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UI elements using the new structs #8

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remoll merged 37 commits from feature/unify-api-frontend into main 2024-06-23 19:21:50 +00:00
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1 changed files with 120 additions and 33 deletions
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backend/src/optimizer.py
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@@ -10,11 +10,44 @@ class landmark :
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
def untangle2(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_landmarks = res.sum() # number of visited landmarks
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
# 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(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_landmarks = resx.sum() # number of edges
visit_order = []
cnt = 0
@@ -40,47 +73,67 @@ def untangle(res: list) :
return visit_order
# Just to print the result
def print_res(res: list, P) :
def print_res(res: list, landmarks: list, P) :
X = abs(res.x)
order = untangle(X)
N = int(np.sqrt(len(X)))
for i in range(N):
print(X[i*N:i*N+N])
order = untangle2(X)
order_ideal = [0, 0, 0, 0, 0, 0, 1, 0]
# 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)
#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 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)
for idx in order :
print('- ' + landmarks[idx].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):
# Constraint to not have d14 and d41 simultaneously
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
for i in range(L) :
for j in range(L) :
if i >= j :
l[j+i*L] = 1
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_eq = np.vstack((A_eq,l))
b_eq.append(0)
A_ub = np.vstack((A_ub,l))
b_ub.append(1)
return A_eq, b_eq
"""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 and have a connected path
@@ -99,6 +152,10 @@ def respect_order(landmarks: list, A_eq, b_eq):
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
@@ -111,12 +168,19 @@ def manhattan_distance(loc1: tuple, loc2: tuple):
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
@@ -135,24 +199,37 @@ def init_ub_dist(landmarks: list, max_steps: int):
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
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) :
@@ -161,27 +238,30 @@ def path_length(P: list, resx: list) :
# 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("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)))
# CONSTRAINT TO RESPECT MAX NUMBER OF STEPS
max_steps = 25
max_steps = 16
# 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 = 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, 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)
@@ -192,10 +272,17 @@ res = linprog(c, A_ub=A_ub, b_ub=b_ub, A_eq=A_eq, b_eq = b_eq, bounds=x_bounds,
# Raise error if no solution is found
if not res.success :
raise ValueError("No solution has been found, please adapt your max steps")
print(f"No solution has been found within given timeframe.\nMinimum steps to visit all must_see is : {H}")
# 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)
# Print result
print_res(res, P)
print_res(res, landmarks, P)