Maurice Tutor

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    Argosy University/ Phoniex University/
    Nov-2005 - Oct-2011

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    Phoniex University
    Oct-2001 - Nov-2016

Category > Computer Science Posted 16 Jul 2017 My Price 11.00

Canadian north

Your friends are planning an expedition to a small town deep in the Canadian north next winter break. They’ve researched all the travel options and have drawn up a directed graph whose nodes represent intermediate destinations and edges represent the roads between them.

In the course of this, they’ve also learned that extreme weather causes roads in this part of the world to become quite slow in the winter and may cause large travel delays. They’ve found an excellent travel Web site that can accurately predict how fast they’ll be able to travel along the roads; however, the speed of travel depends on the time of year. More precisely, the Web site answers queries of the following form: given an edge e = (v, w) connecting two sites v and w, and given a proposed starting time t from location v, the site will return a value fe(t), the predicted arrival time at w. The Web site guarantees that fe(t) ≥ t for all edges e and all times t (you can’t travel backward in time), and that fe(t) is a monotone increasing function of t (that is, you do not arrive earlier by starting later). Other than that, the functions fe(t) may be arbitrary. For example, in areas where the travel time does not vary with the season, we would have fe(t) = t + ℓe, where ℓe is the time needed to travel from the beginning to the end of edge e.

Your friends want to use the Web site to determine the fastest way to travel through the directed graph from their starting point to their intended destination. (You should assume that they start at time 0, and that all predictions made by the Web site are completely correct.) Give a polynomial-time algorithm to do this, where we treat a single query to the Web site (based on a specific edge e and a time t) as taking a single computational step

Answers

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Status NEW Posted 16 Jul 2017 08:07 PM My Price 11.00

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