Target Variable Selection (TVS)

5.4

Target Variable Selection (TVS)

Once the set of potential locations has been identified, it is then necessary to determine the set of variables into which faults will be actually injected. Specifically, there may be locations at which no fault will be injected and other locations where several faults may be injected.

The challenge in selecting target variables set from the fact that when a variable uis overlooked, then it means either that a variablev on which it depends has been selected (and selectinguwill override the effect of propagating error from v tou) or a variablewthat depends onvhas been selected. Thus, the decision of selecting a variable is not a local one. This problem then is very similar to the problem of generating dominating sets. Thus, this chapter proceeds to prove that the problem of target variables selection (TVS) is NP-complete.

5.4.1

Complexity Analysis of TVS

The problem of target variables selection is formally defined as an optimisation problem.

Definition 5.2 (Target Variables Selection (TVS)). Given a graph GD

P =

(U, A, U0_{, W, L), where U}0 _{⊆ U, a positive integer N, a positive integer}

A,

does there exist a setUV ⊆U such that:

• |UV| ≤N • ∀u, u0∈UV :

∀p∈ρu,u_G 0 : UV _∩pˆ=∅:

length(p)≤A

It should be observed here that the amplification factor is carried over from the injection location selection problem (ILS). Specifically, given that no two

5.4. TARGET VARIABLE SELECTION (TVS) 60

successive potential locations are no more than distance _Aapart, the selection of target variables should not violate this requirement.

Lemma 5.2.1 (NP membership). TVS is in NP.

Proof. To prove this, the correctness of the set Uv is shown in polynomial-

time. So, given an instance of TVS as described and a solution set Uv, the verification is performed as follows: The first condition is trivially verified. For the second condition, it is required to verify that, from any vertex u∈Uv_{, all}

paths originating from uwill contain another vertexu0 ∈Uv _{with distance at}

most_Aaway. This is done as follows: It first select a nodeu∈Uv_{, and construct}

a spanning tree of depth_A, rooted atu, by doing a depth-first traversal on G. This tree is denoted byUt_{. Now, given graphU}t_{, it is needed to verify whether}

for every pathporiginating fromuand ending at a leaf has at least one vertex u0 ∈Uv_{. If the answer is negative, thenU}v _{is not a solution for TVS. On the}

other hand, if the answer is true forU , then the process is repeated for all other verticesu∈Uv. The complexity of this verification procedure isO(|U|2_).

Lemma 5.2.2 (NP-hardness). TVS is NP-hard.

Proof. To prove this, MDS problem [48] is reduced to the TVS problem. Before

defining the MDS problem, it is denoted by U1_{, the set of vertices adjacent to} a nodeu∈V. Then, the MDS problem is defined as follows:

MDS: Given a graph G= (V, E), a positive integer K, find a setV0⊆V such that

• |V0_{| ≤K}

• ∀u6∈V0:∃v∈U1_:v∈V0

With this definition of MDS, now the mapping between MDS and TVS is de- veloped.

5.4. TARGET VARIABLE SELECTION (TVS) 61

Mapping

It is assume that the graph for MDS has a set of vertices with in-degree 0, denoted byV0, which does not affect the complexity of MDS.

• U =V • U0=V0 • A=E • N =K • W(a) = 1,∀a∈A • _A= 3. Reduction

It now has to be to shown that a solution to MDS exists if and only if a solution of TVS exists.

(⇒) Let V0 ⊆V be a solution to MDS with graph G= (V, E). Let Uv _{be a}

solution to the instance of TVS as defined under the mapping, for graph G0 = (U, U0_{, A, W) with amplification factor}

A= 3, such thatUv=V0. It is shown that this solutionUv_{is valid for TVS. First, sinceV}0_{is a solution} to MDS andUv _=V0_{, then|U}v_{| ≤K. Secondly, for ∀n6∈V}0 _:∃m∈N1_: m∈V0. Now, assume there is a edge (p, q)∈Asuch thatp, q6∈V0. Then, there are two extreme cases: (i) if∃m∈P1: m6=q∧m6∈Q1:m ∈V0 and ∃n ∈ Q1 : n 6= p∧n 6∈ P1 : n ∈ V0, then the distance between nodesmand nis at most 3, satisfyingA, and (ii)∃m∈P1:m6=q and

∃n∈Q1_{:n6=p, then if m=n, then the distance is 0, which is less than}

A. Hence, the maximum distance between any pair of vertices inV0 is at most 3, thus not violating_A. SinceUl_=V0_,V0 _{is a solution to TVS.}

(⇐) LetUv_{be a solution to the instance of TVS as defined under the mapping,}

for graphG0= (U, U0_{, A, W) with amplification factor}

A= 3. LetV0⊆V be a solution to MDS with graphG= (V, E) such that V0 =Uv_{. Now,}