In today’s Programming Praxis exercise we have to implement a brute-force algorithm for solving the well-known traveling salesman algorithm. The provided Scheme solution clocks in at 28 lines. Let’s see if we can come up with something a tad more compact.
A quick import or two:
import Data.List import qualified Data.List.Key as K
In order to calculate the total distance traveled, we need to calculate the distance between two points.
dist :: Floating a => (a, a) -> (a, a) -> a dist (x1, y1) (x2, y2) = sqrt ((x1 - x2) ** 2 + (y1 - y2) ** 2)
Calculating all possible tours is just a matter of generating all the permutations of the points, making sure to duplicate the first element of a tour at the end in order to get back home. We also add an index to the points to make the result a bit easier to read (and mainly because the Scheme solution does it too).
tours :: [b] -> [[(Int, b)]] tours = map (\(x:xs) -> x:xs ++ [x]) . permutations . zip [0..]
Calculating the total cost of a tour is a simple matter of summing the distances between every consecutive pair of points. This function resembles the typical definition of the Fibonacci sequence, since that works in much the same way.
cost :: Floating a => [(b, (a, a))] -> a cost xs = sum $ zipWith dist xs (tail xs)
Finally, showing the shortest path is a simple matter of generating all the tours, taking the one with the lowest cost, showing the indices of the points and getting rid of the duplicated starting point.
shortestPath :: [(Double, Double)] -> [Int] shortestPath = init . map fst . K.minimum cost . tours
As usual, a test to see if everything is working correctly:
main :: IO () main = print $ shortestPath [(5, 2), (19, 13), (4, 8), (6, 32), (23, 7), (57, 54), (55, 8), (70, 59)]
You’ll notice that the result is different from the Scheme solution. Specifically, it’s reversed and cycled a few positions. Since we’re walking a loop, this means that we’re walking clockwise instead of counterclockwise and starting in a different town. A trivial difference, since the route is the same and can be reversed and cycled at will. The reason is the order in which the permutations are generated.
That brings us to 4 lines total, an 85% reduction compared to the Scheme solution. That’s not half bad in my book.