The k-Canadian Travelers Problem with communication

Abstract

This paper studies a variation of the online k-Canadian Traveler Problem (k-CTP), in which there are multiple travelers who can communicate with each other, to share real-time blockage information of the edges. We study two different communication levels for the problem, complete communication (where all travelers can receive and send blockage information with each other) and limited communication (where only some travelers can both receive and send information while the others can only receive information). The objective is that at least one traveler finds a feasible route from the origin to the destination with as small cost as possible. We give lower bounds on the competitive ratio for both the two communication levels. Considering the urban traffic environment, we propose the Retrace-Alternating strategy and Greedy strategy for both the two communication levels, and prove that increasing the number of travelers with complete communication ability may not always improve the competitive ratio of online strategies.

Appendices

Appendix 1: Competitive analysis of Retrace-Alternating strategy for P ₁

Case 2. L>d. Let X=x ₁+x ₂=min{L,2d}. According to the process of the strategy, there are mainly two cases. One is that the strategy doesn’t execute step 4, the other is that the strategy executes step 4.

Case 2.1: Not execute Step 4. In this case, the two groups travelers go to the destination alternatively when the other group is blocked. All the x ₁/x ₂ travelers will return to O when the selected paths are all blocked and the cost to find a blockage on graph G−E _j is at most $l(\mathit{SP}_{G-E_{j}})$. Step 2 and step 3 are totally repeated at most $i=2\alpha\lfloor\frac{k}{X}\rfloor$ times. The total cost of the online strategy is $c_{\mathit{on}}\leq\alpha\sum_{j=1}^{i}l(\mathit{SP}_{G-E_{j}})+\alpha l(\mathit{SP}_{G-E_{i+1}})+a\alpha l(\mathit{SP}_{G-E_{i+2}})$, where a=0 when some of the x ₁ travelers reach D and a=1 when some of the x ₂ travelers reach D. That is to say

The offline optimal cost is

The ratio in this subcase is

Case 2.2: Execute Step 4. In this case, before the strategy executes the step 4, it continues to return to O when all the selected paths are blocked. Suppose the strategy goes to step 4 on graph G−E _i+1 and the cost at this time is $c_{\mathit{on}}\leq\alpha\sum_{j=1}^{i}l(\mathit{SP}_{G-E_{j}})+\alpha l(\mathit{SP}_{G-E_{i+1}})$. The length of the detouring path in step 4 will be $l(P_{x}) =2x l(e(v_{i},D)\leq2 \alpha l(\mathit{SP}_{G-E_{i+1}}))$. Therefore, the total cost is $c_{\mathit{on}}\leq\alpha\sum_{j=1}^{i} l(\mathit{SP}_{G-E_{j}})+\alpha l(\mathit{SP}_{G-E_{i+1}})+2\alpha l(\mathit{SP}_{G-E_{i+1}})\leq \alpha(2i+3)l(\mathit{SP}_{G-E_{i+1}})$. The offline optimal cost is $c_{\mathit{off}}\geq l(\mathit{SP}_{G-E_{i+1}})$.

The ratio is $c_{2}=\frac{c_{\mathit{on}}}{c_{\mathit{off}}}\leq\alpha(i+3)$.

Obviously if a=0, $i<2\lfloor\frac{k}{X}\rfloor$; else there are must some traveler arrived at D and no need to detour. The worst case is $c_{2}\leq\alpha(2\lfloor\frac{k}{X}\rfloor+2)$.

Else if a=1, $i\leq2\lfloor\frac{k}{X}\rfloor$; else there are must some traveler arrived at D and no need to detour. The worst case is $c_{2}\leq\alpha(2\lfloor\frac{k}{X}\rfloor+3)$.

So the ratio in this subcase is

Above all, the competitive ratio of the Retrace-Alternating strategy when L>d is

Appendix 2: Retrace-Alternating strategy for P ₂

Case 1

L ₁≤d.

Step 1

i=1, E _i =ϕ, x=L, No=0.

Step 2

x travelers go to node D along separate paths without overlap on graph G. If the selected paths are all blocked and no blockage is connected to D, the travelers all go back towards node O, i=i+1, add all the blocked edges to set E _i , G=G−E _i , No=No+1, let x=x+1 when k−No=x, go to step 2; if all the paths are blocked and there is blockage connected to D, go to step 3; if at least one traveler reaches the node D, stop.

Step 3

Just let the traveler who is blocked by the edge connected to D detour until the arrival at node D, stop.

Case 2

L ₁>d.

Step 1

Let i=1, x ₁=d, x ₂=min{d,L ₁−d}, E _i =ϕ, No=0.

Step 2

When there are at least x ₁ travelers at O, the x ₁ travelers go to D along separate paths without overlap on graph. If the x ₁ path are all blocked and no blockage is connected to D, the x ₁ travelers all go back towards O, i=i+1, add all the blocked edges to set E _i , G=G−E _i , No=No+x ₁, let x ₂=x ₂+1 when k−No=x ₂, go to step 3; if all the paths are blocked and there is at least one blockage connected to D, go to step 4; if some traveler reaches D, stop.

Step 3

When there are x ₂ travelers at node D, x ₂ travelers go to D along separate paths without overlap on graph. If all the x ₂ paths are all blocked and no blockage is connected to D, i=i+1, add all the blocked edges to set E _i , G=G−E _i , No=No+x ₂, let x ₂=x ₂+1 when k−No=x ₁, the x ₂ travelers all go back towards O, go to step 2; if all the paths are blocked and there is at least one blockage connected to D, go to step 4; if some traveler reaches D, stop.

Step 4

Just let the traveler who is blocked by the edge connected to D detour until the arrival at node D, stop.