Jong C. Park
Computer Science Division,
Today’s TopicsIntroductionPaths and CyclesHamiltonian Cycles and the
Traveling Salesperson ProblemA Shortest-Path AlgorithmRepresentations of GraphsIsomorphisms of Graphs
– a cycle in a graph G that contains each vertex in G
exactly once, except for the starting and ending
– Determine if the following graphs have a
• the graph of Figure 8.3.4• the graph of Figure 8.3.5• the graph of Figure 8.3.6• the graph of Figure 8.3.7
– Given a weighted graph G, find a minimum-length
– When can an n-cube simulate a ring model with 2n
– Equivalently, when does an n-cube contain a
– The n-cube has a Hamiltonian cycle if and only if n
2 and there is a sequence, s1, s2, ., s2n, where
each si is a string of n-bits, satisfying:
• Every n-bit string appears somewhere in the sequence.
• si and si+1 differ in exactly one bit, i = 1, ., 2n-1.
• s2n and s1 differ in exactly one bit.
– The sequence above is called a Gray code.
– When n 2, a Gray code corresponds to the
Hamiltonian cycle s1, s2, ., s2n, s1.
– Let G1 denote the sequence 0, 1. We define Gn in
(a) Let GRn-1 denote the sequence Gn-1 written in reverse.
(b) Let G’n-1 denote the sequence obtained by prefixing each
(c) Let G’’n-1 denote the sequence obtained by prefixing each
(d) Let Gn be the sequence consisting of G’n-1 followed by
Then Gn is a Gray code for every positive integer n.
• The proof is done by induction on n.
– The n-cube has a Hamiltonian cycle for every
– Construct the Gray code G3 beginning with G1– The Knight’s Tour
• A knight’s tour of an n x n board begins at some square,
visits each square exactly once making legal moves, and
• The problem is to determine for which n a knight’s tour
– Dijkstra’s shortest-path algorithm correctly finds the
length of a shortest path from a to z.
– Find a shortest path from a to z and its length for
– For input consisting of an n-vertex, simple,
connected, weighted graph, Dijkstra’s algorithm has
– Select an ordering of the vertices, say a, b, c, d, e.
– Label the rows and columns of a matrix with the
– The entry in this matrix in row i, column j, i j, is
the number of edges incident on i and j. If i = j, the
entry is twice the number of loops incident on i.
– The degree of a vertex v in a graph G is obtained
by summing row v or column v in G’s adjacency
– If A is the adjacency matrix of a simple graph, the
ijth entry of An is equal to the number of paths of
length n from vertex i to vertex j, n = 1, 2, . .
– We label the rows with the vertices and the columns
with the edges (in some arbitrary order).
– The entry for row v and column e is 1 if e is incident on
– Find the incidence matrix for the graph of Figure 8.5.4.
– In a graph without loops, each column has two 1’s and
the sum of a row gives the degree of the vertex
– Graphs G1 and G2 are isomorphic if there is a one-
to-one, onto function f from the vertices of G1 to
the vertices of G2 and a one-to-one, onto function
g from the edges of G1 to the edges of G2, so that
an edge e is incident on v and w in G1 if and only if
the edge g(e) is incident on f(v) and f(w) in G2.
– The pair of functions f and g is called an
– Define an isomorphism for the graphs G1 and G2 of
– The Mesh Model for Parallel Computation
• The two dimensional mesh model for parallel computation
when described as a graph consists of a rectangular array
• When can an n-cube simulate a two-dimensional mesh?• When does an n-cube contain a subgraph isomorphic to a
• If M is a mesh p vertices by q vertices, where p ≤ 2i and q
≤ 2j, then the (i+j)-cube contains a subgraph isomorphic
– Graphs G1 and G2 are isomorphic if and only if for
some ordering of their vertices, their adjacency
– Let G1 and G2 be simple graphs. The following are
(b) There is a one-to-one, onto function f from the vertex set
of G1 to the vertex set of G2 satisfying the following:
Vertices v and w are adjacent in G1 if and only if the
vertices f(v) and f(w) are adjacent in G2.
– Determine if G1 and G2 in Figure 8.6.1 are
isomorphic by examining their adjacency matrices.
– A property P is an invariant if whenever G1 and G2
If G1 has property P, G2 also has property P.
• “has e edges”• “has n vertices”
– Use the notion of an invariant to determine if the
graphs G1 and G2 in Figure 8.6.3 are isomorphic.
– Show that if k is a positive integer, “has a vertex of
– Suppose G1 and G2 are isomorphic graphs and f (resp., g) is
a one-to-one, onto function from the vertices (resp., edges)
of G1 onto the vertices (resp., edges) of G2.
– Suppose further that G1 has a vertex v of degree k.
– Use the fact that “has a vertex of degree 3” is an
invariant to determine if the graphs G1 and G2 in
– Show that if k is a positive integer, “has a simple
cycle of length k” is an invariant.
– Use the fact that “has a simple cycle of length 3” is
an invariant to determine if the graphs G1 and G2 of
• Paths and Cycles• Hamiltonian Cycles
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Product Datasheet Overview Product name Description Specificity This antibody has a Kd of 3 nM for haloperidol and an IC50 value of 260 nM for the reduced haloperidol metabolite(+/-)-4-(4-chlorophenyl)-a-(4-fluorphenyl)-4-hydroxy-1-peperidinebutanol). Tested applications Immunogen Haloperidol conjugated through the tertiary alcohol at the 4’ position of the piperidine ring