fixed circular symmetry
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		| @@ -40,9 +40,6 @@ def untangle2(resx: list) : | ||||
|  | ||||
|     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 | ||||
| @@ -102,7 +99,49 @@ def print_res(res: list, landmarks: list, P) : | ||||
|     steps = path_length(P, abs(res.x)) | ||||
|     print("\nSteps walked : " + str(steps)) | ||||
|  | ||||
| # Constraint to not have d14 and d41 simultaneously | ||||
|  | ||||
| # 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) | ||||
| @@ -125,6 +164,26 @@ def break_sym(landmarks, A_ub, b_ub): | ||||
|  | ||||
|     return A_ub, b_ub | ||||
|  | ||||
|  | ||||
| def prevent_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 = [] | ||||
| @@ -152,9 +211,9 @@ 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): | ||||
|             """for i in range(7): | ||||
|                 print(l[i*7:i*7+7]) | ||||
|             print("\n") | ||||
|             print("\n")""" | ||||
|  | ||||
|     return A_eq, b_eq | ||||
|  | ||||
| @@ -268,7 +327,19 @@ A_eq, b_eq = respect_order(landmarks, A_eq, b_eq)              # Respect order o | ||||
| 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) | ||||
| circle = has_circle(res.x) | ||||
|  | ||||
| while len(circle) != 0 : | ||||
|     print("The solution has a circular path. Not interpretable.") | ||||
|     print("Need to add constraints until no circle ") | ||||
|  | ||||
|     A_ub, b_ub = prevent_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) | ||||
|  | ||||
|  | ||||
|  | ||||
| # Raise error if no solution is found | ||||
| if not res.success : | ||||
|   | ||||
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