inkstitch/lib/stitches/auto_fill.py

451 wiersze
15 KiB
Python

import sys
import shapely
import networkx
import math
from itertools import groupby
from collections import deque
from .fill import intersect_region_with_grating, row_num, stitch_row
from ..i18n import _
from ..svg import PIXELS_PER_MM
from ..utils.geometry import Point as InkstitchPoint
class MaxQueueLengthExceeded(Exception):
pass
def auto_fill(shape, angle, row_spacing, end_row_spacing, max_stitch_length, running_stitch_length, staggers, starting_point=None):
stitches = []
rows_of_segments = intersect_region_with_grating(shape, angle, row_spacing, end_row_spacing)
segments = [segment for row in rows_of_segments for segment in row]
graph = build_graph(shape, segments, angle, row_spacing)
path = find_stitch_path(graph, segments)
if starting_point:
stitches.extend(connect_points(shape, starting_point, path[0][0], running_stitch_length))
stitches.extend(path_to_stitches(graph, path, shape, angle, row_spacing, max_stitch_length, running_stitch_length, staggers))
return stitches
def which_outline(shape, coords):
"""return the index of the outline on which the point resides
Index 0 is the outer boundary of the fill region. 1+ are the
outlines of the holes.
"""
# I'd use an intersection check, but floating point errors make it
# fail sometimes.
point = shapely.geometry.Point(*coords)
outlines = enumerate(list(shape.boundary))
closest = min(outlines, key=lambda (index, outline): outline.distance(point))
return closest[0]
def project(shape, coords, outline_index):
"""project the point onto the specified outline
This returns the distance along the outline at which the point resides.
"""
outline = list(shape.boundary)[outline_index]
return outline.project(shapely.geometry.Point(*coords))
def build_graph(shape, segments, angle, row_spacing):
"""build a graph representation of the grating segments
This function builds a specialized graph (as in graph theory) that will
help us determine a stitching path. The idea comes from this paper:
http://www.sciencedirect.com/science/article/pii/S0925772100000158
The goal is to build a graph that we know must have an Eulerian Path.
An Eulerian Path is a path from edge to edge in the graph that visits
every edge exactly once and ends at the node it started at. Algorithms
exist to build such a path, and we'll use Hierholzer's algorithm.
A graph must have an Eulerian Path if every node in the graph has an
even number of edges touching it. Our goal here is to build a graph
that will have this property.
Based on the paper linked above, we'll build the graph as follows:
* nodes are the endpoints of the grating segments, where they meet
with the outer outline of the region the outlines of the interior
holes in the region.
* edges are:
* each section of the outer and inner outlines of the region,
between nodes
* double every other edge in the outer and inner hole outlines
Doubling up on some of the edges seems as if it will just mean we have
to stitch those spots twice. This may be true, but it also ensures
that every node has 4 edges touching it, ensuring that a valid stitch
path must exist.
"""
graph = networkx.MultiGraph()
# First, add the grating segments as edges. We'll use the coordinates
# of the endpoints as nodes, which networkx will add automatically.
for segment in segments:
# networkx allows us to label nodes with arbitrary data. We'll
# mark this one as a grating segment.
graph.add_edge(*segment, key="segment")
for node in graph.nodes():
outline_index = which_outline(shape, node)
outline_projection = project(shape, node, outline_index)
# Tag each node with its index and projection.
graph.add_node(node, index=outline_index, projection=outline_projection)
nodes = list(graph.nodes(data=True)) # returns a list of tuples: [(node, {data}), (node, {data}) ...]
nodes.sort(key=lambda node: (node[1]['index'], node[1]['projection']))
for outline_index, nodes in groupby(nodes, key=lambda node: node[1]['index']):
nodes = [ node for node, data in nodes ]
# heuristic: change the order I visit the nodes in the outline if necessary.
# If the start and endpoints are in the same row, I can't tell which row
# I should treat it as being in.
for i in xrange(len(nodes)):
row0 = row_num(InkstitchPoint(*nodes[0]), angle, row_spacing)
row1 = row_num(InkstitchPoint(*nodes[1]), angle, row_spacing)
if row0 == row1:
nodes = nodes[1:] + [nodes[0]]
else:
break
# heuristic: it's useful to try to keep the duplicated edges in the same rows.
# this prevents the BFS from having to search a ton of edges.
min_row_num = min(row0, row1)
if min_row_num % 2 == 0:
edge_set = 0
else:
edge_set = 1
#print >> sys.stderr, outline_index, "es", edge_set, "rn", row_num, inkstitch.Point(*nodes[0]) * self.north(angle), inkstitch.Point(*nodes[1]) * self.north(angle)
# add an edge between each successive node
for i, (node1, node2) in enumerate(zip(nodes, nodes[1:] + [nodes[0]])):
graph.add_edge(node1, node2, key="outline")
# duplicate every other edge around this outline
if i % 2 == edge_set:
graph.add_edge(node1, node2, key="extra")
if not networkx.is_eulerian(graph):
raise Exception(_("Unable to autofill. This most often happens because your shape is made up of multiple sections that aren't connected."))
return graph
def node_list_to_edge_list(node_list):
return zip(node_list[:-1], node_list[1:])
def bfs_for_loop(graph, starting_node, max_queue_length=2000):
to_search = deque()
to_search.appendleft(([starting_node], set(), 0))
while to_search:
if len(to_search) > max_queue_length:
raise MaxQueueLengthExceeded()
path, visited_edges, visited_segments = to_search.pop()
ending_node = path[-1]
# get a list of neighbors paired with the key of the edge I can follow to get there
neighbors = [
(node, key)
for node, adj in graph.adj[ending_node].iteritems()
for key in adj
]
# heuristic: try grating segments first
neighbors.sort(key=lambda (dest, key): key == "segment", reverse=True)
for next_node, key in neighbors:
# skip if I've already followed this edge
edge = (tuple(sorted((ending_node, next_node))), key)
if edge in visited_edges:
continue
new_path = path + [next_node]
if key == "segment":
new_visited_segments = visited_segments + 1
else:
new_visited_segments = visited_segments
if next_node == starting_node:
# ignore trivial loops (down and back a doubled edge)
if len(new_path) > 3:
return node_list_to_edge_list(new_path), new_visited_segments
new_visited_edges = visited_edges.copy()
new_visited_edges.add(edge)
to_search.appendleft((new_path, new_visited_edges, new_visited_segments))
def find_loop(graph, starting_nodes):
"""find a loop in the graph that is connected to the existing path
Start at a candidate node and search through edges to find a path
back to that node. We'll use a breadth-first search (BFS) in order to
find the shortest available loop.
In most cases, the BFS should not need to search far to find a loop.
The queue should stay relatively short.
An added heuristic will be used: if the BFS queue's length becomes
too long, we'll abort and try a different starting point. Due to
the way we've set up the graph, there's bound to be a better choice
somewhere else.
"""
#loop = self.simple_loop(graph, starting_nodes[-2])
#if loop:
# print >> sys.stderr, "simple_loop success"
# starting_nodes.pop()
# starting_nodes.pop()
# return loop
loop = None
retry = []
max_queue_length = 2000
while not loop:
while not loop and starting_nodes:
starting_node = starting_nodes.pop()
#print >> sys.stderr, "find loop from", starting_node
try:
# Note: if bfs_for_loop() returns None, no loop can be
# constructed from the starting_node (because the
# necessary edges have already been consumed). In that
# case we discard that node and try the next.
loop = bfs_for_loop(graph, starting_node, max_queue_length)
#if not loop:
#print >> dbg, "failed on", starting_node
#dbg.flush()
except MaxQueueLengthExceeded:
#print >> dbg, "gave up on", starting_node
#dbg.flush()
# We're giving up on this node for now. We could try
# this node again later, so add it to the bottm of the
# stack.
retry.append(starting_node)
# Darn, couldn't find a loop. Try harder.
starting_nodes.extendleft(retry)
max_queue_length *= 2
starting_nodes.extendleft(retry)
return loop
def insert_loop(path, loop):
"""insert a sub-loop into an existing path
The path will be a series of edges describing a path through the graph
that ends where it starts. The loop will be similar, and its starting
point will be somewhere along the path.
Insert the loop into the path, resulting in a longer path.
Both the path and the loop will be a list of edges specified as a
start and end point. The points will be specified in order, such
that they will look like this:
((p1, p2), (p2, p3), (p3, p4) ... (pn, p1))
path will be modified in place.
"""
loop_start = loop[0][0]
for i, (start, end) in enumerate(path):
if start == loop_start:
break
path[i:i] = loop
def find_stitch_path(graph, segments):
"""find a path that visits every grating segment exactly once
Theoretically, we just need to find an Eulerian Path in the graph.
However, we don't actually care whether every single edge is visited.
The edges on the outline of the region are only there to help us get
from one grating segment to the next.
We'll build a "cycle" (a path that ends where it starts) using
Hierholzer's algorithm. We'll stop once we've visited every grating
segment.
Hierholzer's algorithm says to select an arbitrary starting node at
each step. In order to produce a reasonable stitch path, we'll select
the vertex carefully such that we get back-and-forth traversal like
mowing a lawn.
To do this, we'll use a simple heuristic: try to start from nodes in
the order of most-recently-visited first.
"""
original_graph = graph
graph = graph.copy()
num_segments = len(segments)
segments_visited = 0
nodes_visited = deque()
# start with a simple loop: down one segment and then back along the
# outer border to the starting point.
path = [segments[0], list(reversed(segments[0]))]
graph.remove_edges_from(path)
segments_visited += 1
nodes_visited.extend(segments[0])
while segments_visited < num_segments:
result = find_loop(graph, nodes_visited)
if not result:
print >> sys.stderr, _("Unexpected error while generating fill stitches. Please send your SVG file to lexelby@github.")
break
loop, segments = result
#print >> dbg, "found loop:", loop
#dbg.flush()
segments_visited += segments
nodes_visited += [edge[0] for edge in loop]
graph.remove_edges_from(loop)
insert_loop(path, loop)
#if segments_visited >= 12:
# break
# Now we have a loop that covers every grating segment. It returns to
# where it started, which is unnecessary, so we'll snip the last bit off.
#while original_graph.has_edge(*path[-1], key="outline"):
# path.pop()
return path
def collapse_sequential_outline_edges(graph, path):
"""collapse sequential edges that fall on the same outline
When the path follows multiple edges along the outline of the region,
replace those edges with the starting and ending points. We'll use
these to stitch along the outline later on.
"""
start_of_run = None
new_path = []
for edge in path:
if graph.has_edge(*edge, key="segment"):
if start_of_run:
# close off the last run
new_path.append((start_of_run, edge[0]))
start_of_run = None
new_path.append(edge)
else:
if not start_of_run:
start_of_run = edge[0]
if start_of_run:
# if we were still in a run, close it off
new_path.append((start_of_run, edge[1]))
return new_path
def outline_distance(outline, p1, p2):
# how far around the outline (and in what direction) do I need to go
# to get from p1 to p2?
p1_projection = outline.project(shapely.geometry.Point(p1))
p2_projection = outline.project(shapely.geometry.Point(p2))
distance = p2_projection - p1_projection
if abs(distance) > outline.length / 2.0:
# if we'd have to go more than halfway around, it's faster to go
# the other way
if distance < 0:
return distance + outline.length
elif distance > 0:
return distance - outline.length
else:
# this ought not happen, but just for completeness, return 0 if
# p1 and p0 are the same point
return 0
else:
return distance
def connect_points(shape, start, end, running_stitch_length):
outline_index = which_outline(shape, start)
outline = shape.boundary[outline_index]
pos = outline.project(shapely.geometry.Point(start))
distance = outline_distance(outline, start, end)
num_stitches = abs(int(distance / running_stitch_length))
direction = math.copysign(1.0, distance)
one_stitch = running_stitch_length * direction
#print >> dbg, "connect_points:", outline_index, start, end, distance, stitches, direction
#dbg.flush()
stitches = [InkstitchPoint(*outline.interpolate(pos).coords[0])]
for i in xrange(num_stitches):
pos = (pos + one_stitch) % outline.length
stitches.append(InkstitchPoint(*outline.interpolate(pos).coords[0]))
end = InkstitchPoint(*end)
if (end - stitches[-1]).length() > 0.1 * PIXELS_PER_MM:
stitches.append(end)
#print >> dbg, "end connect_points"
#dbg.flush()
return stitches
def path_to_stitches(graph, path, shape, angle, row_spacing, max_stitch_length, running_stitch_length, staggers):
path = collapse_sequential_outline_edges(graph, path)
stitches = []
for edge in path:
if graph.has_edge(*edge, key="segment"):
stitch_row(stitches, edge[0], edge[1], angle, row_spacing, max_stitch_length, staggers)
else:
stitches.extend(connect_points(shape, edge[0], edge[1], running_stitch_length))
return stitches