*The 2-opt algorithm is highly dependent on the initialization that we pick for our data.I started with using Nearest neighbour based initial path (ie Choose the nearest point as the next point in path), since theoretically, it gives the best result.*

This is intuitively easy to understand and also likely going to give us better results as compared to a random path.Image showing how cyclic order of points is calculated using a reference point " data-medium-file="https://i1com/crondev.blog/wp-content/uploads/2018/08/untitled-diagram.png?fit=300,205&ssl=1" data-large-file="https://i1com/crondev.blog/wp-content/uploads/2018/08/untitled-diagram.png? fit=331,226&ssl=1" class="alignnone size-full wp-image-2042" src="https://i2com/crondev.blog//wp-content/uploads/2018/08/untitled-diagram.png? resize=331,226&ssl=1" alt="Untitled Diagram" width="331" height="226" srcset="https://i1com/crondev.blog/wp-content/uploads/2018/08/untitled-diagram.png? w=331&ssl=1 331w, https://i1com/crondev.blog/wp-content/uploads/2018/08/untitled-diagram.png? resize=300,205&ssl=1 300w" sizes="(max-width: 331px) 100vw, 331px" data-recalc-dims="1" / Intuitively, it makes sense, and it does work for 60% of the cases we encountered.The first attempt was to try and arrange all vertices in a cyclic order to create a Polygon.The algorithm is very easy, we take a reference point and arrange all points in an increasing order of angles relative to that point.Let us compare the cost of the tour ACBDA with the tour A’B’C’D’A’ (Figure 4, bottom).Note that the edges AC and BD meet at a point Q which is not one of the original sites involved in the TSP. However AQ QC = AC and DQ QB = DB and hence, if we replace AC and DB by A’B’ and D’C’, we get a shorter tour. A lot of other solutions can also exist for this problem as well(for example the clockwise order example above).We at Lean Agri believe that software should be adaptive of the user rather than the other way round.We build solutions which allow users to improve their efficiency of working rather than training them to efficiently use the software we build.We know that AQ QB is larger than A’B’ (because in the triangle AQB, the sum of any two sides has Euclidean length greater than the third side). Intuitively for a user, the travelling salesman problem solution is going to make the most sense. TSP is solvable in time which for 15 points is going to be ~10^15 iterations!Please note that humans are extremely good at solving the Travelling Salesman Problem. Fortunately, a lot of research is done on this problem to generate approximate heuristic solutions much faster than the brute force solution.

## Comments Solving Travelling Salesman Problem

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