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Count of graphs formed by changing color of any red-colored node with black parent to black
Introduction
In graph theory, nodes and edges form the fundamental units of connected structures. They are widely used to represent a broad range of relationships and connections between different entities. In this article, we will dive into an interesting problem involving counting graphs formed by changing the color of red-colored nodes with black parent nodes in C++. We will explain the concept behind graph coloring, introduce an algorithmic approach to solve this problem, and provide a detailed C++ code implementation that we can use.
Count of graphs formed by changing color
Graph coloring is a concept that involves assigning colors to various elements within a graph such that no two adjacent elements share the same color. This technique finds applications across several domains like map coloring problems or scheduling conflicts in event management systems. Typically, two-coloring like black and white or multi-coloring schemes are utilized for different scenarios.
We have a graph consisting of nodes where each node has either its default color (black) or an additional color option (red). Our task is to count all the possible resulting distinct graphs generated by switching any red-colored node's color to black if it has a parent with black as its current color.
Example
Let's consider the following binary tree −
(Root)A
/ \
B(Black) C(Red)
/ \ /
D(Black) E(Black) F(Red)
Initially, we start from the root node A. Since both nodes C and F are red with a parent being black, we change their color to black. In the above graph, the root node is only black and all other are red.
The resulting updated graph would be −
Root(A) / \ B C <-(Red turned Black) / \ / D(Black) E F <-(Red turned Black)
Hence, after applying our algorithm, one distinct graph is formed.
Approach 1: C++ code to return the count of graphs formed by changing the color of any red colored node to black parent node using the recursion function
The function is called recursively at the end to get the count value. The graph is initialized and we can traverse through the graph using the method called Depth-first search. The count is nothing but the value formed by changing the red nodes with their black parents to black.
Algorithm
Step 1 − We start by creating a structure for our individual nodes, which consists of two properties − 'color' representing whether it’s 'black' or 'red', and 'child' holding pointers or references to its children.
Step 2 − Next, we construct the sample graph using these defined nodes and set their colors accordingly.
Step 3 − Now comes the crucial part where recursion plays a vital role in traversing through each node and checking if it satisfies our condition mentioned earlier which is changing any red-color coded node connected directly with a black-colored parent to black. We will increment the 'count' value whenever such operations occur.
Step 4 − Finally, we call our recursive function on the root node of our graph to obtain the desired result - counting all distinct graphs formed after modifying the colors.
Step 5 − At last, there are 10 different possible graphs that can be formed.
Example
//Including the required header files
#include <string>
#include <vector>
#include <iostream>
using namespace std;
struct Node {
string color;
vector<Node*> child;
};
Node* newNode(string col){
Node* temp = new Node();
temp -> color = col;
return temp;
}
Node* buildSampleGraph() {
// Initialize Nodes
Node* root = newNode("red");
Node* node1 = newNode("black");
Node* node2 = newNode("red");
Node* node3 = newNode("red");
Node* node4 = newNode("red");
Node* node5 = newNode("red");
// Build Graph
root -> child.push_back(node1);
root -> child.push_back(node2);
node2 -> child.push_back(node3);
node3 -> child.push_back(node4);
node3 -> child.push_back(node5);
root->color="black";
return root;
}
int dfs(Node* node) {
int count = 1;
// Base case: If there are no children, return 1
if (node -> child.empty())
return count;
for (auto child : node->child) {
// Check if the child's color is red and its parent's color is black.
if (child -> color == "red" && node -> color == "black") {
child -> color = "black";
count++;
}
// Continue traversing the graph
count += dfs(child);
}
return count;
}
int main() {
Node* root = buildSampleGraph();
int totalGraphsCount = dfs(root);
cout << "The total number of distinct graphs formed by changing red nodes with their black parents to black is: ";
cout << totalGraphsCount << endl;
return 0;
}
Output
The total number of distinct graphs formed by changing red nodes with their black parents to black is: 10
Conclusion
The problem of counting graphs formed by changing the color of specific nodes in given criteria presents an interesting challenge in graph theory. By utilizing concepts like DFS or BFS algorithms along with recursive function calls, we can efficiently traverse through binary trees while identifying and modifying relevant nodes based on the color conditions provided.
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