ters and to determine the relationship between pen-
etration probability, iterations of ORC SAM, and the
selection of configurations for adding distraction clus-
ters as well as investigate the impact of the length of
distraction clusters on the network and the location of
the point at which the distraction clusters reconnect to
the network.
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APPENDIX
Proof of Theorem 1. The Cluster Addition Problem is NP-
hard and the associated decision problem is NP-Complete
when the number of sequences from (s,ℓ) to (s
′
,ℓ
′
) is a poly-
nomial in the number of nodes in the intruder penetration
network.
Proof. Membership in NP (if the number of sequences from
(s,ℓ) to (s
′
,ℓ
′
) is polynomial can be shown when the certifi-
cate is the set of configuration-cluster pairs.
For NP-hardness, consider the set cover problem (Feige,
1998) where the input consists of as set of elements S,
a family of subsets of S denoted as H, and natural num-
bers K, X. The output of this problem is a subset of H
of size K or less such that their union covers X or more
elements of S. This problem is NP-hard and can be em-
bedded into an instance of the cluster addition problem
as follows: L = {0, 1}, S = {s,t} ∪ {v
w
|w ∈ S}, R =
{(s,v
w
),(v
w
,t)|w ∈ S}, ∀w ∈ S, set π((s, 1),(v
w
,1)) = 1 and
π((v
w
,1), (t,1)) = 1 (otherwise set π by definition), ∀w ∈ S,
set f((s,1),(v
w
,1)) = 1 and f((v
w
,1), (t, 1)) = 1 (otherwise
set f by definition), ℓ
s
,ℓ
s
′
= 1, CL = {cl}, CFG = {cfg
h
|h ∈
H}, for each cfg
h
∈ CFG and cl set π
cfg
h
,cl
(v
w
) = 1 if s ∈ h
and 0 otherwise, for each cfg
h
∈ CFG and cl set f
cfg
h
,cl
(v
w
) =
|S| if s ∈ h and 0 otherwise, x = 1−
|S|−X
|S|
−
X
|S|(|S|+1)
, k = K,
and t = 2. Clearly this construction can be completed in
polynomial time.
Next, we show that a solution to set cover will provide
a solution to the constructed cluster-addition problem. If
H
′
is a solution to set cover, select the set {cfg
h
|h ∈ H
′
}.
Clearly this meets the cardinality constraint. Note that in
the construction, all sequences from (s,1) to (t, 1) are of the
form ⟨(s, 1),(v
w
,1), (t,1)⟩ where w ∈ S. Note that as every
system v
w
∈ S −{ s,t} is now connected to a cluster. Hence,
each cluster now has had its probability reduced from 1/|S|
to at most 1/(|S |(|S| + 1). Hence, Pen
t
IPN∪PCP
((s,ℓ),(s
′
,ℓ
′
)) <
|S|−X
|S|
+
X
|S|(|S|+1)
which completes this claim of the proof.
Going the other way, we show that a solution to the
constructed cluster-addition problem will provide a solu-
tion to set cover. Given cluster-addition solution PCP, con-
sider H
′
= {h|(cfg
h
,cl) ∈ PCP}. Note that, by the construc-
tion, all elements of PCP are of the form (cfg
h
,cl) where
h ∈ H
′
. Clearly, the cardinality constraint is met by the
construction. Suppose, BWOC, H
′
is not a valid solution
to set cover. We note that this must imply that there are
some v
s
that are not attached to a distraction cluster. Let
us assume there are δ number of these systems. Hence,
Pen
t
IPN∪PCP
((s,ℓ),(s
′
,ℓ
′
)) >
|S|−X+δ
|S|
+
X−δ
|S|(k|S|+1)
. Let us now
assume, by way of contradiction, that this quantity is less
than or equal to
|S|−X
|S|
+
X
|S|(|S|+1)
, which is the upper bound
on Pen
t
IPN∪PCP
((s,ℓ),(s
′
,ℓ
′
)) is at least X of the v
s
systems have
KeepingIntrudersatLarge-AGraph-theoreticApproachtoReducingtheProbabilityofSuccessfulNetworkIntrusions
29