2 Art and Geometry

2.1.1 What is Perspective?

Have you ever looked at a painting of a pretty scene of the country side and wondered how the artist was able to capture the beauty and essence of the landscape in the picture. Or perhaps, you wondered how the artist was able to create a three dimensional effect on a piece paper of that is only one dimensional. All these effects can be created by using different types of perspective. Perspective is the art of drawing objects in manner that creates three dimensional effect or depth in the drawing. There are several ways to create a perspective in art. The artist can use light, shading, variance in the size of objects, and lines of sight to create depth perception in a drawing or painting. As it turns out, there are several types of perspective. In this course, we will study four types of perspective.

Types of Perspective

1. Diminishing Sizes

2. Overlapping Shapes

3. Atmospheric Perspective

4. One Point Perspective

As we study each type of perspective, we will look both art work and photographs that exhibit each type of perspective.

2.1.2 Diminishing Sizes

In diminishing sizes depth is created by systematically making objects smaller. Objects that are closer to the eye are naturally larger that objects that are further from the eye. In some drawings or paintings, artist use variances in sizes to create depth perception. Diminishing sizes also are prevalent in photographs. In this first example, you can see how diminishing sizes occurs in a regular photograph. Notice that the first tower of the Golden Gate Bridge is much larger that the second tower. A depth perception is created naturally in the photo due to the different sizes of the objects in the photograph. If we look closer at the photo, we can also see that the suspension cables of the bridge get smaller as the bridge get further away.

58 Here are some examples of actual artwork that make use of diminishing sizes to create perspective.

Example 1: Pieter Bruegal’s The Peasant Wedding

Example 2: Pieter Bruegal’s Village Scene with Dance Around the May Pole

59 Summary of Diminishing Sizes

1. Diminishing sizes makes use of the position of the object and shading to create perspective.

2. Diminishing sizes can be used with other perspectives to create perspective, in particular it can be used with on point perspective.

2.1.3 Overlapping Shapes

In overlapping shapes depth perception is created by using overlapping shapes. Overlapping shapes create a sense of space and depth in a drawing. By placing one object behind another object, an artist can create the space of three dimensions to the ﬂat two-dimensional space of the paper. When shapes overlap each other in a drawing, it creates an illusion called shallow space. This illusion help create depth in the drawing. We strengthen the illusion of objects going back in space by placing them one behind another along the page.

Here is a painting by Rita Auerbach in titled Buﬀalo Waterfront Montage that exhibits perspective using overlapping shapes.

60 Notice how the artist creates perspective by overlapping the boats in the bottom of the painting and the buildings near the top of the building. In the next example, notice how overlapping shapes can occur in a normal photograph. In the photograph, the building naturally overlap each other creating a depth perception.

Summary of Overlapping Shapes

1. Overlapping create illusion called shallow space by overlapping objects to create perspective. 2. Overlapping shapes can be used with other perspectives to create perspective.

2.1.4 Atmospheric Perspective In atmospheric perspective depth is created by making objects that are farther away less clear by diminishing both color and shading. If you have ever looked at objects such as mountains at a distance, you will notice that they seem less vivid than objects that are closer. This eﬀect can be captured in a painting or drawing by using diﬀerent shading on

61 objects closer to the naked eye than object farther from the eye. Again, this perspective can also be clearly seen in a photograph as well. In next photograph, you can see how atmospheric perspective appears to the naked eye in real life. Notice that object closer the camera, the plants and buildings, appear to be much more vivid, where as objects further away from the camera, the mountains and San Francisco Bay Bridge, are less vivid.

The next example shown below is a painting by Albert Bierstadt entitled ”Garden of the Rockies”

Notice that the mountains and sky in the background that are farther from the eye are shaded diﬀerently than object that are closer to the eye.

62 Summary of Atmospheric Perspective

1. Atmospheric perspective uses color, shading, and vividness to create perspective.

2. Atmospheric perspective can be used with other perspectives to create perspective.

2.1.5 One Point Perspective

During the Renaissance period many artists wanted to create drawings and painting that look more realistic. In particular artists had trouble painting figures such as building, bridges, streets, and other man made structures. In search for answer Filippo Brunellechi , an Italian architect, discovered a technique called one point perspective. This technique helped artists paint scenes of man made structures with a more realistic appearance. In one perspective depth is created by using lines of site that all connect to a point of focus called a vanishing point. The first Example of One point perspective we will examine is Vincent Van Gogh’s Flower Beds of Holland. In Van Gogh’s paints painting you can see how the rows of the flowers all meet at vanishing point.

In the next illustration, you see how the rows of the ﬂowers bed all converge to one vanishing point.

63 You can also see one point perspective in a actual photograph. Notice how we can see that the cable car tracks meet at a vanishing point in the photograph. Notice that the vanishing point is located at a point near the end of the street or below the tower of the San Francisco Bay Bridge.

If you have ever walked on a straight stretch of railroad tracks, you usually can look oﬀ in the distance and image where the rails of the tracks would meet at a vanishing point. (See diagram below)

64 Summary of Atmospheric Perpective

1. One Perspective is the only one of the four perspectives we study that uses a focal point or vanishing point. 2. One point perspectives use a vanishing point along with line of site to create perspective. 3. One point perspective can be used with other perspectives to create perspective, espe- cially with diminishing sizes.

2.1.6 Dominant Perspective In many drawings there may be more than one perspective in eﬀect in the drawing. For example a painting or drawing could have elements of one point perspective, diminishing sizes, and atmospheric perspective. If there is a perspective that stands out more than other perspectives, then that perspective would be the dominant perspective. In the next photograph, one point perspective would be the dominant perspective.

Notice how the other perspectives, Diminishing sizes and overlapping shapes, are not as prevalent in the photograph as one point perspective is in the photograph.

65 Art without perspective Some paintings or drawings will not have a visible perspective in the picture. If this is the case, then the artwork will appear ﬂat without any depth perception. Notice how the Egyptian hieroglyphics in the picture below have ﬂat appearance with out any perspective. Often Egyptian hieroglyphics will have a lack of perspective.

2.1.7 Similar Triangles and Perspective

Suppose we asked to make a drawing with a few trees in it and we were asked to draw the trees using diminishing sizes to create a perspective in the drawing. How could we use math to ﬁgure out how to space the trees apart from one another. The answer to the problem is to use similar triangles and proportions to ﬁnd the correct distances. Before we work example like this, let’s review a few things about similar triangles. In the diagram below is a picture of two triangles that are similar.

Since these triangles are similar, the sides of two similar triangles are proportional. Given that the sides of the triangle are proportional, we can set up the following ratio between the sides. ℎ1 = ℎ2 d1 d2 Now, let try using proportions to ﬁnd the missing side of the triangle.

Example 3

Given that the triangles below are proportional, ﬁnd ℎ2

66 Solution Set up a proportion and solve for the missing variable which is ℎ2

ℎ1 = ℎ2 d1 d2 5 ℎ2 8 = 14 Now, ﬁnd the cross-product of the proportion in the problem. 8ℎ2 = 5(14) 8ℎ2 = 70 ℎ2 = 8 .75

Here is an example of using a proportion to ﬁnd the correct distance in drawing.

Example 4

Use the picture below and the values of a,b,c,d, and e to ﬁnd the missing value. If a =5 in, b = 3 in, and e = 12 in, ﬁnd d. Round your answer to 2 decimal places.

Solution: Draw a pair of similar triangles from the diagram above:

67 Set up a proportion and solve for d. a b e = d 5 3 12 = d Now, ﬁnd the cross-product of the proportion in the problem. 5d = 12(3) 5d = 36 d = 7 .2in

68 Example 5

Use the picture below and the values of a,b,c,d, and e to ﬁnd the missing value. If a =10 in, b = 18 in, and e = 30 in, ﬁnd d. Round your answer to 2 decimal places.

Solution: Draw a pair of similar triangles from the diagram above:

Set up a proportion and solve for d. a b e = d 10 b 30 = 18 Now, ﬁnd the cross-product of the proportion in the problem. 30 b = 10(18) 30 b = 180 b = 6 in

Example 6

In the diagram below, the distance between the telephone poles is c, the height of the two telephone pole is a and b respectively, the distance from ﬁrst telephone pole a to the vanishing point is e, and the distance from second telephone pole to the vanishing point is d. Using the diagram to ﬁnd distance d and c, given that a = 3 in, b = 2 in, and e = 12 in.

69 Set up a proportion and solve for d. a b e = d 3 2 12 = d Now, ﬁnd the cross-product of the proportion in the problem. 3d = 12(2) 3d = 24 d = 8 in c = 12 − 8 = 4 in

70 2.1.8 Exercises 1. Name the four types of perspective discussed in this unit. 2. Give a brief description of each of the four types of perspective 3. Describe any perspectives that can be found in the following picture.

4. Find the dominant perspective in the following picture

(A) Atmospheric Perspective (B) One Point Perspective (C) Diminishing Sizes (D) Overlapping Shapes

71 5. In the picture shown below use lines of sight and a vanishing point to draw several telephone poles along the roadside.

6. In the diagram below, the distance between the telephone poles is c, the height of the two telephone pole is a and b respectively, the distance from ﬁrst telephone pole a to the vanishing point is e, and the distance from second telephone pole to the vanishing point is d. Using the diagram to ﬁnd distance d and c, given that a = 5 in, b = 4 in, and e = 10 in.

7. Use the picture below and the values of a,b,c,d, and e to ﬁnd the missing value. If a =10 in, d = 18 in, and e = 30 in, ﬁnd b. Round your answer to 2 decimal places.

72 8. Use the picture below and the values of a,b,c,d, and e to ﬁnd the missing value. If a =5 in, b = 3 in, and e = 12 in, ﬁnd d. Round your answer to 2 decimal places.

9. Use the following diagram and the values of a,b,c,d, and e to ﬁnd the missing value. If a = 6 in, b = 3 in, and e = 8 in, ﬁnd d. Round your answer to 2 decimal places.

73 2.2 Fractals

2.2.1 What is a Fractal?

To start out this section on fractals we will begin by answering several questions. The first question one might ask is what is a fractal? In the 1970s, the mathematical Benoit Mandelbrot discovered a method to create geometric figures with special properties. This method allows figures to be enlarged repeatedly while preserving the detail of the figure. Mandelbrot refer to these figures as fractals. At the current time, we do not have universal agreement on an exact definition of a fractal. Based on the fractals we will study, We will define a fractal as a geometric figure that is divided into smaller versions of itself. A second common question that is asked is what does a fractal look like? This question is not as easy to answer because fractals can take on a wide range of patterns and designs, but the common element to all fractals is that they all contain repeating patterns. Below are some computer generated fractals that will give us an idea what a fractal looks like.

All of these picture where generated by Suzanne Alejandre they can view at this website: http://mathforum.org/alejandre/workshops/fractal/fractal3.html

2.2.2 How are Fractals Made?

A third question might be is how are fractals created? Usually fractals are made by starting with a general shape which is called an initiator. The initiator is then expanded out into diﬀerent shapes by using what is called a generator. Here is an examples of how a fractal can be generated by using an initiator and a generator. This special fractal is known as the Koch curve.

Example 1: The Koch Curve

74 When producing the Koch Curve, we start with the given initiator and develop each step using the given generator. In this problem the generator is a line segment and the generator is created by dividing the segment into three equal segments and replacing the middle segment with a hump as shown in the below ﬁgure. The hump in the middle is formed by segments that equal in length to the outer two segments.

The ﬁrst step or step 0 is the initiator which is a simple line segment. Next, step 1 you will apply the generator to the initiator which will result in a pattern identical to the initiator where there pattern consists of four line segments. (See below)

On the next step (Step 2), you will apply the pattern of the initiator to each of the four line segment in step 1. The resulting pattern will look like the pattern below:

Repeating this process for several step will result in the fractal called the Koch curve.

2.2.3 The Dimension of a fractal

To ﬁnd the dimension of a fractal, we to know three quantities. These quantities are know as the replacement ratio of the fractal, the scaling ratio, and the magniﬁcation ratio. The replacement ratio N is the number objects that replace the original object in the previous

75 step. The magnification ratio s is the ratio between the new object and the original object. The scaling ratio r is the reciprocal of the magnification ratio. The equation to find the log (N) dimension of the fractal is given d = log (r)

Example 2: The Dimension of the Koch Curve

Recall that the Koch curve took one line segment and replaced it with four new line segments, so the replacement ratio would be N = 4. If look at the original generator for the Koch curve, each new segment has a length that is one third the length of the original segment. Thus, the 1 magniﬁcation ratio would be s = 3 . The scaling ratio of the fractal would be the reciprocal 1 log (4) of s. s = 1 = 3 Therefore, the dimension of the Koch curve would be d = log = 1 .26 3 (3)

Example 3: The Koch Snowﬂake

The Koch snowﬂake is similar to the Koch curve, except you use an equilateral triangle as an initiator. To generate the Koch Snowﬂake, you start with an equilateral triangle and use the Koch curve generator on each of the sides of the equilateral triangle.

In step 2, we repeat the pattern of the Koch curve on each side of the ﬁgure in step 1.

If we keep repeating this process will get a ﬁgure that resembles a snowﬂake.

Images courtesy thinkquest.org

Example 4: Sierpinski’s Triangle

To generate Sierpinski’s triangle, we start with a equilateral triangle and connect the midpoints of the three midpoints of the triangle. This will result in four congruent triangles see step 1 below; Equilateral triangle

76 Equilateral triangle with the three midpoints connected by line segments

Now, let’s shade in the outer three triangles and remove the middle triangle.

In step 2, you now take each of the three shaded triangles and repeat the process of divided the triangle into four equal triangles and removing the middle triangle.

Repeating this process will result in Sierpinski’s triangle

Example 5: The dimension of the Sierpinski’s Triangle Recall that the Sierpinski’s triangle took one solid triangle and replaced it with three solid triangles of the next step, so the replacement ratio would be N = 3. If look at step 1 of Sierpinski’s triangle, each new side of new triangle has a length that is one half the length 1 of the original triangle. Thus, the magniﬁcation ratio would be s = 2 . The scaling ratio of 1 the fractal would be the reciprocal of s. r = 1 = 2 Therefore, the dimension of the Koch 2 log (3) curve would be d = log (2) = 1 .46

77 2.2.4 Uses of Fractals Engineer John Chenoweth discovered that fractal antennas are 25 percent more eﬃcient than rubbery ”stubby” antennas. In addition, these types of antenna are cheaper to manufacture and fractal antennas also can operate on multiple bands.

78 2.2.5 Exercises 1. Find the dimension of the fractal given its scaling ratio s and replacement ratio N. (r = 4 and N = 5) 2. Sketch the ﬁrst two iterations of the fractal and then ﬁnd its dimension.

3. Find the dimension of the fractal given its scaling ratio s and replacement ratio N. (r = 5 and N = 7) . 4. Find the dimension of the following fractal given the initiator and generator of the fractal

5. Find the dimension of the fractal with the given initiator and generator.

6. Exploratory Question: Find the dimension of Serpinski’s Carpet. Compare the result to the dimension of the fractal in the above problem (The Hilbert Curve).

79 2.3 The Golden Ratio

2.3.1 Finding the Golden Ratio

In this next section, we will continue study the relation between art and geometry. During the renaissance period, many artist, architects, and sculptors used ratios of distance that were more pleasing to the eye. This special ratio is called the Golden Ratio. The Golden Ratio can be found by dividing a segment into two pieces where the ratio of the longer piece to the shorter is the same as the ratio entire line segment to the longer piece. In the next passage, we will ﬁnd the value of the Golden Ratio

a a+b b = a a2 = b(a + b) a2 = ab + b2 35 a2 − ab − b2 = 0 Now, use the quadratic formula to solve for a

b+√b2+4 b2 a = 2 b+√5b2 a = 2 b+b√5 a = 2 b+b√5 a 2 b+b√5 1 Now, substitute back into the original ratio to the Golden Ratio Á = b = b = 2 b = b(1+ √5) 1 1+ √5 2 b = 2 = 1 .61803 ......

The Golden Ratio was referred to as the divina proportione meaning divide proportion by Luca Parcioli in 1509. The famous Physicist Johann Kelper later called it the sectio divina in 1610 which translates to divine section. The term Golden Ratio or Golden section came into use around 1640. Many artists and composers have used the Golden Ratio in their work including famous artist Leonardo da Vinci and composer Bela Barlok.

2.3.2 The Parthenon and the Golden Rectangle

Typically, the value 1.62 is used for the Golden Ratio to make computations easier. The Golden Rectangle is a rectangle that the ratio between the two sides is 1 to 1.62. The Greeks believe that Golden Rectangles where more pleasing to the eye than other rectangles. The Golden Rectangle can be found in the measurement in many ancient Greece. In Athens, buildings such as the Parthenon on the Arcopolis contain the Golden Ratio.

80 Using the picture of the parthenon above, we a can draw a rectangle around the perimeter of the parthenon as shown in the next diagram. After making this rectangle, we will measure the length and width of the rectangle.

Now, let’s ﬁnd the ratio between the length and width of the rectangle draw over the perimeter of the Parthenon.

LengtℎofRectangle 7.5 W idtℎofRectangle = 4.6 = 1 .6304

2.3.3 The Golden Cross

A Golden Cross is a cross that is constructed using two special ratios. We start by deﬁning length of the upper portion of the cross as T and the length of the lower portion as B. We will also deﬁne the overall height of the cross as T+B and the width of the cross as W. (See illustration below:)

81 A cross with the measurements shown above is a Golden Cross if the following ratios equal the golden ratio. B B+T T = Á = 1 .62 and B = Á = 1 .62

In the next example, we check to see if the cross is a Golden Cross.

Example 1

Is the following cross shown below a Golden Cross?

Solution: Check both ratios. B 4.86 cm T = 3cm = 1 .62

B+t 7.86 cm B = 4.85 cm = 1 .62 Example 2 Find the value of T, H, and W that will make the cross a Golden Cross.

82 Solution: First ﬁnd the value of T by setting the following ratio equal to the Golden Ratio. 8 T = 1 .62 8 T ( T ) = 1 .62( T ) 8 = 1 .62 T 8 T = 1.62 T = 4 .9in

Now, ﬁnd H by using the fact that H = T + B. Therefore, H = 8 + 4.9 = 12.9 inches

Now, set the following ratio equal to the Golden Ratio (1.62) to ﬁnd W. 12 .9 W = 1 .62 12 .9 W ( W ) = 1 .62( W ) 12 .9 = 1 .62 W 12 .9 W = 1.62 W = 8 .0in

2.3.4 The Golden Box The Golden Box is a rectangle solid whose length (L), width (W), and height (H) satisfy the Golden Ratio. These Golden Ratios are deﬁne as: W L H = W = Á = 1 .62

83 Example 3

Find the width and length to the nearest tenth of a Golden Box that has a height of 3 centimeters.

Solution:

First, ﬁnd the value of the width by setting up a ratio using the Golden Ratio.

W 3in = 1 .62 W 3in ( 3 ) = 1 .62(3 in ) W = 4 .9in

Now, ﬁnd the length of the box.

L 4.9in = 1 .62 W 4.9in ( 4.9in ) = 1 .62(4 .9in ) W = 7 .9in

2.3.5 Art and the Golden Ratio

The Golden Ratio can be found in the artwork of many famous artist. In particular, the Golden Ratio can be in some of Leonardo Da Vinci works. The next picture is a photo of Leonardo self portrait.

84 In the next picture, it is shown how a golden rectangle can be constructed around Leonardo face in his self portrait. Notice that the width of the rectangle is 2.49 cm and the height of the rectangle of 4.02 cm.

If you divide the height of the rectangle by width of the rectangle, you will get the Golden Ratio. Heigtℎ 4.02 W idtℎ = 2.49 = 1 .614

85 Other Examples of the Golden Ratio

Another classic example of the Golden Ratio is the Nautilus Spiral. The Nautilus is a seashell that can be divided into several golden rectangles. The illustration below shows how the Nautilus Spiral can be divided into at least 8 golden rectangles.

86 2.3.6 Exercises 1. Using a straightedge or a rule, try to draw a Golden Rectangle around the Mona Lisa’s face. Conﬁrm your results by dividing the longer side by the shorter side.

2. Find the width and length to the nearest tenth of a Golden Box that has a height of 8 centimeters. 3. Find the height and length to the nearest tenth of a Golden Box that has a height of 6 centimeters. 4. Find the width and length to the nearest tenth of a Golden Box that has a height of 12 centimeters. 5. Find the value of T, H, and W that will make the cross a Golden Cross.

87 6. Find the value of T, H, and W that will make the cross a Golden Cross.

88 3 Graph Theory (Optional)

3.1 Introduction to Graph Theory

3.1.1 Seven Bridges of Konigsberg

Mathematician Leonhard Euler visited Konigsberg, Prussia which is now Kaliningrad, Russia in 1736. On his visit, he learned about what is called the seven bridges problem. The Pregel river passed through the city of Konigsberg forming two large islands that are connected by seven bridges. In the seven bridges of Konigsberg problem, the objective is to ﬁnd a path through the city that would cross each bridge once and only once. The islands could not be reached by any route other than the bridges, and every bridge must have been crossed completely every time. (In the next section, we ﬁnd out the solution to this problem.)

3.1.2 What is Graph Theory

Graph theory is a branch of geometry that study graphs or networks. A graph or network is a series of edges and vertices. So, let’s start by learning some key terms.

Key Terms

1. Vertex (Vertices): Each point of a graph

2. Edge: An edge is a segment that connects two vertices.

3. Graph: A graph is a collection of edges and vertices.

In the next diagram, there are some illustrations of graphs. The graph on the left has 5 edges and 4 vertices. The ﬁgure on the right has 4 edges and 4 vertices.

103 3.1.3 The Degree of a Vertex

Each vertex in a graph has a degree. The degree of a vertex is the number of edges that meet at that vertex. In this ﬁrst example, it will be shown how to ﬁnd the degree of each vertex. Example 1

Find the degree of vertex A and vertex E.

Simply count the number of edges that meet at each vertex. Two edges meet at vertex A, so vertex A degree is 2. Three edges meet at vertex E, so vertex E degree is 3.

Example 2

Find the degree of vertex.

104 Simply count the number of edges that meet at each vertex. Two edges meet at vertex P, so vertex P degree is 2. Three edges meet at vertex Q, so vertex Q degree is 3. Two edges meet at vertex R, so vertex R degree is 2. Three edges meet at vertex S, so vertex S degree is 3. Now let’s summarize everything we have covered so far by completing in the next example.

Example 3

Complete the following table

Graph Edges Vertices Degree of Each Vertex Sum of Degrees Graph 1 Graph 2 Graph 3 Solution

To complete this chart simply count the number of vertices and edges in each graph. Count the number of edges at each vertex to get the degree of each vertex. Simply sum up all the degrees of each vertex to get the sum of the degrees. (Note: In the last graph, the edges CA and CA do not meet, so the graph has only 5 edges)

105 Graph Edges Vertices Degree of Each Vertex Sum of Degrees Graph 1 3 3 2,2,2 6 Graph 2 5 4 3,2,3,2 10 Graph 3 5 4 3,2,2,3 10

3.1.4 Exercises

1. Find the degree of vertex A

2. Find the degree of each vertex.

3. Complete the following chart below using the following graphs

106 Graph Edges Vertices Degree of Each Vertex Sum of Degrees Graph 1 Graph 2 Graph 3

107 3.2 Euler Circuits and Hamiltonian Cycles 3.2.1 Path and Circuits A path between vertices formed by traveling along the edges of a graph is called a walk. A walk that starts a vertex and ends at the same vertex is a circuit or cycle.

Degree and Special Paths

The ﬁrst special type of walk we will discuss in this section is called an Euler Walk. An Euler walk is a walk that uses every edge, but does not use any edge more than once. The next type of special path is called an Euler Circuit. An Euler Circuit is an Euler Walk that starts and ﬁnishes at the same vertex. We can actually use the degrees of each vertex to determine if the graph has a Euler Walk or Euler Circuit.

3.2.2 Degree and Special Paths The relationship between the degree of the vertices and Euler Walks and Euler Circuits 1. If the network has no odd vertices, then the network does have both an Euler Walk and Euler Circuit any point is a starting point. The starting point will also turn out to be the ending point. 2. If the network has exactly one odd vertex, then the network does not contain an Euler Walk or Euler Circuit . A network cannot have only one starting point or ending point without the other. 3. If the network has two odd vertices, then the network has an Euler Walk, but does not have an Euler Circuit. One odd vertex must be the starting point and the other odd vertex must be the ending point. 4. If the network has more than two odd vertices, then the network does not contain an Euler Walk or Euler Circuit. A network cannot have more than one starting point and one ending point. Example 1

Determine if the following graph have either an Euler walk or an Euler Circuit.

108 Solution:

Graph 1 has an Euler walk because the graph contains exactly two odd vertices. If you start a vertex and follow the path A → D → C → B → A → B, you travel over every edge with out visiting an edge more that once. Graph 2 has both an Euler Walk and an Euler Circuit because all of the vertices of the graph are even. (Each vertex is degree 2) Note that edges AC and BD do not meet at a vertex, so the graph has only four vertices. One possible path that is both an Euler Circuit and Euler Walk is A → D → C → B → A

Example 2: Revisiting the Seven Bridges of Konigsberg problem

Upon Euler’s return from Konigsberg, he studied the seven bridges more thoroughly in search for a solution to the problem. In the process of trying to ﬁnd a solution, Euler reconstructed the problem in terms of a graph. The notes that he wrote on trying to solve this problem lead to the discovery of graph theory. He divided the city of Konigsberg in four sections. (The area above the rivers, the area below of the rivers, the island, and the area to the right of the island) He then constructed a graph using the sections of the city as vertices and the bridges as edges. Using this process, he constructed a graph similar to the graph shown below:

In this case, a circuit that would start at one point in the city and cross every bridge only once would be an Euler circuit. If you ﬁnd the degree of each vertex, you will discover that the graph has three odd vertices which are degree 3, degree 5 and degree 3. According the rule for graphs that have more than 2 odd vertices, the graph does not have an Euler circuit. Therefore the seven bridges of Konigsberg has no solution.

3.2.3 Hamiltonian Cycles

A another special circuit is called a Hamiltonian cycle. A hamiltonian cycle is a cycle that starts and ﬁnishes a one vertex and visits every vertex only once. Now, let’s look at a few graph and determine if they have a hamiltonian cycle.

Example 3

Determine if the graphs have a Hamiltonian cycle.

109 Solution

Graph 1 has a path A → D → C → B → A that starts at vertex A visits each vertex only once and returns to vertex A. Graph 2 has a path A → B → C → D → E → A that visit that starts at vertex A visits each vertex only once and returns to A.

3.2.4 Weighted Graphs

A graph that has a value assigned to each edge is called a weighted graph.

Nearest Neighbor Algorithm

The nearest neighbor algorithm can be use to ﬁnd a cycle of least weight. To accomplish this you start at a vertex and use the following set of rules.

1. Start at a vertex and chose a path to the nearest vertex or chose the edge of least weight.

2. From that vertex ﬁnd the nearest vertex that has not be visited.

3. Repeat this process until there is no other choice than to return to original vertex.

Example 4: The traveling salesman problem

A salesman wants to visit ﬁve Virginia cities, Arlington, Charlottesville, Radford, Emporia, and Norfolk. Driving distances are shown in the ﬁgure below. What is the shortest trip starting and ending in Arlington?

110 Solution

To find the shortest trip we will use the nearest neighbor algorithm. We will start in Arlington and find the nearest neighboring city which would be Charlottesville. From Charlottesville we would find the nearest neighboring city which would be Radford. From Radford the nearest city that has not be visited would be be Norfolk, and then from Norfolk the nearest city would be Emporia. After visiting Emporia, all the cities would have been visited so you would return to Arlington. Therefore, the Hamiltonian cycle you would follow would be:

−→ −→ −→ −→ −→ Arlington 100 Cℎarlottesville 140 Radford 300 Norfolk 75 Emporia 180 Arlington

The total distance for the trip would be: 100 miles + 140 miles + 300 miles + 75 miles + 180 miles = 795 miles

Here is what the corresponding graph would look like.

111 3.2.5 Connected Graphs and Dirac’s Theorem A connected graph is a graph that is in only one pieces. Graphs that have more than one piece a not connected. Here are some examples of some graphs that are connected and not connected.

Dirac’s Theorem

Consider a connected graph with at least three vertices and no multiple edges. Let n be the n number of vertices in the graph. If every vertex has degree of at least 2 then the graph must be Hamiltonian.

112 3.2.6 Exercises 1. Determine if the following graph have either an Euler walk or an Euler Circuit.

2. Determine if the following graph have either an Euler walk or Euler Circuit.

3. Determine if the graph has a Hamiltonian Cycle.

113