Maximum number of elements in the path of a matrix

ghz 12hours ago ⋅ 2 views

I tried to solve a problem of map (matrix 4x4) using python.

I want to find Maximum number of elements in the path of a map provided the next node must be lesser than the previous node with all possible combinations of elements in the matrix.

4 8 7 3  
2 5 9 3  
6 3 2 5  
4 4 1 6

The movement is like from an element can move to east-west-north-south

For example from m[0][1] can move to m[0][2] and m[1][1] 4-> 8 or 2

Here is the sample code but i have no idea to how to recursively check every element.

#import itertools
n = 4 
matrix = [[4, 8, 7, 3 ], [2, 5, 9, 3 ], [6, 3, 2, 5 ], [4, 4, 1, 6]]
for index,ele in enumerate(matrix):
    vals=[]
    for i2,e2 in enumerate(ele):
        for index2,ele2 in enumerate(ele):
            if index < (n-1):
                if ele2 > matrix[index+1] [index2]:
                    vals.append(matrix[index+1] [index2])
            if index > 0:
                if ele2 > matrix[index-1] [index2]:
                    vals.append(matrix[index-1] [index2])
            if index2 < n-1:
                if ele2 > matrix[index] [index2+1]:
                    vals.append(matrix[index] [index2+1])
            if index2 >0:
                if ele2 > matrix[index] [index2-1]:
                    vals.append(matrix[index] [index2-1])

how to recurse this function to loop till the end

For Example the answer will be like 8-5-3-2-1 (Longest Path with decreasing factor)

Answer

To solve the problem of finding the maximum number of elements in a decreasing path in the matrix, we can use Depth-First Search (DFS) with memoization. The idea is to recursively explore all possible paths from each element while making sure that we only move to neighboring elements with smaller values, and we want to keep track of the longest path we can find for each element.

Approach:

  1. DFS with Memoization: For each matrix element, we will use DFS to explore its neighbors. We need to store (memoize) the longest path starting from each element so that we do not recompute it multiple times.
  2. Directions: We can move to any of the four possible neighbors (north, south, east, west).
  3. Recursive Exploration: We explore each neighboring element that has a smaller value (decreasing path).
  4. Base Case: If a cell has no valid neighbors to move to, the recursion stops, and we return a length of 1 (the element itself).

Code Implementation:

n = 4
matrix = [[4, 8, 7, 3], [2, 5, 9, 3], [6, 3, 2, 5], [4, 4, 1, 6]]

# Directions for moving (north, south, east, west)
directions = [(-1, 0), (1, 0), (0, -1), (0, 1)]

# Memoization table to store the longest path from each cell
memo = [[-1 for _ in range(n)] for _ in range(n)]

# Check if the current position is inside the matrix bounds
def is_valid(x, y):
    return 0 <= x < n and 0 <= y < n

# DFS to find the longest decreasing path starting from (x, y)
def dfs(x, y):
    # If already computed, return the stored result
    if memo[x][y] != -1:
        return memo[x][y]

    max_length = 1  # The path at least contains the current element
    for dx, dy in directions:
        nx, ny = x + dx, y + dy
        # Move to valid neighbors with a smaller value
        if is_valid(nx, ny) and matrix[nx][ny] < matrix[x][y]:
            max_length = max(max_length, 1 + dfs(nx, ny))

    memo[x][y] = max_length
    return memo[x][y]

# Find the longest decreasing path in the entire matrix
def find_longest_decreasing_path():
    longest_path = 0
    start_x, start_y = 0, 0

    # Compute the longest path starting from each cell
    for i in range(n):
        for j in range(n):
            longest_path = max(longest_path, dfs(i, j))

    return longest_path

# Finding the longest path
longest_path = find_longest_decreasing_path()
print("Longest decreasing path length:", longest_path)

Explanation:

  1. is_valid(x, y): A helper function to ensure that the current position is within bounds of the matrix.
  2. dfs(x, y): The main DFS function. It recursively explores all valid moves (in decreasing order) from the current cell. We memoize the results to avoid redundant calculations.
  3. Memoization (memo[x][y]): We store the longest decreasing path from each cell in the memo array to speed up the process.
  4. find_longest_decreasing_path(): This function iterates over each cell in the matrix and computes the longest path using DFS starting from that cell.

Output:

This code will output the length of the longest decreasing path. To track the path itself (not just the length), you can modify the code to store and return the actual path.

Modifying the Code to Return the Path:

To store the actual path, we can slightly modify the DFS function to track the sequence of elements in the longest decreasing path.

# DFS to find the longest decreasing path starting from (x, y)
def dfs(x, y):
    # If already computed, return the stored result
    if memo[x][y] != -1:
        return memo[x][y], []

    max_length = 1  # The path at least contains the current element
    longest_path = [matrix[x][y]]  # Start path with current element

    for dx, dy in directions:
        nx, ny = x + dx, y + dy
        # Move to valid neighbors with a smaller value
        if is_valid(nx, ny) and matrix[nx][ny] < matrix[x][y]:
            path_length, path = dfs(nx, ny)
            if path_length + 1 > max_length:
                max_length = path_length + 1
                longest_path = [matrix[x][y]] + path  # Prepend current element to the path

    memo[x][y] = max_length
    return max_length, longest_path

# Finding the longest decreasing path
def find_longest_decreasing_path():
    longest_path = []
    for i in range(n):
        for j in range(n):
            _, path = dfs(i, j)
            if len(path) > len(longest_path):
                longest_path = path

    return longest_path

# Finding the longest decreasing path
longest_path = find_longest_decreasing_path()
print("Longest decreasing path:", longest_path)

Output (with path):

Longest decreasing path: [8, 5, 3, 2, 1]

Explanation of Path Finding:

  • DFS: As we explore from each matrix element, we keep track of the longest path found, including the path itself.
  • Memoization: After computing the longest path from a given element, we store the result to avoid redundant calculations.
  • Building the Path: During the DFS exploration, when a better path is found (i.e., a longer path), we prepend the current element to the path to build the sequence from start to end.

With this solution, you will be able to both find the length of the longest decreasing path and the actual path itself.

#py