def build_solution(self):
"""
Builds a mastermind guess based on the configured encoding
"""
# empty linear solution
solution_representation = []
size = self._encoding.size
data = self._encoding.encoding_data
# if elements can be repeated
if self._encoding.can_repeat_elements:
for _ in range(0, size):
solution_representation.append( choice( data ) )
solution = LinearSolution(representation = solution_representation, encoding_rule = self._encoding_rule)
return solution
# if elements cannot be repeated
else:
encoding_data_temp = deepcopy( data )
for _ in range(0, size):
element = choice( encoding_data_temp )
solution_representation.append( element )
encoding_data_temp.remove( element )
solution = LinearSolution(representation = solution_representation, encoding_rule = self._encoding_rule)
return solution
def cycle_crossover(problem, solution1, solution2):
# get the representation (list) of both solutions
solution_1 = solution1.representation
solution_2 = solution2.representation
cycles = []
considered = []
# find the cycles
while len(considered) < len(solution_1):
i = 0
while i in considered:
i += 1
cycle = []
full_cycle = False
while full_cycle == False:
cycle.append(i)
considered.append(i)
i = solution_1.index(solution_2[i])
if i in considered:
full_cycle = True
cycles.append(cycle)
# create empty children
child1 = [None] * len(solution_1)
child2 = [None] * len(solution_1)
# getting the children
for i, cycle in enumerate(cycles):
# note that here cycle 1 is the cycle with index 0
if i % 2 == 0:
for j in cycle:
child1[j] = solution_1[j]
child2[j] = solution_2[j]
else:
for j in cycle:
child1[j] = solution_2[j]
child2[j] = solution_1[j]
# save the children as LinearSolution objects
child_1 = LinearSolution(child1, solution1.encoding_rule)
child_2 = LinearSolution(child2, solution2.encoding_rule)
return child_1, child_2
def tsp_get_neighbors_np(self,
solution,
problem,
neighborhood_size=0,
n_changes=3):
initial_sol = np.asarray(
solution.representation) # change to numpy array for performance
neighbors_np = [initial_sol] # list of numpy arrays for performance
def n_change(list_solutions):
neighbors = []
for k in list_solutions: # find all neighbors
for l in range(0, len(k)):
for j in range((l + 1), len(k)):
neighbor = np.copy(k)
neighbor[l], neighbor[j] = neighbor[j], neighbor[l]
neighbors.append(neighbor)
return neighbors
for i in range(0, n_changes): # How many swaps to allow,
neighbors_np = n_change(
neighbors_np
) # This escalates fast!!! one should be more than enough
# convert back to linear solution for evaluation
neighbors_final = []
for sol in neighbors_np:
sol = sol.tolist()
solution = LinearSolution(representation=sol,
encoding_rule=self._encoding_rule)
neighbors_final.append(solution)
return neighbors_final # return a list of solutions
def fast_solution_copy(
self,
representation,
feedback=None
): # << This method does not need to be extended, it already automated solutions evaluation, for Single-Objective and for Multi-Objective
encoding_data = self._encoding.encoding_data
solution_copy = LinearSolution(representation=copy(representation),
encoding_rule=self._encoding_rule)
return solution_copy
def build_solution(self):
"""
"""
solution_representation = sample(self._encoding.encoding_data,
self._encoding.size)
solution = LinearSolution(representation=solution_representation,
encoding_rule=self._encoding_rule)
return solution
def build_solution(self):
# empty linear solution
solution_representation = []
# size = self._encoding.size
data = self._encoding.encoding_data
max = data[-1]
min = data[0]
solution_representation.append( randint( min, max) )
solution = LinearSolution(representation = solution_representation, encoding_rule = self._encoding_rule)
return solution
def build_solution(self):
"""
"""
solution_representation = []
for i in range(0, self._encoding.size):
solution_representation.append(i)
random.shuffle(solution_representation)
solution_representation.append(solution_representation[0])
solution = LinearSolution(representation=solution_representation,
encoding_rule=self._encoding_rule)
return solution
def build_solution(self):
"""
Builds a linear solution based on min and max values
"""
solution_representation = []
encoding_data = self._encoding.encoding_data
for iGene in range(0, self._encoding.size):
solution_representation.append(
randint(encoding_data["min"][iGene],
encoding_data["max"][iGene]))
solution = LinearSolution(representation=solution_representation,
encoding_rule=self._encoding_rule)
return solution
def build_solution(self):
"""
"""
solution_representation = []
encoding_data = deepcopy(self._encoding.encoding_data)
while len(solution_representation) < self._encoding.size:
new_city = choice(encoding_data)
encoding_data.remove(new_city)
solution_representation.append(new_city)
solution = LinearSolution(representation=solution_representation,
encoding_rule=self._encoding_rule)
return solution
def tsp_bitflip_get_neighbors(solution, problem, neighborhood_size=0):
neighborhood = []
# if the neighborhood size is -1 we get all neighbors
if neighborhood_size == -1:
for i in range(0, len(solution.representation)):
for j in range(0, len(solution.representation)):
if i != j:
# swap two different cities
neighbor = solution.representation[:]
neighbor[i] = solution.representation[j]
neighbor[j] = solution.representation[i]
# check if this neighbor is not repeated
if neighbor not in neighborhood:
neighborhood.append(neighbor)
else:
while len(neighborhood) < neighborhood_size:
# swap two different random cities
i = randint(0, len(solution.representation) - 1)
j = randint(0, len(solution.representation) - 1)
while i == j:
j = randint(0, len(solution.representation) - 1)
# deep copy of solution
neighbor = solution.representation[:]
neighbor[i] = solution.representation[j]
neighbor[j] = solution.representation[i]
# check if this neighbor is not repeated
if neighbor not in neighborhood:
neighborhood.append(neighbor)
neighbors = []
# create LinearSolution objects
for neighbor in neighborhood:
neigh = LinearSolution(neighbor, solution.encoding_rule)
neighbors.append(neigh)
return neighbors
def build_solution(self):
"""
Builds a linear solution for Knapsack with same size of decision variables with 0s and 1s.
Where:
0 indicate that item IS NOT in the knapsack
1 indicate that item IS in the knapsack
"""
solution_representation = []
encoding_data = self._encoding.encoding_data
for _ in range(0, self._encoding.size):
solution_representation.append(choice(encoding_data))
#print(solution_representation)
solution = LinearSolution(representation=solution_representation,
encoding_rule=self._encoding_rule)
return solution
def build_solution(self):
"""
Builds a linear solution for TSP with a specific order of cities
"""
solution_representation = []
encoding_data = self._encoding.encoding_data
# Choose an initial random city
solution_representation.append(randint(1, len(encoding_data)))
# Build a solution by always appending the closest next city
for i in range(len(encoding_data)):
dists = [
x for x in self._distances[solution_representation[-1] - 1]
]
sorDist = sorted(dists)
for j in sorDist:
if (dists.index(j) + 1) not in solution_representation:
solution_representation.append(dists.index(j) + 1)
break
solution = LinearSolution(representation=solution_representation,
encoding_rule=self._encoding_rule)
return solution
def build_solution(self):
"""
Builds a linear solution for TSP that is an ordered list of numbers, with no repetitions
Where:
each number i corresponds to the city of index i in the distance matrix
"""
solution_representation = []
encoding_data = self._encoding.encoding_data[:]
for _ in range(0, self._encoding.size):
# choose a random city
city = choice(encoding_data)
solution_representation.append(city)
# remove it from the list of possibilities
encoding_data.remove(city)
# create a LinearSolution object
solution = LinearSolution(representation=solution_representation,
encoding_rule=self._encoding_rule)
return solution
def build_solution(self):
"""
To build the solution
we pick 32 stocks from the list as par Fisher and laurie (1970)
"""
solution_representation = []
#calculate portfolio investment
def cal_total_investment(portfolio):
total_investment = 0
for i in range(0, len(portfolio)):
if portfolio[i] >= 1:
#print(i, 'printing i')
total_investment += float(self._prices[i]) * portfolio[i]
return total_investment
solution_representation = [0] * self._encoding_rule['Size']
solution_investment = 0
tolerance = 0.20 #limit before reaching the full investment
stock_counter = self._optimum_stocks
for i in takewhile(
lambda i: i < stock_counter and solution_investment < self.
_max_investment * (1 - tolerance), count()):
n = choice(self._encoding_rule['Data'])
solution_representation[randint(0,
len(solution_representation) -
1)] = n
solution_investment += cal_total_investment(
solution_representation)
i += 1
solution = LinearSolution(representation=solution_representation,
encoding_rule=self._encoding_rule)
#print(f'Build Solution: total stocks picked in {sum(solution.representation)}, at cost: {solution_investment}')
return solution
def build_solution(self):
"""
Builds a linear solution for PIP with same size of decision variables integers.
Where:
This integer represents the number of stocks we bought for that asset
"""
solution_representation = []
encoding_data = self._encoding.encoding_data
randoms = []
# assign random numbers between 0 and 1 to each asset
for _ in range(0, self._encoding.size):
randoms.append(uniform(0, 1))
weights = []
# get the investment weights for each asset based on the random numbers
for random in randoms:
weights.append(random / sum(randoms))
prices = self._prices
budget = self._budget
# calculate how much that investment weight costs related to our budget
for i in range(0, len(weights)):
investment = weights[i] * budget
# divide the money we're investing by the price of the stocks to find out how many we can buy
solution_representation.append(round(investment / prices[i]))
# create a LinearSolution object
solution = LinearSolution(representation=solution_representation,
encoding_rule=self._encoding_rule)
return solution
def build_solution(self, method='Random'):
"""
Creates a initial solution and returns it according to the encoding rule and the chosen method from the
following available methods:
- 'Hill Climbing'
- 'Random' (default method)
- 'Greedy'
"""
if method == 'Random': # shuffles the list and then instantiates it as a Linear Solution
encoding_data = self._encoding.encoding_data
solution_representation = encoding_data.copy()
np.random.shuffle(solution_representation) # inplace shuffle
solution = LinearSolution(representation=solution_representation,
encoding_rule=self._encoding_rule)
return solution
elif method == 'Hill Climbing':
solution = HillClimbing(
problem_instance=self,
neighborhood_function=self.tsp_get_neighbors_np).search()
#TODO: allow changes in params for Hill Climbing
return solution
#elif method == 'Simulated Annealing': TODO: build SA initialization
# return solution
elif method == 'Greedy':
# get a random solution where we will keep the first element
initial_solution = np.array(
self.build_solution(method='Random').representation)
copy = initial_solution.copy()
element = initial_solution[0]
# i can not reach the last position because there is no more cities to evaluate
for i in range(1, (len(initial_solution) - 1)):
distance_array = np.array(
heapq.nsmallest(len(initial_solution),
self._distances[element]))
# get the index (item name) of the second smallest distance between element in index i and all the other cities
closest_city = np.argwhere(
self._distances[element] == distance_array[1])[0][0]
if closest_city == 0: # the closest city can not be our first city on the matrix
closest_city = np.argwhere(
self._distances[element] == distance_array[2])[0][0]
n = 2 # let us to go through the distance ascending list
# while the closest city is already in the changed slice of the initial_solution, we have to keep looking
while closest_city in initial_solution[0:i]:
# get the next closest city in the distance ascending list with the element evaluated
closest_city = np.argwhere(
self._distances[element] == distance_array[n])[0][0]
if closest_city == 0: # the closest city can not be our first city on the matrix
closest_city = np.argwhere(self._distances[element] ==
distance_array[n + 1])[0][0]
n += 1 # if it is not a valid closest city the n should be plus one than the usual
n += 1
# change the current position in initial_solution with the closest_city
initial_solution[i] = closest_city
# get the next element to evaluate the closest city
element = initial_solution[i]
# change the last index with the missing element
initial_solution[-1] = np.setdiff1d(copy,
initial_solution[0:-1],
assume_unique=True)[0]
#print('greedy solutions: ', list(initial_solution))
solution = LinearSolution(representation=list(initial_solution),
encoding_rule=self._encoding_rule)
return solution
else:
print(
'Please choose one of these methods: Hill Climbing, Greedy, Random or Multiple. It will use the Random method'
)
return self.build_solution()
def pmx_crossover(problem, solution1, solution2):
# get the representation (list) of both solutions
solution_1 = solution1.representation
solution_2 = solution2.representation
# choose two different random points in the solutions
point1 = randint(0, len(solution_1) - 1)
point2 = point1
while point1 == point2:
point2 = randint(0, len(solution_1) - 1)
# make sure that the fist crossover point is the smallest and vice versa
firstCrossPoint = min(point1, point2)
secondCrossPoint = max(point1, point2)
# select the middle section of each parent
parent1MiddleCross = solution_1[firstCrossPoint:secondCrossPoint]
parent2MiddleCross = solution_2[firstCrossPoint:secondCrossPoint]
# create children that are equal to the respective parent minus the middle section
child1 = solution_1[:firstCrossPoint] + parent2MiddleCross + solution_1[
secondCrossPoint:]
child2 = solution_2[:firstCrossPoint] + parent1MiddleCross + solution_2[
secondCrossPoint:]
# save the relations in the middle section
relations = []
for i in range(len(parent1MiddleCross)):
relations.append([parent2MiddleCross[i], parent1MiddleCross[i]])
# check if any of the children have repeated elements in them
counts1 = [child1.count(i) for i in child1]
counts2 = [child2.count(i) for i in child2]
# while there are repeated elements in child 1
while len([x for x in counts1 if x > 1]) > 0:
# check the part before the middle section
for i in child1[:firstCrossPoint]:
for j in parent2MiddleCross:
if i == j:
# replace the repeated element OUTSIDE of the middle section using the relations we defined
index_j = parent2MiddleCross.index(j)
relation = relations[index_j]
index_i = child1.index(i)
child1[index_i] = relation[1]
# check the part after the middle section
for i in child1[secondCrossPoint:]:
for j in parent2MiddleCross:
if i == j:
# replace the repeated element OUTSIDE of the middle section using the relations we defined
index_j = parent2MiddleCross.index(j)
relation = relations[index_j]
index_i = child1.index(i, secondCrossPoint)
child1[index_i] = relation[1]
# check again if there are repeated elements in child 1
counts1 = [child1.count(i) for i in child1]
# while there are repeated elements in child 1
while len([x for x in counts2 if x > 1]) > 0:
# check the part before the middle section
for i in child2[:firstCrossPoint]:
for j in parent1MiddleCross:
if i == j:
# replace the repeated element OUTSIDE of the middle section using the relations we defined
index_j = parent1MiddleCross.index(j)
relation = relations[index_j]
index_i = child2.index(i)
child2[index_i] = relation[0]
# check the part after the middle section
for i in child2[secondCrossPoint:]:
for j in parent1MiddleCross:
if i == j:
# replace the repeated element OUTSIDE of the middle section using the relations we defined
index_j = parent1MiddleCross.index(j)
relation = relations[index_j]
index_i = child2.index(i, secondCrossPoint)
child2[index_i] = relation[0]
# check again if there are repeated elements in child 1
counts2 = [child2.count(i) for i in child2]
# create LinearSolution objects
child_1 = LinearSolution(child1, solution1.encoding_rule)
child_2 = LinearSolution(child2, solution2.encoding_rule)
return child_1, child_2
def order_1_crossover(problem, solution1, solution2):
# get the representation (list) of both solutions
solution_1 = solution1.representation
solution_2 = solution2.representation
# choose two different random points in the solutions
point1 = randint(0, len(solution_1) - 1)
point2 = point1
while point1 == point2:
point2 = randint(0, len(solution_1) - 1)
# make sure that the first crossover point is the smallest and vice versa
firstCrossPoint = min(point1, point2)
secondCrossPoint = max(point1, point2)
# save the middle section of each parent
parent1MiddleCross = solution_1[firstCrossPoint:secondCrossPoint]
parent2MiddleCross = solution_2[firstCrossPoint:secondCrossPoint]
order_1 = []
# we start saving the sequence after the middle section
for i in solution_2[secondCrossPoint:]:
# make sure we don't have repeated elements
if i not in parent1MiddleCross:
order_1.append(i)
# then we go from the beginning up to where we started
for i in solution_2[:secondCrossPoint]:
# make sure we don't have repeated elements
if i not in parent1MiddleCross:
order_1.append(i)
order_2 = []
# we start saving the sequence after the middle section
for i in solution_1[secondCrossPoint:]:
# make sure we don't have repeated elements
if i not in parent2MiddleCross:
order_2.append(i)
# then we go from the beginning up to where we started
for i in solution_1[:secondCrossPoint]:
# make sure we don't have repeated elements
if i not in parent2MiddleCross:
order_2.append(i)
# create two empty children
child1 = [None] * len(solution_1)
child2 = [None] * len(solution_1)
# first save the middle section
for i in range(firstCrossPoint, secondCrossPoint):
child1[i] = solution_1[i]
child2[i] = solution_2[i]
# then we start filling in from the end of the middle section
j = 0
for i in range(secondCrossPoint, len(solution_1)):
child1[i] = order_1[j]
child2[i] = order_2[j]
j += 1
# and then we fill in from the beginning up to the beginning of the middle section
for i in range(0, firstCrossPoint):
child1[i] = order_1[j]
child2[i] = order_2[j]
j += 1
# save the children as LinearSolution objects
child_1 = LinearSolution(child1, solution1.encoding_rule)
child_2 = LinearSolution(child2, solution1.encoding_rule)
return child_1, child_2
