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@yoyolicoris
Last active June 28, 2020 00:16
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import numpy as np
import networkx as nx
def W(x):
return (x + np.pi) % (2 * np.pi) - np.pi
def mcf(x: np.ndarray, capacity=None):
assert x.ndim == 2, "Input x should be a 2d array!"
# construct index for each node
N, M = x.shape
index = np.arange(N * M).reshape(N, M)
if capacity is None:
cap_args = dict()
else:
cap_args = {'capacity': capacity}
# get pseudo estimation of gradients along x and y axis
psi1 = W(np.diff(x, axis=0))
psi2 = W(np.diff(x, axis=1))
G = nx.Graph()
demands = np.round(-(psi1[:, 1:] - psi1[:, :-1] - psi2[1:, :] + psi2[:-1, :]) * 0.5 / np.pi).astype(np.int)
# for convenience let's pad the demands so it match the shape of image
# this add N + M - 1 dummy nodes with 0 demand
demands = np.pad(demands, ((0, 1),) * 2, 'constant', constant_values=0)
G.add_nodes_from(zip(index.ravel(), [{'demand': d} for d in demands.ravel()]))
# set earth node index to -1, and its demand is the negative of the sum of all demands,
# so the total demands is zero
G.add_node(-1, demand=-demands.sum())
# edges along x and y axis
edges = np.vstack((
np.vstack((index[:, :-1].ravel(), index[:, 1:].ravel())).T,
np.vstack((index[:-1].ravel(), index[1:].ravel())).T)
)
# set the edge weight to 1 when its left (upper) node demands is equal to zero
# I found it achieve very stable result
weights = np.concatenate(((demands[:, :-1] == 0).ravel(), (demands[:-1] == 0).ravel()))
G.add_weighted_edges_from(zip(edges[:, 0], edges[:, 1], weights), **cap_args)
# add the remaining edges that connected to earth node
G.add_edges_from(zip([-1] * M, range(M)), **cap_args)
G.add_edges_from(zip([-1] * M, range((N - 1) * M, N * M - 1)), **cap_args)
G.add_edges_from(zip([-1] * N, range(0, N * M, M)), **cap_args)
G.add_edges_from(zip([-1] * N, range(M - 1, N * M, M)), **cap_args)
# make graph to directed graph, so we can distinguish positive and negative flow
G = G.to_directed()
# perform MCF
cost, flowdict = nx.network_simplex(G)
# construct K matrix with the same shape as the gradients
K2 = np.empty((N, M - 1))
K1 = np.empty((N - 1, M))
# add the flow to their orthogonal edge
# the sign of the flow depends on those 4 vectors direction (clockwise or counter-clockwise)
# when calculating the demands
for i in range(N - 1):
for j in range(M):
if j == 0:
K1[i][0] = -flowdict[-1][i * M] + flowdict[i * M][-1]
else:
K1[i][j] = -flowdict[i * M + j - 1][i * M + j] + flowdict[i * M + j][i * M + j - 1]
for i in range(N):
for j in range(M - 1):
if i == 0:
K2[i][j] = flowdict[-1][j] - flowdict[j][-1]
else:
K2[i][j] = flowdict[(i - 1) * M + j][i * M + j] - flowdict[i * M + j][(i - 1) * M + j]
# the boundary node with index = 0 have only one edge to earth node,
# so set one of its edge's K to zero
K2[0, 0] = 0
# derive correct gradients
psi1 += K1 * 2 * np.pi
psi2 += K2 * 2 * np.pi
# integrate the gradients
y = np.full_like(x, x[0, 0])
y[1:, 0] += np.cumsum(psi1[:, 0])
y[:, 1:] = np.cumsum(psi2, axis=1) + y[:, :1]
return y
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