- My laptop is broken, so I can't work on code for this on the plane.
- The delights of Sweden will go less noticed, since the number theorist in me will not let me forget. (It's already saying, "It's such a simple description, how hard can it be? Just give it some thought.")
This was cruel timing. ;-)
---Frank
On Friday, Aug 22, 2003, at 02:25 US/Eastern, Andrew J Bromage wrote:
G'day all.
On Fri, Aug 22, 2003 at 03:41:14PM +1000, [EMAIL PROTECTED] wrote:
Seeing as its thst time of year again and everyone is posting their homework, has anyone got any good puzzles to do? I wouldn't mind having a go at something a bit tricky.
OK, here's a tricky problem.
Take a list S. Delete some elements from the list. What you have left is a subsequence of S. For example, [1,3,2] is a subsequence of [2,1,2,1,3,2,1,3].
Consider the following list:
[1,2,3,1,2,3,1]
This list contains all permutations of the list [1,2,3] as
subsequences. It is also minimal, in the sense that there is no shorter
subsequence which will do (though there are potentially many minimal
subsequences). We will call such a list a shortest supersequence
over the alphabet [1..3].
The challenge is multi-part. You may answer as many or as few questions
as you want.
1. Write a function sss :: Int -> [Int] where sss n is a shortest supersequence over the alphabet [1..n]. Make this as efficient as possible. Prove an upper-bound complexity on your function.
2. Write a function sss_length :: Int -> Int where sss_length n is the length of a shortest supersequence over the alphabet [1..n]. Make this as efficient as possible. Prove an upper-bound complexity on your function.
If you can't solve this problem efficiently, write a function sss_length_bounds :: Int -> (Int,Int) which returns the best upper and lower bounds that you can.
(Hint: n is a trivial lower bound, n^2 is a trivial upper bound. A tighter upper bound is n^2-n+1. Prove this as an exercise.)
3. Write a function sss_count :: Int -> Int where sss_count n is the number of shortest supersequences over the alphabet [1..n]. Make this as efficient as possible. Prove an upper-bound complexity on your function.
(Hint: sss_count n must be a multiple of n factorial. Why?)
If you can't solve this problem efficiently, write a function sss_count_bounds :: Int -> (Int,Int) which returns the best upper and lower bounds that you can.
Incidentally, I don't have sub-exponential answers to any of these questions. You did ask for something "a bit tricky".
Cheers, Andrew Bromage _______________________________________________ Haskell-Cafe mailing list [EMAIL PROTECTED] http://www.haskell.org/mailman/listinfo/haskell-cafe
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