Interview Question for Design Verification & Test Engineer at Marvell Technology:

Trail 1: At random weigh two stones vs. two stones (3 sitting on the side)
             A: Of the 4 on the scales if one side weighs more then the other then weigh one on each side (since one of them must be heavier)
             B. If the 2 vs 2 are equal then at random weigh 2 (one on each side) of the three left on the side. If they are the same then the 3rd one that never got weighed is the heaviest.

Simple case of process of elimination by grouping (Divide and Test)

I would first weigh in one stone, say stone 1, and assume the weight is say 2 lbs(try 1). Then separate the 6 remaining stones into 2 piles, 2,3,4 and 5,6,7. Weigh in either 2,3,4 or 5,6,7, it doesn't matter. Say 2,3,4, if the sum of these 3 is 6 then the lighter stone has to be in the 5,6,7(try 2). Weigh in 5,6, if the sum of the two is 4 then the lighter is stone 7(try3). If sum is less than 4 then weigh in either 5 or 6 to find out. So, the maximum number of tries is 4 and least is 3.

Needs two weighing at most:
1. Put {1, 2, 3} on LHS and {4, 5, 6} on RHS.
2. If LHS and RHS are equal 7 is the lighter one.
    else discard heavier of previously weighed group.
Now we have a group 3 stones left. Lets call them A, B, C.
3. Put A on LHS and B on RHS.
4. If LHS and RHS are equal C is the lighter one.
    else lighter or LHS or RHS is the lighter one.
Voila!