# SPDX-License-Identifier: MIT # Part of https://algorithm.tamnd.com/octree/. The authors are listed in prov/AKGCFN.jsonl. from typing import Optional Point = tuple[int, int, int] Box = tuple[Point, Point] class Node: """A cell of the octree: a cube with corner (x, y, z) and side size. A leaf has children None and holds up to B points in points. An internal node has eight children, indexed by the octant number o = bx + 2*by + 4*bz, where bx is 1 for the upper half in x, and so on. count is the number of points in the subtree. """ __slots__ = ("x", "y", "z", "size", "points", "children", "count") def __init__(self, x: int, y: int, z: int, size: int) -> None: self.x, self.y, self.z, self.size = x, y, z, size self.points: list[Point] = [] self.children: Optional[list[Node]] = None self.count = 0 def octant(self, p: Point) -> int: """The index of the child cell that holds p. O(1).""" h = self.size // 2 return ( (1 if p[0] >= self.x + h else 0) | (2 if p[1] >= self.y + h else 0) | (4 if p[2] >= self.z + h else 0) ) def holds(self, p: Point) -> bool: return ( self.x <= p[0] < self.x + self.size and self.y <= p[1] < self.y + self.size and self.z <= p[2] < self.z + self.size ) class Octree: """A point-region (PR) octree with leaf capacity B over the cube [0, 2^depth)^3. Points have integer coordinates and are stored as a set. Cells split at their centers, so the shape depends only on the set of points, not on the order of insertion. Invariant: a leaf holds at most B points, an internal node holds more than B points in its subtree, and each point lies in the cell of its leaf. With distinct integer points, a cell of side 1 holds at most one point, so the height is at most depth. """ def __init__(self, depth: int, capacity: int = 1) -> None: if depth < 0 or capacity < 1: raise ValueError("need depth >= 0 and capacity >= 1") self.depth = depth self.capacity = capacity self.root = Node(0, 0, 0, 1 << depth) def __len__(self) -> int: return self.root.count def _check_point(self, p: Point) -> Point: q = (int(p[0]), int(p[1]), int(p[2])) if not self.root.holds(q): raise ValueError("the point is outside the root cube") return q def _leaf(self, p: Point) -> tuple[Node, list[Node]]: """The leaf whose cell holds p, and the path of nodes above it. O(depth).""" u = self.root path: list[Node] = [] while u.children is not None: path.append(u) u = u.children[u.octant(p)] return u, path def contains(self, p: Point) -> bool: """Whether p is stored. O(depth + B).""" q = self._check_point(p) u, _ = self._leaf(q) return q in u.points def insert(self, p: Point) -> bool: """Add p; return False if it was already there. O(depth * B) worst.""" q = self._check_point(p) u, path = self._leaf(q) # @line L1 if q in u.points: return False for v in path: # @line L2 v.count += 1 u.points.append(q) # @line L3 u.count += 1 B = self.capacity while len(u.points) > B: # @line L4: overflow; a cell of side 1 never overflows h = u.size // 2 u.children = [ # @line L5 Node(u.x + (o & 1) * h, u.y + (o >> 1 & 1) * h, u.z + (o >> 2 & 1) * h, h) for o in range(8) ] for r in u.points: # @line L6 c = u.children[u.octant(r)] c.points.append(r) c.count += 1 u.points = [] nxt = None for c in u.children: # @line L7: at most one child can overflow if len(c.points) > B: nxt = c if nxt is None: break u = nxt return True # @line L8 def range_search(self, lo: Point, hi: Point) -> list[Point]: """The stored points p with lo[i] <= p[i] <= hi[i] for each axis. Visits each node whose cell meets the box; an explicit stack is used. """ out: list[Point] = [] stack = [self.root] while stack: u = stack.pop() if ( # @line L9 hi[0] < u.x or lo[0] >= u.x + u.size or hi[1] < u.y or lo[1] >= u.y + u.size or hi[2] < u.z or lo[2] >= u.z + u.size ): continue if u.children is None: # @line L10 for p in u.points: if all(lo[i] <= p[i] <= hi[i] for i in range(3)): out.append(p) continue stack.extend(u.children) # @line L11 return out def delete(self, p: Point) -> bool: """Remove p; return False if it is not stored. O(depth + B).""" q = self._check_point(p) u, path = self._leaf(q) # @line L12 if q not in u.points: return False u.points.remove(q) # @line L13 u.count -= 1 for v in path: v.count -= 1 B = self.capacity for v in reversed(path): # @line L14: merge upward while the subtree is small if v.count > B: break assert v.children is not None pts: list[Point] = [] for c in v.children: pts.extend(c.points) # all children are leaves here v.points = pts v.children = None return True # @line L15 def points(self) -> list[Point]: """All stored points. O(number of nodes).""" out: list[Point] = [] stack = [self.root] while stack: u = stack.pop() if u.children is None: out.extend(u.points) else: stack.extend(u.children) return out def height(self) -> int: """The number of levels below the root (0 for a single leaf).""" best = 0 stack = [(self.root, 0)] while stack: u, d = stack.pop() best = max(best, d) if u.children is not None: stack.extend((c, d + 1) for c in u.children) return best def node_count(self) -> int: n = 0 stack = [self.root] while stack: u = stack.pop() n += 1 if u.children is not None: stack.extend(u.children) return n def check_invariant(self) -> None: """Raise AssertionError if the structure is broken. O(number of nodes).""" seen: set[Point] = set() stack = [self.root] while stack: u = stack.pop() if u.children is None: assert len(u.points) <= self.capacity assert u.count == len(u.points) for p in u.points: assert u.holds(p) assert p not in seen seen.add(p) else: assert not u.points and len(u.children) == 8 and u.size >= 2 h = u.size // 2 total = 0 for o, c in enumerate(u.children): assert (c.x, c.y, c.z, c.size) == ( u.x + (o & 1) * h, u.y + (o >> 1 & 1) * h, u.z + (o >> 2 & 1) * h, h) total += c.count stack.append(c) assert u.count == total assert u.count > self.capacity assert len(seen) == self.root.count