1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180 181 182 183 184 185 186 187 188 189 190 191 192 193 194 195 196 197 198 199 200 201 202 203 204 205 206 207 208 209 210 211 212 213 214 215 216 217 218 219 220 221 222 223 224 225 226 227 228 229 230 231 232 233 234 235 236 237 238 239 240 241 242 243 244 245 246 247 248 249 250 251 252 253 254 255 256 257 258 259 260 261 262 263 264 265 266 267 268 269 270 271 272 273 274 275 276 277 278 279 280 281 282 283 284 285 286 287 288 289 290 291 292 293 294 295 296 297 298 299 300 301 302 303 304 305 306 307 308 309 310 311 312 313 314 315 316 317 318 319 320 321 322 323 324 325 326 327 328 329 330 331 332 333 334 335 336 337 338 339 340 341 342 343 344 345 346 347 348 349 350 351 352 353 354 355 356 357 358 359 360 361 362 363 364 365 366 367 368 369 370 371 372 373 374 375 376 377 378 379 380 381 382 383 384 385 386 387 388 389 390 391 392 393 394 395 396 397 398 399 400 401 402 403 404 405 406 407 408 409 410 411 412 413 414 415 416 417 418 419 420 421 422 423 424 425 426 427 428 429 430 431 432 433 434 435 436 437 438 439 440 441 442 443 444 445 446 447 448 449 450 451 452 453 454 455 456 457 458 459 460 461 462 463 464 465 466 467 468 469 470 471 472 473 474 475 476 477 478 479 480 481 482 483 484 485 486 487 488 489 490 491 492 493 494 495 496 497 498 499 500 501 502 503 504 505 506 507 508 509 510 511 512 513 514 515 516 517 518 519 520 521 522 523 524 525 526 527 528 529 530 531 532 533 534 535
// Copyright (c) The Diem Core Contributors
// SPDX-License-Identifier: Apache-2.0
//! This module provides an abstraction for positioning a node in a binary tree,
//! A `Position` uniquely identifies the location of a node
//!
//! In this implementation, `Position` is represented by the in-order-traversal sequence number
//! of the node. The process of locating a node and jumping between nodes is done through
//! in-order position calculation, which can be done with bit manipulation.
//!
//! For example
//! ```text
//! 3
//! / \
//! / \
//! 1 5 <-[Node index, a.k.a, Position]
//! / \ / \
//! 0 2 4 6
//!
//! 0 1 2 3 <[Leaf index]
//! ```
//! Note1: The in-order-traversal counts from 0
//! Note2: The level of tree counts from leaf level, start from 0
//! Note3: The leaf index starting from left-most leaf, starts from 0
use crate::proof::definition::{LeafCount, MAX_ACCUMULATOR_LEAVES, MAX_ACCUMULATOR_PROOF_DEPTH};
use anyhow::{ensure, Result};
use mirai_annotations::*;
use std::fmt;
#[cfg(test)]
mod position_test;
#[derive(Clone, Copy, Debug, Eq, PartialEq, Hash, Ord, PartialOrd)]
pub struct Position(u64);
// invariant Position.0 < u64::max_value() - 1
#[derive(Debug, Eq, PartialEq)]
pub enum NodeDirection {
Left,
Right,
}
impl Position {
/// What level is this node in the tree, 0 if the node is a leaf,
/// 1 if the level is one above a leaf, etc.
pub fn level(self) -> u32 {
(!self.0).trailing_zeros()
}
pub fn is_leaf(self) -> bool {
self.0 & 1 == 0
}
/// What position is the node within the level? i.e. how many nodes
/// are to the left of this node at the same level
#[cfg(test)]
fn pos_counting_from_left(self) -> u64 {
self.0 >> (self.level() + 1)
}
/// pos count start from 0 on each level
pub fn from_level_and_pos(level: u32, pos: u64) -> Self {
precondition!(level < 64);
assume!(1u64 << level > 0); // bitwise and integer operations don't mix.
let level_one_bits = (1u64 << level) - 1;
let shifted_pos = if level == 63 { 0 } else { pos << (level + 1) };
Position(shifted_pos | level_one_bits)
}
pub fn from_inorder_index(index: u64) -> Self {
Position(index)
}
pub fn to_inorder_index(self) -> u64 {
self.0
}
pub fn from_postorder_index(index: u64) -> Result<Self> {
ensure!(
index < !0u64,
"node index {} is invalid (equal to 2^64 - 1)",
index
);
Ok(Position(postorder_to_inorder(index)))
}
pub fn to_postorder_index(self) -> u64 {
inorder_to_postorder(self.to_inorder_index())
}
/// What is the parent of this node?
pub fn parent(self) -> Self {
assume!(self.0 < u64::max_value() - 1); // invariant
Self(
(self.0 | isolate_rightmost_zero_bit(self.0))
& !(isolate_rightmost_zero_bit(self.0) << 1),
)
}
/// What is the left node of this node? Will overflow if the node is a leaf
pub fn left_child(self) -> Self {
checked_precondition!(!self.is_leaf());
Self::child(self, NodeDirection::Left)
}
/// What is the right node of this node? Will overflow if the node is a leaf
pub fn right_child(self) -> Self {
checked_precondition!(!self.is_leaf());
Self::child(self, NodeDirection::Right)
}
fn child(self, dir: NodeDirection) -> Self {
checked_precondition!(!self.is_leaf());
assume!(self.0 < u64::max_value() - 1); // invariant
let direction_bit = match dir {
NodeDirection::Left => 0,
NodeDirection::Right => isolate_rightmost_zero_bit(self.0),
};
Self((self.0 | direction_bit) & !(isolate_rightmost_zero_bit(self.0) >> 1))
}
/// Whether this position is a left child of its parent. The observation is that,
/// after stripping out all right-most 1 bits, a left child will have a bit pattern
/// of xxx00(11..), while a right child will be represented by xxx10(11..)
pub fn is_left_child(self) -> bool {
assume!(self.0 < u64::max_value() - 1); // invariant
self.0 & (isolate_rightmost_zero_bit(self.0) << 1) == 0
}
pub fn is_right_child(self) -> bool {
!self.is_left_child()
}
// Opposite of get_left_node_count_from_position.
pub fn from_leaf_index(leaf_index: u64) -> Self {
Self::from_level_and_pos(0, leaf_index)
}
/// This method takes in a node position and return its sibling position
///
/// The observation is that, after stripping out the right-most common bits,
/// two sibling nodes flip the the next right-most bits with each other.
/// To find out the right-most common bits, first remove all the right-most ones
/// because they are corresponding to level's indicator. Then remove next zero right after.
pub fn sibling(self) -> Self {
assume!(self.0 < u64::max_value() - 1); // invariant
Self(self.0 ^ (isolate_rightmost_zero_bit(self.0) << 1))
}
// Given a leaf index, calculate the position of a minimum root which contains this leaf
/// This method calculates the index of the smallest root which contains this leaf.
/// Observe that, the root position is composed by a "height" number of ones
///
/// For example
/// ```text
/// 0010010(node)
/// 0011111(smearing)
/// -------
/// 0001111(root)
/// ```
pub fn root_from_leaf_index(leaf_index: u64) -> Self {
let leaf = Self::from_leaf_index(leaf_index);
Self(smear_ones_for_u64(leaf.0) >> 1)
}
pub fn root_from_leaf_count(leaf_count: LeafCount) -> Self {
assert!(leaf_count > 0);
Self::root_from_leaf_index((leaf_count - 1) as u64)
}
pub fn root_level_from_leaf_count(leaf_count: LeafCount) -> u32 {
assert!(leaf_count > 0);
let index = (leaf_count - 1) as u64;
MAX_ACCUMULATOR_PROOF_DEPTH as u32 + 1 - index.leading_zeros()
}
/// Given a node, find its right most child in its subtree.
/// Right most child is a Position, could be itself, at level 0
pub fn right_most_child(self) -> Self {
let level = self.level();
Self(self.0 + (1_u64 << level) - 1)
}
/// Given a node, find its left most child in its subtree
/// Left most child is a node, could be itself, at level 0
pub fn left_most_child(self) -> Self {
// Turn off its right most x bits. while x=level of node
let level = self.level();
Self(turn_off_right_most_n_bits(self.0, level))
}
}
// Some helper functions to perform general bit manipulation
/// Smearing all the bits starting from MSB with ones
fn smear_ones_for_u64(v: u64) -> u64 {
let mut n = v;
n |= n >> 1;
n |= n >> 2;
n |= n >> 4;
n |= n >> 8;
n |= n >> 16;
n |= n >> 32;
n
}
/// Turn off n right most bits
///
/// For example
/// ```text
/// 00010010101
/// -----------
/// 00010010100 n=1
/// 00010010000 n=3
/// ```
fn turn_off_right_most_n_bits(v: u64, n: u32) -> u64 {
debug_checked_precondition!(n < 64);
(v >> n) << n
}
/// Finds the rightmost 0-bit, turns off all bits, and sets this bit to 1 in
/// the result. For example:
///
/// ```text
/// 01110111 (x)
/// --------
/// 10001000 (~x)
/// & 01111000 (x+1)
/// --------
/// 00001000
/// ```
/// http://www.catonmat.net/blog/low-level-bit-hacks-you-absolutely-must-know/
fn isolate_rightmost_zero_bit(v: u64) -> u64 {
!v & v.overflowing_add(1).0
}
// The following part of the position implementation is logically separate and
// depends on our notion of freezable. It should probably move to another module.
impl Position {
// Given index of right most leaf, calculate if a position is the root
// of a perfect subtree that does not contain any placeholder nodes.
//
// First find its right most child
// the right most child of any node will be at leaf level, which will be a either placeholder
// node or leaf node. if right most child is a leaf node, then it is freezable.
// if right most child is larger than max_leaf_node, it is a placeholder node, and not
// freezable.
pub fn is_freezable(self, leaf_index: u64) -> bool {
let leaf = Self::from_leaf_index(leaf_index);
let right_most_child = self.right_most_child();
right_most_child.0 <= leaf.0
}
// Given index of right most leaf, calculate if a position should contain
// a placeholder node at this moment
// A node is a placeholder if both two conditions below are true:
// 1, the node's in order traversal seq > max_leaf_node's, and
// 2, the node does not have left child or right child.
pub fn is_placeholder(self, leaf_index: u64) -> bool {
let leaf = Self::from_leaf_index(leaf_index);
if self.0 <= leaf.0 {
return false;
}
if self.left_most_child().0 <= leaf.0 {
return false;
}
true
}
/// Creates an `AncestorIterator` using this position.
pub fn iter_ancestor(self) -> AncestorIterator {
AncestorIterator { position: self }
}
/// Creates an `AncestorSiblingIterator` using this position.
pub fn iter_ancestor_sibling(self) -> AncestorSiblingIterator {
AncestorSiblingIterator { position: self }
}
}
impl fmt::Display for Position {
fn fmt(&self, f: &mut fmt::Formatter) -> fmt::Result {
write!(f, "Pos({})", self.to_inorder_index())
}
}
/// `AncestorSiblingIterator` generates current sibling position and moves itself to its parent
/// position for each iteration.
#[derive(Debug)]
pub struct AncestorSiblingIterator {
position: Position,
}
impl Iterator for AncestorSiblingIterator {
type Item = Position;
fn next(&mut self) -> Option<Position> {
let current_sibling_position = self.position.sibling();
self.position = self.position.parent();
Some(current_sibling_position)
}
}
/// `AncestorIterator` generates current position and moves itself to its parent position for each
/// iteration.
#[derive(Debug)]
pub struct AncestorIterator {
position: Position,
}
impl Iterator for AncestorIterator {
type Item = Position;
fn next(&mut self) -> Option<Position> {
let current_position = self.position;
self.position = self.position.parent();
Some(current_position)
}
}
/// Traverse leaves from left to right in groups that forms full subtrees, yielding root positions
/// of such subtrees.
/// Note that each 1-bit in num_leaves corresponds to a full subtree.
/// For example, in the below tree of 5=0b101 leaves, the two 1-bits corresponds to Fzn2 and L4
/// accordingly.
///
/// ```text
/// Non-fzn
/// / \
/// / \
/// / \
/// Fzn2 Non-fzn
/// / \ / \
/// / \ / \
/// Fzn1 Fzn3 Non-fzn [Placeholder]
/// / \ / \ / \
/// L0 L1 L2 L3 L4 [Placeholder]
/// ```
pub struct FrozenSubTreeIterator {
bitmap: u64,
seen_leaves: u64,
// invariant seen_leaves < u64::max_value() - bitmap
}
impl FrozenSubTreeIterator {
pub fn new(num_leaves: LeafCount) -> Self {
Self {
bitmap: num_leaves,
seen_leaves: 0,
}
}
}
impl Iterator for FrozenSubTreeIterator {
type Item = Position;
fn next(&mut self) -> Option<Position> {
assume!(self.seen_leaves < u64::max_value() - self.bitmap); // invariant
if self.bitmap == 0 {
return None;
}
// Find the remaining biggest full subtree.
// The MSB of the bitmap represents it. For example for a tree of 0b1010=10 leaves, the
// biggest and leftmost full subtree has 0b1000=8 leaves, which can be got by smearing all
// bits after MSB with 1-bits (got 0b1111), right shift once (got 0b0111) and add 1 (got
// 0b1000=8). At the same time, we also observe that the in-order numbering of a full
// subtree root is (num_leaves - 1) greater than that of the leftmost leaf, and also
// (num_leaves - 1) less than that of the rightmost leaf.
let root_offset = smear_ones_for_u64(self.bitmap) >> 1;
assume!(root_offset < self.bitmap); // relate bit logic to integer logic
let num_leaves = root_offset + 1;
let leftmost_leaf = Position::from_leaf_index(self.seen_leaves);
let root = Position::from_inorder_index(leftmost_leaf.to_inorder_index() + root_offset);
// Mark it consumed.
self.bitmap &= !num_leaves;
self.seen_leaves += num_leaves;
Some(root)
}
}
/// Given an accumulator of size `current_num_leaves`, `FrozenSubtreeSiblingIterator` yields the
/// positions of required subtrees if we want to append these subtrees to the existing accumulator
/// to generate a bigger one of size `new_num_leaves`.
///
/// See [`crate::proof::accumulator::Accumulator::append_subtrees`] for more details.
pub struct FrozenSubtreeSiblingIterator {
current_num_leaves: LeafCount,
remaining_new_leaves: LeafCount,
}
impl FrozenSubtreeSiblingIterator {
/// Constructs a new `FrozenSubtreeSiblingIterator` given the size of current accumulator and
/// the size of the bigger accumulator.
pub fn new(current_num_leaves: LeafCount, new_num_leaves: LeafCount) -> Self {
assert!(
new_num_leaves <= MAX_ACCUMULATOR_LEAVES,
"An accumulator can have at most 2^{} leaves. Provided num_leaves: {}.",
MAX_ACCUMULATOR_PROOF_DEPTH,
new_num_leaves,
);
assert!(
current_num_leaves <= new_num_leaves,
"Number of leaves needs to be increasing: current_num_leaves: {}, new_num_leaves: {}",
current_num_leaves,
new_num_leaves
);
Self {
current_num_leaves,
remaining_new_leaves: new_num_leaves - current_num_leaves,
}
}
/// Helper function to return the next set of leaves that form a complete subtree. For
/// example, if there are 5 leaves (..0101), 2 ^ (63 - 61 leading zeros) = 4 leaves should be
/// taken next.
fn next_new_leaf_batch(&self) -> LeafCount {
let zeros = self.remaining_new_leaves.leading_zeros();
1 << (MAX_ACCUMULATOR_PROOF_DEPTH - zeros as usize)
}
}
impl Iterator for FrozenSubtreeSiblingIterator {
type Item = Position;
fn next(&mut self) -> Option<Self::Item> {
if self.remaining_new_leaves == 0 {
return None;
}
// Now we compute the size of the next subtree. If there is a rightmost frozen subtree, we
// may combine it with a subtree of the same size, or append a smaller one on the right. In
// case self.current_num_leaves is zero and there is no rightmost frozen subtree, the
// largest possible one is appended.
let next_subtree_leaves = if self.current_num_leaves > 0 {
let rightmost_frozen_subtree_leaves = 1 << self.current_num_leaves.trailing_zeros();
if self.remaining_new_leaves >= rightmost_frozen_subtree_leaves {
rightmost_frozen_subtree_leaves
} else {
self.next_new_leaf_batch()
}
} else {
self.next_new_leaf_batch()
};
// Now that the size of the next subtree is known, we compute the leftmost and rightmost
// leaves in this subtree. The root of the subtree is then the middle of these two leaves.
let first_leaf_index = self.current_num_leaves;
let last_leaf_index = first_leaf_index + next_subtree_leaves - 1;
self.current_num_leaves += next_subtree_leaves;
self.remaining_new_leaves -= next_subtree_leaves;
Some(Position::from_inorder_index(
(first_leaf_index + last_leaf_index) as u64,
))
}
}
fn children_of_node(node: u64) -> u64 {
(isolate_rightmost_zero_bit(node) << 1) - 2
}
/// In a post-order tree traversal, how many nodes are traversed before `node`
/// not including nodes that are children of `node`.
fn nodes_to_left_of(node: u64) -> u64 {
// If node = 0b0100111, ones_up_to_level = 0b111
let ones_up_to_level = isolate_rightmost_zero_bit(node) - 1;
// Unset all the 1s due to the level
let unset_level_zeros = node ^ ones_up_to_level;
// What remains is a 1 bit set every time a node navigated right
// For example, consider node=5=0b101. unset_level_zeros=0b100.
// the 1 bit in unset_level_zeros at position 2 represents the
// fact that 5 is the right child at the level 1. At this level
// there are 2^2 - 1 children on the left hand side.
//
// So what we do is subtract the count of one bits from unset_level_zeros
// to account for the fact that if the node is the right child at level
// n that there are 2^n - 1 children.
unset_level_zeros - u64::from(unset_level_zeros.count_ones())
}
/// Given `node`, an index in an in-order traversal of a perfect binary tree,
/// what order would the node be visited in in post-order traversal?
/// For example, consider this tree of in-order nodes.
///
/// ```text
/// 3
/// / \
/// / \
/// 1 5
/// / \ / \
/// 0 2 4 6
/// ```
///
/// The post-order ordering of the nodes is:
/// ```text
/// 6
/// / \
/// / \
/// 2 5
/// / \ / \
/// 0 1 3 4
/// ```
///
/// post_order_index(1) == 2
/// post_order_index(4) == 3
pub fn inorder_to_postorder(node: u64) -> u64 {
let children = children_of_node(node);
let left_nodes = nodes_to_left_of(node);
children + left_nodes
}
pub fn postorder_to_inorder(mut node: u64) -> u64 {
// The number of nodes in a full binary tree with height `n` is `2^n - 1`.
let mut full_binary_size = !0u64;
let mut bitmap = 0u64;
for i in (0..64).rev() {
if node >= full_binary_size {
node -= full_binary_size;
bitmap |= 1 << i;
}
full_binary_size >>= 1;
}
let level = node as u32;
let pos = bitmap >> level;
Position::from_level_and_pos(level, pos).to_inorder_index()
}