Path Problems in Acyclic Graphs in Java

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Recall that a directed acyclic graph has no cycles This important class of graphs simplifies the solution to the shortest-path problem For instance, we do not have to worry about negative-cost cycles because there are no cycles Thus we consider the following problem
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WEIGHTED SINGLE-SOURCE, SHORTEST-PATH PROBLEM FOR ACYCLIC GRAPHS FINDTHE SHORTEST PATH (MEASURED BY TOTAL COST) FROM A DESIGNATED VERTEX S TO EVERY VERTEX IN AN ACYCLIC GRAPH EDGE COSTS ARE
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1551 Topological Sorting
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Before considering the shortest-path problem, let us examine a related problem: a topological sort A topological sort orders vertices in a directed acyclic graph such that if there is a path from u to v, then v appears after u in the ordering For instance, a graph is typically used to represent the prerequisite requirement for courses at universities An edge (v, w) indicates that course v must be completed before course w may be attempted A topological order of the courses is any sequence that does not violate the prerequisite requiremerits
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A topo~ogica~ sort Orders vertices in a d~rected acyclic graph that if there is a path from u to V, then "appears afieruin the ordering A graph that has a cycle havea topological order
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Graphs and Paths
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1 / / Single-source negative-weighted shortest-path algorithm & startName ) 3 { 4 vmap::iterator itr = vertexMapfind( startName 1 ; 5 if( itr == vertexMapend( ) j 6 throw Graph~xception( startName + " is not a vertex" ) ; 7 8 clearAll( j ; 9 Vertex *start = (*itr)second; 10 list<Vertex * > q; 11 qpush-back( start ) ; start->dist = 0; start->scratch++; 12 while( !qempty( ) ) 13 14 { 15 Vertex *v = qfront ( ) ; qpop-front ( ) ; 16 if( v->scratch++ > 2 * vertexMapsize( j ) 17 throw GraphExceptionj "Negative cycle detected" ) ; 18 19 for( int i = 0; i < v->adjsize( ) ; i++ ) 20 i 21 Edge e = v->adj[ i I ; 22 Vertex *w = edest; 23 double cvw = ecost; 24 if( w->dist > v->dist + cvw ) 25 26 I 27 w->dist = v-zdist + cvw; 28 w->prev = v; 29 / / Enqueue only if not already on the queue 30 if( w->scratch++ % 2 == 0 ) 31 q push-back ( w ) ; 32 else 33 w->scratch++; / / In effect, adds 2 34 } 35 1 36 } 37 }
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2 void Graph::negative( const string
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Figure 1530 A negative-weighted, shortest-path algorithm: Negative edges are allowed
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Clearly, a topological sort is not possible if a graph has a cycle because, for two vertices v and w on the cycle, there is a path from v to w and w to v Thus any ordering of v and w would contradict one of the two paths A graph may have several topological orders, and in most cases, any legal ordering will do
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Path problems in Acyclic Graphs
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Figure 1531 A topological sort The conventions are the same as those in Figure 1522
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In a simple algorithm for performing a topological sort we first find any vertex v that has no incoming edges Then we print the vertex and logically remove it, along with its edges, from the graph Finally, we apply the same strategy to the rest of the graph More formally, we say that the indegree of a vertex v is the number of incoming edges (u, v) We compute the indegrees of all vertices in the graph In practice, logica11y remove means that we lower the count of incoming edges for each vertex adjacent to v Figure 1531 shows the algorithm applied to an acyclic
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The indegreeof a vertex is the number of incoming edges A topological sort can be performed in linear time by repeatedly and logically removing vertices that have no incoming edges
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The algorithm produces the correct answer and detects cycles if the graph is not acyclic
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The running time is linear if a queue is used
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graph The indegree is computed for each vertex Vertex V, has indegree 0, so it is first in the topological order If there were several vertices of indegree 0, we could choose any one of them When V 2 and its edges are removed from the graph, the indegrees of V,,, V,, and V5 are all decremented by 1 Now V has indegree 0, so it is next in the topological order, and V, and V, , have their indegrees lowered The algorithm continues, and the remaining vertices are examined in the order V1, V3, V4 V6,and Vs To reiterate, we do not physically delete edges from the graph; removing edges just makes it easier to see how the indegree count is lowered Two important issues to consider are correctness and eficiencj Clearly, any ordering produced by the algorithm is a topological order The question is whether every acyclic graph has a topological order, and if so, whether our algorithm is guaranteed to find one The answer is yes to both questions If at any point there are unseen vertices but none of them have an indegree of 0, we are guaranteed that a cycle exists To illustrate we can pick any vertex A,, Because A has an incoming edge, let A, be the vertex connected to it And as A , has an incoming edge, let A , be the vertex connected to it We repeat this process N times, where N is the number of unprocessed vertices left in the graph Among Ao, A ,, ,A,, there must be two identical vertices (because there are N vertices but N + 1 A,'s) Tracing backward between those identical A, and A, exhibits a cycle We can implement the algorithm in linear time by placing all unprocessed indegree 0 vertices on a queue Initially, all vertices of indegree 0 are placed on the queue To find the next vertex in the topological order, we merely get and remove the front item from the queue When a vertex has its indegree lowered to 0, it is placed on the queue If the queue empties before all the vertices have been topologically sorted, the graph has a cycle The running time is clearly linear, by the same reasoning used in the unweighted shortest-path algorithm
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