Dilawar Singh Being Dilawar/ Tools/

  \This implementation is not-very naive but it is not very efficient either. It can generate solutions up to n = 12, then it gets very slow.

This is an extremely horrible solution to generate langford pairs by dilawar@ee.iitb.ac.in succesfully finished on Nov 27, 2011 after two days of labour. Is is advised not to use this program in any real-time application. Its runs slower than Chacha Chaudary brain. One can achieve much better performace by using Arrays in Haskell. This is discussed in Knuth, TAOP - Vol 4(a)

mkZero n = take n [0,0..] genBaseList n      | n < 1 = [] >     | otherwise = n : mkZero n ++ [n] We are producing baselists e.g. for n=4, baselists are [1,0,1], [2,0,0,2], [3,0,0,0,3] and [4,0,0,0,0,4]. allBaseLists n = map genBaseList [n,n-1..1] Following horrible step is the core of our algorithm. In this step we use two base-list. First list is known as topList while the other one is bottomList. To fuse these lists, we first shift the top-list to right by prepending a 0 to it and fuse it with bottom. Two lists are no fusable whenver both of them has non-zero integer at the same place. This process stops when size of top list become more than 2*n we stop. For example, for n = 3 Top   : 1 0 1    | 0 1 0 1 | 0 0 1 0 1 | 0 0 0 1 0 1 | size more than 2n Bot   : 2 0 0 2  | 2 0 0 2 | 2 0 0 2   | 2 0 0 2 Fuse  : X        | 2 1 0 X | 2 0 1 2 1 | 2 0 0 X Nothing    Nothing     Just       Nothing ^                        ^ |                        | Two nozero             This is the only Entries coincide       possible fusion. In second step, we shift the bottom and fuse Top   : 1 0 1    | 1 0 1      | 1 0 1         | 1 0 1 Bot   : 2 0 0 2  | 0 2 0 0 2  | 0 0 2 0 0 2   | size more than 2n Fuse  : X        | 1 2 1 0 2  | 1 0 X Nothing      Just        Nothing fuseBothList [] y = Just y fuseBothList x [] = Just x fuseBothList (x:xs) (y:ys)      | (x /= 0) && (y /= 0) = Nothing      | otherwise = (helper $ fuseBothList xs ys)                          where helper Nothing = Nothing                                helper (Just p) = if y/=0 then Just (y:p)                                                  else Just (x:p)

topRightShiftAndFuse t b n      | length (t) > 2*n = []      | fuseBothList t b == Nothing = topRightShiftAndFuse (0:t) b n      | otherwise = (\(Just p) -> p:(topRightShiftAndFuse (0:t) b n)) (fuseBothList t b)

botRightShiftAndFuse t b n      | length (b) > 2*n = []      | fuseBothList t b == Nothing = botRightShiftAndFuse t (0:b) n      | otherwise = (\(Just p) -> p:(botRightShiftAndFuse t (0:b) n)) (fuseBothList t b)

At each step we get fused lists by above method. These lists are needed to be fused with next list And now we should merge what we have fused. For example we have got [0,2,0,0,2] and [1,2,1,0,2] from previous merge, now we need to merge it with [3,0,0,0,3]. by doing so, we’ll get the solution for n=3. mergeTopBottom t b n = (topRightShiftAndFuse t b n) ++ (botRightShiftAndFuse t b n) Now rest is putting these function together to get the final solution. initList n = allBaseLists n mergeInto n = (\(x:xs) -> mergeTopBottom (head xs) x n) (initList n) mergeFrom n = (\(x:y:zs) -> zs) (initList n) This is my ultimate answer. generateSol mergeFromList mergeIntoList n. It generates duplicate solution. generateSol (b:[]) a n      = foldr (++) [] $ map (\x -> mergeTopBottom b x n) a generateSol (b:bs) (a) n      = generateSol bs (foldr (++) [] (map (\x -> mergeTopBottom b x n) a)) n And now I should combile all of them to write the top-most function. langfordPair n = generateSol (mergeFrom n) (mergeInto n) n totalPairs n = length (langfordPair n) Who needs tests now! test1 = mergeTopBottom [2,0,0,2] [3,0,0,0,3] 3 test2 = map (\x -> mergeTopBottom [1,0,1] x 3) test1 test4 = mergeThisList [2,0,0,2] test5 = mergeThisList