A More Complicated Hardness Proof for Finding Densest Subgraphs in Bounded Degree Graphs

# A More Complicated Hardness Proof for Finding Densest Subgraphs in Bounded Degree Graphs

Manuel Sorge
###### Abstract

We consider the Densest-Subgraph problem, where a graph and an integer  is given and we search for a subgraph on exactly  vertices that induces the maximum number of edges. We prove that this problem is NP-hard even when the input graph has maximum degree three.

## 1 Introduction

We consider the following problem.

Densest-Subgraph:
Input: A graph , and a nonnegative integer .
Task: Find a vertex set  of size exactly  such that  has maximum number of edges.

We call every vertex set  with  a solution for this particular instance. Densest-Subgraph is clearly a fundamental problem and has received much attention in the literature [7, 6, 1, 4]. In particular, this problem has been proven NP-hard in various variants. We prove that Densest-Subgraph remains NP-hard on graphs of degree at most three. While this result has been proven before [5], we show it in a much more complicated manner. More precisely, we first give a reduction from Clique to Densest-Subgraph with maximum degree five, proving its correctness through an intricate replacement argument. Then we successively reduce the maximum degree by replacing each high-degree vertex by gadget graphs. We prove that, for some optimal solution , each gadget graph is either fully contained in  or not at all. This is done using elaborate case-distinctions which are most tedious to verify. We note that the previous proof is essentially contained within two pages. Our proof needs eight pages.

### Preliminaries.

We use standard graph notation as used, for example, by Diestel [2]. Where it is appropriate, we denote  and .

## 2 NP-hardness of Densest-Subgraph in Graphs with Maximum Degree Three

Our proof of NP-hardness is divided into three steps. We first show that Densest-Subgraph is NP-hard in graphs with maximum degree at most five. Then, we proceed to reduce the maximum degree in hard instances. We successively employ two gadgets that replace vertices in the input graph, giving hardness in graphs with degree at most four and then hardness in graphs with degree at most three.

###### Lemma 2.1.

Densest-Subgraph is NP-hard even on graphs with maximum degree five.

###### Proof.

Let  be an instance of Clique and without loss of generality assume that  is an odd square number. The construction is as follows. For every vertex  in  introduce a quadratic grid graph  with  vertices. In order to simplify the analysis, identify vertices with degree three on opposite boundaries, creating a grid model of a torus in which every vertex has degree exactly four. More formally,  is the graph with vertex set

 {vij:0≤i,j,≤n2−1}

and edge set

 {{vij,v(i+1)modn2j},{vij,vi(j+1)modn2}:0≤i,j≤n2−1}.

Let  be a bijection and let

 ψ :V→N:i↦(ϕ(i)n√n)modn2, and σ :V→N:i↦⌊ϕ(i)n√n/n2⌋.

For every edge  connect the tori  and  by an “inter-torus” edge between the vertices  and . Call the graph that is constructed in this way  and set the instance of Densest-Subgraph 11todo: 1Densest--Subgraph!?to . Clearly, this construction can be carried out in polynomial time. We decide  to be a yes-instance if and only if an optimal solution of  contains at least  edges.

For the correctness, first, observe that  contains a subgraph of  with  edges if  contains a clique with  vertices: for every vertex  in the clique choose every vertex of the torus . For the other direction, we prove that every optimal solution to  comprises a collection of complete tori. That is, if an optimal solution contains a vertex of one of the tori , every vertex of  is in the solution. Then it is clear that if the optimal solution contains  edges, the vertices in  that correspond to a torus  in the optimal solution must induce a clique on  vertices.

An optimal solution  to  consists of a number of possibly proper vertex subsets of the tori, altogether containing exactly  vertices. For the sake of contradiction assume that at least one of these vertex subsets is proper, that is, not all of the vertices of some torus are contained in . Call the vertices in  “black” and the vertices in  “white”. Call edges between black and white vertices within one torus “cut”. Call tori  with at most  black vertices “small” and the remaining tori “large”. We prove the following.

Claim: In each small torus , there are at least twice as many cut edges as there are inter-torus edges incident with black vertices of .

Before proving the claim, we note that it implies the lemma. There are at most  large tori, that is, for every sufficiently large111The relation holds if and only if . Since  is strictly monotone ascending for  and equality holds for , we have that  for all . We can assume this without loss of generality. , there are at most  large tori. Consider distributing all black vertices of small tori to large tori, completing them in the process, and combining the remaining black vertices in small tori to complete tori. This procedure cannot decrease the number of edges induced by the black vertices: All inter-torus edges between large tori are preserved and for every inter-torus edge incident with a black vertex of a small torus, there were at least two cut edges, and thus, at least one edge is added to the solution. (Observe that after this procedure, there are no cut edges remaining.) Furthermore, after this procedure, each black vertex has at least four black neighbors within its torus, whereas before it had at most four. Thus, we may assume that there are no incomplete tori and the correctness of our construction follows.

We prove our claim using the following result by Efe and Feng [3].

Fact: In order to remove  vertices from a quadratic grid with  vertices, where  is odd, at least  edges have to be cut.

It is clear that in a torus with the same number of vertices at least as many edges have to be cut.222Consider removing a set of vertices from a quadratic grid and then removing it from a torus on the same vertices as the grid. The number of cut edges incident with each vertex differ only for vertices that are removed from the boundary of the grid. But in the torus, these vertices have higher degree than in the grid. From this statement, our claim follows directly for tori with at least  and at most  vertices. Assume that there is a torus  with  black vertices. The number of cut edges is

 2(n4−nv)n2−1≥2n3n2−1>2n.

Since there are at most  inter-torus edges incident with , there are at least twice as many cut edges as there are inter-torus edges. The same follows analogously for tori  with  black vertices.

It remains to show the claim for tori with at most  black vertices. Consider such a torus  and call a vertex  in  to reside in “row”  and in “column” . We show that incident with each connected component  of black vertices in  there are at least twice as many cut edges as there are inter-torus edges incident with black vertices in . The claim then follows, since there are no cut edges that are incident with two of ’s black connected components. First, assume that  contains each vertex of some row . Since  contains at most  black vertices, the number of columns in  with fewer than  black vertices is at least  and this gives a lower bound on the number of cut edges incident with . Thus, since there are at most  inter-torus edges incident with , for every sufficiently large , there are at least twice as many cut edges as inter-torus edges incident with . Analogously we can prove this, if  contains each vertex of some column. Next, assume that  does not contain complete rows or columns. If there is only one inter-torus edge incident with , clearly there are always more than two cut edges. If there are at least two inter-torus edges incident with , consider a shortest black path between these two edges. By the placement of the inter-torus edges in , this path either touches at least  rows or at least as many columns. Since there are no complete columns or rows in  is then a lower bound on the number of cut edges incident with . Thus, our claim now follows for all  larger than some constant and our construction is correct. ∎

Next, we reduce the maximum degree in hard instances to four, by replacing each vertex with a “solid” gadget, that is, a gadget that can be assumed to be either completely contained in a solution or to be disjoint to it. Thus, in order to maximize the number of edges induced by a solution, one has to choose gadgets correspondingly to vertices in the original graph.

###### Lemma 2.2.

There is a polynomial-time many-one reduction from Densest-Subgraph in graphs with maximum degree six or maximum degree five to Densest-Subgraph in graphs with maximum degree four.

###### Proof.

Let  be an instance of Densest-Subgraph where each vertex in  has degree at most six or at most five. We replace each vertex  in  by the graph shown in Figure 1 (the “fence gadget”) and distribute ’s edges to the outer vertices such that each vertex in the fence gadget gets degree at most four. Call the graph constructed in this way  and set the new instance of Densest-Subgraph 22todo: 2Densest--Subgraph!?to . We argue that from any solution  to  we can construct another solution with at least as many edges such that each fence gadget either is completely contained in  or its vertices are disjoint to . Then, from any solution to  with  edges (there are  edges in a gadget) we can construct a solution for  with at least  edges and vice-versa by completing every gadget and then exchanging gadgets with their corresponding vertices. Thus, since it is easy to achieve at least  edges in a solution for , finding an optimal solution or any solution with a number of edges above a given threshold is equivalent in these instances. It is also not hard to see, that constructing the solution for  from a solution for  can be done in polynomial time, we omit the details.33todo: 3MS: Dürfen wir das? Das ist: höchstens -mal iterieren zum Komplettieren der Gadgets. Darin: Immer wieder den Claim herstellen (mittels iterieren über alle Gadgets und Property 1 und 2 herstellen = ), und beim Komplettieren entsprechende Austauschknoten finden = auf jeden Fall . Wäre unschön, den Text noch mehr vollzustopfen. Oder einen Extraabsatz ganz am Ende dazu?

Let  be a solution for . Call the vertices in  “black” and the remaining ones “white”. The black-degree of a vertex is the number of black vertices adjacent to it. Call edges between two black vertices “black” and the remaining edges “non-black”. We consider fence gadgets that are partially black and argue that we can move black vertices between gadgets without losing edges, ultimately completing all gadgets. To do this, we prove the following.

Claim: Given a solution , we can modify  without losing black edges and without changing the number of black vertices in any fence gadget such that the following holds.

1. For  it holds that in each fence gadget with exactly  white vertices, there are at least  non-black edges.

2. For  it holds that in each fence gadget with at least  black vertices and at least  white vertices, there is a set of  black vertices in the gadget that is incident with at most  black edges (counting both inner-gadget edges and edges between gadgets).

3. Each fence gadget with exactly four white vertices has at least eight non-black edges (within the gadget) and a set of four black vertices that is incident with at most eight black edges (both within and outside of the gadget).

4. Each fence gadget with exactly  black vertices contains a black vertex with black-degree at most two.

44todo: 4MS: Ich finde es besser, erst zu zeigen, wie man den Claim anwendet und ihn dann zu beweisen. Egal wierum man es macht, man muss immer scrollen; entweder ist die Figure mit dem Gadget nicht da, oder der Claim selbst, wenn man ihn anwendet. Aber bei der Variante den Claim erst hinterher zu beweisen, verliert man imho weniger Motivation beim Beweislesen, weil man erst sieht, warum der Claim so aussieht, wie er aussieht, und nicht erst irgendeine Aussage überprüfen muss, von der man keine Ahnung hat, wie sie angewendet wird oder einen weiterbringt.

We first show how this claim can be used to complete all gadgets in . Assume that the claim holds. We show how we can make gadgets completely black that have some minimum number of white vertices other than . Let  be the minimum number of white vertices in gadgets and let  be a gadget with exactly  white vertices.

• Assume that . It follows that  contains a white vertex  that has at least three black neighbors and Part 2 of our claim tells us that we can color a vertex in another gadget white and color  black without losing black edges.

• Assume that . Then, either there are two gadgets  with exactly one black vertex or there is a gadget  with at least two black vertices and at least two white vertices. In the first case, we can color both black vertices of  and  white, losing at most two black edges in the process, and complete , gaining at least five black edges (Part 1 of the claim). In the second case, we use Part 2 to color two black vertices of  white, losing at most five black edges and complete , gaining at least five black edges.

• Assume that . Then, either we can use Part 2 or Part 4 of the claim to find three black vertices such that, if we color them white, we lose at most seven black edges. We color the found vertices white and complete the gadget  without losing black edges by Part 1.

• Assume that . Then, either we can find another gadget with exactly four white vertices and use Part 3 of our claim to color its vertices white and complete  without losing black edges, or, if there is no other gadget with exactly four white vertices, by Part 4 of our claim, we can successively find four vertices each with black-degree at most two, color them white and, after that, we can complete . We lose at most eight black edges, but gain at least eight by Part 3 of the claim.

• Assume that . Observe that in a gadget with exactly three black vertices, the number of non-black edges is at least ten. Thus, we can again apply Part 4 of our claim.

• The case that  is analogous to  and the case  is trivial.

To make each gadget either completely black or completely white, we first ensure that each part of the claim holds, then apply a modification as described above to complete the gadget  and iterate. The modifications that are necessary for each part of the claim to hold do not change the number of black vertices in any gadget, and in each of the above cases, we modify  such that the number of completely black gadgets increases. Thus, after at most  steps, each gadget is either completely black or completely white, and the correctness of our construction follows if the claim holds.

Next, we prove the claim. We first prove that we can make two assumptions about the solution  without loss of generality. We modify  without losing edges and without changing the number of black vertices in any fence gadget such that the following holds.

Property 1: In every gadget, there is no white inner vertex that has at most one outer white neighbor and no white inner neighbor.

Assume that there is such a white inner vertex . If  does not have a white outer neighbor, we may simply color one of its black outer neighbors white, losing at most three black edges and then color  black, gaining at least three black edges. If  has a white outer neighbor, then it has a black outer neighbor  with black-degree at most two. Thus, we can make  white, losing at most two black edges, and make  black, gaining at least two black edges.

Property 2: In every gadget with at most four white vertices, the white vertices induce a connected graph.

Consider a fence gadget with at least two white vertices. If there is a singleton white vertex , we color it black and color white a black neighbor  of one of the other white vertices. In the exchange, if  and  are not neighbors, at most three black edges incident with  are lost and at least three black edges incident with  are gained. If  and  are neighbors, at most two black edges are lost and at least two black edges are gained. In the following, assume that there are no singleton white vertices. If there are exactly three white vertices that do not induce a connected graph, then there is at least one singleton white vertex. Hence, we may assume that there are at least four white vertices. We remove each maximal connected component  with at most two white vertices: Observe that if both inner vertices are white, the white vertices always induce a connected graph. Thus, we may assume that there is only one inner white vertex and, by Property 1,  has no inner vertices. We now know that if the white vertices do not induce a connected graph, there are at least four of them, the minimum size of a white connected component is two, each white connected component of size exactly two contains no inner vertices, and there is at most one white inner vertex. Without loss of generality, there are only two possible configurations left. Either the vertices  or the vertices  are white. In case one, we can make  black and  white, losing at most two black edges and gaining at least two black edges. In case two, we can color  black and  white. Property 2 now follows.

Notice that ensuring either property does not change the number of black vertices in the gadget. Furthermore, since ensuring Property 1 does not destroy Property 2, and since Property 1 does not depend on Property 2, we can modify the solution  such that both statements hold simultaneously.

To prove Part 1 through 4 of the claim, we iterate through the the number  of white vertices in a gadget, and show that each of the relevant statements hold; no further modifications are necessary. Let  be a gadget with exactly  white vertices.

• The claim is trivial for .

• Assume that . Both white vertices of  have at least three incident non-black edges. The neighborhood of the white vertices overlaps in at most one vertex, and, thus, we have at least  non-black edges, matching the given bound of Part 1 of the claim. Part 2 of the claim follows by choosing two black vertices  in the neighborhood of the white vertices such that  are neighbors. It is easily seen that such black vertices must exist.

• Assume that .

• If both inner vertices of  are white, these two vertices alone have seven incident non-black edges and Part 1 of the claim follows. For Part 2 and , we choose three consecutive black vertices on the cycle formed by the outer vertices. Each of these vertices has at most three incident black edges and the neighborhoods overlap in at most one vertex for neighboring vertices. Thus, these three vertices have at most  incident black edges. If we are to choose only two black vertices, that is, , we elect the neighbors of the outer white vertex. Both these black vertices have black-degree at most two. The case  is trivial and Part 2 of the claim follows.

• If exactly one inner vertex is white, then, by Property 1, it has two white outer neighbors. Without loss of generality, let the white vertices be . These three vertices have at least seven incident non-black edges and the first part of the claim follows. Choosing the black vertices , , and  yields at most , , and  incident black edges, respectively. Thus, Part 2 of the claim follows.

• Now, if all three white vertices are outer vertices, by Property 2, they form a connected component. Without loss of generality, there are only two possible configurations: , and , each yielding at least seven non-black edges, fulfilling the first part of the claim. For Part 2, we choose , , and , yielding at most , , and  incident black edges in the first configuration. For the second configuration, we choose , , and , yielding at most , , and  incident black edges and, thus, Part 2 of the claim holds.

• Now assume that .

• If both inner vertices are white, they yield at least seven non-black vertices plus at least two from one of the outer white vertices, satisfying the lower bound of Part 3 of the claim. For the upper bound on the number of black edges incident with some black vertices in Part 3, observe that there is a black vertex with black-degree at most two (incident with one of the outer white vertices) and deleting it makes its black neighbor (if there is any) have black-degree at most two. For the upper bound from Part 2 of our claim, analogously to , we choose three consecutive black vertices or, if this is not possible, neighbors of the white outer vertices. It is easy to check that Part 2 of the claim holds in this case.

• Now assume that exactly one inner vertex is white. Then, by Property 1, it has at least two white outer neighbors. Without loss of generality, the only possible configurations are and . Each configuration yields at least eight non-black edges. Since there are exactly  edges in the gadget and at most three black vertices with edges, the upper bound of Part 3 of our claim follows. For the upper bound of Part 2, we choose , , and  in configuration one, yielding at most , , and  black edges, and , , and , yielding at most , , and  edges.

• Assume that every white vertex is an outer vertex. Then, by Property 2, they induce a connected graph. Thus, without loss of generality, the only possible configurations are  through  and  through . The lower bound on the number of non-black edges in Part 3 follows, since the vertices  are incident with seven edges in the gadget and both  and  contribute at least one further non-black edge. The upper bound on black edges from Part 3 follows, since there are at most  black edges within the gadget and at most two black edges not within the gadget incident to  and  or , respectively. For Part 2 we may choose the vertices , , and , yielding at most , , and  incident black edges in both configurations.

• It remains to prove Part 2 and Part 4 of our claim for . The only nontrivial case is , that is, there are exactly three black vertices. Both Part 2 and Part 4 trivially follow, if the black vertices are not connected. If they are connected, they either form a triangle or a path. In the first case, at most two black vertices are incident with non-gadget edges, and, thus, the remaining vertex has black-degree at most two and the black vertices have at most five incident black edges. In the second case, there are two black vertices with black-degree at most two and the black vertices have incident at most five black edges.

Thus, our claim follows, and our construction is correct. ∎

We now use the same strategy, albeit with a much simpler proof, replacing the vertices by a further gadget to further reduce the maximum degree in the input graph.

###### Lemma 2.3.

There is a polynomial-time many-one reduction from Densest-Subgraph in graphs with maximum degree four to Densest-Subgraph in graphs with maximum degree three.

###### Proof.

Let  be an instance of Densest-Subgraph where each vertex in  has degree at most four. We replace each vertex  in the graph  by a four-vertex cycle (the “cycle gadget”), and distribute ’s edges to the vertices of the cycle such that each vertex in the cycle gets degree at most three. Call the graph constructed in this way . We set the new instance of Densest-Subgraph to . We show 55todo: 5MS: Was spricht gegen argue?that this construction is correct by showing that from any solution to  we can construct a solution with at least as many edges such that the vertices of each cycle gadget are either completely contained in the solution or are disjoint. Thus, from any solution to  with  edges, we can construct a solution for  with at least  edges and vice-versa. It is clear, that solutions for  with at least  edges are achievable in polynomial time, and it will also not be hard to check that we construct the solution for  from a solution for  in polynomial time. Thus, finding an optimal solution or any solution with a number of edges above some given threshold is polynomial-time equivalent in these instances.

Combining the three lemmas of this section, the following is now easily obtained.

###### Theorem 2.1.

Densest-Subgraph is NP-hard even on graphs with maximum degree three.

## 3 Conclusion and Outlook

We have demonstrated that stubbornness not only may lead to the proof of a desired result. It also can be used to obtain much more complicated proofs of previously known ones. We leave it as an open question as to which elegant proofs can be reproved more elaborately and in a nontrivial way.

### Acknowledgement.

I thank Christian Komusiewicz for insightful discussions about the matter of the paper and for proofreading.

## References

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• [2] R. Diestel. Graph Theory, volume 173 of Graduate Texts in Mathematics. Springer Verlag, New York, 4th edition, 2010.
• [3] K. Efe and G.-L. Feng. A proof for bisection width of grids. International Journal of Mathematical and Computer Sciences, 4(3), 2008.
• [4] U. Feige, D. Peleg, and G. Kortsarz. The dense k-subgraph problem. Algorithmica, 29(3):410–421, 2001.
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