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09 B: Graph Algorithms II CS1102S: Data Structures and Algorithms - - PowerPoint PPT Presentation

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Review: Graphs, Shortest Path Unweighted Shortest Paths Dijkstras Algorithm Correctness and Complexity of Dijkstras Algorithm 09 B: Graph Algorithms II CS1102S: Data Structures and Algorithms Martin Henz March 19, 2010 Generated on

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Review: Graphs, Shortest Path Unweighted Shortest Paths Dijkstra’s Algorithm Correctness and Complexity of Dijkstra’s Algorithm

09 B: Graph Algorithms II

CS1102S: Data Structures and Algorithms

Martin Henz

March 19, 2010

Generated on Thursday 18th March, 2010, 00:21 CS1102S: Data Structures and Algorithms 09 B: Graph Algorithms II 1

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Review: Graphs, Shortest Path

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Unweighted Shortest Paths

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Dijkstra’s Algorithm

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Correctness and Complexity of Dijkstra’s Algorithm

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Review: Graphs, Shortest Path

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Unweighted Shortest Paths

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Dijkstra’s Algorithm

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Correctness and Complexity of Dijkstra’s Algorithm

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Graph, Vertices, Edges

Graph A graph G = (V, E) consists of a set of vertices, V, and a set of edges, E.

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Graph, Vertices, Edges

Graph A graph G = (V, E) consists of a set of vertices, V, and a set of edges, E. Edge Each edge is a pair (v, w), where v, w ∈ V.

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Graph, Vertices, Edges

Graph A graph G = (V, E) consists of a set of vertices, V, and a set of edges, E. Edge Each edge is a pair (v, w), where v, w ∈ V. Directed graph If the pairs are ordered, then the graph is directed.

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Graph, Vertices, Edges

Graph A graph G = (V, E) consists of a set of vertices, V, and a set of edges, E. Edge Each edge is a pair (v, w), where v, w ∈ V. Directed graph If the pairs are ordered, then the graph is directed. Weight Sometimes the edges have a third component, knows as either a weight or a cost. Such graphs are called weighted graphs.

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Paths

Path A path in a graph is a sequence of vertices w1, w2, w3, . . . , wN such that (wi, wi+1) ∈ E for 1 ≤ i < N. It is said to lead from w1

  • t wN.

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The Shortest Path Problem

Input Weighted graph: associated with each edge (vi, vj) is a cost ci,j to traverse the edge.

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The Shortest Path Problem

Input Weighted graph: associated with each edge (vi, vj) is a cost ci,j to traverse the edge. Weighted path length Cost of path v1v2 · · · vN is N−1

i=1 ci,i+1.

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Single-Source Shortest-Path Problem

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.

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Example

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Example

Shortest path from v1 to v6

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Example

Shortest path from v1 to v6 has a cost of 6 and goes from v1 to v4 to v7 to v6.

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Unweighted Shortest Paths: Example

Find the shortest path from v3 to all other vertices

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Idea

Level-order traversal Start with s (distance 0) and proceed in phases currDist, each time going through all vertices. If vertex is “known” and has distance currDist, set the distance of its neighbors to currDist+1.

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Implementation

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Inefficiency

Careless loop In each phase, we go through all vertices. We can remember the “known” vertices in a data structure.

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Inefficiency

Careless loop In each phase, we go through all vertices. We can remember the “known” vertices in a data structure. Suitable data structure Queue: will contain the vertices in order of increasing distance

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Implementation

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Dijkstra’s Algorithm: Idea

Idea Treat nodes in the order of shortest distance

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Dijkstra’s Algorithm: Idea

Idea Treat nodes in the order of shortest distance

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Example

Initial configuration:

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Example

After v1 is declared known:

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Example

After v4 is declared known:

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Example

After v2 is declared known:

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Example

After v5 and then v3 are declared known:

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Example

After v7 is declared known:

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Example

After v6 is declared known:

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Pseudocode for Dijkstra’s Algorithm

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Shortest Subpath

Lemma Any subpath of a shortest path must also be a shortest path.

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Shortest Subpath

Lemma Any subpath of a shortest path must also be a shortest path. Proof By contradiction: Assume that a subpath p = vi · · · vj of the shortest path q = v1 · · · p · · · vk is not a shortest path. Then there is a shorter path p′ from vi to vj. Plug p′ into q to get q′ = v1 · · · p′ · · · vk shorter than q!

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Definition of Shortest Distance

Notation We use the notation δ(v, w) to denote the length of the shortest path from v to w.

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Definition of Shortest Distance

Notation We use the notation δ(v, w) to denote the length of the shortest path from v to w. Distance We call δ(v, w) the distance between v and w.

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dist is a Relaxation

Lemma At any point in time and for any vertex v, we have v . dist ≥ δ(s, v)

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dist is a Relaxation

Lemma At any point in time and for any vertex v, we have v . dist ≥ δ(s, v) Proof Idea We show this by proving that whenever we set v . dist to a finite value, there exists a path of that length.

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dist is a Relaxation

Lemma At any point in time and for any vertex v, we have v . dist ≥ δ(s, v)

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dist is a Relaxation

Lemma At any point in time and for any vertex v, we have v . dist ≥ δ(s, v) Proof By induction over the number of iterations of outer loop.

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dist is a Relaxation

Lemma At any point in time and for any vertex v, we have v . dist ≥ δ(s, v) Proof By induction over the number of iterations of outer loop. Start: claim holds for s (distance 0) and all other vertices (distance ∞)

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dist is a Relaxation

Lemma At any point in time and for any vertex v, we have v . dist ≥ δ(s, v) Proof By induction over the number of iterations of outer loop. Start: claim holds for s (distance 0) and all other vertices (distance ∞) Hypothesis: claim holds for previous iterations.

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dist is a Relaxation

Lemma At any point in time and for any vertex v, we have v . dist ≥ δ(s, v) Proof By induction over the number of iterations of outer loop. Start: claim holds for s (distance 0) and all other vertices (distance ∞) Hypothesis: claim holds for previous iterations. Induction step: Updates are done such that an edge weight is added to a previously computed dist value

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Order of Adding Vertices

Observation Vertices are becoming known in order of increasing dist values.

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Order of Adding Vertices

Observation Vertices are becoming known in order of increasing dist values. Observation Once a vertex becomes known, its dist value does not change.

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Correctness of Dijkstra’s Algorithm

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Main Correctness Lemma

Lemma When we set v.known = true then v. dist= δ(s, v).

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Main Correctness Lemma

Lemma When we set v.known = true then v. dist= δ(s, v). Correctness of Dijkstra’s algorithm Each iteration through the outer loop makes one vertex known. At the end every vertex v is known and thus, according to the lemma, its dist value is δ(s, v).

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Proving the Main Lemma

Proof by contradiction Assume that there is a vertex v for which the claim does not hold.

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Proving the Main Lemma

Proof by contradiction Assume that there is a vertex v for which the claim does not hold. Therefore, using relaxation property, when we set v .known = true then v . dist > δ(s, v).

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Proving the Main Lemma

Proof by contradiction Assume that there is a vertex v for which the claim does not hold. Therefore, using relaxation property, when we set v .known = true then v . dist > δ(s, v). Then, there must be a vertex u for which this is the case for the first time in the algorithm.

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Situation

Analysis Situation just before u.known = true: The value u. dist reflects the length of the path given by u.path.

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Situation

Analysis Situation just before u.known = true: The value u. dist reflects the length of the path given by u.path. The real shortest path The real shortest path from s to u is shorter than u. dist. Consider the real shortest path s · · · u.

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Situation

Jumping into “unknown” Since u.known still false, and s.known=true, there must be a pair of two neighboring vertices r and t in the real shortest path such that r .known=true and t .known=false. First pair There must be a first pair x and y where this is the case.

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Analysis of x and y

Processing of x We have processed x, but not yet y. Since y’s dist value is decreased by decrease(...), we know that y . dist ≤ x . dist+cx,y

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Analysis of x and y

Processing of x We have processed x, but not yet y. Since y’s dist value is decreased by decrease(...), we know that y . dist ≤ x . dist+cx,y Using hypothesis Since x becomes known earlier than u, we have x . dist= δ(s, x)

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Analysis of x and y

Processing of x We have processed x, but not yet y. Since y’s dist value is decreased by decrease(...), we know that y . dist ≤ x . dist+cx,y Using hypothesis Since x becomes known earlier than u, we have x . dist= δ(s, x) Shortest subpath s · · · xy is subpath of shortest path, thus δ(s, y) = δ(s, x) + cx,y = x . dist+cx,y

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Recap

We have y . dist ≤ x . dist+cx,y and δ(s, y) = δ(s, x) + cx,y = x . dist+cx,y

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Recap

We have y . dist ≤ x . dist+cx,y and δ(s, y) = δ(s, x) + cx,y = x . dist+cx,y and thus: y . dist ≤ δ(s, y)

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Recap

We have y . dist ≤ x . dist+cx,y and δ(s, y) = δ(s, x) + cx,y = x . dist+cx,y and thus: y . dist ≤ δ(s, y) and therefore y . dist = δ(s, y)

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Finale

Since y . dist = δ(s, y), we have y = u.

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Finale

Since y . dist = δ(s, y), we have y = u. Edge costs between y and u are non-negative, thus δ(s, y) ≤ δ(s, u) and thus y . dist= δ(s, y) ≤ δ(s, u) < u. dist

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Finale

Since y . dist = δ(s, y), we have y = u. Edge costs between y and u are non-negative, thus δ(s, y) ≤ δ(s, u) and thus y . dist= δ(s, y) ≤ δ(s, u) < u. dist If y . dist< u. dist, and since vertices become known in order of increasing distance, y would have become known before u, which contradicts the assumption that u is next vertex to become known!

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