Much much tougher brain teaser

[PorscheNGuns]

Most OOTers are probably not smart enough to figure this one out on their own, but nonetheless, don't spoil it if you know it.

You have 12 rocks and a scale, and 11 of the rocks weigh exactly the same. The remaining rock is either heavier or lighter than the other 11.

Using the scale only 3 times, figure out which one of the rocks is the odd one out, and whether it is heavier or lighter.

You only get one chance, I am holding a .357 mag to your head, and if you screw up, I blow your brains out. Go!

-Matt

[cnfuzzd]

pick up scale. Bash you in head 3 times. Problem solved.

peace

john nickle

[Luzion]

is that even possible? not knowing if that 12th special rock is heavier or lighter really makes it impossible

[ClaytonN]

soab. i can definitely figure it in 4, but not 3. hold on.

[Stuey]

so it's like a puzzle with socks. if you think about it.

[deadmoney98]

Separate the rocks into 3 equal piles of 4. Weigh each, the odd one out contains the different rock (rock X). This also will tell you if it is lighter or heavier and the weight of each individual other rock (Y).

Take two rocks from the group you now know contains rock X and weigh them. If they weigh 2Y, take one of the two remaining rocks and weigh it. It is either X or a Y.

If they don't weigh 2Y, pick one and weigh. It is either X or a Y.

Obviously if the rock weighs Y, the other rock is X and vice versa.

[Matty]

that's more than 3. that's like 5.

[Josh W]

Is it a scale like "this weighs 10 lbs."? Or a balance like "this side weighs more than that side"?

Josh

[Stuey]

you can only use the scale 3 times.

So if you weigh the first 2 piles of 4 rocks and they are not equal your methods fails. If you get lucky and they are equal it works though. Good chance I'm blowing smoke here. /images/graemlins/blush.gif

[Luzion]

[ QUOTE ]
Separate the rocks into 3 equal piles of 4. Weigh each, the odd one out contains the different rock (rock X). This also will tell you if it is lighter or heavier and the weight of each individual other rock (Y).

Take two rocks from the group you now know contains rock X and weigh them. If they weigh 2Y, take one of the two remaining rocks and weigh it. It is either X or a Y.

If they don't weigh 2Y, pick one and weigh. It is either X or a Y.

Obviously if the rock weighs Y, the other rock is X and vice versa.

[/ QUOTE ]

Wait wait wait... I want to clarify something.... What kind of scale are we using? Are we using something that weighs the thing you put in it? Or are we talking about a scale like this... http://teapot.usask.ca/cdn-firearms/Reform/scale.gif

[milliondollaz]

if you know if it's heavier or lighter...

1. you split the group into two and weigh (compare) them.
2. if odd stone is heavier, you take the heavier 6, if it's lighter, you take the lighter 6, and split them into two groups of three, and weigh them.
3. you take the three stones that are heavier (lighter) and you weigh just two of them.

it will be obvious if one is heavier (lighter). if they balance, the one in your hand is heavier (lighter).

still can't figure out how to do it if you don't know if it's supposed to be heavier or lighter. cause you could get the first step wrong and split up a group of evenly weighing stones, and you would need another step fix your wrong initial guess...

[Klepton]

6 on each side = take 6 rocks from heavier side
3 on each side = take 3 rocks from heavier side
put any 2 on the scale = if one is heavier, thats the one, if they weigh the same, one not on scale is

oh...dammit i wrote this and didn't see heavier OR lighter.

oh well i got it if it was heavier

[yellowjack]

Based on my attempts and other replies w/ similar results, I'm beginning to think this was one cruel post.

[chuddo]

this puzzle is only solvable in 3 steps when using a balance scale.

and it is not too tough, but a bit fun.

separate into 3 piles of 4 rocks: A, B, and C

assume odd rock is light.

Place A and B on the balance. if balance = even, odd rock is in pile C. if uneven, odd rock is obviously on the lighter scale.

you now have a group of 4 rocks and one is odd, and can use the balance scale two more times.

Seperate this group of 4 into 2 groups of 2 stones: D and E.

again, lighter side contains the odd rock.

finally take these last 2 rocks, balance them on the scale. lighter rock is obviously the odd rock of the 12.

[Stuey]

3 groups of 4 rocks is right
if they don't weigh the same or in the case of the type of scale I am using "balance" I remove 2 rocks from each side at the same time.

If that causes the scale to balance then I know I got the bad one in one of my mitts.

If the scale stays put then I can use the 4 stones I pulled as standard weighing ones to eliminate the remaining 4 on the scale with the previous method.

Basically I am cheating and using the scale more than 3 times but it does not apear that way as I am just removing the rocks in a certian order to help my deductions.

Makes sence to me but I describe it badly.

[battschr]

[ QUOTE ]
this puzzle is only solvable in 3 steps when using a balance scale.

and it is not too tough, but a bit fun.

separate into 3 piles of 4 rocks: A, B, and C

assume odd rock is light.

Place A and B on the balance. if balance = even, odd rock is in pile C. if uneven, odd rock is obviously on the lighter scale.

you now have a group of 4 rocks and one is odd, and can use the balance scale two more times.

Seperate this group of 4 into 2 groups of 2 stones: D and E.

again, lighter side contains the odd rock.

finally take these last 2 rocks, balance them on the scale. lighter rock is obviously the odd rock of the 12.

[/ QUOTE ]
Why can we assume that the odd rock is light?

[Luzion]

[ QUOTE ]
this puzzle is only solvable in 3 steps when using a balance scale.

and it is not too tough, but a bit fun.

separate into 3 piles of 4 rocks: A, B, and C

assume odd rock is light.

Place A and B on the balance. if balance = even, odd rock is in pile C. if uneven, odd rock is obviously on the lighter scale.

you now have a group of 4 rocks and one is odd, and can use the balance scale two more times.

Seperate this group of 4 into 2 groups of 2 stones: D and E.

again, lighter side contains the odd rock.

finally take these last 2 rocks, balance them on the scale. lighter rock is obviously the odd rock of the 12.

[/ QUOTE ]

What the hell man... the hard thing about this puzzle is NOT KNOWING if the rock is gonna be heavier or lighter then the other 11 rocks...

Anyone couldve solved it if we already knew ahead of time if the rock was heavier or lighter then the other rocks

[chuddo]

blah, im a donkey. didn't take note of the fact you have to devine whether or not the stone is lighter or heavier relative to the other stones while finguring it out.

this makes the problem much more challenging, but you can see that the partial algorithm in my last post is a piece of it.

i just worked it out and now think this is a good puzzle, and will leave the final solution out so others can work it out.

[Stuey]

[ QUOTE ]
3 groups of 4 rocks is right
if they don't weigh the same or in the case of the type of scale I am using "balance" I remove 2 rocks from each side at the same time.

If that causes the scale to balance then I know I got the bad one in one of my mitts.

If the scale stays put then I can use the 4 stones I pulled as standard weighing ones to eliminate the remaining 4 on the scale with the previous method.

Basically I am cheating and using the scale more than 3 times but it does not apear that way as I am just removing the rocks in a certian order to help my deductions.

Makes sence to me but I describe it badly.

[/ QUOTE ]

Am I wrong? here is a longer version

Separate the rocks into 3 equal piles of 4.

Step 1

Put any 2 piles on a balance type scale, if they balance the 3rd pile has rock X in it. And you can use any 2 rocks on the scale as standard weight rocks to find out where X is in the remaining 4 using step 2 below.

If they don’t balance remove 2 from each side at the same time. If that makes it balance then you know you are holding rock x and all other rocks are standard weight so you can use them to find which of the 4 in your hands is X easily with step 2.

If it does not balance when you remove 2 from each side you know rock x is on the scale and all other rocks are standard and you can use step 2 to find it.

It will look like you only use the scale 1 time but you actually use it 2 times as you remove the rocks in a certain order in the case when it does not balance on the first test. It is a cheat but you gota live!

Step 2

Now that you know which group of 4 rocks x is in just take 2 from that pile and weigh them vs 2 standard ones. If it balances take 2 of the remaining ones and weight it vs 2 standard ones and you will have found it. Just remove one of the rocks and watch the scale to find out which is X and if it is heavy or light based on the scale reaction. Then say I hold X and it is heavy or light or say rock X is on the scale and it is heavy or light whatever the case might be.

You only need to use the scale twice in all possible combinations once you narrow it down to 4 rocks and you have 2 standard ones to use. You have to pull the trick from step one here again to find out if the rock is heavy or light though but it will look like you only used the scale twice. The trick is watching what happens when you remove one rock from each side.

[deadmoney98]

[ QUOTE ]
Separate the rocks into 3 equal piles of 4. Weigh each, the odd one out contains the different rock (rock X). This also will tell you if it is lighter or heavier and the weight of each individual other rock (Y).

Take two rocks from the group you now know contains rock X and weigh them. If they weigh 2Y, take one of the two remaining rocks and weigh it. It is either X or a Y.

If they don't weigh 2Y, pick one and weigh. It is either X or a Y.

Obviously if the rock weighs Y, the other rock is X and vice versa.

[/ QUOTE ]

[ QUOTE ]
that's more than 3. that's like 5.

[/ QUOTE ]

Wow, I am dumb at 4 AM, sorry about being an idiot.

[ptmusic]

Ok, the way I solved this was to use Scenarios for each possible outcome.

Step 1 always starts the same for each Scenario; Steps 2 and 3 vary depending on the result of Step 1.

I am calling the rocks by number 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, and 12.

SCENARIO A

Step 1: 1,2,3,4 vs. 5,6,7,8. They are equal.

Step 2: 1,2,3 vs. 9,10,11. They are equal

Step 3: 1 vs. 12. 12 is the odd rock, and it is heavy or light depending on how it weighs against 1.



SCENARIO B:

Step 1: 1,2,3,4 vs. 5,6,7,8. They are equal.

Step 2: 1,2,3 vs. 9,10,11. They are NOT equal. Odd rock is 9, 10, or 11; you know if odd rock is heavy or light now, depending on how this group weighed against 1,2,3.

Step 3a: 9 vs. 10. They are Equal. Odd rock is 11.
Step 3b: 9 vs. 10. They are NOT equal. Odd rock is the heavier rock of the two, if group was heavier in Step 2, or odd rock is the lighter rock if the group was lighter in Step 2.



SCENARIO C:

Step 1: 1,2,3,4 vs. 5,6,7,8. They are NOT equal. Call 1,2,3,4 the side that was lighter. You know the odd rock is one of these 8 rocks. You also know that if it is 1,2,3, or 4 then it is lighter; if it 5,6,7, or 8 then it is heavier. You also know that it is not 9,10,11, or 12.

Step 2: 1,8,9,10,11 vs. 2,3,5,6,12. They are equal. You know that odd is 4 (and light) or 7 (and heavy).

Step 3a: 12 vs. 4. They are equal. 7 is the odd rock and it is heavy.
Step 3b: 12 vs. 4. They are NOT equal. 4 is the odd rock and it is light.



SCENARIO D:

Step 1: 1,2,3,4 vs. 5,6,7,8. They are NOT equal. You know the same information as in Scenario C.

Step 2: 1,8,9,10,11 vs. 2,3,5,6,12. They are NOT equal. If the first group is heavier, 8 is odd rock and it is heavy OR 2 or 3 is odd and light. Go to Step 3a. If 2nd group is heavy, 5 or 6 is odd OR 1 is light; go to Step 3b.

Step 3a: 2 vs. 3. If equal, then 8 is odd rock and heavy. If Not equal, then the lighter of the two is the odd rock.

Step 3b: 5 vs. 6. If equal, then 1 is odd rock and light. If NOT equal, the heavier of the two is the odd rock.


-ptmusic

[sirio11]

Beautiful !

[Matty]

well done

[1800GAMBLER]

You're a sick sick person for posting the unknown weight one first.

[PorscheNGuns]

it is possibe dummy

[Matty]

[ QUOTE ]
it is possibe dummy

[/ QUOTE ]What the [censored] are you talking about?

[PorscheNGuns]

chuddo Im pretty sure you dont have jack [censored] worked out

-Matt

[PorscheNGuns]

Luzion, instead of attempting to solve the riddle, questions its validity:

[ QUOTE ]
is that even possible? not knowing if that 12th special rock is heavier or lighter really makes it impossible

[/ QUOTE ]

thats what the [censored] Im talking about

-Matt

[PorscheNGuns]

Getting closer, but by removing a rock at the end you are essentially using the scale 4 times, making your solution wrong.

-Matt

[PorscheNGuns]

http://www.nasa.gov/images/content/114299main_iss010e25158.jpg

-Matt

[Indiana]

this is a trivial problem.

Do you see why?

Indy

[ptmusic]

[ QUOTE ]
http://www.nasa.gov/images/content/114299main_iss010e25158.jpg

-Matt

[/ QUOTE ]

Thanks for the photo. It almost makes up for the hour I lost working that one out and trying to put it into words!

-ptmusic

[Stuey]

Why do I always have to try and cheat? Even when you are holding a .357 mag to my head. I just got what little brains I have blown out. /images/graemlins/frown.gif

I had fun trying though, nice job ptmusic. /images/graemlins/laugh.gif

Vezzini: Haha.. you fool! You fell victim to one of the classic blunders. The most famous is: Never get involved in a land war in Asia. Only slightly less well know is this: Never go in against a Sicilian when death is on the line!

[RiverFenix]

[ QUOTE ]

Vezzini: Haha.. you fool! You fell victim to one of the classic blunders. The most famous is: Never get involved in a land war in Asia. Only slightly less well know is this: Never go in against a Sicilian when death is on the line!

[/ QUOTE ]

best movie ever

[Slow Play Ray]

I came across this elegant solution here (http://mathforum.org/library/drmath/view/55618.html).

Of 12 coins, one is counterfeit and weighs either more or less than
the other coins. Using an old-fashioned balance with two scales, and
performing only three measurements, you have to find out which coin is
counterfeit, and whether it is lighter or heavier then the other
coins. How would you do that?

The basic idea of this solution is to perform the three measurements,
look at the result of each measurement, and then look at the way each
coin has participated in the measurements to identify the only coin
that could have caused this particular combination of results.

For instance, if the three results are: equal (0), right side heavier
(R), left side heavier (L), or, in short, 0RL, then it must have been
the coin that was

not in the first measurement,
on the right in the second measurement,
on the left in the third measurement,

and that coin is heavier than the others, or (mirrored) the coin that
was

not in the first measurement,
on the left in the second measurement,
on the right in the third measurement,

and that coin is lighter than the others.

Note how the measurement result 0RL and the participation of a heavier
coin in the measurements coincide.

So all we have to do is to distribute the 12 coins over the scales of
the three measurements in such a way that no coin participates in the
three measurements in the same way (or mirrored) as any other coin.
The distribution below is one of many possible distributions that
fulfills this requirement:

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

If the measurements result in 0RL, then it can be seen from this
distribution that coin 8 is lighter than the other coins. No other
coin can explain the outcome 0RL.

DONE.

We have thus found a solution to the problem. But how can such a
solution be found? I.e., how can a distribution of coins over the
measurements such as the one above be found? And why should there be
four coins on each side of the balance in each measurement? The answer
is in the list of logical steps below.

Each measurement has three possible results: equal (0), right side
heavier (R), or left side heavier (L). The three measurements together
can therefore have a total of 3x3x3 = 27 different outcomes, namely
any possible combination of 3 letters out of 0, L, and R, i.e.,

000, 00L, 00R, 0L0, 0LL, 0LR, .... , RRL, RRR.

We are looking to find the right one among 24 different statements,
i.e.,

"coin 1 is heavier", "coin 1 is lighter", ... ,
"coin 12 is lighter".

We need to map each statement to an outcome, so that if that outcome
occurs, the statement to which it maps must be true. Part of such a
mapping table could be:

LR0 means "coin 1 is heavier"
RLR means "coin 2 is heavier"
etc.

This mapping table serves two purposes. It identifies the counterfeit
coin and its weight after having performed the measurements, but it
also tells us where to put each coin in each measurement. For
instance, since we want LR0 to mean "coin 1 is heavier," coin 1 must
participate in measurement 1 on the left side, in measurement 2 on the
right, and not at all in measurement 3.

Outcome 000 is useless to us since it means that the counterfeit coin
was never part of a measurement and we can not tell whether it was
lighter or heavier than the others. This leaves 26 possible outcomes
that must be mapped to 24 statements.

If LR0 means "coin 1 is heavier," then RL0 (its mirror outcome) must
mean the opposite, i.e., "coin 1 is lighter." This is so because
mapping LR0 to "coin 1 is heavier" determines where coin 1 occurs in
the three measurements, and if that coin is lighter instead, the
measurements will result in the mirror outcome.

We can therefore group the 26 outcomes into 13 pairs (each outcome
with its mirror outcome). The table can now be reduced to mapping 12
statements about coins being heavier to 12 outcomes (each outcome
taken from a different pair).

We have one pair of outcomes too many. If we make a list of all
outcomes, and count the numbers of L's and R's (not counting 0's) that
appear in the first position of each of the outcome strings, we will
end up with the number 9 (or 18 when including mirror outcomes). This
means that 9 coins would participate in the first measurement.
Counting L's and R's in the second and third position would also yield
9. Since we can not distribute 9 coins evenly over 2 scales, the pair
of outcomes we are not going to use should have an L or R in each
position, such as LLL and its mirror outcome, RRR. Not using that pair
will leave 8 coins in each measurement. Alternatively, any other pair
of outcomes with three letters, e.g., (LRR, RLL), can be dropped from
the outcomes and a correct distribution can still be found using the
remaining 12 outcome pairs.

We are now left with 12 outcome pairs and 12 statements about each
coin being heavier. Since the numbering of the coins is irrelevant,
the only remaining choice we have is which outcome of each pair we
will use in the table. If we now start to map "heavier" statements to
outcomes arbitrarily, we run the risk of ending up with 6 coins on the
left and 2 coins on the right in a measurement, which isn't going to
tell us much. We must ensure that in each measurement (position in the
outcome string), there are as many coins on the left (symbols L) as
there are on the right (symbols R). We can achieve this by taking a
mirror outcome if the scales get out of balance.

The rest is a bit of rather straightforward trial and error. Let's
decide not to use the pair (LLL, RRR). From each remaining pair, we
pick one outcome in such a way that we have the same number of L's and
R's in each position of the string. We save the pairs of outcomes with
two 0's to the end to make it easier to get the scales back into
balance.

LLR \
LRL > 3 letters, one of which is different
RLL /

we now have two L's in each column against one R, so let's use some
more R's

R0R \
0RR > one 0 and twice the same letter
RR0 /

each column has now one R too many

LR0 \
0LR > one 0 and two different letters
R0L /

we still have one R too many in each column

L00 \
0L0 > two 0's and one letter
00L /

Since we now have an equal number of L's and R's in each of the three
columns of the outcomes (namely 4), we have an equal number of coins
on each side of the scales in each measurement.

To complete the table, we only need to write down the "heavier"
statements next to the outcomes (and the "lighter" statement to the
corresponding mirror outcomes). The numbering of the coins is of
course irrelevant:

LLR "coin 1 is heavier" RRL "coin 1 is lighter"
LRL "coin 2 is heavier" RLR "coin 2 is lighter"
RLL "coin 3 is heavier" LRR "coin 3 is lighter"
R0R "coin 4 is heavier" L0L "coin 4 is lighter"
0RR "coin 5 is heavier" 0LL "coin 5 is lighter"
RR0 "coin 6 is heavier" LL0 "coin 6 is lighter"
LR0 "coin 7 is heavier" RL0 "coin 7 is lighter"
0LR "coin 8 is heavier" 0RL "coin 8 is lighter"
R0L "coin 9 is heavier" L0R "coin 9 is lighter"
L00 "coin 10 is heavier" R00 "coin 10 is lighter"
0L0 "coin 11 is heavier" 0R0 "coin 11 is lighter"
00L "coin 12 is heavier" 00R "coin 12 is lighter"

As explained before, the left part of the table tells us exactly where
each coin participates in each measurement. The resulting distribution
of coins over scales is the one given at the beginning. In addition,
the table above can be used to find the counterfeit coin after having
performed the three measurements.

[TheWorstPlayer]

This is not really elegant. It is effective, but it is a ton of work and writing, etc. Not elegant at all. A more elegant solution that I came up with is to weigh 4 against 4. If it doesn't balance, move three normal ones (which weren't involved in the first weighing) to the heavy side, three from the heavy side to the light side and three from the light side off. Now weigh again.

If the heavy side is still heavy, then either the one heavy left on it is heavy or the one light on the light side is light. Weigh the light one against a normal and if it is light, it is that one that is weird and it is light. If they balance, then the heavy one from the heavy side is the weird one and it is heavy.

If the heavy side from the first weighing is now light on the second weighing, then you know that one of the three heavy ones that you moved to the original light side must be heavy. So weigh one against one and if it is unbalanced it is the heavy one and it is heavy. If it is balanced, the one you didnt weigh is weird and it is heavy.

If the second weighing balances, then one of the three light ones that you took off is light. Repeat the three heavies procedure to find the light one.

If the initial weighing balanced, then you know that it is one of the four that you didn't weigh that is weird. For the second weighing, weigh three of those against three normal ones. If the weird side is heavy, repeat the heavy procedure from above to determine which one it is and it is heavy. If the weird side is light, same procedure to determine which one and it is light. If they balance, then the one you didn't weigh at all yet is weird. Weigh it against a normal one to see if it is light or heavy.

QED.

[Slow Play Ray]

Elegant meaning the weighings are all the same, regardless of the outcome of the previous weighings.

I am still reading it, but your solution seems to be exactly the one a guy i work with just handed me.

(This problem has essentially ceased all work in the engineering department of my company for the past day).

*Edit* Yup, exactly the same. I wish he explained his as clearly as you did, it took me forever to decipher 3 pages of notes.

[TheWorstPlayer]

[ QUOTE ]
Elegant meaning the weighings are all the same, regardless of the outcome of the previous weighings.


[/ QUOTE ]
OK, I can buy that. Seems more 'brute force' than 'elegant' to me, but matter it's just semantics.

[ QUOTE ]

(This problem has essentially ceased all work in the engineering department of my company for the past day).


[/ QUOTE ]
Good to know why we're lagging behind the Japanese.

[ QUOTE ]
I wish he explained his as clearly as you did, it took me forever to decipher 3 pages of notes.

[/ QUOTE ]
Just because the solution is elegant doesn't mean the explanation is, unfortunately.