12 coins problem game

Coin Weighing Game. Embed this code :. Coin Weighing » ProProfs Games. The point is that you need to arrive at a situation where "either the fake coin is lighter and is one of these coins; or the fake coin is heavier and is one of those coins". We do the second weigh as follows: R1, R2, R3 vs U1, U2, U3 we have 2 cases: F the left side is heavier that the right side is heavier is similar G the two sides weights the same for F the fake coin is lighter and is one of U1, U2, U3.

We weigh U1 vs U2 and we can get the answer for G the fake coin is U4, but we don't know whether it it is lighter or heavier than the real ones, we can find this out by weighing this against one of the real coins. This is the solution for a 12 coin problem, the second weigh of case A is the tricky weigh. Featured Logic Games Sudoku. Crazy Taxi. Mahjongg II. Let us denote the left coins as coins 1, 2, 3, and 4 and the right coins as 5, 6, 7, and 8. For the next step, we should divide the 8 potential answers into groups of 3, 3 and 2.

For example, we want to have three answers in the case when the left cup is lighter or equal to the right cup, and 2 answers when the left cup is heavier than the right cup. That means that 3 out of the 8 coins should be left in the same place that they started, 3 should be taken away, and 2 should change places.

We may use coins 9, 10, 11, and 12 to supplement each weighing, in order to have an equal number of coins in each cup. Firstly let me stress that this puzzle is very difficult, and worse the answer is not particularly satisfying. It is not a good way to be introduced to my site. You have been warned. My answer is something of an amalgam between the Dr. Math web-site , myself, and various emails I received, including a correction from Ben Cronin.

It is now rigorous but as easy to follow as is possible. You are given a set of scales and 12 marbles. The scales are of the old balance variety. That is, a small dish hangs from each end of a rod that is balanced in the middle. The device enables you to conclude either that the contents of the dishes weigh the same or that the dish that falls lower has heavier contents than the other.

The 12 marbles appear to be identical. In fact, 11 of them are identical, and one is of a different weight. Your task is to identify the unusual marble and discard it. You are allowed to use the scales three times if you wish, but no more.

Note that the unusual marble may be heavier or lighter than the others. You are asked to both identify it and determine whether it is heavy or light. Before reading the answer can I interest you in a clue? Most people seem to think that the thing to do is weigh six marbles against six marbles, but if you think about it, this would yield you no information concerning the whereabouts of the only different marble or the nature of it's difference as we already know that one side will be heavier than the other.

It's worth mentioning that if this is used as an interview question this observation, by itself, is likely to get you a good portion of the way there. So that the plan can be followed, let us number the marbles from 1 to For the first weighing let us put on the left pan marbles 1,2,3,4 and on the right pan marbles 5,6,7,8. There are two possibilities. Either they balance, or they don't. If they balance, then the different marble is in the group 9,10,11, So for our second one possibility is to weigh 9,10,11 against 1,2,3 1 They balance, in which case you know 12 is the different marble, and you just weigh it against any other to determine whether it is heavy or light.

In this case, you know that the different marble is 9, 10, or 11, and that that marble is heavy. Simply weigh 9 against 10; if they balance, 11 is the heavy marble. Also, the tree is not full tree, middle branch terminated after first weighing. Infact, we can get 27 leaves of 3 level full 3-ary tree, but only we got 11 leaves including impossible cases.

Analysis: Given N coins, all may be genuine or only one coin is defective. Observe the above figure that not all the branches are generating leaves, i.

When possible, we should group the coins in such a way that every branch is going to yield valid output in simple terms generate full 3-ary tree. Problem 4 describes this approach of 12 coins. How many weighing are required in worst case to figure out the odd coin whether it is heavier or lighter? Label the coins as 1, 2, 3, 4 and G genuine. We now have some information on coin purity. We need to make use that in the groupings. We can best group them as [ G1, 23 and 4 ].

The following diagram explains the procedure. This is possible in two ways, either 1 should be lighter or either of 2, 3 should be heavier. The former instance is obvious when next weighing 2, 3 is balanced, yielding 1 as lighter. The leaf nodes on left branch are named to reflect these outcomes. This is possible in two ways, either 1 is heavier or either of 2, 3 should be lighter. The former instance is obvious when the next weighing 2, 3 is balanced, yielding 1 as heavier.

In the above problem, under any possibility we need only two weighing. We are able to use all outcomes of two level full 3-ary tree. Infact we should have 11 outcomes since we stared with 5 coins, where are other 2 outcomes?

These two outcomes can be declared at the root of tree itself prior to first weighing , can you figure these two out comes? This can be solved in one weighing read Problem 1. Note the equality sign. Problem 4: The classic 12 coin puzzle You are given two pan fair balance. You have 12 identically looking coins out of which one coin may be lighter or heavier. How can you find odd coin, if any, in minimum trials, also determine whether defective coin is lighter or heavier, in the worst case?