Coin Challenge: Who's the Odd Ball?

It can be done in 2 weighings.

You place 3 ball combination in the 3 pan scale. (Standard Tri State Scale)

(Arms 120 degrees apart.)

The offending group will be higher or lower than the other 2.

If in balance the 4th group will have the fake.

The offending group can now be weighed with 1 ball in each pan and the lighter or heavier ball can then be isolated.

I'll second that!

Unless dear Liberty has one hidden under her frock, she doesn't have a scale of ANY kind.

I think someone is thinking of Lady Justice.

http://en.wikipedia.org/wiki/Lady_Justice

You are right. I see it in the courts

The balance I mean has only two pans like a see-saw

Make 2 batch's of 6 each

step 1

weigh and mark odd ball batch

Step 2

again divide odd ball batch in to 2 batch's

pick 2 balls from each batch,put on pan and go for step 3 with conditions

step 3

scale is not balanced , weigh the two balls of odd ball batch those are on pan and find out odd one,

OR

scale is balanced,keep them at side and weigh remaining 1 ball from each batch and find out odd one.

is any other way?

One odd ball is heaver or lighter as per original condition. Thus, you can not mark odd ball batch in step 1

How do you know which is the odd batch. The odd ball may be heavier or lighter. you do not know. Only logic will help and this is not a two minute job. It is something you will enjoy solving

Divide into three groups of four.

Scene one:

1) Compare four against four, assume balance, then discard all 8 balls. Odd ball is in next group of 4

2) Weigh 1 and 1 of reamaining, if balance, discard one,

3) Compare against one of remaining. If balance, then remaining is odd.

If at step 2) is not balanced then (3) compare agaianst one of remaining, if balance you just discarded odd ball, if not balance you have odd ball still in pan.

If at step 1) not balanced, 2)remove 3 balls from each pan, if pans balance 3) weigh 2 of off balance group. if balanced, remaining ball is odd. ----Hmm, now we have 4 weighs required.

I haven't seen any rules preventing me from using a 3 arm scale.

Assuming that fake precious metal coins are always lighter one can isolate the fake in 3 weightings between 27 coins.

Divide in 3 groups of 9 and place any 2 groups on the scale (if the scale is in balance the spare group is lighter)

Divide the lighter group in 3 groups of 3 and place any 2 groups in the pans

compare any 2 of the lightest group and name the culprit.

For the 12 (genderless) coins you need to do the following

W1

Take any 3 coins in each pan

If in balance or not you have 6 known good coins

W2

Weigh the 6 unknown against the 6 good and determine if the fake is lighter or not.

W3

Take 2 of the L or H group (as the case may be) and name the fake

I know it is incorrect but it is all I have time for now.

I have asked my assistant to start designing a 12pan balance.

Oh sorry 11 pan should be enough for one go.

I would prefer a 12 pan one. If you use a 11 pan one and the odd ball is nº 12, you have to make any further measurement to know if it's heavier or lighter.

And once starting to design a "more than two pan scale" why stop at 11?

Kind regards

Yea you are right as it has added advantage of using it as 2.3,4,6,8 and 12 pan scale.where as with 11 pans its only 11pan scale,

http://cr4.globalspec.com/blogentry/983/Counterfeit-Coins-Newsletter-Challenge-01-23-07

Right, the quiz has already been answered in the provided link. But let me add on this that one can detect within three efforts a counterfeit coin among 13, not just 12! Let me give a more mathematical twist to the matter:

First of all, a balance - given certain "questions" we pose to her - has a set of "answers" with which can provide useful information. When I say, "questions", I mean certain placement of coins we perform on the plates (provided that we always ensure that there are same number of coins on each plate). When I say "answers", I mean information that the balance passes to us by either staying balanced, unbalanced, or changing the direction of the tilt. Not having information about whether the counterfeit coin is lighter or heavier, there is an asymmetrical way a balance can answer us, depending on whether it is in a balanced position or not, before posing our "question". That is, if the balance is level, and then after some certain placements of coin it tilts, this counts as one piece of information, regardless of whether the tilt is to the left or to the right. If on the other hand, the balance has already been tilted after a previous step, then it can "answer" with three possible ways in the next step: Either return to the balanced position, or stay where it is, or change tilt to the other side. One may ask, how can we excite all possible answers in order to get the most out of the balance's answers? The given link gives some possible coin moves. They boil down to moving coins from outside to the balance, and moving coins from the one plate to the other (this is useful in case the balance is already tilting), taking care to always leaving same number of coins on the plates, by moving non-suspect coins to or from them.

Given all the above, I created a tree showing the possible answers we can get. With "b" I denote "balanced, "ub" means "unbalanced" (no matter which direction), "c" means "change of tilting direction" and "nc" stands for "no change of tilting direction"

I guess it is straightforward. Now, the interesting bit is that at the third level, we have 13 possible answers, therefore it is possible to detect a counterfeit coin among 13! It is left for homework how it is done! A hint is that by viewing the tree, we start by leaving 5 coins off the balance (see the left part of the tree), and placing the rest 8 on the plates (4+4).

One can easily extend the above tree to see what happens in the next levels. Generally, to get to the level i, we note that it holds:

b_i = b_i-1 + ub_i-1 + c_i-1 + uc_i-1

ub_i = b_i-1

c_i = u_i-1 + c_i-1 + uc_i-1

uc_i = c_i

One may use z-transform to find the analytical function, rather than having to use the above recursive formulas step by step. I may do it if I find some free time.

If my calculations are correct, then we have 34 leaves on level 4 (i.e. we can trace 34 coins to detect one counterfeit in four efforts). In this case, we start with 16 coins off the balance and 9 on each plate. Then, we have 92 coins on level 5 (starting with 38 off balance, and 27 on each plate), and so on.

Can you specify term weighings or trials exact. from putting balls on both the pans and unloading both the pans, is this? what if one pan is not unloaded, is that still one trial?

You can put balls on either pan or one pan. Each time you put balls on the pan and remove it is one trial. I do not see your point in putting balls on one pan. What does that give you. All I can say in any trial ,both pans should have equal number of balls to give you a deduction. I am liking the humour in your answers of having twelve pans etc. Sorry guys only two pans and no other weighing device. all you have is a two pan balance Twelve identical Balls one lighter or heavier and your logic. Go for it.

Well....

let's divide the balls into three groups: A, B, and C.

1. Put A and B groups each on one scale arm. Two things can happen: Balance (then both groups A and B are "good balls") or unbalance. In the last case, obviously one group must be heavier than the other. Let's assume the heavier is "A" group. the if the odd ball is in the "A" group, we know it must be heavier. If it's in "B" group, must be lighter. Resume: the first try give us some information:

- If scale is balanced, the odd ball is in group "C" but we don't known if it's heavier or lighter.

- If scale is unbalance, we know group "C" balls are good ones and if the odd one is in group "A", must be heavier and lighter if belongs to group "B".

2.1 Assuming first try is balanced. Then take any three group "C" balls in one arm and another 3 good ones (we've got 8 from groups "A" and "B") and watch the scale: if balanced, the odd ball is the group "C" remaining and by placing it against a good one in the third try we'll know if its heavier or lighter. If the result of second try is unbalanced, the way the 3 "C" balls arm behaves tell us if odd ball is heavier or lighter and one more thing: the odd ball is one of that 3. Then to identify the ball, in the third try just test two of them (one against the other). If balanced, the odd is the third ball. If unbalanced, the ball heavier or lighter as already determined is the goal.

2.2 Let's go with the first try unbalance result. For the second try, take 3 "A" balls and one "B" ball and test them against the fourth "A" ball and 3 good "C" balls. If the scale is balanced, the odd ball must any of the remaining 3 "B" balls. And we know that "B" balls are lighter... so test any two in the third try and if balanced, the odd ball is the remaining "B" and is of course "lighter". If unbalance happens in the third try, the ball in the upper arm (lighter one) is the goal.

2.3 If the second try is unbalanced, we have 3 "A" + 1 "B" balls in one arm and one "A" and good ones in the opposite. If 3"A" + 1"B" arm gets down, the odd ball must be any of that 3 "A" ball (Remember that "B" balls are lighter and cannot get down their arm and on the other arm, the "A" ball cannot get the arm up because are "heavier")

In this case, the third try must be to compare any two of the 3 "A" balls. If balanced, the odd is the third "A" and therefore heavier. If unbalanced, the "A" ball located in the lower arm is the goal.

If in the second try the lower arm is that with one "A" (heavier) and three good ones, the possible solutions are just two: The odd ball need to be either the "A" ball in the lower arm or the "B" ball in the higher arm. To decide which is the odd just compare as the third try any of them with a good one. If balanced is the remaining, and if unbalanced the odd is in the opposite arm to the "good one".

I think I've take into account all possibilities and use just three tries.

Any comment will be appreciated.

Kind regards

If the