a J5d8@sRddlmZddlmZmZmZmZddlZdej ee eee fdddZ dS)) lru_cache)DictListOptionalTupleN)distance_matrixmaxsizereturncsttdjd}it|dtttdfdd d|}d}dg}|rx||f}||||h}qP||fS)a 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 r)r)niNr csLsdfSfddD}t|ddd\}}|f<|S)Nrc s.g|]&}||f||hfqS) difference).0Znj)r distrr r `/var/www/html/django/DPS/env/lib/python3.9/site-packages/python_tsp/exact/dynamic_programming.py isz?solve_tsp_dynamic_programming..dist..cSs|dS)Nr r )xr r rmz=solve_tsp_dynamic_programming..dist..)key)min)r r ZcostsZnminZmin_costrrmemo)r r rrcs  z+solve_tsp_dynamic_programming..dist) frozensetrangeshaperintfloatappendr)rrr Z best_distancer Zsolutionr rrsolve_tsp_dynamic_programmingsX   r )N) functoolsrtypingrrrrnumpynpZndarrayrrr r r r rs