[Tutorial] Diameter of a tree and its applications

Revision en2, by TheScrasse, 2022-03-14 19:05:58

Hello everyone,
finding the diameter is one of the most frequent ways to solve problems about trees. In this tutorial we will see how to find a diameter and some of its properties, and we will use them to solve some problems of increasing difficulty. I tried to put a lot of examples to make the understanding easier.
The first part of the tutorial is quite basic, so feel free to skip it and jump to the problems if you already know the concepts.

Target: rating $[1400, 2100]$ on CF
Prerequisites: basic graph theory, greedy

## The diameter

Given an unweighted tree, let's define $\text{dist}(a, b) =$ the number of edges in the simple path $a \rightarrow b$.

A diameter of the tree is a simple path $a \rightarrow b$ that maximizes $\text{dist}(a, b)$ over all pairs of nodes. If there are multiple diameters, let's pick any of them.

The same definition is valid for a weighted tree with nonnegative weights (with $\text{dist}(a, b) =$ the sum of the weights of the edges in the simple path $a \rightarrow b$).

#### Finding a diameter

Given a tree with $n$ nodes are multiple ways to find a diameter. Here is one of the simplest ways:

Run a DFS from node $1$. Let $a$ be a node whose distance from node $1$ is maximized. Run another DFS from node $a$. Let $b$ be a node whose distance from node $a$ is maximized. $a \rightarrow b$ is a diameter.

#### Proof 1

Let $f(x)$ be the number of inversions after $x$ moves.
In one move, if you swap the values on positions $i, i + 1$, $f(x)$ either increases by $1$ or decreases by $1$. This is because the only pair $(a_i, a_j)$ whose relative order changed is $(a_i, a_{i+1})$. Since the sorted array has $0$ inversions, you need at least $k$ moves to sort the array.
For example, if you have the permutation $[2, 3, 7, 8, 6, 9, 1, 4, 5]$ ($16$ inversions) and you swap two adjacent elements such that $a_i > a_{i+1}$ (getting, for example, $[2, 3, 7, 6, 8, 9, 1, 4, 5]$), the resulting array has $15$ inversions, and if you swap two adjacent elements such that $a_i < a_{i+1}$ (getting, for example, $[3, 2, 7, 8, 6, 9, 1, 4, 5]$), the resulting array has $17$ inversions.

On the other hand, if the array is not sorted you can always find an $i$ such that $a_i > a_{i+1}$, so you can sort the array in $k$ moves.

#### Proof 2

For each $x$, let $f(x)$ be the number of inversions if you consider only the elements from $1$ to $x$ in the permutation.
First, let's put $x$ at the end of the permutation: this requires $x - \text{pos}(x)$ moves. That's optimal (the actual proof is similar to Proof 1; in an intuitive way, if you put the last element to the end of the array, it doesn't interfere anymore with the other swaps).
For example, if you have the permutation $[2, 3, 7, 8, 6, 9, 1, 4, 5]$ and you move the $9$ to the end, you get $[2, 3, 7, 8, 6, 1, 4, 5, 9]$ and now you need to sort $[2, 3, 7, 8, 6, 1, 4, 5]$. Hence, $f(x) = f(x-1) + x - \text{pos}(x)$. For each $x$, $x - \text{pos}(x)$ is actually the number of pairs $(i, j)$ ($1 \leq i < j \leq x$) such that $x = a_i > a_j$. So $f(x)$ is equal to the number of inversions.

#### Counting inversions in $O(n \log n)$

You can use a Fenwick tree (or a segment tree). There are other solutions (for example, using divide & conquer + merge sort), but they are usually harder to generalize.
For each $j$, calculate the number of $i < j$ such that $a_i > a_j$.
The Fenwick tree should contain the frequency of each value in $[1, n]$ in the prefix $[1, j - 1]$ of the array.
So, for each $j$, the queries look like

• $res := res + \text{range_sum}(a_j + 1, n)$
• add $1$ in the position $a_j$ of the Fenwick tree

#### Observations / slight variations of the problem

By using a Fenwick tree, you are actually calculating the number of inversions for each prefix of the array.

You can calculate the number of swaps required to sort an array (not necessarily a permutation, but for now let's assume that its elements are distinct) by compressing the values of the array. For example, the array $[13, 18, 34, 38, 28, 41, 5, 29, 30]$ becomes $[2, 3, 7, 8, 6, 9, 1, 4, 5]$.

You can also calculate the number of swaps required to get an array $b$ (for now let's assume that its elements are distinct) starting from $a$, by renaming the values. For example,
$a = [2, 3, 7, 8, 6, 9, 1, 4, 5], b = [9, 8, 5, 2, 1, 4, 7, 3, 6]$
is equivalent to
$a = [4, 8, 7, 2, 9, 1, 5, 6, 3], b = [1, 2, 3, 4, 5, 6, 7, 8, 9]$

$a^{-1}$ (a permutation such that $(a^{-1})_{a_x} = x$, i.e. $(a^{-1})_x$ is equal to the position of $x$ in $a$) has the same number of inversions as $a$. For example, $[2, 3, 7, 8, 6, 9, 1, 4, 5]$ and $[7, 1, 2, 8, 9, 5, 3, 4, 6]$ have both $16$ inversions. Sketch of a proof: note that, when you swap two elements in adjacent positions in $a$, you are swapping two adjacent values in $a^{-1}$, and the number of inversions in $a^{-1}$ also increases by $1$ or decreases by $1$ (like in Proof 1).

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Solution

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Solution

## arc088_e (rating: 2231)

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Implementation (C++)

## arc097_e (rating: 2247)

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Implementation (C++)

## Conclusions

We've seen that a lot of problems where you have to swap adjacent elements can be tackled with greedy observations, such as looking at the optimal relative positions of the values in the final array; then, a lot of these problems can be reduced to "find the number of inversions" or similar.

Of course, suggestions/corrections are welcome. In particular, please share in the comments other problems where you have to swap adjacent elements.

I hope you enjoyed the blog!

#### History

Revisions

Rev. Lang. By When Δ Comment
en27 TheScrasse 2023-07-13 11:15:23 24 Tiny change: 't turned IGM. Now you ' -> 't turned International Grandmaster. Now you '
en26 TheScrasse 2023-07-12 22:32:10 91 Tiny change: 'Old blog' -> 'I've just turned IGM. Now you can ask me anything in the comments.' (published)
en25 TheScrasse 2022-06-17 15:27:12 12105
en24 TheScrasse 2022-03-26 14:49:37 24 Tiny change: 'u have to swap adjacent elements.\n\nI hop' -> 'u have to use the diameter.\n\nI hop'
en23 TheScrasse 2022-03-26 14:07:32 3 Tiny change: 'mple path of $a \right' -> 'mple path$a \right'
en22 TheScrasse 2022-03-26 14:06:55 173
en21 TheScrasse 2022-03-26 14:04:27 2448
en20 TheScrasse 2022-03-26 13:14:16 935
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en17 TheScrasse 2022-03-26 12:23:20 1163
en16 TheScrasse 2022-03-26 11:56:33 1772
en15 TheScrasse 2022-03-26 11:42:16 779
en14 TheScrasse 2022-03-26 11:37:47 966 Tiny change: 'e diameter.\n</spoil' -> 'e diameter .\n</spoil'
en13 TheScrasse 2022-03-25 00:43:26 1585
en12 TheScrasse 2022-03-25 00:26:04 128
en11 TheScrasse 2022-03-25 00:22:20 1754
en10 TheScrasse 2022-03-24 23:45:35 1716 Tiny change: 'ays works.\n\nProof:' -> 'ays works._\n\nProof:'
en9 TheScrasse 2022-03-24 23:18:42 2333 Reverted to en7
en8 TheScrasse 2022-03-24 23:17:52 2333 Reverted to en6
en7 TheScrasse 2022-03-16 01:17:14 2333
en6 TheScrasse 2022-03-15 00:47:55 330
en5 TheScrasse 2022-03-15 00:29:55 1145 Tiny change: 'eter).\n\n\n\n#### P' -> 'eter).\n\n![ ](https://i.imgur.com/45DIau0.png)\n\n#### P'
en4 TheScrasse 2022-03-14 23:59:01 19
en3 TheScrasse 2022-03-14 19:50:27 842
en2 TheScrasse 2022-03-14 19:05:58 678 Tiny change: ' diameter.\n\n#### P' -> ' diameter._\n\n#### P'
en1 TheScrasse 2022-03-14 18:47:40 14006 Initial revision (saved to drafts)