# Adjacency Matrix

In the last post, we explored how a graph can be represented using an adjacency list. We’ll now look at how an adjacency matrix can also be used.

## Disclaimer

This post is mainly for me to jot down points about Adjacency Matrices. If this is helpful to you in some way, please let me know!

## What is an Adjacency Matrix?

An

adjacency matrixis a square matrix representation of a finite graph. The elements indicate whether a pair of vertices is adjacent to each other.

A graph with `V`

vertices can be represented with a `V x V`

matrix (`V`

rows and `V`

columns) where:

- The row index represents the origin of the edge
- The column index represents the tail of the edge

Therefore, a point `(i, j)`

represents an edge starting from `i`

and ending at `j`

.

If the value at this point `(i, j)`

is:

`1`

then there is an edge from`i`

to`j`

`0`

then there is no edge from`i`

to`j`

## Building an Adjacency Matrix

We have previously looked at the following graph:

In order to build this matrix, we would first need to create a `VxV`

matrix full of zeros.
`V`

is the number of vertices which for this graph is 6 so we need to build a matrix consisting of 6 rows and 6 columns.

The `6x6`

matrix looks like this:

The rows and columns are enumerated from 0 until 5 which corresponds with vertices `A`

to `F`

respectively.

We now go through each vertex and mark each node is connected to with a 1.
Let’s start with vertex `A`

.

### Vertex A

Vertex `A`

is connected to both `B`

and `C`

.

- Find the row corresponding to
`A`

: 0 - Then we find the column for
`B`

: 1 - Mark this point at row 0 and column 1 (
`(0, 1)`

) with a value of 1 - Then we find the column for
`C`

: 2 - Mark this point row 0 and column 2 (
`(0, 1)`

) with a value of 1

The resulting row for `A`

now looks like:

### Vertex B

Vertex `B`

is connected to both `D`

and `E`

.

- Find the row corresponding to
`B`

: 1 - Then we find the column for
`D`

: 3 - Mark this point at row 1 and column 3 (
`(1, 3)`

) with a value of 1 - Then we find the column for
`E`

: 4 - Mark this point row 1 and column 4 (
`(1, 4)`

) with a value of 1

The resulting row for `B`

now looks like:

### Vertex C

Vertex `C`

is only connected to `F`

.

- Find the row corresponding to
`C`

: 2 - Then we find the column for
`F`

: 5 - Mark this point at row 2 and column 5 (
`(2, 5)`

) with a value of 1

The resulting row for `C`

now looks like:

### Vertex D, E, and F

Vertex `D`

, `E`

, and `F`

have no adjacent nodes.

Therefore, their respective rows can remain as zeros.

## The Result

The completed adjacency matrix for the graph looks like:

Recall:

`A`

connects to both`B`

and`C`

so both`(0, 1)`

and`(0, 2)`

are marked with a`1`

- However, the point at
`(1, 0)`

is**not**marked with a point since`B`

does not connect back to`A`

.

## Implementation

The adjacent matrix can be implemented using a 2D array (aka an array of arrays).

- The inner arrays represent the columns
- The outer array represent the rows

The above matrix would be defined in Swift as follows:

```
let matrix: [[Int]] = [
[0, 1, 1, 0, 0, 0],
[0, 0, 0, 1, 1, 0],
[0, 0, 0, 0, 0, 1],
[0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0]
]
```

### Checking for Existence

If we wanted to determine whether there exists an edge between two vertices, we can do a lookup in this array to determine this. Recall that a value of 1 signifies the existence of an edge.

```
func hasEdge(in matrix: [[Int]], from row: Int, to col: Int) -> Bool {
// 1
guard isIndexValid((row, col), in: matrix) else { return false }
// 2
return matrix[row][col] == 1
}
```

- Calls a helper function that takes in a tuple of the row and column pair and returns if it’s valid for the given matrix
- Accesses the value at the given row and column and verifies if it equals to 1
- This is sometimes easier to follow when separated:
`let rows = matrix[row] return rows[col] == 1`

- This is sometimes easier to follow when separated:

### Depth First Search (DFS)

When we did DFS on the graph, we did pre-order traversal that follows these steps:

- Visit node
- Visit left child
- Visit right child

The result of the traversal was a visitation order as follows:

```
A -> B -> D -> E -> C -> F
```

The DFS traversal for a matrix requires that we traverse as follows:

- Start with the first row
- Iterate through all the columns until we’ve found a
`1`

and if we haven’t visited it yet - Given the column, repeat the DFS using this column as the new row
- Repeat these steps until all rows have been traversed

```
// This algorithm is courtesy of my colleague Michelle Lee!
func dfsMatrix(_ matrix: [[Int]], row: Int = 0) {
guard !matrix.isEmpty else { return }
print(edgesLookup[row])
// 1
rowsVisited.insert(row)
// 2
let colCount = matrix[row].count
for col in 0..<colCount {
// 3
if matrix[row][col] == 1 && !rowsVisited.contains(col) {
// 4
dfsMatrix(matrix, row: col)
}
}
}
```

- After printing the node that corresponds to the row, add this row to our set of visited nodes
- Grab the count of columns for this given row and create a loop from
`0`

up to`colCount`

- Check if the value at
`(row, col)`

is equal to`1`

and that we haven’t visited this node before - Using this column
`col`

, repeat the DFS starting at this row.

The output should be the same as before:

```
A -> B -> D -> E -> C -> F
```

This does assume that the first row has at least one edge through which for most cases is a valid assumption.

## Summary

An adjacency matrix is another way to represent a graph that makes lookups very quick.

However, there is often a lot of space required to define the graph. For large graphs, this can become quite inefficient.

A common example where you’ll see adjacency matrices is in this classic interview question: Island Problem. The example code can be found here.

## Tradeoffs

**Pros:**

- The existence of an edge between two vertices can be determined in constant time:
`O(1)`

- To determine if an edge exists can be done as:
`matrix[i][j] == 1`

- To determine if an edge exists can be done as:

**Cons:**

- A
`V * V`

matrix is required even if only a few vertices have edges between them:`O(V^2)`

- Observe how the matrix above is mainly zeros

- Iterating through all the edges would exhaust the whole matrix in the worst case:
`O(V^2)`

- If you need to iterate through all edges you would need to loop through the whole matrix