Post on 21-Dec-2015
Paths
A path in a graph is a continuous way of getting from one vertex to another by using a sequence of edges.
EG: could get from 1 to 3 circuitously as follows: 1-e12-e11-e33-e42-e62-e52-e43
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1 2
3 4
e1
e3
e2
e4e5
e6
e7
Paths
DEF: A path of length n in an undirected graph is a sequence of n edges e1, e2, … ,en such that each consecutive pair ei , ei+1 share a common vertex. In a simple graph, one may instead define a path of length n as a sequence of n+1 vertices v0, v1, v2, … ,vn such that each consecutive pair vi , vi+1 are adjacent. Paths of length 0 are also allowed according to this definition.
Q: Why does the second definition work for simple graphs?
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Paths
A: For simple graphs, any edge is unique between vertices so listing the vertices gives us the edge-sequence as well.
DEF: A simple path contains no duplicate edges (though duplicate vertices are allowed). A cycle (or circuit) is a path which starts and ends at the same vertex.
Note: Simple paths need not be in simple graphs. E.g., may contain loops.
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PathsA: The following path from 1 to 2 is a maximal simple
path because • simple: each of its edges appears exactly once• maximal: because it contains every edge except
the unreachable edge e7
The maximal path: e1,e5,e6,e2,e3,e4
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1 2
3 4
e1
e3
e2e4
e5
e6
e7
Paths in Directed Graphs
One can define paths for directed graphs by insisting that the target of each edge in the path is the source of the next edge:
DEF: A path of length n in a directed graph is a sequence of n edges e1, e2, … ,en such that the target of ei is the source ei+1 for each i. In a digraph, one may instead define a path of length n as a sequence of n+1 vertices v0, v1, v2, … ,vn such that for each consecutive pair vi , vi+1 there is an edge from vi to vi+1 .
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Paths in Directed Graphs
Q: Consider digraph adjacency matrix:
1. Find a path from 1 to 4.2. Is there a path from 4 to 1?
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0010
0011
1010
0101
Paths in Directed Graphs
A:
1. 1324.2. There’s no path from 4 to 1. From 4 must
go to 2, from 2 must stay at 2 or return to 4. In other words 2 and 4 are disconnected from 1. 10
0010
0011
1010
0101
ConnectivityDEF: Let G be a pseudograph. Let u and v be
vertices. u and v are connected to each other if there is a path in G which starts at u and ends at v. G is said to be connected if all vertices are connected to each other.
1. Note: Any vertex is automatically connected to itself via the empty path.
2. Note: A suitable definition for directed graphs will follow later.
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English Connectivity PuzzleCan define a puzzling graph G as follows:V = {3-letter English words}E : two words are connected if can get one
word from the other by changing a single letter.
One small subgraph of G is:
Q: Is “fun” connected to “car” ?14
jobrob jab
Connected ComponentsDEF: A connected component (or just
component) in a graph G is a set of vertices such that all vertices in the set are connected to each other and every possible connected vertex is included.
Q: What are the connected components of the following graph?
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6 2
435
1
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Connected ComponentsA: The components are {1,3,5},{2,4,6},{7} and
{8} as one can see visually by pulling components apart:
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6 2
4
7835
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N-ConnectivityNot all connected graphs are created equal! Q: Rate following graphs in terms of their
design value for computer networks:1)
2)
3)
4) 18
N-Connectivity
A: Want all computers to be connected, even if 1 computer goes down:
1) 2nd best. However, there’sa weak link— “cut vertex”2) 3rd best. Connectedbut any computer can disconnect 3) Worst! Already disconnected4) Best! Network dies only with 2 bad computers
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N-Connectivity
The network is best because it can only become disconnected when 2 vertices are removed. In other words, it is 2-connected. Formally:
DEF: A connected simple graph with 3 or more vertices is 2-connected if it remains connected when any vertex is removed. When the graph is not 2-connected, we call the disconnecting vertex a cut vertex.
Q: Why the condition on the number of vertices?
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N-Connectivity
A: To avoid being 2-connected.
There is also a notion of N-Connectivity where we require at least N vertices to be removed to disconnect the graph.
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Connectivity inDirected Graphs
1) Weakly connected : can get from a to b in underlying undirected graph
2) Semi-connected (my terminology): can get from a to b OR from b to a in digraph
3) Strongly connected : can get from a to b AND from b to a in the digraph
DEF: A graph is strongly (resp. semi, resp. weakly) connected if every pair of vertices is connected in the same sense.
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Euler and Hamilton Paths-Motivation
An pictorial way to motivate the graph theoretic concepts of Eulerian and Hamiltonian paths and circuits is with two puzzles:
• The pencil drawing problem• The taxicab problem
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Pencil Drawing Problem-Euler Paths
Which of the following pictures can be drawn on paper without ever lifting the pencil and without retracing over any segment?
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Pencil Drawing Problem-Euler Paths
Graph Theoretically: Which of the following graphs has an Euler path?
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Euler Paths and CircuitsDefinition
DEF: An Euler path in a graph G is a simple path containing every edge in G. An Euler circuit (or Euler cycle) is a cycle which is an Euler path.
NOTE: The definition applies both to undirected as well as directed graphs of all types.
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Taxicab Problem-Hamilton Paths
Can a taxicab driver milk his hapless customer by visiting every intersection exactly once, when driving from point A to point B ?
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A
B
Taxicab Problem-Hamilton Paths
Graph Theoretically: Is there a Hamilton path from A to B in the following graph?
(NO in this case)
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A
B
Hamilton Paths and CircuitsDefinition
DEF: A Hamilton path in a graph G is a path which visits ever vertex in G exactly once. A Hamilton circuit (or Hamilton cycle) is a cycle which visits every vertex exactly once, except for the first vertex, which is also visited at the end of the cycle.
NOTE: Again, the definition applies both to undirected as well as directed graphs of all types.
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Implications to CS
Finding Hamilton paths is a very important problem in CS.
EG: Visit every city (vertex) in a region using the least trips (edges) as possible.
EG: Encode all bit strings of a certain length as economically as possible so that only change one bit at a time. (Gray codes).
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Implications to CS
Analyzing difficulty of Euler vs. Hamilton paths is a great CS case study.
• Finding Euler paths can be done in O (n) time
• Finding Hamilton paths is NP-complete!Slight change in definition can result in dramatic
algorithmic bifurcation!
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Finding Euler Paths
To find Euler paths, we’ll first give an algorithm for finding Euler cycles and then modify it to give Euler paths.
THM: An undirected graph G has an Euler circuit iff it is connected and every vertex has even degree.
NOTE: for directed graphs the condition is that G be weakly connected and that every vertex has same in-degree as out-degree.
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Finding Euler Circuits
Let’s prove the theorem constructively. Constructive means that the proof will actually contain an algorithm for constructing the Euler path, when it exists.
Part 1) Suppose G is connected and each vertex has even degree. Construct an Euler cycle. We prove this by strong induction on m –the number of edges in G.
Base case m = 0: Since G is connected and contains no edges, it must consist of a single vertex. The empty path is an Euler cycle.1
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Finding Euler Circuits
Induction step for m+1 edges, assuming proved this up to m 0.
CLAIM: G contains a simple cycle. Consider an edge e (since m+1 > 0). If e is a self-loop, then is a simple cycle. So can assume that e is a loopless edge: v v’
Since deg(v’ ) > 1 (all degrees are even), another edge e’ must be incident with v’ :
39v’ v’’
e
Finding Euler Circuits
So can continue adding edges until we find an edge whose new endpoint has already been encountered during the process. This endpoint is a vertex which is seen twice so at which a simple cycle is based!
(This proves claim)
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Simplecycle
Finding Euler Circuits
Gives rise to a recursive proof/algorithm:1) Find a simple cycle in connected graph with m+1
edges.2) Delete all the edges from the cycle and find Euler
cycles in each resulting component3) Amalgamate Euler cycles together using the simple
cycle obtaining wanted Euler cycle.Let’s see how the amalgamation process works:
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Finding Euler CircuitsMohammed’s Scimitars
Found a cycle after starting from middle vertex.
Delete the cycle:
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Finding Euler CircuitsMohammed’s Scimitars
Found a cycle after starting from middle vertex.
Delete the cycle:
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Finding Euler CircuitsMohammed’s Scimitars
Found a cycle after starting from middle vertex.
Delete the cycle:
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Finding Euler CircuitsMohammed’s Scimitars
Found a cycle after starting from middle vertex.
Delete the cycle:
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Finding Euler CircuitsMohammed’s Scimitars
Found a cycle after starting from middle vertex.
Delete the cycle:
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Finding Euler CircuitsMohammed’s Scimitars
This time, found a cycle starting and ending at middle vertex:
Amalgamate these cycles together from a point of intersection, and delete from graph:
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Finding Euler CircuitsMohammed’s Scimitars
Find another cycle from middle vertex:
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123 4 5
6
789
10
Finding Euler CircuitsMohammed’s Scimitars
Find another cycle from middle vertex:
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123 4 5
6
789
10
11
Finding Euler CircuitsMohammed’s Scimitars
Amalgamate it to Euler cycle of deleted graph, and delete it. Need to insert cycle between former edges 10 & 11:
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123 4 5
6
789
10
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Finding Euler CircuitsMohammed’s Scimitars
Finally, need to add the triangle.
Use same naïve approach looking for cycle in remaining component:
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123 4 5
6
789
10
??
Finding Euler CircuitsMohammed’s Scimitars
Finally, need to add the triangle.
Use same naïve approach looking for cycle in remaining component:
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123 4 5
6
789
10
?? 11
Finding Euler CircuitsMohammed’s Scimitars
Finally, need to add the triangle.
Use same naïve approach looking for cycle in remaining component:
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123 4 5
6
789
10
?? 11 12
Finding Euler CircuitsMohammed’s Scimitars
Finally, need to add the triangle.
Use same naïve approach looking for cycle in remaining component:
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123 4 5
6
789
10
?? 11 1213
Finding Euler CircuitsMohammed’s Scimitars
Finally, need to add the triangle.
Use same naïve approach looking for cycle in remaining component:
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123 4 5
6
789
10
?? 11 121314
Finding Euler CircuitsMohammed’s Scimitars
Finally, need to add the triangle.
Use same naïve approach looking for cycle in remaining component:
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123 4 5
6
789
10
15 11 121314
Finding Euler CircuitsMohammed’s Scimitars
Amalgamate the triangle cycle between edges formerly labeled 9 & 10:
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123 4 5
6
789
10
15 11 121314
Finding Euler CircuitsMohammed’s Scimitars
Amalgamate the triangle cycle between edges formerly labeled 9 & 10:
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123 4 5
6
789
??
?? ?? ??????
Finding Euler CircuitsMohammed’s Scimitars
We found the Euler circuit!
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123 4 5
6
789
13
18 14 151617
1011 12
Euler Circuit All Degrees Even
2nd half of theorem says that an Euler circuit in a graph implies that all degrees are even:
In a simple cycle, whenever path enters vertex, must come out on different edge. Thus every visit of v contributes 2 to deg(v). Thus, if keep only edges which were on the simple cycle, degrees of resulting graph are all even. But in an Eulerian graph G, can find a simple cycle containing all vertices and consequently graph resulting from cycle is G itself!
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Generalizing to Euler Paths
A: YES! Because exactly 2 vertices of odd degree.
So can add a phantom edge between odd degree vertices:
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Generalizing to Euler Paths
THM: An undirected connected graph has an Euler path iff there are exactly two vertices of odd degree.
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