Binary Jumping

Binary Jumping

Explanation

Binary lifting consists of calculating the $2^k$-th ancestor of each node for all relevant values of $k$ and storing them in a table.

With this table, we can then efficiently answer queries regarding the $k$-th ancestor of all nodes. This is because any $k$ can be broken down into a sum of powers of $2$ with its binary representation.

This way, instead of directly computing, say, the $13$th ancestor of a node, we can go to the $8$th, then $4$th, then $1$st ancestor of the node. This results in a logarithmic complexity for computing the $k$th parent.

Here's an animation of how we jump if you're still confused:

To actually compute our binary jumping table, we start with the $2^0=1$st parents of each node, which is their direct parent. We then move on and calculate $2^1=2$nd parents, using the fact that the $2$nd parent can be calculated as the $1$st parent of the $1$st parent. Using similar logic, we move on to $2^2$, $2^3$, and so on. We stop when $2^n$ is greater than the size of the tree, since at that point we're definitely at the root.

Implementation

Time Complexity: $\mathcal{O}((N+Q)\log N)$

Copy
#include <cmath>
#include <iostream>
#include <vector>

using std::cout;
using std::endl;
using std::vector;

class Tree {
  private:

Problems

Lowest Common Ancestor

Explanation 1

To find $\textrm{lca}(a, b)$, we can first lift the lower node of $a$ and $b$ to the same depth as the other. Then, we lift both nodes up decrementally. At the end, the parent of either node is the LCA of the two.

Implementation

Time Complexity: $\mathcal{O}((N+Q)\log N)$

Copy
#include <cmath>
#include <iostream>
#include <vector>

using std::cout;
using std::endl;
using std::vector;

class Tree {
  private:

Explanation 2

We can also use an Euler tour of the tree to help us compute the LCAs as well.

Let $\texttt{start}$ and $\texttt{end}$ be the time-in and time-out table for the nodes in the tree. They'll be filled up in the exact same way the Euler tour module does it.

The cool thing is that while we're filling up $\texttt{start}$ and $\texttt{end}$, we can also calculate our binary jumping table in the exact same way we did in the previous solution! We can do this because in a DFS, we're guaranteed to have processed all of a node's parents before the node itself, so all the tables of any node's ancestors will have been filled when we reach the node.

Now, to actually calculate the LCA without the depths of the nodes, we can use the fact that node $a$ is an ancestor of node $b$ if $\texttt{start}[a] \le \texttt{start}[b]$ and $\texttt{end}[b] \le \texttt{end}[a]$.

In our LCA function, we first check if one node is already an ancestor of the other. In that case, we return the ancestor. If it isn't, then we lift up one of the nodes until its an ancestor of the other in a method that's basically the same as our previous binary jumping algorithm. After that, our answer is the parent of the node we lift up.

Implementation

Time Complexity: $\mathcal{O}((N+Q)\log N)$

Copy
#include <cmath>
#include <iostream>
#include <vector>

using std::cout;
using std::endl;
using std::vector;

class Tree {
  private:

Explanation 3

We can also find the LCA of two nodes using Tarjan's Offline LCA algorithm.

By taking advantage of the DFS traversal, we can precompute the answers to the queries through forming subtrees and calculated the common parent with a similar structure as Disjoint-Set Union.

Implementation

Copy
#include <bits/stdc++.h>
using namespace std;

const int MAX = 2e5 + 1;
bool vis[MAX];
int lca[MAX], fa[MAX];
vector<array<int, 2>> adj[MAX], qry[MAX];
int find(int u) { return (fa[u] == u) ? u : fa[u] = find(fa[u]); }
void tarjan(int node) {
	vis[node] = true;

Explanation

Since we have the depth of all the nodes, the distance between two nodes $a$ and $b$ is $\texttt{depth}[a] + \texttt{depth}[b] - 2 \cdot \texttt{depth}[\textrm{lca}(a, b)]$.

Here's some intuition if you're confused about how this formula works. To get from node $a$ to $b$, one way would be to go to the root of the tree and then to $b$. This gives us a distance of $\texttt{depth}[a] + \texttt{depth}[b]$. However, notice that we're passing through all the nodes above the LCA twice. Thus, we have to subtract off twice the depth of the LCA, giving us our final expression.

Implementation

Time Complexity: $\mathcal{O}((N+Q)\log N)$

Copy
#include <cmath>
#include <iostream>
#include <vector>

using std::cout;
using std::endl;
using std::vector;

 Code Snippet: LCA Tree (Click to expand)

Problems

USACO

General