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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.
Example: (Single rotation into AVL tree, while a new node is inserted into the AVL tree (LL Rotation)) Figure: LL Rotation The rectangles marked A, B & C are trees
Worst Fit method:- In this method the system always allocate a portion of the largest free block in memory. The philosophy behind this method is that by using small number of a ve
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Program segment for All pairs shortest paths algorithm AllPairsShortestPaths(int N, Matrix C, Matrix P, Matrix D) { int i, j, k if i = j then C[i][j] = 0 for ( i =
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