Lib/fontTools/varLib/iup.py - third_party/fonttools - Git at Google

 from __future__ import print_function, division, absolute_import
 from __future__ import unicode_literals
 from fontTools.misc.py23 import *


 def iup_segment(coords, rc1, rd1, rc2, rd2):
 	# rc1 = reference coord 1
 	# rd1 = reference delta 1
 	out_arrays = [None, None]
 	for j in 0,1:
 		out_arrays[j] = out = []
 		x1, x2, d1, d2 = rc1[j], rc2[j], rd1[j], rd2[j]


 		if x1 == x2:
 			n = len(coords)
 			if d1 == d2:
 				out.extend([d1]*n)
 			else:
 				out.extend([0]*n)
 			continue

 		if x1 > x2:
 			x1, x2 = x2, x1
 			d1, d2 = d2, d1

 		# x1 < x2
 		scale = (d2 - d1) / (x2 - x1)
 		for pair in coords:
 			x = pair[j]

 			if x <= x1:
 				d = d1
 			elif x >= x2:
 				d = d2
 			else:
 				# Interpolate
 				d = d1 + (x - x1) * scale

 			out.append(d)

 	return zip(*out_arrays)

 def iup_contour(delta, coords):
 	assert len(delta) == len(coords)
 	if None not in delta:
 		return delta

 	n = len(delta)
 	# indices of points with explicit deltas
 	indices = [i for i,v in enumerate(delta) if v is not None]
 	if not indices:
 		# All deltas are None.  Return 0,0 for all.
 		return [(0,0)]*n

 	out = []
 	it = iter(indices)
 	start = next(it)
 	if start != 0:
 		# Initial segment that wraps around
 		i1, i2, ri1, ri2 = 0, start, start, indices[-1]
 		out.extend(iup_segment(coords[i1:i2], coords[ri1], delta[ri1], coords[ri2], delta[ri2]))
 	out.append(delta[start])
 	for end in it:
 		if end - start > 1:
 			i1, i2, ri1, ri2 = start+1, end, start, end
 			out.extend(iup_segment(coords[i1:i2], coords[ri1], delta[ri1], coords[ri2], delta[ri2]))
 		out.append(delta[end])
 		start = end
 	if start != n-1:
 		# Final segment that wraps around
 		i1, i2, ri1, ri2 = start+1, n, start, indices[0]
 		out.extend(iup_segment(coords[i1:i2], coords[ri1], delta[ri1], coords[ri2], delta[ri2]))

 	assert len(delta) == len(out), (len(delta), len(out))
 	return out

 def iup_delta(delta, coords, ends):
 	assert sorted(ends) == ends and len(coords) == (ends[-1]+1 if ends else 0) + 4
 	n = len(coords)
 	ends = ends + [n-4, n-3, n-2, n-1]
 	out = []
 	start = 0
 	for end in ends:
 		end += 1
 		contour = iup_contour(delta[start:end], coords[start:end])
 		out.extend(contour)
 		start = end

 	return out

 # Optimizer

 def can_iup_in_between(deltas, coords, i, j, tolerance):
 	assert j - i >= 2
 	interp = list(iup_segment(coords[i+1:j], coords[i], deltas[i], coords[j], deltas[j]))
 	deltas = deltas[i+1:j]

 	assert len(deltas) == len(interp)

 	return all(abs(complex(x-p, y-q)) <= tolerance for (x,y),(p,q) in zip(deltas, interp))

 def _iup_contour_bound_forced_set(delta, coords, tolerance=0):
 	"""The forced set is a conservative set of points on the contour that must be encoded
 	explicitly (ie. cannot be interpolated).  Calculating this set allows for significantly
 	speeding up the dynamic-programming, as well as resolve circularity in DP.

 	The set is precise; that is, if an index is in the returned set, then there is no way
 	that IUP can generate delta for that point, given coords and delta.
 	"""
 	assert len(delta) == len(coords)

 	forced = set()
 	# Track "last" and "next" points on the contour as we sweep.
 	nd, nc = delta[0], coords[0]
 	ld, lc = delta[-1], coords[-1]
 	for i in range(len(delta)-1, -1, -1):
 		d, c = ld, lc
 		ld, lc = delta[i-1], coords[i-1]

 		for j in (0,1): # For X and for Y
 			cj = c[j]
 			dj = d[j]
 			lcj = lc[j]
 			ldj = ld[j]
 			ncj = nc[j]
 			ndj = nd[j]

 			if lcj <= ncj:
 				c1, c2 = lcj, ncj
 				d1, d2 = ldj, ndj
 			else:
 				c1, c2 = ncj, lcj
 				d1, d2 = ndj, ldj

 			# If coordinate for current point is between coordinate of adjacent
 			# points on the two sides, but the delta for current point is NOT
 			# between delta for those adjacent points (considering tolerance
 			# allowance), then there is no way that current point can be IUP-ed.
 			# Mark it forced.
 			force = False
 			if c1 <= cj <= c2:
 				if not (min(d1,d2)-tolerance <= dj <= max(d1,d2)+tolerance):
 					force = True
 			else: # cj < c1 or c2 < cj
 				if c1 == c2:
 					if d1 == d2:
 						if abs(dj - d1) > tolerance:
 							force = True
 					else:
 						if abs(dj) > tolerance:
 							# Disabled the following because the "d1 == d2" does
 							# check does not take tolerance into consideration...
 							pass # force = True
 				elif d1 != d2:
 					if cj < c1:
 						if dj != d1 and ((dj-tolerance < d1) != (d1 < d2)):
 							force = True
 					else: # c2 < cj
 						if d2 != dj and ((d2 < dj+tolerance) != (d1 < d2)):
 							force = True

 			if force:
 				forced.add(i)
 				break

 		nd, nc = d, c

 	return forced

 def _iup_contour_optimize_dp(delta, coords, forced={}, tolerance=0, lookback=None):
 	"""Straightforward Dynamic-Programming.  For each index i, find least-costly encoding of
 	points 0 to i where i is explicitly encoded.  We find this by considering all previous
 	explicit points j and check whether interpolation can fill points between j and i.

 	Note that solution always encodes last point explicitly.  Higher-level is responsible
 	for removing that restriction.

 	As major speedup, we stop looking further whenever we see a "forced" point."""

 	n = len(delta)
 	if lookback is None:
 		lookback = n
 	costs = {-1:0}
 	chain = {-1:None}
 	for i in range(0, n):
 		best_cost = costs[i-1] + 1

 		costs[i] = best_cost
 		chain[i] = i - 1

 		if i - 1 in forced:
 			continue

 		for j in range(i-2, max(i-lookback, -2), -1):

 			cost = costs[j] + 1

 			if cost < best_cost and can_iup_in_between(delta, coords, j, i, tolerance):
 				costs[i] = best_cost = cost
 				chain[i] = j

 			if j in forced:
 				break

 	return chain, costs

 def _rot_list(l, k):
 	"""Rotate list by k items forward.  Ie. item at position 0 will be
 	at position k in returned list.  Negative k is allowed."""
 	n = len(l)
 	k %= n
 	if not k: return l
 	return l[n-k:] + l[:n-k]

 def _rot_set(s, k, n):
 	k %= n
 	if not k: return s
 	return {(v + k) % n for v in s}

 def iup_contour_optimize(delta, coords, tolerance=0.):
 	n = len(delta)

 	# Get the easy cases out of the way:

 	# If all are within tolerance distance of 0, encode nothing:
 	if all(abs(complex(*p)) <= tolerance for p in delta):
 		return [None] * n

 	# If there's exactly one point, return it:
 	if n == 1:
 		return delta

 	# If all deltas are exactly the same, return just one (the first one):
 	d0 = delta[0]
 	if all(d0 == d for d in delta):
 		return [d0] + [None] * (n-1)

 	# Else, solve the general problem using Dynamic Programming.

 	forced = _iup_contour_bound_forced_set(delta, coords, tolerance)
 	# The _iup_contour_optimize_dp() routine returns the optimal encoding
 	# solution given the constraint that the last point is always encoded.
 	# To remove this constraint, we use two different methods, depending on
 	# whether forced set is non-empty or not:

 	if forced:
 		# Forced set is non-empty: rotate the contour start point
 		# such that the last point in the list is a forced point.
 		k = (n-1) - max(forced)
 		assert k >= 0

 		delta  = _rot_list(delta, k)
 		coords = _rot_list(coords, k)
 		forced = _rot_set(forced, k, n)

 		chain, costs = _iup_contour_optimize_dp(delta, coords, forced, tolerance)

 		# Assemble solution.
 		solution = set()
 		i = n - 1
 		while i is not None:
 			solution.add(i)
 			i = chain[i]
 		assert forced <= solution, (forced, solution)
 		delta = [delta[i] if i in solution else None for i in range(n)]

 		delta = _rot_list(delta, -k)
 	else:
 		# Repeat the contour an extra time, solve the 2*n case, then look for solutions of the
 		# circular n-length problem in the solution for 2*n linear case.  I cannot prove that
 		# this always produces the optimal solution...
 		chain, costs = _iup_contour_optimize_dp(delta+delta, coords+coords, forced, tolerance, n)
 		best_sol, best_cost = None, n+1

 		for start in range(n-1, 2*n-1):
 			# Assemble solution.
 			solution = set()
 			i = start
 			while i > start - n:
 				solution.add(i % n)
 				i = chain[i]
 			if i == start - n:
 				cost = costs[start] - costs[start - n]
 				if cost <= best_cost:
 					best_sol, best_cost = solution, cost

 		delta = [delta[i] if i in best_sol else None for i in range(n)]


 	return delta

 def iup_delta_optimize(delta, coords, ends, tolerance=0.):
 	assert sorted(ends) == ends and len(coords) == (ends[-1]+1 if ends else 0) + 4
 	n = len(coords)
 	ends = ends + [n-4, n-3, n-2, n-1]
 	out = []
 	start = 0
 	for end in ends:
 		contour = iup_contour_optimize(delta[start:end+1], coords[start:end+1], tolerance)
 		assert len(contour) == end - start + 1
 		out.extend(contour)
 		start = end+1

 	return out
	from __future__ import print_function, division, absolute_import
	from __future__ import unicode_literals
	from fontTools.misc.py23 import *


	def iup_segment(coords, rc1, rd1, rc2, rd2):
	# rc1 = reference coord 1
	# rd1 = reference delta 1
	out_arrays = [None, None]
	for j in 0,1:
	out_arrays[j] = out = []
	x1, x2, d1, d2 = rc1[j], rc2[j], rd1[j], rd2[j]


	if x1 == x2:
	n = len(coords)
	if d1 == d2:
	out.extend([d1]*n)
	else:
	out.extend([0]*n)
	continue

	if x1 > x2:
	x1, x2 = x2, x1
	d1, d2 = d2, d1

	# x1 < x2
	scale = (d2 - d1) / (x2 - x1)
	for pair in coords:
	x = pair[j]

	if x <= x1:
	d = d1
	elif x >= x2:
	d = d2
	else:
	# Interpolate
	d = d1 + (x - x1) * scale

	out.append(d)

	return zip(*out_arrays)

	def iup_contour(delta, coords):
	assert len(delta) == len(coords)
	if None not in delta:
	return delta

	n = len(delta)
	# indices of points with explicit deltas
	indices = [i for i,v in enumerate(delta) if v is not None]
	if not indices:
	# All deltas are None. Return 0,0 for all.
	return [(0,0)]*n

	out = []
	it = iter(indices)
	start = next(it)
	if start != 0:
	# Initial segment that wraps around
	i1, i2, ri1, ri2 = 0, start, start, indices[-1]
	out.extend(iup_segment(coords[i1:i2], coords[ri1], delta[ri1], coords[ri2], delta[ri2]))
	out.append(delta[start])
	for end in it:
	if end - start > 1:
	i1, i2, ri1, ri2 = start+1, end, start, end
	out.extend(iup_segment(coords[i1:i2], coords[ri1], delta[ri1], coords[ri2], delta[ri2]))
	out.append(delta[end])
	start = end
	if start != n-1:
	# Final segment that wraps around
	i1, i2, ri1, ri2 = start+1, n, start, indices[0]
	out.extend(iup_segment(coords[i1:i2], coords[ri1], delta[ri1], coords[ri2], delta[ri2]))

	assert len(delta) == len(out), (len(delta), len(out))
	return out

	def iup_delta(delta, coords, ends):
	assert sorted(ends) == ends and len(coords) == (ends[-1]+1 if ends else 0) + 4
	n = len(coords)
	ends = ends + [n-4, n-3, n-2, n-1]
	out = []
	start = 0
	for end in ends:
	end += 1
	contour = iup_contour(delta[start:end], coords[start:end])
	out.extend(contour)
	start = end

	return out

	# Optimizer

	def can_iup_in_between(deltas, coords, i, j, tolerance):
	assert j - i >= 2
	interp = list(iup_segment(coords[i+1:j], coords[i], deltas[i], coords[j], deltas[j]))
	deltas = deltas[i+1:j]

	assert len(deltas) == len(interp)

	return all(abs(complex(x-p, y-q)) <= tolerance for (x,y),(p,q) in zip(deltas, interp))

	def _iup_contour_bound_forced_set(delta, coords, tolerance=0):
	"""The forced set is a conservative set of points on the contour that must be encoded
	explicitly (ie. cannot be interpolated). Calculating this set allows for significantly
	speeding up the dynamic-programming, as well as resolve circularity in DP.

	The set is precise; that is, if an index is in the returned set, then there is no way
	that IUP can generate delta for that point, given coords and delta.
	"""
	assert len(delta) == len(coords)

	forced = set()
	# Track "last" and "next" points on the contour as we sweep.
	nd, nc = delta[0], coords[0]
	ld, lc = delta[-1], coords[-1]
	for i in range(len(delta)-1, -1, -1):
	d, c = ld, lc
	ld, lc = delta[i-1], coords[i-1]

	for j in (0,1): # For X and for Y
	cj = c[j]
	dj = d[j]
	lcj = lc[j]
	ldj = ld[j]
	ncj = nc[j]
	ndj = nd[j]

	if lcj <= ncj:
	c1, c2 = lcj, ncj
	d1, d2 = ldj, ndj
	else:
	c1, c2 = ncj, lcj
	d1, d2 = ndj, ldj

	# If coordinate for current point is between coordinate of adjacent
	# points on the two sides, but the delta for current point is NOT
	# between delta for those adjacent points (considering tolerance
	# allowance), then there is no way that current point can be IUP-ed.
	# Mark it forced.
	force = False
	if c1 <= cj <= c2:
	if not (min(d1,d2)-tolerance <= dj <= max(d1,d2)+tolerance):
	force = True
	else: # cj < c1 or c2 < cj
	if c1 == c2:
	if d1 == d2:
	if abs(dj - d1) > tolerance:
	force = True
	else:
	if abs(dj) > tolerance:
	# Disabled the following because the "d1 == d2" does
	# check does not take tolerance into consideration...
	pass # force = True
	elif d1 != d2:
	if cj < c1:
	if dj != d1 and ((dj-tolerance < d1) != (d1 < d2)):
	force = True
	else: # c2 < cj
	if d2 != dj and ((d2 < dj+tolerance) != (d1 < d2)):
	force = True

	if force:
	forced.add(i)
	break

	nd, nc = d, c

	return forced

	def _iup_contour_optimize_dp(delta, coords, forced={}, tolerance=0, lookback=None):
	"""Straightforward Dynamic-Programming. For each index i, find least-costly encoding of
	points 0 to i where i is explicitly encoded. We find this by considering all previous
	explicit points j and check whether interpolation can fill points between j and i.

	Note that solution always encodes last point explicitly. Higher-level is responsible
	for removing that restriction.

	As major speedup, we stop looking further whenever we see a "forced" point."""

	n = len(delta)
	if lookback is None:
	lookback = n
	costs = {-1:0}
	chain = {-1:None}
	for i in range(0, n):
	best_cost = costs[i-1] + 1

	costs[i] = best_cost
	chain[i] = i - 1

	if i - 1 in forced:
	continue

	for j in range(i-2, max(i-lookback, -2), -1):

	cost = costs[j] + 1

	if cost < best_cost and can_iup_in_between(delta, coords, j, i, tolerance):
	costs[i] = best_cost = cost
	chain[i] = j

	if j in forced:
	break

	return chain, costs

	def _rot_list(l, k):
	"""Rotate list by k items forward. Ie. item at position 0 will be
	at position k in returned list. Negative k is allowed."""
	n = len(l)
	k %= n
	if not k: return l
	return l[n-k:] + l[:n-k]

	def _rot_set(s, k, n):
	k %= n
	if not k: return s
	return {(v + k) % n for v in s}

	def iup_contour_optimize(delta, coords, tolerance=0.):
	n = len(delta)

	# Get the easy cases out of the way:

	# If all are within tolerance distance of 0, encode nothing:
	if all(abs(complex(*p)) <= tolerance for p in delta):
	return [None] * n

	# If there's exactly one point, return it:
	if n == 1:
	return delta

	# If all deltas are exactly the same, return just one (the first one):
	d0 = delta[0]
	if all(d0 == d for d in delta):
	return [d0] + [None] * (n-1)

	# Else, solve the general problem using Dynamic Programming.

	forced = _iup_contour_bound_forced_set(delta, coords, tolerance)
	# The _iup_contour_optimize_dp() routine returns the optimal encoding
	# solution given the constraint that the last point is always encoded.
	# To remove this constraint, we use two different methods, depending on
	# whether forced set is non-empty or not:

	if forced:
	# Forced set is non-empty: rotate the contour start point
	# such that the last point in the list is a forced point.
	k = (n-1) - max(forced)
	assert k >= 0

	delta = _rot_list(delta, k)
	coords = _rot_list(coords, k)
	forced = _rot_set(forced, k, n)

	chain, costs = _iup_contour_optimize_dp(delta, coords, forced, tolerance)

	# Assemble solution.
	solution = set()
	i = n - 1
	while i is not None:
	solution.add(i)
	i = chain[i]
	assert forced <= solution, (forced, solution)
	delta = [delta[i] if i in solution else None for i in range(n)]

	delta = _rot_list(delta, -k)
	else:
	# Repeat the contour an extra time, solve the 2*n case, then look for solutions of the
	# circular n-length problem in the solution for 2*n linear case. I cannot prove that
	# this always produces the optimal solution...
	chain, costs = _iup_contour_optimize_dp(delta+delta, coords+coords, forced, tolerance, n)
	best_sol, best_cost = None, n+1

	for start in range(n-1, 2*n-1):
	# Assemble solution.
	solution = set()
	i = start
	while i > start - n:
	solution.add(i % n)
	i = chain[i]
	if i == start - n:
	cost = costs[start] - costs[start - n]
	if cost <= best_cost:
	best_sol, best_cost = solution, cost

	delta = [delta[i] if i in best_sol else None for i in range(n)]


	return delta

	def iup_delta_optimize(delta, coords, ends, tolerance=0.):
	assert sorted(ends) == ends and len(coords) == (ends[-1]+1 if ends else 0) + 4
	n = len(coords)
	ends = ends + [n-4, n-3, n-2, n-1]
	out = []
	start = 0
	for end in ends:
	contour = iup_contour_optimize(delta[start:end+1], coords[start:end+1], tolerance)
	assert len(contour) == end - start + 1
	out.extend(contour)
	start = end+1

	return out