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Topic 12:Shortest Path

Basics

Dijkstra AlgorithmRelaxationBellman-Ford Alg.

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Shortest Path

Given a graph G = (V, E) with weight function w : E --> R

Shortest path weight:

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Various Problems

Single-source shortest-paths problem Given source node s to all nodes from V

Single-destination shortest-paths problem From all nodes in V to a destination u

Single-pair shortest-path problem Shortest path between u and v

All-pairs shortest-paths problem Shortest paths between all pairs of nodes

Example

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Negative-weight Edges

Some edges may have negative weights

If there is a negative cycle reachable from s: Shortest path is no longer well-defined Example

Otherwise, it is fine

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Cycles

Shortest path cannot have cycles inside Negative cycles : already eliminated Positive cycles: can be removed 0-weight cycles: can be removed

A path with no cycles inside is also called a simple path.

No-Cycle Theorem: Given any two vertices of graph, there exists a shortest path between them with no cycle.

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Optimal Substructure Property

Proof Proof by contradiction. If there is a shorter path from to , then the path is a

shorter path from to Contradiciton.

Optimal Substructure Property (Theorem): If is a shortest path from to ,

then any sub-path is also a shortest path.

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Shortest-paths Tree For every node v V, π[v] is the predecessor of v

in a shortest path from source s to v Nil if does not exist

A shortest-path tree Root is source s Edges are (π[v] , v)

The shortest path between s and v is the unique tree path from root s to v.

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Example: directed graph

Example: undirected graph

Key property: Given shortest path tree rooted at s ( in this example),

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Shortest Path Tree Question:

Given a shortest path tree of G rooted at s, how fast can we report the shortest path distance from s to all other vertices in G?

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Key property: Given shortest path tree rooted at s ( in this example),

one can obtain the shortest path from s to every other vertex connected to it.

O(V)

The output of shortest-path algorithms usually contains both shortest-path tree and distances.

Shortest Path Tree Related to minimum spanning tree?

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Goal:

Input: a weighted graph G = (V, E), with positive weights! and a source node vs V

Output: For every vertex v V,

v.distance = (vs , v) v.parent =

Shortest-paths tree induced by v.parent

Breadth-First Search

Recall BFS: Shortest path for unweighted graph (i.e, each edge has

weight 1).

Would the same idea work for weighted graph?

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Example

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We can no longer guarantee that at the time the algorithm first discovers a node, the distance is shortest.

BFS is only for shortest number of edges to reach a node!

How to guarantee that when we first reach a node, the distance is shortest?

Dijkstra Algorithm

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Intuition: If Then is the shortest distance to reach through . Note, this does not have to be the same as shortest

distance to

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Example

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Correctness

Invariance: At the beginning of the While-loop, all vertices already

discovered have correct shortest distance value. Prove that this invariance is maintained:

Base case: in the first iteration, only is discovered, and this invariance holds.

Inductive step: If the invariance holds at the beginning of the -th iteration, then it holds at the end of -th iteration (i.e, it holds at the beginning of -th iteration).

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Proof of Induction Step Let as identified in Line 5 of the algorithm.

Let P be the shortest path from to . is the weight of the path .

Proof that any other path from to has larger weight. Consider any other path from to . Let be the first edge in that connects a vertex from to a vertex

in . Argue that the weight of subpath from to is at least . Hence the weight of is larger than that of .

Done.

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Why do we require that the weights are all positive?

Example.

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Running Time Analysis

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Running Time Analysis

Naïve implementation: Spend time to identify in Line 5. Total time:

How to identify more efficiently? First improvement: Storing cost at vertices.

.distance: current estimate of shortest path weight from to

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Storing Cost at Vertices

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Running Time Analysis

First improvement: Spend time to identify in Line 5. Total time: Note: the weight stored at is NOT the smallest weight

of edge connecting to any visited vertex That is how Prim’s MST algorithm works.

How to identify more efficiently? Second improvement: Use Priority Queue to store /

maintain vertex costs! .distance: current estimate of shortest path weight from to

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Second Improvement: Priority Queue

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Running Time Analysis Priority Queue: Q;

Total size: # Q.insert:

Total time: # Q.DeleteMin:

Total time: # Q.IsNotEmpty() and # Q.MinKey(): ,

Total time: # Q.DecreaseKey:

Total time:

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Total time:

Remarks Similar idea as breadth first search:

Greedy type of algorithm Guarantees that when we first discover a node, the

distance is the correct shortest path weight Similar to Prim’s Alg for MST:

But the cost at each vertex is defined differently. Also works for directed graphs But require weights to be positive

Can be improved to by using a better implementation of priority queue (Fibonacci heap)

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A more general algorithm

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Shortest Path

Given a graph G = (V, E) with weight function w : E --> R

Shortest path weight:

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Goal:

Input: directed weighted graph G = (V, E), source node s V

Output: For every vertex v V,

d[v] = (s, v) π[v]

Shortest-paths tree induced by π[v]

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Previously

Dijkstra Algorithm: Compute shortest paths from a source node when the

input graph has no negative edge weights!

Now:

A single source shortest path algorithm for arbitrary edge weight!

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Relaxation

Maintain shortest-path estimate d[v] for each node d[v]: initialize

Algorithms will repeatedly apply Relax.Differ in the order of Relax operation

Intuition:Do we have a shorter path

if use edge (u,v) ?

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Shortest-paths Properties Triangle inequality

For (u, v) E, (s, v) ≤ (s, u) + w(u, v)

Equality holds when u is predecessor of v.

Upper-bound property If start with d[v] = and apply Relax operations Always have d[v] ≥ (s, v) Once d[v] = (s, v), never changes

Proof by contradiction

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Properties cont.

Convergence property Suppose s u -> v is the shortest path from s to v Currently d[u] = (s, u) Relax (u,v) will set d[v] = (s, v)

Proof: follow from Relaxation defintion

Path-relaxation property Given a shortest path Apply Relax in order Then, afterwards,

Proof: Follow from above property

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Bellman-Ford Algorithm Allow negative weights Follow the frame work that first, initialize:

Then apply a set of Relax compute d[v] and π[v] Return False if there exists negative cycles

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Pseudo-code

Time complexity:O(VE)

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Example

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Correctness Why d[v] = (s, v) for every node v ?

Proof: Use path-relaxation property

Return True (or False) iff there is (or is no) negative cycles ? If there is no negative cycle: obvious

use triangle inequality If there is negative cycle:

Proof by contradiction