Computers and Technology
Computers and Technology, 27.11.2019 03:31, parbridg

Collaborative problem: you are some security analysts monitor a collection of networked computers, tracking the spread of an online virus. there are n computers in the system, labeled c1,c2 cn, and as input, you are given a collection of trace data indicating the times at which pairs of computers communicated. thus the data is a sequence of ordered triples (ci, cj, tk) such a triple indicates that ci and c, communicated at time tk. assume there are m triples total. now let us assume that the triples are presented to you in sorted order of time. for purposes of simplicity, we will assume that each pair of computers communicates at most once during the inter val you are observing. the security analysts you are working with would like to be able to answer questions of the following form: if the virus was inserted into computer ca at time r, could it pos- sibly have infected computer cb by time y? the mechanics of infection are simple if an infected computer ci communicates with an uninfected computer cj at time t (in other words, if one of the triples (ci, cj, tk) or (cj, ci, tk) appears in the trace data), then computer c, becomes infected as well, starting at time tk. infection can thus spread from one machine to another across a sequence of communications, provided that no step in this sequence involves a move backwards in time. thus, for example, if ci is infected by time tk and the trace data contains triples (ci, cj, tk) and (cj, ct), where tks tr, then c will become infected via cj. (note that it is okay for tk to be equal to tr. this would mean that c, had open connections to both ci and c at the same time, so a virus could move from ci to cq) for example, suppose 4, the trace data consists of the triples (c1,c2,4), (c2,c4,8) (cs, c4,8) (c1, c4, 12) and the virus was inserted into computer ci at time 2. then cs would be infected at time 8 by a sequence of three steps first c2 becomes infected at time 4, then c4 gets the virus from c2 at time 8, and then cs gets the virus from c4 at time 8.

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