1 / 17

§3 Shortest Path Algorithms

Given a digraph G = ( V, E ), and a cost function c ( e ) for e  E( G ). The length of a path P from source to destination is (also called weighted path length ). 2. v 1. v 2. 2. v 1. v 2. 4. 1. –10. 4. 1. 3. 10. 3. 2. 2. v 3. v 4. v 5. 2. 2.

fritz-wells
Download Presentation

§3 Shortest Path Algorithms

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Given a digraph G = ( V, E ), and a cost function c( e ) for e E( G ). The length of a path P from source to destination is (also called weighted path length). 2 v1 v2 2 v1 v2 4 1 –10 4 1 3 10 3 2 2 v3 v4 v5 2 2 v3 v4 v5 8 4 8 6 5 4 6 5 1 v6 v7 1 v6 v7 §3 Shortest Path Algorithms 1. Single-Source Shortest-Path Problem Given as input a weighted graph, G = ( V, E ), and a distinguished vertex, s, find the shortest weighted path from s to every other vertex in G. Negative-cost cycle Note: If there is no negative-cost cycle, the shortest path from s to s is defined to be zero. 1/17

  2. §3 Shortest Path Algorithms v1 v2 v3 v4 v5 v6 v7  Unweighted Shortest Paths  Sketch of the idea 0: v3 1 2 Breadth-first search 1:  v1 and v6 2 0 3 2:  v2 and v4 3:  1 3 v5 and v7  Implementation Table[ i ].Dist ::= distance from s to vi/* initialized to be  except for s */ Table[ i ].Known ::= 1 if vi is checked; or 0 if not Table[ i ].Path ::= for tracking the path /* initialized to be 0 */ 2/17

  3. §3 Shortest Path Algorithms v8 v7 v6 v5 v3 v2 v1 v9 v4 void Unweighted( Table T ) { int CurrDist; Vertex V, W; for ( CurrDist = 0; CurrDist < NumVertex; CurrDist ++ ) { for ( each vertex V ) if ( !T[ V ].Known && T[ V ].Dist == CurrDist ) { T[ V ].Known = true; for ( each W adjacent to V ) if ( T[ W ].Dist == Infinity ) { T[ W ].Dist = CurrDist + 1; T[ W ].Path = V; } /* end-if Dist == Infinity */ } /* end-if !Known && Dist == CurrDist */ } /* end-for CurrDist */ } If V is unknown yet has Dist < Infinity, then Dist is either CurrDist or CurrDist+1.  T = O( |V|2 ) The worst case: 3/17

  4. §3 Shortest Path Algorithms v1 v2 v3 v4 v5 v6 v7 DistPath v1  0 v2  0 v3 0 0 v4  0 v5  0 v6  0 v7  0  Improvement 1 2 void Unweighted( Table T ) { /* T is initialized with the source vertex S given */ Queue Q; Vertex V, W; Q = CreateQueue (NumVertex ); MakeEmpty( Q ); Enqueue( S, Q ); /* Enqueue the source vertex */ while ( !IsEmpty( Q ) ) { V = Dequeue( Q ); T[ V ].Known = true; /* not really necessary */ for ( each W adjacent to V ) if ( T[ W ].Dist == Infinity ) { T[ W ].Dist = T[ V ].Dist + 1; T[ W ].Path = V; Enqueue( W, Q ); } /* end-if Dist == Infinity */ } /* end-while */ DisposeQueue( Q ); /* free memory */ } 0 3 2 3 1 1 v3 v7 2 v1 v5 v4 2 v1 v2 3 v2 1 v3 v6 3 v4 v3 v1 T = O( |V| + |E| ) 4/17

  5. §3 Shortest Path Algorithms Dijkstra’s Algorithm (for weighted shortest paths) Let S = { s and vi’s whose shortest paths have been found } For any u S, define distance [ u ] = minimal length of path { s ( vi S )  u }. If the paths are generated in non-decreasing order, then  the shortest path must go through ONLYvi S ;  u is chosen so that distance[ u ] = min{ wS | distance[ w ] } (If u is not unique, then we may select any of them) ; /* Greedy Method */ Why? If it is not true, then there must be a vertex w on this path that is not in S. Then ...  if distance [ u1 ] < distance [ u2 ] and we add u1 into S, then distance [ u2 ] may change. If so, a shorter path from s to u2 must go through u1 and distance’ [ u2 ] = distance [ u1 ] + length(< u1, u2>). 5/17

  6. §3 Shortest Path Algorithms Dist Path v1 0 0 6 8 9 5 1 3 3 2 v3 v7 v4 v1 v4 v4 v4 v1 v2  0 v3  0 v4  0 v5  0 2 v1 v2 v6  0 4 1 3 10 v7  0 2 2 v3 v4 v5 8 4 6 5 1 v6 v7 void Dijkstra( Table T ) { /* T is initialized by Figure 9.30 on p.303 */ Vertex V, W; for ( ; ; ) { V = smallest unknown distance vertex; if ( V == NotAVertex ) break; T[ V ].Known = true; for ( each W adjacent to V ) if ( !T[ W ].Known ) if ( T[ V ].Dist + Cvw < T[ W ].Dist ) { Decrease( T[ W ].Dist to T[ V ].Dist + Cvw ); T[ W ].Path = V; } /* end-if update W */ } /* end-for( ; ; ) */ } /* O( |V| ) */ /* not work for edge with negative cost */ Please read Figure 9.31 on p.304 for printing the path. 6/17

  7. §3 Shortest Path Algorithms  Implementation 1 V = smallest unknown distance vertex; /* simply scan the table – O( |V| ) */ Good if the graph is dense T = O( |V|2 + |E| )  Implementation 2 Home work: p.339 9.5 Find the shortest paths p.340 9.10 Modify Dijkstra’s algorithm V = smallest unknown distance vertex; /* keep distances in a priority queue and call DeleteMin – O( log|V| ) */ Decrease( T[ W ].Dist to T[ V ].Dist + Cvw ); Good if the graph is sparse /* Method 1: DecreaseKey – O( log|V| ) */ T = O( |V| log|V| + |E| log|V| ) = O( |E| log|V| ) /* Method 2: insert W with updated Dist into the priority queue */ /* Must keep doing DeleteMin until an unknown vertex emerges */ T = O( |E| log|V| ) but requires |E| DeleteMin with |E| space  Other improvements: Pairing heap (Ch.12) and Fibonacci heap (Ch. 11) 7/17

  8. §3 Shortest Path Algorithms Too simple, and naïve… Try this one out: 2 1 2 – 2 1 3 4 2  Graphs with Negative Edge Costs Hey I have a good idea: why don’t we simply add a constant  to each edge and thus remove negative edges? void WeightedNegative( Table T ) { /* T is initialized by Figure 9.30 on p.303 */ Queue Q; Vertex V, W; Q = CreateQueue (NumVertex ); MakeEmpty( Q ); Enqueue( S, Q ); /* Enqueue the source vertex */ while ( !IsEmpty( Q ) ) { V = Dequeue( Q ); for ( each W adjacent to V ) if ( T[ V ].Dist + Cvw < T[ W ].Dist ) { T[ W ].Dist = T[ V ].Dist + Cvw; T[ W ].Path = V; if ( W is not already in Q ) Enqueue( W, Q ); } /* end-if update */ } /* end-while */ DisposeQueue( Q ); /* free memory */ } T = O( |V|  |E| ) /* each vertex can dequeue at most |V| times */ /* no longer once per edge */ /* negative-cost cycle will cause indefinite loop */ 8/17

  9. §3 Shortest Path Algorithms Signals the completion ofai ai ::= activity vj Index of vertex EC Time Lasting Time Slack Time LC Time  Acyclic Graphs If the graph is acyclic, vertices may be selected in topological order since when a vertex is selected, its distance can no longer be lowered without any incoming edges from unknown nodes. T = O( |E| + |V| ) and no priority queue is needed. Application: AOE ( Activity On Edge ) Networks —— scheduling a project  EC[ j ] \ LC[ j ] ::= the earliest \ latest completion time for node vj  CPM ( Critical Path Method ) 9/17

  10. §3 Shortest Path Algorithms 6 6 16 16 a0=6 a3=1 0 0 a6=9 a9=2 4 8 5 1 7 3 6 2 0 start 7 7 a1=4 18 18 a7=7 finish a10=4 4 6 a4=1 14 14 a2=5 a8=4 5 5 7 7 a5=2  Slack Time of <v,w> = 〖Example〗 AOE network of a hypothetical project 2 2 a11=0 3 Dummy activity  Calculation of EC: Start from v0, for any ai = <v, w>, we have  Calculation of LC: Start from the last vertex v8, for any ai = <v, w>, we have  Critical Path ::= path consisting entirely of zero-slack edges. 10/17

  11. §3 Shortest Path Algorithms 2. All-Pairs Shortest Path Problem For all pairs of vi and vj( i j ), find the shortest path between. Method 1 Use single-source algorithmfor |V| times. T = O( |V|3 ) – works fast on sparse graph. Method 2 O( |V|3 ) algorithm given in Ch.10, works faster on dense graphs. 11/17

  12. s 2 3 1 a b 3 2 4 c d Determine the maximum amount of flow that can pass from s to t. 2 3 t §4 Network Flow Problems 〖Example〗 Consider the following network of pipes: source Note: Total coming in (v)  Total going out (v) where v  { s, t } sink 12/17

  13. §4 Network Flow Problems s s s 2 3 1 a b a a b b 3 2 4 c d c d c d 2 3 t t t G Flow Gf Residual Gr 1. A Simple Algorithm Step 1: Find any path s t in Gr; Step 4: If (there is a path s t in Gr ) Goto Step 1; Else End. Step 2: Take the minimum edge on this path as the amount of flow and add to Gf; augmenting path Step 3: Update Grand remove the 0 flow edges; 13/17

  14. §4 Network Flow Problems s 2 3 1 a b 3 2 4 c d 2 3 t G You are right! What if I pick up the path s a  d  t first? It is simple indeed. But I bet that you will point out some problems here… Uh-oh… Seems we cannot be greedy at this point. 14/17

  15. §4 Network Flow Problems For each edge ( v, w ) with flow fv, w in Gf, add an edge ( w, v ) with flow fv, w in Gr. s s 2 3 1 a b a s b 3 2 3 2 3 2 3 2 4 1 c d a b c d 2 2 2 3 4 2 2 3 2 3 2 3 c d t t 2 3 3 2 G 3 Flow Gf 2 t Residual Gr 2. A Solution – allow the algorithm to undo its decisions 3 1 1 1 〖Proposition〗If the edge capabilities are rational numbers, this algorithm always terminate with a maximum flow. Note: The algorithm works for G with cycles as well. 15/17

  16. §4 Network Flow Problems s 1 000 000 1 000 000 1 a b 1 000 000 1 000 000 t 3. Analysis ( If the capacities are all integers )  An augmenting path can be found by an unweighted shortest path algorithm. T = O( ) where f is the maximum flow. f · |E|  Always choose the augmenting path that allows the largest increase in flow. /* modify Dijkstra’s algorithm */ T = Taugmentation * Tfind a path = O( |E| log capmax) * O( |E| log |V| ) = O( |E|2 log |V| ) if capmax is a small integer. 16/17

  17. §4 Network Flow Problems  Always choose the augmenting path that has the least number of edges. Home work: p.341 9.11 A test case. T = Taugmentation * Tfind a path = O( |E| ) * O( |E| · |V| ) /* unweighted shortest path algorithm */ = O( |E|2 |V| ) • Note: • If every v  { s, t } has either a single incoming edge of capacity 1 or a single outgoing edge of capacity 1, then time bound is reduced to O( |E| |V|1/2 ). • The min-cost flow problem is to find, among all maximum flows, the one flow of minimum cost provided that each edge has a cost per unit of flow. 17/17

More Related