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Chapter 9 : Graphs Part I

(Topological Sort & Shortest Path

Algorithms)

CE 221

Data Structures and Algorithms

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Text: Read Weiss, §9.1 – 9.3

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Definitions - I

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

• Each edge is a pair (v, w), where v, w є V.

• If the pair is ordered then G is directed(digraph).

• Vertex w is adjacent to v iff (v, w) є E.

• In an undirected graph with edge (v, w), w is adjacent to v and v is adjacent to w.

• Sometimes and edge has a third component, weight or cost.

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Definitions - II

• A path in a graph is w1, w2,...,wN such that

(wi, wi+1) є E for 1≤i<N. The length of such

a path is the number of edges on the path.

If a path from a vertex to itself contains no

edges, then the path length is zero. If G

contains an edge (v, v), then the path v, v

is called a loop.

• A simple path is a path such that all

vertices are distinct, except that the first

and the last could be the same.

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Definitions - III

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• A cycle in a directed graph is a path of length at

least 1 such that w1=wN. This cycle is simple if the

path is simple. For undirected graphs, the edges are

required to be distinct (Why?).

• A directed graph is acyclic if it has no cycles (DAG).

• An undirected graph is connected if there is a path

from every vertex to every other vertex. A directed

graph with this property is called strongly

connected. If directed graph is not, but underlying

undirected graph is, it is weakly connected. A

complete graph is a graph in which there is an

edge between every pair of vertices.

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Representation of Graphs - I

• One simple way is to use a two-dimensional array (adjacency matrixrepresentation). If vertices are numbered starting at 1, A[u][v]=true if (u, v) є E. Space requirement is Θ(|V|2).

• If the graph is not dense (sparse), adjacency lists may be used. The space requirement is O(|E|+|V|).

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Representation of Graphs - II

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Topological Sort - I

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•A topological sort is an ordering of vertices

in a DAG, such that if there is path from vi to

vj, then vj appears after vi in the ordering.

•A simple algorithm to find a topological

ordering is first to find any vertex with no

incoming edges. We can then print this vertex,

and remove it, along with its edges. Then apply

the same strategy to the rest of the graph. To

formalize this, define the indegree of a vertex

v as the number of edges (u, v).

8Izmir University of Economics

Topological Sort – Initial Attempt

• running time of

the algorithm is

O(|V|2).

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• We can remove the inefficiency by keeping all the unassigned vertices of indegree 0 in a special data structure (queue or stack). When a new vertex with degree zero is needed, it is returned by removing one from the queue, and when the indegrees of adjacent vertices are decremented, they are inserted into the queue if the indegree falls to zero. The running time is O(|E|+|V|)

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Topological Sort – A Better Algorithm

Homework Assignments

• 9.1, 9.2, 9.3, 9.4, 9.38

• You are requested to study and solve the

exercises. Note that these are for you to

practice only. You are not to deliver the

results to me.

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Shortest-Path Algorithms

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• The input is a weighted graph: associated with each edge (vi, vj) is a cost ci,j. The cost of a path v1v2...vN is ∑ci,i+1 for i in [1..N-1]. This is weighted path length, the unweighted path length on the other hand is merely the number of edges on the path, namely, N-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 Cycles

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• In the graph to the left, the shortest path from v1 to v6 has a

cost of 6 and the path itself is v1v4v7v6. The shortest

unweighted path has 2 edges.

• In the graph to the right, we have a negative cost. The path

from v5 to v4 has cost 1, but a shorter path exists by following

the loop v5v4v2v5v4 which has cost -5. This path is still not the

shortest, because we could stay in the loop arbitrarily long.

Shortest Path Length: Problems

We will examine 4 algorithms to solve four versions of

the problem

1.Unweighted shortest path O(|E|+|V|)

2.Weighted shortest path without negative edges

O(|E|log|V|) using queues

3.Weighted shortest path with negative edges

O(|E| . |V|)

4.Weighted shortest path of acyclic graphs linear

time Izmir University of Economics 13

Unweighted Shortest Paths• Using some vertex, s, which is an input

parameter, find the shortest path from s to all other vertices in an unweighted graph. Assume s=v3.

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Unweighted Shortest Paths

• Algorithm: find vertices that are at

distance 1, 2, ... N-1 by processing

vertices in layers (breadth-first search)

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Unweighted Shortest Paths

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Unweighted Shortest Paths

• Complexity O(|V|2)

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Unweighted Shortest Paths - Improvement

• At any point in time there

are only two types of

unknown vertices that

have dv≠∞. Some have

dv = currDist and the rest

have dv = currDist +1.

• We can make use of a

queue data structure.

• O(|E|+|V|)

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Weighted Shortest Path

Dijkstra’s Algorithm• With weighted shortest path,distance dv is

tentative. It turns out to be the shortest path

length from s to v using only known vertices as

intermediates.

• Greedy algorithm: proceeds in stages doing the

best at each stage. Dijkstra’s algorithm selects a

vertex v with smallest dv among all unknown

vertices and declares it known. Remainder of the

stage consists of updating the values dw for all

edges (v, w).

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Dijkstra’s Algorithm - Example

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►

►

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Dijkstra’s Algorithm - Example

• A proof by contradiction will show that this

algorithm always works as long as no

edge has a negative cost.Izmir University of Economics 21

►

►

►

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Dijkstra’s Algorithm - Pseudocode

• If the vertices are

sequentially scanned to

find minimum dv, each

phase will take O(|V|) to

find the minimum, thus

O(|V|2) over the course

of the algorithm.

• The time for updates is

constant and at most

one update per edge for

a total of O(|E|).

• Therefore the total time

spent is O(|V|2+|E|).

• If the graph is dense,

OPTIMAL.

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Dijkstra’s Algorithm-What if the

graph is sparse?• If the graph is sparse |E|=θ(|V|), algorithm is too

slow. The distances of vertices need to be kept

in a priority queue.

• Selection of vertex with minimum distance via

deleteMin, and updates via decreaseKey

operation. Hence; O(|E|log|V|+|V|log|V|)

• find operations are not supported, so you need

to be able to maintain locations of di in the heap

and update them as they change.

• Alternative: insert w and dw with every update.Izmir University of Economics 23

Graphs with negative edge

costs• Dijkstra’s algorithm does not work with

negative edge costs. Once a vertex u is

known, it is possible that from some other

unknown vertex v, there is a path back to

u that is very negative.

• Algorithm: A combination of weighted and

unweighted algorithms. Forget about the

concept of known vertices.

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Graphs with negative edge costs - I

• O(|E|*|V|). Each vertex

can dequeue at most

O(|V|) times. (Why?

Algorithm computes

shortest paths with at

most 0, 1, ..., |V|-1 edges

in this order). Hence, the

result!

• If negative cost cycles,

then each vertex should

be checked to have been

dequeued at most |V|

times.Izmir University of Economics 25

Acyclic Graphs

• If the graph is known to be acyclic, the

order in which vertices are declared

known, can be set to be the topological

order.

• Running time = O(|V|+|E|)

• This selection rule works because when a

vertex is selected, its distance can no

longer be lowered, since by topological

ordering rule it has no incoming edges

emanating from unknown nodes.Izmir University of Economics 26

Homework Assignments

• 9.5, 9.7, 9.10, 9.40, 9.42, 9.44, 9.46, 9.47,

9.52

• You are requested to study and solve the

exercises. Note that these are for you to

practice only. You are not to deliver the

results to me.

Izmir University of Economics 27