP-complete............ ,.... polynomial time....... .....
> Shortest vs. Longest simple paths: 24.. ........... single source.
> shortest path...... O(VE)..
> longest simple paths....:..... ,........ simple path(.....
> path),...... k........:. ..... Hamiltonian path(......
> path)..........,. NPcomplete ....
> 34 NP-Completeness
> 3 334 44 NP NPNP-Co -Co-Com mmp pplet letletene eneenes sss ss
> Euler tour vs. Hamiltonian cycle:.... ..,............. cycle,. . Euler
> tour...,.. O(E)....... .....,.............. cycle,.. Hamiltonian
> cycle...... .....,. NP-complete....
> 34 NP-Completeness
> 3 334 44 NP NPNP-Co -Co-Com mmp pplet letletene eneenes sss ss
> 2-CNF satisfiability vs. 3-CNF satisfiability:
> ........,............ .. true,.. satisfiability....... . .............
> (x1 . -x2) . (-x1
> x3) . (-x2 . -x3)....,.. 2
> CNF(conjuctive normal form) satisfiability
> ... (........ or,......
> ... and,............ );.
> ...............,.. 3-CNF
> satisfiability.... 2-CNFsatisfiability. ..,. 3-CNF satisfiability.....
> NPcomplete .... 34 NP-Completeness
> 3 334 44 NP NPNP-Co -Co-Com mmp pplet letletene eneenes sss ss
> NP-completeness and the classes P and NP
> ....,.......:..... polynomial time.....,.... P......... polynomial
> time.............,. . NP...,.. Hamiltonian cycle, 3-CNF. .. P .
> NP,...: P . NP?.......: P..... NP. . NP-complete... NP.......,... . NP-
> complete...... polynomial time. ....,.. NP....... polynomial time ....
> 34 NP-Completeness
> 3 334 44 NP NPNP-Co -Co-Com mmp pplet letletene eneenes
> sss
> ss
> .. NP-complete........
> polynomial time....,.......
> .. polynomial time....
> ....,...... P.NP,.. NP-
> complete............,.
> 1971. Cook..... NP-complete..
> ..,..................
> NP-complete... polynomial time....
> ..............,......
> .
> 34 NP-Completeness
> 3
> 334
> 44 NP
> NPNP-Co
> -Co-Com
> mmp
> pplet
> letletene
> eneenes
> sss
> ss
> ......... NP-complete....
> ...,............ NPcomplete
> .
> P.... NP......;.. NPcomplete
> ....,.....
> approximation algorithm;......
> special case....,.........
> polynomial time..........
> 34 NP-Completeness
> 3
> 334
> 44 NP
> NPNP-Co
> -Co-Com
> mmp
> pplet
> letletene
> eneenes
> sss
> ss
> Overview of showing problems to be NP-
> complete
> ....... NP-complete........
> .....,................
> .......;........ NPcomplete
> .................
> ........... O(n lg n)......
> 34 NP-Completeness
> 3
> 334
> 44 NP
> NPNP-Co
> -Co-Com
> mmp
> pplet
> letletene
> eneenes
> sss
> ss
> Decision problems vs. Optimization problems
> ...................,.
> ........ (optimization problems),
> ...........: minimum spanning
> tree. shortest path..........
> ..... decision problems,......
> ..,....... (Yes/No)...:...
> . weight..... 20. spanning tree..
> ........ k. simple cycle.
> .... decision.... optimization..
> ......... NP-complete.,...
> decision problem....34 NP-Completeness
> 3
> 334
> 44 NP
> NPNP-Co
> -Co-Com
> mmp
> pplet
> letletene
> eneenes
> sss
> ss
> Decision problems vs. Optimization problems
> . Optimization......... decision
> ........:......... u, v.
> v
> .......... decision...... u,
> .......... k .......,.
> ...........,..........
> ........ k .... decision....
> .,.. decision.........
> optimization......
> .. decision........,.......
> ..,.... polynomial time,....
> decision problem...
> 34 NP-Completeness
> 3
> 334
> 44 NP
> NPNP-Co
> -Co-Com
> mmp
> pplet
> letletene
> eneenes
> sss
> ss
> Reduction
> ........... reduction,.....
> ..... A. NP-complete.
> ........ A. polynomial time..
> ......,...... polynomial time
> ...........:..... A....
> (instance)........ polynomial time
> ...... B ... .,.... .....
> ...........,.. A....
> polynomial time...
> 1....... polynomial time.
> 2. ..... .....34 NP-Completeness
> 14
> 34 NP-Completeness
> 34 NP-Completeness34 NP-Completeness
> Reduction
> We call such a procedure a polynomial-time
> reduction algorithm.
> polynomial-time
> reduction algorithm
> polynomial-time
> algorithm to decide B
> . .
> yes yes
> no no
> Polynomial-time algorithm to decide A
> 3
> 334
> 44 NP
> NPNP-Co
> -Co-Com
> mmp
> pplet
> letletene
> eneenes
> sss
> ss
> Reduction
> ........ NP-complete.......
> ....... polynomial time.....,
> .......... reduction......
> ........ A. NP-complete.....
> . reduction...,......... NPcomplete
> ... B,. polynomial time...
> B. input reduce. A. input,.. A...
> .. B....
> 34 NP-Completeness
> 3
> 334
> 44 NP
> NPNP-Co
> -Co-Com
> mmp
> pplet
> letletene
> eneenes
> sss
> ss
> . reduce......,.........
> .. NP-complete.....
> .....:... NP-complete.....
> .......
> ... NP-complete problem. Cook...
> . satisfiability,........ NP...
> reduce. satisfiability......
> satisfiability.... NP-complete
> problem.
> 34 NP-Completeness
> 3
> 334
> 44.1
> .1.1 P
> PPo
> ooly
> lylyn
> nno
> oom
> mmia
> iaial
> ll tim
> timtime
> ee
> ....... polynomial time.....
> .. tractable,....:
> 1.
> .. polynomial time........
> .....,.. .(n100).......
> ,... polynomial time.......
> .....
> 2. Polynomial time...........
> model.... polynomial time...
> ,. RAM.... Turing machine..
> .....
> 3. Polynomial time
> ..........
> (closure),..............
> ...34 NP-Completeness
> 3
> 334
> 44.1
> .1.1 P
> PPo
> ooly
> lylyn
> nno
> oom
> mmia
> iaial
> ll tim
> timtime
> ee
> We define an abstract problem Q to be a
> binary relation on a set I of problem
> instances and a set S of problem
> solutions.
> ..: shortest path.........
> G... u... v,... G. u. v
> ............. k. path..
> ....,.... yes..... no,.
> . decision problem....
> Optimization problems.........
> .....,........ decision..
> ....
> 34 NP-Completeness
> 3
> 334
> 44.1
> .1.1 P
> PPo
> ooly
> lylyn
> nno
> oom
> mmia
> iaial
> ll tim
> timtime
> ee
> If a computer program is to solve an
> abstract problem, problem instances must
> be represented in a way that the program
> understands. An encoding of a set S of
> abstract objects is a mapping e from S to
> the set of binary strings.
> ...... polygons, graphs, functions,
> ordered pairs, programs..... binary
> strings.
> 34 NP-Completeness
> 3
> 334
> 44.1
> .1.1 P
> PPo
> ooly
> lylyn
> nno
> oom
> mmia
> iaial
> ll tim
> timtime
> ee
> A computer algorithm that "solves" come
> abstract decision problem actually takes an
> encoding of a problem instance as input. We
> of call a problem whose instance set is the set
> binary strings a concrete problem. An
> algorithm solves a concrete problem in time
> O(T(n)) if, when it is provided a problem
> instance i of length n = |i|, the algorithm can
> produce the solution in O(T(n)) time. A
> concrete
> problem is polynomial-time
> solvable, therefore, if there exists an
> algorithm to solve it in time O(nk) for some
> constant k.
> 34 NP-Completeness
> 3
> 334
> 44.1
> .1.1 P
> PPo
> ooly
> lylyn
> nno
> oom
> mmia
> iaial
> ll tim
> timtime
> ee
> The complexity class P is the set of concrete
> decision problems that are polynomial-time
> solvable.
> ............
> k,. O(k)....
> ...,.............
> ..........,.. unary(...
> ).
> ......,.. O(n)...;... binary
> .......,........ O(2n).
> 34 NP-Completeness
> 3
> 334
> 44.1
> .1.1 P
> PPo
> ooly
> lylyn
> nno
> oom
> mmia
> iaial
> ll tim
> timtime
> ee
> We say that a function f: {0,1}* .
> {0,1}* is
> polynomial-time computable if there exists a
> polynomial-time algorithm A that, given any
> that
> input x .
> {0,1}*, produces as output f(x).
> For some set I of problem instances, we say
> two encodings e1 and e2 are
> polynomially related if there exist two
> polynomial-time computable functions f12 and
> ef21 such that for any i .
> I, we have f12(e1(i)) > 2(i) and f21(e2(i)) = e1(i).
> 34 NP-Completeness
> 3
> 334
> 44.1
> .1.1 P
> PPo
> ooly
> lylyn
> nno
> oom
> mmia
> iaial
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> ee
> Lemma 34.1 Let Q be an abstract decision
> problem on an instance set I, and let e1 and
> e2 be polynomially related encodings on I.
> Then, e1(Q) .
> P if and only if e2(Q) .
> P.
> Proof:... e1(i)...... O(nk)...
> e2(i)... e1(i).... O(nc)....
> e2(i)
> .,.....:... e1(i)... e1(i)...
> ,.... O(nc+k),.. polynomial time..
> ...
> 34 NP-Completeness
> 24
> 34.1 Polynomial time34.1 Polynomial time
> ....... polynomial time......
> ....,........... P....
> .............. binary code,.
> ..............
> 34 NP-Completeness
> 3
> 334
> 44.1
> .1.1 P
> PPo
> ooly
> lylyn
> nno
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> iaial
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> A formal-language framework
> .. formal language....
> decision problem
> An alphabet S
> is a finite set of symbols. A
> language L over S
> is any set of strings made up
> of symbols from S. E.g, if S={0,1}, the set
> L={10,11,101,111,1011, 1101, 10001, ...} is
> the language of binary representations of
> prime numbers.
> empty string ., empty language Ø. The
> language of all strings over S
> is denoted by S*.
> 34 NP-Completeness
> 26
> 3
> 334
> 44.1
> .1.1 P
> PPo
> ooly
> lylyn
> nno
> oom
> mmia
> iaial
> ll tim
> timtime
> ee
> Languages. operations: union .,
> intersection n, complement L = S* – L,
> concatenation, closure.
> The concatenation of two languages L1 and L2
> isthelanguage L = {xx: x.
> Land x.
> L}.
> 1211 22
> The closure or Kleene star of a language L is
> the language L* = {.} .
> L .
> L2 .
> L3 .
> ..., where
> Lk is the language obtained by concatenating
> L to itself k times.
> 34 NP-Completeness
> 3
> 334
> 44.1
> .1.1 P
> PPo
> ooly
> lylyn
> nno
> oom
> mmia
> iaial
> ll tim
> timtime
> ee
> From the point of view of language theory,
> the set of instances for any decision problem
> Q is simply the set S*, where S
> = {0,1}. Since
> Q is entirely characterized by those problem
> instances that produce a 1(yes) answer, we
> can view Q as a language L over S
> = {0,1},
> whereL = {x .S* : Q(x) = 1}.
> ..: PATH = {<G,u,v,k
> : G = (V,E) is an
> undirected graph, u, v .
> V, k = 0 is an
> integer, and there exists a path from u to v in
> G consisting of at most k edges}.
> 34 NP-Completeness
> 3
> 334
> 44.1
> .1.1 P
> PPo
> ooly
> lylyn
> nno
> oom
> mmia
> iaial
> ll tim
> timtime
> ee
> We say that an algorithm A accepts a string x
> .
> {0,1}* if, given input x, the algorithm's
> output A(x) is 1. The language accepted by
> {0,1}*
> an algorithm A is the set of strings L = {x .
> : A(x) = 1}, that is, the set of strings
> that the algorithm accepts. An algorithm A
> rejects string x if A(x) = 0.
> 34 NP-Completeness
> 3
> 334
> 44.1
> .1.1 P
> PPo
> ooly
> lylyn
> nno
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> mmia
> iaial
> ll tim
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> ee
> A language L is decided by an algorithm A if
> every binary string in L is accepted by A and
> every binary string not in L is rejected by A.
> A language L is accepted in polynomial time
> by an algorithm A if it is accepted by A and if
> in addition there is a constant k such that for
> any length-n string x .
> L, algorithm A accepts
> x in time O(nk).
> A language L is decided in polynomial time
> an algorithm A if there is a constant kby
> such that for any length-n string x .
> {0,1}*,
> the algorithm correctly decides whether x L
> in time O(nk). 34 NP-Completeness
> 30
> We define
> languages,
> determined by
> 3
> 334
> 44.1
> .1.1 P
> PPo
> ooly
> lylyn
> nno
> oom
> mmia
> iaial
> ll tim
> timtime
> ee
> a complexity class as a set of
> membership in which is
> a complexity measure, such
> as running time, of an algorithm that
> 34 NP-Completeness
> determines whether a given string x belongs
> to language L.
> We can provide an alternative definition of
> the complexity class P: P = {L .{0,1}* :
> there exists an algorithm A that decides L in
> polynomial time}.
> 3
> 334
> 44.1
> .1.1 P
> PPo
> ooly
> lylyn
> nno
> oom
> mmia
> iaial
> ll tim
> timtime
> ee
> In fact, P is also the class of languages that
> can be accepted in polynomial time.
> Theorem 34.2 P = {L: L is accepted by a
> polynomial-time algorithm}.
> 34 NP-Completeness
> Ex
> ExExer
> ererc
> ccis
> isise
> ees
> ss 3
> 334
> 44.1
> .1.1
> for
> 34.1-4 Is the dynamic-programming algorithm
> for the 0-1 knapsack problem that is asked
> in Exercise 16.2-2 a polynomial-time
> 16.2-2 Give
> algorithm ? Explain your answer.
> a dynamic-programming solution
> to the 0-1 knapsack problem that runs in O(n
> W) time, where n is number of items and W is
> the maximum weight of items that the thief
> can put in his knapsack.
> 34 NP-Completeness
> 33
> Exercises34.1Exercises 34.1
> 34.1-6 Show that the class P, viewed as a set of
> languages, is closed under union,
> intersection, concatenation, complement,
> and Kleene star. That is, if L1, L2 .P, then L1 .
> L2 .P, etc.
> 34.1-6 Show that the class P, viewed as a set of
> languages, is closed under union,
> intersection, concatenation, complement,
> and Kleene star. That is, if L1, L2 .P, then L1 .
> L2 .P, etc.
> 34 NP-Completeness
> 34.234.234.234.2 Po
> PoPol
> lly
> yyno
> nonom
> mmi
> iia
> aal
> ll-
> --ti
> titim
> mme
> ee v
> vve
> eer
> rri
> iif
> ffi
> iic
> cca
> aati
> titio
> oon
> nn
> ... polynomial-time verification...
> ..... polynomial-time.......
> .........,..... PATH..
> .,..... G... u... v,..
> .. k,.... path,..... lineartime
> ....... path... u. v.
> path,......... k,....
> polynomial-time verification.
> ..... polynomial-time......,
> ..... polynomial-time verification.
> ............ polynomial-time
> ...,...... polynomial-time
> verification....34 NP-Completeness
> 35
> 34.2 Polynomial-time verification34.2 Polynomial-time verification
> Hamiltonian cycle
> .............. hamiltonian
> cycle,...............
> hamiltonian cycle of an undirectedAgraph G = (V, E) is a simple
> cycle
> that
> contains each vertex in V. A graph that
> contains a hamiltonian cycle is said to be
> hamiltonian; otherwise, it is
> nonhamiltonian.
> Hamiltonian-cycle problem: "Does a graph
> G have a hamiltonian cycle ?"
> 34 NP-Completeness
> 36
> 34.2 Polynomial-time verification34.2 Polynomial-time verification
> .......... hamiltonian-cycle?
> .....
> dodecahedron
> ...... bipartite graph
> 34 NP-Completeness
> 37
> 34.2 Polynomial-time verification34.2 Polynomial-time verification
> . hamiltonian cycle.......
> polynomial time...,.... O(2n).
> Verification algorithms:.....,...
> .......,............
> ....... G. cycle C,.....
> C
> ... hamiltonian cycle........
> cycle,........,........
> ,............ hamiltonian
> cycle,......:..........
> ............ edge?....
> simple cycle?.......... Yes,.
> ... hamiltonian cycle.
> 34 NP-Completeness
> 38
> 34.2 Polynomial-time verification34.2 Polynomial-time verification
> We define a verification algorithm as
> being a two-argument algorithm A, where
> one argument is an ordinary input string x
> certificate.
> and the other is a binary string y called a
> A two-argument algorithm A
> verifies an input string x if there exists a
> certificate y such that A(x, y) = 1. The
> language verified by a verification
> algorithm A is L = {x .
> {0,1}* : there
> exists y .
> {0,1}* such that A(x, y) = 1}.
> 34 NP-Completeness
> 34.234.234.234.2 Po
> PoPol
> lly
> yyno
> nonom
> mmi
> iia
> aal
> ll-
> --ti
> titim
> mme
> ee v
> vve
> eer
> rri
> iif
> ffi
> iic
> cca
> aati
> titio
> oon
> nn
> The complexity class NP
> ........ polynomial time....,..
> hamiltonian-cycle problem.. NP....
> .... NP. NP. nondeterministic
> model
> polynomial ...,.... non-deterministic
> ..,.. polynomial time......,
> .. NP....
> A language L .
> NP iff there exists a two-input
> polynomial-time algorithm A and constant c such
> that L = {x .
> {0,1}* : there exists a certificate y
> with |y| = O(|x|c) such that A(x, y) = 1}.
> The algorithm A verifies language L in polynomial
> time.
> 34 NP-Completeness
> 34.234.234.234.2 Po
> PoPol
> lly
> yyno
> nonom
> mmi
> iia
> aal
> ll-
> --ti
> titim
> mme
> ee v
> vve
> eer
> rri
> iif
> ffi
> iic
> cca
> aati
> titio
> oon
> nn
> ..... hamiltonian cycle..... NP.
> ..,..... L ... P... L....
> . NP,.... L .............
> ..,.. P .
> NP.
> ...... P> NP........,...
> ........ P . NP.
> 34 NP-Completeness
> 34.234.234.234.2 Po
> PoPol
> lly
> yyno
> nonom
> mmi
> iia
> aal
> ll-
> --ti
> titim
> mme
> ee v
> vve
> eer
> rri
> iif
> ffi
> iic
> cca
> aati
> titio
> oon
> nn
> .. P.... NP..,..........
> ...,.. NP...... complement.
> ......... L .
> NP..... L .
> NP.
> NPWe can define the complexity class co-NP as
> the set of languages L such that L .
> NP.
> ............ NP.... coNP
> ....
> P = NP = co-NP NP = co-NP
> P
> co-NP NPP = NP nco-NP co-NP NPNP nco-NP
> P
> .......
> 34 NP-Completeness
> E
> EEx
> xxe
> eer
> rrc
> cci
> iis
> sse
> ees
> ss 34
> 3434.2
> .2.2
> 42
> 34.2-1 Consider
> ISOMORPHISM
> isomorphic graphs}. Prove that GRAPH-
> ISOMORPHISM .NP by describing a
> polynomial-time algorithm to verify the
> language.
> 34.2-2 Prove that if G is an undirected bipartite
> graph with an odd number of vertices, then G
> is nonhamiltonian.
> the language GRAPH={<
> G,G
> :Gand Gare
> 121 2
> 34 NP-Completeness
> E
> EEx
> xxe
> eer
> rrc
> cci
> iis
> sse
> ees
> ss 34
> 3434.2
> .2.2
> 34.2-4 Prove that the class NP of languages is
> closed under union, intersection,
> 34.2-6
> concatenation, and Kleene star. Discuss the
> closure of NP under complement.
> A hamiltonian path in a graph is a
> simple path that visits every vertex exactly
> once. Show that the language HAM-PATH = {
> <G, u, v
> : there is a hamiltonian path from u
> to v in graph G} belongs to NP.
> 34 NP-Completeness
> E
> EEx
> xxe
> eer
> rrc
> cci
> iis
> sse
> ees
> ss 34
> 3434.2
> .2.2
> 34.2-7 Show that the hamiltonian-path
> problem can be solved in polynomial time on
> 34.2-11
> directed acyclic graphs. Give an efficient
> algorithm for the problem.
> Let G be a connected, undirected
> graph with at least 3 vertices, and let G3 be
> the graph obtained by connecting all pairs of
> vertices that are connected by a path in G of
> length
> most 3. Prove that G3 is
> hamiltonian. (Hint: Construct a spanning tree
> at
> for G, and use an inductive argument.)
> 34 NP-Completeness
> 34.3
> 34.334.3 N
> NNP
> PP-
> --c
> cco
> oom
> mmp
> ppl
> lle
> eet
> tte
> een
> nne
> ees
> sss
> ss a
> aand
> ndnd r
> rre
> eed
> dduc
> ucuci
> iib
> bbi
> iil
> lli
> iity
> tyty
> ..... P . NP........ NPcomplete
> ...,.... NP-complete
> ...... polynomial time.....,
> NP....... polynomial time..
> ....,... P = NP.........
> ,..... NP-complete.......
> ..... polynomial time.....
> .. Hamiltonian cycle..... NPcomplete
> ...,..... polynomial
> time.... Hamiltonian cycle...,
> .. NP....... polynomial time
> .....
> 34 NP-Completeness
> 46
> 34.3 NP-completeness and reducibility34.3 NP-completeness and
> reducibility
> ....,.. NP – P.......,..
> Hamiltonian cycle....... NP –
> P.
> .... NP-complete.... NP...
> ...
> ...... polynomial-time reducibility
> ...... NP-complete......
> 34 NP-Completeness
> 47
> 34.3 NP-completeness and reducibility34.3 NP-completeness and
> reducibility
> L1 is polynomial-time reducible to L2,
> written L1 =P L2,.. L2 . P. L1 . P.
> A
> ..: L2. sorting...,. L1....
> ....,. L1 =PL2.
> language L1 is polynomial-time
> reducible to a language L2, written L1 =P
> L2, if there exists a polynomial-time
> computable function f : {0,1}* .
> {0,1}*
> such that for all x .
> {0,1}*, x .
> L1 if and
> only if f(x) .
> L2.
> 34 NP-Completeness
> 48
> Lemma 34.3 If Lsuch that L
> 34.3
> 34.334.3 N
> NNP
> PP-
> --c
> cco
> oom
> mmp
> ppl
> lle
> eet
> tte
> een
> nne
> ees
> sss
> ss a
> aand
> ndnd r
> rre
> eed
> dduc
> ucuci
> iib
> bbi
> iil
> lli
> iity
> tyty
> 1, L2 .
> {0,1}* are languages
> =L,thenL.
> P implies L.
> 1P22 1
> P.
> 34 NP-Completeness
> A1
> A2F
> f(x)x
> yes, f(x) .L2
> no, f(x) .L2
> yes, x .L1
> no, x .L1
> 49
> 34.3 NP-completeness and reducibility34.3 NP-completeness and
> reducibility
> NP-completeness
> Polynomial-time reductions provide a
> formal means for showing that one
> We
> problem is at least as hard as another, to
> within a polynomial-time factor.
> define the set of NP-complete
> languages, which are the hardest
> problems in NP.
> A language L .
> {0,1}* is NP-complete if
> 1. L .
> NP, and
> 2. L' =P L for every L' .
> NP.
> 34 NP-Completeness
> 34.3
> 34.334.3 N
> NNP
> PP-
> --c
> cco
> oom
> mmp
> ppl
> lle
> eet
> tte
> een
> nne
> ees
> sss
> ss a
> aand
> ndnd r
> rre
> eed
> dduc
> ucuci
> iib
> bbi
> iil
> lli
> iity
> tyty
> . L . NP-complete...,. L . NP,
> ..... NP... L' =P L.
> .......... NP... L' =P L..
> ..., L .. NP-hard....
> .... NPC..
> NP-complete
> languages.
> . L1 =PL2.,.. L1. NP-complete.
> L2.... NP-hard.
> 34 NP-Completeness
> 51
> 34.3 NP-completeness and reducibility34.3 NP-completeness and
> reducibility
> Theorem 34.4 If any NP-complete problem
> is polynomial-time solvable, then P = NP.
> Equivalently, if any problem in NP is not
> complete
> solvable.
> polynomial-time solvable, then no NP-
> problem is polynomial-time
> ...... P .NP
> NP
> P
> NP-Complete
> 34 NP-Completeness
> 34.3
> 34.334.3 N
> NNP
> PP-
> --c
> cco
> oom
> mmp
> ppl
> lle
> eet
> tte
> een
> nne
> ees
> sss
> ss a
> aand
> ndnd r
> rre
> eed
> dduc
> ucuci
> iib
> bbi
> iil
> lli
> iity
> tyty
> Circuit satisfiability
> ... NP-complete .......,....
> .. reducibility............ NP-
> Thus,
> complete....
> we now focus on demonstrating the
> existence of an NP-complete problem: the
> circuit-satisfiability problem.
> A boolean combinational element is any
> circuit element that has a constant number
> of boolean inputs and outputs and that
> performs a well-defined function.
> ....... 0.
> 1, 0..
> FALSE. 1..
> TRUE.
> 34 NP-Completeness
> 53
> 34.3 NP-completeness and reducibility34.3 NP-completeness and
> reducibility
> A boolean combinational circuit consists
> of one or more boolean combinational
> elements interconnected by wires.
> not and or
> 34 NP-Completeness
> 54
> 34.3 NP-completeness and reducibility34.3 NP-completeness and
> reducibility
> x1
> x2
> x3
> 34 NP-Completeness
> 3
> 334.3 N
> 4.3 N4.3 NP-
> P-P-c
> cco
> oom
> mmp
> ppl
> lle
> eet
> tte
> eene
> nenes
> sss
> ss a
> aand
> ndndre
> rered
> dduc
> ucuci
> iib
> bbi
> iil
> lli
> iity
> tyty
> •
> A truth assignment for a boolean
> combinational circuit is a set of boolean input
> values. We say that a one-output boolean
> combinational circuit is satisfiable if it has a
> satisfying assignment: a truth assignment
> that causes the output of the circuit to be 1.
> 34 NP-Completeness
> 3
> 334.3 N
> 4.3 N4.3 NP-
> P-P-c
> cco
> oom
> mmp
> ppl
> lle
> eet
> tte
> eene
> nenes
> sss
> ss a
> aand
> ndndre
> rered
> dduc
> ucuci
> iib
> bbi
> iil
> lli
> iity
> tyty
> The circuit-satisfiability problem is, "Given a
> boolean combinational circuit composed of
> AND, OR, and NOT gates, is it satisfiable?"
> ...... size. circuit. combinational
> element......
> CIRCUIT-SAT = {<C : C is a satisfiable
> boolean combinational circuit}.
> ................ true. false
> ..,.....,.............
> . .(2n),...... n...
> 34 NP-Completeness
> 3
> 334.3 N
> 4.3 N4.3 NP-
> P-P-c
> cco
> oom
> mmp
> ppl
> lle
> eet
> tte
> eene
> nenes
> sss
> ss a
> aand
> ndndre
> rered
> dduc
> ucuci
> iib
> bbi
> iil
> lli
> iity
> tyty
> .... CIRCUIT-SAT. NP-Complete... .
> Lemma 34.5 The circuit-satisfiability problem
> ..
> belongs to the class NP.
> lemma............
> circuit,..... truth assignment,...
> . linear time.......
> assignment.
> .... truth assignment... circuitsatisfiability
> . NP....
> 34 NP-Completeness
> Circuit-satisfiability
> problem is NP-complete
> (cont.)
> • Lemma 34.6: (page 991)
> – CIRCUIT-SAT is NP-hard.
> • Proof: Suppose X is any problem in NP
> – construct a poly-time algorithm F maps
> every problem instance x in X to a circuit
> C=f(x) such that the answer to x is YES if
> and only if CÎCIRCUIT-SAT (is satisfiable).
> Circuit-satisfiability problem is NP-hard
> (cont.)
> –
> F runs in poly time.
> •
> Poly space:
> –
> Size of x is n.
> –
> Size of A is constant, independent of x.
> –
> Size of y is O(nk).
> –
> Amount of working storage is poly in n since A runs at
> most O(nk).
> –
> M has size poly in length of configuration, which is poly
> in O(nk), and hence is poly in n.
> –
> C consists of at most O(nk) copies of M, and hence is poly
> in n.
> –
> Thus, the C has poly space.
> •
> The construction of C takes at most O(nk) steps and
> each step takes poly time, so F takes poly time to
> construct C from x.
> Circuit-satisfiability problem is NP-hard
> (cont.)
> •
> Since XÎNP, there is a poly-time
> algorithm A which verifies X.
> •
> Suppose the input length is n and let T(n)
> denote the worst-case running time. Let k
> be the constant such that T(n)=O(nk) and
> the length of the certificate is O(nk).
> Circuit-satisfiability problem is NP-hard
> (cont.)
> •
> Idea is to represent the computation of A
> as a sequence of configurations, c0, c1,
> …,ci,ci+1,…,cT(n), each ci can be broken into
> –
> (program for A, program counter PC, auxiliary machine
> state, input x, certificate y, working storage) and
> –
> ci is mapped to ci+1 by the combinational circuit
> M implementing the computer hardware.
> –
> The output of A: 0 or 1– is written to some
> designated location in working storage. If the
> algorithm runs for at most T(n) steps, the
> output appears as one bit in cT(n).
> –
> Note: A(x,y)=1 or 0.
> 62
> Circuit-satisfiability problem is NP-hard (cont.)
> •
> The reduction algorithm F constructs a
> single combinational circuit C as follows:
> – Paste together all T(n) copies of the circuit
> M.
> –
> The output of the ith circuit, which produces
> ci, is directly fed into the input of the (i+1)st
> circuit.
> –
> All items in the initial configuration, except
> the bits corresponding to certificate y, are
> wired directly to their known values.
> –
> The bits corresponding to y are the inputs to
> C.
> – All the outputs to the circuit are ignored,
> except the one bit of cT(n) corresponding to
> the output of A.
> Circuit-satisfiability problem is NP-hard (cont.)
> • Two properties remain to be proven:
> – F correctly constructs the reduction, i.e., C
> is satisfiable if and only if there exists a
> certificate y, such that A(x,y)=1.
> ÜSuppose there is a certificate y, such
> that A(x,y)=1. Then if we apply the bits
> of y to the inputs of C, the output of C is
> the bit of A(x,y), that is C(y)= A(x,y) =1,
> so C is satisfiable.
> Þ
> Suppose C is satisfiable, then there is a
> y such that C(y)=1. So, A(x,y)=1.
> – F runs in poly time.
> 34.3 NP-completeness and
> reducibility
> Theorem 34.7 The circuit-satisfiability
> problem is NP-complete.
> Exercises 34.3
> 34.3-2 Show that the =P relation is a
> transitive relation on languages. That is,
> showthatifL=LandL=L,thenL> 1P2 2P31P
> L3.
> 34.4 NP-completeness proofs
> ....... NP-complete.......
> circuit-satisfiability problem.......
> NP.... reduce. circuit-satisfiability
> ...,............
> Lemma 34.8 If L is a language such that L' =P
> L for some L' .
> NPC, then L is NP-hard.
> Moreover, if L .
> NP, then L .
> NPC.
> .. =P. transitive...,......
> NP... L=PL' =P L,.. L. NP-hard
> ....
> 34.4 NP-completeness proofs
> .
> Lemma 34.8........... NPcomplete
> ......
> 1. Prove L .
> NP.
> 2. Select a know NP-complete language L'.
> 3. Describe an algorithm that computes a
> function f mapping every instance x .
> {0,1}* of L' to an instance f(x) of L.
> 4.Prove that the function f satisfies x .
> L' if
> and only if f(x) .
> L for all x .
> {0,1}*.
> 5.Prove that the algorithm computing f
> runs in polynomial time.
> 34.4 NP-completeness proofs
> .... 2~5......... NP-hard.
> ............ NP... reduce
> ........... CIRCUIT-SAT. NPcomplete
> .......
> ........... NP-complete..,
> ..............,.......
> .. NP-complete........
> 34.4 NP-completeness proofs
> Formula satisfiability
> ....... formula satisfiability
> problem,...... NP-complete.
> An instance of SAT is a boolean formula .
> composed of
> 1. n boolean variables: x1, x2, ..., xn;
> 2. m boolean connectives: any boolean
> function with one or two inputs and one
> output, such as ., ., ¬, .(imply), .(iff)
> 3.parentheses.
> 34.4 NP-completeness proofs
> Boolean formula ... encode....
> O(m+n). A truth assignment for a boolean
> formula . is a set of values for the variables
> of ., and a satisfying assignment is a truth
> assignment that causes it to evaluate to 1.
> A formula with a satisfying assignment is a
> satisfiable formula.
> SAT = {<. : . is a satisfiable boolean
> formula}.
> 34.4 NP-completeness proofs
> ..: .= ((x1 .
> x2) .
> ¬ ((¬x1 .
> x3) .
> x4)) .
> ¬ x2,.. satisfying assignment. x1 = 0,
> x2 = 0, x3 = 1, x4 = 1,.
> .= ((0 .
> 0) .
> ¬ ((¬0 .
> 1) .
> 1)) .
> ¬0
> = (1 .
> ¬ (1 .
> 1)) .
> 1 = (1 .
> 0) .
> 1 =1
> SAT...... n...,.... 2n..
> .....,............
> formula.. true,........... .
> (2n).
> 34.4 NP-completeness proofs
> Theorem
> 34.9 Satisfiability of boolean
> formulas is NP-complete.
> . Lemma 34.8 ......,... SAT.
> NP...,...... truth
> assignment,..........
> formula
> .... 1,......... polynomial
> time,.. SAT.. NP.
> ...... CIRCUIT-SAT =P SAT,....
> .. CIRCUIT-SAT............
> ,....... size...........
> 34.4 NP-completeness proofs
> . =x.
> (x.
> ¬x) .
> (x.
> (x.
> x)) .
> (x.
> 104 35126
> ¬x) .
> (x.
> (x.
> x.
> x)) .
> (x.
> (x.
> x)) .
> 47124856
> (x.
> (x.
> x)) .
> (x.
> (x.
> x.
> x)).
> 9 6710 789
> x1
> x2
> x3 x4
> x5
> x6
> x7
> x8
> x9
> x10
> 34.4 NP-completeness proofs
> ...................
> polynomial time........ formula
> .. satisfiable..,... CIRCUIT-SAT
> .. satisfiable,.....
> .. SAT..... NP-complete.
> 34.4 NP-completeness proofs
> 3-CNF satisfiability
> .......: A literal in a boolean
> formula is an occurrence of a variable or its
> negation. A boolean formula is in
> conjunctive normal form, or CNF, if it is
> expressed as an AND of clauses, each of
> which is the OR of one or more literals. A
> boolean formula is in 3-conjunctive normal
> form, or 3-CNF, if each clause has exactly
> three distinct literals.
> 77
> 34.4 NP-completeness proofs
> ..: (x1 .
> ¬x1 .
> ¬x2) .
> (x3 .
> x2 .
> x4) .
> (¬x1
> .
> ¬x3 .
> ¬x4).... 3-CNF.
> In 3-CNF-SAT, we are asked whether a
> given boolean formula . in 3-CNF is
> satisfiable.
> Theorem 34.10 Satisfiability of boolean
> formulas in 3-conjunctive normal form is
> NP-complete.
> 34.4 NP-completeness proofs
> Theorem
> 34.10 Satisfiability of boolean
> formulas in 3-conjunctive normal form is
> NP-complete.
> 3-CNF-SAT .
> NP.... SAT .
> NP....
> ........ SAT =P 3-CNF-SAT.
> .. SAT... reduce. 3-CNF-SAT..
> .,.... SAT. input. polynomial
> time.... 3-CNF-SAT. input,...
> ......... 3-CNF,........
> ...
> 34.4 NP-completeness proofs
> .. SAT... reduce. 3-CNF-SAT..
> .,.... SAT. input. polynomial
> time.... 3-CNF-SAT. input,...
> ......... 3-CNF,........
> ...
> ...... SAT input formula. binary
> tree.
> 34.4 NP-completeness proofs
> ........, . > ((x.
> x) .
> ¬((¬x.
> x)
> 12 13
> .
> x4)) .
> ¬x2,...
> binary tree..:
> ..... ... .' = y1
> .
> (y.
> (y.
> ¬x)) .
> (y.
> 1222
> (y.
> y)) .
> (y.
> (x.
> 3431
> x)) .
> (y.
> ¬y) .
> (y.
> 2455
> (y.
> x)) .
> (y.
> (¬x.
> 646 1
> x3))
> y1
> y2
> .
> y.
> .
> ¬
> .
> .
> y¬x2
> 3
> 4
> y5
> xx
> 1 2y6
> x
> 4
> ¬xx
> 13
> 34.4 NP-completeness proofs
> ..... .'.......
> and,....
> ..... and..............
> ...... or.. y.
> (y.
> ¬x)...
> 12
> y1 y2 x2
> . y1 .
> (y2 .
> ¬x2) = 11
> 1110
> ¬((y.
> y.
> x) .
> (y
> 1221 101
> .
> ¬y2 .
> x2) .
> (y1 .
> 100
> 011
> ¬y.
> ¬x) .
> (¬y.
> 221 010
> y.
> ¬x))
> 22001
> 000
> = (¬y.
> ¬y.
> ¬x) .
> (¬y.
> y
> 12212
> (¬y.
> y.
> x) .
> (y.
> ¬y.
> x)
> 1221 22
> 2
> .y1 ..y2.-x2 ..
> 0
> 1
> 0
> 0
> 1
> 0
> 1
> 1
> .
> ¬x2) .
> 82
> 34.4 NP-completeness proofs
> ... .'............,...
> CNF .... .".............
> .................... (xi
> .
> xj)..,... (xi .
> xj .
> p) .
> (xi .
> xj .
> ¬p)............ (x),... (x
> .
> p .
> q) .
> (x .
> ¬p .
> q) .
> (x .
> p .
> ¬q) .
> (x .
> ¬p .
> ¬q).
> ............. 3-CNF formula
> .,.... satisfiable if and only if..
> ...... satisfiable... 3-CNF-SAT.
> NP-complete.
> Exercises 34-4
> 34.4-6 Suppose that someone gives you a
> polynomial-time algorithm to decide
> formula satisfiability. Describe how to use
> this algorithm to find satisfying
> assignments in polynomial time.
> 34.4-7 Let 2-CNF-SAT be the set of satisfiable
> boolean formulas in CNF with exactly 2
> literals per clause. Show that 2-CNF-SAT .
> P. Make your algorithm as efficient as
> possible.(Hint: Observe that x .
> y is
> equivalent to ¬x .
> y. Reduce 2-CNF-SAT to
> a problem on a directed graph that is
> efficiently solvable.)
> 3
> 334.5 N
> 4.5 N4.5 NP-
> P-P-c
> cco
> oom
> mmp
> ppl
> lle
> eet
> tte
> ee p
> ppr
> rro
> oob
> bbl
> lle
> eem
> mms
> ss
> NP-complete problems arise in diverse
> domains: boolean logic, graphs,
> arithmetic, network design, sets and
> storage and retrieval,
> partitions,
> sequencing and scheduling, mathematical
> programming, algebra and number
> theory, games and puzzles, automata and
> language theory, program optimization,
> biology, chemistry, physics, and more.
> ............ NP-complete.
> 34 NP-Completeness
> 3
> 334.5 N
> 4.5 N4.5 NP-
> P-P-c
> cco
> oom
> mmp
> ppl
> lle
> eet
> tte
> ee p
> ppr
> rro
> oob
> bbl
> lle
> eem
> mms
> ss
> CIRCUIT-SAT
> SAT
> 3-CNF-SAT
> SUBSET-SUMCLIQUE
> VERTEX-COVER
> HAM-CYCLE
> TSP
> 34 NP-Completeness
> 86
> 34.5 NP-complete problems34.5 NP-complete problems
> 34.5-1 The clique problem
> A clique in an undirected graph G = (V, E)
> is a subset V' .
> V of vertices, each pair of
> other
> which is connected by an edge in E. In
> words, a clique is a complete
> subgraph of G. The size of a clique is the
> number of vertices it contains. The clique
> problem is the optimization problem of
> finding a clique of maximum size in a
> graph.
> 34 NP-Completeness
> 3
> 334.5 N
> 4.5 N4.5 NP-
> P-P-c
> cco
> oom
> mmp
> ppl
> lle
> eet
> tte
> ee p
> ppr
> rro
> oob
> bbl
> lle
> eem
> mms
> ss
> 34.5-1 The clique problem
> As a decision problem, we ask simply
> whether a clique of a given size k exists in
> graph. The formal definition isthe
> CLIQUE = {<G,k : G is a graph with a
> clique of size k}.
> ............... G. k .
> ..........,.... complete
> graph,.......... .(k2* C(|V|,
> k)),.. C(|V|, k). |V|........
> k .....,.........
> polynomial time.34 NP-Completeness
> 3
> 334.5 N
> 4.5 N4.5 NP-
> P-P-c
> cco
> oom
> mmp
> ppl
> lle
> eet
> tte
> ee p
> ppr
> rro
> oob
> bbl
> lle
> eem
> mms
> ss
> Theorem
> 34.11 The clique problem is NP-
> complete.
> ..... CLIQUE .
> NP,..... V',
> k.
> ..... polynomial time..
> V'...
> ..........,. |V'|......
> ........ 3-CNF-SAT =P CLIQUE,
> ...... 3-CNF-SAT. input...
> CLIQUE. input,.........,.
> .......... k ... clique.,
> 3-CNF-SAT.. satisfiable,......
> ......:34 NP-Completeness
> 3
> 334.5 N
> 4.5 N4.5 NP-
> P-P-c
> cco
> oom
> mmp
> ppl
> lle
> eet
> tte
> ee p
> ppr
> rro
> oob
> bbl
> lle
> eem
> mms
> ss
> . 3-CNF-SAT.... .=(x1 .
> ¬x2 .
> ¬x3)
> .
> (¬x1 .
> x2 .
> x3) .
> (x1 .
> x2 .
> x3),....:
> (¬x.
> (x.
> ¬x.
> ¬x)
> 123
> 2 .
> x3)
> 1
> 34 NP-Completeness
> ¬x1
> x2
> x1 2¬x3
> x1
> x2
> x2 .x3) (x1 .xx3
> ¬x2
> x3
> 3
> 334.5 N
> 4.5 N4.5 NP-
> P-P-c
> cco
> oom
> mmp
> ppl
> lle
> eet
> tte
> ee p
> ppr
> rro
> oob
> bbl
> lle
> eem
> mms
> ss
> ......... k ... clique ...
> 3-CNF-SAT... k . clause.....
> .. satisfiable... CLIQUE....
> NP-complete.
> 34 NP-Completeness
> 91
> 34.5 NP-complete problems34.5 NP-complete problems
> 34.5.2 The vertex-cover problem
> A vertex cover of an undirected graph G
> = (V, E) is a subset V' .
> V such that if (u,
> and
> v) .
> E, then u .
> V' or v .
> V'(or both). That
> is, each vertex "cover" its incident edges,
> a vertex cover for G is a set of
> vertices that covers all the edges in E.
> The size of a vertex cover is the number
> of vertices in it.
> The vertex-cover problem is to find a
> vertex cover of minimum size in a given
> graph.
> 34 NP-Completeness
> 3
> 334.5 N
> 4.5 N4.5 NP-
> P-P-c
> cco
> oom
> mmp
> ppl
> lle
> eet
> tte
> ee p
> ppr
> rro
> oob
> bbl
> lle
> eem
> mms
> ss
> Restating this optimization problem as a
> decision problem, we wish to determine
> whether a graph has a vertex cover of a
> given size k. As a language, we define
> VERTEX-COVER = {<G,k : graph G has a
> vertex cover of size k}.
> Theorem 34.12 The vertex-cover problem is
> NP-complete.
> ......... NP...,... G> (V,E)... k..... V' .
> V,...
> |V'|
> =k....,...
> E..... (u,v),
> .. u .
> V'. v .
> V'........
> 34 NP-Completeness
> 3
> 334.5 N
> 4.5 N4.5 NP-
> P-P-c
> cco
> oom
> mmp
> ppl
> lle
> eet
> tte
> ee p
> ppr
> rro
> oob
> bbl
> lle
> eem
> mms
> ss
> ..... vertex-cover .... NPhard,
> ........ CLIQUE =P
> VERTEX-COVER.
> ............ complement,
> Given an undirected graph G = (V, E), we
> define the complement of G as G = (V, E),
> where E = {(u, v) : u, v .
> V, u . v, and (u,
> v) .
> E}.... G
> ....... G.,.
> .. G..... G...
> 34 NP-Completeness
> 3
> 334.5 N
> 4.5 N4.5 NP-
> P-P-c
> cco
> oom
> mmp
> ppl
> lle
> eet
> tte
> ee p
> ppr
> rro
> oob
> bbl
> lle
> eem
> mms
> ss
> ......... complement.
> u v
> z w
> y x
> u v
> z w
> y x
> CLIQUE =P VERTEX-COVER.....:.
> ... clique problem. instance <G,k,
> .... G. complement G,...
> vertex-cover .. <G,|V|-k...,...
> ... clique......
> 34 NP-Completeness
> 3
> 334.5 N
> 4.5 N4.5 NP-
> P-P-c
> cco
> oom
> mmp
> ppl
> lle
> eet
> tte
> ee p
> ppr
> rro
> oob
> bbl
> lle
> eem
> mms
> ss
> The graph G has a clique of size k if and
> only if the graph G has a vertex cover of
> size |V| -k.
> V'. G..... clique,. G. V'.
> .............,... G..
> ...... V – V'......,.. V–
> V'. vertex cover.
> . G. V – V'. vertex cover,. V'..
> ........,... G.. V'...
> . clique.
> 34 NP-Completeness
> 3
> 334.5 N
> 4.5 N4.5 NP-
> P-P-c
> cco
> oom
> mmp
> ppl
> lle
> eet
> tte
> ee p
> ppr
> rro
> oob
> bbl
> lle
> eem
> mms
> ss
> .. VERTEX-COVER.... NPcomplete,
> ............
> polynomial time.......
> approximation algorithm.,.....
> polynomial time..........,..
> ...........
> ....... NP-complete...
> approximation algorithm.
> 34 NP-Completeness
> 3
> 334.5 N
> 4.5 N4.5 NP-
> P-P-c
> cco
> oom
> mmp
> ppl
> lle
> eet
> tte
> ee p
> ppr
> rro
> oob
> bbl
> lle
> eem
> mms
> ss
> 34.5.3 The hamiltonian-cycle problem
> Theorem
> 34.13 The hamiltonian cycle
> problem is NP-complete.
> ............... NP...,
> ..... G = (V, E),....
> hamiltonian cycle C,..
> C......
> ....,...
> C...........
> E.....
> .... VERTEX-COVER =HAM-CYCLE
> P
> ... hamiltonian cycle.... NPhard
> .
> 34 NP-Completeness
> [u,v,1]
> [u,v,2]
> [u,v,3]
> [u,v,4]
> [u,v,5]
> [u,v,6]
> Wu v
> (a)
> [v,u,1]
> [v,u,6]
> [u,v,1]
> [u,v,6]
> Wuv
> 3
> 334.5 N
> 4.5 N4.5 NP-
> P-P-c
> cco
> oom
> mmp
> ppl
> lle
> eet
> tte
> ee p
> ppr
> rro
> oob
> bbl
> lle
> eem
> mms
> ss
> [v,u,1]
> [v,u,2]
> [v,u,3]
> [v,u,4]
> [v,u,5]
> [v,u,6]
> [v,u,1]
> [v,u,6]
> [u,v,1]
> [u,v,6]
> Wuv
> [v,u,1]
> [v,u,6]
> [u,v,1]
> [u,v,6]
> Wuv
> (b) (c) (d)
> Given an undirected graph G = (V, E) and
> an integer k, we construct an undirected
> graph G' = (V', E') that has a hamiltonian
> cycle if and only if G has a vertex cover of
> size k. G'....:. G.... (u,v)..
> ... 14 ..........: [u,v,1] ~
> [u,v,6],. [v,u,1]~[v,u,6].
> 34 NP-Completeness
> 99
> 34 NP-Completeness
> (a)
> (b)
> w
> z
> x
> y
> [x,w,1]
> [x,w,6]
> [w,x,1]
> [w,x,6]
> Wuv
> [y,x,1]
> [y,x,6]
> [x,y,1]
> [x,y,6]
> Wuv
> [y,w,1]
> [y,w,6]
> [w,y,1]
> [w,y,6]
> Wuv
> [z,w,1]
> [z,w,6]
> [w,z,1]
> [w,z,6]
> Wuv
> 3
> 334.5 N
> 4.5 N4.5 NP-
> P-P-c
> cco
> oom
> mmp
> ppl
> lle
> eet
> tte
> ee p
> ppr
> rro
> oob
> bbl
> lle
> eem
> mms
> ss
> ..............,....
> k .
> selector verteices s1, s2, ..., sk.
> ... u.. degree(u)...,.....
> ......... u(1), u(2), ..., u(degree(u)),.
> ..
> {([u,u(i),6], [u,u(i+1),1]) : 1 = i > degree(u) – 1}...
> .... vertex cover .. hamiltonian
> cycle.,.... vertex cover .....
> ......... si....
> 34 NP-Completeness
> 3
> 334.5 N
> 4.5 N4.5 NP-
> P-P-c
> cco
> oom
> mmp
> ppl
> lle
> eet
> tte
> ee p
> ppr
> rro
> oob
> bbl
> lle
> eem
> mms
> ss
> ........, {(sj, [u,u(1),1]) : u .
> V
> , [u,u(degree(u)),6]) : u .
> V
> and1 =j= k} .
> {(sj
> and 1 = j = k}.
> .........,... |V'| = 12|E| + k
> = 12|E| + |V|,... |E'| = 14|E| + (2|E| |
> V|) + (2k|V|) = 16|E| + (2k-1)|V|
> ......... polynomial time.
> 34 NP-Completeness
> 34.5.4 The traveling-salesman problem
> Traveling-salesman problem(TSP): A
> salesman must visit n cities. We can say
> 34.5 NP-complete problems
> 34.5 NP-complete problems
> visiting
> to
> that the salesman wishes to make a tour,
> each city exactly once and
> finishing at the city he starts from. There
> is an integer cost c(i, j) to travel from city
> i city j, and the salesman wishes to
> make the tour whose total cost is
> minimum, where the total cost is the sum
> of the individual costs along the edges of
> the tour.
> 34 NP-Completeness
> 3
> 334.5 N
> 4.5 N4.5 NP-
> P-P-c
> cco
> oom
> mmp
> ppl
> lle
> eet
> tte
> ee p
> ppr
> rro
> oob
> bbl
> lle
> eem
> mms
> ss
> TSP = {<G,c,k : G = (V, E) is a complete
> graph, c is a function from V × V .
> Z, k .
> Z, and G has a traveling-salesman tour
> Theorem
> with cost at most k}.
> 34.14 The traveling-salesman
> problem is NP-complete.
> .... TSP..... NP,.....
> TSP...,..... polynomial time
> ..................,..
> ......... cost...... k.
> 34 NP-Completeness
> 3
> 334.5 N
> 4.5 N4.5 NP-
> P-P-c
> cco
> oom
> mmp
> ppl
> lle
> eet
> tte
> ee p
> ppr
> rro
> oob
> bbl
> lle
> eem
> mms
> ss
> ..... NP-complete... reduce.
> TSP...,.... Hamiltonian cycle
> =P TSP.
> ..... G = (V, E),.....
> complete.. G' = (V, E'), E' = {(i,j) : i,
> j .
> V and i . j},... (i,j) .
> E. c(i,j) > 0,. (i,j) .
> E. c(i,j) = 1.
> The graph G has a hamiltonian cycle if
> and only if graph G' has a tour of cost at
> most 0.
> 34 NP-Completeness
> 105
> 3
> 334.5 N
> 4.5 N4.5 NP-
> P-P-c
> cco
> oom
> mmp
> ppl
> lle
> eet
> tte
> ee p
> ppr
> rro
> oob
> bbl
> lle
> eem
> mms
> ss
> 34.5.5 The subset-sum problem
> ....
> subset-sum problem, We are
> given a finite set S .
> N and a target t .
> N.
> ..
> We ask whether there is a subset S' .
> S
> whose elements sum to t.
> S = {1, 2, 7, 14, 49, 98, 343, 686,
> 2409, 2793, 16808, 17206, 117705,
> 117993}. t = 138457,. S' = {1, 2, 7,
> 98, 343, 686, 2409, 17206, 117705}..
> ...
> SUBSET-SUM = {<S,t : there exists a
> subset S' .
> S such that t = S
> s}.
> s.S'34 NP-Completeness
> 106
> 3
> 334.5 N
> 4.5 N4.5 NP-
> P-P-c
> cco
> oom
> mmp
> ppl
> lle
> eet
> tte
> ee p
> ppr
> rro
> oob
> bbl
> lle
> eem
> mms
> ss
> Theorem 34.15 The subset-sum problem is
> NP-complete.
> ........ NP...,..... S' ,
> ..
> S'............ t....
> .. polynomial time......,...
> ..... NP.
> .... 3-CNF-SAT =P SUBSET-SUM..
> ...... NP-hard.
> 34 NP-Completeness
> 34.5 NP-complete problems34.5 NP-complete problems
> 107
> Given a 3-CNF formula . over variables
> x, x, ..., x with clauses C, C, ..., C,
> 12n 12k
> each containing exactly three distinct
> the reduction algorithm
> an instance <S,t of the
> problem such that . is
> literals,
> constructs
> subset-sum
> satisfiable if and only if there is a subset
> of S whose sum is exactly t.
> 34 NP-Completeness
> 108
> 3
> 334.5 N
> 4.5 N4.5 NP-
> P-P-c
> cco
> oom
> mmp
> ppl
> lle
> eet
> tte
> ee p
> ppr
> rro
> oob
> bbl
> lle
> eem
> mms
> ss
> ................,. .> C1 .
> C2 .
> C3 .
> C4,.. C1 = (x1 .
> ¬x2 .
> ¬x),C=(¬x.
> ¬x.
> ¬x) C=(¬x.
> 32 1233 1
> .
> x),C=(x.
> x.
> x)
> 234 123
> ¬x
> ...... satisfying assignment. x1
> = 0, x2= 0, x3 = 1.
> ..... subset-sum..:
> 34 NP-Completeness
> 109
> 34 NP-Completeness
> x1 x2 x3 C1 C2 C3 C4
> v1 = 1 0 0 1 0 0 1
> v1’ = 1 0 0 0 1 1 0
> v2 = 0 1 0 0 0 0 1
> v2’ = 0 1 0 1 1 1 0
> v3 = 0 0 1 0 0 1 1
> v3’ = 0 0 1 1 1 0 0
> s1 = 0 0 0 1 0 0 0
> s1’ = 0 0 0 2 0 0 0
> s2 = 0 0 0 0 1 0 0
> s2’ = 0 0 0 0 2 0 0
> s3 = 0 0 0 0 0 1 0
> s3’ = 0 0 0 0 0 2 0
> s4 = 0 0 0 0 0 0 1
> s4’ = 0 0 0 0 0 0 2
> t = 1 1 1 4 4 4 4
> 110
> 3
> 334.5 N
> 4.5 N4.5 NP-
> P-P-c
> cco
> oom
> mmp
> ppl
> lle
> eet
> tte
> ee p
> ppr
> rro
> oob
> bbl
> lle
> eem
> mms
> ss
> .............,.......
> .. vi. vi',.. xi. Cj...., vi
> ....
> Cj..... 1.... 0, ¬xi
> Cj...., vi'....
> Cj......
> 1.... 0.
> .......... si. si',.. si. Ci
> .... 1.... 0,. si'. Ci....
> 2.... 0.
> t..... n. digits. 1.. k .
> digits. 4.
> .........34 NP-Completeness
> 111
> 3
> 334.5 N
> 4.5 N4.5 NP-
> P-P-c
> cco
> oom
> mmp
> ppl
> lle
> eet
> tte
> ee p
> ppr
> rro
> oob
> bbl
> lle
> eem
> mms
> ss
> ............ subset... t
> ..,....... satisfiable.
> .. subset-sum.... NP-complete.
> 34 NP-Completeness
> E
> EEx
> xxe
> eerc
> rcrci
> iis
> sse
> ees
> ss 34.5
> 34.534.5
> 112
> 34.5-1 The subgraph-isomorphism problem
> takes two graphs G1 and G2 and asks
> whether G1 is isomorphic to a subgraph of
> G2. Show that the subgraph-isomorphism
> problem is NP-complete.
> 34.5-5 The set-partition problem takes as
> input a set S of numbers. The question is
> whether the numbers can be partitioned
> 34 NP-Completeness
> into two sets A and S – A such that Sx.Ax > Sx (.S-A)x. Show that the set-partition
> problem is NP-complete.
> E
> EEx
> xxe
> eerc
> rcrci
> iis
> sse
> ees
> ss 34.5
> 34.534.5
> 34.5-6 Show that the hamiltonian-path
> problem is NP-complete.
> 34.5-7 The longest-simple-cycle problem is
> problem of determining a simple the
> cycle(no repeated vertices) of maximum
> length in a graph. Show that this problem
> is NP-complete.
> 34 NP-Completeness
> .NP
> This proof is correct:
> When a snail scrunches does he become a 'snali'?
> No. [1] http://groups.google.com/group/sci.math/msg/c61385970bc34320?
> [2] http://buildasearch.com/meami?start=351&paginas=40&e=%20prove%20%3E%20Cook%27s%20theorem%20for%20a%20unary%20alphabet" rel="nofollow" target="_blank">http://buildasearch.com/meami?start=351&paginas=40&e=%20prove%20%3E%20Cook%27s%20theorem%20for%20a%20unary%20alphabet
> 34 NP-Completeness 34 NP-Completeness 1971 年 Cook 提出第一個 NP-complete 問
> 題. 之後,至今三十幾年、沒有人能夠找到任何一個. NP ... Prove that GRAPH- ISOMORPHISM. NP by
> describing a. polynomial-time algorithm to verify the ...
> csie.dyu.edu.tw/~spring/.../96_1/Algorithms/chapter34.pdf
> Research conducted and based on: http://buildasearch.com/meami http://meami.org
> Posted by M. Michael Musatov 10.22.2009
> 71.130.128.163 (talk) 02:23, 23 October 2009 (UTC)M. Michael Musatov