Today we continue to discuss the paper on 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 saw how iFrames differ from boxes and why boxes are helpful with the Jigsaw design goals. A Jigsaw  application can have multiple boxes, but all of the boxes live in the same frame. Jigsaw code uses access modifiers to expose state to external domains. The Jigsaw runtime uses these modifiers to validate cross-domain operations. A single Jigsaw application may contain multiple  principals. Jigsaw places each of these principals in a separate box.  As with the DOM, a principal can access its parent principal object or the root.Principal objects can synchronize their writes to DOM storage. Each Jigsaw box can potentially contain a DOM tree and an associated visual field. The updates to these are constrained by the parent. A visual field consists of width height and location within the parents visual field and a z-order.  A parent can set these for the child. If a child wants to modify these, it can try but it has to be permitted by the parent. The child fires a  Jigsaw event on any modification and the parent must have a registered handler for the same. Visual fields can overlap but boxes cannot request overlapping  visual region. With the hierarchy, there  is fine control of the visual field and storage but a principal must still trust in its ancestors. DOM storage obstruction allows each origin to maintain a client-side key/value database. Jigsaw partitions this DOM storage.

#google codejam Riverflow problem:
A river receives water from tributaries that can be diverted to farms by farmers. Farmers choose an integer power of 2 from 1 to D both inclusive and toggle their usage everytime that many number of days has passed. Given N days history of water flow in the river, determine if the farmers are cheaters or the smallest number of farmers who could be diverting from the river.

int GetMinFarmers(int D, int N, List<int> flows)
{
bool cheaters = false;
int net = 0;
int farmers = 0;
for (int d = D; d >= 1; d /= 2)
{
    for (int i = 0; i < d; i++) {
         int net = flows[i+d]-flows[i] - flows[i+d-1] + flows[(i+2*d-1)%(2*d)];
                                                                 // i+d-1 stands for previous day to be reduced by next cycle
                                                                                          // (i+2*d-1)%(2*d) stands for the next cycle
         if (net%2) cheaters = true;
         net /= 2;
         if (net > 0){
                     farmers += net;
                     for(int j = i+d; j <= i+2*d;j++){
                           flows[j%(2*d)] -= net;
                     }
         }else{
                     farmers -= net;
                     for(int j = i; j <= i+d;j++){
                           flows[j%(2*d)] += net;
                     }
         }
    }
}
if (cheaters || flows[0] < 0)
   Console.WriteLine("cheaters");
return farmers;
}
#courtesy: Gnurdux