Originally posted by Zuggy
While it is true that, on the first weighing, you cannot avoid having the heavy and the light weight on the same side of the scales that does not mean you cannot tell them apart on subsequent weighings.
The point is that you don't glean muhc infromation from each weighing.
Lets assume you divide the 9 weights into three groups: a1a2a3 b1b2b3 c1c2c3.
Now you weigh group a against b.
If they are the same it could mean:
all a's and b's have the same weight, that means the orther two are in the c-group.
both different weights are in the a group
OR both are ikn the c-group.
Summary: All you learned with that result is that both sought weights are in the same group, but you still have to do a lot mor weighing. Since now you must check through all of the groups.
1 weighing a1 vs. a2 if they are the same
weighing b1 vs. b2 if they are the same
weighing c1 against c2 [you already are at the fourth weighing here]
IF c1> c2 you still don't know which one is the normal (if any)
so weigh c1 against c3 and you still have to weigh c2 against c3 to establish the order [7 weighings]
back to the first weighing:
If a>b the wither one of the a's is the heavy weight OR one of the b's is the light weight. OR one of the a#s is the heavy AND one of the b's is the light.
so you weigh a vs. c a>c you now know that one of the a's is the heavy one. *
If c>a you know that one of the c's is the heavy one and all a's are normal ones.
On the second case you have to weigh c1 vs. c2 and b1 vs. b2 to find the sought weigths. Making it four weighings at the least.
On case * you still have to weigh c aginst b to find out which batch is the normal one. and you have to weigh then the groups with the sought weights to establish which one is they are.
In Sum you have to make 4 weighings if you are lucky and 7 if you are unlucky.
There might be a simpler solution, but I doubt that.