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67 lines
2.8 KiB
Markdown
67 lines
2.8 KiB
Markdown
---
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id: bfs-tree
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title: Distance boundaries from BFS tree on undirected graphs
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description: |
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Short explanation of distance boundaries deduced from a BFS tree.
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tags:
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- graphs
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- bfs
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last_update:
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date: 2022-04-30
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---
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## Introduction
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As we have talked on the seminar, if we construct from some vertex $u$ BFS tree on an undirected graph, we can obtain:
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- lower bound of length of the shortest path between 2 vertices from the _height difference_
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- upper bound of length of the shortest path between 2 vertices from the _path through the root_
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## Lower bound
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Consider the following graph:
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We run BFS from the vertex $a$ and obtain the following BFS tree:
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Let's consider pair of vertices $e$ and $h$. For them we can safely lay, from the BFS tree, following properties:
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- lower bound: $2$
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- upper bound: $4$
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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 ;)_)
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## Proof by contradiction
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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”! ;)
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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!
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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.
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:::tip
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Do we need to run BFS after **every** change?
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I am leaving that as an exercise ;)
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:::
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Oops, we have gotten a new BFS tree, that has a height difference of 1.
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:::tip
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Try to think about a way this can be generalized for shortening of minimal length 3 to minimal length 2 ;)
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:::
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