A Weighty ball problem?

[zoombini]

Okidoki 8 ball bearings, one of which is slightly different from the rest.
You cannot visually see the difference its only in its weight.
You do not know if its heavier or lighter.

You have a pair of balance scales, you are only allowed to make 3 balance tests.

How do you tell which is the odd bearing?

Something else thats being discussed in the office. :)

Quick replies wanted so I can be the one to answer it :D :D

[altis]

2^3 = 8 so, yes, it can be done in three tests.

1) ABC vs DEF
2) ABF vs DEC
3) AEF vs DBC

then just follow the letters...

Edit
Just read the bit about not knowing if it's heavier of lighter :dunce:
mmmm... thinking cap on, instead.

[danielf]

Put 2 bearings on each side of the scale. If they are not equal in weight, the odd one is in this lot. Now check two of these, if these are equal in weight, check one of these against one of the other two. If these are equal in weight, the remaining one is the odd one out. If not, it's the 'new one'. If the first 4 are equal in weight, do the last steps with the other 4.

Edit: Or, in letters. Check AB vs CD. If unequal, check A v B, and the B v C.
If AB = CD repeat last steps with E, F ,G and H.

[zoombini]

eh?
what about g & h?

[zoombini]

Put 2 bearings on each side of the scale. If they are not equal in weight, the odd one is in this lot. Now check two of these, if these are equal in weight, check one of these against one of the other two. If these are equal in weight, the remaining one is the odd one out. If not, it's the 'new one'. If the first 4 are equal in weight, do the last steps with the other 4.

You only get 3 goes.

[danielf]

You only get 3 goes.

This is three goes. In the first go, you narrow it down to four (say ABCD). Then you check two of these to see if they are equal (say A and B). If they are, you only have two left, one of which includes the odd one out. If you now check if A and C are equal (and they are), the odd one out must be D. If A and C are not equal, it must be C.

If the first 4 are equal, you do step 2 and 3 with EFGH

[MetaWraith]

Okidoki 8 ball bearings, one of which is slightly different from the rest.
You cannot visually see the difference its only in its weight.
You do not know if its heavier or lighter.

You have a pair of balance scales, you are only allowed to make 3 balance tests.

How do you tell which is the odd bearing?

Something else thats being discussed in the office. :)

Quick replies wanted so I can be the one to answer it :D :D

This is an OLD,OLD Problem

Big hint here : Start with weighing 2 vs 2


Full answer here
http://www.mathnerds.com/mathnerds/best/counterfeit/solution.asp

[danielf]

This is an OLD,OLD Problem

Big hint here : Start with weighing 2 vs 2

Put 2 bearings on each side of the scale. <snip? :)

[altis]

Too much haste - ignore me I'm talking balls!

I think Danielf has the answer:

1) AB vs CD, if they balance goto 5
2) A vs B, if they balance goto 4
3) A vs C, if they are unbalanced then answer is A, balanced it's B
4) A vs C, if they are unbalanced then answer is C, balanced it's D
5) AB vs EF, if they balance goto 7
6) A vs E, if they are unbalanced then answer is E, balanced it's F
7) A vs G, if they are unbalanced then answer is G, balanced it's H.

[Matth]

1: Weigh 2 against 2 - if this group are equal, they become your reference, the other group the sample, otherwise reverse it.
2: Weigh 2 of the sample against 2 of the reference - if equal, they move to the reference, otherwise the other two move to the reference.
3. Now out of the remaing two left in the sample, weigh ONE against one of the reference - if unequal, you've found it, otherwise it's the other one

[danielf]

Too much haste - ignore me I'm talking balls!

I think Danielf has the answer:

1) AB vs CD, if they balance goto 5
2) A vs B, if they balance goto 4
3) A vs C, if they are unbalanced then answer is A, balanced it's B
4) A vs C, if they are unbalanced then answer is C, balanced it's D
5) AB vs EF, if they balance goto 7
6) A vs E, if they are unbalanced then answer is E, balanced it's F
7) A vs G, if they are unbalanced then answer is G, balanced it's H.

Sort of. If AB and CD balance, the odd one out must be in EFGH, if not it must be in ABCD. If ABCD contains the odd one out, compare A with B. If they balance, the odd one out must be C or D. If A and C balance, it must be D. If A and C don't balance, it must be C since A balances with B. If A and B don't balance, one of them is the odd one out, so you compare A (or B with C. Three steps.

[danielf]

1: Weigh 2 against 2 - if this group are equal, they become your reference, the other group the sample, otherwise reverse it.
2: Weigh 2 of the sample against 2 of the reference - if equal, they move to the reference, otherwise the other two move to the reference.
3. Now out of the remaing two left in the sample, weigh ONE against one of the reference - if unequal, you've found it, otherwise it's the other one

Slightly different from my solution, but it works just the same. :)

[zoombini]

The links given are for a 12 ball similar problem.

[zoombini]

1: Weigh 2 against 2 - if this group are equal, they become your reference, the other group the sample, otherwise reverse it.
2: Weigh 2 of the sample against 2 of the reference - if equal, they move to the reference, otherwise the other two move to the reference.
3. Now out of the remaing two left in the sample, weigh ONE against one of the reference - if unequal, you've found it, otherwise it's the other one

This is the easiest to understand for me & it seems to work.

I'll have to see what the lads here make of it. cheers.

[MetaWraith]

Now I've had time to take lunch, and write it up here's a solution
I think It's more or less what Danielf had, but with more detail.

STEP 1
Weigh 2 v 2
Possible results
A) pairs are unequal (heavier or lighter not known)
B) pairs are equal
Thus we have 4 balls known to be "good" and 4 balls still "suspect".

STEP 2)
Weigh 2 "suspect" balls v 2 "known good" balls
Possible results
C) pairs are unequal (heavier or lighter is now known)
D) pairs are equal (discard these)

STEP 3C
Weigh one of the "suspect" balls tested in STEP 2 against a "known good" ball
Possible Results
E) Balance equal (the other suspect is guilty)
F) Balance unequal (the chosen suspect is guilty)

STEP 3D
Weigh one of the "suspect" balls not tested in STEP 2 against a "known good" ball
G) Balance equal (the other suspect is guilty)
H) Balance unequal (the chosen suspect is guilty)

NOTE
In G) it is impossible to tell if the guilty ball is heaver or lighter without a further weighing

[zoombini]

Now to make it worse, how do you tell if the odd one out is heavier or lighter?

[zoombini]

Supposing the first 4 are balanced.
Then the 2 ref & 2 sample are balanced.

its one of the last 2.

you weigh one ref & 1 sample, they balance, its the last one - which is it heavy or light?

I think there is some swapping from sides of the scale involved...

[danielf]

Like Metawraith said, in his E and G it is impossible to tell without a further weighing.

[altis]

Now to make it worse, how do you tell if the odd one out is heavier or lighter?
Follow my 'basic' program and you will be able to tell if it's heavier or lighter in every case except the last one. To arrive at ball H you will have had three balances so you'll need to compare it with one of the others to find out if it's heavier or lighter.

On balance, perhaps it would have been clearer if I'd labelled the stages 1, 2a, 3a, 3b, 2b, 3c and 3d instead of 1 to 7.

[zoombini]

Not according to one of the technical magazines we have at work.
Someone reckons to have figured it out but he's keping stum..

[zoombini]

I can do it in 1 weigh in.

Pick them all up, put into little bag, find the Mensa headquaters, go through "way in" & hand to them to solve it :D

[altis]

Lets set up some nomencleture(sp?)

test => left side heavier / balanced / right side heavier

1) start here
AB vs CD => goto 2a / goto 2b / goto 2c

2a) A or B is heavy, or, C or D is light
A vs B => A is heavy / goto 3a / B is heavy

3a) C or D is light
C vs D => D is light / uh? / C is light

2c) A or B is light, or, C or D is heavy
A vs B => B is light / goto 3b / A is light

3b) C or D is heavy
C vs D => C is heavy / uh? / D is heavy

2b) E,F,G or H is light or heavy
ABC vs EFG => goto 3c / goto 3d / goto 3e

3c) E,F or G is light
E vs F => F is light / G is light / E is light

3d) H is light or heavy
A vs H => H is light / uh? / H is heavy

3e) E,F or G is heavy
E vs F => E is heavy / G is heavy / F is heavy

All 8 balls resolved heavy or light in 3 tests.

Can I go to bed now?

[zoombini]

OK, you go to bed & I'll pass it on to the lads to see what they make of it.

[zoombini]

The written solution has arrived today. (The Engineer).

Number the balls 1 to 8.
Denote weighings W1, W2, W3.


W1 Put aside balls 7 & 8. Weigh balls 1, 2, 3 against 4, 5, 6.
If they balance, 7 or 8 is light/heavy.

W2 (a) Weigh 7 against any one of 1 to 6: this will show if 7 is heavy, light or the same.

W3 (a) If the same repeat with 8.

If 1, 2,3 is heavier than 4, 5, 6 on the first weighing then 1, 2 or 3 is heavy or 4, 5, or 6 is light.

W2 (b) Weigh 1, 2, 4 against 3, 7, 8.
If 1, 2, 4 are heavy then 1 or 2 is heavy.

W3 (b) Weigh 1 and any ball except 2. Either one will be the heavier, or, if the same, 2 is heavy.

If in W2 (b) 1,2,4 is light then either 4 is light or 3 is heavy.

W3 (c) Weigh 4 against anything but 3. If the same then 3 is heavy.

Hmm, too heavy for me, but I understand it now.

[altis]

The written solution has arrived today. (The Engineer).

Number the balls 1 to 8.
Denote weighings W1, W2, W3.


W1 Put aside balls 7 & 8. Weigh balls 1, 2, 3 against 4, 5, 6.
If they balance, 7 or 8 is light/heavy.

W2 (a) Weigh 7 against any one of 1 to 6: this will show if 7 is heavy, light or the same.

W3 (a) If the same repeat with 8.

If 1, 2,3 is heavier than 4, 5, 6 on the first weighing then 1, 2 or 3 is heavy or 4, 5, or 6 is light.

W2 (b) Weigh 1, 2, 4 against 3, 7, 8.
If 1, 2, 4 are heavy then 1 or 2 is heavy.

W3 (b) Weigh 1 and any ball except 2. Either one will be the heavier, or, if the same, 2 is heavy.

If in W2 (b) 1,2,4 is light then either 4 is light or 3 is heavy.

W3 (c) Weigh 4 against anything but 3. If the same then 3 is heavy.

Hmm, too heavy for me, but I understand it now.
Hmm... indeed - surely that's only part of it.

Yes, it resolves whether 7 is heavy or light in stage 2a, whether 8 is heavy or light in stage 3a, whether 1 or 2 is heavy in stage 3b and whether 3 is heavy or 4 is light in stage 3c.

But what about the cases when 1 is light, 2 is light, 3 is light, 4 is heavy or anything to do with balls 5 or 6?

[zoombini]

I dunno, thats the given answer in the magazine that posed the question.

I guess that you get to swap some of them about somewhere.

I think you swap 123 with 456 in step something... I'll ask the guys.

??