#### Tabu Search

p
```"""
This is pure Python implementation of Tabu search algorithm for a Travelling Salesman
Problem, that the distances between the cities are symmetric (the distance between city
'a' and city 'b' is the same between city 'b' and city 'a').
The TSP can be represented into a graph. The cities are represented by nodes and the
distance between them is represented by the weight of the ark between the nodes.

The .txt file with the graph has the form:

node1 node2 distance_between_node1_and_node2
node1 node3 distance_between_node1_and_node3
...

Be careful node1, node2 and the distance between them, must exist only once. This means
in the .txt file should not exist:
node1 node2 distance_between_node1_and_node2
node2 node1 distance_between_node2_and_node1

For pytests run following command:
pytest

For manual testing run:
python tabu_search.py -f your_file_name.txt -number_of_iterations_of_tabu_search \
-s size_of_tabu_search
e.g. python tabu_search.py -f tabudata2.txt -i 4 -s 3
"""

import argparse
import copy

def generate_neighbours(path):
"""
Pure implementation of generating a dictionary of neighbors and the cost with each
neighbor, given a path file that includes a graph.

:param path: The path to the .txt file that includes the graph (e.g.tabudata2.txt)
:return dict_of_neighbours: Dictionary with key each node and value a list of lists
with the neighbors of the node and the cost (distance) for each neighbor.

Example of dict_of_neighbours:
>>) dict_of_neighbours[a]
[[b,20],[c,18],[d,22],[e,26]]

This indicates the neighbors of node (city) 'a', which has neighbor the node 'b'
with distance 20, the node 'c' with distance 18, the node 'd' with distance 22 and
the node 'e' with distance 26.
"""

dict_of_neighbours = {}

with open(path) as f:
for line in f:
if line.split()[0] not in dict_of_neighbours:
_list = []
_list.append([line.split()[1], line.split()[2]])
dict_of_neighbours[line.split()[0]] = _list
else:
dict_of_neighbours[line.split()[0]].append(
[line.split()[1], line.split()[2]]
)
if line.split()[1] not in dict_of_neighbours:
_list = []
_list.append([line.split()[0], line.split()[2]])
dict_of_neighbours[line.split()[1]] = _list
else:
dict_of_neighbours[line.split()[1]].append(
[line.split()[0], line.split()[2]]
)

return dict_of_neighbours

def generate_first_solution(path, dict_of_neighbours):
"""
Pure implementation of generating the first solution for the Tabu search to start,
with the redundant resolution strategy. That means that we start from the starting
node (e.g. node 'a'), then we go to the city nearest (lowest distance) to this node
(let's assume is node 'c'), then we go to the nearest city of the node 'c', etc.
till we have visited all cities and return to the starting node.

:param path: The path to the .txt file that includes the graph (e.g.tabudata2.txt)
:param dict_of_neighbours: Dictionary with key each node and value a list of lists
with the neighbors of the node and the cost (distance) for each neighbor.
:return first_solution: The solution for the first iteration of Tabu search using
the redundant resolution strategy in a list.
:return distance_of_first_solution: The total distance that Travelling Salesman
will travel, if he follows the path in first_solution.
"""

with open(path) as f:
start_node = f.read(1)
end_node = start_node

first_solution = []

visiting = start_node

distance_of_first_solution = 0
while visiting not in first_solution:
minim = 10000
for k in dict_of_neighbours[visiting]:
if int(k[1]) < int(minim) and k[0] not in first_solution:
minim = k[1]
best_node = k[0]

first_solution.append(visiting)
distance_of_first_solution = distance_of_first_solution + int(minim)
visiting = best_node

first_solution.append(end_node)

position = 0
for k in dict_of_neighbours[first_solution[-2]]:
if k[0] == start_node:
break
position += 1

distance_of_first_solution = (
distance_of_first_solution
+ int(dict_of_neighbours[first_solution[-2]][position][1])
- 10000
)
return first_solution, distance_of_first_solution

def find_neighborhood(solution, dict_of_neighbours):
"""
Pure implementation of generating the neighborhood (sorted by total distance of
each solution from lowest to highest) of a solution with 1-1 exchange method, that
means we exchange each node in a solution with each other node and generating a
number of solution named neighborhood.

:param solution: The solution in which we want to find the neighborhood.
:param dict_of_neighbours: Dictionary with key each node and value a list of lists
with the neighbors of the node and the cost (distance) for each neighbor.
:return neighborhood_of_solution: A list that includes the solutions and the total
distance of each solution (in form of list) that are produced with 1-1 exchange
from the solution that the method took as an input

Example:
>>> find_neighborhood(['a', 'c', 'b', 'd', 'e', 'a'],
...                   {'a': [['b', '20'], ['c', '18'], ['d', '22'], ['e', '26']],
...                    'c': [['a', '18'], ['b', '10'], ['d', '23'], ['e', '24']],
...                    'b': [['a', '20'], ['c', '10'], ['d', '11'], ['e', '12']],
...                    'e': [['a', '26'], ['b', '12'], ['c', '24'], ['d', '40']],
...                    'd': [['a', '22'], ['b', '11'], ['c', '23'], ['e', '40']]}
...                   )  # doctest: +NORMALIZE_WHITESPACE
[['a', 'e', 'b', 'd', 'c', 'a', 90],
['a', 'c', 'd', 'b', 'e', 'a', 90],
['a', 'd', 'b', 'c', 'e', 'a', 93],
['a', 'c', 'b', 'e', 'd', 'a', 102],
['a', 'c', 'e', 'd', 'b', 'a', 113],
['a', 'b', 'c', 'd', 'e', 'a', 119]]
"""

neighborhood_of_solution = []

for n in solution[1:-1]:
idx1 = solution.index(n)
for kn in solution[1:-1]:
idx2 = solution.index(kn)
if n == kn:
continue

_tmp = copy.deepcopy(solution)
_tmp[idx1] = kn
_tmp[idx2] = n

distance = 0

for k in _tmp[:-1]:
next_node = _tmp[_tmp.index(k) + 1]
for i in dict_of_neighbours[k]:
if i[0] == next_node:
distance = distance + int(i[1])
_tmp.append(distance)

if _tmp not in neighborhood_of_solution:
neighborhood_of_solution.append(_tmp)

index_of_last_item_in_the_list = len(neighborhood_of_solution[0]) - 1

neighborhood_of_solution.sort(key=lambda x: x[index_of_last_item_in_the_list])
return neighborhood_of_solution

def tabu_search(
first_solution, distance_of_first_solution, dict_of_neighbours, iters, size
):
"""
Pure implementation of Tabu search algorithm for a Travelling Salesman Problem in
Python.

:param first_solution: The solution for the first iteration of Tabu search using
the redundant resolution strategy in a list.
:param distance_of_first_solution: The total distance that Travelling Salesman will
travel, if he follows the path in first_solution.
:param dict_of_neighbours: Dictionary with key each node and value a list of lists
with the neighbors of the node and the cost (distance) for each neighbor.
:param iters: The number of iterations that Tabu search will execute.
:param size: The size of Tabu List.
:return best_solution_ever: The solution with the lowest distance that occurred
during the execution of Tabu search.
:return best_cost: The total distance that Travelling Salesman will travel, if he
follows the path in best_solution ever.
"""
count = 1
solution = first_solution
tabu_list = []
best_cost = distance_of_first_solution
best_solution_ever = solution

while count <= iters:
neighborhood = find_neighborhood(solution, dict_of_neighbours)
index_of_best_solution = 0
best_solution = neighborhood[index_of_best_solution]
best_cost_index = len(best_solution) - 1

found = False
while not found:
i = 0
while i < len(best_solution):
if best_solution[i] != solution[i]:
first_exchange_node = best_solution[i]
second_exchange_node = solution[i]
break
i = i + 1

if [first_exchange_node, second_exchange_node] not in tabu_list and [
second_exchange_node,
first_exchange_node,
] not in tabu_list:
tabu_list.append([first_exchange_node, second_exchange_node])
found = True
solution = best_solution[:-1]
cost = neighborhood[index_of_best_solution][best_cost_index]
if cost < best_cost:
best_cost = cost
best_solution_ever = solution
else:
index_of_best_solution = index_of_best_solution + 1
best_solution = neighborhood[index_of_best_solution]

if len(tabu_list) >= size:
tabu_list.pop(0)

count = count + 1

return best_solution_ever, best_cost

def main(args=None):
dict_of_neighbours = generate_neighbours(args.File)

first_solution, distance_of_first_solution = generate_first_solution(
args.File, dict_of_neighbours
)

best_sol, best_cost = tabu_search(
first_solution,
distance_of_first_solution,
dict_of_neighbours,
args.Iterations,
args.Size,
)

print(f"Best solution: {best_sol}, with total distance: {best_cost}.")

if __name__ == "__main__":
parser = argparse.ArgumentParser(description="Tabu Search")
parser.add_argument(
"-f",
"--File",
type=str,
help="Path to the file containing the data",
required=True,
)
parser.add_argument(
"-i",
"--Iterations",
type=int,
help="How many iterations the algorithm should perform",
required=True,
)
parser.add_argument(
"-s", "--Size", type=int, help="Size of the tabu list", required=True
)

# Pass the arguments to main method
main(parser.parse_args())
```