The first step to solving this quiz is to come up with some graphs that

you can

test. I started by thinking of some trivial patterns I could easily

hand figure

the answers for. A good example is the graph:

```
A -> B -> C -> D
```

Finding the longest simple path for that is not hard, it is always all

of the

edges. That’s just a straight line basically and it will include all

the nodes.

Building these kinds of graphs programatically for any size is also

easy:

```
>> Array.new(3) { |i| [(?A + i).chr, (?A + i + 1).chr] }
=> [["A", "B"], ["B", "C"], ["C", "D"]]
```

Himadri ChoudHury took a different approach, generating completely

random

graphs. This has the advantage of testing far more cases including

isolated

nodes and circular paths. Here’s the generation code:

```
# @dg is the generated DiGraph. Store the paths in a hash called @paths
@dg
# @paths is a hash. Each element of the hash is an array that contains
# all the paths from that node
@paths
# @nodes is an array which contains a list of all the nodes
@nodes
# Randomly generate @dg and the corresponding @paths
def generate_dg
nodes = Array.new
# 10 nodes total
10.times do
nodes << rand(10)
end
nodes.uniq!
@paths = Hash.new
nodes.each do |n|
num_paths_from_each_node = rand(3) + 1
next_nodes = Array.new
num_paths_from_each_node.times do
next_nodes << nodes[rand(nodes.length)]
end
next_nodes.uniq!
@paths[n] = next_nodes
end
arr = Array.new
@paths.each do |key,vals|
@paths[key].each do |val|
arr << [key,val]
end
end
@dg = DiGraph.new(*arr)
@nodes = @paths.keys
end
```

(Note: those bare instance variables nixed in with the comments

reference

variables on the Class object and not the instances themselves. They

read and

discard nil values having no effect.)

The code begins by generating up to ten (uniq!() can reduce the count)

random

nodes as simple Integers. After that, each node is randomly connected

to

between one and three other nodes. A Hash is created during this

process to

represent the connections. This turns out to be very helpful later on.

Finally, the paths are converted into the Array representation DiGraph

expects

and the graph is created. The last line also stores the nodes for easy

access.

Here’s a random graph (viewed from the paths Hash) created from this

code, so

you can see how they come out:

```
{ 5 => [8, 6],
0 => [0, 6],
6 => [1, 7, 6],
1 => [8],
7 => [7, 9, 4],
8 => [6, 1],
9 => [4, 8, 0],
4 => [5] }
```

Now, in order to use these random graphs, we really need code that can

tell us

the right answer for the data. Essentially, this means that we require

our own

implementation of the test methods, to prove that we get the same

answers as the

DiGraph implementation. Here is one of those methods, used to test

max_length_of_simple_path_including_node():

```
# Depth first search for the longest simple path starting from 'node'
# Simple path means a path that doesn't contain any duplicate edges
# Note: I'm not using the definition of simply connected based on no
# duplicate nodes
def search(node)
longest_path = 0
if (@paths[node])
@paths[node].each_index do |next_idx|
next_node = @paths[node][next_idx]
@paths[node].delete_at(next_idx)
tmp = 1 + search(next_node)
@paths[node].insert(next_idx,next_node)
if (longest_path < tmp)
longest_path = tmp
end
end
end
return longest_path
end
```

This method is a simple depth-first search, as the comment says,

counting the

steps taken to reach the farthest node without crossing a single edge

twice.

The implementation is just a recursive search from each node reachable

from the

start node. At each step, the last edge crossed is removed from the

paths Hash,

the recursive search is performed, and then the edge is restored. The

returned

result here is just a count, not the actual path.

The tests using that are pretty trivial:

```
def test_03_max_length_of_simple_path_including_node
generate_dg
@nodes.each do |node|
longest_path = search(node)
# ...
assert_equal( longest_path,
@dg.max_length_of_simple_path_including_node(node)
```

)

end

end

Unfortunately, this is where we get to execution problems with the quiz.

The

above code passes some tests, due to bugs in the DiGraph implementation,

but it

does not correctly implement the

max_length_of_simple_path_including_node()

method described in the quiz.

Here’s an example of where it gets wrong answers, using my trivial

one-way path

described at the beginning of this quiz:

```
>> @paths = {"A" => ["B"], "B" => ["C"], "C" => ["D"]}
=> {"A"=>["B"], "B"=>["C"], "C"=>["D"]}
>> search("A")
=> 3
>> search("B")
=> 2
>> search("C")
=> 1
>> search("D")
=> 0
```

Those are the longest simple paths *starting* from a given node, but not

the

longest given paths *including* the given node (which would all be 3).

These errors made it tricky to correctly evaluate the expected results

of the

methods and I apologize for this. I ran into the same issue with my own

tests,

as anyone following my posts to Ruby T. saw in painful detail.

Luckily, it’s easy to get to a real solution of the first method from

here.

First, we can modify Himadri’s search() method to return the path

instead of

just a count:

```
def search(node, longest_path = [node])
if (@paths[node])
@paths[node].each_index do |next_idx|
next_node = @paths[node][next_idx]
@paths[node].delete_at(next_idx)
tmp = search(next_node, longest_path + [next_node])
@paths[node].insert(next_idx,next_node)
if (longest_path.size < tmp.size)
longest_path = tmp
end
end
end
return longest_path
end
```

Then, using that method, we can exhaustively search all the long paths

for the

biggest one containing our node:

```
def max_length_of_simple_path_including_node(n)
all_paths = @paths.keys.map { |node| search(node) }
all_paths = all_paths.select { |path| path.include? n }
all_paths = all_paths.sort_by { |path| -path.size }
all_paths.empty? ? 0 : all_paths.shift.size - 1
end
```

Line by line that makes a list of the longest paths starting from each

node,

reduces that list to those including the desired node, sorts the

remaining paths

biggest to smallest, and finally returns an edge count of the largest

path or 0

if there aren’t any paths left.

The other method is left as an exercise for the interested reader.

Many, many thanks to the submitters who braved the waters and came up

with some

basic tests, right or wrong.

Tomorrow, we will programatically generate 1,000 monkeys in the hopes

that they

can recreate the works of Shakespeare…