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Permafix-optimization-refiner #9

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kscheidecker merged 9 commits from Permafix-optimization-refiner into main 2024-07-17 10:32:33 +00:00
Conversation 0 Commits 9 Files Changed 18 +80 -62
3 changed files with 80 additions and 62 deletions
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2
backend/src/amenities/nature.am
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@@ -8,5 +8,5 @@ geological
'tourism'='alpine_hut'
'tourism'='viewpoint'
'tourism'='zoo'
#'tourism'='artwork'
'tourism'='artwork'
'waterway'='waterfall'

136
backend/src/optimizer_v4.py
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@@ -31,7 +31,7 @@ def print_res(L: List[Landmark], L_tot):
# Prevent the use of a particular solution
def prevent_config(resx, A_ub, b_ub):
def prevent_config(resx):
for i, elem in enumerate(resx):
resx[i] = round(elem)
@@ -52,13 +52,10 @@ def prevent_config(resx, A_ub, b_ub):
if i in vertices_visited :
h[i*L:i*L+L] = ones
A_ub = np.vstack((A_ub, h))
b_ub.append(len(vertices_visited)-1)
return A_ub, b_ub
return h, [len(vertices_visited)-1]
def prevent_circle(circle_vertices: list, L: int, A_eq: list, b_eq: list) :
def prevent_circle(circle_vertices: list, L: int) :
l1 = [0]*L*L
l2 = [0]*L*L
@@ -74,12 +71,7 @@ def prevent_circle(circle_vertices: list, L: int, A_eq: list, b_eq: list) :
l1[g*L + s] = 1
l2[s*L + g] = 1
A_eq = np.vstack((A_eq, l1))
b_eq.append(0)
A_eq = np.vstack((A_eq, l2))
b_eq.append(0)
return A_eq, b_eq
return np.vstack((l1, l2)), [0, 0]
# Prevent the possibility of a given solution bit
def break_circle(circle_vertices: list, L: int, A_ub: list, b_ub: list):
@@ -340,14 +332,16 @@ def init_ub_dist(landmarks: List[Landmark], max_steps: int):
# Constraint to respect only one travel per landmark. Also caps the total number of visited landmarks
def respect_number(L: int, A_ub, b_ub, max_landmarks):
def respect_number(L: int, max_landmarks: int):
ones = [1]*L
zeros = [0]*L
for i in range(L) :
h = zeros*i + ones + zeros*(L-1-i)
A_ub = np.vstack((A_ub, h))
b_ub.append(1)
A = ones + zeros*(L-1)
b = [1]
for i in range(L-1) :
h_new = zeros*i + ones + zeros*(L-1-i)
A = np.vstack((A, h_new))
b.append(1)
if max_landmarks is None :
# Read the parameters from the file
@@ -355,29 +349,35 @@ def respect_number(L: int, A_ub, b_ub, max_landmarks):
parameters = json.loads(f.read())
max_landmarks = parameters['max landmarks']
A_ub = np.vstack((A_ub, ones*L))
b_ub.append(max_landmarks+1)
A = np.vstack((A, ones*L))
b.append(max_landmarks+1)
return A_ub, b_ub
return A, b
# Constraint to not have d14 and d41 simultaneously. Does not prevent circular symmetry with more elements
def break_sym(L, A_ub, b_ub):
# Constraint to not have d14 and d41 simultaneously. Does not prevent cyclic paths with more elements
def break_sym(L):
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) :
A = [0]*L*L # useless row to prevent overhead ? better solution welcomed
# A[up_ind_x[0]*L + up_ind_y[0]] = 1
# A[up_ind_y[0]*L + up_ind_x[0]] = 1
b = [1]
for i, _ in enumerate(up_ind_x[1:]) :
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)
A = np.vstack((A,l))
b.append(1)
return A_ub, b_ub
return A, b
# Constraint to not stay in position. Removes d11, d22, d33, etc.
@@ -395,22 +395,24 @@ def init_eq_not_stay(L: int):
# Go through the landmarks and force the optimizer to use landmarks where attractiveness is set to -1
def respect_user_mustsee(landmarks: List[Landmark], A_eq: list, b_eq: list) :
def respect_user_mustsee(landmarks: List[Landmark]) :
L = len(landmarks)
for i, elem in enumerate(landmarks) :
A = [0]*L*L
b = [0]
for i, elem in enumerate(landmarks[1:]) :
if elem.must_do is True and elem.name not in ['finish', 'start']:
l = [0]*L*L
l[i*L:i*L+L] = [1]*L # set mandatory departures from landmarks tagged as 'must_do'
A_eq = np.vstack((A_eq,l))
b_eq.append(1)
A = np.vstack((A,l))
b.append(1)
return A_eq, b_eq
return A, b
# Constraint to ensure start at start and finish at goal
def respect_start_finish(L: int, A_eq: list, b_eq: list):
def respect_start_finish(L: int):
l_start = [1]*L + [0]*L*(L-1) # sets departures only for start (horizontal ones)
l_start[L-1] = 0 # prevents the jump from start to finish
l_goal = [0]*L*L # sets arrivals only for finish (vertical ones)
@@ -421,32 +423,33 @@ def respect_start_finish(L: int, A_eq: list, b_eq: list):
l_goal[k*L+L-1] = 1
A_eq = np.vstack((A_eq,l_start))
A_eq = np.vstack((A_eq,l_goal))
A_eq = np.vstack((A_eq,l_L))
b_eq.append(1)
b_eq.append(1)
b_eq.append(0)
A = np.vstack((l_start, l_goal))
b = [1, 1]
A = np.vstack((A,l_L))
b.append(0)
return A_eq, b_eq
return A, b
# Constraint to tie the problem together. Necessary but not sufficient to avoid circles
def respect_order(N: int, A_eq, b_eq):
for i in range(N-1) : # Prevent stacked ones
if i == 0 or i == N-1: # Don't touch start or finish
def respect_order(L: int):
A = [0]*L*L # useless row to reduce overhead ? better solution is welcome
b = [0]
for i in range(L-1) : # Prevent stacked ones
if i == 0 or i == L-1: # Don't touch start or finish
continue
else :
l = [0]*N
l = [0]*L
l[i] = -1
l = l*N
for j in range(N) :
l[i*N + j] = 1
l = l*L
for j in range(L) :
l[i*L + j] = 1
A_eq = np.vstack((A_eq,l))
b_eq.append(0)
A = np.vstack((A,l))
b.append(0)
return A_eq, b_eq
return A, b
# Computes the time to reach from each landmark to the next
@@ -514,16 +517,27 @@ def solve_optimization (landmarks :List[Landmark], max_steps: int, printing_deta
L = len(landmarks)
# SET CONSTRAINTS FOR INEQUALITY
c, A_ub, b_ub = init_ub_dist(landmarks, max_steps) # Add the distances from each landmark to the other
A_ub, b_ub = respect_number(L, A_ub, b_ub, max_landmarks) # Respect max number of visits (no more possible stops than landmarks).
A_ub, b_ub = break_sym(L, A_ub, b_ub) # break the 'zig-zag' symmetry
c, A_ub, b_ub = init_ub_dist(landmarks, max_steps) # Add the distances from each landmark to the other
A, b = respect_number(L, max_landmarks) # Respect max number of visits (no more possible stops than landmarks).
A_ub = np.vstack((A_ub, A))
b_ub += b
A, b = break_sym(L) # break the 'zig-zag' symmetry
A_ub = np.vstack((A_ub, A))
b_ub += b
# SET CONSTRAINTS FOR EQUALITY
A_eq, b_eq = init_eq_not_stay(L) # 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 = respect_start_finish(L, A_eq, b_eq) # Force start and finish positions
A_eq, b_eq = respect_order(L, A_eq, b_eq) # Respect order of visit (only works when max_steps is limiting factor)
A_eq, b_eq = init_eq_not_stay(L) # Force solution not to stay in same place
A, b = respect_user_mustsee(landmarks) # Check if there are user_defined must_see. Also takes care of start/goal
A_eq = np.vstack((A_eq, A))
b_eq += b
A, b = respect_start_finish(L) # Force start and finish positions
A_eq = np.vstack((A_eq, A))
b_eq += b
A, b = respect_order(L) # Respect order of visit (only works when max_steps is limiting factor)
A_eq = np.vstack((A_eq, A))
b_eq += b
# SET BOUNDS FOR DECISION VARIABLE (x can only be 0 or 1)
x_bounds = [(0, 1)]*L*L
@@ -541,10 +555,14 @@ def solve_optimization (landmarks :List[Landmark], max_steps: int, printing_deta
i = 0
timeout = 80
while circles is not None and i < timeout:
A_ub, b_ub = prevent_config(res.x, A_ub, b_ub)
A, b = prevent_config(res.x)
A_ub = np.vstack((A_ub, A))
b_ub += b
#A_ub, b_ub = prevent_circle(order, len(landmarks), A_ub, b_ub)
for circle in circles :
A_ub, b_ub = prevent_circle(circle, len(landmarks), A_ub, b_ub)
A, b = prevent_circle(circle, len(landmarks))
A_eq = np.vstack((A_eq, A))
b_eq += b
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)
order, circles = is_connected2(res.x)
#nodes, edges = is_connected2(res.x)

4
backend/src/tester.py
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@@ -99,8 +99,8 @@ def test4(coordinates: tuple[float, float]) -> List[Landmark]:
landmarks_short.insert(0, start)
landmarks_short.append(finish)
max_walking_time = 120 # minutes
detour = 30 # minutes
max_walking_time = 50 # minutes
detour = 0 # minutes
# First stage optimization
base_tour = solve_optimization(landmarks_short, max_walking_time, True)