GSF: Triangular storage allocation

• More or less things are described in the longish comments added in the code (I repeat them below).

• The main difference wrt before is that in reality one does need the diagonal distance (i,i) i.e the distance of one element with it self.

• Frozen tier-0 is here : RunTier0Tests.log

**
 *   For pairwise distance comparisons assuming 0..... N-1 (N total elements )
 *   The pairwise distance matrix  can be represented in a triangular array:
 *   [ (1,0)
 *   [ (2,0), (2,1),
 *   [ (3,0), (3,1), (3,2)
 *   [ (4,0), (4,1), (4,2) , (4,3)
 *   [.......................
 *   [.............................
 *   [(N-1,0),(N-1,1),(N-1,2),(N-1,3).......(N-1,N-2)]
 *
 *   With size 
 *   1+2+3+ .... (N-1) = N*(N-1)/2
 *
 *   The lexicographical storage allocation function is
 *   Loc(i,j) = i*(i-1)/2 + j 
 *   e.g
 *   (1,0) --> 1 *(1-1)/2 + 0 --> 0 
 *   (2,0) --> 2 *(2-1)/2 + 0 --> 1
 *   (2,1) --> 2 *(2-1)/2 + 1 --> 2
 *   (3,0) --> 3 * (3-1)/2 +0 --> 3
 *   Leading to  
 *   [(1,0),(2,0),(2,1),(3,0),(3,1),(3,2)...... (N-1,N-2)]
 * 
 *
 *   The N-1 Rows  map to value K of the 1st element in the pair.
 *   K=1,2,3,...N-1  and each one has size K. 
 *   Each Row starts at array positions K*(K-1)/2
 *   e.g 
 *   The row for element 1 starts at array position 0.
 *   The row for element 2 starts at array position 1.
 *   The row for element N-1  starts at array positon (N-1)*(N-2)/2
 *   
 *   The N-1 Columns map to the value K of the second  element in the pair 
 *   0,1,2 ... N-2
 *   The array position follow (i-1)*i/2+K;
 *   where i : K+1 .... N-1 [for(i=K+1;i<N;++i)
 *   e.g 
 *   0 appears as 2nd element in the pair at array positions 
 *   [0,1,3,6...]
 *   1 appears as 2nd element in the pair at array positions
 *   [2,4,7...] 
 *   2 appears as 2nd element in the pair at array positions
 *   [5,8,12....] 
 *   N-2 appears as 2nd element once at position [N(N-1)/2-1]
 *
 */