- #1
Carl1
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Can You Explain the Puzzle Toad Proof?
- Context: MHB
- Thread starter Carl1
- Start date
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- Explanation Proof Puzzle Web
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SUMMARY
The discussion focuses on the Puzzle Toad proof, specifically the process of walkers exchanging positions along edges in a graph. The procedure involves traversing edges in increasing order of length, starting from the shortest edge $e_1$ to the longest edge $e_m$. At each stage $k$, two walkers swap positions along edge $e_k$, resulting in a total of $2m$ edges traveled. Consequently, the mean number of edges traveled per walker is calculated as $2m/n$, establishing that at least one walker must have traversed $2m/n$ or more edges.
PREREQUISITES- Understanding of graph theory concepts, particularly edges and vertices.
- Familiarity with the concept of walkers in graph traversal.
- Basic knowledge of mathematical proofs and mean calculations.
- Experience with the Puzzle Toad proof and its implications.
- Study advanced graph theory, focusing on edge traversal algorithms.
- Explore the implications of mean value calculations in combinatorial proofs.
- Research the Puzzle Toad proof in detail to understand its applications.
- Learn about different types of graph traversal techniques, such as depth-first and breadth-first search.
Mathematicians, computer scientists, and students interested in graph theory and combinatorial proofs will benefit from this discussion.
- #2
Opalg
Gold Member
MHB
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It would be easier to understand that proof if the second sentence was made more explicit:
At stage $k$ of the procedure, two walkers travel along the edge $e_k$. So the total number of edges traveled during the whole procedure is $2m$. The number of walkers is $n$. So the mean number of edges traveled by a walker is $2m/n$, and therefore at least one of the travellers must have walked along $2m/n$ or more edges.
As the Puzzle Toad site mentions, it is indeed a beautiful proof.
Then go through the edges of $E$ one by one in order, starting with the shortest edge $\color{red}e_1$ and finishing with the longest edge $\color{red}e_m$[/color] ... .
At the $k$th stage of this procedure, when you come to the edge $e_k = (u,v)$, the walkers currently at its endpoints $u$ and $v$ change place by walking along $e_k$. If they have not previously moved then that will be the first leg of their walk. But if one or both of them have already moved during a previous stage of the procedure, they will have moved along edges shorter than $e_k$. When the procedure ends (with the walkers at the endpoints of $e_m$ changing places), each walker will have traveled along some increasingly long sequence of edges.At stage $k$ of the procedure, two walkers travel along the edge $e_k$. So the total number of edges traveled during the whole procedure is $2m$. The number of walkers is $n$. So the mean number of edges traveled by a walker is $2m/n$, and therefore at least one of the travellers must have walked along $2m/n$ or more edges.
As the Puzzle Toad site mentions, it is indeed a beautiful proof.
- #3
Carl1
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- 0
Thank you so much. Now I understand the proof. It is beautiful.