Non-continuous subsequences

Non-continuous subsequences
You are encouraged to solve this task according to the task description, using any language you may know.

Consider some sequence of elements.

A subsequence contains some, but not all elements of this sequence, in the same order.

A continuous subsequence is missing elements only before its begin and after its end.

The task is to find all non-continuous subsequences for a given sequence. Example: For the sequence 1,2,3,4, there are five non-continuous subsequences, namely 1,3; 1,4; 2,4; 1,3,4 and 1,2,4.

D

(I may be misunderstood...) The list of non-continuous subsequences is constructed from some what different from task description, but should be equivalent way:

1. from the original sequence, select some continuous subsequences of length >= 3,
2. from each continuous subsequences in (1), remove some in-between elements beside first and last element,
3. each subsequences some elements removed in (2) is a non-continuous subsequences.

Notes:assumes continuous subsequences of length < 3 does not contain non continuous subsequences, <d>module ncsub ; import std.stdio ;

struct Combi {

 private int n_, m_ ;
 alias int delegate(inout int[]) dg_type ;
 int opApply(dg_type dg) { return combinate([], 0, m_, dg) ; }
 int combinate(int[] fixed, int next, int left, dg_type dg) {
   if (left <= 0) dg(fixed) ;
   else
     for(int i = next ; i < n_ - left + 1; i++)
       combinate(fixed ~ [i+1], i + 1, left - 1, dg) ;
   return 0 ;
 }

}

T[] takesome(T)(T[] seq, int[] rmIndexs) {

 T[] res = seq.dup ;
 foreach(idx ; rmIndexs.reverse) 
   res = res[0..idx] ~ res[idx+1..$] ;
 return res ;

}

T[][] ncsub(T)(T[] seq) {

 if(seq.length < 3) return [] ;
 T[][] ncset ;
 for(int head = 0 ; head < seq.length - 2 ; head++)
   for(int tail = head + 2 ; tail < seq.length ; tail++) {
     T[] contseq = seq[head..tail+1] ;
     for(int takeNum = 1 ; takeNum < contseq.length - 1 ; takeNum++)
       foreach(removeIndexs ; Combi(contseq.length - 2, takeNum))
         ncset ~= takesome(contseq, removeIndexs) ;
   }
 return ncset ;

}

void main() {

 writefln(ncsub([1,2,3])) ;
 writefln(ncsub([1,2,3,4])) ;
 writefln(ncsub([1,2,3,4,5])) ;  // results match Haskell examples
 writefln(ncsub([1,1,1,1,1])) ;  // should this return [] ? that is, no non-cont subseq.

}</d>

Haskell

action p x = if p x then succ x else x

fenceM p q s []     = guard (q s) >> return []
fenceM p q s (x:xs) = do
  (f,g) <- p 
  ys <- fenceM p q (g s) xs
  return $ f x ys

ncsubseq = fenceM [((:), action even), (flip const, action odd)] (>= 3) 0 

Output:

*Main> ncsubseq [1..3]
[[1,3]]
*Main> ncsubseq [1..4]
[[1,2,4],[1,3,4],[1,3],[1,4],[2,4]]
*Main> ncsubseq [1..5]
[[1,2,3,5],[1,2,4,5],[1,2,4],[1,2,5],[1,3,4,5],[1,3,4],[1,3,5],[1,3],[1,4,5],[1,4],[1,5],[2,3,5],[2,4,5],[2,4],[2,5],[3,5]]