Chapter 7 Mathematics of Networks
ESSENTIAL QUESTIONS:
Section 7.1: How is a tree different from other graphs?
Section 7.2: What is a spanning tree?
Section 7.3: How do you use Kruskal’s Algorithm to find Minimum Spanning Tree (MST)?
Section 7.4: How can you find the shortest network connecting 3 points?
WORD WALL:
INTERIOR JUNCTION RULE
INTERIOR JUNCTION POINT
JUNCTION POINT
KRUSKAL’S ALGORITHM
MINIMUM NETWORK PROBLEMS
MINIMUM SPANNING TREE (MST)
NETWORK
REDUNDANCY
SHORTEST NETWORK
SHORTEST NETWORK RULE
SPANNING TREE
STEINER POINT
STEINER TREE
SUBGRAPH
TREE
HOMEWORK:
7.1: p. 259 # 1, 2, 3 – 7 (odd)
7.2: p. 260 #11 – 12 (all), 16 – 18 (all)
7.3: pp. 260 – 1 # 19 – 25 (odd)
7.4: DAY ONE: Triangle Worksheet and DAY TWO: p. 262 # 27-34
Section 7.1: Tree
What is a NETWORK? Any CONNECTED GRAPH
Vertices of networks are often called nodes or terminals and represent “objects”
cities, subway stations, computer servers, people, etc
o Edges of networks are called links and indicated connections among objects
airline routes, rail lines, internet connections, social connections, etc
CHAPTER THEME: Finding optimal networks that connect a set of points the cheapest
OPTIMAL NETWORK GOALS:
1) Connect all the “TERMINALS” = Vertices
o “CONNECT” DOESN’T HAVE TO MAKE an euler/ hamilton path or circuit
A
2) Total cost (length) of the network as small as possible (optimal)
Example Network Problem: The weighted graph represents a system of
roads connecting buildings of a company’s headquarters. The company
is planning to upgrade its fiber optic cable through underground lines
connecting its servers. The weights represent cost in millions of dollars
to upgrade these cables between each building. The goal for the
company is to find a network that utilizes the existing network of roads;
connects all the buildings together; has the least cost.
3.1
E
2.6
2.2
2.5
2.8
B
1.8
2.9
2.7
2.4
D 2.1 C
Key Term Requirements of network in the example:
1) SUBGRAPH: edges come from the original graph
2) SPAN: must include all vertices of the original graph
3) MINIMAL: total weight of the network must be as small as possible “OPTIMAL”
o Cannot have a circuit
o Information, cars, people, etc are allowed to travel along network in both directions
EXAMPLE: Based on the original graph, draw different examples of subgraphs and spans.
A
Original
SUBGRAPH of 4 vertices + 6 edges
C
B
B
D
E
SUBGRAPH of 5 vertices + 4 edges
A
A
B
D
C
E
F
A
D
A
C
F
D
E
SPAN of 5 edges
F
C
B
B
E
C
B
F
SPAN of 6 edges
SPAN of 5 edges
E
F
D
E
F
SPECIFIC TYPE OF NETWORK:
TREE: a network (connected graph) with no circuits of any length
Which of the following graphs a – h represents a tree: Why or why not?
(a)
(b)
NO
YES
(e)
YES
YES
NO
(g)
(f)
(d)
(c)
(h)
YES
NO
NO
Observation Questions for Graphs that are TREES:
1) Is there a relationship between edges and vertices?
ONE MORE VERTEX THAN EDGE or EDGES ONE LESS THAN VERTICES
2) Is there a special term you can use to describe the edges used?
EVERY EDGE IS A BRIDGE
SPANNING TREE:
A subgraph of a network that connects all the vertices and has no circuits
“tree” inside of a graph using all vertices
EXAMPLE: For the given graph, find a spanning tree.
A
C
B
A
D
C
E
B
A
D
C
E
B
A
D
C
E
B
A
D
C
E
B
D
E
PROPERTIES # 1 – 3 OF TREES:
PROPERTY #1:
o In a tree, there is one and only one PATH joining any two vertices.
o If there is one and only one path joining any two vertices of a graph, the graph must be
a tree.
Path joining any two vertices = connected graph
one and only one = no circuits exist
PROPERTY #2: IE no circuits
o In a tree, every edge is a BRIDGE.
o If every edge of a graph is a bridge, then the graph must be a tree.
Every edge is bridge = no circuits because you can’t get back to any vertices to complete a
circuit.
PROPERTY #3:
o A tree with N vertices has N-1 EDGES.
o If a NETWORK has N vertices and N-1 edges, then it must be a tree.
TREE/ NETWORK = connected
“Number of Edges is one less than the number of vertices in a tree”
Practice Problems: A graph G has no loops or multiple edges. Identify if the graph (1)
ALWAYS is a tree, (2) NEVER is a tree, or (3) SOMETIMES is a tree.
1) G has 4 vertices and 3 edges.
SOMETIMES
5) G has 10 vertices and 9 edges.
SOMETIMES
2) G is connected with 4 vertices and 3
edges
ALWAYS
6) G has 10 vertices and 11 edges.
NEVER
7) G has 12 vertices and 11 bridges.
ALWAYS
3) G has 5 vertices and 5 edges.
NEVER
8) G is connected with 8 vertices and
every vertex has degree 7.
NEVER
4) G has 5 vertices and 4 bridges.
ALWAYS
HW 7.1: p. 259 # 1, 2, 3 – 7 (odd)
Section 7.2: Spanning Tree
SPANNING TREE: a subgraph of a network that connects all the vertices and has no circuits
1A. Is the following graph a tree? Why?
A
B
2A. Is the following graph a tree? Why?
A
C
B
D
E
C
D
F
G
E
F
G
1b. Find a spanning tree in the graph.
A
E
B
D
2b. Find a spanning tree in the graph.
The graph itself is a spanning tree because it is a
tree
C
G
F
PROPERTY #4 of Trees:
o If a network (CONNECTED GRAPH) has V vertices and E edges, then E ≥ V - 1.
In a connected graph, number of edges is greater than or equal to number of vertices minus 1.
REDUNDANCY = Difference of EDGES and VERTICES plus ONE
R=E–V+1
Number of edges to remove from network to create tree
o If E = V - 1, then the network is a tree; if M > V - 1, the network has circuits and is
not a tree.
If R = 0; network is tree
If R > 0; network is not a tree
If Redundancy is positive, (R > 0), then many SPANNING TREES exist in network
However, It doesn’t identify which edges must be removed.
How do you find spanning trees?.
1) Your goal is to remove R edges to create a tree.
(Break Up Circuits)
2) DO NOT remove edges that are Bridges.
EXAMPLE #1:
1) How many edges do you need to remove?
R=1 = 8 – (8 – 1)
Circuit = CDHC; CD, HC, HD
B
C
D
A
H
E
2) How many possible spanning trees exist in this graph?
3  remove one edge from circuit of 3 lines
G
F
EXAMPLE #2: How many edges do you need to remove? R = 2 = 9 – 8 + 1
2 EDGES TO REMOVE
Notice there are two separate circuits to break up CDHC and
DEFGD which you need to remove one edge from each to
make R =0
METHOD #1: FIND BY DRAWING
REMOVE CD
REMOVE CH
REMOVE HD
CD, DE
CH, DE
HD, DE
CD, EF
CH, EF
HD, EF
CD, FG
CH, FG
HD, FG
CD, GD
CH, GD
HD, GD
B
C
D
A
H
E
F
G
METHOD #2: How might we know when we’ve found all spanning trees?
Circuit #1 = 3 edges
Circuit #2 = 4 edges
Multiplication Rule: Remove 1 edge from Circuit #1 and Remove 1 edge from Circuit #2
= 3 * 4 = 12
EXAMPLE #3: How many edges do you need to remove? R = 2 = 9 – 8 + 1
2 EDGES TO REMOVE
Notice there are two separate circuits to break up CDHC and DEFGHD which you need to
remove one edge from each of those circuits to make R =0
PROBLEM: HD is shared by both circuits
B
METHOD #1: FIND BY DRAWING
REMOVE CD
REMOVE CH
CD, DE
CH, DE
REMOVE HD
HD, DE
CD, EF
CH, EF
HD, EF
CD, FG
CH, FG
HD, FG
CD, GD
CH, GD
HD, GD
CD, DH
CH, HD
C
D
A
H
E
G
F
METHOD 2: How might we know when we’ve reached finding all spanning trees?
METHOD 2A: REMOVE THE ONE CASE OF HD CHOSEN TWICE: 3 * 5 – 1 = 14
METHOD 2B: Break it Into Cases:
Case #1: Exclude HD and then can exclude any of other 6 edges from circuits = 1 *6 = 6
Case #2: Must Keep HD; Circuit #1 has 2 edges and Circuit #2 has 4 edges = 2 * 4 = 8
SPANNING TREE PRACTICE
For each of the below graphs, find the total number of possible spanning trees.
1) 4
2) 4*5 = 20
3) 4*3 – 1 = 11
4) 4*5*(3*3 – 1) = 160
5) (3*5 – 1)(3*4 – 1) =
154
6) 3 * 4 * 6 = 72
7) 4*5*(4*4 – 1) = 300
8) Advanced: 8 + 15 + 12 + 16 = 39 (4 Cases)
Cases: Keep Both, Remove Both, Remove One
and Keep Other (Vice Versa)
HW 7.2: p. 260 #11 – 12 (all), 16 – 18 (all)
Section 7.3: Kruskal’s Algorithm
MINIMUM SPANNING TREE (MST):
a spanning tree of a weighted network with the least total weight
How many spanning trees exist in each of the given networks?
Network #1:
Network #2:
Network #3:
9
5
7
3
8
6
(3*5 – 1) = 14
19
7
10
11
36
21
9
23
32
23
15
17
45
8 4
11
14
14
8
18
13
42
(3*3 – 1) (4) = 32
29
5
(3*4-1)(3*4-1) = 121
How can we find the minimum spanning tree without checking all of the
possible spanning trees?
Kruskal’s Algorithm for N vertices on a Network.
Step #1: Pick the cheapest edge available.
o In case of a tie, randomly pick one edge.
o Mark it or somehow identify you chose it.
REPEAT: Pick the next cheapest link or edge available and mark it.
 RULE #1: Cannot make a Circuit by choice
 RULE #2: Stop once you’ve chosen N-1 Edges
 RULE #3: Vertices are allowed to have any degree > 0
14
12
Example: The following graph represents roads connecting seven towns in a developing
jungle region that is planning to lay fiber-optic cable. The cost of laying the fiber optic
cable is represented by the weights of graph. Let’s find the minimum spanning tree (MST)
using Kruskal’s Algorithm.
GF = 42, BD = 45, AD = 49, DG = 51, CAN’T USE AB, CD = 53, CE = 59
Positives of Kruskal’s Algorithm:
It’s easy to implement – it doesn’t require a lot of complex steps
It’s an efficient algorithm – it doesn’t take long to complete.
It’s an optimal algorithm – will always find a minimum spanning tree
EXAMPLE PROBLEMS:
A
2.4
2.3
45
3.5
2.6
3.1
C
45
25
40
C
1.2
1.4
5
B
2.8
D
B
35
2.2
2.1
E
A
D
15
20
20
40
E
5
10
F
7.3: pp. 260 – 261 # 19 – 25 (odd)
7.4 Introduction Triangles
GEOMETRY - TRIANGLES REVIEW:
BASIC TRIANGLE RELATIONSHIPS: ANGLES v. SIDE LENGTHS
LONGEST Side of a triangle is opposite (across from) the BIGGEST angle
SHORTEST Side of a triangle is opposite (across from) the SMALLEST angle
MIDDLE Side of triangle is opposite (across from) the MIDDLE angle.
Triangles are Not Drawn to Scale!!
B
B
R
6
7
47º
P
10
35º
C
A
10
Which angle is the largest?
What is measure of angle Q?
B: Opposite 10
98 = 180 – (35 + 47)
Which angle is the smallest?
Which side is the shortest?
A: Opposite 6
RQ: opposite 35
Which angle is the middle?
Which side is the longest?
C: Opposite 7
PR: opposite 98
C
Which angle is larger A or C?
Why?
C < A because 10 < 14
Write an inequality to describe the
length of AC.
AC > 14 longest side
Y
E
27º
25
19
18
14
A
Q
D
105º
What is of angle X or Z larger? Why?
X > Z because 25 > 19
18
33º
Z
X
F
What is the largest total of angle X and Z?
What is measure of angle E and F?
X + Z < 66 b/c X = 33 and Z < 33
E = 27 and F = 126
What value must angle Y be larger than?
Is DE longer or shorter than 18? Why?
Y > 147 because X + Z < 66 Z < 33
DE > 18 because longest
RIGHT TRIANGLES:
State the Pythagorean Theorem
Label the Right Triangle based on your statement.
a2 + b2 = c2
PRACTICE: Find the value of the missing side x. (Triangles not drawn to scale)
5
x
x
5
x
3
X=4
4
6
12
X = 13
X = 8.49
= 6√2
6
X = 6.93
= 4√3
x
8
SPEICAL RIGHT TRIANGLES: 300 – 600 – 900
ALL 300 – 600 – 900 triangles are SIMILAR and the relationship of the sides are proportional.
Hint: Place a 1 and 2 on a triangle and use Pythagorean Theorem
Hypotenuse = Opposite the 900 = 2
300
Short Leg = Opposite the 300 = √3 = Pythagorean Theorem
2
Long Leg = Opposite the 600 = 1
√3
To move between sides because sides are always proportional.
Bigger to Smaller = Divide
Smaller to Bigger = Multiply
600
900
1
Always go through the “1” side otherwise 2 to √3 requires using a fraction value.
PRACTICE: Use your knowledge of 300 – 600 – 900 to find all side lengths.
20
10*2
10√3
10*√3
15
15√3
15*√3
10
60÷√3
2*30÷√3
60
15÷√3
300
300
600
30
30÷√3
2*15÷√3
30
15*2
25
4
8÷2
600
100÷√3
100
0
30÷√3
30
4√3
8
0
200÷√3
2*100÷√3
50
100÷√3
2*50÷√3
40
300
600
600
50÷√3
20
40 ÷ 2
20√3
20*√3
GENERAL TRIANGLE PRACTICE
IDENTIFY missing information as largest, smallest, or middle angles and lengths.
 If possible, provide an exact value for missing measurement.
 If not possible, then provide approximate angle or side values with inequalities.
A
A
18
C
200
15
36
450
C
B
20
LARGEST
A ________________
MIDDLE
B ________________
B
C
>74 = 180 – (53+53
A _______________
< 53;Smallest
C ________________
=115 = 180 – (20+45)
CB_______________
A
A
B
800
600
530
B
120
89
110
> 36; Longest
AC_______________
< 36; Shortest
CB_______________
=115 = 180 – (20+45)
B ________________
SMALLEST
C ________________
A
A
10
0
70
0
35
35
C
B
65
420
B
C
C
>
BC
>
120
AC_______________
> 120
CB_______________
= 10 (C = B)
AC_______________
>10
CB_______________
> 70
BC_______________
< 42
C _______________
=40 = 180 – (60+80)
C ________________
=110 = 180 – (35+35)
A ________________
>96= 180 – (42+42)
A ________________
A
57
A
100
1300
C
41
B
23
C
< 41
AC_______________
= 40 = 180 – (130 + 10)
A_______________
800
23
B
> 23
CB_______________
= 50 = (180 – 80)/2
B_______________
= 50
C ________________
Section 7.4: The Shortest Network Connecting Three Points
Minimum Spanning Tree (MST): The spanning tree of a given network that has the least weight
 Subgraph
 No Circuits
 Connected Graph
MST is the OPTIMAL(shortest) way to connect a set of vertices based on the assumption that all
CONNECTIONS (EDGES) should belong to the links of the ORIGINAL NETWORK. .
 SPANNING part of the definition
A
Example: Consider the network (right) that describes 3 isolated towns (A,
B, C) which are equidistant from each other and located in the heart of the
Australian outback. The towns are connected by three unpaved straight
roads as described by the edges. What is the minimum spanning tree
solution to this network?
B
Pick any 2 sides and they add up to be 1000 miles.
A
A
500
B
C
B
500
500
miles
500 miles
C
A
500
500
500
miles
C
500
B
500
C
Could there be a shorter network than the Minimum Spanning Tree (MST) in a graph?
What would we need to do to the graph to check for shorter answer?
 You must be able to create new vertices and new edges to use.
 You need to be able to measure these new edges.
 This new network must have less weight than the MST. (Shorter than 1000)
SHORTEST NETWORK = the network of all vertices that has the SMALLEST total weight
without RESTRICTION on the edges you can use or create.
 Example: Do power cables have to follow a roads OR highways?
It might be possible to do this in the outback because of the flat land and not necessary to follow along the
current road system. Or in other situations where construction might not be limited by existing systems.
EXAMPLE #1 OF ADDING NEW VERTEX, T, AND EDGE: Vertex T is at the midpoint of the edge BC
and creates edge TA.. (T-NETWORK)
What is the length of TA?
A
AC = 500
500
miles
500
miles
TC = 250 = 500/2
TA = 250(√3) = 433
B
T
500 miles
C
What is the total weight of the T-Network (AT, BT, CT)?
Total Weight = 250 + 250 + 433 = 933
Is this T-Network shorter than the MST?
T-network is shorter than the MST (933 v. 1000)
EXAMPLE #2 OF ADDING NEW VERTEX, Y, AND EDGES: Put a new vertex, Y, in
the exact middle of the triangle and create edges YA, YB, and YC. (Y-NETWORK).
A
What is the length of YA, YB, and YC?
YA = YB = YC each one Is the same value
XC = 250
500
YX = 250 ÷ (√3) = 144.34
500
YC = 2•YX = 288.68
Y
B
C
500
X
What is the total miles
weight of the Y-Network (YA, YB, and YC)? 3•288.68 = 866.04
Is this Y-Network shorter than the MST and the T-Network? Y-network is shorter than both
What is important about the angles the edges of the Y-network form? __1200_______
Shortest Network Key Terms:

Junction Point: any point where two or more edges of the network come together
MST = A
T-NETWORK = T
Y-NETWORK = Y

Native Junction Point: a junction point that is located at an original vertices of the network
A

Interior Junction Point: a nonnative junction point located somewhere other than at one of the
original vertices of the network
T, Y; INTERIOR = MUST BE CREATED OR ADDED TO NETWORK BY YOU

Steiner Point: A specific interior junction point consisting of three line segments (edges) coming
together forming equal 1200 angles (a y-network)
Y
7.4 Continued
TRUE STORY EXAMPLE: in 1989, a consortium of several of the world’s biggest telephone companies
(AT&T, MCI, Spring, and British Telephone) completed the Third Trans-Pacific Cable (TPC-3) line, a network
of submarine fiber-optic lines linking Japan and Guam to the US (via Hawaii). The approximate straight line
distances between the three locations was 1620 Japan to Guam, 3910 Japan to Hawaii, and 3820 Guam to
Hawaii (in miles). The solution to these companies problem was to find the SHORTEST NETWORK between
the Japan-Guam-Hawaii Triangle.
The SOLUTION: use a Steiner Point of the Triangle to create the network of actual length 5690 miles.
Note that 5690 miles is longer than the MST but that is a result of uneven contour of the ocean floor. The MST
would have increased length besides the projected 5440 miles straight distance.
TEACHER NOTE: Students will need to calculate the steiner point for simple triangle involving 30-60-90
Is the SHORTEST NETWORK for THREE POINTS always a STEINER POINT?
GEOMETER’S SKETCHPAD INVESTIGATION???
Not every triangle has a steiner point. Any angle is 1200 or more => no more steiner points
Therefore the shortest Network can involve either a Steiner Point or the MST.
Use the transparency of 120o angles to check if a Steiner Point Exists in each triangle.
YES or NO
YES or NO
YES or NO
YES or NO
YES or NO
SHORTEST NETWORK CONNECTING THREE POINTS:
 If the LARGEST angle of the triangle is greater than or equal to 1200, the shortest
network linking the three vertices consists of the two shortest sides of the triangle. In this
situation, the shortest network is the minimum spanning tree.

If LARGEST angle of the triangle is less than 1200, the shortest network is obtained by
finding a STEINER POINTS inside the triangle and joining S to each of the vertices.
7.4 Continued
INTERIOR JUNCTION RULE for Shortest Networks:
In a shortest network the interior junction points are all Steiner Points.
Any other junction point in a triangle when the largest angle is less than 1200 will
create a longer network than the Steiner Point network.
Consider the triangle ABC with interior junction points of P, Q, and R. We know that
triangle ABC has no angle larger than 120o. The table below gives the lengths of all the
created edges from each junction point to the vertices of the triangle. In each table, which
junction point must be the STEINER POINT?
P
Q
R
A
380
390
700
87
B
620
680
260
77
104
C
300
190
550
243
232
Total
1300
1260
1510
P
Q
R
A
54
72
41
B
61
94
C
125
Total
240
SUMMARY OF TYPE OF SHORTEST NETWORK
Angles
Shortest
Network
Weight
of
Network
Example
Largest Angle ≥ 120
Largest Angle < 120
Minimum Spanning Tree (MST)
STEINER POINT (Y – NETWORK)
Pick the two smallest sides of the original
triangle.
Draw the y-network and find the 3 weight of the
edges drawn. (Normally requires trig)
Special Case: For an equilateral triangle, use the
30-60-90 triangle knowledge.
Apply the Shortest Network for Connecting Three Points
For each triangle, identify the shortest network as a Minimum Spanning Tree
(MST) or a Steiner Point.
Hint: Determine exact or approximate size of all angles in the triangle.
TRIANGLES ARE NOT DRAWN TO SCALE!!
83º
22º
118º
38º
59º
19º
MST:
139 > 120
STEINER:
83 < 120
STEINER:
118 < 120
27º
25
25
37
30º
37
25
STEINER:
60 < 120
MST:
126 > 120
STEINER:
90 < 120
25º
> 150
42
15º
63
30
< 15
MST:
(>150) > 120
< 25
> 130
35
MST:
(>130) > 120
120º
120º
120º
120º
120º
120º
120º
120º
120º
120º
120º
120º
120º
120º
120º
120º
120º
120º