Dynamic Programming: Optimal Canoe Rental Planning

Problem Description

Given a sequence of trading posts along a river, design an efficient algorithm to minimize the total cost of canoe rentals for a river trip. You start at post 1 and must reach post n, with the ability to rent and drop off canoes at any intermediate posts.

Problem Constraints

1. You start at post 1 and must end at post n
2. For any posts i and j where i < j ≤ n, you can:
   • Rent a canoe at post i
   • Drop it off at post j
3. Each rental from post i to j has an associated cost R[i,j]
4. The goal is to minimize the total cost of reaching post n

Example

For a river with 4 posts, the rental costs R[i,j] might be:

Understanding the Solution

Dynamic Programming Approach

This is a classic optimization problem that can be solved efficiently using dynamic programming. Let’s break down the solution step by step.

1. State Definition

Let dp[j] represent the minimum cost to reach post j from post 1. This gives us our subproblem structure.

2. Recurrence Relation

For any post j, we need to consider all possible previous posts i where we could have rented a canoe to reach j. The recurrence relation is:

3. Base Cases

• $dp[1] = 0$ (starting point)
• $dp[j] = \infty$ initially for j > 1

Example Walkthrough

Let’s solve our example step by step:

1. Initialize: $dp = [0, \infty, \infty, \infty]$

2. For j = 2:

   • Can only rent from post 1
   • $dp[2] = dp[1] + R[1,2] = 0 + 10 = 10$

3. For j = 3:

   • From post 1: $dp[1] + R[1,3] = 0 + 15 = 15$
   • From post 2: $dp[2] + R[2,3] = 10 + 40 = 50$
   • $dp[3] = \min(15, 50) = 15$

4. For j = 4:

   • From post 1: $dp[1] + R[1,4] = 0 + 50 = 50$
   • From post 2: $dp[2] + R[2,4] = 10 + 20 = 30$
   • From post 3: $dp[3] + R[3,4] = 15 + 35 = 50$
   • $dp[4] = \min(50, 30, 50) = 30$

Solution Implementation

def min_canoe_cost(R, n):
    dp = [float('inf')] * (n + 1)
    dp[1] = 0

    # prev[j] stores the post from which we rented the canoe to reach j
    prev = [0] * (n + 1)
    
    # For each destination post
    for j in range(2, n + 1):
        # Try all possible starting points i
        for i in range(1, j):
            if R[i][j] != '-':  # If there's a valid rental cost
                cost = dp[i] + R[i][j]
                if cost < dp[j]:
                    dp[j] = cost
                    prev[j] = i
    
    # Reconstruct the path
    path = []
    curr = n
    while curr != 1:
        path.append((prev[curr], curr))
        curr = prev[curr]
    path.reverse()
    
    return dp[n], path

Complexity Analysis

Time Complexity

• $O(n^2)$ where n is the number of posts
• We need to consider all possible (i,j) pairs

Space Complexity

• $O(n)$ for the dp array
• Additional $O(n)$ for path reconstruction

Why is this Solution Efficient?

1. Optimal Substructure

   • The optimal solution to the problem incorporates optimal solutions to subproblems
   • Once we find the minimum cost to reach a post j, we never need to recalculate it

2. Overlapping Subproblems

   • The same subproblems (reaching intermediate posts) are used multiple times
   • We store solutions in the dp array to avoid redundant calculations

3. Greedy Choice Property

   • At each step, we can make a locally optimal choice without affecting future decisions
   • This is because the cost of each rental is independent of other rentals