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Fix typos #1150

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Sep 16, 2023
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Fix typos #1150

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4 changes: 2 additions & 2 deletions src/graph/bridge-searching-online.md
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Expand Up @@ -79,7 +79,7 @@ We will now consistently disassemble every operation that we need to learn to im
For example you can re-root the tree of vertex $a$, and then attach it to another tree by setting the ancestor of $a$ to $b$.

However the question about the effectiveness of the re-rooting operation arises:
in order to re-root the tree with the root $r$ to the vertex $v$, it is necessary to necessary to visit all vertices on the path between $v$ and $r$ and redirect the pointers `par[]` in the opposite direction, and also change the references to the ancestors in the DSU that is responsible for the connected components.
in order to re-root the tree with the root $r$ to the vertex $v$, it is necessary to visit all vertices on the path between $v$ and $r$ and redirect the pointers `par[]` in the opposite direction, and also change the references to the ancestors in the DSU that is responsible for the connected components.

Thus, the cost of re-rooting is $O(h)$, where $h$ is the height of the tree.
You can make an even worse estimate by saying that the cost is $O(\text{size})$ where $\text{size}$ is the number of vertices in the tree.
Expand All @@ -103,7 +103,7 @@ We will now consistently disassemble every operation that we need to learn to im

* Searching for the cycle formed by adding a new edge $(a, b)$.
Since $a$ and $b$ are already connected in the tree we need to find the [Lowest Common Ancestor](lca.md) of the vertices $a$ and $b$.
The cycle will consist of the paths from $b$ to the LCA, from the LCA to $b$ and the edge $a$ to $b$.
The cycle will consist of the paths from $b$ to the LCA, from the LCA to $a$ and the edge $a$ to $b$.

After finding the cycle we compress all vertices of the detected cycle into one vertex.
This means that we already have a complexity proportional to the cycle length, which means that we also can use any LCA algorithm proportional to the length, and don't have to use any fast one.
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