An Euler Circuit is a cycle of an undirected graph, that traverses every edge of the graph exactly once, and ends at the same node from which it began.

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An Euler Circuit is a cycle of an undirected
graph, that traverses every edge of the graph
exactly once, and ends at the same node from
which it began. Euler's Theorem A connected
graph G possesses an Euler circuit if and only if
G does not contain any nodes of odd
degree. Proof of Euler's theorem Assume that
G has zero nodes of odd degree. It can then be
shown that this is a necessary and a sufficient
condition for an Euler circuit to exist. Part
1 It is necessary because any Euler circuit
drawn on the graph must always enter a node
through some edge and leave through another and
all edges on the graph must be used exactly once.
Thus, an even number of incident edges is
required for every node on the graph.
2

• Proof of Euler's theorem Part 2
• Sufficiency, on the other hand, can be shown
  through the following tour construction argument.
  We begin at some initial node k0 and draw a
  circuit through G (thus eventually returning to
  k0). Let this circuit be denoted C0. If C0
  happens to be an Euler circuit, this is fine we
  stop. If C0 is not an Euler circuit, then if we
  remove from G all edges used by circuit C0, there
  must be some edges left over. Moreover, at least
  two of these edges must be incident on some node
  k1 through which circuit C0 has passed. This must
  be so since, by assumption, G is, first,
  connected and, second, all its nodes are of even
  degree (and C0 has only used up an even number of
  edges which are incident on k1). Thus, it is
  possible to draw another circuit C1 originating
  and terminating at k1, which uses only edges of
  G', the graph left after we eliminate the edges
  of C0 from G.
• This procedure may now be continued until
  eventually, say after the nth step, there will be
  no edges left uncovered. At that time, an Euler
  circuit will also have been obtained which will
  be a combination of circuits C0, C1, C2, . . .,
  Cn.

3

• Euler's theorem is a proof by construction.
• However, the algorithm could be O(E2) or O(E log
  E) where E is the number of edges in the graph.
• Note both of these are O(E2)

for i 1 to n for k 1 to n
for i 1 to n for i 1 to n . .
(linear function of n times) . for i 1 to n
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Fleury's Algorithm O(E)?

• Pick any vertex to start.
• From that vertex pick an edge to traverse,
  considering following rule never cross a bridge
  of the reduced graph unless there is no other
  choice.
• Darken that edge, as a reminder that you can't
  traverse it again.
• Travel that edge, coming to the next vertex.
• Repeat 2-4 until all edges have been traversed,
  and you are back at the starting vertex.
• By reduced graph we mean the original graph minus
  the darkened (already used) edges.
• A bridge of a graph G is an edge whose deletion
  increases the number of components of G.

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Fleury's Algorithm in Action
Pick any vertex (e.g. F) Take F to C  (arbitrary)
Take C to D  (arbitrary)
Take D to A  (arbitrary)
A bridge is not a local property (i.e. if edge EF
existed then AB would not be a bridge).
How can we recognize a bridge efficiently? In the
original graph, AB was not a bridge. Can we
preprocess the graph in O(E) time identifying
bridges and building a structure that can be
updated in constant time with each reduction?
Take A to C. Can't go to B that edge is a bridge
of the reduced graph, and there are two other
choices.
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Scrub Tile Puzzle A Hamiltonian Cycle

• Puzzle Applet

Given 10 equally sized index cards each with one
of the following words HEN, HUT, WIT, SAW, CAR,
CUB, MOB, DIM, RED, SON. Arrange the cards in a
rectangular closed chain such that any two
adjacent words must share a common letter.
The above may serve to demonstrate a possible
arrangement. However, the words SON and RED,
although adjacent, do not have a letter in
common. Therefore, it's not a valid
configuration. Repeat the same problem with the
cards SON and HUT replaced by SUN and HOT.
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Scrub Tile Puzzle
Of the two variants of the puzzle, the second is
solvable, the first is not. The two variants are
represented by different graphs. A solution to
the puzzle constitutes a Hamiltonian circuit on
its graph. The first of the two is the famous
Petersen graph that is known not to house any
Hamiltonian circuits.
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Dodecahedral Graph Is it Hamiltonian? If so,
find the Hamiltonian Cycle
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Knight's Tour Problem

• The "Knight's Tour" is a sequence of moves done
  by a knight on a chessboard. The knight is placed
  on an empty chessboard and, following the rules
  of chess, must visit each square exactly once.
  There are several billion solutions to the
  problem, of which about 122,000,000 have the
  knight finishing on a square which is just a move
  away from the starting square. Such a tour is
  described as closed.
• The Knight's Tour problem is an instance of the
  more general Hamiltonian path problem in graph
  theory. The problem of getting a closed Knight's
  Tour is similarly an instance of the Hamiltonian
  cycle problem.

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Find a Closed Knight's Tour
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State as Hamiltonian Path Problem

• For a given n value, find an arrangement of the
  first n natural values such that, at the same
  time
• The sum of every two adjacent numbers is a prime.
• The sum of every two adjacent squares is a prime.
• Find such a sequence for n 10.