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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.
Thus far, we have been considering sorting depend on single keys. However, in real life applications, we may desire to sort the data on several keys. The simplest instance is that
what algorithms can i use for the above title in my project desing and implmentation of road transport booking system
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