Help with Ramsey's Theorem Homework

[RyozKidz]

Homework Statement

Ramsey's theorem. Let G be a graph . A clique in G is a subgraph in which every two nodes are connected by an edge . An anti-clique , also called an independent set , is a subgraph in which every two nodes are not connected by an edge . Show that every graph with n nodes contains either a clique or an anti-clique with at least 1/2log₂n nodes.

Homework Equations

3. The Attempt at a Solution [/b

Make space for two piles of nodes , A and B . Then , starting with the entire graph , repeatedly add each remaining node x to A if its degree is greater than one half the number of remaining nodes and to B otherwise , and discard all nodes to which x isn't(is) connected if it was added to A(B) . Continue until no nodes are left. At most half of the nodes are discarded at each of these steps , so at least log₂ n steps will occur before the process terminates . Each step adds a node to one of the piles , so one of the piles ends up with at least 1/2log₂ n nodes. The pile contains the nodes of a clique and the B pile contains the nodes of an anti-clique.

I can't interpret this solution ..~ can anyone help me out..~ pls...

 

[KataKoniK]

~

Sure, let me try to explain the solution in simpler terms.

The problem is asking us to show that in any graph with n nodes, there will always be either a clique (a subgraph where all nodes are connected) or an anti-clique (a subgraph where no nodes are connected) with at least 1/2log2n nodes. In other words, there will always be a group of at least 1/2log2n nodes that are either all connected or all not connected.

To prove this, we can use a strategy called "divide and conquer." Imagine we have two piles of nodes, A and B. We will start with the entire graph and go through each node one by one. For each node, we will add it to pile A if its degree (the number of edges it has) is greater than half of the number of remaining nodes. Otherwise, we will add it to pile B.

For example, let's say we have a graph with 10 nodes. We start with all 10 nodes in one pile. We look at the first node and see that it has 6 edges. Since 6 is greater than half of the remaining 9 nodes (which is 4.5, but we round up to 5), we add it to pile A. Then we look at the next node and see that it only has 2 edges. Since 2 is less than 5, we add it to pile B. We continue this process until we have gone through all 10 nodes.

Now, at each step, we are either adding a node to pile A or pile B, and we are discarding all nodes that are not connected to the node we are currently looking at. This means that at most half of the nodes will be discarded at each step. So at least half of the nodes will still be left after each step.

Since we are adding a node to either pile A or pile B at each step, and at least half of the nodes will still be left after each step, it will take at least log2n steps before we run out of nodes. This is because log2n is the number of times you can divide n in half before you get to a number less than 1.

By the end of this process, one of the piles (either A or B) will have at least 1/2log2n nodes. This pile will represent either a clique or an

 

[Mene]

I would suggest breaking down the solution step by step and providing clear explanations for each step. This will help the reader understand the solution better. Also, providing diagrams or examples can make it easier to visualize the process. Additionally, you can provide alternative ways to approach the problem and explain the logic behind each method. It may also be helpful to provide further resources or references for better understanding of Ramsey's theorem.