permafixed the optimizer ???
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@ -8,5 +8,5 @@ geological
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'tourism'='alpine_hut'
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'tourism'='viewpoint'
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'tourism'='zoo'
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#'tourism'='artwork'
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'tourism'='artwork'
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'waterway'='waterfall'
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@ -31,7 +31,7 @@ def print_res(L: List[Landmark], L_tot):
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# Prevent the use of a particular solution
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def prevent_config(resx, A_ub, b_ub):
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def prevent_config(resx):
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for i, elem in enumerate(resx):
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resx[i] = round(elem)
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@ -52,13 +52,10 @@ def prevent_config(resx, A_ub, b_ub):
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if i in vertices_visited :
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h[i*L:i*L+L] = ones
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A_ub = np.vstack((A_ub, h))
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b_ub.append(len(vertices_visited)-1)
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return A_ub, b_ub
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return h, [len(vertices_visited)-1]
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def prevent_circle(circle_vertices: list, L: int, A_eq: list, b_eq: list) :
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def prevent_circle(circle_vertices: list, L: int) :
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l1 = [0]*L*L
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l2 = [0]*L*L
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@ -74,12 +71,7 @@ def prevent_circle(circle_vertices: list, L: int, A_eq: list, b_eq: list) :
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l1[g*L + s] = 1
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l2[s*L + g] = 1
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A_eq = np.vstack((A_eq, l1))
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b_eq.append(0)
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A_eq = np.vstack((A_eq, l2))
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b_eq.append(0)
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return A_eq, b_eq
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return np.vstack((l1, l2)), [0, 0]
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# Prevent the possibility of a given solution bit
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def break_circle(circle_vertices: list, L: int, A_ub: list, b_ub: list):
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@ -340,14 +332,16 @@ def init_ub_dist(landmarks: List[Landmark], max_steps: int):
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# Constraint to respect only one travel per landmark. Also caps the total number of visited landmarks
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def respect_number(L: int, A_ub, b_ub, max_landmarks):
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def respect_number(L: int, max_landmarks: int):
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ones = [1]*L
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zeros = [0]*L
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for i in range(L) :
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h = zeros*i + ones + zeros*(L-1-i)
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A_ub = np.vstack((A_ub, h))
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b_ub.append(1)
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A = ones + zeros*(L-1)
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b = [1]
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for i in range(L-1) :
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h_new = zeros*i + ones + zeros*(L-1-i)
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A = np.vstack((A, h_new))
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b.append(1)
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if max_landmarks is None :
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# Read the parameters from the file
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@ -355,29 +349,35 @@ def respect_number(L: int, A_ub, b_ub, max_landmarks):
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parameters = json.loads(f.read())
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max_landmarks = parameters['max landmarks']
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A_ub = np.vstack((A_ub, ones*L))
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b_ub.append(max_landmarks+1)
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A = np.vstack((A, ones*L))
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b.append(max_landmarks+1)
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return A_ub, b_ub
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return A, b
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# Constraint to not have d14 and d41 simultaneously. Does not prevent circular symmetry with more elements
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def break_sym(L, A_ub, b_ub):
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# Constraint to not have d14 and d41 simultaneously. Does not prevent cyclic paths with more elements
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def break_sym(L):
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upper_ind = np.triu_indices(L,0,L)
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up_ind_x = upper_ind[0]
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up_ind_y = upper_ind[1]
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for i, _ in enumerate(up_ind_x) :
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A = [0]*L*L # useless row to prevent overhead ? better solution welcomed
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# A[up_ind_x[0]*L + up_ind_y[0]] = 1
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# A[up_ind_y[0]*L + up_ind_x[0]] = 1
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b = [1]
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for i, _ in enumerate(up_ind_x[1:]) :
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l = [0]*L*L
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if up_ind_x[i] != up_ind_y[i] :
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l[up_ind_x[i]*L + up_ind_y[i]] = 1
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l[up_ind_y[i]*L + up_ind_x[i]] = 1
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A_ub = np.vstack((A_ub,l))
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b_ub.append(1)
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A = np.vstack((A,l))
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b.append(1)
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return A_ub, b_ub
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return A, b
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# Constraint to not stay in position. Removes d11, d22, d33, etc.
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@ -395,22 +395,24 @@ def init_eq_not_stay(L: int):
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# Go through the landmarks and force the optimizer to use landmarks where attractiveness is set to -1
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def respect_user_mustsee(landmarks: List[Landmark], A_eq: list, b_eq: list) :
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def respect_user_mustsee(landmarks: List[Landmark]) :
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L = len(landmarks)
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for i, elem in enumerate(landmarks) :
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A = [0]*L*L
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b = [0]
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for i, elem in enumerate(landmarks[1:]) :
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if elem.must_do is True and elem.name not in ['finish', 'start']:
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l = [0]*L*L
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l[i*L:i*L+L] = [1]*L # set mandatory departures from landmarks tagged as 'must_do'
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A_eq = np.vstack((A_eq,l))
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b_eq.append(1)
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A = np.vstack((A,l))
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b.append(1)
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return A_eq, b_eq
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return A, b
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# Constraint to ensure start at start and finish at goal
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def respect_start_finish(L: int, A_eq: list, b_eq: list):
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def respect_start_finish(L: int):
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l_start = [1]*L + [0]*L*(L-1) # sets departures only for start (horizontal ones)
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l_start[L-1] = 0 # prevents the jump from start to finish
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l_goal = [0]*L*L # sets arrivals only for finish (vertical ones)
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@ -421,32 +423,33 @@ def respect_start_finish(L: int, A_eq: list, b_eq: list):
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l_goal[k*L+L-1] = 1
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A_eq = np.vstack((A_eq,l_start))
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A_eq = np.vstack((A_eq,l_goal))
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A_eq = np.vstack((A_eq,l_L))
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b_eq.append(1)
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b_eq.append(1)
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b_eq.append(0)
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A = np.vstack((l_start, l_goal))
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b = [1, 1]
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A = np.vstack((A,l_L))
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b.append(0)
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return A_eq, b_eq
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return A, b
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# Constraint to tie the problem together. Necessary but not sufficient to avoid circles
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def respect_order(N: int, A_eq, b_eq):
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for i in range(N-1) : # Prevent stacked ones
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if i == 0 or i == N-1: # Don't touch start or finish
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def respect_order(L: int):
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A = [0]*L*L # useless row to reduce overhead ? better solution is welcome
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b = [0]
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for i in range(L-1) : # Prevent stacked ones
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if i == 0 or i == L-1: # Don't touch start or finish
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continue
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else :
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l = [0]*N
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l = [0]*L
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l[i] = -1
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l = l*N
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for j in range(N) :
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l[i*N + j] = 1
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l = l*L
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for j in range(L) :
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l[i*L + j] = 1
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A_eq = np.vstack((A_eq,l))
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b_eq.append(0)
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A = np.vstack((A,l))
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b.append(0)
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return A_eq, b_eq
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return A, b
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# Computes the time to reach from each landmark to the next
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@ -514,15 +517,26 @@ def solve_optimization (landmarks :List[Landmark], max_steps: int, printing_deta
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L = len(landmarks)
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# SET CONSTRAINTS FOR INEQUALITY
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c, A_ub, b_ub = init_ub_dist(landmarks, max_steps) # Add the distances from each landmark to the other
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A_ub, b_ub = respect_number(L, A_ub, b_ub, max_landmarks) # Respect max number of visits (no more possible stops than landmarks).
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A_ub, b_ub = break_sym(L, A_ub, b_ub) # break the 'zig-zag' symmetry
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c, A_ub, b_ub = init_ub_dist(landmarks, max_steps) # Add the distances from each landmark to the other
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A, b = respect_number(L, max_landmarks) # Respect max number of visits (no more possible stops than landmarks).
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A_ub = np.vstack((A_ub, A))
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b_ub += b
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A, b = break_sym(L) # break the 'zig-zag' symmetry
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A_ub = np.vstack((A_ub, A))
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b_ub += b
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# SET CONSTRAINTS FOR EQUALITY
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A_eq, b_eq = init_eq_not_stay(L) # Force solution not to stay in same place
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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
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A_eq, b_eq = respect_start_finish(L, A_eq, b_eq) # Force start and finish positions
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A_eq, b_eq = respect_order(L, A_eq, b_eq) # Respect order of visit (only works when max_steps is limiting factor)
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A_eq, b_eq = init_eq_not_stay(L) # Force solution not to stay in same place
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A, b = respect_user_mustsee(landmarks) # Check if there are user_defined must_see. Also takes care of start/goal
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A_eq = np.vstack((A_eq, A))
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b_eq += b
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A, b = respect_start_finish(L) # Force start and finish positions
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A_eq = np.vstack((A_eq, A))
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b_eq += b
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A, b = respect_order(L) # Respect order of visit (only works when max_steps is limiting factor)
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A_eq = np.vstack((A_eq, A))
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b_eq += b
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# SET BOUNDS FOR DECISION VARIABLE (x can only be 0 or 1)
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x_bounds = [(0, 1)]*L*L
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@ -541,10 +555,14 @@ def solve_optimization (landmarks :List[Landmark], max_steps: int, printing_deta
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i = 0
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timeout = 80
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while circles is not None and i < timeout:
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A_ub, b_ub = prevent_config(res.x, A_ub, b_ub)
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A, b = prevent_config(res.x)
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A_ub = np.vstack((A_ub, A))
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b_ub += b
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#A_ub, b_ub = prevent_circle(order, len(landmarks), A_ub, b_ub)
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for circle in circles :
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A_ub, b_ub = prevent_circle(circle, len(landmarks), A_ub, b_ub)
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A, b = prevent_circle(circle, len(landmarks))
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A_eq = np.vstack((A_eq, A))
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b_eq += b
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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)
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order, circles = is_connected2(res.x)
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#nodes, edges = is_connected2(res.x)
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@ -99,8 +99,8 @@ def test4(coordinates: tuple[float, float]) -> List[Landmark]:
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landmarks_short.insert(0, start)
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landmarks_short.append(finish)
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max_walking_time = 120 # minutes
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detour = 30 # minutes
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max_walking_time = 50 # minutes
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detour = 0 # minutes
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# First stage optimization
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base_tour = solve_optimization(landmarks_short, max_walking_time, True)
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