Abstract
Let \(G\) be a random geometric graph formed by \(n\) nodes with adjacency distance \(r_n\) and let each edge of \(G\) be assigned an independent exponential passage time with mean that depends on the graph size \(n.\) We connect \(G\) to two nodes source \(s_A\) and destination \(s_B\) at deterministic locations spaced \(d_n\) apart in the unit square and find upper and lower bounds on the minimum passage time between \(s_A\) and \(s_B\) through paths in \(G\) having constant stretch, i.e., whose length is constrained to be proportional to the Euclidean distance between \(s_A\) and \(s_B.\)
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Acknowledgements
I thank Professors Rahul Roy, C. R. Subramanian and the referees for crucial comments that led to an improvement of the paper. I also thank IMSc for my fellowships.
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Appendices
Appendix 1: Edge weight assigments
Let \(\{W_{j_1,j_2,j_3}\}_{j_1,j_2,j_3 \ge 1}\) be i.i.d. random variables with cumulative distribution function (cdf) \(F_n\) that are also independent of the node locations \(\{X_i\}.\) We first describe edge weight assignments and then discuss the conditional independence property.
Edge weight assignments To assign weights to the edges of the random graph \(G\) from the set \(\{W_{j_1,j_3,j_4}\},\) we follow an analogous procedure as in Chapter \(1\) of Meester and Roy [10]. The basic idea is to first assign each node \(X_j,\) a unique pair of positive integers \(m(X_j) := (q,l_j).\) Next if \(X_i\) and \(X_j\) are joined by an edge \((X_i,X_j)\) in \(G,\) then we define the passage time \(t(X_i,X_j) := W_{q,l_i,l_j}.\) The integer pair \((q,l_j)\) is obtained by dividing the unit square \(S\) into successively finer subsquares of size \(\frac{1}{2^{q}}, q \ge 1\) each, until each subsquare contains at most one node of \(\{X_i\}.\)
Formally, for \(k \ge 1\) divide the unit square \(S\) into disjoint \(\frac{1}{2^{k}} \times \frac{1}{2^{k}}\) squares \(\{S_{k,l}\}_{1 \le l \le 2^{2k}}.\) Let \(q\) be the smallest integer such that each square in \(\{S_{q,l}\}\) contains at most one node of \(\{X_i\}.\) If node \(X_j\) is present in square \(S_{q,l_j},\) we then set \(m(X_j) := (q,l_j).\)
We illustrate the above procedure in Fig. 3, where the unit square \(ABCD\) is first subdivided into \(4\) subsquares each of side length \(\frac{1}{2}.\) This partition is not fine enough since the square \(AEFG\) contains two nodes \(x\) and \(y.\) Further subdividing \(\frac{1}{2} \times \frac{1}{2}\) subsquare into \(4\) subsquares each, we get a fine enough partition so that each subsquare contains at most one node.
Conditional independence property We use the following conditional independence property throughout. For any measurable function \(g = g(X_1,\ldots ,X_n)\) and any \(i \ne j,\) we have
We then say that given the node locations \(\{X_i\},\) the passage times \(\{t(X_i,X_j)\}\) are independent in the sense of (A1.1).
Appendix 2: Deviation estimates
We use the following deviation estimate throughout. Let \(\{R_j\}_{1 \le j \le m}\) be independent Bernoulli random variables with
and let \(R = \sum _{j=1}^{m} R_j.\) If \(p_{low} \le p_j \le p_{up}\) for all \(1 \le j \le m\) then then
for all \(m \ge 1.\)
Proof of (A2.1): Letting \(\mu = {\mathbb {E}}R\) and \(0 < \epsilon \le \frac{1}{2},\) we have from Corollary A.1.14, pp. 312 of [1] that
where \(C(\epsilon ) = \min \left( \frac{\epsilon ^2}{2}, -\epsilon + (1+\epsilon )\log (1+\epsilon )\right) \ge \frac{\epsilon ^2}{4}\) because \(\log (1+\epsilon ) > \epsilon - \frac{\epsilon ^2}{2}\) and so
since \(0 < \epsilon \le \frac{1}{2}.\) Setting \(\epsilon = \frac{1}{2}\) we therefore get
and an analogous analysis holds for the lower bound. \(\square \)
Let \(R_1,\ldots ,R_m\) be independent exponential random variables with mean \(\lambda \) and let \(R = \sum _{j=1}^{m} R_j\) so that \({\mathbb {E}}R = m\lambda .\) We then have
and
Proof of (A2.3): For \(s < \frac{1}{\lambda },\) we have \({\mathbb {E}}e^{sR_i} = \lambda ^{-1}\int _{0}^{\infty }e^{sx}e^{-x\lambda ^{-1}} = \frac{1}{1-s\lambda }.\) Therefore
where \(\delta = 2s\lambda + \log \left( 1-s\lambda \right) .\) If \(s\lambda = \frac{1}{3},\) then
and so \(\delta = \frac{1}{6}\) and consequently, \({\mathbb {P}}(R \ge 2m\lambda ) \le e^{-m/6}.\)
For the lower bound, we let \(s > 0\) and get
For \(x > 0\) we therefore have
where \(\theta = \log (1+s\lambda ) - sx.\) Setting \(s=1\) and \(x = \frac{\log (1+\lambda )}{2}\) we get (A2.4). \(\square \)
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Ganesan, G. Constrained Minimum Passage Time in Random Geometric Graphs. Algorithmica 83, 576–588 (2021). https://doi.org/10.1007/s00453-020-00766-8
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DOI: https://doi.org/10.1007/s00453-020-00766-8