This code is a framework to solve a transportation problem...

April 3, 2025 at 03:45 PM

import sys
import util
sys.setrecursionlimit(10000)

### Model (search problem)

class TransportationProblem(object):
    def __init__(self, N):
        # N = number of blocks
        self.N = N
    def startState(self):
        return 1
    def isEnd(self, state):
        return state == self.N
    def succAndCost(self, state):
        # Returns list of (action, newState, cost) triples
        result = []
        if state + 1 <= self.N:
            result.append(('walk', state + 1, 1))
        if 2 * state <= self.N:
            result.append(('tram', 2 * state, 2))
        return result

### Inference (search algorithms)

def printSolution(solution):
    totalCost, history = solution
    print 'totalCost:', totalCost
    for item in history:
        print item

def backtrackingSearch(problem):
    # Best found so far
    # (technicality: using array because of Python scoping)
    bestTotalCost = [float('+inf')]
    bestHistory = [None]
    def recurse(state, history, totalCost):
        # At |state| having undergone |history|, accumulated |totalCost|.
        # Explore the rest of the subtree under |state|.
        if problem.isEnd(state):
            # Update the best solution so far
            if totalCost < bestTotalCost[0]:
                bestTotalCost[0] = totalCost
                bestHistory[0] = history
            return

        # Recurse on children
        for action, newState, cost in problem.succAndCost(state):
            recurse(newState, history + [(action, newState, cost)], totalCost + cost)

    recurse(problem.startState(), history=[], totalCost=0)
    return (bestTotalCost[0], bestHistory[0])

def dynamicProgramming(problem):
    cache = {}  # state => futureCost(state), action, newState, cost
    def futureCost(state):
        # Base case
        if problem.isEnd(state):
            return 0
        if state in cache:  # Exponential savings!
            return cache[state][0]
        # Actually do work
        result = min((cost + futureCost(newState), action, newState, cost) \
            for action, newState, cost in problem.succAndCost(state))
        cache[state] = result
        return result[0]

    state = problem.startState()
    totalCost = futureCost(state)

    # Recover history
    history = []
    while not problem.isEnd(state):
        _, action, newState, cost = cache[state]
        history.append((action, newState, cost))
        state = newState

    return (totalCost, history)

def uniformCostSearch(problem):
    frontier = util.PriorityQueue()
    frontier.update(problem.startState(), 0)
    while True:
        # Move from frontier to explored
        state, pastCost = frontier.removeMin()
        if problem.isEnd(state):
            # Note: don't compute history
            return (pastCost, [])
        # Push out on the frontier
        for action, newState, cost in problem.succAndCost(state):
            frontier.update(newState, pastCost + cost)

### Main

problem = TransportationProblem(N=40)
print problem.succAndCost(3)
print problem.succAndCost(9)
printSolution(backtrackingSearch(problem))
printSolution(dynamicProgramming(problem))
printSolution(uniformCostSearch(problem))

This code is a framework to solve a transportation problem using three search techniques: backtracking search, dynamic programming, and uniform cost search. The goal of the problem is to get from the starting state (block 1) to the ending state (block N) at the minimal cost. Below is a breakdown of what each part of the code does:

1. TransportationProblem Class:

Models the transportation problem as a search problem.
__init__(N): Initializes the problem with N blocks.
startState():
- Returns the starting state, which is always 1 (first block).
isEnd(state):
- Checks if the given state is the goal state (state == N).
succAndCost(state):
- Defines the possible actions (walk or tram) that can transition to new states:
  - walk: Move to the next block with a cost of 1.
  - tram: Jump to double the current block number with a cost of 2.
- Returns a list of tuples for possible (action, newState, cost) transitions.

2. Search Algorithms:

This section implements three search techniques to find the optimal solution:

a. Backtracking Search:

Performs an exhaustive depth-first search to find the optimal solution.
Recursively explores all paths from startState() to isEnd(state), keeping track of the accumulated cost (totalCost) and history of actions.
Updates the best solution found so far in terms of cost (bestTotalCost) and history if the current path has a lower cost.
Returns a tuple (totalCost, history) for the optimal solution.

b. Dynamic Programming (DP):

Uses memoization (via the cache dictionary) to avoid recomputing results for already-visited states (state) recursively.
futureCost(state) computes the minimal future cost to reach the goal from the current state, storing results in a cache for performance.
Recovers the path (action history) by backtracking through the states stored in the cache after computing the minimal cost.
Returns the optimal solution (totalCost, history).

c. Uniform Cost Search (UCS):

Uses a priority queue from the util module to implement Dijkstra's algorithm.
Expands states in order of cost, ensuring it always explores the least-cost path first.
Only returns the minimal cost (pastCost) without reconstructing the history.

3. Execution and Outputs:

Creates a problem instance with N=40 blocks.
Prints the successor states and costs for states 3 and 9 using the succAndCost() method.
- For example, succAndCost(3) would return actions like [('walk', 4, 1), ('tram', 6, 2)].
Runs each of the three search algorithms (backtrackingSearch, dynamicProgramming, uniformCostSearch), prints their solutions, and includes:
- totalCost (minimum cost to reach the goal).
- history (sequence of actions, new states, and costs taken to reach the goal).

Key Notes:

The code solves a specific search problem where the goal is to minimize cost while transitioning from block 1 to block N.
sys.setrecursionlimit(10000) ensures recursive backtracking can handle large values of N by increasing the recursion depth limit.
The util.PriorityQueue is used in UCS, but the util module is assumed to provide this utility (not shown in the code).
Differences among algorithms:
- Backtracking is brute-force and slow.
- DP optimizes by caching results, avoiding redundant calculations.
- UCS guarantees the optimal solution using a cost-priority exploration.

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