subject

You’re helping 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’re 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 Cj exchanged bits at time tk. There are m triples total. We’ll assume that the triples are presented to you in sorted order of time. For purposes of simplicity, we’ll assume that each pair of computers communicates at most once during the interval you’re observing.
The security analysts you’re working with would like to be able to answer questions of the following form: If the virus was inserted into computer Ca at time x, could it possibly 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 tk (in other words, if one of the triples (Ci, Cj, tk) or (Cj, Ci, tk) appears in the trace data), then computer Ci 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 backward in time. Thus, for example, If Ci is infected by time tk, and the trace data contains triples (Ci, Cj, tk) and (Cj, Cq, tr), where tk <= tr, then Cq will become infected via Cj. (Note that it is okay for tk to be equal to tr; this would mean that Cj had open connections to both Ci and Cq at the same time, and so a virus could move from Ci to Cq.) For example, suppose n =
4, the trace data consists of the triples (C1, C2, 4), (C2, C4, 8), (C3, C4, 8), (Cl, C4, 12)
and the virus was inserted into computer C1 at time 2. Then C3 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 G3 gets the virus from C4 at time 8. On the other hand, if the trace data were (C2, C3, 8), (C1, C4, 12), (C1, C2, 14)
and again the virus was inserted into computer C1 at time 2, then C3 would not become infected during the period of observation: although C2 becomes infected at time 14, we see that C3 only communicates with C2 before C2 was infected. There is no sequence of communications moving forward in time by which the virus could get from C1 to C3 in this second example.
Aim:
Design an algorithm that answers questions of this type: given a collection of trace data, the algorithm should decide whether a virus introduced at computer Ca at time x could have infected computer Cb by time y. If yes, give the trace of infection.
Hint for building graph:
For each triple (Ci, Cj, tk) we can see in our scan, we create nodes (Ci, tk) and (Cj, tk), we create directed edges joining these two nodes in both directions. If this is not the first node involving Cx (for all x from 1 to n), then we include a directed edge from (Cx, tx) to (Cx, tk), where tx, tk are the timestamps and tx < tk.
For implementation notes and format of input file and Program please check the attached pdf file.

ansver
Answers: 1

Another question on Computers and Technology

question
Computers and Technology, 22.06.2019 13:30
Write lines of verse that rhyme to remember the following information: acid rain is a type of air pollution caused by chemicals in the air.
Answers: 1
question
Computers and Technology, 25.06.2019 05:10
What is the output of the following program? #include using namespace std; class bclass { public: void print() const; bclass(int a = 0, int b = 0); //postcondition: x = a; y = b; private: int x; int y; }; class dclass: public bclass { public: void print() const; dclass(int a = 0, int b = 0, int c = 0); //postcondition: x = a; y = b; z = c; private: int z; }; int main() { bclass bobject(2, 3); dclass dobject(3, 5, 8); bobject.print(); cout < < endl; dobject.print(); cout < < endl; return 0 ; } void bclass: : print() const { cout < < x < < " " < < y < < endl; } bclass: : bclass(int a, int b) { x = a; y = b; } void dclass: : print() const { bclass: print(); cout < < " " < < z < < endl; } dclass: : dclass(int a, int b, int c) : bclass(a, b) { z = c; }
Answers: 3
question
Computers and Technology, 25.06.2019 08:00
The bus width and bus speed together determine the bus's or bandwidth; that is, the amount of data that can be transferred via the bus in a given period of time.
Answers: 1
question
Computers and Technology, 25.06.2019 08:20
The binary numbering system uses only two symbols—the digits 0 and 1—to represent all possible numbers. - true or false
Answers: 3
You know the right answer?
You’re helping some security analysts monitor a collection of networked computers, tracking the spre...
Questions
question
Mathematics, 09.03.2021 21:10
question
Mathematics, 09.03.2021 21:10
question
Mathematics, 09.03.2021 21:10
question
Chemistry, 09.03.2021 21:10
question
Advanced Placement (AP), 09.03.2021 21:10
Questions on the website: 13722362