Docstrings and formatting improvements (TheAlgorithms#2418)

* Fix spelling in docstrings

* Improve comments and formatting

* Update print statement to reflect doctest change

* improve phrasing and apply black

* Update rat_in_maze.py

    This method is recursive starting from (i, j) and going in one of four directions:
    up, down, left, right.
    If a path is found to destination it returns True otherwise it returns False.

Co-authored-by: Christian Clauss <cclauss@me.com>

Author: Hasenn

backtracking/hamiltonian_cycle.py

@@ -51,16 +51,16 @@ def util_hamilton_cycle(graph: List[List[int]], path: List[int], curr_ind: int)
     """
     Pseudo-Code
     Base Case:
-    1. Chceck if we visited all of vertices
+    1. Check if we visited all of vertices
         1.1 If last visited vertex has path to starting vertex return True either
             return False
     Recursive Step:
     2. Iterate over each vertex
         Check if next vertex is valid for transiting from current vertex
             2.1 Remember next vertex as next transition
             2.2 Do recursive call and check if going to this vertex solves problem
-            2.3 if next vertex leads to solution return True
-            2.4 else backtrack, delete remembered vertex
+            2.3 If next vertex leads to solution return True
+            2.4 Else backtrack, delete remembered vertex
 
     Case 1: Use exact graph as in main function, with initialized values
     >>> graph = [[0, 1, 0, 1, 0],

backtracking/rat_in_maze.py

@@ -1,13 +1,13 @@
 def solve_maze(maze: list) -> bool:
     """
-    This method solves rat in maze algorithm.
-    In this problem we have n by n matrix and we have start point and end point
-    we want to go from source to distination. In this matrix 0 are block paths
-    1 are open paths we can use.
+    This method solves the "rat in maze" problem.
+    In this problem we have some n by n matrix, a start point and an end point.
+    We want to go from the start to the end. In this matrix zeroes represent walls
+    and ones paths we can use.
     Parameters :
         maze(2D matrix) : maze
     Returns:
-        Return: True is maze has a solution or False if it does not.
+        Return: True if the maze has a solution or False if it does not.
     >>> maze = [[0, 1, 0, 1, 1],
     ...         [0, 0, 0, 0, 0],
     ...         [1, 0, 1, 0, 1],
@@ -47,13 +47,13 @@ def solve_maze(maze: list) -> bool:
     ...         [0, 1, 0],
     ...         [1, 0, 0]]
     >>> solve_maze(maze)
-    Solution does not exists!
+    No solution exists!
     False
 
     >>> maze = [[0, 1],
     ...         [1, 0]]
     >>> solve_maze(maze)
-    Solution does not exists!
+    No solution exists!
     False
     """
     size = len(maze)
@@ -63,16 +63,15 @@ def solve_maze(maze: list) -> bool:
     if solved:
         print("\n".join(str(row) for row in solutions))
     else:
-        print("Solution does not exists!")
+        print("No solution exists!")
     return solved
 
 
 def run_maze(maze, i, j, solutions):
     """
-    This method is recursive method which starts from i and j
-    and goes with 4 direction option up, down, left, right
-    if path found to destination it breaks and return True
-    otherwise False
+    This method is recursive starting from (i, j) and going in one of four directions:
+    up, down, left, right.
+    If a path is found to destination it returns True otherwise it returns False.
     Parameters:
         maze(2D matrix) : maze
         i, j : coordinates of matrix

boolean_algebra/quine_mc_cluskey.py

@@ -146,7 +146,7 @@ def main():
     minterms = [
         int(x)
         for x in input(
-            "Enter the decimal representation of Minterms 'Spaces Seprated'\n"
+            "Enter the decimal representation of Minterms 'Spaces Separated'\n"
         ).split()
     ]
     binary = decimal_to_binary(no_of_variable, minterms)

cellular_automata/one_dimensional.py

@@ -32,8 +32,9 @@ def new_generation(cells: List[List[int]], rule: List[int], time: int) -> List[i
     next_generation = []
     for i in range(population):
         # Get the neighbors of each cell
-        left_neighbor = 0 if i == 0 else cells[time][i - 1]  # special: leftmost cell
-        right_neighbor = 0 if i == population - 1 else cells[time][i + 1]  # rightmost
+        # Handle neighbours outside bounds by using 0 as their value
+        left_neighbor = 0 if i == 0 else cells[time][i - 1]
+        right_neighbor = 0 if i == population - 1 else cells[time][i + 1]
         # Define a new cell and add it to the new generation
         situation = 7 - int(f"{left_neighbor}{cells[time][i]}{right_neighbor}", 2)
         next_generation.append(rule[situation])