Brain teaser

sharky86

You have the following:

• 12 red pool balls
• 1 balancing scales

All the balls weight is equal with one exception, one ball is heaver or lighter you are not sure which, but just slightly. You are allowed to place the balls on the balancing scales 3 times in any combination. No more. How you find the one ball?

Give it a good lads it is possible

gerry87

Break the 12 into 3 groups of 4. A B and C. Then number each group A1, A2, A3, A4 etc...

     [U]X[/U]              [U]Y[/U]             [U]Z[/U]
1| A1, A2         B1, B2        C1, C2
 | A3, A4         B3, B4        C3, C4

Keep one from each group where it is (1) and rotate the rest of them to the group to the left.

2| A1, B2         B1, C2        C1, A2
 | B3, B4         C3, C4        A3, A4

Then weigh X1 and B1, after that weigh X2 and B2. Now consider what happens when you go from X1 to X2. There are a few scenarios.

1) X1 is heavier(lighter) and X2 heavier(lighter). Then either A1 is heavy(light) or B1 is light(heavy). Weigh A1 and C1 to see which it is.

2) X1 is heavier(lighter) and X2 is lighter(heavier). Then either B2, B3 or B4 is light(heavy). Weigh any 2 to find the light one.

3) X1 and X2 both balance with Y1 and Y2. Then the ball is C1. Weigh with any to find if it's heavy or light.

4) X1 balances with Y1 but X2 is heavier (lighter). This tells us C2, C3 or C4 is lighter (heavier). Weigh any 2 of them to find the light (heavy) one.

5) X1 is heavier (lighter), X2 and Y2 balance. Then A2, A3 or A4 is heavier (lighter). Weight any 2 to find the heavy one.

sharky86

gerry87 wrote: »

Break the 12 into 3 groups of 4. A B and C. Then number each group A1, A2, A3, A4 etc...

     [U]X[/U]              [U]Y[/U]             [U]Z[/U]
1| A1, A2         B1, B2        C1, C2
 | A3, A4         B3, B4        C3, C4

Keep one from each group where it is (1) and rotate the rest of them to the group to the left.

 
2| A1, B2         B1, C2        C1, A2
 | B3, B4         C3, C4        A3, A4

Then weigh X1 and B1, after that weigh X2 and B2. Now consider what happens when you go from X1 to X2. There are a few scenarios.

1) X1 is heavier(lighter) and X2 heavier(lighter). Then either A1 is heavy(light) or B1 is light(heavy). Weigh A1 and C1 to see which it is.

2) X1 is heavier(lighter) and X2 is lighter(heavier). Then either B2, B3 or B4 is light(heavy). Weigh any 2 to find the light one.

3) X1 and X2 both balance with Y1 and Y2. Then the ball is C1. Weigh with any to find if it's heavy or light.

4) X1 balances with Y1 but X2 is heavier (lighter). This tells us C2, C3 or C4 is lighter (heavier). Weigh any 2 of them to find the light (heavy) one.

5) X1 is heavier (lighter), X2 and Y2 balance. Then A2, A3 or A4 is heavier (lighter). Weight any 2 to find the heavy one.

Shiny gold star for you gerry

My maths "professor" in portabello couldnt figure it out

Mellor

sharky86 wrote: »

My maths "professor" in portabello couldnt figure it out

The man must really be an idiot. any decent leaving cert student should solve this.

JoeB-

I don't understand the solution... but I don't think it's right to be honest.. it's not an easy puzzle...


What I mean is that the solution given is confusing, and I can't really follow it... I have solved this puzzle myself and know the answer.. it just isn't the one given above.. and I don't think the one above works..


the first instruction is 'weigh x1 and b1'... x1 is a group of balls, b1 is a single ball??? etc etc


Gold star awarded too hastily methinks...


hmmm..
edited to add..
maybe I was a little hasty.. he he. (sorry)

The solution does seem to be accurate, it's the typos that are confusing... I have an alternative solution, but maybe it's the same one, just written differently... confused...

red_fox

Call the scales A and B.

Place any 5 on A, any other 5 on B and leave 2 off.
1) the scales balance - the odd ball is one of the 2 not on the scales, place one on each and you have your answer
2) the scales tip - you know that the odd ball is one of 5 so for the second weighing place 2 of those 5 on A, another 2 on B and 1 off.

2a) The scales balance - the odd ball is the 1 not on either.
2b) The scales tip - you know the ball is one of two, weigh those against each other (third use of scales) and you win.

Follow the arrow where you find the lightest:

                                       (A=5  B=5           Off=2)
                                              |              |                            
                                              V              V
                                   (A=2 B=2   Off=1)   (A=1 B=1)
                                         |       |         |
                                         V       V         V
                                   (A=1 B=1)    WIN       WIN
                                         |
                                         V
                                       WIN

Or, as I clearly have nothing better to do, here is the above tree in pictorial form:

Fremen

I think you misread the question red fox, the odd ball can be heavier or lighter. That way, using the 5/5/2 strategy if the scales tip then all you know is that the odd ball is one of the ten.

I have a feeling Gerry87's is the only solution.

I'd have to disagree with Mellor, it's a reasonably difficult problem.

Wasn't able to solve it myself, though I only tried for fifteen minutes or so. I knew of the problem before: it's in "Heard on the street" by timothy crack. The book has a list of all the brainteasers you're likely to be asked in a job interview. This is actually a bit annoying since competitors at interviews memorise lists of these things in the hopes of appearing bright. The only thing (unoriginal) brainteasers in interviews test is how many brainteasers an applicant has studied.

red_fox

Awh but I drew a picture and everything, you're right though, my method only works if you know if its weight realative to the others.

sharky86

Mellor wrote: »

The man must really be an idiot. any decent leaving cert student should solve this.

Completely disagree

gerry87

That first instuction was supposed to be weight X1 and Y1 then X2 and Y2, sry bout that.

I saw it before in heard on wall street , it still took a good while to solve after i remembered something about rotating parts of the groups.

Definitely not an easy one though, there are so many combinations of ways to try before you realise you're wrong, and it's almost impossible to keep track of them in a tree or anything like that. It would take some luck to stumble on the right method if you hadn't heard it before.

JoeB-

yep, I agree.. I have known many people attempt this puzzle and fail..

I wouldn't even expect all A grade leaving cert students to solve this in an hour.. I'd say only 20% or less of B grade leaving students would solve it in 30 mins...

Why would people rotate the balls as in step 2??? This is the special step, and one that might seem easy in hindsight but not easy otherwise..my solution is slightly different, I gave it on these boards about two years ago.. I feel it's slightly easier but more or less equivilent, it does seem to be a slightly different sequence with the same results

Mellor

Fremen wrote: »

I'd have to disagree with Mellor, it's a reasonably difficult problem.

sharky86 wrote: »

Completely disagree

Actually you're right, I didn't read it carefully and missed a key piece of info.

But I would still expected a maths prof to solve it. And a top LC students. I remember doing similar ones in school, scales, water puzzles etc

jolt

Here's an alternative pseudocode type answer. It's pretty similar to Gerry87's actually just presented differently:

Number the balls 1 to 12.
put 8 random balls on the scales, keep 4 off:

E.g. (1,2,3,4) vs. (5,6,7,8)

Case 1 : balance

Take three of the remaining and one known equal weight ball. e.g. (9,10) vs. (11,1)
Case 1 : balance

Remaining ball (12) is the odd one out.

Case 2 : Imbalance

3_Unknown_1_Known Method:

Scale will either tip in favour or against the known equal weight ball (1)
Take ball 11 off the scale
Switch ball 9 over to the right hand side.
Put another known equal weight ball on the other side to give e.g. (2,10) vs. (9,1)

Case 1 : scales stay the same - ball on the other side (10) is the odd one out
Case 2 : scale balances - Removed ball (11) is the odd one out.
Case 3 : scales tip over - Switched ball (9) is the culprit

Case 2 : imbalance

You now have 8 balls, say 1,2,3,4 on the left-hand of the scale and 5,6,7,8 on the right-hand side.

Take balls 3,4 and 8 off. Swap 2,6, and 7 to their opposite sides and add a known equal weight ball to the right-hand side
to give say (1,6,7) vs. (5,2,9)

Case 1 : Scale stays the same - either 1 or 5 is the culprit : Swap one for an equal weight ball to find which
Case 2 : Scale balances - Either 3,4 or 8 is the culprit. use 3_Unknown_1_Known method to find out which
Case 3 : Scale tips to the opposite side - Either 2,6 or 7 is the culprit. use 3_Unknown_1_Known method to find out which.

JoeB-

Hi..

My answer given earlier is here..
http://www.boards.ie/vbulletin/showpost.php?p=52649150&postcount=69

it's different to Gerry87's answer... from step 2 onwards.. but only a little...

Cheers so...