Krausz dimension and its generalizations in special graph classes

A {\it krausz $(k,m)$-partition} of a graph $G$ is the partition of $G$ into cliques, such that any vertex belongs to at most $k$ cliques and any two cliques have at most $m$ vertices in common. The {\it $m$-krausz} dimension $kdim_m(G)$ of the graph $G$ is the minimum number $k$ such that $G$ has a krausz $(k,m)$-partition. 1-krausz dimension is known and studied krausz dimension of graph $kdim(G)$. In this paper we prove, that the problem $"kdim(G)\leq 3"$ is polynomially solvable for chordal graphs, thus partially solving the problem of P. Hlineny and J. Kratochvil. We show, that the problem of finding $m$-krausz dimension is NP-hard for every $m\geq 1$, even if restricted to (1,2)-colorable graphs, but the problem $"kdim_m(G)\leq k"$ is polynomially solvable for $(\infty,1)$-polar graphs for every fixed $k,m\geq 1$.


Introduction
In this paper we consider finite undirected graphs without loops and multiple edges.The vertex and the edge sets of a graph (hypergraph) G are denoted by V (G) and E(G) respectively.N (v) = N G (v) is the neighborhood of a vertex v in G and deg(v) is the degree of v. Let G(X) denote the subgraph of G induced by a set X ⊆ V (G) and ecc G (v) is the eccentricity of a vertex v ∈ V (G).
A Krausz partition of a graph G is a decomposition of G into cliques, which are called clusters of the partition, such that every edge of G belongs to exactly one cluster.If every vertex of G belongs to at most k clusters, then the partition is called Krausz k-partition.The Krausz dimension kdim(G) of the graph G is a minimal k such that G has Krausz k-partition.
Krausz k-partitions are closely connected with the representation of a graph as an intersection graph of a hypergraph.The intersection graph L(H) of a hypergraph H = (V (H), E(H)) is defined as follows: 1. the vertices of L(H) are in a bijective correspondence with the edges of H; 2. two vertices are adjacent in L(H) if and only if the corresponding edges have a nonempty intersection.
Hypergraph H is called linear if any two of its edges have at most one common vertex; k-uniform if every edge contains k vertices.
The multiplicity of the pair of vertices u, v of the hypergraph H is the number m(u, v) = |{E ∈ E(H) : u, v ∈ E}|; the multiplicity m(H) of the hypergraph H is the maximum multiplicity of the pairs of its vertices.So, linear hypergraphs are the hypergraphs with the multiplicity 1.
Denote by H * the dual hypergraph of H and by H [2] the 2-section of H, i.e., the simple graph obtained from H by the transformation of each edge into a clique.It follows immediately from the definition that L(H) = (H * ) [2] . (1) First this relation was implicitly formulated by C. Berge in [3].(1) implies that a graph G has Krausz k-partition if and only if it is an intersection graph of linear k-uniform hypergraph.
A graph G is called polar [7,21] if there exists a partition of its vertex set such that all connected components of the graphs G(A) and G(B) are complete graphs.The partition (A, B) is called bipartition.If, in addition, α and β are fixed integers and the orders of connected components of the graphs G(A) and G(B) are at most α and β respectively, then the polar graph G with bipartition (2) is called (α, β)-polar [5].Given a polar graph G with bipartition (2), if the order of connected components of the graph G(A) (the graph G(B)) is not restricted above, then the parameter α (respectively β) is supposed to be equal ∞.Thus an arbitrary polar graph is (∞, ∞)-polar, and (1, 1)polar graphs are exactly well-known split graphs.
Denote by KDIM (k) the problem of determining whether kdim(G) ≤ k and by KDIM the problem of finding the Krausz dimension.
Graphs with Krausz dimension at most 2 are exactly line graphs.The class of line graphs has been studied for a long time.It is characterized by a finite list of forbidden induced subgraphs [1], efficient algorithms for solving the problem KDIM (2) and constructing the corresponding Krausz 2-partition are also known [6,12,18,19].
The situation changes radically if one takes k = 3 instead of k = 2 : the problem KDIM (k) is NP-complete for every fixed k ≥ 3 [9].The case k = 3 was studied in the different papers [10,15,16,17,20], and several graph classes, where the problem KDIM (3) is polynomially solvable or remains NP-complete, were found.
In [9] P. Hlineny and J. Kratochvil studied the computational complexity of the Krausz dimension in detail.Besides other results, the following findings were obtained in their paper: R1) The problem KDIM is polynomially solvable for graphs with bounded treewidth.In particular, it is polynomially solvable for chordal graphs with bounded clique size.
R2) For the whole class of chordal graphs, the problem KDIM (k) is NP-complete for every k ≥ 6.

R3
) For every fixed k the problem KDIM (k) is polynomially solvable in the class of split graphs.Note, that this result was independently obtained in [13].
Note, that the study of Krausz dimension of split graphs were motivated by results of L. Beineke and I. Broere [2], who studied the Krausz dimension of complete graph without one edge, which appeared to be non-trivial problem.
Based on their results, P. Hlineny and J. Kratochvil posed the following open problems: P1) decide the complexity of KDIM (k) restricted to chordal graphs for k = 3, 4, 5; P2) decide the complexity of KDIM for split graphs and complements of bipartite graphs.
In this paper we solve Problem P2 and partially solve Problem P1.In Section 2 we prove that the problem KDIM (3) is polynomially solvable in the class of chordal graphs.In Section 3 we show that the problem KDIM is NP-hard for both classes of split graphs and complements of bipartite graphs.
In Section 4 we consider a natural generalization of the Krausz dimension.A Krausz (k, m)-partition of a graph G is a decomposition of G into cliques, which are called clusters of the partition, such that any vertex belongs to at most k clusters, and any two clusters have at most m vertices in common.As shown above, the relation (1) implies the following statement: k the class of graphs with a Krausz (k, m)partition.It was proved in [11] that the class L m 3 could not be characterized by a finite set of forbidden induced subgraphs for every m ≥ 2, but the complexity of the problem KDIM m for an arbitrary m has not been established yet.We prove that the problem KDIM m is NP-hard for every m ≥ 1.
The class L m k is hereditary, i.e., closed with respect to deleting the vertices.Therefore it can be characterized in terms of forbidden induced subgraphs.We prove that for every fixed integers m, k such finite characterization of the class exists when restricted to (∞, 1)-polar graphs.In particular, it follows that the problem KDIM m (k) is fixed-parameter tractable when parameterized by k and m.It is the strong generalization of the result R3.Note also, that the class of (∞, 1)-polar graphs could be recognized in polynomial time [5].

Krausz 3-partitions of chordal graphs
Let F be a family of cliques of graph G.The cliques from F are called clusters of F. Denote by l F (v) the number of clusters from F covering the vertex v and by C(v 1 , . . ., v r ) the cluster containing vertices v 1 , . . ., v r .
A maximal clique with at least k 2 − k + 2 vertices is called a k-large clique.For such cliques the following statement holds: Lemma 2 ( [9,10,17]) Any k-large clique of a graph G belongs to every Krausz k-partition of G.
Further in this section 3-large clique will be called simply large clique.Let G be a graph with kdim(G) ≤ 3 and Q be some its Krausz 3-partition.Any subset F ⊆ Q is called a fragment of the Krausz 3-partition Q or simply a fragment.
Let F be some fragment of Krausz 3-partition Q and H be the subgraph of G obtained by deleting edges covered by F .Further we will use the notation H = G − E(F ).The set F could be empty.Fix some vertex a ∈ V (H) and positive integer k.Denote by B k [a] the kth neighborhood of a in H, i.e. the set of vertices at distance at most k from a.In particular, fragment for some a and k.In particular, by Lemma 2 large cliques are special.
The following statement is evident.
Then the following propositions hold: 4. For every a ∈ V (H) and every k there exists at least one (a, k)-local fragment; 5) If the clique C is special, then F ∪ {C} is a fragment. Proof: implies, that at least one of the clusters C 1 , C 2 , C 3 has more than 7 vertices.
2. If F k (a) is a local fragment, then by the definition v belongs to at most one cluster

The family of cliques
Lemma 4 Let G be a chordal graph and S be its cycle with length at least 4. Then for every two consecutive vertices of S at least one of them belongs to some chord.
Without lost of generality consider two adjacent vertices a k , a 1 of S.
If the statement of lemma is not true, then one can choose the chord a p a q , 1 < p < q < k such, that (p − 1) + (k − q) is minimal.But then (a 1 , . . ., a p , a q , . . ., a k ) is a chordless cycle. 2 Denote by lc(H) the length of a longest induced cycle of the graph H.
Lemma 5 Let G be a chordal graph with a Krausz 3-partition Q, F be a fragment of Q and H = G − E(F ).Let further there be no (a, 2)-local fragments, which contain special cliques in H. Then lc(H) ≤ 6.
Proof: Suppose the contrary, i.e., let a 1 , . . ., a k form the induced cycle Since each a i has two nonadjacent neighbors in H, in every local fragment with center in a i it is covered by at least 2 clusters.It implies l F (a i ) ≤ 1 for every i = 1, . . ., k.
As G is a chordal graph, there exist chords of the cycle S covered by the fragment F .As l F (a i ) ≤ 1, for every vertex a i chords incident to this vertex are covered by exactly one cluster of F .It implies that no pair of chords has the form {a i a j , a i a j+1 }, since in this case the vertices a i , a j , a j+1 are covered by one cluster of F and thus the edge a j a j+1 should be covered by F .
Assume without loss of generality that a chord of S contains a 1 .Let us show that all chords of S are covered by the cluster C chord ⊇ {a 1 , a 3 , . . ., a k−1 }.Indeed, suppose that some chords of S are covered by the cluster ) is a cycle of length at least 4, in which at least one of the vertices a ip , a ip+1 , say a ip , belongs to some chord according to Lemma 4. That chord should be covered by a cluster C ∈ F , C = C. So, we have l F (a ip ) ≥ 2, the contradiction.
In particular, this proposition implies the following: 1. k is even; 2. for any odd i and even j such that a i a j is not the edge of S, the vertices a i and a j are nonadjacent in G.It follows from the fact, that otherwise l F (a i ) ≥ 2.
Since there are no special cliques in H there exist two different local fragments therefore by the definition of local fragment all edges of H adjacent to a 3 are covered by 1).Note that since k ≥ 7, we have a 4 a k−1 ∈ E(H).
Recall that the vertex v in the local fragment F 2 (a 1 ) is covered by the cluster C (v, a 1 , a k ).All other edges of H incident to v should be covered by at most two clusters of F 2 (a 1 ).But this is impossible since the vertices a 2 , a 4 , a k−1 are pairwise nonadjacent.This contradiction proves Lemma 5. 2 The considerations above suggest the following algorithm which reduces the problem of recognition chordal graphs with Krausz dimension at most 3 to the same problem for graphs with bounded maximum degree and maximum induced cycle length.
Proof: Let us first prove the correctness of Algorithm 1. Once the "while" loop of Algorithm 1 is completed, we obtain the graph H and the set of cliques F .By the construction F contains only special cliques.Therefore if the Krausz 3-partition of G exists then it has the form F ∪ R, where R is a clique partition of H such that every vertex v ∈ V (H) belongs to at most 3 − l F (v) clusters of R and every edge of H belongs to exactly one cluster of R. By the construction of Algorithm 1 l F (v) ∈ {0, 1} for every non-isolated vertex v ∈ V (H).Therefore R is a clique partition of H with required properties if and only if R ∪ {vp v : l F (v) = 1} is a Krausz 3-partition of H * .Now we will show that the Algorithm 1 is polynomial.Indeed, the procedure of finding large clique which contains the fixed vertex v ∈ V (H) has the complexity O(m).We start to generate all possible (v, e)-local fragments for a vertex v ∈ V (H) only when deg(v) ≤ 18.This implies |B e [v]| ≤ const and thus the complexity of this procedure is constant.The outer loop of Algorithm 1 is performed at most m times.
After performing the Algorithm 1, we obtain the graph H * with bounded maximum degree and the length of the longest induced cycle.By Theorem 6 H * has bounded treewidth.For such a graph the problem of determining its Krausz dimension is polynomially solvable [9]. 2

Krausz dimension of split graphs and complements of bipartite graphs
Theorem 8 The problem KDIM is NP-hard in the class of split graphs.
Proof: We will reduce the problem KDIM for general graphs to the problem KDIM for split graphs.
Let G be a connected graph with n vertices and m edges and k ≤ n − 2 be an integer.Modify G by adding additional vertices and edges as follows: 1.For every u, v ∈ V (G) such that uv ∈ E(G), u = v add vertices x uv , y uv and edges uv, x uv u, x uv v, y uv u, y uv v, x uv y uv .Let X = {x uv : uv ∈ E(G), u = v}, |X| = n 2 − m. 2. For every u, v, w ∈ V (G) such that uv ∈ E(G) and every w ∈ V (G) \ {u, v} add a vertex y uvw and edges x uv w, x uv y uvw , wy uvw ; 3. For every two vertices x uv , x sw ∈ X add a vertex z uvsw and edges x uv z uvsw , x sw z uvsw , x uv x sw ; Let H be obtained graph (see an example of graph H for the graph G = K 1 + P 3 on Figure 2).Note, that A = V (G) ∪ X and B = V (H) \ A is a clique and a stable set in H, respectively.So, H is a split graph.We will show, that G has Krausz k-partition if and only if H has Krausz k -partition, where Let Q uv ⊆ Q be the set of clusters containing x uv .N uv is a stable set, and Therefore every cluster of Q uv contains exactly one vertex from N uv , i.e.
Every vertex of H, which is adjacent to x uv , belongs to one of those clusters.For every vertex from N (x uv ) \ N uv = A \ {x uv } we have the following.
1.The only vertex from N uv , which is adjacent to u and v, is y uv .Moreover, N (y uv ) = {u, v, x uv }.
For every v ∈ V (G) let Q (v) be the set of clusters of Q which contain v.We have So, The graph obtained from H by deleting edges covered by Q and isolated vertices is exactly the graph G.
Figure 3: The property (*) implies that M = {(a i , b i , c i ) : i = 1, . . ., q} ⊆ M and, by the consideration above, M is the 3-dimensional matching. 2 Now we turn to the complexity of the recognition problem KDIM m (k) in the class of (∞, 1)-polar graphs.
In [14] the following two statements were proved.Since they were published only in Russian in a journal that is of difficult access for general readers, we repeat their proofs here.Now we will prove that if . Consider a vertex u ∈ B i \ B j .Any edge of the form ux, where x ∈ B j \ B i (if such one exists) is contained neither in B i , nor in B j .Besides, no cluster of B contains more than m of such edges by definition of Krausz (k, m)-partition.Taking into account that u belongs to at most k − 1 clusters of B different from B i , we obtain the inequality Consider an arbitrary vertex v of the clique C. Let, without loss of generality, it belong to the clusters . The following equality is obvious then by the statement proved above, each term in the right part of the equality (10), starting from the second, does not exceed m(k − 1).Hence we have . Let, on the contrary, i ∈ {2, . . ., s} be the maximal number such that (B 1 ∪ . . . and the sum of the first i terms in the right part of ( 10) is equal to Each of the other terms does not exceed m(k − 1) by the maximality of i. Hence So, in any case we obtain that the inequality The obtained contradiction proves the lemma. 2 Theorem 12 For every fixed k, m ≥ 1, there exists a finite set F 0 of forbidden induced subgraphs such that a split graph G belongs to the class L m k if and only if no induced subgraph of G is isomorphic to an element of F 0 .
Proof: Denote by R p the graph obtained from the complete graph H ∼ = K f (k,m) by adding a new vertex and connecting it with exactly p vertices of H. Put F 0 = {R p : km + 1 ≤ p ≤ f (k, m) − 1} ∪ {K 1,k+1 }.Using Theorem 11 one can immediately verify that no graph from F 0 belongs to L m k .Let, without loss of generality, G be a connected graph, and V (G) = C ∪ S be a bipartition of V (G) into clique C and stable set S such that C is a maximal clique.Let also no induced subgraph of G be isomorphic to an element of F 0 .Put S = {v 1 , . . ., v s }.Consider two cases: 1) |C| > (km − 1)k + 1.
In this case we have Then, since no induced subgraph of G is isomorphic to a graph R p , km Hence, there exists a vertex u from C, which is not adjacent to any vertex from v 1 , . . ., v k .But then G(u, u , v 1 , . . ., v k ) ∼ = K 1,k+1 , a contradiction.Now we can construct a Krausz (k, m)-partition of G. Since deg(v i ) ≤ km for any i = 1, 2, . . ., s, then there exists a partition N ( i. e., the order of graph G is bounded above by a value, depending on k and m.Add to the list F 0 all such split graphs H, that H ∈ L m k and |H| ≤ ((km − 1)k + 1)(k + 1).Obviously, the constructed in the cases 1) and 2) finite list F 0 is a required list of forbidden induced subgraphs. 2 Since K 1,k+1 ∈ L m k , the heredity of L m k immediately implies Lemma 13 A bipartite graph G belongs to the class L m k if and only if no induced subgraph of G is isomorphic to K 1,k+1 .Theorem 14 For every fixed k, m ≥ 1, there exists a finite set F of forbidden induced subgraphs such that an (∞, 1)-polar graph G belongs to the class L m k if and only if no induced subgraph of G is isomorphic to an element of F.
Proof: Without loss of generality, we can suppose that (∞, 1)-polar graph G is connected.Let G have bipartition (A, B); A i , i = 1, 2, . . ., t, be the vertex sets of connected components of G(A); F 0 be the set of split graphs from Theorem 12. Denote by F 1 the set of (∞, 1)-polar graphs which do not belong to the class L m k and have order at most k .Now suppose that G(A) is neither complete nor empty graph.Then 2 ≤ t ≤ |A| − 1.Since K 1,k+1 ∈ F, then |A i | ≤ k for any i = 1, 2, . . ., t.Now we will prove that since K f (k,m)+1 − e ∈ F, then t ≤ f (k, m)−1.Let, to the contrary, t ≥ f (k, m).As G(A) is not a complete graph, there exists an index i 0 ∈ {1, 2, . . ., t} such that |A i0 | ≥ 2. Consider the set S = {a 1 , a 2 , . . ., a i0−1 , a i0 , a i0 , a i0+1 , . . ., a t }, where a i ∈ A i for any i ∈ {1, 2, . . ., t} \ {i 0 } and a i0 , a i0 ∈ A i0 .Then G(S) contains K f (k,m)+1 − e as an induced subgraph, a contradiction.Therefore

It follows from the inclusion
Corollary 15 The problem KDIM m (k) is fixed parameter tractable in the class of (∞, 1)-polar graphs, when parameterized by k, m ≥ 1.
Proof: The proofs of Theorems The complexity of the algorithm in lines 1-14 is O(n 2 ) at worst, where n = |V (G)|.Since the order of the graph G in line 15 is bounded above by value g(k, m) = max{((km − 1)k + 1)(k + 1), (k + 1)k(f (k, m) − 1)} the problem "kdim m (G) ≤ k" in this line could be solved in time O(h(k, m)), where h is some function.
So, the total complexity of Algorithm 2 is O(n 2 + h(k, m)). 2

Proposition 1 A
graph G has Krausz (k, m)-partition if and only if it is the intersection graph of a k-uniform hypergraph with the multiplicity at most m.The m-Krausz dimension kdim m (G) of the graph G is the minimum k such that G has a Krausz (k, m)partition.The Krausz dimension in those terms is the 1-Krausz dimension.Denote by KDIM m the problem of determining the m-Krausz dimension of graph, by KDIM m (k) the problem of determining whether kdim m (G) ≤ k and by L m Figure 2:

Theorem 11
Any (k, m)-large clique C of a graph G belongs to every Krausz (k, m)-partition of G. Proof: Let A be a Krausz (k, m)-partition of graph G, A 1 , A 2 , . . ., A t be those clusters of A which have common vertices with C. Assume that C ∈ A. Then the family B = (B 1 , B 2 , . . ., B t ), where B i = A i ∩ C, is a Krausz (k, m)-partition of the graph G(C), and by maximality of C B i = C for every i = 1, 2, . . ., t.Let us show that |B i | ≤ mk for any i = 1, 2, . . ., t.Consider a cluster of B, say B 1 , and a vertex u ∈ C \ B 1 .No edge of the form ux, where x ∈ B 1 , is contained in B 1 .Moreover, by the definition of Krausz (k, m)-partition each cluster of B different from B 1 contains at most m such edges.Taking into account that the vertex u belongs to at most k clusters of B, we obtain the inequality |B 1 | ≤ mk.
where K f (k,m)+1 − e is the graph obtained from the complete graph K f (k,m)+1 after deleting an edge.The set F is finite, since F 0 and F 1 are finite.According to Theorem 11, there is no Krausz (k, m)-partition for K f (k,m)+1 −e.Therefore, K f (k,m)+1 − e ∈ L m k .Thus, F ∩ L m k = ∅.The necessity of the statement follows from the heredity of the class L m k .Now let G contain no induced subgraph isomorphic to an element from F. If G(A) is complete, then G is a split graph and by Theorem 12 G ∈ L m k .If G(A) is empty, then G is a bipartite graph and by Lemma 13 G ∈ L m i | ≤ k(f (k, m) − 1).Since |N (a) ∩ B| ≤ k for any vertex a ∈ A and G is connected, we have|G| ≤ |A| + |B| ≤ |A| + a∈A |N (a) ∩ B| ≤ k(f (k, m) − 1) + k 2 (f (k, m) − 1) = (k + 1)k(f (k, m) − 1).
For every non-isolated vertex v ∈ V (H) generate all (v, e)-local fragments, e = min{ecc H (v), 2}; if there exists a vertex v ∈ V (H) such that there are no (v, e)-local fragments the answer is "kdim(G) > 3"; stop; if there exists a special clique C F := F ∪ {C}; continue to the next iteration of the cycle; isContinue := f alse endwhile; construct a graph H * by adding a pendant edge vp to the next iteration of the cycle; else the answer is "kdim(G) > 3"; stop; if there exists a vertex v ∈ V (H) with deg(v) ≥ 19 if v is contained in a clique C with |C| ≥ 8 extend C to a maximal clique; F := F ∪ {C}; continue to the next iteration of the cycle; else the answer is "kdim(G) > 3"; stop; we have deg(v i ) ≤ km for any i = 1, 2, . . ., s.Since G contains no induced K 1,k+1 , we have |N (u) ∩ S| ≤ k for any vertex u from C.Moreover, we prove that for any vertex u from C the inequality |N (u) ∩ S| ≤ k − 1 holds.Assume this is not true.Let, without loss of generality, some vertex u from C be adjacent to the vertices v 1 , . . ., v k from S, k ≤ s.Since deg(v i ) ≤ km, i = 1, 2, . . ., k, and u ∈ 12, 14 and Lemma 13 imply the following algorithm of recognition whether a given (∞, 1)-polar graph belong to the class L m k .Algorithm 2 Input: connected (∞, 1)-polar graph G with bipartition (A, B), A = A 1 ∪ A 2 ∪ . . .∪ A t where A i , i = 1, 2, . . ., t, be the vertex sets of connected components of G(A), integers k, m ≥ 1. Output: 1 if kdim m (G) ≤ k, and 0 otherwise.1. begin 2. if |N (a) ∩ B| ≤ k for every a ∈ A 3. if t = 1, i.e. graph G is bipartite, 4. if deg(b) ≤ k for every b ∈ B 5.