# Pacific Island Hopping using R and iGraph

Last month I enjoyed a relaxing holiday in the tropical paradise of Vanuatu. One rainy day I contemplated how to go island hopping across the Pacific ocean visiting as many island nations as possible. The Pacific ocean is a massive body of water between, Asia and the Americas, which covers almost half the surface of the earth. The southern Pacific is strewn with island nations from Australia to Chile. In this post, I describe how to use R to plan your next Pacific island hopping journey.

The Pacific Ocean.

## Listing all airports

My first step was to create a list of flight connections between each of the island nations in the Pacific ocean. I am not aware of a publically available data set of international flights so unfortunately, I created a list manually (if you do know of such data set, then please leave a comment).

My manual research resulted in a list of international flights from or to island airports. This list might not be complete, but it is a start. My Pinterest board with Pacific island airline route maps was the information source for this list.

The first code section reads the list of airline routes and uses the `ggmap` package to extract their coordinates from Google maps. The data frame with airport coordinates is saved for future reference to avoid repeatedly pinging Google for the same information.

```# Init
library(tidyverse)
library(ggmap)
library(ggrepel)
library(geosphere)

# Read flight list and airport list
flights <- read.csv("Geography/PacificFlights.csv", stringsAsFactors = FALSE)
f <- "Geography/airports.csv"
if (file.exists(f)) {
} else airports <- data.frame(airport = NA, lat = NA, lon = NA)

# Lookup coordinates for new airports
all_airports <- unique(c(flights\$From, flights\$To))
new_airports <- all_airports[!(all_airports %in% airports\$airport)]
if (length(new_airports) != 0) {
coords <- geocode(new_airports)
new_airports <- data.frame(airport = new_airports, coords)
airports <- rbind(airports, new_airports)
airports <- subset(airports, !is.na(airport))
write.csv(airports, "Geography/airports.csv", row.names = FALSE)
}

# Add coordinates to flight list
flights <- merge(flights, airports, by.x="From", by.y="airport")
flights <- merge(flights, airports, by.x="To", by.y="airport")
```

## Create the map

To create a map, I modified the code to create flight maps I published in an earlier post. This code had to be changed to centre the map on the Pacific. Mapping the Pacific ocean is problematic because the -180 and +180 degree meridians meet around the date line. Longitudes west of the antemeridian are positive, while longitudes east are negative.

The `world2` data set in the borders function of the `ggplot2` package is centred on the Pacific ocean. To enable plotting on this map, all negative longitudes are made positive by adding 360 degrees to them.

```# Pacific centric
flights\$lon.x[flights\$lon.x < 0] <- flights\$lon.x[flights\$lon.x < 0] + 360
flights\$lon.y[flights\$lon.y < 0] <- flights\$lon.y[flights\$lon.y < 0] + 360
airports\$lon[airports\$lon < 0] <- airports\$lon[airports\$lon < 0] + 360

# Plot flight routes
worldmap <- borders("world2", colour="#efede1", fill="#efede1")
ggplot() + worldmap +
geom_point(data=airports, aes(x = lon, y = lat), col = "#970027") +
geom_text_repel(data=airports, aes(x = lon, y = lat, label = airport),
col = "black", size = 2, segment.color = NA) +
geom_curve(data=flights, aes(x = lon.x, y = lat.x, xend = lon.y,
yend = lat.y, col = Airline), size = .4, curvature = .2) +
theme(panel.background = element_rect(fill="white"),
axis.line = element_blank(),
axis.text.x = element_blank(),
axis.text.y = element_blank(),
axis.ticks = element_blank(),
axis.title.x = element_blank(),
axis.title.y = element_blank()
) +
xlim(100, 300) + ylim(-40,40)
```

## Pacific Island Hopping

This visualisation is aesthetic and full of context, but it is not the best visualisation to solve the travel problem. This map can also be expressed as a graph with nodes (airports) and edges (routes). Once the map is represented mathematically, we can generate travel routes and begin our Pacific Island hopping.

The igraph package converts the flight list to a graph that can be analysed and plotted. The `shortest_path` function can then be used to plan routes. If I would want to travel from Auckland to Saipan in the Northern Mariana Islands, I have to go through Port Vila, Honiara, Port Moresby, Chuuk, Guam and then to Saipan. I am pretty sure there are quicker ways to get there, but this would be an exciting journey through the Pacific.

```library(igraph)
g <- graph_from_edgelist(as.matrix(flights[,1:2]), directed = FALSE)
par(mar = rep(0, 4))
plot(g, layout = layout.fruchterman.reingold, vertex.size=0)
V(g)
shortest_paths(g, "Auckland", "Saipan")
```

# Euler Problem 18 & 67: Maximum Path Sums

An example of a pedigree. Source: Wikimedia.

Euler Problem 18 and 67 are exactly the same besides that the data set in the second version is larger than in the first one. In this post, I kill two Eulers with one code.

These problems deal with binary trees, which is a data structure where each node has two children. A practical example of a binary tree is a pedigree chart, where each person or animal has two parents, four grandparents and so on.

## Euler Problem 18 Definition

By starting at the top of the triangle below and moving to adjacent numbers on the row below, the maximum total from top to bottom is 23.

3
7 4
2 4 6
8 5 9 3

That is, 3 + 7 + 4 + 9 = 23. Find the maximum total from top to bottom of the triangle below:

75
95 64
17 47 82
18 35 87 10
20 04 82 47 65
19 01 23 75 03 34
88 02 77 73 07 63 67
99 65 04 28 06 16 70 92
41 41 26 56 83 40 80 70 33
41 48 72 33 47 32 37 16 94 29
53 71 44 65 25 43 91 52 97 51 14
70 11 33 28 77 73 17 78 39 68 17 57
91 71 52 38 17 14 91 43 58 50 27 29 48
63 66 04 68 89 53 67 30 73 16 69 87 40 31
04 62 98 27 23 09 70 98 73 93 38 53 60 04 23

As there are only 16,384 routes, it is possible to solve this problem by trying every route. However, Problem 67, is the same challenge with a triangle containing one-hundred rows; it cannot be solved by brute force, and requires a clever method! ;o)

## Solution

This problem seeks a maximum path sum in a binary tree. The brute force method, as indicated in the problem definition, is a very inefficient way to solve this problem. The video visualises the quest for the maximum path, which takes eleven minutes of hypnotic animation.

A more efficient method is to define the maximum path layer by layer, starting at the bottom. The maximum sum of 2+8 or 2+5 is 10, the maximum sum of 4+5 or 4+9 is 13 and the last maximum sum is 15. These numbers are now placed in the next row. This process cycles until only one number is left. This algorithm solves the sample triangle in four steps:

Step 1:

3
7 4
2 4 6
8 5 9 3

Step 2:

3
7 4
10 13 15

Step 3:

3
20 19

Step 4:

23

In the code below, the data is triangle matrix. The variables rij (row) and kol (column) drive the search for the maximum path. The triangle for Euler Problem 18 is manually created and the triangle for Euler Problem 67 is read from the website.

```path.sum <- function(triangle) {
for (rij in nrow(triangle):2) {
for (kol in 1:(ncol(triangle)-1)) {
triangle[rij - 1,kol] <- max(triangle[rij,kol:(kol + 1)]) + triangle[rij - 1, kol]
}
triangle[rij,] <- NA
}
return(max(triangle, na.rm = TRUE))
}

# Euler Problem 18
triangle <- matrix(ncol = 15, nrow = 15)
triangle[1,1] <- 75
triangle[2,1:2] <- c(95, 64)
triangle[3,1:3] <- c(17, 47, 82)
triangle[4,1:4] <- c(18, 35, 87, 10)
triangle[5,1:5] <- c(20, 04, 82, 47, 65)
triangle[6,1:6] <- c(19, 01, 23, 75, 03, 34)
triangle[7,1:7] <- c(88, 02, 77, 73, 07, 63, 67)
triangle[8,1:8] <- c(99, 65, 04, 28, 06, 16, 70, 92)
triangle[9,1:9] <- c(41, 41, 26, 56, 83, 40, 80, 70, 33)
triangle[10,1:10] <- c(41, 48, 72, 33, 47, 32, 37, 16, 94, 29)
triangle[11,1:11] <- c(53, 71, 44, 65, 25, 43, 91, 52, 97, 51, 14)
triangle[12,1:12] <- c(70, 11, 33, 28, 77, 73, 17, 78, 39, 68, 17, 57)
triangle[13,1:13] <- c(91, 71, 52, 38, 17, 14, 91, 43, 58, 50, 27, 29, 48)
triangle[14,1:14] <- c(63, 66, 04, 68, 89, 53, 67, 30, 73, 16, 69, 87, 40, 31)
triangle[15,1:15] <- c(04, 62, 98, 27, 23, 09, 70, 98, 73, 93, 38, 53, 60, 04, 23)

```

## Euler Problem 67

The solution for problem number 67 is exactly the same. The data is read directly from the Project Euler website.

```# Euler Problem 67
triangle.67 <- matrix(nrow = 100, ncol = 100)
for (i in 1:100) {
triangle.67[i,1:i] <- as.numeric(unlist(strsplit(triangle.file[i,], " ")))
}
```

# Euler Problem 15: Pathways Through a Lattice

Euler Problem 15 analyses taxicab geometry. This system replaces the usual distance function with the sum of the absolute differences of their Cartesian coordinates. In other words, the distance a taxi would travel in a grid plan. The fifteenth Euler problem asks to determine the number of possible routes a taxi can take in a city of a certain size.

## Euler Problem 15 Definition

Starting in the top left corner of a 2×2 grid, and only being able to move to the right and down, there are exactly 6 routes to the bottom right corner. How many possible routes are there through a 20×20 grid?

## Solution

The defined lattice is one larger than the number of squares. Along the edges of the matrix, only one pathway is possible: straight to the right or down. We can calculate the number of possible pathways for the remaining number by adding the number to the right and below the point.

$p_{i,j}=p_{i,j-1}+p_{{i+1},j}$

For the two by two lattice the solution space is:

6  3  1
3  2  1
1  1  0

The total number of pathways from the upper left corner to the lower right corner is thus 6. This logic can now be applied to a grid of any arbitrary size using the following code.

The code defines the lattice and initiates the boundary conditions. The bottom row and the right column are filled with 1 as there is only one solution from these points. The code then calculates the pathways by working backwards through the matrix. The final solution is the number is the first cell.

```# Define lattice
nLattice <- 20
lattice = matrix(ncol=nLattice + 1, nrow=nLattice + 1)

# Boundary conditions
lattice[nLattice + 1,-(nLattice + 1)] <- 1
lattice[-(nLattice + 1), nLattice + 1] <- 1

# Calculate Pathways
for (i in nLattice:1) {
for (j in nLattice:1) {
lattice[i,j] <- lattice[i+1, j] + lattice[i, j+1]
}
}