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Docstrings and formatting improvements (#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>
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@ -51,7 +51,7 @@ def util_hamilton_cycle(graph: List[List[int]], path: List[int], curr_ind: int)
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"""
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Pseudo-Code
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Base Case:
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1. Chceck if we visited all of vertices
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1. Check if we visited all of vertices
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1.1 If last visited vertex has path to starting vertex return True either
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return False
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Recursive Step:
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@ -59,8 +59,8 @@ def util_hamilton_cycle(graph: List[List[int]], path: List[int], curr_ind: int)
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Check if next vertex is valid for transiting from current vertex
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2.1 Remember next vertex as next transition
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2.2 Do recursive call and check if going to this vertex solves problem
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2.3 if next vertex leads to solution return True
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2.4 else backtrack, delete remembered vertex
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2.3 If next vertex leads to solution return True
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2.4 Else backtrack, delete remembered vertex
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Case 1: Use exact graph as in main function, with initialized values
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>>> graph = [[0, 1, 0, 1, 0],
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@ -1,13 +1,13 @@
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def solve_maze(maze: list) -> bool:
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"""
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This method solves rat in maze algorithm.
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In this problem we have n by n matrix and we have start point and end point
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we want to go from source to distination. In this matrix 0 are block paths
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1 are open paths we can use.
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This method solves the "rat in maze" problem.
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In this problem we have some n by n matrix, a start point and an end point.
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We want to go from the start to the end. In this matrix zeroes represent walls
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and ones paths we can use.
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Parameters :
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maze(2D matrix) : maze
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Returns:
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Return: True is maze has a solution or False if it does not.
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Return: True if the maze has a solution or False if it does not.
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>>> maze = [[0, 1, 0, 1, 1],
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... [0, 0, 0, 0, 0],
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... [1, 0, 1, 0, 1],
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@ -47,13 +47,13 @@ def solve_maze(maze: list) -> bool:
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... [0, 1, 0],
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... [1, 0, 0]]
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>>> solve_maze(maze)
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Solution does not exists!
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No solution exists!
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False
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>>> maze = [[0, 1],
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... [1, 0]]
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>>> solve_maze(maze)
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Solution does not exists!
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No solution exists!
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False
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"""
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size = len(maze)
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@ -63,16 +63,15 @@ def solve_maze(maze: list) -> bool:
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if solved:
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print("\n".join(str(row) for row in solutions))
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else:
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print("Solution does not exists!")
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print("No solution exists!")
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return solved
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def run_maze(maze, i, j, solutions):
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"""
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This method is recursive method which starts from i and j
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and goes with 4 direction option up, down, left, right
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if path found to destination it breaks and return True
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otherwise False
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This method is recursive starting from (i, j) and going in one of four directions:
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up, down, left, right.
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If a path is found to destination it returns True otherwise it returns False.
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Parameters:
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maze(2D matrix) : maze
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i, j : coordinates of matrix
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@ -146,7 +146,7 @@ def main():
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minterms = [
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int(x)
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for x in input(
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"Enter the decimal representation of Minterms 'Spaces Seprated'\n"
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"Enter the decimal representation of Minterms 'Spaces Separated'\n"
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).split()
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]
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binary = decimal_to_binary(no_of_variable, minterms)
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@ -32,8 +32,9 @@ def new_generation(cells: List[List[int]], rule: List[int], time: int) -> List[i
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next_generation = []
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for i in range(population):
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# Get the neighbors of each cell
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left_neighbor = 0 if i == 0 else cells[time][i - 1] # special: leftmost cell
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right_neighbor = 0 if i == population - 1 else cells[time][i + 1] # rightmost
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# Handle neighbours outside bounds by using 0 as their value
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left_neighbor = 0 if i == 0 else cells[time][i - 1]
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right_neighbor = 0 if i == population - 1 else cells[time][i + 1]
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# Define a new cell and add it to the new generation
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situation = 7 - int(f"{left_neighbor}{cells[time][i]}{right_neighbor}", 2)
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next_generation.append(rule[situation])
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