Ramsey Numbers¶
In graph theory, the Ramsey number R(m, n) represents the smallest number such that any graph with this many vertices contains either a complete subgraph with m vertices or an independent subgraph with n vertices.
Example: R(3, 3) = 6. This means that in any graph with 6 vertices, there must be either a triangle (complete 3-vertex subgraph) or an independent set of 3 vertices.
Note: Calculating Ramsey numbers is extremely challenging, and only a few exact values have been determined to date.
Sample code for visualizing a graph with 5 vertices:
# Importing the NetworkX graph library
import networkx as nx
import matplotlib.pyplot as plt
# Creating a graph with 5 vertices
G = nx.Graph()
# اضافه کردن یالها
G.add_edges_from([(1,2), (1,3), (2,3), (2,4), (3,5), (4,5)])
# نمايش گراف
nx.draw(G, with_labels=True, node_color='lightblue')
plt.show()
Note: This is a simplified example of modeling a Ramsey problem in Python. In practice, investigating higher Ramsey numbers requires more sophisticated algorithms and intensive computations.
Introduction¶
A tree is a connected acyclic graph. Below is an example of tree representation in code:
class TreeNode:
def __init__(self, data):
self.data = data
self.children = []
def print_tree(root): # Prints the tree
if not root:
return
print(root.data)
for child in root.children:
print_tree(child)
Trees have wide applications in data structures and algorithms. A tree with n nodes has exactly n-1 edges. For example, the tree in the image above has 5 nodes and 4 edges. Key properties include:
There is exactly one unique path between any two nodes
Adding any edge creates a cycle
Removing any edge disconnects the graph
First Problem¶
We have a group of 6 people where every two are either friends or enemies. Prove that there exist three people who are either all mutual friends or all mutual enemies.
Definition¶
The function \(R(a,b)\) is equal to the smallest \(n\) such that no matter how we color the edges of a \(K_n\) with two colors, blue and red, there will either be a blue \(K_a\) or a red \(K_b\).
You can easily prove that the above problem is equivalent to demonstrating the statement \(R(3,3) \leq 6\).
Bounds¶
Obtaining the exact value of \(R(a,b)\) is not possible, but we can provide bounds for it.
First, we must prove that \(R(a, b)\) exists. (It might be that for no \(n\), the desired property holds for \(K_n\))
The solution idea is to find a vertex with sufficiently many blue or red adjacents, then consider only its neighbors and find a smaller blue or red clique there.
More precisely, suppose there exists a vertex \(u\) with at least \(R(a-1,b)\) blue neighbors. In this case, considering its neighbors leads to two scenarios:
There exists a blue \(K_{a-1}\) among them, in which case adding vertex \(u\) creates a blue \(K_a\).
There exists a red \(K_b\) among them.
Therefore, if there exists a vertex with at least \(R(a-1,b)\) blue neighbors, the problem is solved. Similarly, if there exists a vertex with at least \(R(a, b-1)\) red neighbors, the problem is also solved.
Thus, we conclude that \(R(a,b) \leq R(a-1,b) + R(a,b-1)\). Because if our graph has at least \(R(a-1,b) + R(a,b-1)\) vertices, any arbitrary vertex will either have sufficiently many blue neighbors or sufficiently many red neighbors.
This inequality reminds us of Pascal's identity ( \(\binom{n}{k} = \binom{n-1}{k} + \binom{n-1}{k-1}\) ).
Furthermore, we can prove by induction that \(R(a, b) \leq \binom{a+b}{a}\).
Generalization to k Dimensions
------------------------------
Suppose we have an n-element set, and we have colored each of its k-element subsets with either blue or red. We call a subset :math:`A` a **k-dimensional cluster** if all k-element subsets of :math:`A` have the same color (if blue, we call it a blue cluster; if red, a red cluster).
We define :math:`R_k(a,b)` as the minimum :math:`n` such that no matter how we color the k-element subsets of an n-element set with blue and red, there will either be a blue cluster of size :math:`a` or a red cluster of size :math:`b`.
**Proof idea**: Similar to the previous approach. Assume :math:`n = R_k(a,b)`. Consider a specific element :math:`u` from the n-element set :math:`A`. Let :math:`B = A - \{u\}`. Color each :math:`(k-1)`-element subset :math:`S` of :math:`B` with the color of the k-element subset :math:`S \cup \{u\}`. If the size of :math:`B` is at least :math:`R_{k-1}( R_k(a-1,b), R_k(a,b-1) )`, then one of the following must occur:
- :math:`B` contains a :math:`(k-1)`-dimensional blue cluster :math:`S` of size at least :math:`R_k(a-1,b)`. In this case, either :math:`S` contains a :math:`k`-dimensional red cluster of size :math:`b` (solving the problem), or :math:`S` contains a :math:`k`-dimensional blue cluster of size :math:`a-1`. We can then add :math:`u` to this cluster to form a :math:`k`-dimensional blue cluster of size :math:`a`.
- :math:`B` contains a :math:`(k-1)`-dimensional red cluster :math:`S` of size at least :math:`R_k(a,b-1)`. In this case, either :math:`S` contains a :math:`k`-dimensional blue cluster of size :math:`a` (solving the problem), or :math:`S` contains a :math:`k`-dimensional red cluster of size :math:`b-1`. We can then add :math:`u` to this cluster to form a :math:`k`-dimensional red cluster of size :math:`b`.
Based on this reasoning, we can prove that :math:`R_k(a,b) \leq R_{k-1}(R_k(a-1,b),R_k(a,b-1))`.
Generalization to k Dimensions and c Colors¶
Similarly, we can formulate and prove the problem for more than two colors. We leave finding bounds on \(R_k(a_1,a_2,...,a_c)\) to the reader.
This book is made by The Shaazzz contributors.
It is publicly available with cc-by-sa license.
Last updated at آوریل 19, 2025.
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