Try out the ODDBALL logic test?

Here is my solution I hope it is correct, if not please inform me why

Good morning people,

Here is the solution(s) to the 12 odd ball problem!

You could use chalk to identify ball potially lighter or heaver or both

Before I explain some conventions in setting out the solution to the Twelve Balls problem, here again is a statement of the task.

There are twelve balls, all the same size, shape and color. All weigh the same, except that one ball is minutely different in weight, but not noticeably so in the hand. Moreover, the odd ball might be lighter or heavier than the others.

Your challenge was to discover the odd ball and whether it is lighter or heavier.

You must use a beam balance only, and you are restricted to three weighing operations.

Note: by starting with 6 /6 or, 5/ 5 or 3/ 3. or 2/2 or 1/1 it is impossible to solve the problem

“The only way to solve this rather complex problems is by starting by weighing 4 balls against 4 as you will see by the solution below”

Conventions:

At every weighing one of three things theoretically can happen: the pans can balance, the left pan can go down or the left pan can go up.

It will be necessary to refer to a given ball as definitely normal (N), potentially “heavier” (H) or potentially “lighter” (L). Often our identification of a ball in this way will be as part of a group (= “This group contains a heavier/lighter ball”), and will depend on what we learn from a previous weighing. At the start, all balls have a status of unknown (U).

To show at each weighing what is being placed in each pan, represent the situation as per the following examples:
.
First Weighing UUUU ——— UUUU

Pans balance All these U’s are now known to be N’s; the odd ball is one of the remaining unweighed four (call them UUUU from now on).

Proceed to Second Weighing: Case 1

Left pan down

One of the four balls in the left pan might be heavier (call them HHHH from now on) or one of the four balls in the right pan might be lighter (call them LLLL from now on).

Proceed to Second Weighing — Case 2

Left pan up

One of the four balls in the left pan might be lighter (call them LLLL from now on) or one of the four balls in the right pan is heavier (call them HHHH from now on).

Proceed to Second Weighing — Case 2

Second Weighing

Case 1 UUU ——— NNN
Pans balance All these U’s are now known to be N’s; the odd ball is the remaining unweighed U, but we don’t yet know if it’s heavier or lighter than normal.

Proceed to Third Weighing — Case 1

Left pan down

One of these U’s is heavier than normal, but we don’t yet know which one (call them HHH from now on).


Proceed to Third Weighing — Case 2

Left pan up

One of these U’s is lighter than normal, but we don’t yet know which one (call them LLL from now on).

Proceed to Third Weighing — Case 3

Case 2 HHL ——— HLN
Pans balance

All these H’s and L’s are now known to be N’s; the odd ball is one of the remaining unweighed H or two L’s.

Proceed to Third Weighing — Case 4

Left pan down

The odd ball is one of the left two H’s or the right L.

Proceed to Third Weighing — Case 5

Left pan up

The odd ball is either the right H or the left L.

Proceed to Third Weighing — Case 6

Third Weighing
Case 1 U ——— N
Pans balance Not possible

Left pan down The odd ball is this U, and it’s heavier

Left pan up The odd ball is this U, and it’s lighter

Case 2 H ——— H
Pans balance The odd ball is the remaining unweighed H (heavier)

Left pan down The odd ball is the left H (heavier)

Left pan up The odd ball is the right H (heavier)

Case 3 L ——— L

Pans balance The odd ball is the remaining unweighed L (lighter)

Left pan down The odd ball is the right L (lighter)

Left pan up The odd ball is the left L (lighter)

Case 4 L ——— L

Pans balance The odd ball is the remaining unweighed H (heavier)

Left pan down The odd ball is the right L (lighter)

Left pan up The odd ball is the left L (lighter)
Case 5 H ——— H

Pans balance The odd ball is the remaining unweighed L (lighter)


Left pan down The odd ball is the left H (heavier)

Left pan up The odd ball is the right H (heavier)

Case 6 H ——— N
Pans balance The odd ball is the remaining unweighed L (lighter)

Left pan down The odd ball is this H (heavier)

Left pan up Not possible

 

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This is the solution I found online.
It is very similar to Stryders attempt, but it works in every case.
http://www.primepuzzle.com/leeslatest/12_ball_solution.html

A Solution To The 12 Ball Problem

Statement of the problem: You have 12 balls, one of which is heavy or
light. In 3 weighings, determine which ball is odd and whether it is light
or heavy.

Label the 12 balls 1, 2, etc. to 12. Perform the following 3 weighings:

1 2 3 4 ^5 6 7 8
1 4 8 9 ^2 3 11 12
3 7 9 12 ^1, 2 5 10

If the left side goes

Up, Up, Down, then 1 is light
Up, Down, Down, then 2 is light
Up, Down, Up, then 3 is light
Up, Up, Even, then 4 is light
Up, Even, Up, then 5 is heavy
Even, Up, Up, then 9 is light
Up, Even, Even, then 6 is heavy
Even, Even, Up, then 10 is heavy
Even, Up, Even, then 11 is heavy
Down, Up, Even, then 8 is light
Down, Even, Up, then 7 is light
Even, Down, Up, then 12 is light

Changing all words Up to Down and Down to Up (and leaving Even alone,) the
12 other possibilities are given by changing light to heavy and heavy to
light.


It is worth taking a look at the link.
In it Lee wonders if there is a connection between this and the triple coding used in DNA. Interesting idea.

The strange thing about this puzzle is how difficult the solution is to find.
I bet the cryptography people are looking at higher numbers of balls for code purposes.

 

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Correct and ingenious solution.

 

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It's possible that I didn't put the lettering across correctly.

You see I was working on spreadsheets and coding at the same time, I didn't copy the intial conclusion, however you are right the "Pairing" of I and L occurs too often, if you look in the solution you found it is only suppose to occur once. I even knew that but for some reason it got into my coded version.

The main thing to look at however is a decent matrix array, as with it you can identify the duplications easier.

[edit]
I've updated the HTML version to use the conclusion that Kremmen found. Incidentally the conclusion being right or wrong doesn't really matter, it's semantics, at the end of the day any such test is actually more about presentation and method.

 

I did give you a bit of a clue that you were 99% there, but at the time Alan had not given his answer and I felt it wrong to butt in on his thread.

My own answer only went wrong in one case, and I was sure it worked for ages.

How did you arrive at three weighing of four balls as the best method?

@Alan
It is going to be difficult for anyone to spend the time going through your solution to find errors, because they would really have to try it out with every possible case. You can only really do that yourself. I did it with mine, and it broke my heart :bawl:
I would guess however, that you will find a fault if you persevere, and that this matrix solution (if that is what it is), is the only one that works.

I think you could use this puzzle to make a physical lock that would be very secure.

 

I identified that it's a logic problem. I looked at it from a "programmers" reasoning in the sense that looking for a "simplified" solution made more sense than having multiple case scenario's. After all, it's one method to rule them all.

I obviously didn't have the lettering the right way round, but I reasoned that juggling them would eventually reveal the solution. I just quit juggling them prior to the weekend otherwise my solution would of eventually been corrected.

The 4 x 3 method was because obviously any weighing had to be equal in number of balls, reducing the number to 3 x 4 or increasing to 5 x 2 or 6 x 2 would of allowed for too many unknowns. There was also the point that one of the golden rules was "You only have three weighing's". Now on a case by case method, you might equate "I only weigh three times" but you might plot the case scenario to weigh say twenty different weigh-in's based upon different case scenario's. So technically although you are only weighing three times you are actually weighing a lot more.

There was one other reason I was looking at the "moving the balls" method, back some years ago in the infancy of Answerphones. The systems use to use a four digit number as a password that could be plugged in externally to listen to recorded messages or even record a new message. The answer phones of that period weren't very secure, they didn't test for a number of failed attempts or create a failed attempt when you reached the forth digit. Instead they would sit listening for the correct four digits to be applied. This led to the creation of a "Rainbow Array", which was basically a string of numbers that would systematically attempt every 4 digit combination. For instance the string 123456 would cover 1234, 2345, 3456, alternating the numbers allowed for the creation of a smaller number to type in rather than typing sequentially (0000 0001 0002 0003 etc).

A lot of people lost messages and potentially customers because of the ability to prank(Phrack) with it.

 

method I used was to make twelve pieces of paper which I took for the twelve balls I drew an imaginary scale and imagined it in the various positions up or down etc.

You can only get to the answer by first weighing 4 balls against another 4 balls .as you have noticed it is very difficult to test another persons logic, . the main mistake made by people is to wrongly, at the beginning of the problem, assume the oddball is heavier or lighter

Great effort by all you guys, who am I say if you were right or wrong. But this problem can and has been solved by simple logic that we have all being using. My clever daughter solved it mathematically

In the thirty years or so ago that I was given this problem only one of my friends managed to solve, share it with your friends as well if you want to

 

He never siad you could label the balls.

He said,

the puzzle is about 12 steel balls, identical in almost everything, color, size and shape, but as for weight they are all identical except for one,

 

It didn't need saying, it was obvious, otherwise you couldn't solve it. He didn't say that you couldn't label them.

 

Yes you should label all the balls (Chalk) to keep track on them during this test of logic. You must start with 4 balls in both pans of the weighing scale. If they do not balance you already know something vital towards solving the puzzle

The upper end pan might contain the lighter ball, or the upper or the lower pan might have contain a heavier ball. So label all the balls in the lower pan HB= (Heavy or Balanced) Then mark the balls in the upper pan LB=(light or balanced)

If the don't balance the odd ball is amongst the 4 balls put aside leading to an easy solution if this happens. But you must work through all possibilities and must not let luck be a part of your solution

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can't you just pick them up in your hand and see which one's the oddball? why do you need a scale?

 

How sensitive do you think a human is to minute differences in weight?
The only way you could fulfil the conditions of the OP - only three weighings to identify the different ball - would be through luck.

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Exactly and luck should play no part in the solution

 

some people are more sensitive.

i'd like to try. where are these balls and where can i get them?

what kinda scale is it? is it like a see saw? cuz then you could stack them all on it and see what side of the see saw gives, and then take the lighter side and weigh that and then weigh the lighter side. at least would give you a clue.

 

Have you read the thread?
There's a photo of the scale in question. And a specification of the problem.

 

i wanted to think of it on my own. so no i didn't read the thread, dywy.

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is it like a see saw then?

 

So if you didn't read the thread how do you know what question you're answering?

No, it's like a balance scale.

 

a balance scale is a see saw.

i'm answering the OP.

 

Go to; http://www.google.co.za/imgres?imgu...a=X&ei=VQNaTKCsOsT68AaJ-I2cCw&ved=0CBgQ9QEwAA

 

A balance scale can be referred to as "A pair of scales", the pair referencing that there are two pans being used to "scale" against one another.

A see saw, while operating similarly be having an equal lateral projection applied across a pivotal point is a child's toy. To my knowledge I haven't seen a school class attempting to weigh 12 similar looking kids to find out which one is fatter using a see saw.

The overall problem was very simple, in fact one of the golden rules was simple as well. "Find the oddball and whether it's heavier or lighter than the other in 3 weighings" (This means when you have alternative methods based upon outcome of the first outcome, you are in fact plotting more than 3 weighings)