Traveling Salesman Problem: Difference between revisions
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=Algorithm= | =Algorithm= | ||
<font size=-1> | |||
Let A = 2D array, indexed by subsets S ⊆ {1,2,...,n} that contain 1, and destinations j ∈ {1,2,...,n} | |||
<font color=teal># Initialize base case</font> | |||
│ 0 if S={1} | |||
A[S,1] = │ | |||
│ +∞ otherwise | |||
for m=2,3,4,...,n: | |||
for each set S ⊆ {1,2,...,n} of size m that contains 1: | |||
for each j∈S, j≠1: | |||
A[S,j] = min { A[S-{j},k] + C<sub>kj</sub> | |||
k∈S | |||
k≠j | |||
n | |||
return min { A[{1,2,3,...,n},j] + C<sub>j1</sub> } | |||
j=1 | |||
</font> | |||
==Playground Implementation== | ==Playground Implementation== |
Revision as of 02:44, 27 November 2021
External
- https://www.coursera.org/learn/algorithms-npcomplete/lecture/49MkW/the-traveling-salesman-problem
- https://www.coursera.org/learn/algorithms-npcomplete/lecture/uVABz/a-dynamic-programming-algorithm-for-tsp
Internal
Overview
The input is a complete undirected graph with non-negative edge costs. The output is a minimum cost tour (permutation of the vertices, a cycle that visits every vertex exactly once that minimize the sum of the edges). The brute force search running time is O(n!). The dynamic programming approach described here has a running time of O(n22n).
Algorithm
Let A = 2D array, indexed by subsets S ⊆ {1,2,...,n} that contain 1, and destinations j ∈ {1,2,...,n} # Initialize base case │ 0 if S={1} A[S,1] = │ │ +∞ otherwise for m=2,3,4,...,n: for each set S ⊆ {1,2,...,n} of size m that contains 1: for each j∈S, j≠1: A[S,j] = min { A[S-{j},k] + Ckj k∈S k≠j n return min { A[{1,2,3,...,n},j] + Cj1 } j=1