from functools import lru_cache
from typing import Dict, List, Optional, Tuple

import numpy as np


def solve_tsp_dynamic_programming(
    distance_matrix: np.ndarray,
    maxsize: Optional[int] = None,
) -> Tuple[List, float]:
    """
    Solve TSP to optimality with dynamic programming.

    Parameters
    ----------
    distance_matrix
        Distance matrix of shape (n x n) with the (i, j) entry indicating the
        distance from node i to j. It does not need to be symmetric

    maxsize
        Parameter passed to ``lru_cache`` decorator. Used to define the maximum
        size for the recursion tree. Defaults to `None`, which essentially
        means "take as much space as needed".

    Returns
    -------
    permutation
        A permutation of nodes from 0 to n that produces the least total
        distance

    distance
        The total distance the optimal permutation produces

    Notes
    -----
    Algorithm: cost of the optimal path
    ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

    Consider a TSP instance with 3 nodes: {0, 1, 2}. Let dist(0, {1, 2}) be the
    distance from 0, visiting all nodes in {1, 2} and going back to 0. This can
    be computed recursively as:

        dist(0, {1, 2}) = min(
            c_{0, 1} + dist(1, {2}),
            c_{0, 2} + dist(2, {1}),
        )

    wherein c_{0, 1} is the cost from going from 0 to 1 in the distance matrix.
    The inner dist(1, {2}) is computed as:

        dist(1, {2}) = min(
            c_{1, 2} + dist(2, {}),
        )

    and similarly for dist(2, {1}). The stopping point in the recursion is:

        dist(2, {}) = c_{2, 0}.

    This process can be generalized as:

        dist(ni, N) =   min   ( c_{ni, nj} + dist(nj, N - {nj}) )
                      nj in N

    and

        dist(ni, {}) = c_{ni, 0}

    With starting point as dist(0, {1, 2, ..., tsp_size}). The notation
    N - {nj} is the difference operator, meaning set N without node nj.


    Algorithm: compute the optimal path
    ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
    The previous process returns the distance of the optimal path. To find the
    actual path, we need to store in a memory the following key/values:

        memo[(ni, N)] = nj_min

    with nj_min the node in N that provided the smallest value of dist(ni, N).
    Then, the process goes backwards starting from
    memo[(0, {1, 2, ..., tsp_size})].

    In the previous example, suppose memo[(0, {1, 2})] = 1.
    Then, look for memo[(1, {2})] = 2.
    Then, since the next step would be memo[2, {}], stop there. The optimal
    path would be 0 -> 1 -> 2 -> 0.


    Reference
    ---------
    https://en.wikipedia.org/wiki/Held%E2%80%93Karp_algorithm#cite_note-5
    """
    # Get initial set {1, 2, ..., tsp_size} as a frozenset because @lru_cache
    # requires a hashable type
    N = frozenset(range(1, distance_matrix.shape[0]))
    memo: Dict[Tuple, int] = {}

    # Step 1: get minimum distance
    @lru_cache(maxsize=maxsize)
    def dist(ni: int, N: frozenset) -> float:
        if not N:
            return distance_matrix[ni, 0]

        # Store the costs in the form (nj, dist(nj, N))
        costs = [
            (nj, distance_matrix[ni, nj] + dist(nj, N.difference({nj})))
            for nj in N
        ]
        nmin, min_cost = min(costs, key=lambda x: x[1])
        memo[(ni, N)] = nmin
        return min_cost

    best_distance = dist(0, N)

    # Step 2: get path with the minimum distance
    ni = 0  # start at the origin
    solution = [0]
    while N:
        ni = memo[(ni, N)]
        solution.append(ni)
        N = N.difference({ni})

    return solution, best_distance

from functools import lru_cache
from typing import Dict, List, Optional, Tuple

import numpy as np


def solve_tsp_dynamic_programming(
    distance_matrix: np.ndarray,
    maxsize: Optional[int] = None,
) -> Tuple[List, float]:
    """
    Solve TSP to optimality with dynamic programming.

    Parameters
    ----------
    distance_matrix
        Distance matrix of shape (n x n) with the (i, j) entry indicating the
        distance from node i to j. It does not need to be symmetric

    maxsize
        Parameter passed to ``lru_cache`` decorator. Used to define the maximum
        size for the recursion tree. Defaults to `None`, which essentially
        means "take as much space as needed".

    Returns
    -------
    permutation
        A permutation of nodes from 0 to n that produces the least total
        distance

    distance
        The total distance the optimal permutation produces

    Notes
    -----
    Algorithm: cost of the optimal path
    ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

    Consider a TSP instance with 3 nodes: {0, 1, 2}. Let dist(0, {1, 2}) be the
    distance from 0, visiting all nodes in {1, 2} and going back to 0. This can
    be computed recursively as:

        dist(0, {1, 2}) = min(
            c_{0, 1} + dist(1, {2}),
            c_{0, 2} + dist(2, {1}),
        )

    wherein c_{0, 1} is the cost from going from 0 to 1 in the distance matrix.
    The inner dist(1, {2}) is computed as:

        dist(1, {2}) = min(
            c_{1, 2} + dist(2, {}),
        )

    and similarly for dist(2, {1}). The stopping point in the recursion is:

        dist(2, {}) = c_{2, 0}.

    This process can be generalized as:

        dist(ni, N) =   min   ( c_{ni, nj} + dist(nj, N - {nj}) )
                      nj in N

    and

        dist(ni, {}) = c_{ni, 0}

    With starting point as dist(0, {1, 2, ..., tsp_size}). The notation
    N - {nj} is the difference operator, meaning set N without node nj.


    Algorithm: compute the optimal path
    ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
    The previous process returns the distance of the optimal path. To find the
    actual path, we need to store in a memory the following key/values:

        memo[(ni, N)] = nj_min

    with nj_min the node in N that provided the smallest value of dist(ni, N).
    Then, the process goes backwards starting from
    memo[(0, {1, 2, ..., tsp_size})].

    In the previous example, suppose memo[(0, {1, 2})] = 1.
    Then, look for memo[(1, {2})] = 2.
    Then, since the next step would be memo[2, {}], stop there. The optimal
    path would be 0 -> 1 -> 2 -> 0.


    Reference
    ---------
    https://en.wikipedia.org/wiki/Held%E2%80%93Karp_algorithm#cite_note-5
    """
    # Get initial set {1, 2, ..., tsp_size} as a frozenset because @lru_cache
    # requires a hashable type
    N = frozenset(range(1, distance_matrix.shape[0]))
    memo: Dict[Tuple, int] = {}

    # Step 1: get minimum distance
    @lru_cache(maxsize=maxsize)
    def dist(ni: int, N: frozenset) -> float:
        if not N:
            return distance_matrix[ni, 0]

        # Store the costs in the form (nj, dist(nj, N))
        costs = [
            (nj, distance_matrix[ni, nj] + dist(nj, N.difference({nj})))
            for nj in N
        ]
        nmin, min_cost = min(costs, key=lambda x: x[1])
        memo[(ni, N)] = nmin
        return min_cost

    best_distance = dist(0, N)

    # Step 2: get path with the minimum distance
    ni = 0  # start at the origin
    solution = [0]
    while N:
        ni = memo[(ni, N)]
        solution.append(ni)
        N = N.difference({ni})

    return solution, best_distance

Solve TSP to optimality with dynamic programming.      Parameters     ----------     distance_matrix         Distance matrix of shape (n x n) with the (i, j) entry indicating the         distance from node i to j. It does not need to be symmetric      maxsize         Parameter passed to ``lru_cache`` decorator. Used to define the maximum         size for the recursion tree. Defaults to `None`, which essentially         means "take as much space as needed".      Returns     -------     permutation         A permutation of nodes from 0 to n that produces the least total         distance      distance         The total distance the optimal permutation produces      Notes     -----     Algorithm: cost of the optimal path     ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~      Consider a TSP instance with 3 nodes: {0, 1, 2}. Let dist(0, {1, 2}) be the     distance from 0, visiting all nodes in {1, 2} and going back to 0. This can     be computed recursively as:          dist(0, {1, 2}) = min(             c_{0, 1} + dist(1, {2}),             c_{0, 2} + dist(2, {1}),         )      wherein c_{0, 1} is the cost from going from 0 to 1 in the distance matrix.     The inner dist(1, {2}) is computed as:          dist(1, {2}) = min(             c_{1, 2} + dist(2, {}),         )      and similarly for dist(2, {1}). The stopping point in the recursion is:          dist(2, {}) = c_{2, 0}.      This process can be generalized as:          dist(ni, N) =   min   ( c_{ni, nj} + dist(nj, N - {nj}) )                       nj in N      and          dist(ni, {}) = c_{ni, 0}      With starting point as dist(0, {1, 2, ..., tsp_size}). The notation     N - {nj} is the difference operator, meaning set N without node nj.       Algorithm: compute the optimal path     ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~     The previous process returns the distance of the optimal path. To find the     actual path, we need to store in a memory the following key/values:          memo[(ni, N)] = nj_min      with nj_min the node in N that provided the smallest value of dist(ni, N).     Then, the process goes backwards starting from     memo[(0, {1, 2, ..., tsp_size})].      In the previous example, suppose memo[(0, {1, 2})] = 1.     Then, look for memo[(1, {2})] = 2.     Then, since the next step would be memo[2, {}], stop there. The optimal     path would be 0 -> 1 -> 2 -> 0.       Reference     ---------     https://en.wikipedia.org/wiki/Held%E2%80%93Karp_algorithm#cite_note-5 # Get initial set {1, 2, ..., tsp_size} as a frozenset because @lru_cache # requires a hashable type # Step 1: get minimum distance # Store the costs in the form (nj, dist(nj, N)) # Step 2: get path with the minimum distance # start at the origin