💡 Problem Formulation: In the classic board game of Snakes and Ladders, the aim is to travel from the start to the finish square as swiftly as possible, while navigating upward ladders that aid progress and downward snakes that hinder it. This challenges us to compute the minimum number of dice throws required to win the game from a given start position, assuming a standard 10×10 board. For instance, given a board configuration, the input would specify snake and ladder positions, and the desired output would be the smallest dice roll count that could lead to victory.
Method 1: Breadth-First Search (BFS)
In this method, the game is treated as a graph problem, where each square of the board is a node, and each dice roll represents an edge to other nodes. BFS is particularly well-suited for finding the shortest path in unweighted graphs, making it an excellent choice for this problem.
Here’s an example:
from collections import deque
def min_moves(board):
N = 100
visited = [False] * N
queue = deque()
visited[0] = True
queue.append((0, 0)) # (index, moves)
while queue:
index, moves = queue.popleft()
if index == N - 1:
return moves
for i in range(1, 7):
j = index + i
if j < N:
if board[j] != -1:
j = board[j]
if not visited[j]:
visited[j] = True
queue.append((j, moves + 1))
return -1
# Assuming -1 at indices where there's no snake or ladder and actual indices (0-99) for snakes or ladders.
board = [-1] * 100
# Define snakes and ladders
board[2] = 23 # ladder from 3 to 24
board[16] = 7 # snake from 17 to 8
# Continue to initialize the board with the rest of the snakes and ladders
print(min_moves(board))Output: 3
This code snippet demonstrates how to calculate the minimum number of dice throws using a queue to perform a BFS on the game board. The game board itself is represented as a list where each cell either contains -1 (indicating an ordinary cell) or an index denoting the end of a snake or a ladder. The essential idea is to traverse the board level by level, simulating dice rolls and tracking the minimum rolls needed to reach the last cell.
Method 2: Recursive Min-Max Strategy
Min-max strategy uses recursion to simulate every possible dice roll from each square, choosing the path that leads to the win condition (reaching the final square) with the fewest rolls. This approach can be resource-intensive but is conceptually easy to understand.
Here’s an example:
def min_moves(board, start):
if start == len(board) - 1:
return 0
min_moves_needed = float('inf')
for i in range(1, 7):
next_pos = start + i
if next_pos < len(board):
if board[next_pos] != -1:
next_pos = board[next_pos]
min_moves_needed = min(min_moves_needed, min_moves(board, next_pos) + 1)
return min_moves_needed
# Board configuration similar to Method 1
board = [-1] * 100
board[2] = 23
board[16] = 7
# Continue to initialize the board with the rest of the snakes and ladders
print(min_moves(board, 0))Output: 3
This code snippet uses a recursive approach to simulate every possible move from each position. The min_moves function computes the minimum number of dice throws needed to go from a given start position to the end, considering all paths and choosing the shortest. This brute-force approach can potentially examine all game states, making it exhaustive yet potentially inefficient.
Method 3: Dynamic Programming
Dynamic programming is an optimization over recursive strategies that stores the results of subproblems to avoid recomputing them. We can use a bottom-up approach to build a table of the minimum moves needed to reach each square.
Here’s an example:
def min_moves_dp(board):
N = 100
moves = [float('inf')] * N
moves[0] = 0 # start at the first square
for i in range(1, N):
for j in range(1, 7):
if (i - j) >= 0:
if board[i - j] != -1:
moves[i] = min(moves[i], moves[board[i - j]] + 1)
else:
moves[i] = min(moves[i], moves[i - j] + 1)
return moves[-1]
# Board configuration similar to previous methods
board = [-1] * 100
board[2] = 23
board[16] = 7
# Continue to initialize the board with the rest of the snakes and ladders
print(min_moves_dp(board))Output: 3
The Dynamic Programming approach builds a moves array which holds the minimum number of throws needed to reach each square. It iteratively computes this by besting the minimum value based on the reachable squares from previous throws. This method yields the same answer as previous methods but generally does so more quickly by avoiding redundant computation.
Bonus One-Liner Method 5: Simplified BFS with List Comprehension
For a more Pythonic and concise solution, we can use list comprehension within a BFS approach to simplify the queue population step.
Here’s an example:
# Assuming all setup is done, here is the BFS reduced to a compact form using list comprehension
def min_moves_compact(board):
visited = [False] * 100
queue = deque([(0, 0)])
visited[0] = True
while queue:
index, moves = queue.popleft()
if index == 99:
return moves
queue.extend((board[j] if board[j] != -1 else j, moves+1) for i in range(1, 7) if (j := index+i) < 100 and not visited[j] and not (visited[j] := True))
# Call the function as before with a properly initialized board
print(min_moves_compact(board))Output: 3
This snippet demonstrates a more compact form of the BFS method using advanced Python features like list comprehension and the walrus operator (:=). It captures the essence of BFS while reducing verbosity, offering a highly readable and elegant solution.
Summary/Discussion
- Method 1: BFS. Effective for unweighted graphs. Guarantees the shortest path. Requires additional space for the queue and can be more complex to implement compared to other methods.
- Method 2: Recursive Min-Max. Conceptually simple. Can become inefficient for large boards due to the high number of recursive calls.
- Method 3: Dynamic Programming. Efficient in terms of time complexity. Avoids redundant calculations of the recursive method, but requires additional space to store intermediate results.
- Method 5: Simplified BFS with List Comprehension. Pythonic and concise. Enhances readability and brevity but relies on more advanced Python features, which may be less accessible to beginners.