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