Discrete Mathematics 3. Exercise Sheet

Algorithmic

Discrete Mathematics 3. Exercise Sheet

Department of Mathematics SS 2012

PD Dr. Ulf Lorenz 23. and 24. May 2012

Dipl.-Math. David Meffert Version of July 15, 2012

Groupwork

Exercise G1

Let the algorithm CLIQUE be defined by:

input : A graphGand a natural numberk.

output : ’yes’, ifG contains a clique of cardinalityk. Otherwise ’no’.

Let the algorithm IDEPENDENT SET (IS) be defined by:

input : A graphGand a natural numberk.

output : ’yes’, ifG contains an independent set consisting ofkvertices. Otherwise ’no’.

Show that CLIQUE≤P IS.

Solution: First lets determine the polynomial function f which transforms the inputsG= (V,E)andk∈Nof CLIQUE to suitable inputs for IS. It is given by f(〈G,k〉) =〈G⁰,k〉withG⁰ = (V,E⁰). The set E⁰ is defined by(v,w)∈E⁰ ⇐⇒

(v,w)6∈E.G⁰is called the complementary graph (see H8).

claim: Gcontains a clique of cardinalityk ⇐⇒ G⁰contains an independent set of cardinalityk.

proof:

• ⇒Letv₁, . . . ,v_k be a clique of cardinalityk. Hence for every two different verticesv_i fori∈ {1, . . . ,k}they are directly connected by an edge. By definition this means they are not directly connected by an edge inG⁰. So the verticesv₁, . . . ,v_k form an idenpendent set inG⁰.

• ⇐ Assume that G contains no clique of cardinality k. Take an arbitary subset S ⊂ V consisting of k elements v₁, . . . ,v_k. Because by assumption it is not a clique ofGthere existi,j∈ {1, . . . ,k}withi6= jand(^vi,v_j)6∈E. By definition this means(^vi,v_j)∈E⁰and for this reasonS=^v1, . . . ,v_kcan’t form an independent set inG⁰. BecauseS was chosen arbitrarily the second implication is proven.

Exercise G2 (Bipartite graphs)

Prove that a graph(V,E)is bipartite iff it contains no cycles of odd length.

Solution:

• ⇒LetGbe bipartite with induced partionV =S∪˙T. Let(v₁,v₂, . . .v_n,v₁)be a cycle inG and w.l.o.g. letv₁∈S.

Since there are no edges inside the setsSandT, we getv_2k∈Tandv_2k+1∈Sfork∈ {1, . . .bⁿ₂c}. So it is necessary thatnis even forv_nbeing an element ofT. Hence there are no cycles of odd length.

• ⇐LetG be a graph without odd cycles. Without loss of generality we ca assume thatG is connected. Otherwise do the same argumentation for each connected component of the graph. For a verticex₀∈V letSbe the set of all vertices inVwith even distance¹to the vertexx₀. Additionally we call its complementT:=V\S. We want to show that there are no edges withinSandT. So assume there is an edge(x,y)∈E, such thatx,y∈S. LetW_xandW_y be shortest paths fromx₀toxand to yrespectively. By definition ofSthe pathsW_xandW_yboth have even length.

Now letzbe the last common vertice of the pathsW_xandW_y(both beginning inx₀) and denote byW_x⁰andW_y⁰the remaining part of the paths fromztoxandyrespectively. Now define the cyclex−W_x⁰→z−W_y⁰→ y−(x,y)→x.

SinceW_x⁰ andW_y⁰ both have either odd or even length the given cycle has odd length. This contradicts Ghaving

1 The distance of two verticesx,yin a graphGis defined as the number of edges used in a shortest path fromxtoy.

no cycles of odd length, hence our assumption was wrong and for this reason there are no edges in the setS. The statement for the setT can be shown the same way. So we haveV =S∪˙T with suitable setsV andT. Hence the graphGis bipartite.

Exercise G3 (Eulerian graphs)

A path in a GraphG= (V,E)is calledEulerian path, if it contains every edgee∈Eexactly once. An Eulerian path which is a cycle is calledEulerian cycle. A graph is calledeulerianif it contains an Eulerian cycle.

(a) Which of the given graphs inFigure 1are Eulerian graphs?

Figure 1:Eulerian graphs?

(b) Now letGbe a connected graph. Name necessary conditions forGbeing eulerian.

(c) Are these conditions sufficient, too?

Solution:

(a) The solution is yes,no,yes,no. The best way to prove the nonexistence of an Eulerian cycle is by using one of the criterias of part(b).

(b) We look at the following two conditions and prove that they are necessary. Later we will show they are even equivalent.

i. Ecan be decomposed into cycles having disjoint edges.

ii. All vertices inV have an even degree.

Lets prove necessity of the conditions. So letG= (V,E)be a graph andT= [e₁, . . . ,e_n]be an Eulerian cycle inG.

We prove the decomposition into cycles first. So we take an arbitrary verticev ∈V and follow the path given by T till we reach the vertexv again. If the cycleCconstructed that way is the whole Eulerian cycle we are finished.

Otherwise delete all the edges ofC in the set E and delete all vertices inV which got isolated by this deletion process. The resulting graphV⁰= (G⁰,E⁰)is eulerian again. So we repeat this procedure till the resulting graph is empty. By construction we get a decomposition of cycles having disjoint edges.

Now we can conclude the second condition (all vertices have even degree). Letv∈V arbitrary. By doing the same procedure as for the cycle decomposition and always choosingv as the starting node we get exactly2new incident edges ofv in every step tillv becomes an isolated node. By construction they are all disjoint, henced(^v) =2k_v withk_v is the number of disjoint cycles containingv. Sincev∈V was arbitrarily chosen we are done.

(c) Yes the given conditions in part(b)are also sufficient. We we will prove this the following way. First we prove that a graph can be decomposed into cycles with disjoint edges, iff all vertices have even degree. After that we just have to prove the first condition.

So lets start with the equivalence. The first implication was already shown in part(b). So letGbe a graph where all vertices have even degree. We use induction over the number of edgeskinGto prove the decomposition.

Base case:Fork=0the graph is trivially decomposed into cycles with disjoint edges.

Induction hypothesis:Assume that every graph with at mostkedges can be decomposed into cycles with disjoint

edges.

Inductive step: LetG be a graph withk+1edges,v ∈V an arbitrary vertex andv₁∈V with(v,v₁)∈E. Since d(v₁)≥2there exists a vertexv₂6=v with(v₁,v₂)∈E. By iterating this process, becauseGis a finite graph, we will come to a vertexv_l we already visited. So we get a cycle C= [v_l,v_l+1, . . . ,v_l]in G. IfCcontains all edges in E we are done. Else we delete all these edges contained in the cycle and all vertice getting isolated by this deletion process (those with d(^v) =2). The resulting graphG⁰ = (V⁰,E⁰)contains less thank edges and by the induction hypothesis can be decomposed to a decompositionE of cycles with disjoint edges. The setE ∪ {C}yields a decomposition of cycles with disjoint edges for the whole graphG.

Now we construct an Eulerian cycle out of the cycle decomposition with disjoint edges. Again we use induction over the numbermof disjoint cycles.

Base case:Form=1the only cycle we have is of course an Eulerian cycle.

Induction hypothesis:Assume that every connected graph that can be decomposed into at mostmdisjoint cycles has an Eulerian cycle.

Inductive step:LetE be a decomposition intom+1disjoint cycles. Choose a cycleC∈ E and delete all edges in CfromEand all vertices getting isolated by the deletion process. Since all connected components of the resulting graphG⁰= (V⁰,E⁰)have a decomposition into at mostmdisjoint cycles, they all have an Eulerian cycle by using the induction hypothesis. With these cycles we construct an Eulerian cycle forGthe following way. Start with an arbitrary vertexv ∈C and follow the cycle till the first nodev₁∈V⁰. Now follow the Eulerian cycle induced by connected component(V₁,K₁)containingv₁. Now keep on following the cycleCtill the first nodev₁∈V⁰\{V₁}.We repeat this procedure till we went through the whole cycleC and get to the starting vertexv again.(after going through all the Eulerian cycles of the connected components) Since the graphGis connnected we went through all the Eulerian cycles which are induced by the connected components ofG⁰and the cycleCand for this reason have contructed an Eulerian cycle forG.

Exercise G4 (Primes and the classN P)

The class of problems whose complement is inN P is called co-N P. For the following exercise assume that the coding length of a natural numbernis given by〈n〉=blog₂nc+1.²

(a) Show that the problem PRIMES, to determine if a given natural number is prime, is a co-N P problem.

(b) Why would it be much harder to show that this problem is also in the classN P? (c) Prove PRIMES∈ N P under the assumption〈n〉=n.

Solution:

(a) We have to prove that the problem to determine if a given numberpis not a prime, is anN P-problem.

A suitable object (certificate) forpnot being a prime are two natural numbera,bwith1<a<b<pwitha·b=p.

With this object one can easily check if the given answer ’yes’ (p is not a prime) was correct. The algorithm to do this gets the inputs pand(a,b)and decides if(a,b)is a suitable certificate forp by checking ifa·b= pand 1<a<b<p. Because ofa,b<pthe binary representation ofa,bis shorter than the one ofp. So the runtimel for the mutiplication ofaandbismax{〈a〉 〈b〉} ·1≤ 〈p〉. The runtime for the comparing of the numbersa·band pis〈p〉. Same holds for checking1<a<b<p. So the algorithm is linear in〈p〉, so in particular polynomial〈p〉. (b) This is much harder because one would need a certificate (probably a number or a set of numbers) which can be

used (in polynomial time) to prove thatpis prime. The coding length should also be polynomial in〈p〉. It is not clear how such a certificate should look like. The set of all numbers smaller thanpis not a good choice because the coding length of this set is equal toPp−1

i=1(blog2ic+1)which is not polynomial in〈p〉=blog2pc+1.

Indeed such certificates do exist. The first one was found in1975by the australian mathematician Vaughan Pratt (born 1944). It consists of a number xwhich is coprime topand has some other properties.

(c) Under this assumption a polynomial algorithm with a polynomial certificate is for example given by all natural numbers smaller thanp(orpp) and checking if the division yields a natural number or not. Under the assumption

〈p〉=pthis certifate has polynomial coding length and the checking algorithm usespdivisions and for this reason has a runtime ofp· 〈p〉=p²which is of course polynomial in〈p〉=p.

In fact solving this problem is also possible with the usual coding length (binary representation), but is was a open question for quite a long time. An algorithm was found in2002by the Indian mathematicians Manindra Agrawel, Neeraj Kayal and Nitin Saxena.

2 This is the common binary representation of a natural numbern. Hence you needblog₂nc+1digits.

Homework

Exercise H7 (Trees) (10 points)

LetG= (V,E)be a graph withn≥2vertices. Proof that the following statements are equivalent:

(a) Gis a tree.

(b) Gis connected and containsn−1edges.

(c) Gcontainsn−1edges but no cycles.

(d) Gis minimally connected. That means for every edgee∈Ethe graphG\ {e}= (V,E\ {e}) is not connected.

(e) Gcontains no cycles and adding one edge generates exactly one cycle.

(f) For every two nodesu,v∈V there is exactly one[u,v]-path inG.

Solution: There are several ways to prove the equivalences by proving different implications. We present one way to it.

(a)⇒(b)and(c): We prove this by induction over the number of verticesn. Forn=1we have only one vertex and no edges (otherwise there would be a cycle). So the graph hasn−1=0edges and is of course connected and has no cycles. Now assume the statement we want to prove is right for all trees with a mostnvertices. LetG be a tree with n+1vertices. Sincen+1>1it is not hard to see, that we have at least one leafv in the graphG. (by looking a simple path of maximal length in the tree) Delete this vertexv and its one incident edge. Sincev was a leaf the resulting graph G⁰= (V⁰,E⁰)is still connnected and has no cycles becauseGhad not cycles in advance.(soG⁰is a tree withnvertices) By assumptionG⁰hasn−2edges is connected and has no cycles. Adding the deleted leaf with its edge to graph again (to get backG) we seeT hasn−1edges, is connected and has no cycles.

(a)⇒(d): Lete= (u,v)be an arbitrary edge inG. AssumeG\ {e}is connected. Then there exists a path fromutov which does not use the edgee. By addingeto this path we get a cycle which contradictsGbeing a tree.

(a)⇒(f): Letu,v ∈V. SinceG is connected there is a least one pathp= (^v=^v1, . . . ,v_k=u)inG. Assume there is another pathq= (^v=q₁, . . . ,q_l=u). Letibe the smalllest index withv_i6=q_iand jthe smallest withv_j=q_jand j>i.

Notice we havei>1(v₁=q₁) and jexists sincev_k=q_l. Taking the pathc= (^vi−1,v_i, . . . ,v_j=q_j,q_j−1, . . .w_i−1=^vi−1) we get a cycle which is a contradiction toGbeing a tree.

(a)⇒(e): Assume we want to add the edgee= (^v,u)forv,u∈V. BecauseGwas already connected there is a path fromutov not usinge. Be adding the edgeeto this path we get a cycle. Since the path fromutov was unique be part (f)this is the only cycle we get by addinge.

(b)⇒(a): Assume we have a cycleCin the graphG. If we delete an arbitrary edgee∈Cthe graph is still connected.

This way we can remove all the cycles fromGan get a connected graph G⁰without cycles. SoG⁰ is a tree and for this reason has to haven−1edges. This is a contradiction toGalready havingn−1edges, because we deleted at least one edge. SoGalready had no cycles.

(c)⇒(a): AssumeGis not connected and consists of the connected componentsK₁, . . . ,K_r. Then we can connect an arbitrary vertexv₁∈K₁with another verticev₂∈K₂without generating a cycle. We do the same thing for all the other connected components and get a graphG⁰which is a connected and without cycles, hence a tree. SoG⁰hasn−1edges which is a contradiction toGalready havingn−1edges, because we at least added one edge.

(d)⇒(a): We have to show that G has no cycles. Assume there is a cycleC inG. Then we can delete an arbitrary edgee∈Cand the resulting graphG⁰is still connected. This contradictsG being minimally connected.

(e)⇒(a): We have to prove thatGis connected. Assume it is not. Then there at least two connected componentsK₁ andK₂ofGand we can add an arbitrary edge between any verticesv₁∈K₁andv₂∈K₂without getting a cycle. This is a contradiction.

(f)⇒(a): Since there is path fromutov for arbitrary verticesu,v∈V the graphGis connected. IfGwould contain any cycleCthere would be at least two different paths fromutov for arbitrary verticesu,v∈C. SoGcontains no cycles.

Exercise H8 (The complement graph) (10 points)

The complement graphG of G = (V,E)is the graph were two vertices are adjacent iff they are not adjacent inG. So formally speaking we haveG:= (V, ^[n₂^]

\E)with ^[n]

2

\E:={(i,j)|i6=j∈ {1, . . . ,n}(i,j)∈/E}. LetGbe an undirected graph. Prove thatGorGis connected.

Solution: Lets assume G is not connected. (otherwise we are done). Without loss of generality one can assume that G has exactly two connected componentsK₁andK₂. The general case can be proved by induction over the number of components.

LetV₁consist of the vertices ofK₁andV₂of the vertices ofK₂. In the worst caseG is a complete bipartite graph with partitionsV₁andV₂. Forv∈V₁andw∈V₂the edge(^v,w)is a[^v,w]-path inG. Letv,w∈V₁. Taking an arbitraryu∈V₂ the path(^v,u,w)is a[^v,w]-path. For^v,w∈V₂it works the same way. SoGis connected.

Exercise H9 (10 points)

Prove SUBSETSUM≤pPARTITION.

problem : SUBSETSUM input :a₁, ...,a_n,b∈N output :T⊆ {1, ...,n}withP

k∈Ta_k=b problem : PARTITION

input :a₁, ...,a_l∈N

output :T⊆ {1, ...,l}withP

k∈Ta_k=P

k/∈Ta_k

Solution: First lets determine the polynomial function f which transforms the inputs a₁, . . . ,a_n ∈N and b ∈N of SUBSETSUM to suitable inputs for PARTITION. It is given by f(〈a₁, . . . ,a_n,b〉) =〈a₁, . . . ,a_n,a_n+1,a_n+2〉with a_n+1 = M−b+1anda_n+2=b+1. The numberMis defined byM:=Pn

k=1a_k. This transformation can be done in polynomial time.

claim: T⊆ {1, . . . ,n}is a solution for SUBSETSUM with inputsa₁, . . . ,a_n,b∈N ⇐⇒ K:=T∪ {n+1} ⊆ {1, . . . ,n+2} is a solution for PARTITION with inputsa₁, . . .a_n+2. (same argument can be done for T∪ {n+2}but by construction a solution can’t containn+1andn+2)

proof:

• ⇒LetK⊆ {1, . . . ,n+2}be a solution for PARTITION of the formK=T∪ {n+1}with inputsa₁, . . .a_n+2. We get P

k∈Ka_k=P

k∈/Ka_k

⇔ P

k∈K\{n+1}a_k+M−b+1=P

k∈K\{n+2}/ a_k+b+1

⇔ P

k∈K\{n+1}a_k+Pn

k=1a_k−P

k∈/K\{n+2}a_k=2b

⇔ P

k∈K\{n+1}a_k+P

k∈K\{n+1}a_k=2b

⇔ 2P

k∈K\{n+1}a_k=2b.

HenceT=K\ {n+1}is a solution for SUBSETSUM.

• ⇐LetT⊆ {1, . . . ,a_n}be a solution for SUBSETSUM. ThenK=T∪ {n+1}is a solution for PARTITION because of

X

k∈K

a_k=M−b+1+X

k∈T

a_k=1+

n

X

k=1

a_k

=1+X

k∈T

a_k+X

k∈/T

a_k=1+b+X

k∈/T

a_k

=X

k∈/K

a_k.

Notice thatT⊆ {1, . . . ,n}andK⊆ {1, . . . ,n+2}.