Today we continue to discuss jigsaw.Efficient, low effort mashup isolation by James Mickens and Matthew Finifter. Mashups are web applications that contain code from different principals.
Jigsaw lets mashups be isolated with its framework  and lets them selectively expose private state. We were discussing the porting effort involved with Jigsaw and performance improvements. We now review end to end performance study of Jigsaw. Jigsaw performs checks on property accesses. It does bookkeeping of all objects be setting their box id and object id. It checks access to virtualized browser resources for compliance to security policies. All this might affect the end to end performance. Consequently three different libraries were ported over to Jigsaw and their performance evaluated. These libraries include  JSON-RPC a library that performs RPC over Ajax connections.  A local host server is chosen over a remote server to avoid network delays in performance evaluation. A DOM-SQL library that uses SQL as the storage for DOM was another library used. The AES encryption function I'm the standard crypto library was also used for the study. Mousemove is another simple benchmark library that registers handlers on mode moves. In each case, the library code was put in a box separate from that of the integrator. All cross box communication with the integrator took place through virtualized resources or via a principal object.
#codejam exercise  Campinatorics 
A park ranger has to distribute families to tents in a N×N grid subject to the following rules:
Families can be 1 or 2 or 3 members 
The number of 3 member families is X
Thereally are plenty of 1 or 2 member families as fillers 
A cell can be occupied by only one tent
The number of members in any row or column must be exactly 3
The number of tents in any row or column cannot exceed two.
Find how many arrangements are possible if two arrangements are different if any of their cells are different or if the members in the same cell are different.
Solution : we are looking at different arrangements of 0, 1, 2 or 3 in an N×N grid subject to the above combinations. Therefore one way to do this would be to combine those values for each N×N cell . After all the cells have been given a number, we can quickly reject or accept an arrangement by running the checks above. We return the count of all accepted arrangements.
If we list the cells row wise as a string then the ith element belongs to the i/N row and i%N column.
Void arrange (StringBuilder a, int start)
{
  If (start == N×N){ print a.toString (); return; }
  For ( int m = 0; m <= 3; m++)
  {

      a [start] = m;
      arrange (a, start +1);
      a [start] = '/0';
}
}

Bool isValid (string s, int N, int X) // representation for 
{
Int [,] matrix = s.toMatrix ();
Int threes = 0;
For ( int I = 0; I < N; i++)
{
Int sum = 0;
Int count = 0;
For (int j=0; j < N; j++)
{
If (matrix [i][j] != 0) count++;
If (count >2) return False;
If (matrix [i][j] == 3) threes++;
sum += matrix [i][j];
}
If (sum != 3) return False;
}
For ( int j = 0; j< N; j++)
{
Int sum = 0;
Int count = 0;
For (int i=0; i< N; i++)
{
If (matrix [i][j] != 0) count++;
If (count >2) return False;
If (matrix [i][j] == 3) threes++;
sum += matrix [i][j];
}
If (sum != 3) return False;
}
If (threes != X) return False;
Return True;
}