diff --git a/algorithms/backtracking/Knight'stour.cpp b/algorithms/backtracking/Knight'stour.cpp
new file mode 100644
index 00000000..a0d8efb8
--- /dev/null
+++ b/algorithms/backtracking/Knight'stour.cpp
@@ -0,0 +1,82 @@
+//
+// C/C++ program of Knight's Tour using Backtracking
+//
+// The All ▲lgorithms Project
+
+// https://github.com/allalgorithms/cpp
+//
+// Contributed by: Prayag Thakur
+// Github: @PrAyAg9
+//
+
+#include <iostream>
+#include <vector>
+#define N 8 // Chessboard is N x N (8 x 8 in this case)
+using namespace std;
+
+// Utility function to check if the move is valid
+bool isSafe(int x, int y, vector<vector<int>> &board) {
+    return (x >= 0 && x < N && y >= 0 && y < N && board[x][y] == -1);
+}
+
+// Utility function to print the board
+void printSolution(const vector<vector<int>> &board) {
+    for (int x = 0; x < N; ++x) {
+        for (int y = 0; y < N; ++y) {
+            cout << board[x][y] << "\t";
+        }
+        cout << endl;
+    }
+}
+
+// A recursive utility function to solve Knight Tour problem
+bool solveKnightTour(int x, int y, int movei, vector<vector<int>> &board,
+                     const vector<int> &moveX, const vector<int> &moveY) {
+    int nextX, nextY;
+    // Base case: If all squares are visited, return true
+    if (movei == N * N)
+        return true;
+
+    // Try all next moves from the current coordinate x, y
+    for (int k = 0; k < 8; ++k) {
+        nextX = x + moveX[k];
+        nextY = y + moveY[k];
+        if (isSafe(nextX, nextY, board)) {
+            board[nextX][nextY] = movei;
+            if (solveKnightTour(nextX, nextY, movei + 1, board, moveX, moveY))
+                return true;
+            else
+                // Backtrack: Undo the current move
+                board[nextX][nextY] = -1;
+        }
+    }
+    return false;
+}
+
+// This function solves the Knight Tour problem using Backtracking.
+// This function mainly uses solveKnightTour() to solve the problem.
+bool knightTour() {
+    // Initialize the chessboard with -1 (unvisited squares)
+    vector<vector<int>> board(N, vector<int>(N, -1));
+
+    // The knight's possible moves in the x and y directions
+    vector<int> moveX = {2, 1, -1, -2, -2, -1, 1, 2};
+    vector<int> moveY = {1, 2, 2, 1, -1, -2, -2, -1};
+
+    // Knight starts at the first block (0, 0)
+    board[0][0] = 0;
+
+    // Start the Knight's Tour from the first position
+    if (!solveKnightTour(0, 0, 1, board, moveX, moveY)) {
+        cout << "Solution does not exist." << endl;
+        return false;
+    } else
+        printSolution(board);
+
+    return true;
+}
+
+int main() {
+    knightTour();
+    return 0;
+}
diff --git a/algorithms/backtracking/M_ColoringProblem.cpp b/algorithms/backtracking/M_ColoringProblem.cpp
new file mode 100644
index 00000000..39b09767
--- /dev/null
+++ b/algorithms/backtracking/M_ColoringProblem.cpp
@@ -0,0 +1,89 @@
+// C/C++ M-colouring Problem using Backtracking
+//
+// The All ▲lgorithms Project
+
+// https://github.com/allalgorithms/cpp
+//
+// Contributed by: Prayag Thakur
+// Github: @PrAyAg9
+
+// Problem Statement:
+// Given an undirected graph and a number M, determine if the graph can be colored using at most M different colors such that no two adjacent vertices have the same color.
+
+#include <iostream>
+#include <vector>
+using namespace std;
+
+// Function to check if the current color assignment is safe for vertex v
+bool isSafe(int v, const vector<vector<int>> &graph, const vector<int> &color, int c) {
+    // Check for each adjacent vertex
+    for (int i = 0; i < graph.size(); i++) {
+        // If the vertex is adjacent and has the same color, return false
+        if (graph[v][i] == 1 && color[i] == c) {
+            return false;
+        }
+    }
+    return true;
+}
+
+// A recursive utility function to solve the M-coloring problem
+bool solveMColoring(int v, const vector<vector<int>> &graph, vector<int> &color, int m) {
+    // Base case: If all vertices are assigned a color, return true
+    if (v == graph.size()) {
+        return true;
+    }
+
+    // Try assigning each color from 1 to m
+    for (int c = 1; c <= m; c++) {
+        // Check if assigning color c to vertex v is safe
+        if (isSafe(v, graph, color, c)) {
+            color[v] = c; // Assign the color
+
+            // Recursively assign colors to the rest of the vertices
+            if (solveMColoring(v + 1, graph, color, m)) {
+                return true;
+            }
+
+            // If assigning color c doesn't lead to a solution, backtrack
+            color[v] = 0;
+        }
+    }
+    return false; // If no color can be assigned, return false
+}
+
+// Function to solve the M-coloring problem
+bool graphColoring(const vector<vector<int>> &graph, int m) {
+    int V = graph.size();
+    vector<int> color(V, 0); // Initialize all vertices with no color (0)
+
+    // Call the solveMColoring function starting from the first vertex
+    if (!solveMColoring(0, graph, color, m)) {
+        cout << "Solution does not exist." << endl;
+        return false;
+    }
+
+    // Print the solution
+    cout << "Solution exists: Following are the assigned colors" << endl;
+    for (int i = 0; i < V; i++) {
+        cout << "Vertex " << i << " ---> Color " << color[i] << endl;
+    }
+    return true;
+}
+
+int main() {
+    // Example: A graph represented as an adjacency matrix
+    // (1 means an edge between two vertices, 0 means no edge)
+    vector<vector<int>> graph = {
+        {0, 1, 1, 1},
+        {1, 0, 1, 0},
+        {1, 1, 0, 1},
+        {1, 0, 1, 0}
+    };
+    
+    int m = 3; // Number of colors
+
+    // Solve the M-Coloring problem
+    graphColoring(graph, m);
+
+    return 0;
+}