Tutorial: Tree Rerooting approach for Path XOR Sums

Hello everyone! I recently solved this problem and wanted to share a Tree Rerooting (In-Out DP) solution based on the contribution technique. Problem Link: 766E - Mahmoud and a xor trip

If you're looking for a structured way to handle path XOR queries across a tree, breaking it down into bits and using Tree DP is the way.

1. The Contribution Technique

Instead of computing the XOR sum for every single path $$$d(x, y)$$$, we can process the answer bit by bit. Since, $$$a[i] \lt = 1e6$$$, 20 bits are enough.

For a specific bit $$$b \in [0, 20]$$$, we can treat the node values $$$a_i$$$ as a binary sequence (either $$$0$$$ or $$$1$$$ depending on the $$$b$$$-th bit). Our goal for each bit $$$b$$$ reduces to: Find the number of paths in the tree where the sum of the $$$b$$$-th bit is odd (i.e., XOR sum is $$$1$$$).

If we can find this count, let's call it $$$res(b)$$$, then our final answer is simply:

2. Downwards Tree DP (In-DP)

Let's root the tree arbitrarily (e.g., at node $$$1$$$). For a given bit $$$b$$$, let $$$a(u)$$$ be the value of the $$$b$$$-th bit of node $$$u$$$. We define our states: $$$c(u, 0) =$$$ number of paths starting at $$$u$$$ and going down into its subtree with an XOR sum of $$$0$$$. $$$c(u, 1) =$$$ number of paths starting at $$$u$$$ and going down into its subtree with an XOR sum of $$$1$$$.

Base Case: A single node forms a valid path by itself.

Transitions: For every child $$$v$$$ of $$$u$$$, if a path from $$$v$$$ has a sum of $$$S$$$, extending it to $$$u$$$ makes the sum $$$S \oplus a(u)$$$. To get a total sum of $$$0$$$, we need $$$S \oplus a(u) = 0 \implies S = a(u)$$$. To get a total sum of $$$1$$$, we need $$$S \oplus a(u) = 1 \implies S = 1 \oplus a(u)$$$.

Therefore, we sum over all children $$$v$$$:

3. Upwards Tree Rerooting (Out-DP)

To get the paths that go up from $$$u$$$ (towards its parent and into other branches), we define: $$$up(u, 0) =$$$ number of paths starting at $$$u$$$ and going up with an XOR sum of $$$0$$$. $$$up(u, 1) =$$$ number of paths starting at $$$u$$$ and going up with an XOR sum of $$$1$$$.

When transitioning from $$$u$$$ to its child $$$v$$$, the paths going "up" from $$$v$$$ consist of all valid paths starting at $$$u$$$ except the ones that go down into $$$v$$$'s subtree.

Let $$$x = a(v)$$$ and $$$y = a(u)$$$. The rerooting transitions are:

Explanation: $$$up(u, x) + c(u, x)$$$ gives all paths starting at $$$u$$$ with XOR sum $$$x$$$. We subtract $$$c(v, x \oplus y)$$$ to remove the paths that strictly lie in $$$v$$$'s subtree. Adding $$$v$$$ to these paths toggles their sum by $$$x$$$, giving us the required $$$0$$$ or $$$1$$$ sum for $$$up(v, \dots)$$$.

4. Assembling the Answer for Bit $$$b$$$

Once both DP passes are complete, the total number of paths starting at any node $$$u$$$ with an XOR sum of $$$1$$$ is $$$c(u, 1) + up(u, 1)$$$.

However, if we simply sum this over all nodes:

We will double-count every path of length $$$ \gt 1$$$ (since it is counted once from each endpoint). Single-node paths (where $$$a(u) = 1$$$) are only counted once.

To fix this, we subtract the single-node paths, divide by 2, and then add the single-node paths back:

Finally, add $$$res(b) \cdot 2^b$$$ to your global answer!

Hope this makes the In-Out DP logic clear.