Posts Tagged ‘brute’

Programming Praxis – Traveling Salesman: Brute Force

March 12, 2010

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.

Programming Praxis – Brute Force

August 21, 2009

Today’s Programming Praxis post is another easy one. We have to implement a brute-force string search. Let’s get started.

Our import:

import Data.List

Since it’s no extra effort, we’ll just make our search function generic.

bruteSearch :: Eq a => [a] -> Maybe Int -> [a] -> Maybe Int
bruteSearch n o = fmap fst . find (isPrefixOf n . snd) .
                  drop (maybe 0 id o) . zip [0..] . tails

With that out of the way, let’s define our test suite:

test :: (String -> Maybe Int -> String -> Maybe Int) -> IO ()
test f = do print $ f ""   Nothing  "Hello World" == Just 0
            print $ f "He" Nothing  "Hello World" == Just 0
            print $ f "od" Nothing  "Bonsai Code" == Just 8
            print $ f "ef" Nothing  "Bonsai Code" == Nothing
            print $ f ""   (Just 1) "Hello World" == Just 1
            print $ f "He" (Just 1) "Hello World" == Nothing
            print $ f "od" (Just 1) "Bonsai Code" == Just 8
            print $ f "ef" (Just 1) "Bonsai Code" == Nothing

And let’s run it to see if everything’s in order.

main :: IO ()
main = test bruteSearch

It works fine, but obviously it’s not the most efficient search algorithm. For that you’ll have to wait for the upcoming other search algorithms.