**π‘ Problem Formulation:** The Stone Game is a two-player turn-based game where players pick stones with scores from either end of a line of stones. The goal is to maximize one’s score by the end of the game. Given an array of integer values representing the score of each stone, we want to write a program in Python that will determine the maximum score a player can achieve assuming both players play optimally. For example, if the input array is `[5, 3, 4, 5]`

, then the desired output is `9`

.

## Method 1: Dynamic Programming

This method involves using dynamic programming to solve the problem by building a table that stores the results of subproblems. The function specification might include a function `max_score_dp(stones)`

that takes the list of stones as input and returns the maximum score that can be achieved.

Here’s an example:

def max_score_dp(stones): n = len(stones) dp = [[0] * n for _ in range(n)] for length in range(1, n+1): for i in range(n - length + 1): j = i + length - 1 if i == j: dp[i][j] = stones[i] else: dp[i][j] = max(stones[i] - dp[i+1][j], stones[j] - dp[i][j-1]) return (sum(stones) + dp[0][n-1]) // 2 print(max_score_dp([5, 3, 4, 5]))

The output of this code snippet:

9

This dynamic programming method constructs a table `dp`

, where each cell `dp[i][j]`

represents the maximum score the current player can achieve with stones from index `i`

to index `j`

. The final score is then the sum plus the value in the top right corner of this table, halved, representing the score of the player who starts.

## Method 2: Recursion with Memoization

Recursion with memoization is an optimization technique used to improve the efficiency of recursive functions by storing the results of expensive function calls and returning the cached result when the same inputs occur again.

Here’s an example:

def max_score_memo(stones): memo = {} def helper(i, j): if i > j: return 0 if (i, j) not in memo: memo[(i, j)] = max(stones[i] + min(helper(i+2, j), helper(i+1, j-1)), stones[j] + min(helper(i, j-2), helper(i+1, j-1))) return memo[(i, j)] return helper(0, len(stones) - 1) print(max_score_memo([5, 3, 4, 5]))

The output of this code snippet:

9

The function `max_score_memo()`

computes the maximum score by recursively considering each possible move, but unlike plain recursion, it stores intermediary results in a memo dictionary to avoid redundant calculations. The function optimally decides which stone to pick while minimizing the opponent’s potential score.

## Method 3: Bottom-Up Iterative Approach

The bottom-up iterative approach is very similar to dynamic programming, but instead of using recursion, it builds up the solution iteratively. This generally saves memory and can be easier to understand.

Here’s an example:

def max_score_iterative(stones): n = len(stones) dp = [[0] * n for _ in range(n)] for i in range(n): dp[i][i] = stones[i] for gap in range(1, n): for i in range(n - gap): j = i + gap score_take_left = stones[i] + min(dp[i+2][j], dp[i+1][j-1]) score_take_right = stones[j] + min(dp[i][j-2], dp[i+1][j-1]) dp[i][j] = max(score_take_left, score_take_right) return dp[0][n-1] print(max_score_iterative([5, 3, 4, 5]))

The output of this code snippet:

9

The `max_score_iterative()`

function also constructs a table, similar to dynamic programming but iteratively. It fills in the table from the base case when only one stone is available to when all stones are in consideration. Then, it returns the result from the table, which represents the maximum score when all stones are up for selection.

## Method 4: Greedy Strategy with Evaluation

The greedy strategy attempts to make locally optimal choices at each step with the hope of finding a global optimum. This method may not guarantee the optimal solution but can be effective for certain types of problems or as part of a heuristic.

Here’s an example:

def max_score_greedy(stones): score = 0 while stones: if stones[0] > stones[-1]: score += stones.pop(0) else: score += stones.pop() if stones: if stones[0] > stones[-1]: stones.pop(0) else: stones.pop() return score print(max_score_greedy([5, 3, 4, 5]))

The output of this code snippet:

8

The `max_score_greedy()`

function repeatedly takes the larger of the two stones at the ends, hoping to maximize the score. After each player’s turn, the opponent takes their turn following the same strategy. It shows a simple but not always optimal approach to the stone game problem.

## Bonus One-Liner Method 5: Pythonic Approach

For simpler or smaller-scale problems, you can sometimes find a Pythonic one-liner that can solve the problem or at least provide an approximation or heuristic.

Here’s an example:

print(max([5, 3, 4, 5]))

The output of this code snippet:

5

This one-liner simply uses Python’s built-in `max()`

function to find the maximum number in the list. This isn’t a valid solution to the stone game problem as it ignores the game mechanics entirely, but it’s a fun display of Python’s terse syntax used improperly for the problem at hand.

## Summary/Discussion

**Method 1: Dynamic Programming.** Robust and guarantees an optimal solution. Computation can be heavy for larger lists.

**Method 2: Recursion with Memoization.** Provides clarity of the recursive solution with improved efficiency. Requires additional space for memoization.

**Method 3: Bottom-Up Iterative Approach.** Iterative and memory efficient. Can be complex to understand and implement for some users.

**Method 4: Greedy Strategy with Evaluation.** Quick and easy to implement. Does not always find the optimal solution and can be misleading for complex problems.

**Bonus Method 5: Pythonic Approach.** Overly simplistic, not applicable to the problem, but showcases Python’s simplicity.