Merge branch 'feature/optimizer-use-fastapi'

backend/.gitignore

@@ -0,0 +1,164 @@
+# 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
+.ipynb_checkpoints
+
+# 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
+#poetry.lock
+
+# pdm
+#   Similar to Pipfile.lock, it is generally recommended to include pdm.lock in version control.
+#pdm.lock
+#   pdm stores project-wide configurations in .pdm.toml, but it is recommended to not include it
+#   in version control.
+#   https://pdm.fming.dev/latest/usage/project/#working-with-version-control
+.pdm.toml
+.pdm-python
+.pdm-build/
+
+# PEP 582; used by e.g. github.com/David-OConnor/pyflow and github.com/pdm-project/pdm
+__pypackages__/
+
+# Celery stuff
+celerybeat-schedule
+celerybeat.pid
+
+# SageMath parsed files
+*.sage.py
+
+# Environments
+.env
+.venv
+env/
+venv/
+ENV/
+env.bak/
+venv.bak/
+
+# Spyder project settings
+.spyderproject
+.spyproject
+
+# Rope project settings
+.ropeproject
+
+# mkdocs documentation
+/site
+
+# mypy
+.mypy_cache/
+.dmypy.json
+dmypy.json
+
+# Pyre type checker
+.pyre/
+
+# pytype static type analyzer
+.pytype/
+
+# Cython debug symbols
+cython_debug/
+
+# PyCharm
+#  JetBrains specific template is maintained in a separate JetBrains.gitignore that can
+#  be found at https://github.com/github/gitignore/blob/main/Global/JetBrains.gitignore
+#  and can be added to the global gitignore or merged into this file.  For a more nuclear
+#  option (not recommended) you can uncomment the following to ignore the entire idea folder.
+#.idea/

backend/Pipfile

@@ -7,8 +7,6 @@ name = "pypi"
 numpy = "*"
 scipy = "*"
 fastapi = "*"
+osmpythontools = "*"
 
 [dev-packages]
-
-[requires]
-python_version = "3.12"

backend/app/dependencies.py

@@ -0,0 +1 @@
+import app.src

backend/app/main.py

@@ -0,0 +1,34 @@
+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()"""

backend/app/src/landmarks_manager.py

@@ -0,0 +1,57 @@
+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)

backend/app/src/optimizer.py

@@ -0,0 +1,323 @@
+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)
+
+
+
+
+
+

backend/src/optimizer.py

@@ -1,206 +0,0 @@
-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)
-
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