mirror of
https://github.com/mfocko/blog.git
synced 2024-11-10 08:19:07 +01:00
68 lines
2.8 KiB
Markdown
68 lines
2.8 KiB
Markdown
---
|
||
id: bfs-tree
|
||
title: Distance boundaries from BFS tree on undirected graphs
|
||
description: |
|
||
Short explanation of distance boundaries deduced from a BFS tree.
|
||
tags:
|
||
- graphs
|
||
- bfs
|
||
last_update:
|
||
date: 2022-04-30
|
||
---
|
||
|
||
## Introduction
|
||
|
||
As we have talked on the seminar, if we construct from some vertex $u$ BFS tree on an undirected graph, we can obtain:
|
||
|
||
- lower bound of length of the shortest path between 2 vertices from the _height difference_
|
||
- upper bound of length of the shortest path between 2 vertices from the _path through the root_
|
||
|
||
## Lower bound
|
||
|
||
Consider the following graph:
|
||
|
||
|
||
|
||
|
||
We run BFS from the vertex $a$ and obtain the following BFS tree:
|
||
|
||
|
||
|
||
|
||
Let's consider pair of vertices $e$ and $h$. For them we can safely lay, from the BFS tree, following properties:
|
||
|
||
- lower bound: $2$
|
||
- upper bound: $4$
|
||
|
||
By having a look at the graph we started from, we can see that we have a path ‹$e, j, h$› that has a length 2. Apart from that we can also notice there is another path from $e$ to $h$ and that is ‹$e, a, c, i, d, h$›. And that path has a length of $5$. Doesn't this break our statements at the beginning? (_I'm leaving that as an exercise ;)_)
|
||
|
||
## Proof by contradiction
|
||
|
||
Let's keep the same graph, but break the lower bound, i.e. I have gotten a lower bound $2$, but “there must be a shorter path”! ;)
|
||
|
||
Now the more important question, is there a shorter path in that graph? The answer is no, there's no shorter path than the one with length $2$. So what can we do about it? We'll add an edge to have a shorter path. Now we have gotten a lower bound of $2$, which means the only shorter path we can construct has $1$ edge and that is ‹$e, h$› (no intermediary vertices). Let's do this!
|
||
|
||
|
||
|
||
|
||
|
||
Okay, so we have a graph that breaks the rule we have laid. However, we need to run BFS to obtain the new BFS tree, since we have changed the graph.
|
||
|
||
:::tip
|
||
|
||
Do we need to run BFS after **every** change?
|
||
|
||
I am leaving that as an exercise ;)
|
||
|
||
:::
|
||
|
||
|
||
|
||
|
||
Oops, we have gotten a new BFS tree, that has a height difference of 1.
|
||
|
||
:::tip
|
||
|
||
Try to think about a way this can be generalized for shortening of minimal length 3 to minimal length 2 ;)
|
||
|
||
:::
|