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Expert

31-05-2009, 12:18 PM

There are 12 stones, 11 of equal weight, 1 of unequal weight (either heavier or lighter).

Use 3 weighings on a balance scale to tell what stone has a different weight and if it is heavier or lighter.

Can anyone solve this?

mayank

06-06-2009, 11:08 AM

Step 1 - weigh 6 + 6 on both the sides. Pick the 6 stones on the lighter side.

Step 2 - Then weight 3+3 on both the Sides ( picked from Step 1) . pick the 3 stones on the lighter side.

Step 3 - Then weigh any two randomly from the stones picked from Step 2. If they weigh equal the other stone will be lighter. If not the Scale will itself give you an answer.

CopyCat

13-06-2009, 05:39 PM

There are 12 stones, 11 of equal weight, 1 of unequal weight (either heavier or lighter).

Use 3 weighings on a balance scale to tell what stone has a different weight and if it is heavier or lighter.

Can anyone solve this?

Make 4 sets of 3 each. A,B,C,D

compare A & B. If they are equal, take one of each stone out from C and D.

now compare C & D. If they are equal... Add the 2 stone that you took out back... you will know which one is odd and you will if it was heavy or light.

IF A&B were not equal... take one stone each out from A&B.. if they the weight level, you know which one is odd and also know heavy or light ( by looking at wheather the light side went down or heavy side went up).

If they still don't level... take one stone each out again from A&B.

Natraj

14-06-2009, 06:33 PM

Here is my solution. It's long because I explain in detail.

Given:

12 identical looking stones and a balance scale.

1 of the stones is an oddball, and does not weigh the sames as the other 11.

Problem:

Using three weighings on the scale, determine the oddball stone and

whether it is heavier or lighter than the others.

Analysis:

Any solution will require breaking the 12 stones an N piles of equal size.

(I can't prove why, but it seems reasonable.)

Candidate starts:

2 piles of 6

3 piles of 4

4 piles of 3

6 piles of 2

The initial measurement must deliver the most amount of information possible.

2 piles of 6 (divide and conquer) is not a solution:

While a simple divide and conquer appears most obvious, it can only be

applied if the type of the oddball (L or H) is known beforehand.

Divide and Conquer:

Assume the oddball is light.

Make two piles of six and compare

- The oddball must be in the light side

Make two piles of three from the light side and compare

- The oddball must be in the light side

Take two stones from the pile of three on the light side and compare

- If equal, the oddball is the one not on the scale

otherwise, the oddball is the stone on the light side

While the approach above is commendable, it will not bear truth in 3 weighings

if the oddball is in fact heavy.

When the oddball is heavy, the comparison of three will show equal weights

implying the heavy stone is in the heavy pile of 6. Now you are

'one measurement behind'. When you split the heavy side into two piles of 3,

you do not know which one is heavier (the one that would necessarily contain

the heavy stone).

4 piles of 3 is not a solution:

After much empirical doodling I find four piles of three can only be solved

if the initial comparison is unequal.

When the first comparison of piles is equal, the oddball must be in one of the other

two piles. Two more measures is insufficient for determining both which and type

of oddball is amongst the remaining six stones.

3 piles of 4 is the solution

Only initial piles of four offers the most amount of logical inference.

Notation:

W1, W2, W3 mean weighing 1, 2 or 3

The weight of the 11 same stones will be called unit weight

Divide the stones into three piles of four

Compare any two piles (name the piles A and B)

Either they are equal or unequal

W1: A = B

The eight stones on the scale are all the same weight

The oddball must be in the pile C (c1 c2 c3 c4) not on the scales.

Compare any two stones from C (c1 c2) with

one more stone from C and one unit stone (c3 1)

Either the are equal or unequal

W2: (c1 c2) equal (c3 1)

The odd ball must be c4 and it must be heavy or light

Compare c4 to a unit stone

They must be unequal

W3: c4 < 1

The oddball is light and is on the scale.

W3: c4 > 1

The oddball is heavy and is on the scale.

W2: unequal, (c1 c2) < (c3 1)

Either c1 or c2 is a light oddball, or c3 is a heavy oddball.

Compare c1 to c2

Either they are equal or unequal

W3: c1 <> c2

The oddball is light and is on the scale.

W3: c1 = c2

The oddball is heavy and is stone c3.

W2: unequal, (c3 1) < (c1 c2)

Either c1 or c2 is a heavy oddball, or c3 is a light oddball.

Compare c1 to c2

Either they are equal or unequal

W3: c1 <> c2

The oddball is heavy and is on the scale.

W3: c1 = c2

The oddball is light and is stone c3.

W1: A <> B

The oddball is in one of the piles on the scale.

The four stones in C, not on the scale, are all unit weight.

Call the light side pile L and the heavy side pile H.

The piles must contain an oddball in one of the following states

Light Heavy

------- -------

Either 1 1 1 L < 1 1 1 1

Or 1 1 1 1 < H 1 1 1

New information can be obtained by manipulating the states,

changing them from either/or to if/then inferences.

State changes that afford information are:

- scale balance in-equality is reversed on next weighing

- scale balance equal on the next weighing

- scale balance in-equality is the same on next weighing

These state changes are accomplished by moving stones

from one side to another, or replacing stones with stones

of unit weight.

After trying many configurations, the following manipulation

yields inference that guarantees determination.

Let's break the manipulation into steps

s1. Remove two stones from L, call them L-off. Call the other two L-on.

s2. Remove three stones from H, call them H-off. Call the other one H-on.

s3. Move two H-Off stones to L.

s4. Move one stone of L-off to H.

s5. Move two stones of unit to H.

Let look at the possible states after each step.

s1. If oddball is light, L-off either contains it or not. (1 new state)

If oddball is heavy, L-off will never contain it.

L 1 . . 1 1 1 1 1 1

1 1 . . 1 L 1 1 1 1

1 1 . . 1 1 H 1 1 1

s2. If oddball is light, H-off will never contain it.

If oddball is heavy, H-off either contains it or not. (1 new state)

L 1 . . 1 1 1 . . . 1 1 1

1 1 . . 1 L 1 . . . 1 1 1

1 1 . . 1 1 1 . . . 1 1 H

1 1 . . 1 1 H . . . 1 1 1

s3. When moving a stone from H-off (1 1 H) to L

either the possible heavy is or is not moved. (1 new state)

Put a * to the right to indicate stones from H-off

L 1 1* 1* 1 1 1 . . . 1 . .

1 1 1* 1* 1 L 1 . . . 1 . .

1 1 1* H* 1 1 1 . . . 1 . .

1 1 1* 1* 1 1 1 . . . H . .

1 1 1* 1* 1 1 H . . . 1 . .

s4. When moving a stone from L-off (L 1) to H

either the possible light is or is not moved. (1 new state)

Put a * to the left to indicate stones from L-off

. 1 1* 1* 1 1 1 *L . . 1 . .

. L 1* 1* 1 1 1 *1 . . 1 . .

. 1 1* 1* 1 L 1 *1 . . 1 . .

. 1 1* H* 1 1 1 *1 . . 1 . .

. 1 1* 1* 1 1 1 *1 . . H . .

. 1 1* 1* 1 1 H *1 . . 1 . .

s5. Show unit stones as =

Necessary scale balance of next step is also shown

. 1 1* 1* 1 1 > 1 *L = = 1 . .

. L 1* 1* 1 1 = 1 *1 = = 1 . .

. 1 1* 1* 1 L < 1 *1 = = 1 . .

. 1 1* H* 1 1 > 1 *1 = = 1 . .

. 1 1* 1* 1 1 = 1 *1 = = H . .

. 1 1* 1* 1 1 < H *1 = = 1 . .

Now compare the two piles

Either they will be equal or unequal

W2: equal

Only two possible states can be equal

. L 1* 1* 1 1 = 1 *1 = = 1 . .

. 1 1* 1* 1 1 = 1 *1 = = H . .

Note the mutually exclusive states of the remaining L-off and H-off stones.

L

H

Compare the remaining L-off stone to a unit stone

Either they are equal or unequal

W3: unequal

The oddball is light and is on the scale.

W3: equal

The oddball is heavy and is the one remaining H-off.

W2: unequal - L is still the light pile

Only two possible states have L still light

. 1 1* 1* 1 L < 1 *1 = = 1 . .

. 1 1* 1* 1 1 < H *1 = = 1 . .

Note the mutually exclusive states of the remaining L-on and H-on stones.

1 L 1

1 1 H

Compare the two L-on stones

Either they are equal or unequal

W3: unequal

The oddball is light and is on the scale.

W3: equal

The oddball is heavy and is H-on.

W2: unequal - L is now the heavy pile

Only two possible states have a 'sign change'

. 1 1* 1* 1 1 > 1 *L = = 1 . .

. 1 1* H* 1 1 > 1 *1 = = 1 . .

Note the mutually exclusive states of the remaining L-swap and H-swap stones.

1* 1* *L

1* H* *1

Compare the two H-swap stones

Either they are equal or unequal

W3: unequal

The oddball is heavy and is on the scale.

W3: equal

The oddball is light and is L-swap.

Now after this long answer Its tea time :tea:

Master

18-06-2009, 07:42 PM

Expert - Answer Please

Expert

20-06-2009, 06:17 PM

Mayank & Copycat your answers are not correct.

Natraj - You have hit the Jackpot. :peace:

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