hw2_sol

# hw2_sol - CSE 202 Homework 2 Solutions 1 Kleinberg Tardos...

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1 Connected graph G = ( V, E ) . Each edge e E has a time-varying edge cost f e = a e t 2 + b e t + c e such that f e > 0 for all t . Denote n = | V | and m = | E | . Given the graph G and the values ( a e , b e , c e ) : e E , we want to find the time t at which the minimum spanning tree has minimum cost. We first give intuition into the problem. By reviewing the Kruskal’s algorithm to construct a minimum spanning tree of the graph G , we note that the algorithm output only depends on the sorted order of edge cost values in E . So even though the edge cost values change with time, the minimum spanning tree will not change if the sorted order of edge costs does not change. Hence, our key idea is to find the time instants at which the relative order of edge cost values may change. For that purpose, we rely on the parabola shape of the quadratic edge cost functions. In particular, we note that two distinct parabolas have at most two intersections, and the relative order of the two parabolas can only change at those crossing points. We describe the details of our algorithm as follows. 1. Since edge costs vary with time, the set of edges constituting the mininum spanning tree also changes with time. Nevertheless, we can find a finite number of such minimum spanning trees as follows. 2. Consider the Kruskal’s algorithm to construct minimum spanning tree. We sort the edges in E by costs, and successively iterate through E in order of increasing cost. Then an edge e is inserted into the minimum spanning tree only if it does not create a cycle; otherwise it is just discarded. We observe that if the sorted order of E does not change, then the Kruskal’s algorithm will not change the set of edges constituting the mininum spanning tree. This fact holds true even though the edge costs vary with time. 3. Therefore, the next step is to investigate how to enumerate all possible sorted orders of E . Consider two edges e 1 and e 2 in E . If a e 1 = a e 2 , b e 1 = b e 2 , c e 1 = c e 2 , then f e 1 f e 2 for all time t . Otherwise, by simple geometric characteristics of parabolas, f e 1 and f e 2 vary over time as two separate parabolas and have at most two intersections. In particular, we have the following three cases. f

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hw2_sol - CSE 202 Homework 2 Solutions 1 Kleinberg Tardos...

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