Classification of Scheduling Problems

NP-completeness

NP-complete problems

Homework

Vertex Cover

•Instance. A graph G and an integer k.

•Question. Is there a vertex cover of cardinality k?

Clique

•Instance. A graph G and an integer k.

•Question. Has G a clique of cardinality k?

Prove NP-completeness of Vertex cover and Clique.

Vertex cover, stable set, clique,…(2)

Proposition 2.2.

Let G be a graph and X⊆V(G). Then thefollowing three statements are equivalent:

(a) X is a vertex cover in G,

(b) V(G)\X is a stable set in G,

NP-completeness of Vertex Cover

•Transform Stable Set to Vertex Cover.

•Let G = (V, E) and k constitute any instance ofStable Set.

•The corresponding instance of Vertex Cover isprovided simply by the graph G′ = G and theinteger k′ = n − k, where |V |= n.

Hamiltonian Circuit

•Instance: A graph G.

•Question: Is there a Hamiltonian circuit?

Hamiltonian Circuit

Theorem 11.1 (Karp 1972)

Hamiltonian Circuit is NP-complete.

Sketch of Proof

•«Vertex Cover» → «Hamiltonian Circuit»

•«Vertex Cover»: G = (V,E), an integer k ≥ 0.

•We must construct a graph G′ = (V′,E′) such that G′has a Hamiltonian circuit if and only if G has a vertexcover H of size k or less.

•Let |E| = m.

Construction of G′

•|V′| = 12m+k

•For each edge (vi, vj) ∊ E, G′ contains a “cover-testing”component that will be used to ensure that at least oneendpoint of that edges is among the selected k vertices.

•The component has 12 vertices uij1, uij2, uij3, uij4, uij5,uij6, uji1, uji2, uji3, uji4, uji5, uji6 and 14 edges.

•Additionally G′ has k “selector” vertices a1, a2,…, ak,which will be used to select k vertices from the vertexset V for G.

Component (vi, vj)

uij1

uij2

uij3

uij4

uij5

uij6

uji1

uji2

uji3

uji4

uji5

uji6

In the completed construction, the onlyvertices from this component that willbe involved in any additional edges are

uij1, uji1, uij6, uji6.

This implies that any Hamiltoniancircuit of G′ has to pass through thesevertices by the exactly one of the threefollowing ways.

Component (vi, vj)

uij1

uij2

uij3

uij4

uij5

uij6

uji1

uji2

uji3

uji4

uji5

uji6

vi ∊ H,

vj ∉ H

vi ∉ H,

vj ∊ H

vi ∊ H,

vj ∊ H

Component vi

•Additional edges in our construction will serve to joinpairs of cover-testing components or to join a covertesting component to a selector vertex.

•For each vertex vi let r denote a degree of vi in G.Arbitrarily order the edges incident on vi as(vi, vj1), (vi, vj2),…, (vi, vjr).

•All the cover testing components corresponding tothese edges are joined together by the followingconnecting edges:

Vertex Component

uij11

uij16

uij21

uij26

uijr1

uijr6

uij31

uij36

Construction of G′

•|V′| = 12m+k

•For each edge (vi, vj) ∊ E, G′ contains a “cover-testing”component that will be used to ensure that at least oneendpoint of that edges is among the selected k vertices.

•The component has 12 vertices uij1, uij2, uij3, uij4, uij5,uij6, uji1, uji2, uji3, uji4, uji5, uji6 and 14 edges.

•Additionally G′ has k “selector” vertices a1, a2,…, ak,which will be used to select k vertices from the vertexset V for G.

Selector vertices

•Each selector vertex al join with the first and lastvertices of every component vi.

•It is easy to see that G′ can be constructed from G inpolynomial time.

Component of G′

uij11

uij16

uij21

uij26

uijr1

uijr6

uij31

uij36

Proof (1)

•We claim that G′ has a Hamiltonian circuit if andonly if G has a vertex cover of size k or less.

•Suppose there is a Hamiltonian circuit for G′.Consider any portion of this circuit that begins at aselector vertex, ends at another selector vertex andthat encounters no such vertex internally.

Component of G′

uij11

uij16

uij21

uij26

uijr1

uijr6

uij31

uij36

Proof (2)

•We claim that G′ has a Hamiltonian circuit if andonly if G has a vertex cover of size k or less.

•Suppose there is a Hamiltonian circuit for G′.Consider any portion of this circuit that begins at aselector vertex, ends at another selector vertex andthat encounters no such vertex internally.

•This portion of the circuit must pass through a set ofcover-testing components corresponding to exactlythose edge from E that are incident on some oneparticular vertex vi ∊ V.

Proof(3)

•Each of the cover-testing components istraversed in one of the three modes, and novertex from any other cover-testing componentis encountered.

Proof(4)

•So the k selector vertices divide the Hamiltonian circuit into kpaths, each path corresponding to a distinct vertex vi ∊ V.

•Since the Hamiltonian circuit must include all vertices fromevery one of the cover-testing components, and since verticesfrom the cover-testing component for edge e ∊ E can betraversed only by a path corresponding to an endpoint of e,every edge in E must have at least one endpoint among those kselected vertices.

•Therefore, this set of k vertices forms the desired vertex coverfor G.

Proof(5)

•Suppose V* is a vertex cover for G with |V*| ≤ k.

•Let us suppose that |V*| = k.

•Denote the elements of V* by v1, v2,…, vk.

•Choose the edges in the cover-testing componentrepresenting each edge e = {vi, vj} depending onwhether {vi, vj} ∩ V* equals, respectively, {vi}, {vj},or {vi, vj}.

Component (vi, vj)

uij1

uij2

uij3

uij4

uij5

uij6

uji1

uji2

uji3

uji4

uji5

uji6

vi ∊ H,

vj ∉ H

vi ∉ H,

vj ∊ H

vi ∊ H,

vj ∊ H

Optimization problem

Definition 11.2.

•A (discrete) optimization problem is a quadrupleΠ = (X, (Sx) x∈X , c, goal), where

– X is a language over {0, 1} decidable in polynomial time;

–Sx is a subset of {0,1}∗ for each x ∈ X; there exists a polynomial p withsize(y) ≤ p(size(x)) for all y ∈ Sx and all x ∈ X, and the languages{(x, y) : x ∈ X, y ∈ Sx} and {x∈X: Sx =∅}are decidable in polynomialtime;

–c:{(x,y): x ∈ X, y ∈ Sx} → Q is a function computable in polynomialtime; and

–goal ∈ {max, min}.

Optimization problem

•The elements of X are called instances of Π.

•For each instance x, the elements of Sx are called feasiblesolutions of x.

•We write OPT(x) := goal{c(x, y) : y ∈ Sx }.

•An optimum solution of x is a feasible solution y of x withc(x, y) = OPT(x).

Algorithm

•An algorithm for an optimization problem(X, (Sx ) x∈X , c, goal) is an algorithm A whichcomputes for each input x ∈ X with non-empty Sx a feasible solution y ∈ Sx.

•We sometimes write A(x) := c(x, y).

•If A(x) = OPT(x) for all x ∈ X with non-emptyS, then A is an exact algorithm.

Polynomial reduction

•The concept of polynomial reductions easilyextends to optimization problems: a problempolynomially reduces to an optimization problemΠ = (X, (Sx )x∈X , c, goal) if it has an exactpolynomial-time oracle algorithm using anyfunction f with f(x)∈{y∈Sx :c(x,y) = OPT(x)}for all x∈X with non-empty Sx .