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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.
This notation bounds a function to in constant factors. We say f(n) = Θ(g(n)) if there presents positive constants n 0 , c 1 and c 2 such that to the right of n 0 the value of f
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12345 SOLVE BY USING FOLDING METHOD
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