How to solve flow game using Google OR tools? - algorithm

Question

I tried to make solver for flow game using google-OR tools.
I made a few rules for the corner to only contains corner pipes, but other than that, i can not figure out how to make the pipe connected to each other nor how to tell the model to make a pipe that is connecting to each other.
A few snippet
pipe_types = {
0: " ",
1: "-",
2: "|",
3: "┗" ,
4: "┛" ,
5: "┓",
6: "┏",
7: "●"
}
model = cp_model.CpModel()
filled_map = [[0,0,0,0],
[0,0,7,0],
[0,0,0,0],
[0,7,0,0]]
mesh_size = int(np.sqrt(len(np.array(filled_map).flatten())))
target_map = [[model.NewIntVar(1, 6, 'column: %i' % i) for i in range(mesh_size)] for j in range(mesh_size)]
flow_map = init_map(model, target_map, filled_map)
for i in range(len(flow_map)):
for j in range(len(flow_map[0])):
# check if top or bottom side
if (i == 0) or (i == len(flow_map)-1):
model.Add(flow_map[i][j] != 2)
# check if left or right side
if (j == 0) or (j == len(flow_map[0])-1):
model.Add(flow_map[i][j] != 1)
# left up corner
if (i == 0) & (j == 0):
model.Add(flow_map[i][j] != 3)
model.Add(flow_map[i][j] != 4)
model.Add(flow_map[i][j] != 5)
# right up corner
if (i == 0) & (j == len(flow_map[0])-1):
model.Add(flow_map[i][j] != 3)
model.Add(flow_map[i][j] != 4)
model.Add(flow_map[i][j] != 6)
# left bottom corner
if (i == len(flow_map)-1) & (j == 0):
model.Add(flow_map[i][j] != 4)
model.Add(flow_map[i][j] != 5)
model.Add(flow_map[i][j] != 6)
# right bottom corner
if (i == len(flow_map)-1) & (j == len(flow_map[0])-1):
model.Add(flow_map[i][j] != 3)
model.Add(flow_map[i][j] != 5)
model.Add(flow_map[i][j] != 6)
# Solving
status = solver.Solve(model)
res = []
if status == cp_model.OPTIMAL or status == cp_model.FEASIBLE:
for i in range(len(flow_map)):
for j in range(len(flow_map[0])):
res.append(solver.Value(flow_map[i][j]))
print(solver.Value(flow_map[i][j]), end=" ")
print()
This would results horizontal pipes on the center of the mesh. Later on, i would have to figure out how to add color and such on this too.
Is there any pointer on how to make this on OR tools?
Edit 1:
Based on David Eisenstat's answer, I can find solution. Visualizing this solution based on JohanC's answer, I get this result.
Can I get the pathing made from google-OR tools?
Edit 2:
Using hamilton path from "Hamiltonian" path using Python
I could generate somewhat correct pathing.
But it feels so weird since OR tools already calculate the pathing, and I have to recalculate the path. The path generated from "Hamiltonian" path using Python doesn't show all possible combinations. If I can take the path from OR tools, I think that would be my best interest.

Answer 1

As I don't have experience with OR-tools, here is an approach with Z3.
The initial board is represented by numbers for the end points, one number for each color. The idea is a bit similar to how Sudoku is represented.
Each other cell on the board will get either a value for zero, or a number. This number should be equal to exactly two of its neighbors.
The initial endpoints should have exactly one neighbor with its color.
from z3 import Solver, Sum, Int, If, And, Or, sat
def plot_solution(S):
import matplotlib.pyplot as plt
ax = plt.gca()
colors = plt.cm.tab10.colors
for i in range(M):
for j in range(N):
if board[i][j] != 0:
ax.scatter(j, i, s=500, color=colors[board[i][j]])
if S[i][j] != 0:
for k in range(M):
for l in range(N):
if abs(k - i) + abs(l - j) == 1 and S[i][j] == S[k][l]:
ax.plot([j, l], [i, k], color=colors[S[i][j]], lw=15)
ax.set_ylim(M - 0.5, -0.5)
ax.set_xlim(-0.5, N - 0.5)
ax.set_aspect('equal')
ax.set_facecolor('black')
ax.set_yticks([i + 0.5 for i in range(M - 1)], minor=True)
ax.set_xticks([j + 0.5 for j in range(N - 1)], minor=True)
ax.grid(b=True, which='minor', color='white')
ax.set_xticks([])
ax.set_yticks([])
ax.tick_params(axis='both', which='both', length=0)
plt.show()
board = [[1, 0, 0, 2, 3],
[0, 0, 0, 4, 0],
[0, 0, 4, 0, 0],
[0, 2, 3, 0, 5],
[0, 1, 5, 0, 0]]
M = len(board)
N = len(board[0])
B = [[Int(f'B_{i}_{j}') for j in range(N)] for i in range(M)]
s = Solver()
s.add(([If(board[i][j] != 0, B[i][j] == board[i][j], And(B[i][j] >= 0, B[i][j] < 10))
for j in range(N) for i in range(M)]))
for i in range(M):
for j in range(N):
same_neighs_ij = Sum([If(B[i][j] == B[k][l], 1, 0)
for k in range(M) for l in range(N) if abs(k - i) + abs(l - j) == 1])
if board[i][j] != 0:
s.add(same_neighs_ij == 1)
else:
s.add(Or(same_neighs_ij == 2, B[i][j] == 0))
if s.check() == sat:
m = s.model()
S = [[m[B[i][j]].as_long() for j in range(N)] for i in range(M)]
print(S)
plot_solution(S)
Solution:
[[1, 2, 2, 2, 3],
[1, 2, 4, 4, 3],
[1, 2, 4, 3, 3],
[1, 2, 3, 3, 5],
[1, 1, 5, 5, 5]]
As mentioned in the comments, a possible requirement is that all cells would need to be colored. This would need a more complicated approach. Here is an example of such a configuration for which the above code could create a solution that connects all end points without touching all cells:
board = [[0, 1, 2, 0, 0, 0, 0],
[1, 3, 4, 0, 3, 5, 0],
[0, 0, 0, 0, 0, 0, 0],
[0, 2, 0, 4, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0],
[0, 5, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0]]

Answer 2

The best way is probably with AddCircuit. This constraint takes a directed graph where each arc is labeled with a literal and requires that the arcs labeled true form a subgraph where each node has in- and out-degree 1, and further that there is at most one cycle that is not a self-loop. By forcing an arc from the end to the beginning, we can use this constraint type to require that there is a single path from the beginning to the end.
The documentation is somewhat poor, so here's a working code sample. I'll leave the drawing part to you.
import collections
from ortools.sat.python import cp_model
def validate_board_and_count_colors(board):
assert isinstance(board, list)
assert all(isinstance(row, list) for row in board)
assert len(set(map(len, board))) == 1
colors = collections.Counter(square for row in board for square in row)
del colors[0]
assert all(count == 2 for count in colors.values())
num_colors = len(colors)
assert set(colors.keys()) == set(range(1, num_colors + 1))
return num_colors
def main(board):
num_colors = validate_board_and_count_colors(board)
model = cp_model.CpModel()
solution = [
[square or model.NewIntVar(1, num_colors, "") for (j, square) in enumerate(row)]
for (i, row) in enumerate(board)
]
true = model.NewBoolVar("")
model.AddBoolOr([true])
for color in range(1, num_colors + 1):
endpoints = []
arcs = []
for i, row in enumerate(board):
for j, square in enumerate(row):
if square == color:
endpoints.append((i, j))
else:
arcs.append(((i, j), (i, j)))
if i < len(board) - 1:
arcs.append(((i, j), (i + 1, j)))
if j < len(row) - 1:
arcs.append(((i, j), (i, j + 1)))
(i1, j1), (i2, j2) = endpoints
k1 = i1 * len(row) + j1
k2 = i2 * len(row) + j2
arc_variables = [(k2, k1, true)]
for (i1, j1), (i2, j2) in arcs:
k1 = i1 * len(row) + j1
k2 = i2 * len(row) + j2
edge = model.NewBoolVar("")
if k1 == k2:
model.Add(solution[i1][j1] != color).OnlyEnforceIf(edge)
arc_variables.append((k1, k1, edge))
else:
model.Add(solution[i1][j1] == color).OnlyEnforceIf(edge)
model.Add(solution[i2][j2] == color).OnlyEnforceIf(edge)
forward = model.NewBoolVar("")
backward = model.NewBoolVar("")
model.AddBoolOr([edge, forward.Not()])
model.AddBoolOr([edge, backward.Not()])
model.AddBoolOr([edge.Not(), forward, backward])
model.AddBoolOr([forward.Not(), backward.Not()])
arc_variables.append((k1, k2, forward))
arc_variables.append((k2, k1, backward))
model.AddCircuit(arc_variables)
solver = cp_model.CpSolver()
status = solver.Solve(model)
if status == cp_model.OPTIMAL:
for row in solution:
print("".join(str(solver.Value(x)) for x in row))
if __name__ == "__main__":
main(
[
[1, 0, 0, 2, 3],
[0, 0, 0, 4, 0],
[0, 0, 4, 0, 0],
[0, 2, 3, 0, 5],
[0, 1, 5, 0, 0],
]
)