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criss.dev

Posted on • Updated on • Originally published at cserb.net

# Build your first blockchain from scratch (#3 Finding the longest chain)

In the previous articles (Probabilistic Finality, DAG) we looked into how Probabilistic Finality works and how DAGs can help to find the longest chain.

Today we are going have a look at implementing a basic DAG in Crystal. We will focus on implementing a `DAG::Vertex` class and finding the vertex with the longest distance from a given starting vertex.
Each vertex will have a unique name and a Hash to represent its edges pointing to its children.

Here is the complete finished implementation:

``````module DAG
record Tip,
vertex : DAG::Vertex,
distance : Int32,
branch_root : (DAG::Vertex | Nil)

class Vertex
alias Name = String

getter name : Name
getter edges : Hash(Name, Vertex)

def initialize(@name, @edges = {} of Name => Vertex)
end

def add(edge_to vertex : Vertex) : Nil
@edges[vertex.name] = vertex
end

def children : Array(Vertex)
@edges.values
end
end

extend self

def tips(
from vertex : Vertex,
visited = Hash(Vertex::Name, Bool).new(false),
distance = Hash(Vertex, Int32).new(0),
tips = Array(DAG::Tip).new,
start : (Vertex | Nil) = nil,
current_branch_root : (Vertex | Nil) = nil
) : Array(DAG::Tip)
# remember the starting vertex
start = vertex if distance.empty?

# mark current vertex as visited
visited[vertex.name] = true

# sort children just to make it easier to test
sorted_children = vertex.children.sort_by { |c| c.name }
# if it has no children it's the tip of a branch
if !sorted_children[0]?
tips << DAG::Tip.new(
vertex: vertex,
distance: distance[vertex],
branch_root: current_branch_root
)
else
# it has children so let's continue the traverse
sorted_children.each do |child|
if !visited[child.name]
# we mark the child of the starting vertex as the current root
# of the branch we are exploring
if start == vertex
current_branch_root = child
end
# we calculate the distance of the child
distance[child] = distance[vertex] + 1

tips(
from: child,
visited: visited,
distance: distance,
tips: tips,
start: start,
current_branch_root: current_branch_root
)
end
end
end

tips
end

def tip_of_longest_branch(from vertex : DAG::Vertex) : DAG::Tip
tips = self.tips(from: vertex)

tip_of_longest_branch from: tips
end

def tip_of_longest_branch(from tips : Array(DAG::Tip)) : DAG::Tip
# Tip of longest branch (chain)
tips.max_by { |t| t.distance }
end
end
``````

Let's break the code apart to understand how it works.

1. DAG::Vertex Class

``````module DAG
...

class Vertex
alias Name = String

getter name : Name
getter edges : Hash(Name, Vertex)

def initialize(@name, @edges = {} of Name => Vertex)
end

def add(edge_to vertex : Vertex) : Void
@edges[vertex.name] = vertex
end

def children : Array(Vertex)
@edges.values
end
end

...

end
``````

This should be straight forward. We build a class `Vertex` and pass along the `Name` when we initialize it. Afterwards we create its edges:

``````v1 = DAG::Vertex.new("A")
v2 = DAG::Vertex.new("B")

pp v1.name #=> "A"
pp v1.children #=> [#<DAG::Vertex:0x7f0f4866ee60 @edges={}, @name="B">]
# same as
pp v1.edges.values #=> [#<DAG::Vertex:0x7f0f4866ee60 @edges={}, @name="B">]
``````

You can play with the code here

2. Distances and the Tips of branches

``````module DAG

...

extend self

def tips(
from vertex : Vertex,
visited = Hash(Vertex::Name, Bool).new(false),
distance = Hash(Vertex, Int32).new(0),
tips = Array(DAG::Tip).new,
start : (Vertex | Nil) = nil,
current_branch_root : (Vertex | Nil) = nil
) : Array(DAG::Tip)
# remember the starting vertex
start = vertex if distance.empty?

# mark current vertex as visited
visited[vertex.name] = true

# sort children just to make it easier to test
sorted_children = vertex.children.sort_by { |c| c.name }
# if it has no children it's the tip of a branch
if !sorted_children[0]?
tips << DAG::Tip.new(
vertex: vertex,
distance: distance[vertex],
branch_root: current_branch_root
)
else
# it has children so let's continue the traverse
sorted_children.each do |child|
if !visited[child.name]
# we mark the child of the starting vertex as the current root
# of the branch we are exploring
if start == vertex
current_branch_root = child
end
# we calculate the distance of the child
distance[child] = distance[vertex] + 1

tips(
from: child,
visited: visited,
distance: distance,
tips: tips,
start: start,
current_branch_root: current_branch_root
)
end
end
end

tips
end

``````

To find out the distances starting from say block `B1` to `Bn` we use depth first search (DFS), that explores each branch to the end and than moves on to the next one. If you think you need to better understand DFS you should watch this video.

``````  def tips(
from vertex : Vertex,
visited = Hash(Vertex::Name, Bool).new(false),
distance = Hash(Vertex, Int32).new(0),
tips = Array(DAG::Tip).new,
start : (Vertex | Nil) = nil,
current_branch_root : (Vertex | Nil) = nil
) : Array(DAG::Tip)

...
end
``````

The `tips` method is a recursive method that initially only needs the `from vertex` parameter and all other parameters are passed along with each step.

`visited` is an empty hash, which has been initialized with a default value of `false` for each key that is unknown.

`distance` contains a `Hash` of `DAG::Vertex` with its value being the distance from the first block passed as `from vertex` parameter

`tips` contains an `Array` of `DAG::Tip` which is a Struct defined using the record macro

``````module DAG
record Tip,
vertex : DAG::Vertex,
distance : Int32,
branch_root : (DAG::Vertex | Nil)

...
end
``````

We will examine the missing `start` and `current_branch_root` parameters later. First let's refresh our memory with an example of a DAG:

If we pass block `3000.1` as `from vertex` we should receive an Array of `[ 3001.2, 3004.2, 3005.1 ]` as the `tips` of this DAG.

Let's step into the method now:

``````    ...

# remember the starting vertex
start = vertex if distance.empty?

# mark current vertex as visited
visited[vertex.name] = true

# sort children just to make it easier to test
sorted_children = vertex.children.sort_by { |c| c.name }

...
``````

`start` is set to the first vertex that we actually passed to the method and it's not altered afterwards as `distance` will never be empty once we move on to the next step of the traverse.

`visited` is marking the vertex as visited and this is repeated each step

`sorted_children` is only set for convenience

``````    # if it has no children it's the tip of a branch
if sorted_children.empty?
tips << DAG::Tip.new(
vertex: vertex,
distance: distance[vertex],
branch_root: current_branch_root
)
else
# it has children so let's continue the traverse
sorted_children.each do |child|
if !visited[child.name]
# we mark the child of the starting vertex as the current root
# of the branch we are exploring
if start == vertex
current_branch_root = child
end
# we calculate the distance of the child
distance[child] = distance[vertex] + 1

tips(
from: child,
visited: visited,
distance: distance,
tips: tips,
start: start,
current_branch_root: current_branch_root
)
end
end
end
``````

`sorted_children.empty?` if this returns true it means we are at the end of a branch. So it's a `DAG::Tip`

`if !visited[child.name]` otherwise we continue traversing unvisited children.

``````if start == vertex
current_branch_root = child
end
``````

Remember the DAG example above? You have to imagine that once the traverse reaches a tip it steps back until the the branch splits off again and starts traversing unvisited vertices. If we start from `3000.1` all possible `current_branch_roots` are `3001.1` and `3001.2`. We will need this later on to determine which is the branch root of the longest chain.

`distance[child] = distance[vertex] + 1` now we calculate the distance and in the next step call `tips` again to make another step.

``````  def tip_of_longest_branch(from vertex : DAG::Vertex) : DAG::Tip
tips = self.tips(from: vertex)

tip_of_longest_branch from: tips
end

def tip_of_longest_branch(from tips : Array(DAG::Tip)) : DAG::Tip
# Tip of longest branch (chain)
tips.max_by { |t| t.distance }
end
``````

`tips.max_by { |t| t.distance }` to find the tip of the longest branch and its branch root we filter for the one with the highest distance value.
If there are multiple vertices with the same distance we don't really care which one we pick. Why this is not important will become apparent later in the series.

Now let's give it a try:

``````v1 = DAG::Vertex.new("A")
v2 = DAG::Vertex.new("B")
v3 = DAG::Vertex.new("C")
v4 = DAG::Vertex.new("D")

pp DAG.tips from: v1
#[DAG::Tip(
#  @branch_root=
#   #<DAG::Vertex:0x7fce8d002e60
#    @edges=
#     {"C" =>
#       #<DAG::Vertex:0x7fce8d002e40
#        @edges={"D" => #<DAG::Vertex:0x7fce8d002e20 @edges={}, #@name="D">},
#        @name="C">},
#    @name="B">,
#  @distance=3,
#  @vertex=#<DAG::Vertex:0x7fce8d002e20 @edges={}, @name="D">)]

pp DAG.tip_of_longest_branch from: v1
# @vertex=#<DAG::Vertex:0x7f3acc219e20 @edges={}, @name="D">)

``````

You can try it out yourself here and create more complex structures resembling real life scenarios.

3. Conclusion
We found an easy way to create a DAG and using a simple depth first search we were able to find the tip of the longest chain.

Integrating the code into a working blockchain will require some more work, but we'll get to that later on.

4. Cocol Project
The Cocol Project is an effort to implement a basic, but fully working, blockchain in Crystal — a "Minimum Viable Blockchain"