Hi all,
Here's my attempt to convert a list of integers to a list of range tuples -
Given [1,2,3,6,8,9,10], I need [(1,3),(6,6),8,10)]
My attempt using foldl yields me the output in reverse. I can ofcourse
reverse the result, but what would be a better way?
f xs = foldl ff [] xs
where
On Thursday 23 December 2010 18:27:43, C K Kashyap wrote:
Hi all,
Here's my attempt to convert a list of integers to a list of range
tuples -
Given [1,2,3,6,8,9,10], I need [(1,3),(6,6),8,10)]
My attempt using foldl yields me the output in reverse. I can ofcourse
reverse the result, but
On Thu, 23 Dec 2010, C K Kashyap wrote:
Here's my attempt to convert a list of integers to a list of range tuples -
Given [1,2,3,6,8,9,10], I need [(1,3),(6,6),8,10)]
That's an interesting problem!
My first attempt:
List.unfoldr (\xs - case xs of [] - Nothing; y:ys - case span (uncurry
On Thu, 23 Dec 2010, Daniel Fischer wrote:
On Thursday 23 December 2010 18:27:43, C K Kashyap wrote:
Hi all,
Here's my attempt to convert a list of integers to a list of range
tuples -
Given [1,2,3,6,8,9,10], I need [(1,3),(6,6),8,10)]
My attempt using foldl yields me the output in
On Thursday 23 December 2010 18:27:43, C K Kashyap wrote:
Hi all,
Here's my attempt to convert a list of integers to a list of range
tuples -
Given [1,2,3,6,8,9,10], I need [(1,3),(6,6),8,10)]
My attempt using foldl yields me the output in reverse. I can ofcourse
reverse the result, but
On Thu, Dec 23, 2010 at 10:57:43PM +0530, C K Kashyap wrote:
Here's my attempt to convert a list of integers to a list of range tuples -
Given [1,2,3,6,8,9,10], I need [(1,3),(6,6),8,10)]
import Data.Function
import Data.List
ranges ns =
[(head gp, last gp) |
gp - map
I'd go with direct recursion for this one - the pattern of consumption
and production that generates the answer doesn't seem to neatly match
any of the standard recursion combinators (map, unfold, fold,
mapAccum, ...) nor exotic ones (skipping streams c.f. the Stream
fusion paper, apomorphisms,
On 23 December 2010 21:12, Stephen Tetley stephen.tet...@gmail.com wrote:
I'd go with direct recursion for this one - the pattern of consumption
and production that generates the answer doesn't seem to neatly match
any of the standard recursion combinators (map, unfold, fold,
mapAccum, ...)
On Thu, 23 Dec 2010, Stephen Tetley wrote:
On 23 December 2010 21:12, Stephen Tetley stephen.tet...@gmail.com wrote:
I'd go with direct recursion for this one - the pattern of consumption
and production that generates the answer doesn't seem to neatly match
any of the standard recursion
On 23 December 2010 22:01, Henning Thielemann
lemm...@henning-thielemann.de wrote:
This could be seen as type Step st a = (Maybe a, st). I have thought about
mapping from [Int] to [Maybe (Int, Int)] by mapAccumL, then compressing the
result with catMaybes. However we need to append a final
This one looks somewhat symmetrical:
f xs =
let xys = filter ( \ (x,y) - y - x 1 )
$ zip xs ( tail xs )
in zip ( [ head xs ] ++ map snd xys )
( map fst xys ++ [ last xs ] )
___
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All the previous solutions seem to assume that the list of numbers is already
sorted. In cases where this assumption cannot be made, an alternative solution
is to simply insert the numbers into a diet.
eecs.oregonstate.edu/~erwig/papers/abstracts.html#JFP98
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