Counterfeit Coin

You have 12 marbles, one of them is either heavier or lighter than the rest. 
In three weighs on a balance scale, you must determine which is the odd marble.
How do you do this?

This one's a great classic, known as the Counterfeit Coin Problem. 
The solution tabulated below:
It's also possible to detect an odd marble among 13. 
If we are given an extra marble known to be standard, detection among
as many as 14 other marbles becomes possible.

The weighing procedure for 12 marbles: ABCDEFGHIJKL.

First Weighing	Second Weighing	Third Weighing
ABCD = EFGH	AI = JK	A = L   is not possible. A < L  Þ  L is heavy. A > L  Þ  L is light.
	AI < JK	J = K  Þ  I is light. J < K  Þ  K is heavy. J > K  Þ  J is heavy.
	AI > JK	J = K  Þ  I is heavy. J < K  Þ  J is light. J > K  Þ  K is light.
ABCD > EFGH	ABE = CFL	G = H  Þ  D is heavy. G < H  Þ  G is light. G > H  Þ  H is light.
	ABE < CFL	C = L  Þ  E is light. C < L   is not possible. C > L  Þ  C is heavy.
	ABE > CFL	A = B  Þ  F is light. A < B  Þ  B is heavy. A > B  Þ  A is heavy.
ABCD < EFGH	ABE = CFL	G = H  Þ  D is light. G < H  Þ  H is heavy. G > H  Þ  G is heavy.
	ABE < CFL	A = B  Þ  F is heavy. A < B  Þ  A is light. A > B  Þ  B is light.
	ABE > CFL	C = L  Þ  E is heavy. C < L  Þ  C is light. C > L   is not possible.