Medicamentsen-ligne vous propose les traitements dont vous avez besoin afin de prendre soin de votre santé sexuelle. Avec plus de 7 ans d'expérience et plus de 90.000 clients francophones, nous étions la première clinique fournissant du
acheter cialis original en France à vente en ligne et le premier vendeur en ligne de Kamagra dans le monde. Pourquoi prendre des risques si vous pouvez être sûr avec Medicamentsen-ligne - Le service auquel vous pouvez faire confiance.
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
Abstract: Intractable hiccups in transplanted patients may be caused by varioushiccups; transplantation; adverse e¡ects;medical conditions including infections.We report a case of a 44 -year-old man whocorticosteroids; esophagitis; pulmonary abscesssu¡ered from intractable hiccups after cadaveric kidney transplantation.Weidenti¢ed 3 di¡erent hiccup periods with di¡erent causes: 1) steroi
Leseprobe aus: Wilfried Schütz – Ganzheitliche Astromedizin Pathophysiologie medizinisch — astrologisch (Pathophysiologie: gr . Lehre von den Krankheitsvorgängen) Um das Gleichnishafte des Krankheitsgeschehens besser erfassen zu können und zu unserer gewohnten naturwissenschaftlichen Betrachtungsweise eine Brücke zu schlagen, beschäftigen wir uns im Folgenden beispielhaft mit der