Reliable critical nodes detection for Internet of Things (IoT)

Abstract

The 3D critical node (C-N) detection can play a vital role in algorithm development of security, surveillance, monitoring, topology detection, and situation-aware emergency navigation for the Internet of Things (IoT). However, 3D C-N detection problem in IoT raises some issues and also introduces new challenges. The existing state of the art in 3D C-N detection shows that rely on prior known anchor node, known coordinate, embedding of the 3D situation on a 2D geometrical structure like circles and presence of unreliable node and ignores the energy constraint in Low Power and Lossy Networks IoT. In this paper, we present a practical, distributed, and energy-efficient algorithm for reliable 3D C-N detection. The goal of the proposed mechanism is twofold, firstly a 3D critical nodes (C-N) detection algorithm is proposed which uses only Received Signal Strength Indicator information of neighbor. Secondly, a correlation-based algorithm for the reliability approach is proposed to increases the node resilience against malicious IoT nodes. The complexity of our proposed algorithms has a time complexity of $\mathcal {O}(\log (N))$ and computation cost $\mathcal {O}(\delta (\log N))$ where N is the number of nodes in networks, and $\delta$ is the total number of forward and the backward message from an individual node. To validate our work, we implemented our proposed approach with the IPv6 Routing Protocol for Low-Power and Lossy Networks (RPL) based IoT routing protocol compare it with RPL and cryptographic approach Version Number and Rank Authentication (VeRa). The result shows that the proposed approach can detect 10–15% more C-N nodes. Result also shows that our proposed algorithm has better PDR than RPL based approach by 12% and less than VeRa (cryptographic approach) by 8% however our proposed approach consumes almost 50% less power than the VeRa.

Appendix

In this Appendix, we derive the 3D critical node (C-N) detection rule for 3D topology. Let us consider a tetrahedron with vertices ABCDE, as illustrated in the Fig. 2(b). The angle between the any two planes of ABCDE can be derived using Lemma 1

Lemma 1

For a given a tetrahedron ABCD, the angle between the two planes i.e, $\mathcal {P}$(CAB) versus $\mathcal {P}$ (CAD) is:

$$\begin{aligned} \cos ^{-1}\left( \frac{\cos (\angle BAD) -\cos (\angle CAB)*\cos (\angle CAD) }{\sin (\angle CAB)\sin (\angle CAD)} \right) \end{aligned}$$

(1)

where $\angle BAD$= $\cos ^{-1}\left( \frac{dis(\varvec{(}B,A))^2 + dis(\varvec{(}A,D))^2-dis(\varvec{(}B,D))^2}{2dis(\varvec{(}A,B))*dis(\varvec{(}A,D))^2} \right)$,

$\angle CAB=\cos ^{-1}\left( \frac{dis(\varvec{(}C,A))^2 +dis(\varvec{(}A,B))^2 -dis(\varvec{(}C,B))^2}{2dis(\varvec{(}C,A))*dis(\varvec{(}A,B))}\right)$,

$\angle CBD=\cos ^{-1}\left( \frac{dis(\varvec{(}C,A))^2 + dis(\varvec{(}A,D))^2-dis(\varvec{(}C,D))^2}{2dis(\varvec{(}C,A))*dis(\varvec{(}A,D))}\right)$ and

$dis(\mathbf {B},\mathbf {A})$,$dis(\mathbf {A},\mathbf {D})$,$dis(\mathbf {B},\mathbf {D})$,$dis(\mathbf {C},\mathbf {A})$, $dis(\mathbf {A},\mathbf {B})$, $dis(\mathbf {C},\mathbf {D})$,$dis(\mathbf {C}, \mathbf {D)}$ are the distance between the nodes $\overline{\mathrm{BA}}$,$\overline{\mathrm{AD}}$,$\overline{\mathrm{BD}}$,$\overline{\mathrm{CA}}$,$\overline{\mathrm{AB}}$,$\overline{\mathrm{CB}}$ and $\overline{\mathrm{CD}}$.

Proof

Viewing the tetrahedron from vertices A, we get three triangular planes (Fig. 3). The angle: $\angle BCA= \cos ^{-1}\left( \frac{dis(\varvec{(}B,C))^2 + dis(\varvec{(}C,A))^2-dis(\varvec{(}B,A))^2}{2dis(\varvec{(}B,C))*dis(\varvec{(}C,A))^2} \right)$,

$\angle CAD\cos ^{-1}\left( \frac{dis(\varvec{(}C,A))^2 + dis(\varvec{(}A,D))^2-dis(\varvec{(}C,D))^2}{2dis(\varvec{(}A,D))*dis(\varvec{(}C,D))^2} \right)$ and

$\angle DAB=\cos ^{-1}\left( \frac{dis(\varvec{(}D,A))^2 + dis(\varvec{(}A,B))^2-dis(\varvec{(}D,B))^2}{2dis(\varvec{(}D,A))*dis(\varvec{(}A,B))^2} \right)$ ( by using inverse Cosine Formula for triangle). Now, label the 3 edges associated with node A as $\overline{\mathrm{AB}}$,$\overline{\mathrm{AC}}$ and $\overline{\mathrm{AD}}$ and arrange them in counterclockwise manner. Let $\vec {v}_{ab}$, $\vec {v}_{ac}$ and $\vec {v}_{ad}$ be the unit vector pointing away from the vertex A along the direction of the edge $\overline{\mathrm{AB}}$,$\overline{\mathrm{AC}}$ and $\overline{\mathrm{AD}}$ respectively. Any of the edges can be seen as the contact point of two planes or faces. Lets say edge $\overline{\mathrm{AC}}$, can be seen as the juncture of vectors bounded by $\vec {v}_{ab}$ and $\vec {v}_{ac}$ with another vector bounded by $\vec {v}_{ac}$ and $\vec {v}_{ad}$. The inward normal vectors of these two faces are given by: $\vec {n}_{ab,ac}$ = $\frac{\vec {v}_{ab} \times \vec {v}_{ac}}{|\vec {v}_{ab} \times \vec {v}_{ac}|}$ $\,\,\,\text { and }\,\,\, \vec {n}_{ac,ad}$ = $\frac{\vec {v}_{ac} \times \vec {v}_{ad}}{|\vec {v}_{ac} \times \vec {v}_{ad}|}$

$$\begin{aligned}&= -\vec {n}_{ab,ac}\cdot \vec {n}_{ac,ad} \end{aligned}$$

(2)

$$\begin{aligned}&= -\frac{( \vec {v}_{ab} \times \vec {v}_{ac} )\cdot (\vec {v}_{ac} \times \vec {v}_{ad})}{ |\vec {v}_{ab} \times \vec {v}_{ac}||\vec {v}_{ac} \times \vec {v}_{ad}|} \end{aligned}$$

(3)

$$\begin{aligned}&=-\frac{(\vec {v}_{ab}\cdot \vec {v}_{ac})(\vec {v}_{ac}\cdot \vec {v}_{ad}) - (\vec {v}_{ab}\cdot \vec {v}_{ad})|\vec {v}_{ac}|^2}{|\vec {v}_{ab} \times \vec {v}_{ac}||\vec {v}_{ac} \times \vec {v}_{ad}|} \end{aligned}$$

(4)

$$\begin{aligned}&= \frac{\vec {v}_{ab}\cdot \vec {v}_{ad} - (\vec {v}_{ab}\cdot \vec {v}_{ac})(\vec {v}_{ac}\cdot \vec {v}_{ad})}{|\vec {v}_{ab} \times \vec {v}_{ac}||\vec {v}_{ac} \times \vec {v}_{ad}|} \end{aligned}$$

(5)

without loss of generality we will get the dihedral angle between $\mathcal {P}$(CAB) versus $\mathcal {P}$ (CAD) as: $\cos ^{-1}\left( \frac{\cos (\angle BAD) -\cos (\angle CAB)*\cos (\angle CAD) }{\sin \angle CAB)*\sin (\angle CAD)} \right)$ Similarly, dihedral angles between other planes can be obtained as: $\mathcal {P}_{\infty }$(ABC) versus $\mathcal {P}_{\in }$ (ABD) can be derived as: $\cos ^{-1}\left( \frac{\cos (\angle CBD) -\cos (\angle ABC)*\cos (\angle ABD) }{\sin (\angle ABC))*\sin (\angle ABD)} \right)$ and corresponding dihedral angles between other planes are derived. $\square$

For 3D B-N, assume a smaller subgraph of four nodes $\{A, B, C, E\}\in \{A, B, C, D, E\}$, as illustrated in Figs. 2 and 3. Suppose node E wants to test whether it is B-N or I-N (Fig. 2b shows the placement of node E as I-N). Node E tests the coverness information (according to Definition-2), i.e., at least four nodes covers it. If node E lies inside the tetrahedral, then it is I-N else it is B-N. The stated Lemma 2 is used to derive the relation and to detect the node as I-N or B-N as illustrated in (Fig. 9).

Fig. 9

Tetrahydron $\{\hbox {A,B,C,D,E}\}$ is sub divided into sub tetrahydron() considering E as top vertex

Lemma 2

Node ’E’ is interior node (I-N) if it satisfy all the six relationship in terms of angularity.

• 1. $\mathcal {P}$(ABC) versus $\mathcal {P}$ (ABD) < $\mathcal {P}$(ABC) versus $\mathcal {P}$ (ABE) + $\mathcal {P}$(ABD) versus $\mathcal {P}$ (ABE)

• 2. $\mathcal {P}$(BDA) versus $\mathcal {P}$ (BDC) < $\mathcal {P}$(BDA) versus $\mathcal {P}$ (BDE) + $\mathcal {P}$(BDC) versus $\mathcal {P}$ (BDE)

• 3. $\mathcal {P}$(DCA) versus $\mathcal {P}$ (DCB) < $\mathcal {P}$(DCA) versus $\mathcal {P}$ (DCB) + $\mathcal {P}$(DCB) versus $\mathcal {P}$ (DCE)

• 4. $\mathcal {P}$(BCA) versus $\mathcal {P}$ (BCD) < $\mathcal {P}$(BCA) versus $\mathcal {P}$ (BCE) + $\mathcal {P}$(BCD) versus $\mathcal {P}$ (BCE)

• 5. $\mathcal {P}$(ADB) versus $\mathcal {P}$ (ADC) < $\mathcal {P}$(ADB) versus $\mathcal {P}$ (ADE) + $\mathcal {P}$(ADC) versus $\mathcal {P}$ (ADE)

• 6. $\mathcal {P}$(ACB) versus $\mathcal {P}$ (ACD) < $\mathcal {P}$(ACB) versus $\mathcal {P}$ (ACE) + $\mathcal {P}$(ACD) versus $\mathcal {P}$ (ACE)

Proof

“without loss of generality” angle divided into two-part will always be greater than the individual split parts (direct proof by triangle inequality). Hence if the equation fails then it is a boundary node (B-N). $\square$

1.1 Correctness for coverage and connectivity of 3D critical nodes (C-N)

Proving optimality in coverage and connectivity in 3D topology is surprisingly difficult, even though similar problems in 2D can be proved easily. As it has at least 14 different directions when assumed $R_c =R_s$ (Kepler’s sphere-packing problem has been around since 1611 and proof of Kepler’s conjecture has only been found in 1998). Similarly, no optimality proof for Kelvin’s 1887 conjecture is known yet. For coverage, we have proposed a K-Hop distance-vector algorithm. Each Critical Node (C-N) has (node id) so they will extend their Nbr table only and we use an extended k-Hop distance vector routing [43]. Each node transmits its Nbr table each receives the same the most recently received vector from the neighbor is changed according to the distance vector formula. Once this initial helper node were identified then the final Helper nodes [43] are identified using Lemma-III. Figure 10 shows the lattice pattern to achieve full coverage with 14-connectivity (nodes are $\frac{R_c}{2}$ distance apart). Solid circular dots denote sensors, solid lines represent connected links with other sensors and the dashed lines construct a cube. The graph also represents the Voronoi polyhedron generated by each sensing sphere is a truncated octahedron [44].

Fig. 10

Figure shows the lattice pattern to achieve full coverage with 14-connectivity [44]

Lemma 3

Node(u) becomes a interior node if $deg(u)<14$ and all neighbours are $\frac{R_c}{2}$ distance apart.

Proof

Paper [44] proves that patterns for full coverage requires 14 connectivity illustrated in Fig. 10. $\square$