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Unformatted text preview: Subhash Suri UC Santa Barbara CS230b Advanced Algorithms and Applications Subhash Suri Computer Science Department UC Santa Barbara Fall Quarter 2004. Subhash Suri UC Santa Barbara Shortest Paths u v s 10 1 5 7 2 3 x y 2 9 6 • Find shortest length path from s to v ? • s → x → u → v has length 5 + 3 + 1 = 9 . • Many applications. Discussed later. • A network G = ( V, E ) . • Vertices (nodes) V = { 1 , . . . , n } . • Edges (links) E = { e 1 , e 2 , . . . , e m } . Edge e ij = ( i, j ) is directed from i to j . • Edge e ij has cost (weight) c ij . The costs can be positive or negative! Subhash Suri UC Santa Barbara Negative Cost Shortest Paths v s x y 7 6 2 5 9342 8 7 u • What’s the shortest path from s to y ? • s → x → v → u → y has length 2 . • How can costs be negative? • Examples later. (Arbitrage trading, scientific simulations, matching algorithms, min cost network flows). • More general the formulation, the better. • Any simple way to eliminate negative edges? Adding a constant to all edges? Subhash Suri UC Santa Barbara Getting Started u v s 10 1 5 7 2 3 x y 2 9 6 • Source node s = v . • Compute SP distances from s to every node v j . • The paths themselves can be recovered from predecessors. • Distance labels d ( j ) : best path length to j found so far. • Initially, d (0) = 0 , and d ( j ) = ∞ for others. • The algorithm improves estimates for all d ( j ) until SP distances become known. Subhash Suri UC Santa Barbara Basic Idea • How to improve the distance estimate? s c ij i j 50 75 10 • Suppose there is an edge ( i, j ) such that d ( j ) > d ( i ) + c ij then we can improve the estimate of d ( j ) : d ( j ) = d ( i ) + c ij • Previously, d ( i ) = 50 and d ( j ) = 75 . The relabeling step finds a better path to j via i of cost 60 . Subhash Suri UC Santa Barbara Optimality Condition Theorem: Suppose each d ( j ) is the length of some feasible path from s to j . Then, these d () distances are shortest path distance if and only if d ( j ) ≤ d ( i ) + c ij , for all ( i, j ) ∈ E u v s 10 1 5 7 2 3 x y 2 9 6 8 9 5 7 Necessity Proof: Suppose ∃ an edge ( i, j ) violating the condition. Then, d ( j ) > d ( i ) + c ij . But then we can reach j via i at cost d ( i ) + c ij , which is smaller than d ( j ) , contradicting the d ( j ) is shortest path distance. Subhash Suri UC Santa Barbara Sufficiency s 1 2 k1 k c c c 01 12 k1,k 1. Suppose d ( j ) ≤ d ( i ) + c ij holds. 2. Let k be a node with incorrect distance. 3. s → 1 → 2 ··· k 1 → k be actual SP. 4. So, c 01 + c 12 + ··· + c k 1 ,k < d ( k ) . ( * ) 5. By optimality condition we have d ( k ) ≤ d ( k 1) + c k 1 ,k d ( k 1) ≤ d ( k 2) + c k 2 ,k 1 ....
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 Fall '09
 Vandam
 Graph Theory, Shortest path problem, UC Santa Barbara, Subhash Suri

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