We provide next several examples of well-conditioned α() functions. In the first section, we focus on functions concerning the r-Set Packing with α()-Overlap problem that by Transformation 15 could be used to restrict the overlap for the graph version as well. After that in Section9.3.2, we provide functions that consider properties exclusively from graphs.
9.3.1
Restricting the Overlap Between Sets
Constant Overlap. Upper-bounding the overlap size by a constant t (as in the r-Set Packing with t-Overlap problem) where 0 ≤ t ≤ r − 1 is a well-conditioned α() function. More precisely, let α- Constant(si, sj) be a function that returns 0 (no-conflict ) if |si∩ sj| ≤ t, otherwise returns 1 (α-conflict ).
Function α-Constant(si, sj)
Result =no-conflict
if |si∩ sj| > t then Result =α-conflict end if
Return Result EndFunction
Lemma 77. The function α-Constant is well-conditioned.
Proof. (i) α-Constant is hereditary. For any pair of sets si,sj with α(si, sj) = 0, |si∩ sj| ≤ t. Thus, there
cannot be a pair of subsets s0i⊆ si, s0j⊆ sj with |s0i∩ s0j| > t.
(ii) Since t ≥ 0, for any pair si, sj with α(si, sj) = 1, |si∩ sj| ≥ t + 1 ≥ 1. Let s0i⊆ siand s0j ⊆ sj be a
pair of subsets with α(s0i, s0j) = 0. Note that at most one s0i= sj or s0j= sj; otherwise α(s0i, s0j) = 1. Since
α(s0i, s0j) = 0, |s0i∩ s0
j| ≤ t. Therefore, ((si∩ sj)\(s0i∩ s0j)) 6= ∅.
Weighted Overlap. Let us assume that each u ∈ U has associated a non-negative weight w(u). We could restrict the overlap region by its weight. The function α-Weight(si, sj) returns no-conflict if w(si∩ sj) =
(P
u∈(si∩sj)w(u)) ≤ wtwhere wt≥ 0 is a constant, else returns α-conflict.
Function α-Weight(si, sj)
Result =no-conflict
if w(si∩ sj) = (Pu∈si∩sjw(u)) > wtthen Result =α-conflict end if
Return Result EndFunction
Lemma 78. The function α-Weight is well-conditioned.
Proof. (i) α-Weight is hereditary. For any pair of sets si,sj with w(si∩ sj) ≤ wt(α(si, sj) = 0), there is no
pair of subsets s0
i⊆ si, s0j⊆ sj with w(si0∩ s0j) > wt(α(si, sj) = 1). For the sake of contradiction, suppose
otherwise. Notice that (s0i∩ s0j) ⊆ (si∩ sj). Thus, if w(si∩ sj) ≤ wtbut w(s0i∩ s0j) > wt, then there must
be some elements in (si∩ sj)\(s0i∩ s0j) with negative weights, a contradiction because the weights in U are
non-negative.
(ii) If α(si, sj) = 1, then w(si∩ sj) > wt and (si∩ sj) 6= ∅. Let s0i ⊆ si and s0j ⊆ sj be any pair of
subsets with α(s0i, s0j) = 0, (i.e., w(s0i∩ s0
j) ≤ wt). Note that at most one s0i = sj or s0j = sj; otherwise
α(s0i, s0j) = 1. Since w(si∩ sj) > wt, w(si∩ sj) − w(s0i∩ s0j) > 0. Therefore, ((si∩ sj)\(s0i∩ s0j)) 6= ∅.
(iii) Finally, we can determine in O(r) time, if w(si∩ sj) > wt.
Notice that we could use α-Weight(si, sj) to upper-bound the overlap size by a constant t. To this end,
we make w(u) = 1 for each u ∈ U , and wt = t. However, we have defined these functions separately to
combine them into a well-conditioned α() function that restricts the overlap by size and weight. Function α-ConstantAndWeight(si, sj)
Result =no-conflict
if |si∩ sj| > t OR w(si∩ sj) > wt then Result =α-conflict end if
Return Result EndFunction
Lemma 79. The function α-ConstantAndWeight is well-conditioned.
Measures Overlap. A measure of a set S is a function µ that satisfies the following conditions: i. (non- negative) µ(S) ≥ 0, ii. (zero for the empty set) µ(S) = 0 if S = ∅, and iii. (additive) µ(S) = µ(S1) + µ(S2)
for any disjoint sets of S. The last property also implies that for any S0 ⊆ S, µ(S0) ≤ µ(S). Let µ be a
measure on each set si∩ sj, and t ≥ 0 a constant value. The function α-Measure(si, sj) returns no-conflict
if µ(si∩ sj) ≤ t otherwise returns α-conflict.
Function α-Measure(si, sj)
Result =no-conflict
if µ(si∩ sj) > t then Result =α-conflict end if
Return Result EndFunction
Proof. (i) α-Measure is hereditary. Assume for contradiction that µ(si∩ sj) ≤ t but there is a pair of
subsets s0i ⊆ si, s0j ⊆ sj with µ(s0i∩ s0j) > t. Let S = (s0i∩ s0j). First S 6= ∅, otherwise, µ(S) = 0 and
since t ≥ 0 there would be a contradiction. Second S ⊆ (si∩ sj), thus by the additive property of µ,
µ(S) ≤ µ(si∩ sj). Since µ(si∩ sj) ≤ t, the claim holds.
(ii) If µ(si∩ sj) > t, then |si∩ sj| ≥ 1. This follows because µ(∅) = 0 and t ≥ 0. Let s0i⊆ siand s0j ⊆ sj
be a pair of subsets with α(s0i, s0j) = 0, (i.e., µ(s0i∩ s0
j) ≤ t). Note that at most one s0i = si or s0j = sj;
otherwise α(s0i, s0j) = 1. Since µ(si∩ sj) > t, µ(si∩ sj) − µ(s0i∩ s0j) > 0. In this way, ((si∩ sj)\(s0i∩ s0j)) 6= ∅.
(iii) The function µ is computed in polynomial time; thus, we can verify in constant time whether µ(si∩ sj) > t or not.
Metric Overlap. Let us assume that U is a metric space. That is, there is a metric or a distance function that defines a distance between each pair of elements u, v of U , subject to the following conditions: distU(u, v) ≥ 0, distU(u, v) = 0 if (u = v), distU(u, v) = distU(v, u) and distU(u, w) ≤ distU(u, v) +
distU(v, w). For a constant dt > 0, we define the function α-Metric(si, sj) which returns no-conflict if
|si∩ sj| ≤ 1 or distU(u, v) ≤ dt for each pair u, v (u 6= v) in si∩ sj, else it returns α-conflict.
Function α-Metric(si, sj)
if |si∩ sj| ≤ 1 OR (distU(u, v) ≤ dtfor each pair u, v in (si∩ sj), u 6= v) then
Result =no-conflict else Result =α-conflict end if Return Result EndFunction
Lemma 81. The function α-Metric is well-conditioned.
Proof. (i) α-Metric is hereditary. For any pair of sets si,sj that does not α-conflict (i.e., α(si, sj) = 0),
there is no pair of subsets s0i ⊆ si, s0j ⊆ sj with α(s0i, s0j) = 1. Assume the opposite by contradiction.
|s0
i∩ s0j| ≥ 1, otherwise s0iand s0j would not α-conflict. Observe that (s0i∩ s0j) ⊆ (si∩ sj). In addition, since
U is a metric-space and α(si, sj) = 0, distU(u, v) ≤ t for each pair u, v (u 6= v) in (si∩ sj). Given that we
are using distU and not dists0
i∩s0j, there is no pair of elements in (s
0
i∩ s0j) with distU(u, v) > t.
(ii) Since dt > 0, for any pair si, sj with α(si, sj) = 1, |si∩ sj| > 1. Let s0i ⊆ si and s0j ⊆ sj be a
pair of subsets with α(s0i, s0j) = 0. Note that at most one s0i = si or s0j = sj; otherwise α(s0i, s0j) = 1.
distU(u, v) ≤ t for each pair u, v (u 6= v) in (s0i∩ s0j). Since α(si, sj) = 1 but α(s0i, s0j) = 0, then it must
exists at least one element u in ((si∩ sj)\(s0i∩ s0j)) such that distU(u, v) > t for some v in (si∩ sj). In
this way, ((si∩ sj)\(s0i∩ s0j)) 6= ∅.
(iii) Assuming as input a metric space U , α-Metric is verified in O(r2) time.
9.3.2
Restricting the Overlap Between Subgraphs
Prescribed Pattern. It has been observed in social networks that the overlap region is often more densely connected than the rest of the community [138]. Inspired by this, we will allow pairwise-overlap in
a (k, α())-Π-packing, if the overlap region has a specific pattern, for example, it is a clique. More precisely, we say that a pair of Π-subgraphs Hi,Hj does not α-conflict, if G[V (Hi) ∩ V (Hj)] is isomorphic to a graph
F in a class F . To define a well-conditioned α(), F is a graph class that is hereditary (i.e., it is closed under taking induced subgraphs). To preserve our FPT results, any graph in F should be polynomial time verifiable. Examples of F are cliques, planar and chordal graphs. Indeed this applies to any graph class that is closed under minors, since this is hereditary and by the Robertson-Seymour theorem the graph is polynomially testable by checking for the forbidden minors [125]. We define the function α-Pattern(si, sj)
that returns no-conflict if |si∩ sj| = 0 or if G[si∩ sj] is isomorphic to a graph F in F ; otherwise, it returns
α-conflict.
Function α-Pattern(si, sj)
if |si∩ sj| = 0 OR G[si∩ sj] is isomorphic to some F ∈ F then
Return no-conflict else
Return α-conflict end if
EndFunction
Lemma 82. The function α-Pattern is well-conditioned.
Proof. (i) α-Pattern is hereditary. Assume for contradiction that there is pair si,sj with α(si, sj) = 0 but
there is a pair of subsets s0i ⊆ si and s0j ⊆ sjwith α(s0i, s0j) = 1. If α(s0i, s0j) = 1, this implies that G[s0i∩ s0j]
is not isomorphic to a graph F in F . Notice that (s0
i∩ s0j) ⊆ (si∩ sj). In addition, G[si∩ sj] is isomorphic
to a graph F ∈ F (otherwise, α(si, sj) = 1). Since F is a graph class that is hereditary, G[s0i∩ s0j] is also
isomorphic to F and s0i and s0j do not α-conflict.
(ii) It follows by definition of α-Pattern(si, sj) that for any pair si, sj that α-conflict (i.e., α(si, sj) = 1)
|si∩ sj| ≥ 1. Let s0i⊆ siand s0j ⊆ sjbe a pair of non-empty subsets where α(s0i, s0j) = 0. This implies that
G[s0i∩s0
j] is isomorphic to a graph F ∈ F . Recall that (s0i∩s0j) ⊆ (si∩sj). Since G[si∩sj] is not isomorphic
to a graph in F but G[s0i∩ s0
j] is and F is closed under taking induced subgraphs, ((si∩ sj)\(s0i∩ s0j)) 6= ∅.
(iii) Finally, α-Pattern is computed in polynomial time as it is a constraint of the class F .
Distance. In [139], overlapping nodes occur only in the boundary regions of overlapping communities in sensor networks. Motivated by this, we consider in the overlap region only nodes that are “closer” to each other. In this way, two Π-subgraphs Hi,Hj do not α-conflict if the distance in G between any
pair of vertices u, v in V (Hi) ∩ V (Hj) is at most a constant dt> 0, i.e., distG(u, v) ≤ dt. Recall that a
Π-subgraph Hiis represented by a set Si= V (Hi) in S. Since the graph distance is a metric on V (G), we
use the function α-Metric defined previously (Lemma81). Notice that using distG[Hi∩Hj] instead is not a
well-conditioned α() function. Specifically, it does not follow the hereditary property.
Property. There are several vertex properties that are relevant in the analysis of real networks: vertex strength [19,110], vertex weight [99], and disparity [110], among others. On the other hand, weights in the edges of a network may better reflect the strength of a relationship in social and communication networks [70]. Moreover, Onnela et al., [118] suggest that edges in the overlapping regions have large weights.
Hence, we suggest considering only overlapping nodes or edges that present the same property ξ (or properties). For example, overlapping nodes or edges each with at least a specific strength or weight. We assume however that the properties values for each vertex or edge are given as part of the input.
Recall that by Transformation15, we have a set S = V (H) per each induced Π-subgraph H in G. To consider edges properties in the overlap region, we would need to slightly modify that transformation and make U = E(G) and S = E(H) instead. This only increases the size of the sets in S by a factor of r and |S| = O(nr2).
We define α-Property(si, sj) which simply returns no-conflict either if |si∩ sj| = 0 or if each element
u in (si∩ sj) satisfies ξ. Otherwise, it returns α-conflict.
Function α-Property(Si, Sj)
if |Si∩ Sj| = 0 OR (for each element u in (Si∩ Sj) u satisfies ξ ) then
Return no-conflict else
Return α-conflict end if
EndFunction
Lemma 83. The function α-Property is well-conditioned.
Proof. (i) α is hereditary. Assume by contradiction that there is pair si,sj with α(si, sj) = 0 but there is
a pair of subsets α(s0i, s0j) = 1 where s0i⊆ si and s0j⊆ sj. If α(si, sj) = 0, every element in si∩ sj satisfies
the property ξ. Since (s0
i∩ s0j) ⊆ (si∩ sj), every element in s0i∩ s0j satisfies ξ as well.
(ii) By definition of α-Property any pair of disjoint sets do not α-conflict. Let s0i⊆ si and s0j ⊆ sj be a
pair of subsets with α(s0i, s0j) = 0. If α(si, sj) = 1 but α(s0i, s0j) = 0, there must exists at least one element
in ((si∩ sj)\(s0i∩ s0j)) that does not follow ξ and therefore ((si∩ sj)\(s0i∩ s0j)) 6= ∅.
(iii) The property ξ for each element of U is given as part of the input. Thus, we can verify in constant time whether si, sj α-conflict or not.
Dense Overlap. We design another α() function to model the behavior that the overlap region is densely connected. To that end, we define α-DenseOverlap(si, sj) that returns no-conflict if |si∩ sj| = 0 or
|E(G[si∩ sj])| ≥ O(O−1)2 − c, where O = |si∩ sj| and c ≥ 0 is a constant; otherwise, it returns α-conflict.
Function α-DenseOverlap(si, sj)
Set O = |si∩ sj|.
if |si∩ sj| = 0 or |E(G[si∩ sj])| ≥O(O−1)2 − c then
Return no conflict end if
Return α-conflict EndFunction
Lemma 84. The function α-DenseOverlap is well-conditioned.
Proof. (i) α-DenseOverlap is hereditary. Assume by contradiction that there is pair of sets si, sj with
α(si, sj) = 0 but there is a pair of subsets s0i⊆ siand s0j⊆ sj with α(s0i, s0j) = 1. Notice that (s0i∩ s0j) 6= ∅;
otherwise, α(s0i, s0j) = 0. Therefore, if α(si0, s0j) = 1, then |E(G[s0i∩s0 j])| < O0(O0−1) 2 −c, where O 0= |s0 i∩s0j|. However since G[s0i∩ s0
j] is an induced subgraph of G[si∩ sj], |E(G[si∩ sj])| < O(O−1)
(ii) If α(si, sj) = 1, |si ∩ sj| ≥ 1. Furthermore, for any pair of subsets s0i ⊆ si and s0j ⊆ sj with
α(s0i, s0j) = 0, (si∩ sj)\(s0i∩ s0j) 6= ∅. Assume otherwise by contradiction. If α(s0i, s0j) = 0 then |E(G[s0i∩
s0j])| ≥ O0(O20−1)− c, where O0 = |s0
i∩ s0j|. Thus, if (si∩ sj)\(s0i∩ s0j) = ∅, then |E(G[si∩ sj])| ≥ O(O−1)2 − c,
where O = |si∩ sj|, a contradiction to our assumption that α(si, sj) = 1.
(iii) We can verify in polynomial time this condition.