# Classroom exercise with networks

In the previous post I looked at graphs created from representing geographic regions with nodes and connecting nodes with edges if the corresponding regions share a border.

It’s an interesting exercise to recover the geographic regions from the network. For example, take a look at the graph for the continental United States.

It’s easy to identify Alaska in the graph. The node on the left represents Maine because Maine is the only state to border exactly one other state. From there you can bootstrap your way to identifying the rest of the states.

## Math class

This could make a fun classroom exercise in a math class. Students will naturally come up with the idea of the degree of a node, the number of edges that meet that node, because that’s a handy way to solve the puzzle: the only possibilities for a node of degree n are states that border n other states.

This also illustrates that networks preserve topology, not geometry. That is, the connectivity information is retained, but the shape is dramatically different.

## Geography class

Someone asked me on Twitter to make a corresponding graph for Brazil. Mathematica, or at least my version of Mathematica, doesn’t have data on Brazilian states, so I made an adjacency graph using GraphViz.

Labeling the blank nodes is much easier for Brazil than for the US because Brazil has about half as many states, and the topology of the graph gives you more to work with. Three nodes connect to only one other node, for example.

Here the exercise doesn’t involve as much logic, but the geography is less familiar, unless of course you’re more familiar with Brazil than the US. Labeling the graph will require staring at a map of Brazil and you might accidentally learn a little about Brazil.

## GraphViz

The labeled version of the graph above is available here. And here are the GraphViz source files that make the labeled and unlabeled versions.

The layout of a GraphViz file is very simple. The file looks like this:

    graph G {

layout=sfdp

AC [label="Acre"]
AL [label="Alagoas"]
...
AC -- AM
AC -- RO
...
}


There are three parts: a layout, node labels, and connections.

GraphViz has several layout engines, and the sfdp one matched what I was looking for in this case. Other layout options lead to overlapping edges that were confusing.

The node names AC, AL, etc. do not appear in the output. They’re just variable names for your convenience. The text inside the label is what appears in the final output. I’ll give an example in the next post in which it’s very convenient for the variables to be different from the labels. The order of the labels doesn’t matter, only which variables are associated with which labels.

Finally, the lines with variables separated by dashes are the connection data. Here we’re telling GraphViz to connect node AC to nodes AM and RO. The order of these lines doesn’t matter.

## Related posts

Suppose you want to color a map with no two bordering regions having the same color. If this is a map on a plane, you can do this using only four colors, but maybe you’d like to use more.

You can reduce the problem to coloring the nodes in a graph. Each node corresponds to a region, and there is an edge between two nodes if and only if their corresponding regions share a border.

Here is a sort of topologists’s or graph theorist’s view of the continental United States.

This was created using the following sample code from the Mathematica documentation.

    RelationGraph[MemberQ[#2["BorderingStates"], #1] &,
EntityList[


You can recognize Maine in the graph because it’s the only state that only borders one other state. Alaska is also easy to locate. Exercise for the reader: mentally add Hawaii to the graph.

The analogous graph for Texas counties took much longer to draw: there are 49 continental US states but 254 Texas counties.

This was created with the following code.

    RelationGraph[MemberQ[#2["BorderingCounties"], #1] &,


You can find El Paso county in the top left; it only borders one county just as Maine only borders one state.

# Shortest tours of Eurasia and Oceania

This is the final post in a series of three posts about shortest tours, solutions to the so-called traveling salesmen problem.

The first was a tour of Africa. Actually two tours, one for the continent and one for islands. See this post for the Mathematica code used to create the tours.

The second was about the Americas: one tour for the North American continent, one for islands, and one for South America.

This post will look at Eurasia and Oceania. As before, I limit the tours to sovereign states, though there are disputes over which regions are independent nations. I first tried to do separate tours of Europe and Asia, but this would require arbitrarily categorizing some countries as European or Asian. The distinction between Asia and Oceania is a little fuzzy too, but not as complicated.

## Oceania

Here’s a map of the tour of Oceania.

Here’s the order of the tour:

1. Australia
2. East Timor
3. Indonesia
4. Palau
5. Papua New Guinea
6. Micronesia
7. Marshall Islands
8. Nauru
9. Solomon Islands
10. Vanuatu
11. Fiji
12. Tuvalu
13. Kiribati
14. Samoa
15. Tonga
16. New Zealand

The total length of the tour is 28,528 kilometers or 17,727 miles.

## Eurasia

Here’s a map of the the Eurasian tour.

Here’s the order of the tour:

1. Iceland
2. Norway
3. Sweden
4. Finland
5. Estonia
6. Latvia
7. Lithuania
8. Belarus
9. Poland
10. Czech Republic
11. Slovakia
12. Hungary
13. Romania
14. Moldova
15. Ukraine
16. Georgia
17. Armenia
18. Azerbaijan
19. Turkmenistan
20. Uzbekistan
21. Afghanistan
22. Pakistan
23. Tajikistan
24. Kyrgyzstan
25. Kazakhstan
26. Russia
27. Mongolia
28. China
29. North Korea
30. South Korea
31. Japan
32. Taiwan
33. Philippines
34. East Timor
35. Indonesia
36. Brunei
37. Malaysia
38. Singapore
39. Cambodia
40. Vietnam
41. Laos
42. Thailand
43. Myanmar
45. Bhutan
46. Nepal
47. India
48. Sri Lanka
49. Maldives
50. Yemen
51. Oman
52. United Arab Emirates
53. Qatar
54. Bahrain
55. Saudi Arabia
56. Kuwait
57. Iran
58. Iraq
59. Syria
60. Lebanon
61. Jordan
62. Israel
63. Cyprus
64. Turkey
65. Bulgaria
66. North Macedonia
67. Serbia
68. Bosnia and Herzegovina
69. Montenegro
70. Albania
71. Greece
72. Malta
73. Italy
74. San Marino
75. Croatia
76. Slovenia
77. Austria
78. Liechtenstein
79. Switzerland
80. Monaco
81. Andorra
82. Spain
83. Portugal
84. France
85. Belgium
86. Luxembourg
87. Germany
88. Netherlands
89. Denmark
90. United Kingdom
91. Algeria

The total length of the tour is 61,783 kilometers or 38,390 miles.

# Three tours of the Americas

The previous post looked at an optimal tour of continental Africa. This post will give analogous tours of continental North America, North American Islands, and South America. The next post looks at Eurasia and Oceania.

## North American Continent

Here’s the North American continental tour.

The order of the tour is as follows.

2. United States
3. Mexico
4. Guatemala
6. Costa Rica
7. Panama
8. Nicaragua
9. Honduras
10. Belize

## North American Islands

Here’s a tour of the North American islands.

Trinidad and Tabago is about ten miles from the South American continent, but the country is classified as being part of North America, at least for some purposes.

Here is the order of the tour.

1. Cuba
2. Jamaica
3. Haiti
4. Dominican Republic
8. Saint Vincent and the Grenadines
9. Saint Lucia
10. Dominica
11. Antigua and Barbuda
12. Saint Kitts and Nevis
13. Bahamas

## South American tour

Here’s the tour of South America.

Here’s the order of the tour:

1. Venezuela
2. Guyana
3. Suriname
4. French Guiana
5. Brazil
6. Paraguay
7. Uruguay
8. Falkland Islands
9. Argentina
10. Chile
11. Bolivia
12. Peru
14. Colombia

# A traveling salesman tour of Africa

Suppose you’d like to tour Africa, visiting each country once, then returning to your starting point, minimizing the distance traveled.

Here’s my first attempt at a solution using Mathematica, based on an example in the documentation for FindShortestTour.

    africa = CountryData["Africa"]
FindShortestTour[africa]
GeoGraphics[{Thick, Red, GeoPath[africa[[%[[2]]]]]}]


This produced the following map:

Hmm. Maybe I should have been more specific about what I mean by “Africa.” My intention was to find a tour of continental Africa, i.e. not including islands. This means I needed to remove several items from Mathematica’s list of African countries. Also, I had in mind sovereign states, not territories of overseas states and not disputed territories.

After doing this, the map is more like what I’d expect.

The tour is then

1. Algeria
2. Tunisia
3. Libya
4. Egypt
6. Central African Republic
7. Democratic Republic of the Congo
8. Burundi
9. Rwanda
10. Uganda
11. South Sudan
12. Sudan
13. Eritrea
14. Djibouti
15. Somalia
16. Ethiopia
17. Kenya
18. Tanzania
19. Malawi
20. Zambia
21. Mozambique
22. Zimbabwe
23. Eswatini
24. Lesotho
25. South Africa
26. Botswana
27. Namibia
28. Angola
29. Republic of the Congo
30. Gabon
31. Equatorial Guinea
32. Cameroon
33. Nigeria
34. Niger
35. Mali
36. Burkina Faso
37. Benin
38. Togo
39. Ghana
40. Ivory Coast
41. Liberia
42. Sierra Leone
43. Guinea
44. Guinea-Bissau
45. Gambia
46. Senegal
47. Mauritania
48. Morocco

The initial tour, including islands, foreign territories, and Western Sahara, was 23,744 miles or 38,213 kilometers. The second tour was 21,074 miles or 33915 kilometers.

Here’s a tour of just the islands, excluding foreign territories.

The order of the tour is

1. Cape Verde
2. Seychelles
3. Mauritius
5. Comoros
6. São Tomé and Príncipe

This tour is 13,034 miles or 20,976 kilometers.

Update: See the next two posts for tours of the Americas and Eurasia and Oceania.

# Direction between two cities

This post is the third in a series of posts on spherical trigonometry. We first looked at the analog of the Pythagorean theorem on a sphere, then the analog of the law of cosines on a sphere. Now we look at the analog of the law of sines on a sphere.

As before we denote the sides of a triangle with a, b, and c and denote the angle opposite a side by the capital version of that letter. So A is the angle opposite side a and so forth.

The law of sines on a sphere says

sin a / sin A = sin b / sin B = sin c / sin C.

## Small triangles

Suppose for a moment that we’re looking at a relatively small triangle, one for which the length of the sides is small relative to the radius of the sphere. Then we can apply the small angle approximation

sin xx

to each side [1] and find

a / sin Ab / sin Bc / sin C.

In other words, the law of sines from plane trigonometry holds approximately in spherical trigonometry for small triangles. This is not surprising since we know we can ignore the curvature of the earth when working over relatively small areas, but it is reassuring.

## Example: LAX to IAH

In the previous post we found the air distance between LAX (Los Angeles) and IAH (Houston) airports by making a spherical triangle with vertices at LAX, IAH, and the north pole. We labeled the side from LAX to the north pole a and the side from IAH to the north pole b. We found the length of side c using the law of cosines.

Now suppose we’re at LAX and about to fly to IAH. What direction should we fly? Since Los Angeles and Houston are approximately at the same latitude, and they’re not terribly far apart relative to the size of the earth, we know that we’ll roughly fly due east. Let’s find the flight direction more precisely.

The angle B in our triangle gives us the angle between our heading and a heading of due north. We know b from the latitude of IAH, we know C from the difference in longitude of the two airports, and we found c in the previous post. Therefore we can find B from the law of sines:

sin b / sin B = sin c / sin C

and we know everything but B. So

sin B = sin b sin C / sin c = sin 60.02° sin 23.07° / sin 19.92°.

This tells us sin B = 0.9960. So what is B? If we ask software for the arcsine of 0.9960 we will get 84.88°. This would say we head 5.12° north of due east to get from Los Angeles to Houston. But Houston is at a lower latitude than Los Angeles and so that doesn’t seem right.

In the first post in this series I mentioned that a flight between Houston and Wuhan would fly over Alaska even though the two cities are at essentially the same latitude. Wuhan is at a slightly higher latitude than Houston, but a flight leaving Wuhan for Houston would head somewhat north. So it is possible that a flight from a higher latitude to a lower latitude would head north [2].

But Los Angeles is much closer to Houston than Wuhan is, and the difference in latitude is greater. A flight from LA to Houston would head mostly east and a little south. So where did we go wrong?

The arcsine function in any software package returns a solution to the equation

sin(x) = y,

but it might not return the solution we need. The angle 84.88° is a valid solution to the equation

sin(B) = 0.9960

but it’s not the appropriate solution in our context. We need the solution 95.12°. Our flight heada 5.12° south of east. This assumes our flight takes a great circle path; airlines may have reasons not to take the shortest possible path between two airports.

***

[1] The approximation assumes x is measured in radians. If we measure angles in degrees or any other units, the right side of the approximation will need to be multiplied by a constant. But the same constant will appear in all the numerators and denominators of the law of sines and so the approximation holds in any units.

[2] Here’s another illustration of the same point. Draw a circle around the north pole, say a circle a mile in diameter, and suppose you want to walk from longitude 0 to longitude 180°. Following a path of constant latitude is walking around the circle. Heading north instead is walking across the diameter of the circle. If your destination is a little bit outside the circle on the other side, it’s still shorter to walk across the circle, i.e. head north, even though your destination is at a lower latitude, i.e. south.

# Numeric distance vs geographic distance in zip codes

If two zip codes numbers are close, are the regions they represent close? How much can you tell about how far apart two regions are by comparing their zip codes? (Zip codes are US postal codes. The name is an acronym for “Zone Improvement Plan” and was introduced in 1963.)

To investigate this, I looked at some data on zip codes that I purchased last year. I selected the zip codes with latitude and longitude coordinates [1] and computed the distance between consecutive zip codes, consecutive in the sense of being the next item in the list of assigned zip codes, not necessarily consecutive integers.

The closest consecutive zip codes are 10171 and 10172 in Manhattan. These zip codes are neighboring blocks and so you could say their distance is zero. The distance using the latitude and longitude values in my data is 0.05 miles.

The furthest consecutive zip codes are 96863 in Honolulu, Hawaii and 97001 in Antelope, Oregon. These locations are 2,658 miles apart.

The mean distance between consecutive zip codes is 34.4 miles and the median distance is 24.9 miles. So consecutive zip codes are often fairly close geographically, but not always.

Next I went back and divided the geographic distance between consecutive pair of zip codes by the numerical difference between the codes, taking a sort of “derivative”.

By this measure, the furthest consecutive zip codes are 99685 and 99686, which happen to also be consecutive integers. The both are in Alaska, the form former in Unalaska in the Aleutian Islands and the latter in Valdez on the mainland.

The mean and median distance in terms of miles per integer difference is 26.4 and the median is 16.9.

The previous post shows how you could use Hilbert curves to assign zip codes to a grid so that consecutive codes always correspond to neighboring squares.

## Related posts

[1] A lot of zip codes don’t have coordinates because they serve an administrative function more than strictly representing a geographic region.

# Zip codes, geocodes, and Hilbert curves

You might think that if zip codes are close, then the regions they represent are close. Or that if zip codes are consecutive, then their regions touch. Neither of these are true. I explore how far they are from being true in the next post.

But these statements could have been true [1]. It’s possible to lay a grid on a region and number the squares in the grid in such a way that consecutive numbers correspond to contiguous squares, and nearby numbers correspond to nearby squares. There are geocoding systems that accomplish this using Hilbert curves. For example, Google’s S2 system is based on Hilbert curves.

A Hilbert curve is a curve that winds through a square, coming arbitrarily close to every point in the square. A Hilbert curve is a fractal, defined as the limit of an iterative process. We aren’t concerned with the limit because we only want to carry out a finite number of steps in the construction. But the fact that the limit exists tells us that with enough steps we can get any resolution we want.

To illustrate Hilbert curves and how they could be used to label grids, we will use a Hilbert curve to tour a chessboard. We will number the squares of a chess board so that consecutive numbers correspond to neighboring squares.

Here’s Python code to produce our numbering.

    from hilbertcurve.hilbertcurve import HilbertCurve

p = 3 # 2^3 squares on a side
n = 2 # two dimensional cube, i.e. square

hc = HilbertCurve(p, n)

for row in range(2**p):
for col in range(2**p):
d = hc.distance_from_point([row, col])
print(format(d,"2d"), " ", end="")
print()


This produces the following output.

     0   1  14  15  16  19  20  21
3   2  13  12  17  18  23  22
4   7   8  11  30  29  24  25
5   6   9  10  31  28  27  26
58  57  54  53  32  35  36  37
59  56  55  52  33  34  39  38
60  61  50  51  46  45  40  41
63  62  49  48  47  44  43  42


Here’s Python code to draw lines rather than print numbers.

    N = (2**p)**n
points = hc.points_from_distances(range(N))

for i in range(1, N):
x0, y0 = points[i-1]
x1, y1 = points[i]
plt.plot([x0, x1], [y0, y1])
plt.gca().set_aspect('equal')
plt.show()


This produces the following image.

There’s a small problem. Following the square numberings above produces a Hilbert curve, and the plotting code produces and Hilbert curve, but they’re not the same Hilbert curve. The implicit axes created by our sequence of print statements doesn’t match the geometry in the plotting code. To get the curves to match we need to do a sort of change of coordinates. We can do this by changing the call to plt.plot to

    plt.plot([y0, y1], [8-x0, 8-x1])

This produces the following plot, matching the numbering above.

## Related posts

[1] It could have been true in the sense that it was theoretically possible. It might not have been practically possible. Postal codes have different design constraints than geocodes.

# Decoding a grid square

I saw a reference last night to the grid square EL29fx and wanted to figure out where that is. There are many programs that will do this for you, but I wanted to do it by hand.

I wrote about how grid squares work a year ago, but I was rusty on the details, so this morning I went back and reread my earlier post.

## Parsing a locator

The system used to create grid squares in amateur radio is the Maidenhead Locator System. Last year I wrote about why the system works the way it does. Here I will recap how the system works but refer you to the earlier post for the motivation behind the design choices.

First of all, the Maidenhead system interleaves the longitude and latitude information. So in the square EL29fx, the longitude information is (E, 2, f) and the latitude information is (L, 9, x). The first two characters, EL, gives a rough location, a “field.” Then the next two characters, 29, refine the location to a “square.” The last two, fx, narrow the square down to a “subsquare.”[1]

## Finding the field

The first character in the grid square specifies a range of longitude east of the International Date Line [2]. Each letter represents a range of 20° longitude: A for 0 to 20° longitude, B for 20° to 40°, etc. So E represents 80° to 100° longitude east of the antemeridian or 260° to 280° east of the prime meridian.

The second character in the grid square represents latitude north of the south pole in increments of 10°. So L represents 110° to 120° north of the south pole, or 20° to 30° north of the equator.

Together, EL tells us we’re looking at some place 260° to 280° E and 20° to 30° N, i.e. somewhere in Mexico or the US Gulf Coast.

## Finding the square

The two digits in the middle of a grid square divide the field into 10 parts in both directions. Since a field is 20° wide and 10° tall, the first digit represents multiples of 2° longitude and the second digit represent multiples of 1° latitude.

So if we look at EL29, we know our location is between 264° and 266° E and between 29° and 30° N, which is somewhere around Houston, Texas

## Finding the subsquare

A subsquare divides a square into 24 parts in each direction. So a subsquare is 5 minutes of longitude wide and 2.5 minutes of latitude high.

The f in EL29fx refines the longitude is 264° 25′ E (95° 35′ W) and the x refines the latitude is 29° 55′ N. This narrows the location to somewhere in northwest Houston.

[1] There’s no significance to the case of the letters other than as a reminder that the first and second pair of letters are interpreted differently.

[2] Technically, degrees east of 180th meridian, a.k.a. the antemeridian, which mostly coincides with the International Date Line. The latter zigzags a little to avoid splitting island groups.

# Information theory and coordinates

The previous post explains how the Maidenhead location system works. The first character in a location code specifies the longitude in 20 degree increments and the second character specifies the latitude in 10 degree increments. Both are a letter from A through R that breaks the possible range down into 18 parts. (The longitude range is wider because longitude ranges over 360 degrees but latitude ranges over 180 degrees.)

If you divide a range into 16 equally probable parts, you get four bits of information because 24 = 16. Since we’re dividing longitude and latitude into 18 parts, we get around 4 bits of information from each. But the longitude component carries more information than the latitude component. This post will explain why.

## Defining information

The Shannon entropy of a random variable with N possible values, each with probability pi, is defined to be

where the base b is usually 2. Occasionally b might be another base, such as 2 or 10. (More on that here.) The statement above about 16 equal parts giving 4 bits of information assumes b = 2.

The expected information gain from observing the value of a random variable is measured by the random variable’s Shannon entropy.

## Even vs uneven allocation

Suppose we pick a point at random on a sphere. The first component of the point’s Maidenhead location narrows the location of the point to one of 18 equally probable sectors. So each of the ps in the equation above is 1/18. That means the information content in the first component is log2(1/18) = 4.17 bits.

The second component of the location also divides the sphere into 18 parts, but not evenly. A 10 degree band of latitude near the equator covers more area than a 10 degree band of latitude near the polls.

Shannon entropy is maximized when all the ps are equal. Said another way, the amount of surprise from observing a random variable is greatest when all possibilities are equally likely.

Knowing that a random point on a sphere is within 10 degrees of the equator is not as surprising as knowing that it is within 10 degrees of the North Pole because there’s less land within 10 degrees of the North Pole.

## Archimedes and Shannon

To calculate the information content in the latitude band, we need to know the area of each band.

Archimedes (c. 287 – 212 BC) discovered that if you have a sphere sitting inside a cylinder, the area of a band on the sphere is proportional to the area of its projection onto the cylinder. That means the area of a band of latitude is proportional to the difference in the z coordinates of the band.

The 18 bands of latitude run from -90° to 90° in increments of 10°. If we number the bands starting with 1 at the South Pole, the probability of landing in the nth band is

(sin(−90° + 10n°) − sin(-90° + 10(n − 1)°)/2

and you can calculate from this that the Shannon entropy is 3.97.

So the first component of the location, the longitude, carries 4.17 bits of information, and the second component, the latitude, carries 3.97 bits.