Re: [Haskell-cafe] Improving *> and >> for Data.Sequence

23 Nov 2014


      OK, sorry for the flood of posts, but I think I've found a way to make that
work. Specifically, I think I can write a three-Seq append that takes the
total size and uses it to be as lazy as possible in the second of the three
Seqs. I'm still working out the details, but I think it will work. It does
the (possibly avoidable) rebuilding, but I *think* it's at least
asymptotically optimal. Of course, if Ross Paterson can find something more
efficient, that'd be even better.

On Sat, Nov 22, 2014 at 10:10 PM, David Feuer  wrote:
...
OK, so I've thought about this some more. I think the essential *concept*
I want is close to this, but it won't quite work this way:
fs <*> xs = equalJoin $ fmap (<$> xs) fs
equalJoin :: Int -> Seq (Seq a) -> Seq a
equalJoin n s
  | length s <= 2*n = simpleJoin s
  | otherwise       = simpleJoin pref ><
                      equalJoin (2*n) mid ><
                      simpleJoin suff
  where (pref, s')  = splitAt n s
        (mid, suff) = splitAt (length s - 2*n) s'
simpleJoin :: Seq (Seq a) -> Seq a
simpleJoin s
  | null s = empty
  | length s == 1 = index s 0
  | otherwise = simpleJoin front >< simpleJoin back
  where
    (front,back) = splitAt (length s `quot` 2) s
I think the reason this doesn't work is that >< is too strict. I believe
the only potential way around this is to dig into the FingerTree
representation and build the thing top-down. I still don't understand how
(if at all) this can be done.
On Sat, Nov 22, 2014 at 12:57 PM, David Feuer 
wrote:
...
...
To be precise, I *think* using the fromList approach for <*> makes us
create O
(n) thunks in order to extract the last element of the result. If we
build the
result inward, I *think* we can avoid this, getting the last element of
The ideal goal, which has taken me forever to identify and which may well
be unattainable, is to get O(log(min{i,mn-i})) access to each element of
the result, while maintaining O(mn) time to force it entirely. Each of
these is possible separately, of course. To get them both, if it's
possible, we need to give up on the list-like approach and start splitting
Seqs instead of lists. As we descend, we want to pass a single thunk to
each element of each Digit to give it just enough to do its thing.
Representing the splits efficiently and/or memoizing them could be a bit of
a challenge.
On Fri, Nov 21, 2014 at 02:00:16PM -0500, David Feuer wrote:
the
...
result in O(1) time and space. But my understanding of this data
structure
remains primitive.
This modification of the previous should do that.
mult :: Seq (a -> b) -> Seq a -> Seq b
mult sfs sxs = fromTwoLists (length sfs * length sxs) ys rev_ys
  where
    fs = toList sfs
    rev_fs = toRevList sfs
    xs = toList sxs
    rev_xs = toRevList sxs
    ys = [f x | f <- fs, x <- xs]
    rev_ys = [f x | f <- rev_fs, x <- rev_xs]
-- toRevList xs = toList (reverse xs)
toRevList :: Seq a -> [a]
toRevList = foldl (flip (:)) []
-- Build a tree lazy in the middle, from a list and its reverse.
--
-- fromTwoLists (length xs) xs (reverse xs) = fromList xs
--
-- Getting the kth element from either end involves forcing the lists
-- to length k.
fromTwoLists :: Int -> [a] -> [a] -> Seq a
fromTwoLists len_xs xs rev_xs =
    Seq $ mkTree2 len_xs 1 (map Elem xs) (map Elem rev_xs)
-- Construct a fingertree from the first n elements of xs.
-- The arguments must satisfy n <= length xs && rev_xs = reverse xs.
-- Each element of xs has the same size, provided as an argument.
mkTree2 :: Int -> Int -> [a] -> [a] -> FingerTree a
mkTree2 n size xs rev_xs
  | n == 0 = Empty
  | n == 1 = let [x1] = xs in Single x1
  | n <  6 = let
            nl = n `div` 2
            l = Data.List.take nl xs
            r = Data.List.take (n - nl) rev_xs
        in Deep totalSize (mkDigit l) Empty (mkRevDigit r)
  | otherwise = let
            size' = 3*size
            n' = (n-4) `div` 3
            digits = n - n'*3
            nl = digits `div` 2
            (l, xs') = Data.List.splitAt nl xs
            (r, rev_xs') = Data.List.splitAt (digits - nl) rev_xs
            nodes = mkNodes size' xs'
            rev_nodes = mkRevNodes size' rev_xs'
        in Deep totalSize (mkDigit l) (mkTree2 n' size' nodes rev_nodes)
(mkRevDigit r)
  where
    totalSize = n*size
mkDigit :: [a] -> Digit a
mkDigit [x1] = One x1
mkDigit [x1, x2] = Two x1 x2
mkDigit [x1, x2, x3] = Three x1 x2 x3
mkDigit [x1, x2, x3, x4] = Four x1 x2 x3 x4
-- length xs <= 4 => mkRevDigit xs = mkDigit (reverse xs)
mkRevDigit :: [a] -> Digit a
mkRevDigit [x1] = One x1
mkRevDigit [x2, x1] = Two x1 x2
mkRevDigit [x3, x2, x1] = Three x1 x2 x3
mkRevDigit [x4, x3, x2, x1] = Four x1 x2 x3 x4
mkNodes :: Int -> [a] -> [Node a]
mkNodes size (x1:x2:x3:xs) = Node3 size x1 x2 x3:mkNodes size xs
-- length xs `mod` 3 == 0 =>
--    mkRevNodes size xs = reverse (mkNodes size (reverse xs))
mkRevNodes :: Int -> [a] -> [Node a]
mkRevNodes size (x3:x2:x1:xs) = Node3 size x1 x2 x3:mkRevNodes size xs
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