Computing Fundamentals 2
Lecture 2
A theory of Graphs
Lecturer: Patrick Browne
http://www.comp.dit.ie/pbrowne/
Lecture Room K320, Labs A117
Based on Chapter 19.
A Logical approach to Discrete Math
By David Gries and Fred B. Schneider
A Theory of Graphs
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In this lecture we look at:
Graph Morphisms
Hamilton Circuits
Planar Graphs
Shortest Path
Travelling Sales-person Problem (TSP)
Similar Graphs
• The two graphs below look different.
• Are from a graph theoretic point of view
structurally the ‘same’?
2
b
1
a
3
c
e
d
5
4
Graph Morphisms
• The two graphs below look different but are
f.c
structurally the ‘same’.
b
2(c)
f.a
1(a)
a
c
e
3(e)
f.d
5(d)
d
f.b
4(b)
f.e
• They are the ‘same’ up to the renaming of
the vertices. These graphs are isomorphic.
Graph Isomorphism
•Given two graphs G and G’ with vertices V and
V’. The graphs are isomorphic if there exists a
1-to-1 function f:V->V’ such that:
•An isomorphism f “preserves edges” thus
preserving properties of graphs that
depend on edges e.g. cycles and degrees
are preserved.
Isomorphism Invariant
• An isomorphic invariant for simple graphs
is the existence of a simple circuit of length
k ,where k is an integer > 2. Example:
G1 and G2 are isomorphic since we have
the invariants, similarity in degree of
nodes, number of edges, length of circuits.
• Checking for an isomorphism in large
complex graphs is difficult computationally
and visually.
G1
G2
Graph Isomorphism
• Sometimes it is easy to check whether two
graphs are not isomorphic. The left graph below
has two vertices of degree 3 while the right one
has none.
• It is much harder to prove that two graphs are
isomorphic. The computation is time is O(cn)
exponential wrt. the number of vertices .
Web search tasks
• Why should we want to know whether two
graphs are isomorphic? What sort of problems
can be solved?
Complexity O notation
•
Big-O notation can be used to express complexity of
an algorithm or computation. For a problem of size N:
• a constant-time is "order 1": O(1)
• a linear-time is "order N": O(N)
• a quadratic-time is "order N squared": O(N2)
• a exponential-time is "order constant to the ": O(CN)
Complexity O notation
• The time to compute Dijkstra’s Shortest path algorithm
takes O(|E|*log(|V|)) where:
– |V| is the number of vertices
– |E| is the total number of edges
• The travelling salesman problem takes exponential
time . As the size of the problem increases linearly, the
time to solve the problem increases exponentially
(O(C to the power N)). The shortest route A to A,
which visits all points and returns to point A.
Homeomorphic Graphs
• Two graphs are homeomorphic if they can both be
obtained from the same graph by inserting or
removing vertices of degree 2.
1
2
3
b
e
c
d
• Adding b to 1 gives 2
• Adding c,d,e, to 1 gives 3
• Hence 2 and 3 are homeomorphic because they were
obtained from 1 by adding vertices of degree 2.
Recall: Euler and Hamiltonian
Cycles
• An Euler cycle is a cycle in a graph that includes each
edge exactly once. Examples: Designing and optimizing
routes refuse trucks, snow ploughs, or postmen. In all of
these applications, every edge in a graph must be
traversed at least once.
• A Hamiltonian cycle is a cycle in a graph G that
contains each vertex exactly once except for the starting
and ending vertex that appears twice. Examples:
travelling sales man who wishes to visit every city and
also minimize his route to each city and return to his
home city.
Hamilton Path & Circuit
• A Hamilton path of a graph or digraph is a
path that contains each vertex exactly
once, except that the end vertices may be
the same.
• A Hamilton circuit (or cycle) is a Hamilton
path that is a cycle.
• Contrast this with an Euler circuit which
contains each edge exactly once.
Hamilton Path and Circuit
• Graphs with Hamilton path but no
Hamilton circuit (or cycle).
Start
Adjacent and Incident
• Two vertices v and w of a graph G are
adjacent if there is an edge e=<v,w>
joining them. The vertices v and w are
then incident with such an edge e, also e
is incident on v and w . Two distinct edges
e and f are adjacent if they have a vertex
in common
Hamiltonian cycles
Traveling salesperson problem (TSP)
– To visit every vertex of a graph G
only once by a simple cycle.
– Such a cycle is called a Hamilton
cycle.
– If a connected graph G has a
Hamilton cycle, G is called a
Hamiltonian graph.
– A Hamilton cycle or path in G will be
a subgraph of G.
Hamiltonian cycles (HC)
Traveling salesperson problem (TSP)
– To visit every vertex of a graph G
only once by a simple cycle.
– Such a cycle is called a Hamilton
cycle.
– If a connected graph G has a
Hamilton cycle, G is called a
Hamiltonian graph.
– if G has ‘enough’ edges, then G
has HC (add pecked red edge)
– (n=7,deg.b + deg.c ≥ n-1)
Properties of Hamilton Circuit (HC)
• If G has a HC, then deg.v >= 2 for all v
on HC.
• If deg.v = 2 then both edges incident on
v are in HC.
• If deg.v > 2 and two of the edges
incident on v are in a HC, then the other
edges incident on v are not in that HC.
• A complete digraph (v>2) has a Hamilton
Path.
Recall Euler Cycle(EC)
• Graph G has EC if and only if every
vertex of G has even degree.
– Necessary Condition If G has EC, then every
vertex of G has even degree.
– Sufficient Condition If every vertex of G has
even degree, then G has EC.
Properties of Hamilton Circuit (HC)
• Establishing whether or not a graph has a
HC is nontrivial. No elegant (convenient)
characterization of HC exists, although
several necessary or sufficient conditions
are known. Sufficient conditions usually
require ‘enough edges’.
• Dirac’s and Or’s Theorems provide
sufficient conditions for HC, but not a
necessary conditions.
Dirac’s Theorem
• In a graph G, two vertices are adjacent if
they are joined by an edge. If each vertex
of a connected simple graph with n
vertices (n>3) is adjacent to at least n/2
vertices then the graph has a HC.
n = 7, n/2 = 3.5
Since each vertex is adjacent to at
least 4 other vertices the graph has a
Hamilton Cycle by Dirac’s Theorem.
Dirac’s Theorem gives no information
about being non-Hamiltonian. It does
not tell us how to find a HC.
Ore's theorem
• Ore's theorem states that:
• if deg v + deg w ≥ n – 1 for every
pair of non-adjacent vertices v and w of G
then G has a HC.
Different Hamilton Circuits in
Complete Graph
The number of different Hamiltonian cycles
in a complete undirected graph on n vertices
is (n − 1)!/2 and in a complete directed
graph on n vertices is (n − 1)!
Hamilton Circuit
• Find a Hamilton circuit in the G.
b
a
d
c
e
i
f
g
j
h
Hamilton Circuit
• Find a Hamilton circuit in the G.
b
a
d
c
e
i
Solution: d,a
f
g
j
h
Hamilton Circuit
• Find a Hamilton circuit in the G.
b
a
d
c
e
i
Solution: d,a,e
f
g
j
h
Hamilton Circuit
• Find a Hamilton circuit in the G.
b
a
d
c
e
i
Solution: d,a,e,b
f
g
j
h
Hamilton Circuit
• Find a Hamilton circuit in the G.
b
a
d
c
e
i
Solution: d,a,e,b,g
f
g
j
h
Hamilton Circuit
• Find a Hamilton circuit in the G.
b
a
d
c
e
i
Solution: d,a,e,b,g,c
f
g
j
h
Hamilton Circuit
• Find a Hamilton circuit in the G.
b
a
d
c
e
i
Solution: d,a,e,b,g,c,h
f
g
j
h
Hamilton Circuit
• Find a Hamilton circuit in the G.
b
a
d
c
e
i
Solution: d,a,e,b,g,c,h,j
f
g
j
h
Hamilton Circuit
• Find a Hamilton circuit in the G.
b
a
d
c
e
f
i
Solution: d,a,e,b,g,c,h,j,f
g
j
h
Hamilton Circuit
• Find a Hamilton circuit in the G.
b
a
d
c
e
f
i
Solution: d,a,e,b,g,c,h,j,f,i
g
j
h
Hamilton Circuit
• Find a Hamilton circuit in the G.
b
a
d
c
e
f
i
Solution: d,a,e,b,g,c,h,j,f,i,d
g
j
h
Hamilton Circuit
• Find another Hamilton circuit in the same
graph G.
b
a
d
c
e
i
f
g
h
j
•Note that while verticies a and b were on
previous Hamilton circuit the edge <a.b> was
not. This is significant for TSP.
Hamilton Circuit
• Find another Hamilton circuit in the G.
b
a
d
c
e
i
f
g
j
h
Hamilton Circuit
• Find another Hamilton circuit in the G.
b
a
d
c
e
i
f
g
j
h
Hamilton Circuit
• Find another Hamilton circuit in the G.
b
a
d
c
e
i
f
g
j
h
Hamilton Circuit
• Find another Hamilton circuit in the G.
b
a
d
c
e
i
f
g
j
h
Hamilton Circuit
• Find another Hamilton circuit in the G.
b
a
d
c
e
i
f
g
j
h
Hamilton Circuit
• Find another Hamilton circuit in the G.
b
a
d
c
e
i
f
g
j
h
Hamilton Circuit
• Find another Hamilton circuit in the G.
b
a
d
c
e
i
f
g
j
h
Hamilton Circuit
• Find another Hamilton circuit in the G.
b
a
d
c
e
i
f
g
j
h
Hamilton Circuit
• Find another Hamilton circuit in the G.
• Solution a,b,c,g,h,j,f,e,i,d,a
Is HC unique?
• There are graphs with unique Hamiltonian
cycles:
Two classes of Hamiltonian Paths
1. A complete digraph (v>2) has a
Hamiltonian path.
2. Let G be an undirected graph of n
vertices*. If deg.b + deg.c ≥ n-1 for
each pair of vertices b,c then G has a
Hamiltonian path. Note direction of
implication
b,c(deg.b + deg.c ≥ n-1)
HP
If 2 does not hold we do not know about HP
*Note: every vertex must be on
the Hamilton path
Showing there is no HC
• The HC itself must contain exactly two edges
incident on each vertex. Hence each vertex v on the
HC has degree(v)= 2.
• In any graph sum(degree(v))=2*|E|. In the HC case
|V|*2=2*|E|, which reduces to |V|=|E|. Since G
(below) has 5 vertices then the HC should have 5
edges. The graph G (not HC) has 6 edges. Can we
v1
delete an edge from G to form a HC?
• Answer No.
e2
Graph G
e3
v5
v2
v4 e6 e5
e4
e1
v3
Adjacent and Incident
• First we need definitions of adjacent and
incident. Two vertices v and w of a graph G are
adjacent if there is an edge e=<v,w> joining
them. The vertices v and w are then incident
with such an edge e, also e is incident on v and
w . Two distinct edges e and f are adjacent if
they have a vertex in common
Edges can be incident on a vertex
Showing there is no HC
Graph G
v1
• For a HC in G, each vertex on the HC
should have degree two. We can eliminate v4
edges incident on vertices with degree
greater than two.
• The edges <v4,v5> and <v2, v5> are
incident on v4, v2, v5.
• If we eliminate <v4,v5> and <v2, v5> are
left with only 4 edges not enough to
v4
complete a HC, because |V|=|E| should
hold. No HC.
v5
v2
v3
v1
v5
v2
v3
Showing there is no HC
v1
• Using this method we must avoid
v5
double counting. We must not v4
v2
count an eliminated edges more
v3
than once. In the previous case,
v1
when we eliminated one edge
incident on v2 and one edge
v5
incident on v4 these edges are v4
v2
distinct, and hence the argument
v3
of no HC is valid.
Graph G; valid reasoning
Incorrect claim: G has no HC.
In G2 there are 5 vertices, hence HC must
have 5 edges. Following the method from
previous slide we eliminate two edges
incident at v5 and one edge incident at
each v1,v2,v3, and v4, leaving just 2 v4
edges, hence no HC. This is incorrect
reasoning, because the red edges are
counted twice,
When we eliminate 2 edges incident on v5
(which we must) we also eliminate edges
incident on another vertices. Then we only
need to remove 1 other edge. Now we
have 5 edges and 5 vertices, hence a
there is a HC –correct reasoning.
G2
v1
v5
v2
v3
v1
v5
v4
v2
v3
Showing there is no HC
• How could we show that there is no HC in the
graph below?
a
c
b
e
d
g
f
i
h
l
m
j
k
Showing there is no HC
• Suppose HC=H,
then edges
<a,b>, <a,g>,
<b,c>,<c,k>
must be in H, since
each of these has
at least one vertex
with degree two.
• Thus <b,d> and
<b,f> are not in H
since then Deg(b)
>2 would be true.
a
c
b
e
d
g
f
i
h
l
m
j
k
Showing there is no HC
• Therefore edges
<g,d>, <d,e>,
<e,f>, <f, k> are
in H (because d and
f are now degree 2).
• The edges now form
a cycle C. Adding
additional edges to C
will give some vertex
in H a degree greater
that 2. This is a
contradiction,
showing G does not
have HC.
a
c
b
e
d
g
f
i
h
l
m
j
k
Hamilton Cycles & Paths
Hamilton Cycles in K4
The number of Hamilton circuits in KN is
(N - 1) × (N - 2) × ... × 3 × 2 × 1 = (N - 1)!
Hamilton Cycles in K4
Summary of Euler and
Hamilton cycles
• A Euler cycle is a cycle in a graph that includes all of
the edges and all of the vertices in that cycle.
• A Hamilton cycle is a cycle in a graph G the contains
each vertex exactly once except for the starting and
ending vertex that appears twice.
• Recall:
• A path is a sequence of vertices and edges but with no
edge traversed more than once
• A path with at least one edge and with the first and last
vertices the same is called a cycle.
Summary of Euler and
Hamilton cycles
• Whether a graph does or does not have a Hamilton
cycle tells you nothing about whether it has an Euler
cycle , and vice versa. The same is true for Hamilton
and Euler paths (rather than cycles).
Summary of Euler and
Hamilton cycles
B
D
F
C
A
B
E
• G has a Euler cycle but no Hamilton
cycle
Summary of Euler and
Hamilton cycles
Test for Euler cycle
• If a graph G has an Euler cycle, then G is
connected and every vertex has even
degree.
• If G is a connected graph and every vertex
has even degree, then G has a Euler
cycle.
a
b
f
d
c
This graph G does not have
a Euler cycle, since there are
vertices of odd degree, yet G
has a Hamiltonian cycle of:
(a,b,c,d,e,f,g,a).
g
e
Unlike Euler cycles, no easily verified necessary and
sufficient conditions are known to exist for a Hamilton
cycle in a graph
edge(a,b), edge(a,f), edge(a,g), edge(a,e), edge(b,f),
edge(b,d), edge(b,c), edge(c,d), edge(c,e), edge(e,d),
edge(e,f), edge(e,g), edge(g,f)
This graph has a Euler cycle but no Hamiltonian cycle.
Cycle or Path?
If a graph has a Hamilton cycle does it have
to have a Hamilton path?
Answer Yes: HC implies HP.
If <v1…vn-1,vn>, with v1=vn, is a Hamilton
cycle then <v1…vn-1>, is a Hamilton path
. A Hamiltonian cycle exists on a
complete graph
• Recall, a Hamilton cycle is a cycle in a graph G the
contains each vertex exactly once except for the
starting and ending vertex that appears twice.
• Recall, a complete graph on n vertices, Kn, is the
simple graph (no loops or parallel edges) with n
vertices in which there is an edge between every pair
of distinct vertices.
• Since there is an edge between every pair of vertices
and this edge may be traversed in either direction, it
follows that any permutation of vertices is a Hamilton
path.
A Hamilton cycle exists on a
complete graph.
• Moreover, a Hamilton cycle is obtained by
including the edge from the final to the initial
vertex of such a path.
• So in summary
• Every complete graph has a Hamilton path
and a Hamilton cycle.
Hamiltonian cycles can take a
long time to compute!
• For a general graph, every known algorithm for finding
Hamiltonian cycles requires factorial time in the worst
case. For a complete undirected graph with n vertices,
there are a possible (n-1)!/2 essentially different
Hamiltonian cycles. The number of addition operations
required to obtain the lengths of these cycles is O(n!).
For example for n=12 it will take 5 seconds to perform
the computation, for n=50 it will take 1049 years
(assuming 108 (100,000,000, or 100 million) additions
per second). For this reason, methods that produce nearminimum-length cycles are often used.
Planar Graphs
• A graph is called planar if it can be drawn
on the plane in a way that no two of its
edges cross each other
Planar graphs
A graph is planar if it can be drawn in the
plane without crossing edges
Planar graphs
K4 can be drawn with no edges crossing.
A planar graph the points in the plane into
regions or faces. Any planar embedding of
a given graph has the same numbers of
faces.
Planar graphs
Two points (not vertices) are in the
same region iff it is possible to draw an
edge (curved) between them without
touching an edge of the region.
Each interior region (e.g. C and B) is
characterized by a cycle that forms its boundary.
The exterior region (D) contains all the points not
bounded by some cycle.
Euler’s Planar Graph Theorem
Euler’s Characteristic
Theorem : Euler's planar graph theorem
For all connected planar graph or multigraph:
v–e+r=2
number
of vertices
number
of edges
Not(v – e + r = 2) implies Not(Planar)
number
of regions
or faces
Planar implies v – e + r = 2
This theorem provides necessary conditions for planarity is
not a sufficient condition, and therefore can only be used
to prove a graph is not planar, not that it is planar
Recall the Handshaking
Theorem
Theorem: Let G=(V,E) be an undirected graph. Then
Example of Handshaking Theorem
• Draw a graph with 4 edges and 4 vertices , having
degrees 1,2,3,4.
• Solution: No such graph exists. The sum of the
degrees of all the vertices should be equal to twice
the number of edges. Half of the sum of the degrees
is 5, which is not equal to actual number of edges
required in the question (4).
• (1+2+3+4)/2 = 5, 5 =/= 4.
Planar Graphs
Example where Euler’s theorem holds
A planar graph divides the plane
into several regions (faces), one
of them is the infinite external
region.
• v=4,e=5,r=3,
• v-e+r=2
•4-5+3=2
Planar graphs
A graph (or multigraph) G is called planar if G can be
drawn in the plane with its edges intersecting only at
vertices of G, such a drawing of G is called an embedding
of G in the plane. Used in VLSI design, where
overlapping edges requires extra layers, wires cannot
overlap on the circuit board. Representation examples:
K1,K2,K3,K4 are planar, Kn for n>4 are non-planar
Edges in series
Edges in series:
• If vV(G) has degree 2 and there are
edges (v, v1), (v, v2) with v1  v2,
• we say the edges (v, v1) and (v, v2)
are in series.
Series reduction
• A series reduction consists
of deleting the vertex v 
V(G) and replacing the
edges (v,v1) and (v,v2)
by the edge (v1,v2)
• The new graph G’ has one
vertex and one edge less
than G and is said to be
obtained from G by series
reduction
Homeomorphic graphs
• Two graphs G and G’ are said to be
homeomorphic if G’ is obtained from G by
a sequence of series reductions.
– By convention, G is said to be obtainable from
itself by a series reduction, i.e. G is
homeomorphic to itself.
• Define a relation R on graphs: GRG’ if G
and G’ are homeomorphic.
• R is an equivalence relation on the set of
all graphs.
Kuratowski’s theorem
• G is a planar graph if
and only if G does not
contain a sub-graph
homeomorphic to
either K 5 or K 3,3
Utility Problem(again)
• There are three houses and three utility companies--say, gas,
electric, and water--and asks if each utility can be connected to
each house without having any of the gas-water-electric lines
pass over any other. This is equivalent to the equation "Can a
planar graph be constructed from each of three nodes
('houses') to each of three other nodes ('utilities')?" No such
planar graph exists.
Not planar
Here is an example of a
graph which doesn't have
K5 or K3,3 as its
subgraph. However, it has
a subgraph that is
homeomorphic to K3,3
(replace red lines with
yellow line) and is
therefore not planar.
A finite graph is planar if and only if it does not
contain a subgraph that is homeomorphic to K5 or
K3,3.
Euler’s formula
• We can test if a graph is non-planar, but there is
no simple check for planarity. If v is the number
of vertices (at least 3) and e is the number of
edges, then (e  3v−6 ) is a necessary (but not
sufficient) condition for planarity.
• That is, every connected planar graph satisfies (e
 3v−6 ), but there are graphs that satisfy that
constraint which are not planar.
• Planar Û (e  3v−6 )
• not(e  3v−6 ) Û not(Planar)
• Contrapositive holds.
Euler’s formula
• Euler’s formula forms the basis of two theorems
that provide conditions for non-planarity :
• Theorem 1: If v ≥ 3 then e ≤ 3v − 6;
• Theorem 2: If v > 3 and there are no cycles of
length 3, then e ≤ 2v − 4.
• If either of these theorems fail to hold then the
graph is not planar.
Euler’s formula
• The graph K3,3 has 6 vertices, 9 edges,
and no cycles of length 3. Using Theorem
2 we get the implication.
•
If (v > 3) and not(cycle(length(3)) then e ≤ 2v − 4.
•
RHS = 9 ≤ (2*6) – 4 = 9 ≤ 8 = false
• The LHS of the implication is true, but
RHS is false (T=>F is false1) .
• By theorem 2, K3,3 cannot be planar.
Satisfies Euler’s formula but not
planar.
• By itself, Euler's formula does not provide us
with a tool for showing that some graphs are not
planar, because it refers to the set of regions in
a potential planar drawing of the graph. We may
not have information on regions.
• There are graphs that satisfy Euler’s formula but
are not planar
Planar Example
• Euler Characteristic: r = e – v + 2, (4 = 8 – 6 + 2) (true)
• r = regions or faces
a
b
C
f
B
e
D
A
d
c
• For all planar graphs (e  3v − 6), 8 > 12 = true
• Note, this test cannot be used to prove
planarity.
Planar Example
Planar embedding for K2,2, K2,3 ,
& K2,4
Kuratowski’s theorem
•
•
Kuratowski’s Theorem: A graph is non-planar if and only if it contains a
subgraph homeomorephic to K3,3 or K5 .
The graph G on the left contain K,3,3. Deleting vertex h and its incident
edges gives the middle graph. We can replace c-g-f by c-f (homeomorphic
graphs). Then the graph on the left is K3,3 therefore G contains a subgraph
that is homeomorphic to K3,3. (Page 447, course text)
Theorem on vertices, edges,
regions
• If the planar representation divides the
plane then, v – e + r = 2
• If v≥3 then, e ≤ 3v – 6
• If v≥3 and there are no cycles of length 3
then, e ≤ 2v – 4
Proof that K5 is not planar
• A planar graph is a graph that can be drawn on the plane
in such a way that its edges intersect only at their
endpoints.
• Solution: K5 has 5 vertices each with degree 4, so by
the handshaking theorem it has 5*4/2 = 10 edges. By
theorem 1 of Euler’s formula, every connected planar
graph must satisfy E ≤ 3V – 6, gives 10 ≤ 3*5 – 6 or 10 ≤
9, which is false. So K5 cannot be a connected planar
graph.
• Or more simply, if K5 was planar then m ≤ 15 - 6 = 9, but
this is not the case because m = 10
Trees
• A tree is a loop-free connected graph that
contains no cycles.
• Each pair of vertices of a tree is connected
by a unique simple path.
A Tree
Not a tree
Trees
• A tree with at least two vertices has at
least two vertices of degree 1.
• For every tree <V,E>, #V = 1 + #E.
A Tree
Not a tree
Trees
• Let G=<V,E> be a loop free graph, then the following are
equivalent statements:
• G is a tree
• G is connected and the removal of any edge yields two
trees
•
G contains no cycles and #V = 1 + #E
• G is connected and #V = 1 + #E
• G has no cycles and adding one edge introduces one
cycle
Trees
• Every tree has the following properties:
• Any connected subgraph is a tree.
• There is a unique simple path between every pair of
vertices.
• Adding an edge between two vertices creates a cycle.
• Removing any edge disconnects the graph.
• If it has at least two vertices, then it has at least two
leaves.
• The number of vertices is one larger than the number of
edges.
Trees : File Directories
• A graph may contain a cycle. There is the possibility that
deleting a sub-directory could delete the directory from
which the delete command was issued. A tree structure
prohibits the sharing of files, but an acyclic graph would
permit file sharing
• A tree structure prohibits the sharing of files or
directories. Acyclic graphs allow directories to have
shared subdirectories and files, but deletion of file is
difficult.
Trees : File Directories
• Acyclic graphs provide an extension of tree-structured
directories. It is possible for two directories to contain the
same file or subdirectory.
Spanning Tree Example
A spanning tree of a connected graph is a tree that
is a sub-graph and that contains all the vertices of
the graph. In the middle below is a graph; on
either side of it is a spanning tree of that graph.
A spanning tree of a graph contains the minimum
number of edges to keep the graph connected.
Spanning Tree Example
An algorithm used to construct a spanning tree is
non-deterministic in two ways: 1) the choice of
vertex u is arbitrary, 2) the choice of the edge
incident on u to add to the tree is arbitrary.
A spanning tree of a graph contains the minimum
number of edges to keep the graph connected.
Spanning Tree Example
We can limit the choice of vertex by either breath first search
(BFS) or depth first search (DFS).
BFS
Original Graph
1
3
6
DFS
1
2
5
4
7
2
4
3
The edges are traversed in the order specified by the
numbers above.
A spanning tree of a graph contains the minimum
number of edges to keep the graph connected.
7
5
6
Shortest path
Permanent Labels
Temporary Labels
V the set of vertices in the graph, xpaths are next step
F set of vertices whose shortest-path lengths remain to be calculated
V – F (permanent labels) set of vertices whose shortest-path lengths have
been calculated
Each iteration moves one vertex from F to (V – F)
Invariant : PF : (V – F)    b ∊ (V – F)
Dijkstra’s Shortest Path Algorithm
• Dijkstra’s algorithm for finding the shortest
from a to v, involves assigning labels to
vertices. Let L(v) denote the label of
vertex v. At any point, some vertices have
temporary labels and the rest have
permanent labels, these are shown in red
in the following slides. We let T denote the
set of vertices having temporary labels.
Dijkstra’s Shortest Path Algorithm
• If L(v) is a permanent label of v the
L(v) is the shortest path to v.
• Initially all vertices have temporary labels.
• Each iteration of the algorithm changes
the status of one of the labels from
temporary to permanent .
• Terminate when z is not an element of T.
Dijkstra’s Shortest Path Algorithm
•
•
•
•
Initialize:
L(a):= 0
all vn (except a) are labelled .
T := set of all temporary vertices
while z ∊ T [
choose v ∊ T with min L(v)
T := T – {v}
for each x ∊ T adjacent to v [
L(x) := min{(L(x), L(v) + w(v,x)}]]
Where w(v,x) is distance from v to x.
Example Graph G(*)
2
b
c
2
1
2
a
3
4
z
4
d
1
6
7
3
f
e
5
g
Initialized Graph G(*)

2
b
c 
2
1

4
3
2
a

z

e
4
d
1
7
3
f

6
5
g


After First Iteration Graph G(*)
For each x in T adjacent to v: L(x) = min{Lold(x),L(v)+w(v,x)}
2 b
c 
2
T := set of all temporary vertices
2
1

Labels
L(a)=min(,0)
L(f) = min{,0+1} sorted
L(b) = min{,0+2}
z

e
4
d
f
6
7
3
1
3
4
2
a
1
Where w(v,x)
is distance
from v to x.
5
g

Chose (v= f ) ∊ T as minimum L(v)

After Second Iteration Graph G(*)
For each x in T adjacent to v: L(x) = min{Lold(x),L(v)+w(v,x)}
2
T := set of all temporary vertices
c 
2
b
2
Where w(v,x)
is distance
from v to x.
1
2
a
Labels
L(a)=min(,0)
1
L(f) = min{,0+1}
L(b) = min{,0+2} sorted
L(d) = min{,0+1+3}
L(g) = min{,0+1+5}
4
z

e
4
d
f
6
7
3
1
3
4
5
g
6
Chose (v= b ) ∊ T as minimum L(v)

After Third Iteration(*)
For each x in T adjacent to v: L(x) = min{Lold(x),L(v)+w(v,x)}
2
T := set of all temporary vertices
c 4
2
b
2
1
Labels
L(a)=min(,0)
1
4
L(f) = min{,0+1}
L(b) = min{,0+2}
L(c) = min{,0+2+2} sorted
L(d) = min{,0+1+3}
L(d) = min{4,0+2+2} --dif. path
L(e) = min{,0+2+4}
L(g) = min{,0+1+5}
3
4
2
a
Where w(v,x)
is distance
from v to x.
z
6
e
4
d
7
3
f
1
6
5
g
Chose (v= c ) ∊ T as minimum L(v)
6

Fourth Iteration(*)
For each x in T adjacent to v: L(x) = min{Lold(x),L(v)+w(v,x)}
2
T := set of all temporary vertices
c 4
2
b
2
a
Labels
L(a)=min(,0)
1
L(f) = min{,0+1}
4
L(b) = min{,0+2}
L(c) = min{,0+2+2}
L(d) = min{,0+1+3} sorted
L(d) = min{4,0+2+2} dif. Path
L(e) = min{,0+2+4}
L(z) = min{,0+2+2+1}
L(g) = min{,0+1+5}
L(e) = min{6,0+2+2+3} dif. Path
Where w(v,x)
is distance
from v to x.
1
2
d
4
3
4
6
e
z
6
7
3
f
1
5
g
6
Chose (v= d ) ∊ T as minimum L(v)
5
Fourth Iteration(*)
For each x in T adjacent to v: L(x) = min{Lold(x),L(v)+w(v,x)}
2
T := set of all temporary vertices
c 4
2
b
2
a
Labels
L(a)=min(,0)
L(f) = min{,0+1}
1
4
L(b) = min{,0+2}
L(c) = min{,0+2+2}
L(d) = min{4,0+2+2}
L(z) = min{,0+2+2+1} sorted
L(e) = min{,0+2+4}
L(g) = min{,0+1+5}
L(e) = min{6,0+2+2+3} dif. Path
b c e
Where w(v,x)
is distance
from v to x.
1
2
d
4
3
4
6
e
z
6
7
3
f
1
5
5
g
6
Chose (v= z ) ∊ T as minimum L(v)
Fourth Iteration(*)
For each x in T adjacent to v: L(x) = min{Lold(x),L(v)+w(v,x)}
2
T := set of all temporary vertices
c 4
2
b
2
a
1
Labels
4
L(a)=min(,0)
L(f) = min{,0+1}
L(b) = min{,0+2}
L(c) = min{,0+2+2}
L(d) = min{,0+1+3}
L(d) = min{4,0+2+2} dif. Path
L(z) = min{,0+2+2+1}
L(e) = min{,0+2+4} sorted
L(g) = min{,0+1+5}
L(e) = min{6,0+2+2+3} dif. Path
Where w(v,x)
is distance
from v to x.
1
2
d
4
3
4
6
e
z
6
7
3
f
1
5
5
g
6
Terminate zT
Path to Z shorter than any unexplored path
Another Example: 1
a,2
b
c 
3
2
2
0
1
a
z 
1
Labels
L(b) = min{,0+2}
L(d) = min{,0+1}
2
d
a,1
1
e

Choose v ∊ T with minimum L(v)=d
Another Example: 2
a,2
b
c 
3
2
2
0
1
a
z 
1
Labels
L(b) = min{,0+2}
L(d) = min{,0+1}
L(e)=min{,0+1+1}
2
d
a.1
1
e
d.2
Choose v ∊ T with minimum L(v)=d
Another Example: 3
a,2
b
3
c b,5
2
2
0
1
a
z 
1
Labels
L(b) = min{,0+2}
L(d) = min{,0+1}
L(e)=min{,0+1+1}
L(c)=min{,0+2+3}
L(e)=min{2,0+2+1} diff path
2
d
a.1
1
e
d.2
Choose v ∊ T with minimum L(v) =e
Another Example: 3
A,2
B
3
C
2
2
0
1
A
Z 
1
2
D
A,1
1
E
Another Example: 4
a,2
b
3
c b,5
2
2
0
1
a
z e,4
1
Labels
L(b) = min{,0+2}
L(d) = min{,0+1}
L(e)=min{,0+1+1}
L(c)=min{,0+2+3}
L(e)=min{2,0+2+1} diff path
L(z)=min{,0+1+1+2}
Each node is labelled with its
source node and the distance
from the start node
2
d
1
e
a.1
d.2
Shortest path: a,d,e,z
Choose v ∊ T with minimum L(v)=z
Find shortest path a-f
7
d
b
2
10
5
2
4
3
e
a
6
7
c
f
Find shortest path a-f

7
d
b

2
10
5
2

a
4
3
e
6
7
c

All distances set to infinity 

f

Find shortest path a-f
7
d
b
2
10
5
2
4
3
e
a(0,a)
a
6
7
c
f
Labels
L(a) = min{,0}
L(b) = min{,0+5}
L(c) = min{,0+6}
Each node is labelled with its
source node and the distance
from the start node
Find shortest path a-f
7
d
b
2
10
5
2
4
3
e
a(0,a)
a
6
7
c
f
Labels
L(a) = min{,0}
L(b) = min{,0+5}
L(c) = min{,0+6}
L(d) = min{,0+5+7}
L(f) = min{,0+5+10}
L(e) = min{,0+5+2}
L(c) = min{,0+5+3}
Each node is labelled with its
source node and the distance
from the start node
Find shortest path a-f
7
d
b
2
10
5
2
4
3
e
a(0,a)
a
6
7
c
f
Labels
L(a) = min{,0}
L(b) = min{,0+5}
L(c) = min{,0+6}
L(d) = min{,0+5+7}
L(f) = min{,0+5+10}
L(e) = min{,0+5+2}
L(c) = min{,0+5+3}
Each node is labelled with its
source node and the distance
from the start node
Find shortest path a-f
7
d
b
2
10
5
2
4
3
e
a(0,a)
a
6
7
c
f
Labels
L(a) = min{,0}
L(b) = min{,0+5}
L(c) = min{,0+6}
L(e) = min{,0+5+2}
L(d) = min{,0+5+7}
L(f) = min{,0+5+10}
L(c) = min{,0+5+3}
L(e) = min{,0+6+7}
Each node is labelled with its
source node and the distance
from the start node
Find shortest path a-f
7
d
b
2
10
5
2
4
3
a
f
e
a(0,a)
6
7
c
Labels
L(a) = min{,0}
L(b) = min{,0+5}
L(c) = min{,0+6}
L(e) = min{,0+5+2}
L(f) = min{, 0+5+2+4}
L(d) = min{,0+5+7}
L(f) = min{,0+5+10}
L(c) = min{,0+5+3}
L(e) = min{,0+6+7}
Find shortest path a-f
7
d
b
Minimum
Paths
2
10
5
2
4
3
a
f
e
a(0,a)
6
7
c
Shortest path: a,b,e,f
Length of path =11
Labels
L(b) = min{,0+5} –from a
L(c) = min{,0+6} –from a
L(d) = min{,0+5+7} –from b
L(f) = min{,0+5+10} – from b
L(e) = min{,0+5+2} –from b
L(c) = min{6,0+5+3} –from b
L(c) = min{6,0+5+2 +7} –from e
L(f) = min{, 0+5+2 +4} – from e
d not explored because (5+2+4)<12
Shortest Path Example
Shortest Path Example

v3 
5
v2
3
6
3
0
1
v1
v6 
2
1
6
3
v4

4
v5


v3 
5
v2
3
6
3
1
0 v1
v6 
2
1
6
3

Step
Vertex to be
marked
4
v4
v5 
Distance to vertex
Marked vertices
v1
v2
v3
v4
v5
v6

0
v1
0
3

6

1
v2
0
3
,8
6,4
,6 
2
v4
0
3
8,5
4
6,8

v1,v2,v4
3
v3
0
3
5
4
6,8
,11
v1,v2,v4,v3
4
v5
0
3
5
4
6
11,9
v1,v2,v4,v3,v5
5
v6
0
3
5
4
6
9
v1,v2,v4,v3,v5,v6
v1
v1,v2

v3 
5
v2
3
6
3
1
0 v1
v6 
2
1
6
3

Step
Vertex to be
marked
4
v4
v5 
Distance to vertex
Marked vertices
v1
v2
v3
v4
v5
v6

0
v1
0
3

6

1
v2
0
3
,8
6,4
,6 
2
v4
0
3
8,5
4
6,8

v1,v2,v4
3
v3
0
3
5
4
6,8
,11
v1,v2,v4,v3
4
v5
0
3
5
4
6
11,9
v1,v2,v4,v3,v5
5
v6
0
3
5
4
6
9
v1,v2,v4,v3,v5,v6
vi
v1,v2
Path = <A,D,E,F>
Step
Vertex
to be
marked
Distance to vertex
Unmarked vertices
A
B
C
D
E
0
A
0
2

3
  B,C,D,E,F
1
B
0
2
3
3
6
 C,D,E,F
2
D
0
2
3
3
5
 C,E,F
3
C
0
2
3
3
5
8 E,F
4
E
0
2
3
3
5
6 F
5
F
0
2
3
3
5
6
F
Len = 6
Path = <A,D,E,F>
Step
Vertex
to be
marked
Distance to vertex
Unmarked vertices
A
B
C
D
E
0
A
0
2

3
  B,C,D,E,F
1
B
0
2
3
3
6
 C,D,E,F
2
D
0
2
3
3
5
 C,E,F
3
C
0
2
3
3
5
8 E,F
4
E
0
2
3
3
5
6 F
5
F
0
2
3
3
5
6
F
Len = 6
Find path a to f
UML for Dijkstra
Petri Nets
• A Petri Net is a directed bipartite graph where nodes are:
Transitions are events denoted by boxes
Places are conditions denoted by circles.
Arcs (arrows) describe which places are pre-conditions or
postconditions for which transitions.
Petri Nets
• The states of the net is called markings. A place can
hold tokens usually denoted by black dots. The places
represent types of states. An instance of a state is
represented by a token inside the circle. The distribution
of tokens over the places of a net corresponds to a state
of the system. Petri Nets can be expressed in rewriting
logic (RWL). A token passing through the places
represents a process, and a record of the passage is a
history.
Petri Nets in CafeOBJ
• Marking is represented as an element of the finite multiset sort
Marking. The constant null represents the empty marking and _ _
is the corresponding multiset union operator. Associativity,
commutativity and identity laws are specified as structural equations by
the operator attributes in curled brackets. For each place p there is a
constant p, called token constructor, representing a single token
residing in that place. For each transition t there is a rule, called
transition rule, labelled by t and stating that its left marking may be
replaced by its right marking
.
Petri Net : Definition
• A net is a triple: N=(P,T,F)
• P and T are disjoint finite sets of places
and transitions, respectively.
• F is a set of arcs (or flow relations).
• F ¹ (P ² T) ¼ (T ² P) º
• Given a net N = (P,T,F ), a
configuration is a set C so that C º P.
Example of Rewriting Logic1
A Petri net of
the vending
machine
mod VENDING-MACHINE {
[ Coin Item < Place < Marking ]
op null : -> Marking -- empty marking
ops $ q : -> Coin
ops a c : -> Item
var M : Marking
op _ _ : Marking Marking -> Marking {assoc comm id: null}
rl [buy-c] : $ => c .
rl [buy-a] : $ => a q .
rl [change] : q q q q => $ . }
A Petri net is a directed bipartite
graph, in which the nodes represent
1) Transitions (events) denoted by
boxes
2) Places (conditions) denoted by
circles. The directed arcs (arrows)
describe which places are
pre/postconditions for which
transitions.
Example of Rewriting Logic
-- Show one way to spend 3 dollars ($ $ $)
exec $ $ $ .
-- Show all ways to spend 3 dollars ($ $ $)
red $ $ $ =(*)=> m:Marking .
-- How can I get 2 apples with 3 dollars?
red $ $ $ =(*)=> a a M:Marking .
Example of Rewriting Logic
If I have a dollar and three quarters, can I get a cake
and an apple? Yes.
red $ q q q =(*,*)=>! a c M:Marking .
Transitions
River Puzzle1
mod! RIVER-CROSSING {
[ Side Group ]
ops west east : -> Side
op change : Side -> Side
ops farmer wolf goat cabbage : Side -> Group
op __ : Group Group -> Group { assoc comm }
ops initial final : -> Group
vars S S' : Side
eq change(west) = east .
eq change(east) = west .
River Puzzle1
-- constraint for goat eaten
ceq wolf(S) goat(S) farmer(S') =
wolf(S) farmer(S') if S =/= S' .
-- constraint for cabbage eaten
ceq cabbage(S) goat(S) wolf(S') farmer(S') =
goat(S) wolf(S') farmer(S') if S =/= S' .
-- initial configuration all elements on the west.
eq initial = farmer(west) wolf(west) goat(west) cabbage(west) .
-- final configuration all elements on the east
eq final =
wolf(east) farmer(east) goat(east) cabbage(east) .
River Puzzle1
trans [farmer-alone] : farmer(S) => farmer(change(S)) .
trans [wolf] : farmer(S) wolf(S) => farmer(change(S))
wolf(change(S)) .
trans [goat] : farmer(S) goat(S) => farmer(change(S))
goat(change(S)) .
trans [cabbage] : farmer(S) cabbage(S) =>
farmer(change(S))cabbage(change(S)) . }
open RIVER-CROSSING
-- Evaluate using the rewrite predicate =(*)=>
red initial =(*)=>
wolf(east) farmer(east) goat(east) cabbage(east) .
-- The following command will show the states
show path 27
Starts with Farmer, Wolf, Goat, Cabbage on all West (W) bank
[state 1]
farmer(west) wolf(west) goat(west) cabbage(west)
[state 2]
farmer(east) wolf(west) goat(east) cabbage(west)
[state 3]
farmer(west) wolf(west) goat(east) cabbage(west)
[state 4] farmer(east) wolf(east) goat(east) cabbage(west)
[state 5] farmer(west) wolf(east) goat(west) cabbage(west)
[state 6] farmer(east) wolf(east) goat(west) cabbage(east)
[state 7] farmer(west) wolf(east) goat(west) cabbage(east)
[state 8] farmer(east) wolf(east) goat(east) cabbage(east)
Farmer, Wolf, Goat,
Cabbage on all West
(W) bank
River-Crossing
River-Crossing possible states & safe states
Safe states
A directed graph or digraph g1
c
e
a
b
d
Fig. 19.1
Sample Graph g2
v1
Graph g2
v4
v2
v5
v3
Sample Graph g3
v3
v6
Graph g2
v1
v5
v2
v4
v7
[('a','c',2), ('a','d',6), ('b','a',3) ,('b','d',8), ('c','d',7), ('c','e',5) ,('d','e',10)]
Sample Graph g3
8
b
d
10
6
3
2
a
buildGraph
[('a','c',2), ('a','d',6),
('b','a',3)
,('b','d',8), ('c','d',7),
('c','e',5)
,('d','e',10)]
7
c
5
e
fromList [('a',[('b',3.0),('d',6.0),('c',2.0)]),
('b',[('d',8.0),('a',3.0)]),
('c',[('e',5.0),('d',7.0),('a',2.0)]),
('d',[('e',10.0),('c',7.0),('b',8.0),('a',6.0)]),
('e',[('d',10.0),('c',5.0)])
]
build graphs makes edges bidirectional
Review
• Definitions for path, simple path, degree of
vertex.
• Understand real world examples where
these concept could be used.
• Find path, simple path, cycle, or simple
cycle.
Review
• Handshaking Theorem
• Holds for a particular graph
Review
• Showing Euler Cycle exists in a graph
• Finding an Euler Cycle in a particular
graph.
• Finding a Hamilton Cycle in a particular
graph.
Review
• Definitions complete graph and planar
graph.
• Prove that the complete graph K5 is not
planar.
Review
• What is a bipartite graph.
• The complete K3 is not bipartite.
• Shortest path a-f
Shortest Path in C#
• A=0(A) B=2(A) C=3(D) D=1(A) E=3(D) F=6(G)
G=5(D)
x
2
B
C
2
x
1
2
A
1
Z
4
D
F
E
6
7
3
x
3
4
5
G

Graph Summary
• A complete digraph (v>2) has a Hamilton
Path.
Application of Graph Theory to
Programming
• Control flow graph use graph notation to show
all paths that might be traversed through a
program during its execution.
Application of Graph Theory to
Programming
•
Cyclomatic complexity is a software metric . It is used to indicate the
complexity of a program. It directly measures the number of linearly
independent paths through a program's source code. Cyclomatic complexity
is computed using the control flow graph of the program. The nodes of the
graph correspond to indivisible groups of commands of a program, and a
directed edge connects two nodes if the second command might be
executed immediately after the first command. Cyclomatic complexity may
also be applied to individual functions, modules, methods or classes within
a program.
•
It can also be used to test each linearly independent path through the
program, the number of test cases will equal the cyclomatic complexity of
the program
Order Sorted Algebra and
Preorder Algebra
• CafeOBJ includes two distinct algebras:
• OSA is good for representing abstract data type
and values (a being ontology)
• POA representing states and reasoning about
transitions between states (a doing ontology).
• The intuitive distinction is the difference between
you as an individual and your life as a process.
The Blocks World Definition - Actions
• Blocks are picked up and put down by the arm
• Blocks can be picked up only if they are clear, i.e.,
without any block on top
• The arm can pick up a block only if the arm is empty, i.e.,
if it is not holding another block, i.e., the arm can be pick
up only one block at a time
• The arm can put down blocks on blocks or on the table
A problem from blocks world
Initial State
Goal State
• Find a sequence of robot moves to
rearrange blocks from the initial state to
the goal or final state.
Blocks world state space
Start
Goal
Block Worlds
• In general, if there are n places, there
should be n*(n-1) possible moves.
• In this case 3*2=6 possible moves
Block Worlds
• There are 9 possible positions for each of
the 3 blocks but we cannot use a
permutation calculation because many
configurations such as
are not allowed.
• There are 10 possible arrangements of
blocks if the letters are disregarded. Three
block can be arranged in 6 ways, there are
10x6=60 possible configurations.
Block Worlds
• There are permCalc(9,3) = 504 possible
ways to distinctly configure 3 items from 9.
But in this case the rules of the block world
only allows 10*6 = 60 possible states
Block Worlds
• Holding as a state
Block Worlds
• In general, if there are n places, there
should be n*(n-1) possible moves.
• In this case the first set of moves 3*2=6
Blocks World Complexity
• An optimal solution performs the
transformation using as few steps as
possible. Any solution: linear complexity
• Optimal solution: exponential complexity
(NP hard: see later).
Complexity
• Linear time O(N) : Finding an item in an unsorted list.
• Polynomial time O(NK) : computing square roots,
powers, and logarithms (k is a constant ).
• Exponential Time O(KN): The shortest route A to A,
which visits all points and returns to point A. This is
called the travelling salesman problem. As the size of the
problem increases linearly (O(N)), the time to solve the
problem increases exponentially (O(K to the power N)).
P and NP
• The time of efficient algorithms of type P
grow as a polynomial function (such as n,
n2, or n3) of the size of the input.
• NP problems could be solved in
exponential time by checking through all
candidate solutions. But such an algorithm
whose running time is 2n or above is
useless in practice.
Blocks world state space
(Maude: specification and programming in rewriting logic
M. Clavel, F. Duran , S. Eker , P. Lincoln , N. Marti-Oliet,J. Meseguer , J.F. Quesada
,)
Instead of equations (denoted =)we use user-defined transition rules
denoted => .
Blocks world state space
let Start = empty & clear('b) &
table('a) & on('b,'a)
.
let Goal = empty & clear('a) &
table('b) & on('a,'b)
.
-- Start & goal configurations
let Start = empty & clear('b) &
table('a) & on('b,'a)
.
let Goal = empty & clear('a) &
table('b) & on('a,'b)
.
-- Search for solution
red
Start =(1,*)=>* Goal .
-- Examine solution path
show path 4
empty&clear('b)&table('a)& on('b,'a)
[state 1] unstack
(hold('b)&clear('a)& table('a))
[state 2] putdown
empty&clear('b)&table('b)&clear('a)
&table('a)
[state 3] pickup
hold('a) & clear('b) & table('b)
[state 4] stack
empty&clear('a)&on('a,'b)&table('b)
Could be written as transition name
followed by argument as follows:
unstack(b) ; putdown(b) ; pickup(a)
; stack(a,b)
Sorting
• An algorithm for sorting N distinct elements could potentially
have N! states. We will ignore duplicates for the moment.
• In this case 3! = 3 * 2 * 1 = 6 states.
• [0] 3 . 2 . 1 =swap=> [1] 2 . 3 . 1
• [1] 2 . 3 . 1 =swap=> [3] 2 . 1 . 3
• [3] 2 . 1 . 3 =swap=> [5] 1 . 2 . 3
• Sorting involves finding a path through the states <0,1,3,4>
• Sorting a list involves finding one of permutation p from N!
such that p will be in sorted order, in this case state 5 is such
a permutation. It can be reached in two distinct but equivalent
ways.
How many states?
• If there are N elements then there are N!=M
potential states. To find a solution path use the
command: show path M. CafeOBJ numbers the
states from 0. Termination can be proved.
• Disorder becomes lower along transitions between
some of the states. Using CafeOBJ Diaconescu[1]
proved that, s ==> s’ implies disorder
(s’) < disorder(s).
Preorder algebra (POA)
(A Methodological Guide to the CafeOBJ Logic: Razvan Diaconescu)
Note this is a transition rule.
ctrans N . N’ => N’ . N if (N’ <= N) and (N =/= N’)
This transition equation says swap elements if they are in the wrong
order. We can use exec or the search command (next slide)to
execute transition rules.
CafeOBJ’s search command
(Futatsugi )*
• The search command searches the state space of POA
models. Written red (t1 =(m,n)=>* t2) it returns
true if t1 can be translated (or rewritten), via more than
0 times transitions, to some term which matches to t2.
Otherwise, it returns false . Possible transitions are
searched in breadth first fashion. n is upper bound of
the depth of the search, m is upper bound of the number
of terms which match to t2. If either of the depth of the
search or the number of the matched terms reaches to
the upper bound, the search stops.
* Lecture from Kokichi Futatsugi: http://lipn.univ-paris13.fr/~choppy/IFIP/WINCHESTER-2011/WINCH-DATA/Futatsugi.pdf
CafeOBJ’s search command
(Futatsugi)
Permutation & Combination
• Fruit salad is a combination of apples,
grapes and bananas. We don't care what
order the fruits are in.
• The permutation that will open the lock is
942, we do care about the order.
• There are fewer combinations than
permutations. See the Calculations.cafe.
State Space
• State space = Directed graph
• Nodes = Problem situations
• Arcs = Actions, legal moves
• Problem = ( State space, Start, Goal)
• Note: several nodes may satisfy goal condition
• Solving a problem = Finding a path
• Problem solving = Graph search
• Problem solution = Path from start to a goal