[Guest]

there are 10 balls, two of which are special. one of which is light and the other heavy.

you have a balance scale. if both the light and heavy are on the same side the will equal the weight of two normal balls.

can you identify the heavy and the light in 4 weighings? if so how?

[Guest]

Weigh all 10 balls with 5 on each side:

If initially they didnt weigh equally, you know which 5 balls have the heavy ball and which 5 balls have the light ball

Then, move 3 balls from each pan to the other

if they weigh equally, then 3 balls moved from lighter have the light ball and the 2 balls in the heavier have the heavy ballweigh the 2 heavy side balls and find which one is heavier

Weigh 2 of the lighter side balls and find the light ball

If 3 balls remain and one has an odd weight, weigh any 2 balls. If they weigh equally, the third ball weighs odd

If they dont weigh equally, according to the tilt of balance, you know which ball is odd weighted

If lighter side is still lighter then, the 2 lighter side balls have light ball and the 2 heavy ones have the heavy ball

1 weighing each for these sides can determine the odd weighing balls

If lighter side is now heavier then the 3 balls moved from heavier side have the heavy ball and the 3 balls moved from lighter side have the lighter ball

1 wieghing each can determine the odd balls

If initially they weigh equally, you know that one set of 5 balls has both light and heavy balls

Move 4 balls across pans

If they weigh equally, you now have 2 sets of 4 which contain the odd weight balls

If not, you know which ball is lighter or heavier

Move 3 of these 4 balls across

If they weigh equally, you now have 2 sets of 3 balls containing odd weight balls

If not, you know which ball is lighter or heavier

Move 2 of these 3 balls across

If they weigh equally, you now have 2 sets of 2 balls containing odd weight balls

If not, you know which ball is lighter or heavier

Take one from each set and weigh against normal balls

You find one of the odd weighing balls (and you know whether it is heavy or light) and the other balls in the previous set is then the other odd wieghing ball

Edited August 5, 2009 by DeeGee

[Guest]

Weigh all 10 balls with 5 on each side:

If initially they didnt weigh equally, you know which 5 balls have the heavy ball and which 5 balls have the light ball

Then, move 3 balls from each pan to the other

if they weigh equally, then 3 balls moved from lighter have the light ball and the 2 balls in the heavier have the heavy ballweigh the 2 heavy side balls and find which one is heavier

I suppose if they weigh equally, then 3 balls moved from lighter may be normal. Instead, 3 balls moved from heavier may be heavier and 2 light side balls may contain

an odd (lighter) ball.

Did I missed something?

I intentioanally ignore spoilers if I give no hint

[plainglazed]

For my answer I say it cant be done.

Would think that four weighings gives a total of 81 possible results and there initially seem to be 90 possible combinations. But with the parameters given in the OP, the example below defines as normal one ball after two weighings (or otherwise defines the heavy and light balls). I think the total possible combinations needed with the given parameters is less than 90 but as far as I can tell, still greater than 81.

Breaking the 10 balls into three sets of three with one extra remaining:

1,2,3 and

A,B,C and

X,Y,Z and

O

First weigh 1,2,3 and A,B,C

Then weigh A,B,C and X,Y,Z

If 1,2,3 > A,B,C

and A,B,C = X,Y,Z then 1,2,or 3 is heavy and O is light compare 1 and 2, if 1=2, 3 heavy, 1<2, 2 is heavy; or 1>2, 1 is heavy.

The same method applies for the other three scenerios where one weighing is equal and the other is greater than or less than.

If 1,2,3 > A,B,C

and A,B,C > X,Y,Z then 1,2,or 3 is heavy and X,Y, or Z is light and O is normal. compare 1 and 2 and compare X and Y as above.

The same method applies for the other three scenerios where both weighings are either greater than or less than.

If 1,2,3 = A,B,C

and A,B,C = X,Y,Z then both heavy and light are in either 1,2,3 or A,B,C or X,Y,Z and O is normal.

I can only seem to solve this for every scenerio using three more weighings for a total of five.

Unless I've missed something, I think for all but six cases (instead of 90-81=nine) it can be solved in four weighings by weighing 1,A,O to X,Y,Z for the third trial.

And you may be able to do better. And by that I mean "you" as in not me....

[Guest]

i tend to agree that its unsolvable, but i can solve the equal case and not solve the unequal one.

wiegh 3 agaisnt 3. call the left group A the right group B and the remaining group C.

if they are equal. then you are in one of 3 cases.

A = HLN B = NNN C = NNNN

A = NNN B = HLN C = NNNN

A = NNN B = NNN C = HLNN

wiegh 2A 1C vs 2B 1C

if equal then either the 2A are heavy and light or the 2B are heavy and light, and can solve in 2 wieghs.

if unequal, say left heavy. then you are in 1 of four cases.

case1: 2A = HN Aremain =L

case2: 2B = LN Bremain = H

case3: Cl = H, one of the 3C light,

case4: Cr = L, one of the 3C heavy.

wiegh 1 of the 2A and 1 of the 2B against the 2 C you didn't weigh.

if equal they must all be normal, and you can be in any one of the 4 cases.

wiegh the other 1A against the Cl. if equal, your in case 2. if Cl heavy, the Cr must be light, and if A heavy then a remain light.

if 1A and 1B are heavy, you are in case 1 or case 3. weigh the 2 C's on the right against one another. if equal then the A from the left must be heavy, if unequal then that one is light, and the Cl is heavy.

if 1A and 1B are light, then you are in case 2 or case 4, and can solve the same way.

if unequal (left side heavy say.) then you are in one of 3 cases.

A = HNN B = NNN C= LNNN

A= HNN B = LNN C = NNNN

A = NNN B = LNN C = HNNN

if you now weigh 3 B agianst 3 C as you suggest, then you run into a problem.

if the C side is light, then the 3Bs are normal, and can solve.

if equal, then they are all normal and can solve.

however, if the B side is light, you are still in case 2 or 3.

[plainglazed]

i tend to agree that its unsolvable, but i can solve the equal case and not solve the unequal one.

wiegh 3 agaisnt 3. call the left group A the right group B and the remaining group C.

if they are equal. then you are in one of 3 cases.

A = HLN B = NNN C = NNNN

A = NNN B = HLN C = NNNN

A = NNN B = NNN C = HLNN

wiegh 2A 1C vs 2B 1C

if equal then either the 2A are heavy and light or the 2B are heavy and light, and can solve in 2 wieghs.

if equal then either the 2A are heavy and light or the 2B are heavy and light, and can solve in 2 wieghs.

Seems to me that the heavy and light balls could still be in group C in this case (A = NNN B = NNN C = HLNN).

[plainglazed]

Compare two sets of three balls labeled A₁,A₂,A₃ and B₁,B₂,B₃ with C₁,C₂,C₃,C₄ remaining:

If A₁,A₂,A₃ > B₁,B₂,B₃ then either the heavy ball is in the A's or the light ball is in the B's. Compare B₁,B₂,B₃ and C₁,C₂,C₃:

If B

1

,B

2

,B

3

= C

1

,C

2

,C

3

then B

1

,B

2

,B

3

,C

1

,C

2

,C

3

are normal and either A

1

,A

2

,or A

3

is heavy and C

4

is light. Compare A

1

and A

2

:

If A

1

=A

2

then A

3

is heavy.

If A

1

<A

2

then A

2

is heavy.

If A

1

>A

2

then A

1

is heavy.

The same method applies when A

1

,A

2

,A

3

< B

1

,B

2

,B

3

and B

1

,B

2

,B

3

= C

1

,C

2

,C

3

.

If B

1

,B

2

,B

3

> C

1

,C

2

,C

3

then B

1

,B

2

,B

3

,C

4

are normal and either A

1

,A

2

,or A

3

is heavy and either C

1

,C

2

, or C

3

is light.

Compare A

1

and A

2

and compare C

1

and C

2

as above.

The same method applies for the other three scenerios where A

1

,A

2

,A

3

> or < B

1

,B

2

,B

3

and B

1

,B

2

,B

3

> or < C

1

,C

2

,C

3

.

If A₁,A₂,A₃ = B₁,B₂,B₃ then both the heavy and light balls are in either the A's, the B's, or the C's. Compare A₁,B₁,C₁ and C₂,C₃,C₄

If A

1

,B

1

,C

1

> C

2

,C

3

,C

4

then the possibilities are: A

1

>A

2

, A

1

>A

3

, B

1

>B

2

, B

1

>B

3

, C

1

>C

2

, C

1

>C

3

, C

1

>C

4

. Compare A

1

and B

1

:

If A

1

>B

1

then either A

1

>A

2

or A

1

>A

3

compare A

2

and A

3

the lighter of which is light and A

1

is heavy.

If A

1

<B

1

then either B

1

>B

2

or B

1

>B

3

compare B

2

and B

3

the lighter of which is light and B

1

is heavy.

If A

1

=B

1

then either C

1

>C

2

, C

1

>C

3

, or C

1

>C

4

. Compare C

2

and C

3

:

If C

2

=C

3

then C

1

is heavy C

4

is light; if C

2

>C

3

then C

1

is heavy C

3

is light; if C

2

<C

3

then C

1

is heavy C

2

is light.

If A

1

,B

1

,C

1

< C

2

,C

3

,C

4

the same method as above applies.

If A

1

,B

1

,C

1

= C

2

,C

3

,C

4

then the possibilities are A

2

,A

3

; B

2

,B

3

; C

2

,C

3

; C

2

,C

4

; C

3

,C

4

and A

1

,B

1

,C

1

are normal (N). Compare C

2

and C

3

:

If C

2

>C

3

(likewise if C

2

<C

3

) then compare C

4

and N:

If C

4

=N then C

2

is heavy C

3

is light.

If C

4

>N then C

4

is heavy C

3

is light.

If C

4

<N then C

2

is heavy C

4

is light.

If C

2

=C

3

then possibilities are A

2

><A

3

and B

2

><B

3

; compare A

2

and A

3

If A

2

>A

3

then A

2

is heavy, A

3

is light.

If A

2

<A

3

then A3 is heavy, A

2

is light.

If A

2

=A

3

then B

2

and B

3

are special.

Question: Can you identify the heavy and light balls in four weighings? Yes, and in 98.8% of the time you can even define which is which. Granted a semantic interpretation of the question in terms of the validity of the solution, but am pretty confident of the above being the optimum solution with the given parameters. Have come to the same answer using several different methods. While all these methods involved purely intuitive reasoning/brute strengh, would love to see some math confirming or refuting this solution as optimal.

[Guest]

once again you don't seem to be considering the case of B1,B2,B3 < C1,C2,C3, after A1,A2,A3 > B1,B2,B3.

here, either 3B is light and 3C is heavy, or 3B is light and 3C is normal, with either one of the 3A or the remaining C heavy.

if you simply weigh 1A against 1A, if one is heavy, then your good, but if they are equal, then your in deep.

[plainglazed]

rethinking

Edited August 14, 2009 by plainglazed