Finding Connected-Dense-Connected (CDC) subgraphs from Triple Networks is NP-Hard. finding One-Connected-Dense (OCD) sub- graphs from Triple Networks is also NP-Hard. We present formal proofs of these theorems hereby

Prasad, Sushil

Shah, Dhara

Wu, Yubao

English

Georgia State University

Georgia State University
ScholarWorks @ Georgia State University
Computer Science Technical Reports Department of Computer Science
Spring 4-3-2019
Finding Connected-Dense-Connected Subgraphs
and variants is NP-Hard
Dhara Shah
dshah8@student.gsu.edu
Sushil Prasad
sprasad@gsu.edu
Yubao Wu
ywu28@gsu.edu
Follow this and additional works at: https://scholarworks.gsu.edu/
computer_science_technicalreports
Part of the Theory and Algorithms Commons
This Working Paper is brought to you for free and open access by the Department of Computer Science at ScholarWorks @ Georgia State University. It
has been accepted for inclusion in Computer Science Technical Reports by an authorized administrator of ScholarWorks @ Georgia State University.
For more information, please contact scholarworks@gsu.edu.
Recommended Citation
Shah, Dhara; Prasad, Sushil; and Wu, Yubao, "Finding Connected-Dense-Connected Subgraphs and variants is NP-Hard" (2019).
Computer Science Technical Reports. 2.
https://scholarworks.gsu.edu/computer_science_technicalreports/2
finding Connected-Dense-Connected Subgraphs
and variants is NP-Hard
Dhara Shah1, Yubao Wu1, and Sushil Prasad1
Department of Computer Science, Georgia State University, Atlanta 30303, USA
{dshah8, ywu28, sprasad}@gsu.edu
Abstract. Finding Connected-Dense-Connected (CDC) subgraphs from
Triple Networks is NP-Hard. finding One-Connected-Dense (OCD) sub-
graphs from Triple Networks is also NP-Hard. We present formal proofs
of these theorems hereby.
Keywords: Triple Networks · Connected-Dense-Connected subgraphs · One-
Connected-Dense subgraphs · NP-Hard
Theorem 1. Finding a CDC subgraph in a Triple Network is NP Hard.
Proof. We prove that finding a CDC subgraph is a reduction of set-cover prob-
lem. Let R = {r1, · · · , rp} be a set and and C = {C1, · · · , Cq} be its cover
with R = ∪qi=1Ci. The aim of this set cover problem is to find minimum subset
Copt ⊂ C, known as optimal set-cover, such that each rj ∈ R belongs to at least
one set of Copt. This problem is proved to be NP complete.
Let T = {t1, · · · , tp} be a set of points, having the same cardinality as R.
Let D = {D1, · · · , Dq} be a set-cover of T , analogous to C, such that if ri ∈ Cj ,
then ti ∈ Dj . Hence, T,D can be considered as a copy of R,C.
We construct the Triple Network as follows. Let Va = {h, r1, · · · , rp, C1, · · · , Cq},
where node h is connected to every Ci ∈ C and node ri is connected to node Cj
if ri ∈ Cj in the set-cover problem. Similarly, let Vb = {k, t1 · · · , tp, D1, · · · , Dq}
be the analogous set to Va. We connect Va and Vb by connecting all nodes
{r1, · · · , rp, h} to all nodes {t1, · · · , tp, k}.
Construction of this Triple Network is illustrated in figure 1 from an instance
of set-cover problem C1 = {r1, r2}, C2 = {r1}, C3 = {r2, r4}, C4 = {r2, r3}, C5 =
{r4}.
Let Copt ⊂ C be an optimal solution to the set-cover problem of C and
|Copt| = q∗ ≤ q. Similarly, let Dopt be the analogous optimal solution to D and
|Dopt| = q∗ ≤ q. Let H = {h, r1, · · · , rp} and J = {k, t1, · · · , tp}. The subgraph
induced by Sa = H∪Copt is connected in Va, and similarly, the subgraph induced
by Sb = J ∪ Dopt is connected in Vb. Hence, the sub Triple Network G[Sa, Sb]
has density ρ(Sa, Sb) =
(p+1)2
(p+q∗+1) .
Let S1 and S2 be any nonempty node sets where Ga[S1] and Gb[S2] are
connected. In general, S1 = H
′ ∪ C ′ where H ′ ⊂ H and C ′ ⊂ C. Similarly,
S2 = J
′ ∪ D′ where J ′ ⊂ J and D′ ⊂ D. We show that ρ(S1, S2) ≤ ρ(Sa, Sb),
2 Shah et al.
Fig. 1. Triple Network from set-cover
making G[Sa, Sb] a CDC subgraph. Let |H ′| = p1, |C ′| = q1, |J ′| = p2 and
|D′| = q2. Hence, ρ(S1, S2) = p1p2√
(p1+q1)(p2+q2)
.
First, we consider the case when S1 contains all the nodes of H and S2
contains all the nodes of J . In this case, p1 = p2 = p+ 1. Also, by definition of
optimal set-cover, q∗ ≤ q1 and q∗ ≤ q2. Hence, ρ(S1, S2) = (p+1)
2√
(p+q1+1)(p+q2+1)
≤
(p+1)2
(p+q∗+1) = ρ(Sa, Sb).
Second, we consider the case when S1 contains a subset of nodes H
′ ⊂ H. In
this case, we first show that adding elements from H \H ′ to S1 will only increase
its density.
If h 6∈ S1, then after adding h to S1, the resulting subgraph has density
(p1+1)p2√
(p1+q1+1)(p2+q2)
> p1p2√
(p1+q1)(p2+q2)
= ρ(S1, S2). This subgraph is also connected
in Ga, since h is connected to every Ci ∈ C. To add a node rj ∈ H \ H ′ and
making it still connected, we need to add at most one node Ci to C
′ with
rj ∈ Ci. Hence, the density of this resulting subgraph is (p1+1)p2√
(p1+q1+2)(p2+q2)
>
p1p2√
(p1+q1)(p2+q2)
= ρ(S1, S2). We can repeat this process by adding remaining
nodes of H \H ′ to S1, while density of the resulting subgraphs keeps increasing.
Similarly, adding elements from J \J ′ to S2 increases density of the resulting
subgraphs. Since we proved in the first case that the density ρ(S1, S2) when
H ⊂ S1 and J ⊂ S2, we have hence completed the proof of the second case.
In summary, we proved that for any nonempty sets S1 ⊂ Va and S2 ⊂ Vb,
ρ(S1, S2) ≤ ρ(Sa, Sb), making G[Sa, Sb] a CDC subgraph. Also,G[Sa, Sb] is the
solution inducted by optimal set covers, an instance being Sa = {r1, r2, r3, r4, h,
C1, C3, C4} and Sb = {s1, s2, s3, s4, k,D1, D3, D4} hence proving that finding a
CDC subgraph is NP hard.
Lemma 1. Finding OCD subgraph in triple network is NP hard
Proof. We prove that finding OCD subgraph is also reduction of the set cover
problem. We first construct the triple network same as in theorem 1. Let Sa =
H and Sb = J ∪ Dopt. The subgraph G[sa, Sb] hence has density ρ(Sa, Sb) =
finding Connected-Dense-Connected Subgraphs and variants is NP-Hard 3
(p+1)2√
(p+1)(p+q∗+1)
We claim that G[Sa, Sb] is an OCD subgraph. We observe that
G[Sb] is connected.
Let S1 and S2 be any nonempty node sets where either G[S1] or G[S2] is
connected. In general, S1 = H
′ ∪ C ′ where H ′ ⊂ H. Similarly, S2 = J ′ ∪ D′
where J ′ ∪ J . We show that ρ(S1, S2) ≤ ρ(Sa, sb).
First, we consider the case when S1 contains all the nodes of H and S2
contains all the nodes of J . In this case, p1 = p2 = p+ 1. Also, by definition of
optimal set-cover, q∗ ≤ q1 and q∗ ≤ q2. Hence, ρ(S1, S2) = (p+1)
2√
(p+q1+1)(p+q2+1)
≤
(p+1)2√
(p+q∗+1)(p+1)
= ρ(Sa, Sb).
Second, we consider the case when S1 contains a subset of nodes H
′ ⊂ H.
In this case, we first show that adding elements from H \ H ′ to S1 will only
increase its density. Suppose, Ga[S1] is not connected and Gb[S2] is connected.
Then, after adding element from H \ H ′, the resulting subgraph has density
(p1+1)p2√
(p1+q1)(p2+q2)
> p1p2(p1+q1)(p2+q2) = ρ(S1, S2). This includes adding h to S1 if
h 6∈ H ′, making resultant subgraph connected in Va. Now suppose Ga[S1] is
connected. Then, following the same case of theorem 1, we first add h if it is
not in H ′ and then add element from H \H ′ and still show that the resultant
subgraph is connected in Va and its density increases. Similarly, we conclude
that when S2 contains a subset of nodes in J
′ ⊂ J , adding elements from J ′ \ J
also increases the density of the resultant subgraph.
At last, we observe that if Ga[S2] is connected, then the resultant sub-
graph obtained by removing elements from C ′ has density p1p2√
(p1+q1−1)(p2+q2)
>
ρ(S1, S2).
In summary, we have proved that for any nonempty sets S1 ⊂ Va and S2 ⊂ Vb
with either Ga[S1] or Gb[S2] connected has density ρ(S1, S2) ≤ ρ(Sa, Sb), making
G[Sa, Sb] an OCD subgraph. Also, G[Sa, Sb] is the solution induced by optimal
set cover, an instance being Sa = {r1, r2, r3, r4, h}, Sb = {s1, s2, s3, s4, k,D1, D3, D4}
hence proving that finding OCD subgraphs is NP hard.


Finding Connected-Dense-Connected Subgraphs and variants is NP-Hard

ScholarWorks @ Georgia State University

Georgia State UniversityScholarWorks @ Georgia State UniversityComputer Science Technical Reports Department of Computer ScienceSpring 4-3-2019Finding Connected-Dense-Connected Subgraphsand variants is NP-HardDhara Shahdshah8@student.gsu.eduSushil Prasadsprasad@gsu.eduYubao Wuywu28@gsu.eduFollow this and additional works at: https://scholarworks.gsu.edu/computer_science_technicalreportsPart of the Theory and Algorithms CommonsThis Working Paper is brought to you for free and open access by the Department of Computer Science at ScholarWorks @ Georgia State University. Ithas been accepted for inclusion in Computer Science Technical Reports by an authorized administrator of ScholarWorks @ Georgia State University.For more information, please contact scholarworks@gsu.edu.Recommended CitationShah, Dhara; Prasad, Sushil; and Wu, Yubao, "Finding Connected-Dense-Connected Subgraphs and variants is NP-Hard" (2019).Computer Science Technical Reports. 2.https://scholarworks.gsu.edu/computer_science_technicalreports/2finding Connected-Dense-Connected Subgraphsand variants is NP-HardDhara Shah1, Yubao Wu1, and Sushil Prasad1Department of Computer Science, Georgia State University, Atlanta 30303, USA{dshah8, ywu28, sprasad}@gsu.eduAbstract. Finding Connected-Dense-Connected (CDC) subgraphs fromTriple Networks is NP-Hard. finding One-Connected-Dense (OCD) sub-graphs from Triple Networks is also NP-Hard. We present formal proofsof these theorems hereby.Keywords: Triple Networks · Connected-Dense-Connected subgraphs · One-Connected-Dense subgraphs · NP-HardTheorem 1. Finding a CDC subgraph in a Triple Network is NP Hard.Proof. We prove that finding a CDC subgraph is a reduction of set-cover prob-lem. Let R = {r1, · · · , rp} be a set and and C = {C1, · · · , Cq} be its coverwith R = ∪qi=1Ci. The aim of this set cover problem is to find minimum subsetCopt ⊂ C, known as optimal set-cover, such that each rj ∈ R belongs to at leastone set of Copt. This problem is proved to be NP complete.Let T = {t1, · · · , tp} be a set of points, having the same cardinality as R.Let D = {D1, · · · , Dq} be a set-cover of T , analogous to C, such that if ri ∈ Cj ,then ti ∈ Dj . Hence, T,D can be considered as a copy of R,C.We construct the Triple Network as follows. Let Va = {h, r1, · · · , rp, C1, · · · , Cq},where node h is connected to every Ci ∈ C and node ri is connected to node Cjif ri ∈ Cj in the set-cover problem. Similarly, let Vb = {k, t1 · · · , tp, D1, · · · , Dq}be the analogous set to Va. We connect Va and Vb by connecting all nodes{r1, · · · , rp, h} to all nodes {t1, · · · , tp, k}.Construction of this Triple Network is illustrated in figure 1 from an instanceof set-cover problem C1 = {r1, r2}, C2 = {r1}, C3 = {r2, r4}, C4 = {r2, r3}, C5 ={r4}.Let Copt ⊂ C be an optimal solution to the set-cover problem of C and|Copt| = q∗ ≤ q. Similarly, let Dopt be the analogous optimal solution to D and|Dopt| = q∗ ≤ q. Let H = {h, r1, · · · , rp} and J = {k, t1, · · · , tp}. The subgraphinduced by Sa = H∪Copt is connected in Va, and similarly, the subgraph inducedby Sb = J ∪ Dopt is connected in Vb. Hence, the sub Triple Network G[Sa, Sb]has density ρ(Sa, Sb) =(p+1)2(p+q∗+1) .Let S1 and S2 be any nonempty node sets where Ga[S1] and Gb[S2] areconnected. In general, S1 = H′ ∪ C ′ where H ′ ⊂ H and C ′ ⊂ C. Similarly,S2 = J′ ∪ D′ where J ′ ⊂ J and D′ ⊂ D. We show that ρ(S1, S2) ≤ ρ(Sa, Sb),2 Shah et al.Fig. 1. Triple Network from set-covermaking G[Sa, Sb] a CDC subgraph. Let |H ′| = p1, |C ′| = q1, |J ′| = p2 and|D′| = q2. Hence, ρ(S1, S2) = p1p2√(p1+q1)(p2+q2).First, we consider the case when S1 contains all the nodes of H and S2contains all the nodes of J . In this case, p1 = p2 = p+ 1. Also, by definition ofoptimal set-cover, q∗ ≤ q1 and q∗ ≤ q2. Hence, ρ(S1, S2) = (p+1)2√(p+q1+1)(p+q2+1)≤(p+1)2(p+q∗+1) = ρ(Sa, Sb).Second, we consider the case when S1 contains a subset of nodes H′ ⊂ H. Inthis case, we first show that adding elements from H \H ′ to S1 will only increaseits density.If h 6∈ S1, then after adding h to S1, the resulting subgraph has density(p1+1)p2√(p1+q1+1)(p2+q2)> p1p2√(p1+q1)(p2+q2)= ρ(S1, S2). This subgraph is also connectedin Ga, since h is connected to every Ci ∈ C. To add a node rj ∈ H \ H ′ andmaking it still connected, we need to add at most one node Ci to C′ withrj ∈ Ci. Hence, the density of this resulting subgraph is (p1+1)p2√(p1+q1+2)(p2+q2)>p1p2√(p1+q1)(p2+q2)= ρ(S1, S2). We can repeat this process by adding remainingnodes of H \H ′ to S1, while density of the resulting subgraphs keeps increasing.Similarly, adding elements from J \J ′ to S2 increases density of the resultingsubgraphs. Since we proved in the first case that the density ρ(S1, S2) whenH ⊂ S1 and J ⊂ S2, we have hence completed the proof of the second case.In summary, we proved that for any nonempty sets S1 ⊂ Va and S2 ⊂ Vb,ρ(S1, S2) ≤ ρ(Sa, Sb), making G[Sa, Sb] a CDC subgraph. Also,G[Sa, Sb] is thesolution inducted by optimal set covers, an instance being Sa = {r1, r2, r3, r4, h,C1, C3, C4} and Sb = {s1, s2, s3, s4, k,D1, D3, D4} hence proving that finding aCDC subgraph is NP hard.Lemma 1. Finding OCD subgraph in triple network is NP hardProof. We prove that finding OCD subgraph is also reduction of the set coverproblem. We first construct the triple network same as in theorem 1. Let Sa =H and Sb = J ∪ Dopt. The subgraph G[sa, Sb] hence has density ρ(Sa, Sb) =finding Connected-Dense-Connected Subgraphs and variants is NP-Hard 3(p+1)2√(p+1)(p+q∗+1)We claim that G[Sa, Sb] is an OCD subgraph. We observe thatG[Sb] is connected.Let S1 and S2 be any nonempty node sets where either G[S1] or G[S2] isconnected. In general, S1 = H′ ∪ C ′ where H ′ ⊂ H. Similarly, S2 = J ′ ∪ D′where J ′ ∪ J . We show that ρ(S1, S2) ≤ ρ(Sa, sb).First, we consider the case when S1 contains all the nodes of H and S2contains all the nodes of J . In this case, p1 = p2 = p+ 1. Also, by definition ofoptimal set-cover, q∗ ≤ q1 and q∗ ≤ q2. Hence, ρ(S1, S2) = (p+1)2√(p+q1+1)(p+q2+1)≤(p+1)2√(p+q∗+1)(p+1)= ρ(Sa, Sb).Second, we consider the case when S1 contains a subset of nodes H′ ⊂ H.In this case, we first show that adding elements from H \ H ′ to S1 will onlyincrease its density. Suppose, Ga[S1] is not connected and Gb[S2] is connected.Then, after adding element from H \ H ′, the resulting subgraph has density(p1+1)p2√(p1+q1)(p2+q2)> p1p2(p1+q1)(p2+q2) = ρ(S1, S2). This includes adding h to S1 ifh 6∈ H ′, making resultant subgraph connected in Va. Now suppose Ga[S1] isconnected. Then, following the same case of theorem 1, we first add h if it isnot in H ′ and then add element from H \H ′ and still show that the resultantsubgraph is connected in Va and its density increases. Similarly, we concludethat when S2 contains a subset of nodes in J′ ⊂ J , adding elements from J ′ \ Jalso increases the density of the resultant subgraph.At last, we observe that if Ga[S2] is connected, then the resultant sub-graph obtained by removing elements from C ′ has density p1p2√(p1+q1−1)(p2+q2)>ρ(S1, S2).In summary, we have proved that for any nonempty sets S1 ⊂ Va and S2 ⊂ Vbwith either Ga[S1] or Gb[S2] connected has density ρ(S1, S2) ≤ ρ(Sa, Sb), makingG[Sa, Sb] an OCD subgraph. Also, G[Sa, Sb] is the solution induced by optimalset cover, an instance being Sa = {r1, r2, r3, r4, h}, Sb = {s1, s2, s3, s4, k,D1, D3, D4}hence proving that finding OCD subgraphs is NP hard.

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Finding Connected-Dense-Connected Subgraphs and variants is NP-Hard

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