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Merge branch 'feature/optimizer-use-fastapi'

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Remy Moll
2024-05-29 18:23:34 +02:00
parent f292b2a375 ae9860cc8e
commit 80b04905de
15 changed files with 2099 additions and 1127 deletions
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backend/.gitignore vendored Normal file
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# osm-cache
cache/
# Byte-compiled / optimized / DLL files
__pycache__/
*.py[cod]
*$py.class
# C extensions
*.so
# Distribution / packaging
.Python
build/
develop-eggs/
dist/
downloads/
eggs/
.eggs/
lib/
lib64/
parts/
sdist/
var/
wheels/
share/python-wheels/
*.egg-info/
.installed.cfg
*.egg
MANIFEST
# PyInstaller
# Usually these files are written by a python script from a template
# before PyInstaller builds the exe, so as to inject date/other infos into it.
*.manifest
*.spec
# Installer logs
pip-log.txt
pip-delete-this-directory.txt
# Unit test / coverage reports
htmlcov/
.tox/
.nox/
.coverage
.coverage.*
.cache
nosetests.xml
coverage.xml
*.cover
*.py,cover
.hypothesis/
.pytest_cache/
cover/
# Translations
*.mo
*.pot
# Django stuff:
*.log
local_settings.py
db.sqlite3
db.sqlite3-journal
# Flask stuff:
instance/
.webassets-cache
# Scrapy stuff:
.scrapy
# Sphinx documentation
docs/_build/
# PyBuilder
.pybuilder/
target/
# Jupyter Notebook
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# IPython
profile_default/
ipython_config.py
# pyenv
# For a library or package, you might want to ignore these files since the code is
# intended to run in multiple environments; otherwise, check them in:
# .python-version
# pipenv
# According to pypa/pipenv#598, it is recommended to include Pipfile.lock in version control.
# However, in case of collaboration, if having platform-specific dependencies or dependencies
# having no cross-platform support, pipenv may install dependencies that don't work, or not
# install all needed dependencies.
#Pipfile.lock
# poetry
# Similar to Pipfile.lock, it is generally recommended to include poetry.lock in version control.
# This is especially recommended for binary packages to ensure reproducibility, and is more
# commonly ignored for libraries.
# https://python-poetry.org/docs/basic-usage/#commit-your-poetrylock-file-to-version-control
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#pdm.lock
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# in version control.
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# option (not recommended) you can uncomment the following to ignore the entire idea folder.
#.idea/

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backend/Pipfile
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[[source]]
url = "https://pypi.org/simple"
verify_ssl = true
name = "pypi"
[packages]
numpy = "*"
scipy = "*"
fastapi = "*"
[dev-packages]
[requires]
python_version = "3.12"
[[source]]
url = "https://pypi.org/simple"
verify_ssl = true
name = "pypi"
[packages]
numpy = "*"
scipy = "*"
fastapi = "*"
osmpythontools = "*"
[dev-packages]

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import app.src

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from .src.optimizer import solve_optimization
from .src.landmarks_manager import Landmark
from fastapi import FastAPI
app = FastAPI()
@app.get("/optimize/{max_steps}/{print_details}")
def main(max_steps: int, print_details: bool):
# CONSTRAINT TO RESPECT MAX NUMBER OF STEPS
#max_steps = 16
# Initialize all landmarks (+ start and goal). Order matters here
landmarks = []
landmarks.append(Landmark("départ", -1, (0, 0)))
landmarks.append(Landmark("tour eiffel", 99, (0,2))) # PUT IN JSON
landmarks.append(Landmark("arc de triomphe", 99, (0,4)))
landmarks.append(Landmark("louvre", 99, (0,6)))
landmarks.append(Landmark("montmartre", 99, (0,10)))
landmarks.append(Landmark("concorde", 99, (0,8)))
landmarks.append(Landmark("arrivée", -1, (0, 0)))
visiting_order = solve_optimization(landmarks, max_steps, print_details)
#return visiting_order
return("max steps :", max_steps, "\n", visiting_order)
"""if __name__ == "__main__":
main()"""

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backend/app/src/landmarks_manager.py Normal file
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from OSMPythonTools.api import Api
from OSMPythonTools.overpass import Overpass
from dataclasses import dataclass
# Defines the landmark class (aka some place there is to visit)
@dataclass
class Landmarkkkk :
name : str
attractiveness : int
id : int
@dataclass
class Landmark :
name : str
attractiveness : int
loc : tuple
# Converts a OSM id to a landmark
def add_from_id(id: int, score: int) :
try :
s = 'way/' + str(id) # prepare string for query
obj = api.query(s) # object to add
except :
s = 'relation/' + str(id) # prepare string for query
obj = api.query(s) # object to add
return Landmarkkkk(obj.tag('name:fr'), score, id) # create Landmark out of it
# take a lsit of tuples (id, score) to generate a list of landmarks
def generate_landmarks(ids_and_scores: list) :
L = []
for tup in ids_and_scores :
L.append(add_from_id(tup[0], tup[1]))
return L
api = Api()
l = (7515426, 70)
t = (5013364, 100)
n = (201611261, 99)
a = (226413508, 50)
m = (23762981, 30)
ids_and_scores = [t, l, n, a, m]
landmarks = generate_landmarks(ids_and_scores)
for obj in landmarks :
print(obj)

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backend/src/main.py → backend/app/src/main_example.py
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import fastapi
from dataclasses import dataclass
@dataclass
class Destination:
name: str
location: tuple
attractiveness: int
d = Destination()
def get_route() -> list[Destination]:
return {"route": "Hello World"}
endpoint = ("/get_route", get_route)
end
if __name__ == "__main__":
fastapi.run()
import fastapi
from dataclasses import dataclass
@dataclass
class Destination:
name: str
location: tuple
attractiveness: int
d = Destination()
def get_route() -> list[Destination]:
return {"route": "Hello World"}
endpoint = ("/get_route", get_route)
end
if __name__ == "__main__":
fastapi.run()

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backend/app/src/optimizer.py Normal file
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from scipy.optimize import linprog
import numpy as np
from scipy.linalg import block_diag
# landmarks = [Landmark_1, Landmark_2, ...]
# Convert the solution of the optimization into the list of edges to follow. Order is taken into account
def untangle(resx: list) :
N = len(resx) # length of res
L = int(np.sqrt(N)) # number of landmarks. CAST INTO INT but should not be a problem because N = L**2 by def.
n_edges = resx.sum() # number of edges
order = []
nonzeroind = np.nonzero(resx)[0] # the return is a little funny so I use the [0]
nonzero_tup = np.unravel_index(nonzeroind, (L,L))
indx = nonzero_tup[0].tolist()
indy = nonzero_tup[1].tolist()
vert = (indx[0], indy[0])
order.append(vert[0])
order.append(vert[1])
while len(order) < n_edges + 1 :
ind = indx.index(vert[1])
vert = (indx[ind], indy[ind])
order.append(vert[1])
return order
# Just to print the result
def print_res(res, landmarks: list, P) :
X = abs(res.x)
order = untangle(X)
things = []
"""N = int(np.sqrt(len(X)))
for i in range(N):
print(X[i*N:i*N+N])
print("Optimal value:", -res.fun) # Minimization, so we negate to get the maximum
print("Optimal point:", res.x)
for i,x in enumerate(X) : X[i] = round(x,0)
print(order)"""
if (X.sum()+1)**2 == len(X) :
print('\nAll landmarks can be visited within max_steps, the following order is suggested : ')
else :
print('Could not visit all the landmarks, the following order is suggested : ')
for idx in order :
print('- ' + landmarks[idx].name)
things.append(landmarks[idx].name)
steps = path_length(P, abs(res.x))
print("\nSteps walked : " + str(steps))
return things
# Checks for cases of circular symmetry in the result
def has_circle(resx: list) :
N = len(resx) # length of res
L = int(np.sqrt(N)) # number of landmarks. CAST INTO INT but should not be a problem because N = L**2 by def.
n_edges = resx.sum() # number of edges
nonzeroind = np.nonzero(resx)[0] # the return is a little funny so I use the [0]
nonzero_tup = np.unravel_index(nonzeroind, (L,L))
indx = nonzero_tup[0].tolist()
indy = nonzero_tup[1].tolist()
verts = []
for i, x in enumerate(indx) :
verts.append((x, indy[i]))
for vert in verts :
visited = []
visited.append(vert)
while len(visited) < n_edges + 1 :
try :
ind = indx.index(vert[1])
vert = (indx[ind], indy[ind])
if vert in visited :
return visited
else :
visited.append(vert)
except :
break
return []
# Constraint to not have d14 and d41 simultaneously. Does not prevent circular symmetry with more elements
def break_sym(landmarks, A_ub, b_ub):
L = len(landmarks)
upper_ind = np.triu_indices(L,0,L)
up_ind_x = upper_ind[0]
up_ind_y = upper_ind[1]
for i, _ in enumerate(up_ind_x) :
l = [0]*L*L
if up_ind_x[i] != up_ind_y[i] :
l[up_ind_x[i]*L + up_ind_y[i]] = 1
l[up_ind_y[i]*L + up_ind_x[i]] = 1
A_ub = np.vstack((A_ub,l))
b_ub.append(1)
"""for i in range(7):
print(l[i*7:i*7+7])
print("\n")"""
return A_ub, b_ub
# Constraint to not have circular paths. Want to go from start -> finish without unconnected loops
def break_circle(landmarks, A_ub, b_ub, circle) :
N = len(landmarks)
l = [0]*N*N
for index in circle :
x = index[0]
y = index[1]
l[x*N+y] = 1
A_ub = np.vstack((A_ub,l))
b_ub.append(len(circle)-1)
"""print("\n\nPREVENT CIRCLE")
for i in range(7):
print(l[i*7:i*7+7])
print("\n")"""
return A_ub, b_ub
# Constraint to respect max number of travels
def respect_number(landmarks, A_ub, b_ub):
h = []
for i in range(len(landmarks)) : h.append([1]*len(landmarks))
T = block_diag(*h)
"""for l in T :
for i in range(7):
print(l[i*7:i*7+7])
print("\n")"""
return np.vstack((A_ub, T)), b_ub + [1]*len(landmarks)
# Constraint to tie the problem together. Necessary but not sufficient to avoid circles
def respect_order(landmarks: list, A_eq, b_eq):
N = len(landmarks)
for i in range(N-1) : # Prevent stacked ones
if i == 0 :
continue
else :
l = [0]*N
l[i] = -1
l = l*N
for j in range(N) :
l[i*N + j] = 1
A_eq = np.vstack((A_eq,l))
b_eq.append(0)
"""for i in range(7):
print(l[i*7:i*7+7])
print("\n")"""
return A_eq, b_eq
# Compute manhattan distance between 2 locations
def manhattan_distance(loc1: tuple, loc2: tuple):
x1, y1 = loc1
x2, y2 = loc2
return abs(x1 - x2) + abs(y1 - y2)
# Constraint to not stay in position
def init_eq_not_stay(landmarks):
L = len(landmarks)
l = [0]*L*L
for i in range(L) :
for j in range(L) :
if j == i :
l[j + i*L] = 1
l[L-1] = 1 # cannot skip from start to finish
#A_eq = np.array([np.array(xi) for xi in A_eq]) # Must convert A_eq into an np array
l = np.array(np.array(l))
"""for i in range(7):
print(l[i*7:i*7+7])"""
return [l], [0]
# Initialize A and c. Compute the distances from all landmarks to each other and store attractiveness
# We want to maximize the sightseeing : max(c) st. A*x < b and A_eq*x = b_eq
def init_ub_dist(landmarks: list, max_steps: int):
# Objective function coefficients. a*x1 + b*x2 + c*x3 + ...
c = []
# Coefficients of inequality constraints (left-hand side)
A = []
for i, spot1 in enumerate(landmarks) :
dist_table = [0]*len(landmarks)
c.append(-spot1.attractiveness)
for j, spot2 in enumerate(landmarks) :
dist_table[j] = manhattan_distance(spot1.loc, spot2.loc)
A.append(dist_table)
c = c*len(landmarks)
A_ub = []
for line in A :
#print(line)
A_ub += line
return c, A_ub, [max_steps]
# Go through the landmarks and force the optimizer to use landmarks where attractiveness is set to -1
def respect_user_mustsee(landmarks: list, A_eq: list, b_eq: list) :
L = len(landmarks)
H = 0 # sort of heuristic to get an idea of the number of steps needed
for i in landmarks :
if i.name == "départ" : elem_prev = i # list of all matches
for i, elem in enumerate(landmarks) :
if elem.attractiveness == -1 :
l = [0]*L*L
if elem.name != "arrivée" :
for j in range(L) :
l[j +i*L] = 1
else : # This ensures we go to goal
for k in range(L-1) :
l[k*L+L-1] = 1
H += manhattan_distance(elem.loc, elem_prev.loc)
elem_prev = elem
"""for i in range(7):
print(l[i*7:i*7+7])
print("\n")"""
A_eq = np.vstack((A_eq,l))
b_eq.append(1)
return A_eq, b_eq, H
# Computes the path length given path matrix (dist_table) and a result
def path_length(P: list, resx: list) :
return np.dot(P, resx)
# Main optimization pipeline
def solve_optimization (landmarks, max_steps, printing_details) :
# SET CONSTRAINTS FOR INEQUALITY
c, A_ub, b_ub = init_ub_dist(landmarks, max_steps) # Add the distances from each landmark to the other
P = A_ub # store the paths for later. Needed to compute path length
A_ub, b_ub = respect_number(landmarks, A_ub, b_ub) # Respect max number of visits.
# TODO : Problems with circular symmetry
A_ub, b_ub = break_sym(landmarks, A_ub, b_ub) # break the symmetry. Only use the upper diagonal values
# SET CONSTRAINTS FOR EQUALITY
A_eq, b_eq = init_eq_not_stay(landmarks) # Force solution not to stay in same place
A_eq, b_eq, H = respect_user_mustsee(landmarks, A_eq, b_eq) # Check if there are user_defined must_see. Also takes care of start/goal
A_eq, b_eq = respect_order(landmarks, A_eq, b_eq) # Respect order of visit (only works when max_steps is limiting factor)
# Bounds for variables (x can only be 0 or 1)
x_bounds = [(0, 1)] * len(c)
# Solve linear programming problem
res = linprog(c, A_ub=A_ub, b_ub=b_ub, A_eq=A_eq, b_eq = b_eq, bounds=x_bounds, method='highs', integrality=3)
# Raise error if no solution is found
if not res.success :
# Override the max_steps using the heuristic
for i, val in enumerate(b_ub) :
if val == max_steps : b_ub[i] = H
# Solve problem again :
res = linprog(c, A_ub=A_ub, b_ub=b_ub, A_eq=A_eq, b_eq = b_eq, bounds=x_bounds, method='highs', integrality=3)
if not res.success :
s = "No solution could be found, even when increasing max_steps using the heuristic"
return s
#raise ValueError("No solution could be found, even when increasing max_steps using the heuristic")
# If there is a solution, we're good to go, just check for
else :
circle = has_circle(res.x)
i = 0
# Break the circular symmetry if needed
while len(circle) != 0 :
A_ub, b_ub = break_circle(landmarks, A_ub, b_ub, circle)
res = linprog(c, A_ub=A_ub, b_ub=b_ub, A_eq=A_eq, b_eq = b_eq, bounds=x_bounds, method='highs', integrality=3)
circle = has_circle(res.x)
i += 1
if printing_details is True :
if i != 0 :
print(f"Neded to recompute paths {i} times because of unconnected loops...")
X = print_res(res, landmarks, P)
return X
else :
return untangle(res.x)

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from scipy.optimize import linprog
import numpy as np
from scipy.linalg import block_diag
# Defines the landmark class (aka some place there is to visit)
class landmark :
def __init__(self, name: str, attractiveness: int, loc: tuple):
self.name = name
self.attractiveness = attractiveness
self.loc = loc
# Convert the result (edges from j to k like d_25 = edge between vertex 2 and vertex 5) into the list of indices corresponding to the landmarks
def untangle(res: list) :
N = len(res) # length of res
L = int(np.sqrt(N)) # number of landmarks. CAST INTO INT but should not be a problem because N = L**2 by def.
n_landmarks = res.sum() # number of visited landmarks
visit_order = []
cnt = 0
if n_landmarks % 2 == 1 : # if odd number of visited checkpoints
for i in range(L) :
for j in range(L) :
if res[i*L + j] == 1 : # if index is 1
cnt += 1 # increment counter
if cnt % 2 == 1 : # if counter odd
visit_order.append(i)
visit_order.append(j)
else : # if even number of ones
for i in range(L) :
for j in range(L) :
if res[i*L + j] == 1 : # if index is one
cnt += 1 # increment counter
if j % (L-1) == 0 : # if last node
visit_order.append(j) # append only the last index
return visit_order # return
if cnt % 2 == 1 :
visit_order.append(i)
visit_order.append(j)
return visit_order
# Just to print the result
def print_res(res: list, P) :
X = abs(res.x)
order = untangle(X)
# print("Optimal value:", -res.fun) # Minimization, so we negate to get the maximum
# print("Optimal point:", res.x)
# N = int(np.sqrt(len(X)))
# for i in range(N):
# print(X[i*N:i*N+N])
# print(order)
if (X.sum()+1)**2 == len(X) :
print('\nAll landmarks can be visited within max_steps, the following order is most likely not the fastest')
else :
print('Could not visit all the landmarks, the following order could be the fastest but not sure')
print("Order of visit :")
for i, elem in enumerate(landmarks) :
if i in order : print('- ' + elem.name)
steps = path_length(P, abs(res.x))
print("\nSteps walked : " + str(steps))
# Constraint to use only the upper triangular indices for travel
def break_sym(landmarks, A_eq, b_eq):
L = len(landmarks)
l = [0]*L*L
for i in range(L) :
for j in range(L) :
if i >= j :
l[j+i*L] = 1
A_eq = np.vstack((A_eq,l))
b_eq.append(0)
return A_eq, b_eq
# Constraint to respect max number of travels
def respect_number(landmarks, A_ub, b_ub):
h = []
for i in range(len(landmarks)) : h.append([1]*len(landmarks))
T = block_diag(*h)
return np.vstack((A_ub, T)), b_ub + [1]*len(landmarks)
# Constraint to tie the problem together and have a connected path
def respect_order(landmarks: list, A_eq, b_eq):
N = len(landmarks)
for i in range(N-1) : # Prevent stacked ones
if i == 0 :
continue
else :
l = [0]*N
l[i] = -1
l = l*N
for j in range(N) :
l[i*N + j] = 1
A_eq = np.vstack((A_eq,l))
b_eq.append(0)
return A_eq, b_eq
# Compute manhattan distance between 2 locations
def manhattan_distance(loc1: tuple, loc2: tuple):
x1, y1 = loc1
x2, y2 = loc2
return abs(x1 - x2) + abs(y1 - y2)
# Constraint to not stay in position
def init_eq_not_stay(landmarks):
L = len(landmarks)
l = [0]*L*L
for i in range(L) :
for j in range(L) :
if j == i :
l[j + i*L] = 1
#A_eq = np.array([np.array(xi) for xi in A_eq]) # Must convert A_eq into an np array
l = np.array(np.array(l))
return [l], [0]
# Initialize A and c. Compute the distances from all landmarks to each other and store attractiveness
# We want to maximize the sightseeing : max(c) st. A*x < b and A_eq*x = b_eq
def init_ub_dist(landmarks: list, max_steps: int):
# Objective function coefficients. a*x1 + b*x2 + c*x3 + ...
c = []
# Coefficients of inequality constraints (left-hand side)
A = []
for i, spot1 in enumerate(landmarks) :
dist_table = [0]*len(landmarks)
c.append(-spot1.attractiveness)
for j, spot2 in enumerate(landmarks) :
dist_table[j] = manhattan_distance(spot1.loc, spot2.loc)
A.append(dist_table)
c = c*len(landmarks)
A_ub = []
for line in A :
A_ub += line
return c, A_ub, [max_steps]
# Go through the landmarks and force the optimizer to use landmarks where attractiveness is set to -1
def respect_user_mustsee(landmarks: list, A_eq: list, b_eq: list) :
L = len(landmarks)
for i, elem in enumerate(landmarks) :
if elem.attractiveness == -1 :
l = [0]*L*L
if elem.name != "arrivée" :
for j in range(L) :
l[j +i*L] = 1
else : # This ensures we go to goal
for k in range(L-1) :
l[k*L+L-1] = 1
A_eq = np.vstack((A_eq,l))
b_eq.append(1)
return A_eq, b_eq
# Computes the path length given path matrix (dist_table) and a result
def path_length(P: list, resx: list) :
return np.dot(P, resx)
# Initialize all landmarks (+ start and goal). Order matters here
landmarks = []
landmarks.append(landmark("départ", -1, (0, 0)))
landmarks.append(landmark("concorde", -1, (5,5)))
landmarks.append(landmark("tour eiffel", 99, (1,1))) # PUT IN JSON
landmarks.append(landmark("arc de triomphe", 99, (2,3)))
landmarks.append(landmark("louvre", 70, (4,2)))
landmarks.append(landmark("montmartre", 20, (0,2)))
landmarks.append(landmark("arrivée", -1, (0, 0)))
# CONSTRAINT TO RESPECT MAX NUMBER OF STEPS
max_steps = 25
# SET CONSTRAINTS FOR INEQUALITY
c, A_ub, b_ub = init_ub_dist(landmarks, max_steps) # Add the distances from each landmark to the other
P = A_ub # store the paths for later. Needed to compute path length
A_ub, b_ub = respect_number(landmarks, A_ub, b_ub) # Respect max number of visits.
# SET CONSTRAINTS FOR EQUALITY
A_eq, b_eq = init_eq_not_stay(landmarks) # Force solution not to stay in same place
A_eq, b_eq = respect_user_mustsee(landmarks, A_eq, b_eq) # Check if there are user_defined must_see. Also takes care of start/goal
A_eq, b_eq = break_sym(landmarks, A_eq, b_eq) # break the symmetry. Only use the upper diagonal values
A_eq, b_eq = respect_order(landmarks, A_eq, b_eq) # Respect order of visit (only works when max_steps is limiting factor)
# Bounds for variables (x can only be 0 or 1)
x_bounds = [(0, 1)] * len(c)
# Solve linear programming problem
res = linprog(c, A_ub=A_ub, b_ub=b_ub, A_eq=A_eq, b_eq = b_eq, bounds=x_bounds, method='highs', integrality=3)
# Raise error if no solution is found
if not res.success :
raise ValueError("No solution has been found, please adapt your max steps")
# Print result
print_res(res, P)