Euler and hamilton paths

Gazi Zahirul Islam, Assistant Professor, Department of CSE, Daffodil International University, Dhaka
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Euler and Hamilton Paths:
DEFINITION 1:
An Euler circuit in a graph G is a simple circuit containing every edge of G. An Euler path in G is a
simple path containing every edge of G.
Examples 1 and 2 illustrate the concept of Euler circuits and paths.
Example 1:
Which of the undirected graphs in Figure 1 have an Euler circuit? Of those that do not, which have
an Euler path?
Figure 1: The undirected graphs G1, G2 and G3
Solution:
The graph G1 has an Euler circuit, for example, a, e, c, d, e, b, a. Neither of the graphs G2 or G3 has
an Euler circuit. However, G3 has an Euler path, namely, a, c, d, e, b, d, a, b. G2 does not have an
Euler path.
Example 2:
Which of the directed graphs in Figure 2 have an Euler circuit? Of those that do not, which have an
Euler path?
Figure 2: The directed graphs H1, H2 and H3
Solution:
The graph H2 has an Euler circuit, for example, a, g, c, b, g, e, d, f, a. Neither H1 nor H3 has an Euler
circuit. H3 has an Euler path, namely, c, a, b, c, d, b, but H1 does not.
Theorem 1:

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A connected multigraph with at least two vertices has an Euler circuit if and only if each of its
vertices has even degree.
Theorem 2:
A connected multigraph has an Euler path but not an Euler circuit if and only if it has exactly two
vertices of odd degree.
Example 3:
Which graphs shown in Figure 3 have an Euler path?
Figure 3: Three undirected graphs
Solution:
G1 contains exactly two vertices of odd degree, namely, b and d. Hence, it has an Euler path that
must have b and d as its endpoints. One such Euler path is d, a, b, c, d, b. Similarly,G2 has exactly two
vertices of odd degree, namely, b and d. So it has an Euler path that must have b and d as endpoints.
One such Euler path is b, a, g, f, e, d, c, g, b, c, f, d. G3 has no Euler path because it has six vertices of
odd degree.
DEFINITION 2:
A Hamilton circuit in a graph G is a closed path that visits every vertex in G exactly once. Note that
an Euler circuit traverses every edge exactly once, but may repeat vertices, while a Hamilton circuit
visits each vertex exactly once but may repeat edges.
Example 4:
Which of the simple graphs in Figure 4 have a Hamilton circuit or, if not, a Hamilton path?

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Figure 4: Three simple graphs
Solution:
G1 has a Hamilton circuit: a, b, c, d, e, a. There is no Hamilton circuit in G2. But G2 does have a
Hamilton path, namely, a, b, c, d. G3 has neither a Hamilton circuit nor a Hamilton path.
Example 5:
Show that neither graph displayed in Figure 5 has a Hamilton circuit.
Figure 5: Two graphs that do not have a Hamilton circuit
Solution:
There is no Hamilton circuit in G because G has a vertex of degree one, namely, e. However, G has a
Hamilton path, namely, e,d,a,b,c.
Now consider H. Because the degrees of the vertices a, b, d, and e are all two, every edge incident
with these vertices must be part of any Hamilton circuit. It is now easy to see that no Hamilton
circuit can exist in H, for any Hamilton circuit would have to contain four edges incident with c,
which is impossible. H has a Hamilton path: a,b,c,d,e.

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Shortest-Path Problems:
Dijkstra’s Algorithm:
Example 1:
Use Dijkstra’s algorithm to find the length of a shortest path between the vertices a and z in the
weighted graph displayed in Figure 1(a).

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Figure 1: Using Dijkstra’s Algorithm to Find a Shortest Path from a to z.
Solution:
The steps used by Dijkstra’s algorithm to find a shortest path between a and z are shown in Figure 1.

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Introduction to Trees:
DEFINITION 1:
A tree is a connected undirected graph with no simple circuits.
Example 1:
Which of the graphs shown in Figure 1 are trees?
Figure 1: Example of trees and non-trees
Solution:
G1 and G2 are trees, because both are connected graphs with no simple circuits. G3 is not a tree
because e, b, a, d, e is a simple circuit in this graph. Finally, G4 is not a tree because it is not
connected.
Forests:
forests have the property that each of their connected components is a tree. Figure 2 displays a
forest.

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Figure 2: Example of a forest
DEFINITION 2:
A rooted tree is a tree in which one vertex has been designated as the root and every edge is
directed away from the root.
Figure 3: A tree and rooted trees formed by designating two different roots.
For instance, Figure 3 displays the rooted trees formed by designating a to be the root and c to be
the root, respectively, in the tree T.
Tree Terminology:
Suppose that T is a rooted tree. If v is a vertex in T other than the root, the parent of v is the unique
vertex u such that there is a directed edge from u to v. When u is the parent of v, v is called a child of
u. Vertices with the same parent are called siblings. The ancestors of a vertex other than the root
are the vertices in the path from the root to this vertex, excluding the vertex itself and including the
root. The descendants of a vertex v are those vertices that have v as an ancestor. A vertex of a
rooted tree is called a leaf if it has no children. Vertices that have children are called internal
vertices.
If a is a vertex in a tree, the subtree with a as its root is the subgraph of the tree consisting of a and
its descendants and all edges incident to these descendants.

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Example 2:
In the rooted tree T (with root a) shown in Figure 4, find the parent of c, the children of g, the
siblings of h, all ancestors of e, all descendants of b, all internal vertices, and all leaves. What is the
subtree rooted at g?
Figure 4: A rooted tree T
Solution:
The parent of c is b. The children of g are h, i, and j. The siblings of h are i and j. The ancestors of e
are c, b, and a. The descendants of b are c, d, and e. The internal vertices are a, b, c, g, h, and j. The
leaves are d, e, f , i, k, l, and m. The subtree rooted at g is shown in Figure 5.
Figure 5: The subtree rooted at g
DEFINITION 3:

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A rooted tree is called an m-ary tree if every internal vertex has no more than m children. The tree is
called a full m-ary tree if every internal vertex has exactly m children. An m-ary tree with m = 2 is
called a binary tree.
Example 3:
Are the rooted trees in Figure 6 full m-ary trees for some positive integer m?
Solution:
Figure 6: Four rooted trees
T1 is a full binary tree because each of its internal vertices has two children. T2 is a full 3-ary tree
because each of its internal vertices has three children. In T3 each internal vertex has five children,
so T3 is a full 5-ary tree. T4 is not a full m-ary tree for any m because some of its internal vertices
have two children and others have three children.
Example 4:
What are the left and right children of d in the binary tree T shown in Figure 7(a)? What are the left
and right subtrees of c?
Solution:
Figure 7: A binary tree T and left and right subtrees of the vertex c

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The left child of d is f and the right child is g. We show the left and right subtrees of c in Figures 7(b)
and 7(c), respectively.
More about trees:
The level of a vertex v in a rooted tree is the length of the unique path from the root to this vertex.
The level of the root is defined to be zero. The height of a rooted tree is the maximum of the levels
of vertices. In other words, the height of a rooted tree is the length of the longest path from the root
to any vertex.
A rooted m-ary tree of height h is balanced if all leaves are at levels h or h − 1.
Example 5:
Find the level of each vertex in the rooted tree shown in Figure 8. What is the height of this tree?
Solution:
Figure 8: A rooted tree
The root a is at level 0. Vertices b, j, and k are at level 1. Vertices c, e, f, and l are at level 2.Vertices d,
g, i, m, and n are at level 3. Finally, vertex h is at level 4. Because the largest level of any vertex is 4,
this tree has height 4.

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Example 6:
Which of the rooted trees shown in Figure 9 are balanced?
Solution:
Figure 9: Some rooted trees
T1 is balanced, because all its leaves are at levels 3 and 4. However, T2 is not balanced, because it
has leaves at levels 2, 3, and 4. Finally, T3 is balanced, because all its leaves are at level 3.

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Spanning Trees:
DEFINITION 1:
Let G be a simple graph. A spanning tree of G is a subgraph of G that is a tree containing every vertex
of G.
Example 1:
Find a spanning tree of the simple graph G shown in Figure 1.
Figure 1: The simple graph G
Solution:
The graph G is connected, but it is not a tree because it contains simple circuits. Remove the edge {a,
e}. This eliminates one simple circuit, and the resulting subgraph is still connected and still contains
every vertex of G. Next remove the edge {e, f} to eliminate a second simple circuit. Finally, remove
edge {c, g} to produce a simple graph with no simple circuits. This subgraph is a spanning tree,
because it is a tree that contains every vertex of G. The sequence of edge removals used to produce
the spanning tree is illustrated in Figure 2.
Figure 2: Producing a spanning tree for G by removing edges that form simple circuits
The tree shown in Figure 2 is not the only spanning tree of G. For instance, each of the trees shown
in Figure 3 is a spanning tree of G.

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Figure 3: Spanning trees of G
DEFINITION 2:
A minimum spanning tree in a connected weighted graph is a spanning tree that has the smallest
possible sum of weights of its edges.
Prim’s Algorithm:
Example 2:
Use Prim’s algorithm to design a minimum-cost communications network connecting all the
computers represented by the graph in Figure 4.
Figure 4: A weighted graph showing monthly lease costs for lines in a computer network

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Solution:
We solve this problem by finding a minimum spanning tree in the graph in Figure 4. Prim’s algorithm
is carried out by choosing an initial edge of minimum weight and successively adding edges of
minimum weight that are incident to a vertex in the tree and that do not form simple circuits. The
edges in color in Figure 5 show a minimum spanning tree produced by Prim’s algorithm, with the
choice made at each step displayed.
Figure 5: A minimum spanning tree for the weighted graph in figure 4
Example 3:
Use Prim’s algorithm to find a minimum spanning tree in the graph shown in figure 6.
Figure 6: A weighted graph
Solution:

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A minimum spanning tree constructed using Prim’s algorithm is shown in Figure 7. The successive
edges chosen are displayed.
Figure 7: A minimum spanning tree produced using prim’s algorithm
Kruskal’s Algorithm:
Example 4:
Use Kruskal’s algorithm to find a minimum spanning tree in the weighted graph shown in figure 6.
Solution:
A minimum spanning tree and the choices of edges at each stage of Kruskal’s algorithm are shown in
figure 8.

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Figure 8: A minimum spanning tree produced by kruskal’s algorithm

Euler and hamilton paths

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Euler and hamilton paths