Solving the NP-Completeness of Travelling Salesman Problem

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The Travelling Salesman Problem (TSP) is often misunderstood regarding its classification; it is NP-hard, but the decision version of TSP is NP-complete. Verifying if a given route is the shortest can be done in polynomial time, which is a key aspect of NP-completeness. The distinction lies in the fact that a problem is NP-hard if it can be used to solve an NP-complete problem in O(n) time. The discussion emphasizes the importance of understanding the decision version of TSP to grasp its complexity classification. Combinatorial optimization remains a compelling area of study due to these complexities.
Dragonfall
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I don't understand how travenlling salesman is NP-complete. In particular, how can you possibly verify that a given route is the shortest one in polynomial time?
 
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The TSP isn't NP-complete - it's NP-hard. But the question: Is there a path shorter than x is NP-complete. A problem is NP-hard iff it can be used to solve an NP-complete problem in O(n) time.
 
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TSP is only NP-complete if you use the decision version of the problem.

Side note, TSP is one reason why combinatorial optimization is one of my favorite subjects.
 
This makes a lot more sense.
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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