| 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 |