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CSE 202 - Algorithms

Shortest Paths Problems. CSE 202 - Algorithms. Shortest Paths Problems. 1. v. t. Given a weighted directed graph, <u,v,t,x,z> is a path of weight 29 from u to z. <u,v,w,x,y,z> is another path from u to z; it has weight 16 and is the shortest path from u to z. 5. 9. 4. 13. 8. y.

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CSE 202 - Algorithms

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  1. Shortest Paths Problems CSE 202 - Algorithms CSE 202 - Shortest Paths

  2. Shortest Paths Problems 1 v t • Given a weighted directed graph, <u,v,t,x,z> is a path of weight 29 from u to z. <u,v,w,x,y,z> is another path from u to z; it has weight 16 and is the shortest path from u to z. 5 9 4 13 8 y 4 u x 3 2 10 1 2 1 z w 6 CSE 202 - Shortest Paths

  3. Variants of Shortest Paths Problems A. Single pair shortest path problem • Given s and d, find shortest path from s to d. B. Single source shortest paths problem • Given s, for each d find shortest path from s to d. C. All-pairs shortest paths problem • For each ordered pair s,d, find shortest path. • (A) and (B) seem to have same asymptotic complexity. • (C) takes longer, but not as long as repeating (B) for each s. CSE 202 - Shortest Paths

  4. More Shortest Paths Variants • All weights are non-negative. • Weights may be negative, but no negative cycles (A cycle is a path from a vertex to itself.) • Negative cycles allowed. Algorithm reports “- ” if there is a negative cycle on path from source to destination • (2) and (3) seem to be harder than (1). v v 5 5 u 2 -3 u 2 3 6 w -6 w CSE 202 - Shortest Paths

  5. Single Source Shortest Paths • Non-negative weights (problem B-1): From source, construct tree of shortest paths. • Why is this a tree? • Is this a Minimum Spanning Tree? Basic approach: label nodes with shortest path found so far. Relaxation step: choose any edge (u,v) . If it gives a shorter path to v, update v ’s label. 4 4 9 4 9 4 5 5 5 5 s s 9 9 1 2 1 2 7 9 relaxing this edge improves this node CSE 202 - Shortest Paths

  6. Single Source Shortest Paths • Simplest strategy: • Keep cycling through all edges • When all edges are relaxed, you’re done. What is upper bound on work? • Improvement: Mark nodes when we’re sure we have best path. Key insight: if you relax all edges out all marked nodes, the unmarked node that is closest to source can be marked. You can relax edge (u,v) only once - just after u is marked. O(V2) algorithm: mark node; relax its edges; find new min to mark. Dijkstra’s algorithm: Keep nodes on min-priority queue – need E Decrease-Key’s. Gives O(E lg V) - when is this faster? (Aside: can also get O(E + V lg V) using Fibonacci heap.) CSE 202 - Shortest Paths

  7. Single Source Shortest Paths • Negative weights allowed (problem B-2 and B-3) Why can’t we just add a constant to each weight? Why doesn’t Dijkstra’s algorithm work for B-2 ? Bellman-Ford: Cycle through edges V-1 times. O(VE) time. • PERT chart: Arbitrary weights, graph is acyclic: Is there a better algorithm than Bellman-Ford? CSE 202 - Shortest Paths

  8. All-pairs Shortest Paths • Naïve #1: Solve V single-source problems. • O(V2 E) for general problem. • O(V3) or O(V E lg V) for non-negative weights. • Naïve #2: Let dk(i,j) = shortest path from i to j • involving  k edges. d1(i,j) = is original weight matrix. Compute dk+1’s from dk’s by seeing if adding edge helps: dk+1(i,j) = Min { dk(i,m) + d1(m,j) } Hmmm ...this looks a lot like matrix multiplication If there are no negative cycles, dV-1(i,j) is solution. If there is a negative cycle, then dV-1  dV Complexity is O(V4) CSE 202 - Shortest Paths

  9. All-pairs Shortest Paths • No negative cycles – Divide and Conquer says: “Shortest path from a to b with at most 2k+1 hops is a shortest path from a to c with at most 2k hops followed by one from c to b with at most 2k hops.” T(k+1) = T(k) + V3 so T(lg V) is O(V3 lg V). This also looks like matrix multiplication, using d2k = dk x dk instead of dk+1= dk x d1 CSE 202 - Shortest Paths

  10. All-pairs Shortest Paths • No negative cycles – Dynamic Programming approach: Number cities 1 to V. Let S(a,b,k) be length of shortest path from a to b that uses only cities {1,2,...,k} as intermediate points. Intuition: At first (k=0), there are only direct flights between cities. Then (k=1) they allow city 1 to be an intermediate connection point. Then (k=2) they add city 2 as a possible connection point. Etc. Claim: S(a,b,k) = min{ S(a,b,k-1), S(a,k,k-1) + S(k,b,k-1) }. Intuition: either adding k as a possible connection point doesn’t shorten the trip, or it does. “No negative cycles”  needn’t consider paths going through k twice. Complexity: T(k+1)= T(k) + cV2, and T(0) is 0. Thus, T(V) is (V3). This is the Floyd-Warshall algorithm CSE 202 - Shortest Paths

  11. Johnson’s All-Pairs Shortest Paths • Motivation: for sparse non-negative weights, “O(V E lg V)” is better than “O(V3)” We’ll convert “arbitrary weights” (C3) to “non-negative” (C1) problem via reweighting. Dijkstra V times Floyd-Warshall 0 0 -1 -3 1 -3 4 8 -9 8 5 -10 Starting bonus or finishing penalty CSE 202 - Shortest Paths

  12. Johnson’s Algorithm • Reweighting can make all edges non-negative • Make node-weight(x) = shortest path to x from anywhere. • Sounds like all-pairs problem Slick trick use Bellman-Ford (only O(VE) time) 5 4 0 -1 -3 10 2 0 8 3 0 5 -10 0 0 s 0 -1 -3 0 0 8 5 -10 CSE 202 - Shortest Paths

  13. Glossary (in case symbols are weird) •        subset  element of infinity  empty set  for all  there exists  intersection  union  big theta  big omega  summation  >=  <=  about equal  not equal  natural numbers(N)  reals(R)  rationals(Q)  integers(Z) CSE 202 - Shortest Paths

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