Aargh! those balls....

https://en.wikipedia.org/wiki/Balance puzzle#The twelve-coin_problem

Now, My answer and explanation:- There are 12 virtually identical stones except one stone has a weight that is slightly different (could be more or less) than the other 11 equally weighted stones. Now, let us start the game:- We have 12 stones and a balance (beam balance).

Explanation:-

Now, in the above shown balance there are, naturally two weighing pans, namely PAN A and PAN B on which we will weigh the stones. Then, the first operation we perform is that we divide the twelve stones into two equal sets each set containing 6 stones. Now let set p and set q be the two sets- it is obviously implied that our wanted stone (either slightly less or slightly more in weight) is in either of these sets and let’s say we will place set p on PAN A and set q on PAN B. After this, let us assume that our wanted stone is w is greater in weight than the other 11 stones. Now, after placement of each of stones of the sets on PAN A and PAN B respectively, let us assume that our observation is that PAN A which being heavy, settles down below than PAN B .This obviously implies that w is in set p placed on PAN A .Now we carefully pick up those 6 stones LYING IN PAN A in set p and then divide into further two sets of each set containing 3 stones each (set H and set R). Hence, we will place each of the 3 stones on PANA and PAN B of the balance. We are bound to have an observation- i.e. our wanted stone is amongst the stones currently in either of PAN A or PAN B- but let’s say we see that the PAN B (containing set R of stones) goes below than PAN A – which implies that PAN B contains our wanted stone w (since we had assumed that w is slightly heavier )-or to simplify our resultant observation implies that our wanted stone is among the 3 stones of PAN B. Finally we take those particular 3 stones (previously set R)- we place one of those stones on PAN A and the remaining 2 on PAN B-PANB will be obviously heavier (we don’t need a scale to calculate that!)-this is followed by our main/decisive operation- WE REMOVE ONE STONE FROM THE 2 ON PANB- which can actually yield two cases/observations:- ONE OBSERVATION MIGHT BE THAT THE TWO PANS BECOME EQUALLY BALANCED, WHICH IMPLIES THAT THE STONE WE REMOVED WAS OUR REQUIRED STONE (since all the other 11 stones are of equal weight) or another case/solution might be;- THAT THE STONE WE REMOVED WAS ACTUALLY ONE OF THE 11 EQUALLY WEIGHTED ONES- IN THAT CASE THE PAN CONTAING OUR REQUIRED STONE GOES DOWN (since we had assumed it to be slightly heavier) THAN THE OTHER PAN-BEACAUSE WE HAD ALREADY PROVED THAT OUR WANTED STONE MUST BE AMONGST THE 3 CONCERNED STONES (PREVIOUSLY SET R)-therefore, in either of these two cases we are able to identify our required stone. N.B.- WE HAVE DONE THE PROCESS IN 3 WEIGHS-(OR SO TO SAY 3 MEASURMENTS). AND ANOTHER IMPORTANT FACT IS THAT IF WE HAD ASSUMED THAT OUR WANTED STONE IS COMPARATIVELY SLIGHTLY LIGHTER THAN THE OTHER 11 EQUALLY WEIGHTED ONES,THEN THE ONLY CHANGE IN OUR PROOF WILL BE THAT THE BALANCE WILL GO UP(WE WILL ONLY HAVE TO REPLACE THE MOTION OF THE BALANCE-THAT IS IF LIGHT IT WILL BE UP-IF HEAVY,IT WILL GO DOWN/DESCEND).

Therefore, the required algorithm is :-

STEP I:- DIVIDE THE 12 STONES INTO TWO SETS OF 6 EACH AND PLACE EACH SET OF 6 ON PAN A AND PANB. STEP II:- WE WILL OBSERVE WHICH PAN WEIGHS HEAVIER (OR LIGHTER) AND TAKE THE PARTICULAR SET OF 6 STONES FROM THE PAN AND THEN WE PICK UP THOSE 6 STONES TO DIVIDE THEM INTO TWO EQUAL SETS OF 3 STONES STEP III:- WE THEN PLACE EACH SET OF 3 STONES ON PAN A AND PAN B AND THEN OBSERVE WHICH PAN WEIGHS HEAVIER (OR LIGHTER) AND TAKE THE PARTICULAR SET OF 3 STONES FROM THAT DIFFERENTLY WEIGHING PAN AND DIVIDE THEM INTO A SET OF ONE STONE AND ANOTHER SET OF 2 STONES (since obviously 3=2 +1) . STEP IV:- (we obviously know that a set of two stones placed on PAN B will definitely weigh more than PAN A containing only one stone- but we will still place them on the balance.)NOW FOR OUR THIRD(3rd ) and FINAL MEASUREMENT WE REMOVE ONE STONE FROM THE TWO STONES ON PAN B. STEP V:- (IF THE PAN A AND PAN B EQUALLY BALANCE EACH OTHER); THE ALGORITHM TERMINATES:THAT IS THE STONE WE REMOVED IS OUR REQUIRED STONE ( w). STEP VI:- (IF PAN A AND PAN B DON’T BALANCE EACH OTHER,WEIGH DIFFERENTLY), THE REQUIRED STONE CAN BE FOUND TO BE LYING ON ANY ONE OF THE PANS DEPENDING ON THE FACT WHETHER THE REQUIRED STONE IS COMPARATIVELY LIGHTER OR COMPARATIVELY HEAVIER). N.B.-WE HAVE USED ONLY 3 WEIGHS.