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Example 3: Travelling Salesman problem
Given: n associated cities and distances among them
Find: tour of minimum length that visits all of city.
Solutions: How several tours are possible?
n*(n -1)...*1 = n!
Because n! > 2(n-1)
Therefore n! = ? (2n) (lower bound)
As of now, there is no algorithm that determines a tour of minimum length plus covers all of the cities in polynomial time. But, there are many very good heuristic algorithms.
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Normally a potential y satisfies y r = 0 and 0 ³ y w - c vw -y v . Given an integer K³0, define a K-potential to be an array y that satisfies yr = 0 and K ³ y w - c vw -y v
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