chinese puzzle

Put four balls on one side and four balls on the other.

You will discover one pile is heavier or, if they are equal, the other unweighed pile has the heavier ball.

Take the four balls that contain the heavy ball and weigh two of them on each side.

From that weighing, weigh the two remaining balls to find the heaviest.

Re: chinese puzzle

So which one is it, heavier or lighter?

And the question is supposed to have 27 balls (weighing 3 times can have a maximum of 27 balls).

This is a different (more difficult) problem. You don't know if the odd ball is heavier or lighter.

punit_mg

The 12 ball puzzle has been asked and answered in this forum several times. We have even discussed the elegance of the writing of the solution !

but buddy u don't know if the odd ball is heavy or light,so u can't choose heavy four straight away.

Start with six balls on each side. You will know which six has the lighter ball, because it will go up. Now remove two balls from the six, and set them aside, and weigh the four balls with two on each side. The one that goes up has the lighter ball. If it's balanced, it means the two balls you set aside has the lighter ball which you can determine with the third weighing. Otherwise, one of the two balls weighed has the lighter ball, and you can determined the weight with your third weighing .

The trick is to keep track of all information, including potential heavy, potential light information, as this information will come in handy in later weigh-ins.

Here is the naming convention:
balls are 1-12,
N means neutral,
L means maybe light,
H means maybe Heavy,
X means unknown
W1-W3 refers to the 3 weighings:

Start with X1-X12 (all balls unknown)

After W1 (measure 4 against 4), they either balance or don't:

-----------------------------W1A they balance-----------------------------
outcome W1A: balanced: N1-N8 and X9-X12 (8 of them are neutral and 4 unknown)

W2:weigh the following: x9,x10 versus x11,n1

You have 3 outcomes:
2a: balance
2b: right side up --> L9,L10 and H11
2c: left side up --> H9,H10 and L11

2.a: balance means 9-11 are also neutral (N9,N10,N11,X12) --> X12 is the oddball; use W3 with X12 and any other ball to see if it is lighter or heavier

2b use W3: weigh 10 and 11 against 2 neutral: 3 outcomes
balance means oddball is 9 and you already know it is L9
10,11 lighter means L10 is the oddball (lighter)
10,11 heavier means H11 is the oddball (heavier)

2c is same procedure resulting in 3 outcomes and each leads you to the oddball (H9, L11, H10 respectively being the oddball and their L or h is known)

------------------------W1B they don't balance---------------------------------

outcome W1B: First weigh in is not balanced : H1-H4, L5-L8, N9-N12 (the ones not measured are neutral, and you have two sets of possibly heavy and possibly light ones)

W2: H1,H2,L5 versus H3,H4,L6 (set aside L7 L8)
a: balance? --> W3 with L7 and an N --> if balance L8 , if not L7 is oddball (and the L denotes the lightness)

b: right-side heavy? narrows oddballs to H1,H2,L6 --> W3: H1vH2 and get a heavy oddball, or L6 (in case of balance)

c: left side heavy? narrows oddballs to L5, H3, H4, similar to b.....

solution to chinese problem
ok... a simple solution to this problem is....

[/B]number the balls 1,2,3....12
now weigh 1,2,3,4 against 5,6,7,8
then weigh 1,5,9,4 against 2,6,11,12
then weigh 3,7,9,11 against 4,6,10,2

now taking into consideration which pan goes up or which comes down it can easily determined which ball is the odd one.