# 2D Peak Finding Algorithm in Python

## Introduction to 2D Peak Finding

The 2D peak discovery problem in computer science is finding a local peak in a 2D array or matrix. An element that is greater than or equal to its neighbouring elements in the left, right, up, and down directions is referred to as a local peak. This problem can be solved using a variety of algorithms, such as a divide-and-conquer strategy, an optimized method, and a brute-force method. It is an essential example of a 2D variant of the 1D peak-finding problem. This explanation will go over a productive divide-and-conquer strategy for handling the Python 2D peak finding challenge.

## 2D Peak Finding Algorithm

### Basic Idea

One popular and effective method for resolving the 2D peak finding problem is the divide and conquer strategy. The fundamental principle is to break the problem down into smaller subproblems and search in the direction most likely to contain a peak recursively. The binary search algorithm and this are comparable.

### Algorithm Steps

- We have two subproblems to work with when we begin in the centre column of the matrix: one on the left and one on the right.
- Determine which element in the middle column is the maximum (we'll refer to it as mid_element) and compare it to its neighbours up and down (above and below).
- We can return mid_element if it is a 2D peak, at which point we are done.
- The top half of the matrix is where the peak must be if the mid-element is less than up.
- The peak must be in the lower half of the matrix if the mid-element is less than down.
- Apply the same reasoning recursively to the subproblem that has a higher probability of having a peak.

## Implementation of 2D Peak Finding Algorithm in Python

Here is a divide-and-conquer implementation of the 2D peak discovery algorithm in Python:

**Code:**

def find_peak(matrix): rows, cols = len(matrix), len(matrix[0]) mid_col = cols // 2 mid_row = max(range(rows), key=lambda i: matrix[i][mid_col]) mid_element = matrix[mid_row][mid_col] if (mid_col == 0 or mid_element >= matrix[mid_row][mid_col - 1]) and (mid_col == cols - 1 or mid_element >= matrix[mid_row][mid_col + 1]): # If mid_element is a 2D peak, return it return mid_element if mid_col > 0 and matrix[mid_row][mid_col - 1] > mid_element: return find_peak([row[:mid_col] for row in matrix]) return find_peak([row[mid_col + 1:] for row in matrix]) matrix = [ [1, 2, 3], [4, 5, 6], [7, 8, 9] ] peak = find_peak(matrix) print(f"The 2D peak is: {peak}")

**Output:**

The 2D peak is: 9

**Explanation:**

This code shows how to apply the divide and conquer strategy to discover a 2D peak effectively. It begins in the middle column, does a neighbour comparison on the middle element, and then recursively reduces the search space. The temporal complexity of the algorithm is O(m * log(n)), where m and n are the matrix's row and column counts, respectively.

Let's examine the 2D peak discovery technique in more detail, along with a complexity analysis and some other points to think about.

## Complexity Analysis

The temporal complexity of the divide and conquer strategy previously discussed for 2D peak detection is O(m * log(n)), where denotes the number of rows and 'n' is the number of columns in the matrix. It is a succinct description of the temporal complexity:

- The first step, which takes O(m) time, is to determine the maximum element in the middle column.
- Subsequently, we either shrink the search space to the left or right submatrix. In the worst scenario, a logarithmic phrase results from repeatedly halving the search space.

Assuming we operate with slices of the original matrix rather than creating new ones, the division of the matrix results in O(log(m)) space complexity for the recursive function calls.

## Edge Cases

The algorithm handles various cases effectively, including:

- The procedure functions correctly in the case of a single-row matrix since there is no horizontal division.
- Since there is just one column to take into account in a matrix with a single column, the peak will be found.
- The method will find a peak in one of the middle rows or columns if the matrix is completely sorted in that row or column.

## Optimizations

The code that was previously shown is simple and illustrates the fundamental idea of the divide-and-conquer strategy. Still, there are a few adjustments that can be done to increase its effectiveness:

**1. Choosing the Middle Element:** You can start by choosing the middle row and then discover the maximum there, as opposed to starting with the middle column. In certain instances, this might be more effective.

**2.**** ****Randomized Starting Point:** Instead of always beginning in the centre column, you can start at any random place. This method introduces some unpredictability, which helps lower the worst-case temporal complexity.

**3.**** ****Early Exit:** You can choose to restrict your search to the left or right submatrix during the recursive calls if you determine that this submatrix is more likely to contain the peak. It will cut down on the number of recursive calls.

**4.**** ****Parallelization:** To expedite the operation in large matrices, you can parallelize the left and right submatrices' peak search.

## Conclusion

In conclusion, the divide and conquer strategy used by the 2D peak identification algorithm offers a quick and practical solution to the issue of detecting local peaks in a two-dimensional matrix. This technique effectively finds peaks in the matrix by recursively reducing the possibilities and intelligently partitioning the search space. It is a viable option for many applications due to its O(m * log(n)) time complexity, which can handle a variety of edge cases and be further improved to enhance speed. The 2D peak finding algorithm is an essential tool for discovering important features in two-dimensional spaces, with applications ranging from computer graphics and image processing to terrain study.