There are 12 pieces of radioactive metal that look identical. 11 of th

I have a very dumb question here -- Under what constraint should we go ahead solving this question "What is the minimum number of comparisons?"

Should it be least number of comparisons in which case every comparison yields something (optimistic approach) that helps us identify the defective piece or should it be the smallest number of comparisons required to be absolutely sure of finding the defective piece every time (pessimistic approach)?

If I consider the least number of comparisons wherein every comparison is optimistic ... and it yields the defective piece then the answer should be 4 comparisons.

The question is to find out the least number in the worst case ... or in other words the "pessimistic approach"

The answer is 3

Its a fairly difficult question to get right, but in solving it you don't need to know any advanced math; just basic logic

I knew this question already (but with balls with different weights instead), see that the difficulty is that we don't know whether the radiation is greater or smaller, so it's not that obvious when you measure them.

However, the right approach is to divide them the groups as you did.

In the Case 2 you can mix them up using the "neutral" remaining 4, so that can help you to discover the "different one".

You can do it in two comparisons. Call the 11 identical elements x, and the counterfeit y.

On the first weighing, you put two on one side of the scale and two on the other side, and one of the sides has the counterfeit. You'll get a different reading, but you won't know which side has the counterfeit yet. You do have the comparison of x+x and x+y, though.

On the second weighing, take one of the sides off and leave the other on the scale. Then take two more random elements and weigh them against the pair that remained. If the pair that you removed from the scale had the counterfeit, then you're going to get an equal weighing. So now you know that the pair you removed had the counterfeit, and the first weighing told you whether x+y was greater than or less than x+x.

(A)

This cannot work, because you are picking a total of 6 elements across the two measurements, there is guarantee that you even pick the counterfeit one in all this

All they asked for was the minimum amount of measurements it would take. This is the best-case scenario, and therefore the minimum amount.

Well that implicitly means worst case scenario actually. Minimum measurements it takes to find out with probablity 1, you can't pick the case where we just happen to select the counterfeit one, that would be too simple ... its not a trick question

Here is the full solution; divide the 12 into sets of 4 -- X={1,2,3,4} ; Y={5,6,7,8} ; Z={9,10,11,12}

Comparison 1 : X v/s Y

If X=Y, we know the offender is in Z

Comparison 2 : {9,10,11} with {5,6,7}

If they are equal, we know offender is 12. To find out if it is more or less than others, takes one more comparison with any normal sample

If they are not equal we immediately know if counterfeit one is more or less by seeing if {9,10,11} is more or less. Then compare 9 & 10 in measurement 3, if equal answer is 11 else we know the answer between 10 & 11 since we now know if the offender is less or more


If X>Y, we know the offender is in either one of these 2 sets and Z is clean


Comparison 2 : {1,5,9} with {2,7,6}

If they are equal means offender is one of {4,5,8} ... to find out which one compare 4 & 8 with 9 & 10 (both of which are normal). If equal answer is 5. If not, we know lesser or more, hence we know which one of X & Y had the counterfeit, and so we also know if answer is 4 or 8. So done

If {1,5,9} > {6,7,2}, and since X>Y, we know either 1 is more active or one of 6 or 7 is less active. In the third go, weigh 1 & 7 with 11 & 12 (both genuine). If this last measurement comes equal, answer is 6. If not, we know immediately if its 1 or 7 depending on whether answer is lesser or more.

If {1,5,9} < {6,7,2} and since X>Y, we know either 2 is more active or 5 is less active. Compare 2 with 12, and we will know the answer immediately. So done


If X<Y, the case is exactly symmetric to the one above.

Or we can do it this way..
Divide the given into 2 equal sets (of 6 each)
Measure these 2 sets. One set will give an erroneous value.
Divide that very group again into 2 sets (of 3 each)
Measure them. Again one set will be erroneous.

Pick up this erroneous set with 3 members.
Pick out any 2 and measure them. Here there can be 2 possibilities -
a) If the measurement is equal, then the one member left behind is the one we seek.
b) if the measurement is erroneous, then Voila!!!

All in all, it can be done in 3 mesurements.!!!

R J

That does not work, because you don't know if erroneous is higher or lower, so you won't be able to pick the correct lot after the first measurement

Oh Yesss ...you are right ...hmm...with your explanation...the first case where X=Y is pretty simple...but the second case wherein X>Y, tht is something....the most peculiar thing is the way u pick up the numbers in tht case...{1,5,9} and {2,7,6}...any reason behind tht...i guess the explanation requires further expalantion...

Here is the trick to this :

First of all if you are down to just 3 pieces and you know that if the offending piece is less or more active, then it takes exactly 1 measurement to find out the offending piece. So you know you have to reduce the problem to three.

Now when you are down to either A or B after measurement 1, you need the next measurement to (a) reduce the problem set to 3 and (b) to know whether anser is more or less. Now you cannot compare a group of 4 to 4, as in the best case it will only reduce the problem to 4 elements which is not good enough.
If you have to choose a set of 3 to compare, you cannot pick any 3 on the same side from the same set (A or B) because if you do this, a quick check will show you that in every choice there is a case where you can only get down to 4 elements. Eg. If you weighed {1,2,3} v/s {5,9,10} and they were equal you're problem would only reduce to {4,6,7,8}
The easiest way to solve this then is to compare 3 to 3, and make sure each side has elements from both A & B such that whatever the measurement outcome in the worst case the problem reduces to 3 elements only. Which is why the sets {1,5,9} and {2,6,7} OR {A,B,C} & {A,B,B}. The extra element from C is just taken to make the problem symmetric so to say, we have 8 elements and we make it 9, to compose 3 sets of 3 each.

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