# 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.

# Master / Apprentice relationship graph in Star Wars

Here’s a graph of master / apprentice relationships for Jedi and Sith in the Star Wars movies. It’s not exhaustive, but it covers the main relationships in Episodes I through VI.

Here’s what you get if you add a dashed arrow for who killed whom.

Here’s the dot (GraphVis) code that created the diagrams.

```digraph G {

Dooku [label="Dooku/\nDarth Tyranus"];
Luke  [label="Luke Skywalker"];
Ben   [label="Obi-Wan Kenobi"];
Liam  [label="Qui-Gon Jinn"];
DS    [label="Palpatine/\nDarth Sidious"]
DP    [label="Darth Plagueis"];
DM    [label="Darth Maul"];

Yoda  -> Dooku;
Yoda  -> Luke;
Dooku -> Liam;
Liam  -> Ben;
Ben   -> Luke;
DS    -> Dooku;
DP    -> DS;
DS    -> DM;

Ben   -> DM [style=dashed];
DM    -> Liam [style=dashed];
DS    -> DP [style=dashed];
}
```

# Rook graphs and Paley graphs

An m by n rook graph is formed by associating a node with each square of an m by n chess board and connecting two nodes with an edge if a rook can legally move from one to the other. That is, two nodes are connected if they are either in the same row or in the same column.

A Paley graph of order q is a graph with q nodes where q is a prime power and congruent to 1 mod 4. Label each node with an element of the finite field F with q elements and connect two nodes with an edge if the difference between their labels is a quadratic residue. That is, for nodes labeled a and b, connect these nodes if there exist an x in F such that ab = x².

We’ll look at a graph which is both a rook graph and a Paley graph: the 3 by 3 rook graph is the same as the Paley graph of order 9.

## Rook graph 3 x 3

Constructing the rook graph is easy. If we number our 3 by 3 chess board as follows

```    1 2 3
4 5 6
7 8 9
```

then we connect 1 to 2, 3, 4, and 7. We connect 2 to 1, 3, 5, and 8. Etc. This gives us the following graph.

(The graph was made with Sketchviz introduced here.)

## Paley graph of order 9

They Paley graph is a little more complicated to construct. First we need to explain what a field with 9 elements is.

A field with 3 elements is just integers mod 3. We can think of the field with 9 elements as the analog of complex numbers: numbers of the form abi where a and b are integers mod 3. As with the complex numbers, i² = -1.

When we square each of the nine elements of our field, we get four distinct non-zero results: 1, 2, i, and 2i. For example, i is one of our squares because

(1 + 2i)(1 + 2i) = (1 – 4) + 4i = i

Here we used -3 = 0 mod 3 and 4 = 1 mod 3.

Since our squares are 1, 2, i, and 2i, that says 0 is connected to these four values. Then we connect 1 to 2, 0, 1 + i and 1 + 2i because each of these numbers minus 1 is a square. If we keep doing this for all nine nodes, we get the following graph.

## Isomorphism

It’s clear that the two graphs above are relabelings of each other. We can see this explicitly by relabeling our chess board as follows:

```    0    i    2i
1  1+i  1+2i
2  2+i  2+2i
```

Moving horizontally adds i, which is a square, and moving vertically adds 1, which is a square. So the Paley connections are rook connections.

## Uniqueness

Are there more rook graphs that are isomorphic to Paley graphs? No, this is the only one.

Every pair of vertices in a rook graph that are not neighbors have two common. If the points with coordinates (a, b) and (c, d) are on different rows and columns, then both are connected to (a, d) and (c, b).

In a Paley graph of order q, every pair of nodes that aren’t neighbors have (q – 1)/4 common neighbors. For a Paley graph to be isomorphic to a rook graph, we must have (q – 1)/4 = 2, and so q = 9.

# Six degrees of Kevin Bacon, Paul Erdos, and Wikipedia

I just discovered the website Six Degrees of Wikipedia. It lets you enter two topics and it will show you how few hops it can take to get from one to the other.

Since the mathematical equivalent of Six Degrees of Kevin Bacon is Six degrees of Paul Erdős, I tried looking for the distance between Kevin Bacon and Paul Erdős and found this:

Kevin Bacon and Paul Erdős are both known for their prolific collaborations. Many actors have acted with Kevin Bacon, or acted with actors who have acted with him, etc. Many mathematicians wrote papers with Erdős, or have written papers with people who have written papers with him, etc. I’m four degrees of separation from Paul Erdős last I checked.  [1]

Here’s a more complex graph showing the three degrees of separation between Thomas Aquinas and Thomas the Tank Engine.

Note that the edges are directed, links from the starting page to pages that lead to the target page. If you reverse the starting and ending page, you get different results. For example, there are still three degrees of separation if we reverse our last example, going from Thomas the Tank Engine to Thomas Aquinas, but there are about twice as many possible paths.

## More network posts

[1] Your Erdős number, the degrees of separation between your and Paul Erdős, can decrease over time because his collaborators are still writing papers. Even Erdős himself continued to publish posthumously because people who began papers with him finished them years later.

# Spectral sparsification

The latest episode of My Favorite theorem features John Urschel, former offensive lineman for the Baltimore Ravens and current math graduate student. His favorite theorem is a result on graph approximation: for every weighted graph, no matter how densely connected, it is possible to find a sparse graph whose Laplacian approximates that of the original graph. For details see Spectral Sparsification of Graphs by Dan Spielman and Shang-Hua Teng.

You could view this theorem as a positive result—it’s possible to find good sparse approximations—or a negative result—the Laplacian doesn’t capture very well how densely a graph is connected.

Related: Blog posts based on my interview with Dan Spielman at the 2014 Heidelberg Laureate Forum.

# Random minimum spanning trees

I just ran across a post by John Baez pointing to an article [the link has gone away] by Alan Frieze on random minimum spanning trees.

Here’s the problem.

1. Create a complete graph with n nodes, i.e. connect every node to every other node.
2. Assign each edge a uniform random weight between 0 and 1.
3. Find the minimum spanning tree.
4. Add up the edges of the weights in the minimum spanning tree.

The surprise is that as n goes to infinity, the expected value of the process above converges to the Riemann zeta function at 3, i.e.

ζ(3) = 1/1³ + 1/2³ + 1/3³ + …

Incidentally, there are closed-form expressions for the Riemann zeta function at positive even integers. For example, ζ(2) = π² / 6. But no closed-form expressions have been found for odd integers.

## Simulation

Here’s a little Python code to play with this.

```    import networkx as nx
from random import random

N = 1000

G = nx.Graph()
for i in range(N):
for j in range(i+1, N):

T = nx.minimum_spanning_tree(G)
edges = T.edges(data=True)

print( sum( e[2]["weight"] for e in edges ) )
```

When I ran this, I got 1.2307, close to ζ(3) = 1.20205….

I ran this again, putting the code above inside a loop, averaging the results of 100 simulations, and got 1.19701. That is, the distance between my simulation result and ζ(3) went from 0.03 to 0.003.

There are two reasons I wouldn’t get exactly ζ(3). First, I’m only running a finite number of simulations (100) and so I’m not computing the expected value exactly, but only approximating it. (Probably. As in PAC: probably approximately correct.) Second, I’m using a finite graph, of size 1000, not taking a limit as graph size goes to infinity.

My limited results above suggest that the first reason accounts for most of the difference between simulation and theory. Running 100 replications cut the error down by a factor of 10. This is exactly what you’d expect from the central limit theorem. This suggests that for graphs as small as 1000 nodes, the expected value is close to the asymptotic value.

You could experiment with this, increasing the graph size and increasing the number of replications. But be patient. It takes a while for each replication to run.

## Generalization

The paper by Frieze considers more than the uniform distribution. You can use any non-negative distribution with finite variance and whose cumulative distribution function F is differentiable at zero. The more general result replaces ζ(3) with ζ(3) / F ‘(0). We could, for example, replace the uniform distribution on weights with an exponential distribution. In this case the distribution function is 1 − exp(−x), at its derivative at the origin is 1, so our simulation should still produce approximately ζ(3). And indeed it does. When I took the average of 100 runs with exponential weights I got a value of 1.2065.

There’s a little subtlety around using the derivative of the distribution at 0 rather than the density at 0. The derivative of the distribution (CDF) is the density (PDF), so why not just say density? One reason would be to allow the most general probability distributions, but a more immediate reason is that we’re up against a discontinuity at the origin. We’re looking at non-negative distributions, so the density has to be zero to the left of the origin.

When we say the derivative of the distribution at 0, we really mean the derivative at zero of a smooth extension of the distribution. For example, the exponential distribution has density 0 for negative x and density exp(−x) for non-negative x. Strictly speaking, the CDF of this distribution is 1 − exp(−x) for non-negative x and 0 for negative x. The left and right derivatives are different, so the derivative doesn’t exist. By saying the distribution function is simply exp(−x), we’ve used a smooth extension from the non-negative reals to all reals.