[Tech] node swapping algorithm

Hi list,

On LocationManager#shouldSwap(), I wonder if we should use a geometric
mean instead of just multiplying them together.
Let's see the current implement:

748 double A = 1.0;
749 for(int i=0;i<friendLocs.length;i++) {
750 if(Math.abs(friendLocs[i] - myLoc) <=
Double.MIN_VALUE*2) continue;

#1. some friends are skipped.
this number should be small -- my friends shouldn't be sooo close to me.
rare, but not impossible (?)

751 A *= Location.distance(friendLocs[i], myLoc);
752 }
753 for(int i=0;i<hisFriendLocs.length;i++) {
754 if(Math.abs(hisFriendLocs[i] - hisLoc) <=
Double.MIN_VALUE*2) continue;

#2. again some friends are skipped. (same as #1)

755 A *= Location.distance(hisFriendLocs[i], hisLoc);
756 }
757
758 // B = the same, with our two values swapped
759 double B = 1.0;
760 for(int i=0;i<friendLocs.length;i++) {

( never mention his number of friends may not be the same as mine
should we use for(int i=0;i<friendLocs.length &&
;i<hisFriendLocs.length ; i++) ? )

761 if(Math.abs(friendLocs[i] - hisLoc) <=
Double.MIN_VALUE*2) continue;

#3. skip yet other friends...
but this time, we compare my friend's location with his location ..
I have him as my friend, so this is skipped _at least_ once.

The number of friends skipped here is not the same as those in #1 and #3..
i.e. A and B are calculated using different number of samples.

as Location.distance() < 1.0 , the more round you multiply,
the less the value.


762 B *= Location.distance(friendLocs[i], hisLoc);
763 }
764 for(int i=0;i<hisFriendLocs.length;i++) {
765 if(Math.abs(hisFriendLocs[i] - myLoc) <=
Double.MIN_VALUE*2) continue;
766 B *= Location.distance(hisFriendLocs[i], myLoc);
767 }

[...]

771 if(A>B) return true;

#4 And now, we compare A and B. This is unfair!


Regards,
Daniel Cheng