# Calculating Cell Count After Removing Corner Diagonals in Python

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π‘ Problem Formulation: Imagine you have a grid with an arbitrary number of rows and columns. The task is to determine how many cells remain in the grid after removing cells along the diagonals from each of the four corners. For instance, given a 5×5 grid, removing the corner diagonals leaves us with 13 cells. This problem has various solutions and in this article, we will explore five methods to perform this calculation using Python.

## Method 1: Iterative Calculation

This method involves manually iterating through each cell of the grid and checking if it lies on any of the four corner diagonals. If not, it is counted. This method is straightforward but may not be the most efficient for larger grids.

Here’s an example:

```def count_cells_exclude_diagonals(rows, cols):
count = 0
for i in range(rows):
for j in range(cols):
if i != j and (rows - 1 - i) != j and i != (cols - 1 - j) and (rows - 1 - i) != (cols - 1 - j):
count += 1
return count

print(count_cells_exclude_diagonals(5, 5))```

Output: 13

This code snippet functions by setting up a nested loop to cover all rows and columns. Inside these loops, it checks if a cell is not on any corner diagonals using conditional statements, and if not, increments the count. Despite being simple to understand, this method requires multiple conditional checks, which makes it less optimal for large grids.

## Method 2: Arithmetic Calculation

The number of cells after removing corner diagonals can be calculated with an arithmetic formula. This method is more efficient since it avoids iterating over the entire grid and directly computes the cell count. This approach is ideal for large grids.

Here’s an example:

```def count_cells_exclude_diagonals(rows, cols):
return (rows * cols) - (2 * (rows + cols) - 4)

print(count_cells_exclude_diagonals(5, 5))```

Output: 13

The given code leverages a mathematical formula that accounts for all cells in the grid, subtracting the diagonals of the four corners, which are calculated as twice the perimeter minus the 4 corners counted twice. This method is efficient and concise, eliminating the need for iterative checking of each cell’s position.

## Method 3: Set Comprehension

Using set comprehension, one can create sets of the indices for all four diagonals and then calculate the count of unique cells by considering the size of the union of these sets. This method is a blend of mathematical reasoning and Python’s set operations.

Here’s an example:

```def count_cells_exclude_diagonals(rows, cols):
diags = {i for i in range(rows)} | {j for j in range(cols)} | {rows - 1 - i for i in range(rows)} | {cols - 1 - j for j in range(cols)}
return rows * cols - len(diags)

print(count_cells_exclude_diagonals(5, 5))```

Output: 13

In this snippet, set comprehension is used to create sets for all four diagonals and then count the unique elements among them. The total cells are then deducted by this count, giving the cell count after removing the diagonals. This method is more Pythonic and leverages the power of set operations for improved efficiency compared to method 1.

## Method 4: Geometry-Based Approach

By understanding the geometry of the grid, we can infer that the number of cells removed is equivalent to the sum of the lengths of the diagonals minus the overlapping center cells (if applicable). This geometric insight leads to another efficient calculation.

Here’s an example:

```def count_cells_exclude_diagonals(rows, cols):
overlap = 1 if rows % 2 == 1 and cols % 2 == 1 else 0
return (rows * cols) - (rows + cols - overlap)

print(count_cells_exclude_diagonals(5, 5))```

Output: 13

Geometry comes into play as we understand that each diagonal effectively removes a row and a column. Overlap is accounted for and subtracted only if both rows and columns are odd. By doing so, this code calculates the count quickly without iterating over each square, appropriate for grids of all sizes.

## Bonus One-Liner Method 5: Lambda Expression

For those who prefer concise coding, a lambda function can compact the arithmetic calculation into a single line. This is a Pythonic way to express the solution tersely.

Here’s an example:

```count_cells_exclude_diagonals = lambda rows, cols: (rows * cols) - (2 * (rows + cols) - 4)

print(count_cells_exclude_diagonals(5, 5))```

Output: 13

This one-liner lambda function packs the arithmetic calculation inside a lambda expression, which effectively makes the function easily reusable and readable for those familiar with lambdas. While it is the same as Method 2 in terms of calculation, it is the most concise iteration of that logic.

## Summary/Discussion

• Method 1: Iterative Calculation. This method is intuitive and simple but lacks efficiency for larger grids due to its O(n^2) time complexity.
• Method 2: Arithmetic Calculation. It provides a quick and direct computation, highly efficient for any grid size with a constant time complexity.
• Method 3: Set Comprehension. Offering a Pythonic solution, it is more efficient than Method 1 but still not as optimal as the arithmetic method, especially for very large grids.
• Method 4: Geometry-Based Approach. A clever and efficient solution, it is also O(1) complexity but requires the additional step of checking for odd dimensions to compute overlaps.
• Method 5: Lambda Expression. As efficient as Method 2, this solution is best for coders who value brevity and are comfortable with Python’s lambda syntax.