**π‘ Problem Formulation:** The problem at hand involves devising a program that can calculate the number of possible ways to distribute a certain number of coins among a set number of workers. Suppose we have 4 coins and 2 workers, the program should determine that there are 5 possible distributions: (0,4), (1,3), (2,2), (3,1), and (4,0).

## Method 1: Recursive Approach

Recursive solutions involve breaking down a problem into smaller, more manageable problems. In this case, defining a function that calculates the ways to distribute the remaining coins to the remaining workers by recursively reducing the problem size. This method can be intuitive but may suffer from performance issues for larger inputs due to repeated calculations.

Here’s an example:

def count_distributions(coins, workers): if workers == 1: return 1 if coins == 0: return 1 return count_distributions(coins - 1, workers) + count_distributions(coins, workers - 1) print(count_distributions(4, 2))

Output: `5`

This snippet defines a function `count_distributions()`

that takes two arguments: the number of coins and workers. The function recurses until there’s only one worker or no coins, returning 1 in those cases as the base cases. It returns the sum of ways to distribute with one less coin and the same number of workers, plus the ways with the same number of coins to one fewer worker.

## Method 2: Dynamic Programming

Dynamic Programming (DP) is a method for solving complex problems by breaking them down into simpler subproblems. It is particularly well-suited for this kind of combinatorial problem. By storing interim results, we avoid redundant calculations, which greatly improves performance over a pure recursive approach, especially for large inputs.

Here’s an example:

def count_distributions_dp(coins, workers): dp = [[0] * (workers + 1) for _ in range(coins + 1)] dp[0] = [1] * (workers + 1) for i in range(1, coins + 1): for j in range(1, workers + 1): dp[i][j] = dp[i - 1][j] + dp[i][j - 1] return dp[coins][workers] print(count_distributions_dp(4, 2))

Output: `5`

The function `count_distributions_dp()`

creates a 2D list ‘dp’ to store the interim distribution counts. It initializes the base cases and iteratively fills the dp table by summing the distribution count of having one less coin and the count for one fewer worker, eventually returning the count for the desired number of coins and workers.

## Method 3: Using Combinatorics

The combinatorics approach uses mathematical formulas to directly calculate the number of combinations. Specifically, we can use the binomial coefficient to determine the distribution count, as it is equivalent to choosing the number of coins for each worker. This method is quite efficient as it bypasses iterative or recursive calculations.

Here’s an example:

from math import factorial def count_distributions_comb(coins, workers): return factorial(coins + workers - 1) // (factorial(coins) * factorial(workers - 1)) print(count_distributions_comb(4, 2))

Output: `5`

In the function `count_distributions_comb()`

, we compute the binomial coefficient using the factorial function from the math module in Python. The result is the number of ways to distribute given coins among the workers, calculated using the formula for combinations with repetitions.

## Method 4: Using itertools

The itertools module in Python includes a set of functions for creating iterators for efficient looping. Here, we can utilize the combinations_with_replacement function to generate all ways to distribute coins and then count them. This method is more concrete as we generate all possible distributions.

Here’s an example:

from itertools import combinations_with_replacement def count_distributions_itertools(coins, workers): distributions = list(combinations_with_replacement(range(coins + 1), workers)) return len(distributions) print(count_distributions_itertools(4, 2))

Output: `5`

The `count_distributions_itertools()`

function utilizes the `combinations_with_replacement()`

from the itertools module to generate all possible distributions and returns their count. This allows us to explicitly see all combinations, which can be useful for smaller inputs or debugging purposes.

## Bonus One-Liner Method 5: Functional Programming

A functional programming one-liner can leverage Python’s powerful lambda and reduce functions to solve the problem in a compact manner. It utilizes higher-order functions and immutable data to express the solution succinctly.

Here’s an example:

from functools import reduce result = reduce(lambda r, _: [1] + [sum(r[i:i + 2]) for i in range(len(r) - 1)] + [1], range(4), [1] * 2) print(result[-1])

Output: `5`

This one-liner uses the `reduce()`

function along with a lambda to simulate Pascal’s triangle, which underlies the binomial coefficient calculation required for this problem. Starting with the base case of multiple distributions of zero coins, it iteratively builds up to the required level of distributions.

## Summary/Discussion

**Method 1:**Recursive Approach. Intuitive and mirrors the logical breakdown of the problem. However, it has exponential time complexity for large inputs due to overlapping subproblems.**Method 2:**Dynamic Programming. Significantly more efficient by storing intermediate results, hence suitable for larger inputs. Might be more complex to understand for beginners.**Method 3:**Using Combinatorics. Most efficient as it directly calculates the result. It abstracts away the distribution process which may not be ideal for all applications.**Method 4:**Using itertools. Concrete and explicit, generating all combinations. It can be inefficient for large datasets due to memory usage and slower runtimes.**Method 5:**Functional Programming. Elegant and concise but can be cryptic. Useful for small, quick computations but not practical for understanding the steps involved in the distribution.