cleaned up folders and defined proper structs

backend/src/landmarks_manager.py

@@ -0,0 +1,57 @@
+from OSMPythonTools.api import Api
+from OSMPythonTools.overpass import Overpass
+from dataclasses import dataclass
+from pydantic import BaseModel
+
+
+# Defines the landmark class (aka some place there is to visit)
+@dataclass
+class Landmarkkkk :
+    name : str
+    attractiveness : int
+    id : int
+    
+class Landmark(BaseModel) :
+    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/src/main.py

@@ -0,0 +1,42 @@
+from optimizer import solve_optimization
+from .structs.landmarks import LandmarkTest
+from .structs.preferences import Preferences
+from fastapi import FastAPI
+
+app = FastAPI()
+
+# This should become main at some point
+@app.post("optimizer/{longitude}/{latitude}")
+def get_data(longitude: float, latitude: float, preferences: Preferences) :
+    # From frontend get longitude, latitude and prefence list
+
+    return
+
+@app.get("optimizer/{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(LandmarkTest("départ", -1, (0, 0)))
+    landmarks.append(LandmarkTest("tour eiffel", 99, (0,2)))                           # PUT IN JSON
+    landmarks.append(LandmarkTest("arc de triomphe", 99, (0,4)))
+    landmarks.append(LandmarkTest("louvre", 99, (0,6)))
+    landmarks.append(LandmarkTest("montmartre", 99, (0,10)))
+    landmarks.append(LandmarkTest("concorde", 99, (0,8)))
+    landmarks.append(LandmarkTest("arrivée", -1, (0, 0)))
+
+
+    visiting_order = solve_optimization(landmarks, max_steps, print_details)
+
+    #return visiting_order
+
+    # should return landmarks = the list of Landmark (ordered list)
+    return("max steps :", max_steps, "\n", visiting_order)
+
+
+"""if __name__ == "__main__":
+    main()"""

backend/src/main_example.py

@@ -0,0 +1,23 @@
+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()

backend/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/structs/landmarks.py

@@ -0,0 +1,24 @@
+from pydantic import BaseModel
+from .landmarktype import LandmarkType
+from .preferences import Preferences
+
+class LandmarkTest(BaseModel) :
+    name : str
+    attractiveness : int
+    loc : tuple
+
+# Output to frontend
+class Landmark(BaseModel) :
+    name : str
+    type: LandmarkType      # De facto mapping depending on how the query was executed with overpass. Should still EXACTLY correspond to the preferences
+    location : tuple
+
+
+    def score(preferences: Preferences):
+        # loop through the preferences and assign a score
+
+
+        return 29
+
+
+

backend/src/structs/landmarktype.py

@@ -0,0 +1,4 @@
+from pydantic import BaseModel
+
+class LandmarkType(BaseModel):
+    landmark_type: str

backend/src/structs/preferences.py

@@ -0,0 +1,28 @@
+from pydantic import BaseModel
+from .landmarktype import LandmarkType
+
+class Preference(BaseModel) :
+    name: str
+    type: LandmarkType
+    score: int
+
+# Input for optimization
+class Preferences(BaseModel) :
+    # Sightseeing / History & Culture (Musées, bâtiments historiques, opéras, églises)
+    sightseeing : Preference
+
+    # Nature (parcs, jardins, rivières, plages)
+    nature: Preference
+
+    # Shopping (diriger plutôt vers des zones / rues commerçantes)
+    shopping : Preference
+
+    # Food (price low or high. Combien on veut dépenser pour manger à midi/soir)
+    food_budget : Preference
+    
+    # Tolérance au détour (ce qui détermine (+ ou -) le chemin emprunté)
+    detour_tol : Preference
+
+
+
+