kopia lustrzana https://github.com/inkstitch/inkstitch
Merge pull request #268 from inkstitch/lexelby-auto-fill-run
avoid cutting corners in auto-fill running stitchpull/243/head^2
commit
ece81d0c91
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@ -105,9 +105,12 @@ class Fill(EmbroideryElement):
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last_pt = pt
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else:
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last_pt = pt
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if point_ary:
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if len(point_ary) > 2:
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poly_ary.append(point_ary)
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if not poly_ary:
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self.fatal(_("shape %s is so small that it cannot be filled with stitches. Please make it bigger or delete it.") % self.node.get('id'))
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# shapely's idea of "holes" are to subtract everything in the second set
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# from the first. So let's at least make sure the "first" thing is the
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# biggest path.
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@ -6,9 +6,10 @@ from itertools import groupby, izip
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from collections import deque
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from .fill import intersect_region_with_grating, row_num, stitch_row
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from .running_stitch import running_stitch
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from ..i18n import _
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from ..svg import PIXELS_PER_MM
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from ..utils.geometry import Point as InkstitchPoint
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from ..utils.geometry import Point as InkstitchPoint, cut
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class MaxQueueLengthExceeded(Exception):
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@ -437,58 +438,86 @@ def collapse_sequential_outline_edges(graph, path):
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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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def connect_points(shape, start, end, running_stitch_length, row_spacing):
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"""Create stitches to get from one point on an outline of the shape to another.
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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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An outline is essentially a loop (a path of points that ends where it starts).
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Given point A and B on that loop, we want to take the shortest path from one
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to the other. Due to the way our path-finding algorithm above works, it may
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have had to take the long way around the shape to get from A to B, but we'd
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rather ignore that and just get there the short way.
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"""
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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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# We may be on the outer boundary or on on of the hole boundaries.
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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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# First, figure out the start and end position along the outline. The
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# projection gives us the distance travelled down the outline to get to
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# that point.
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start = shapely.geometry.Point(start)
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start_projection = outline.project(start)
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end = shapely.geometry.Point(end)
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end_projection = outline.project(end)
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direction = math.copysign(1.0, distance)
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one_stitch = running_stitch_length * direction
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# If the points are pretty close, just jump there. There's a slight
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# risk that we're going to sew outside the shape here. The way to
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# avoid that is to use running_stitch() even for these really short
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# connections, but that would be really slow for all of the
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# connections from one row to the next.
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#
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# This seems to do a good job of avoiding going outside the shape in
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# most cases. 1.4 is chosen as approximately the length of the
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# stitch connecting two rows if the side of the shape is at a 45
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# degree angle to the rows of stitches (sqrt(2)).
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if abs(end_projection - start_projection) < row_spacing * 1.4:
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return [InkstitchPoint(end.x, end.y)]
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stitches = [InkstitchPoint(*outline.interpolate(pos).coords[0])]
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# The outline path has a "natural" starting point. Think of this as
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# 0 or 12 on an analog clock.
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for i in xrange(num_stitches):
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pos = (pos + one_stitch) % outline.length
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# Cut the outline into two paths at the starting point. The first
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# section will go from 12 o'clock to the starting point. The second
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# section will go from the starting point all the way around and end
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# up at 12 again.
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result = cut(outline, start_projection)
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stitches.append(InkstitchPoint(*outline.interpolate(pos).coords[0]))
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# result will be None if our starting point happens to already be at
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# 12 o'clock.
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if result is not None and result[1] is not None:
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before, after = result
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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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# Make a new outline, starting from the starting point. This is
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# like rotating the clock so that now our starting point is
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# at 12 o'clock.
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outline = shapely.geometry.LineString(list(after.coords) + list(before.coords))
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# Now figure out where our ending point is on the newly-rotated clock.
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end_projection = outline.project(end)
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# Cut the new path at the ending point. before and after now represent
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# two ways to get from the starting point to the ending point. One
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# will most likely be longer than the other.
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before, after = cut(outline, end_projection)
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if before.length <= after.length:
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points = list(before.coords)
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else:
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# after goes from the ending point to the starting point, so reverse
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# it to get from start to end.
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points = list(reversed(after.coords))
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# Now do running stitch along the path we've found. running_stitch() will
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# avoid cutting sharp corners.
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path = [InkstitchPoint(*p) for p in points]
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return running_stitch(path, running_stitch_length)
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return stitches
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def trim_end(path):
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while path and path[-1].is_outline():
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path.pop()
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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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@ -498,6 +527,6 @@ def path_to_stitches(graph, path, shape, angle, row_spacing, max_stitch_length,
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if edge.is_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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stitches.extend(connect_points(shape, edge[0], edge[1], running_stitch_length, row_spacing))
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return stitches
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@ -9,21 +9,22 @@ def cut(line, distance):
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"""
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if distance <= 0.0 or distance >= line.length:
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return [LineString(line), None]
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coords = list(line.coords)
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for i, p in enumerate(coords):
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# TODO: I think this doesn't work if the path doubles back on itself
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pd = line.project(ShapelyPoint(p))
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if pd == distance:
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coords = list(ShapelyPoint(p) for p in line.coords)
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traveled = 0
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last_point = coords[0]
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for i, p in enumerate(coords[1:], 1):
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traveled += p.distance(last_point)
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last_point = p
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if traveled == distance:
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return [
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LineString(coords[:i+1]),
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LineString(coords[i:])]
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if pd > distance:
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if traveled > distance:
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cp = line.interpolate(distance)
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return [
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LineString(coords[:i] + [(cp.x, cp.y)]),
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LineString([(cp.x, cp.y)] + coords[i:])]
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def cut_path(points, length):
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"""Return a subsection of at the start of the path that is length units long.
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13
messages.po
13
messages.po
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@ -8,7 +8,7 @@ msgid ""
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msgstr ""
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"Project-Id-Version: PROJECT VERSION\n"
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"Report-Msgid-Bugs-To: EMAIL@ADDRESS\n"
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"POT-Creation-Date: 2018-08-17 15:47-0400\n"
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"POT-Creation-Date: 2018-08-17 16:07-0400\n"
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"PO-Revision-Date: YEAR-MO-DA HO:MI+ZONE\n"
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"Last-Translator: FULL NAME <EMAIL@ADDRESS>\n"
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"Language-Team: LANGUAGE <LL@li.org>\n"
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@ -154,6 +154,13 @@ msgid ""
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"they fall in the same column position."
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msgstr ""
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#: lib/elements/fill.py:112
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#, python-format
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msgid ""
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"shape %s is so small that it cannot be filled with stitches. Please make"
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" it bigger or delete it."
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msgstr ""
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#: lib/elements/satin_column.py:10
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msgid "Satin Column"
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msgstr ""
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@ -601,13 +608,13 @@ msgstr ""
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msgid "Stitch #"
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msgstr ""
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#: lib/stitches/auto_fill.py:167
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#: lib/stitches/auto_fill.py:168
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msgid ""
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"Unable to autofill. This most often happens because your shape is made "
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"up of multiple sections that aren't connected."
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msgstr ""
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#: lib/stitches/auto_fill.py:392
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#: lib/stitches/auto_fill.py:393
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msgid ""
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"Unexpected error while generating fill stitches. Please send your SVG "
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"file to lexelby@github."
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