The odd stone logic problem

@Alan McDougall,

To MJA:

You can't deduce that it is heavier. Volume is just L x W x H. So the odd ball could be the same size though just hollow.

@Alan McDougall,

nvm found a flawgonna rethink

@click here,

-----------SPOILER-------------

Now I have it. I figured it out after walking home from a friends after having some beer and movie. yum. but i couldn't get on the computer till this morning. :-(

I did use the 4 v 4 clue. kinda wish I didn't have that so I could have figured it out without help.

nonetheless

3 Measures if at first weigh the scale is balanced:

Start off with any four balls on the left side of the scale. Then put any 4 of the other 8 on the scale. Now if the scale is balanced then we know the odd ball is in the 4 that are not on the scale at all. To solve from there. You take 2 of the 4 that were left over and put them on the left side of the scale. On the right side of the scale you place 2 that we measured earlier and they balanced as these are normal. If the scale is balanced then it is one of the 2 that we didn't put on the scale and we can take the ones on the left off and the ones on the right off. Take the 2 left and put one on the scale with a normal ball. If the scale is balanced then it is the ball in your hand that you didn't use. If it is unbalanced then it is the one you put on the left.

If the scale is not balanced then do this for the 2nd weigh.

Remember this situation has 4 balls on the left and 4 on the right and the scale is unbalanced. So the odd ball out is either on the left or the right of the scale. We still have 4 balls on the side that we can use for accurate measurement.
Before weighing the 2nd time:
take 1 on the left and switch it with one on the right it doesn't matter which you would just have to remember which you did switch.
take 3 on the left (excluding the 1 that you switched from the right) and switch them off the scale with 3 that we know are normal (of the 4 that we ignored in the first measure.)

Now we weigh:

If the after 2nd weigh the scale changes which side is up and which side is down:
Then we know it is one of the 2 we switched sides with. (Remember we switched 2 balls. One from the left with one from the right)

Your third weigh in this situation would be:
Take one of the 2 balls that we moved and put it by itself on the left. Weigh it with a normal ball. If the scale is unbalanced then it is the ball on the left. If the scale is balanced it is the other ball of the 2 that we didn't weigh.

If when after 2nd weigh and the scale stays in the same position. (doesn't matter if left is high or right is high just if it stays the same):
Then we know it is one of the 3 that we took off from the left.

We then have to remember when we weighed the first 4 (for the first weigh) whether the left side was higher or lower. If it was lower. Then we know the odd ball is heavier. If higher then lighter.

Third weigh:
We then pick 2 of the 3 that we had taken off from the left earlier and put 2 (doesn't matter which) on opposite sides. If balanced we know its the one we didn't weigh.
Now we determined right above whether or not the ball is lighter or heavier.
So if the scale is unbalanced then we just pick the side that has the lighter/heavier ball in it. It is that simple.

If when after 2nd weigh and the scale does not change its position (meaning it is still unbalanced but hasn't moved)
Then we know that the ball is one of the 3 on the right side.

We then have to remember when we weighed the first 4 (for the first weigh) whether the right side was higher or lower. If it was lower. Then we know the odd ball is heavier. If higher then lighter.

Third weigh:
We then pick 2 of the 3 that we have on the right and put 2 (doesn't matter which) on opposite sides. If balanced we know its the one we didn't weigh.
Now we determined right above whether or not the ball is lighter or heavier.
So if the scale is unbalanced then we just pick the side that has the lighter/heavier ball in it. It is that simple.

Come on guys isn't this clear enough?

This has to be right. let me know if you see an error.

--------

my original plan after 20 or so min of thought.

take 2 on the left off, put 2 on the right on the left and add 2 normal balls to the right.

you can solve it from that if either the scale changes sides or balances but not if if stays the same. :-(

After thinking about it for a while I realized you can't end up having to determine from 2 balls left for the 3rd weigh. but i thought that solving from 3 balls was impossible. then i figured out that you can find out which one is heavier and use that.

@click here,

Click and ZRIS

Quote:

[CENTER]Now I have it. I figured it out after walking home from a friends after having some beer and movie. yum. but i couldn't get on the computer till this morning. :-( [/CENTER]

[CENTER]I did use the 4 v 4 clue. kinda wish I didn't have that so I could have figured it out without help. [/CENTER]

[CENTER]nonetheless[/CENTER]

[CENTER]4 on 4 first weigh with 4 on the side sitting out.[/CENTER]

[CENTER]take 1 on the left and switch it with one on the right.
take 3 on the left and switch them off the scale with 3 that we know are normal (of the 4 left)[/CENTER]

[CENTER]2nd weigh
So if the scale changes which side is the highest from when we weighed the 4. then we know it is one of the 2 we switched sides with. So then just weigh one with a normal ball. done[/CENTER]

[CENTER]2nd weigh
If the scale balances. Then we know it is one of the 3 that we took off one the left. We then have to remember when we weighed the first 4 whether the left side was higher or lower. If it was lower. Then we know the odd ball is heavier. if higher then lighter. we then pick 2 of those 3 and put them on opposite sides. if balanced we know its the one we didn't weigh. If unbalanced then we pick the lighter or heavier side if we know from what we just found out.[/CENTER]

[CENTER]2nd weigh
If the scale does not change from the position it was with the 4. then we know that the ball is one of the 3 on the right side. We then do the same thing we did above remembering which direction the right side was (up or down) when we weighed the first 4.[/CENTER]

[CENTER]This has to be right. let me know if you see an error.[/CENTER]

This might happen in the first weigh :bigsmile: :perplexed:

OOOOVOOOO (BALANCES) :perplexed:

OR

OOOOVOOOO (DOES NOT BALANCE) :perplexed::perplexed: :bigsmile:

"Click you assumed the balls will balance

By the way the balls are of indentical appearance and volume. The odd ball differes so minutely that only a sensitive balance scale will reveal where it is amongst the twelve balls

This is a fun problem and to reveal the solution will spoil it

I know you guys will solve it, but I am perplexed that a philosophical forum which deals in logic thought that so few have taken up the challenge

Remember guys I said in the unfortunate clue that the only way to get to a solution is to weigh four against four!! :detective:

I did not say the four would balance they might not and and they might,there is your dilemma

@Alan McDougall,

1256 on one side then 3478 on otherside.if it goes down on left again either 1or 2 is heavier or 7 or8 is lighter.ummm

@Alan McDougall,

Alan McDougall wrote:

Click and ZRIS

This might happen in the first weigh :bigsmile: :perplexed:

OOOOVOOOO (BALANCES) :perplexed:

OR

OOOOVOOOO (DOES NOT BALANCE) :perplexed::perplexed: :bigsmile:

"Click you assumed the balls will balance

By the way the balls are of indentical appearance and volume. The odd ball differes so minutely that only a sensitive balance scale will reveal where it is amongst the twelve balls

This is a fun problem and to reveal the solution will spoil it

I know you guys will solve it, but I am perplexed that a philosophical forum which deals in logic thought that so few have taken up the challenge

Remember guys I said in the unfortunate clue that the only way to get to a solution is to weigh four against four!! :detective:

I did not say the four would balance they might not and and they might,there is your dilemma

no then your reading my explanation wrong.

my explanation for the 2nd weigh is IF the scale is not balanced.

if the scale was balanced then you know it is in the 4 on the side.
take 2 of them and weigh with 2 normal ones. if balanced then weigh one with a normal one. if not balanced then weigh 1 of the other 2 with a normal one. its easy if at the first weigh things balance. my explanation is for if they DON'T balance.

there is no dilema. I just made my explanation to unclear. but it works.

read it again.

@click here,

click here wrote:

no then your reading my explanation wrong.

my explanation for the 2nd weigh is IF the scale is not balanced.

if the scale was balanced then you know it is in the 4 on the side.
take 2 of them and weigh with 2 normal ones. if balanced then weigh one with a normal one. if not balanced then weigh 1 of the other 2 with a normal one. its easy if at the first weigh things balance. my explanation is for if they DON'T balance.

there is no dilema. I just made my explanation to unclear. but it works.

read it again.

I think you are weighing them more than three times.I can do it easy with four or if one was heavier than the rest but you dont know if one is heavier or lighter remember.

@xris,

xris wrote:

I think you are weighing them more than three times.I can do it easy with four or if one was heavier than the rest but you dont know if one is heavier or lighter remember.

no only 3 times. I swear it works. and i find out which one is heavier or lighter. i'm gonna just make it really pedantic

@Alan McDougall,

read it again. i edited

@click here,

click here wrote:

no then your reading my explanation wrong.

my explanation for the 2nd weigh is IF the scale is not balanced.

if the scale was balanced then you know it is in the 4 on the side.
take 2 of them and weigh with 2 normal ones. if balanced then weigh one with a normal one. if not balanced then weigh 1 of the other 2 with a normal one. its easy if at the first weigh things balance. my explanation is for if they DON'T balance.

there is no dilema. I just made my explanation to unclear. but it works.

read it again.

To get to this stage you have had one weigh..you say you take two of them and weigh with normal ones thats two weighs..if balanced or not balanced you have two that you dont know so what do you do with them?I got that far..

@xris,

xris wrote:

To get to this stage you have had one weigh..you say you take two of them and weigh with normal ones thats two weighs..if balanced or not balanced you have two that you dont know so what do you do with them?I got that far..

no its not. 2 balls on the left. 2 on the right. that is one weigh:brickwall:

@click here,

click here wrote:

read it again. i edited

Number them one to twelve and show me the sequence of weighing.

@xris,

xris wrote:

Number them one to twelve and show me the sequence of weighing.

This is for you xris

urgh... I don't think like this.

First Weigh:
start off with
Left: 1234
Right: 5678
on the table: 9 10 11 12

VERY IMPORTANT: Make a mental note as to which side is higher left or right.

2nd Weigh
If first weigh was balanced then 2nd weigh:
Left: 12
Right: 78
Third Weigh
If 2nd weigh was balanced then:
Left: 3
Right: 8
If after 3rd weigh. Is balanced then odd ball is 4. If not balanced then odd ball is 3.

-------

If first weigh is unbalanced then set up the scale like this:
Left: 9 10 11 5 (9-10 are normal as per the first weigh)
Right: 4 6 7 8
On table: 1 2 3 (12 which we know is normal as per the first weigh

Now we measure

If scale is balanced:
Then the only logical option is that it is either 1, 2, or 3. That we had removed from the left. Since we know (12) is normal.
We only have one weigh left to find out which of the 1 2 or 3 it is.
We then have to remember during the first weigh which direction the left side was. Was it up or down. If it was down then the odd ball of 1 2 or 3 is heavier then all 12. If left side was up then odd ball is lighter then all 12.

So here is our 3rd weigh.
Left: 1
Right: 2
Table: 3
If scale is unbalanced we pick the side that has the lighter or heavier ball in it. (We can do this because we know from the beginning what the weight difference is) If scale is balanced then the correct ball is 3.

Now if after the 2nd weigh and the scale changed positions in height:

Then we know it was one of the 2 that we moved. (5 and 4)
So for our 3rd weigh:
Right: 5
Left: 8
Table: 4
If scale is balanced then 4 is odd ball. If scale is unbalanced then 5 is odd ball.

If after the 2nd weigh the scale stayed in the same unbalanced position. Apply the same logic that we used above for 1 2 3 except with balls (6 7 8).

Please tell me you understand now.

@click here,

If the first weigh is unbalanced how do you know anything about 123 ?

@xris,

xris wrote:

If the first weigh is unbalanced how do you know anything about 123 ?

Because 1 2 3 were on the left side. SO IF we are SURE that the odd ball is 1 of those 3. Then we have to remember which way the scale was leaning at the first weigh. If the left was down. The ball was heavier then the rest. If the scale was up it was lighter then the rest.

Do you have skype or something? lol all this explaining is :perplexed:

@click here,

click here wrote:

Because 1 2 3 were on the left side. SO IF we are SURE that the odd ball is 1 of those 3. Then we have to remember which way the scale was leaning at the first weigh. If the left was down. The ball was heavier then the rest. If the scale was up it was lighter then the rest.

Do you have skype or something? lol all this explaining is :perplexed:

Ill have to think about it later my heads hurting and i have had a hangover all bleeding day..but thanks..

@Alan McDougall,

I bet you it was adding all those numbers. :-D

That drove me nuts. I think its just easier without complicating things with numbers. Though I'm pretty good at visualizing stuff in my head.

@Alan McDougall,

Click Here

Sorry your is not a fool proof solution!! Which ball is the heavier ball? or which ball is the lighter ball??

There are two possible outomes or answers.

You must define both paths to your solution

@Alan McDougall,

Did you not read my explanation?

It states how to find out which ball is heavier or lighter.

In my solution you either narrow it down to 2. Or you narrow it down to 3 on the left and 3 on the right. So you have to then remember which side all of those 3 were on and when they were being weighed which side the scale was on.

Don't you know the answer so this?

It works I don't know how to make my explanation clearer. Please read it again. (preferably the written version on i think page 3.)

@Alan McDougall,

click here

I know the answer, this problem was given to me by a friend of mine over thirty years ago

Last clue, then I will give the solution

Say OOOOVOOOO Balanced then you can take a hypothetical pen and write

BBBBVBBBB all these balls balance.

And now you know the odd ball is amongst the other non-weighed balls OOOO identified that the odd ball so identified these balls with question marks ?????

But if OOOOVOOOO DO NOT BALANCE THEN IF THE LEFT PAN IS HIGHER IT CAN INDICATE THAT THE ODD BLLS IS AMONST THEM AND MAYBE LIGHTER

OR THE RIGHT PAN HAS A HEAVIER ODD BALL. BUT REMEMBER YOU HAVE ONLY ONE ODD BALL

SO PUT LLLLLVHHHH

AND CONTINUE FROM THERE

I will put you out of your pain if you still can not solve it

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