Riddle:12 objects and a balance

Factual Questions
1

You’ll excuse me if this has been covered before elsewhere.
In front of you are twelve objects, identical except for weight. One of the objects is either heavier or lighter, you don’t know which. You have a double-pan balance, but can only use three weighings to determine the object of different weight. How do you do it?

I thought the first weighing was 4 against 4, but that doesn’t seem to get you very far unless the scale balances as “even”.
If this was covered in a column, please link or post the search words to access it, as I couldn’t get it to come up.

2

1st weighing: put 4 on one side, 4 on the other side, 4 dont get weighed.
2nd weighing: if both batches of 4 on the scale are even, the heavier one is in the group you set aside. Put two of those on either side of the scale. (after removing the others.) If one of the groups on the scale is heavier, divide it into two groups of two and weigh those.
3rd weighing: take the heavier group from the 2nd weighing, put one on one side of the scale, one on the other. The heavier one is the mystical object. Voila.

Hmm, whoops, now I see that you said it could be heavier or lighter… oops, this way only works of you know which it is.

::scratches head, goes to sleep::

3

I think you are on the right track

Divide into 3 groups of 4 (group a b c)
Weigh a against b
weigh b against c

if both comparisons are equal the odd block is in b
if a=b then odd block is c
if a=c then odd block is a

take the four from the odd block (A B C D)
weigh A and B against 2 from another set

(*)if equal then then weigh C against one from another set
if equal answer = d if unequal answer =c

if (*) not equal then weigh A against oone from another set
if equal answer =b else answer = a
(I bet somebody has beaten me to it)

4

oops typo

I think you are on the right track

Divide into 3 groups of 4 (group a b c)
Weigh a against b
weigh b against c

if both comparisons are equal the odd block is in b
if a=b then odd block is c
if b=c then odd block is a (typo here)

take the four from the odd block (A B C D)
weigh A and B against 2 from another set

(*)if equal then then weigh C against one from another set
if equal answer = D if unequal answer = C

if (*) not equal then weigh A against oone from another set
if equal answer =B else answer = A

methinks
(I bet somebody has beaten me to it)

5

I don’t know the answer, but Messiah…you’re thing works, but takes 4 weighings to accomplish…you only get 3.

Jman

6

Divide coins into 3 groups.
reference coins by number. Any coin found to be real
will be indicated by “".
First weighing
1,2,3,4 against 5,6,7,8
If 1,2,3,4=5,6,7,8 counterfeit coin must be one of 9,10,11,12 (coins 1-8 are authentic i.e. "”)
weigh 9,* against 10,11
if 9,=10,11, counterfeit is 12. Weigh 12 against "" to determine if it’s heavy or light
if 9,* < 10,11 then either 9 is light or (10 or 11) is heavy. Weigh 10 against 11. If those two are equal, then 9 is light. If 10<11 then 11 is counterfeit, and it’s heavy. If 10>11 then 10 is counterfeit and it’s heavy.

If 1,2,3,4<5,6,7,8 then one of (1,2,3,4) is light or one of (5,6,7,8) is heavy
Weigh 1,2,5 against 3,6,*
If 1,2,5=3,6,* then 4 is light or (7 or 8) is heavy.
Weigh 7 against 8. If 7=8 then 4 is light, otherwise heavier coin is the counterfeit one.
If 1,2,5<3,6,* then (1 or 2) is light or 6 is heavy. Weigh 1 against 2 to determine
If 1,2,5>3,6,* then either 5 is heavy or 3 light. Weigh either against “*” to determine.

If 1,2,3,4>5,6,7,8 then one of (1,2,3,4) is heavy or one of (5,6,7,8) is light.
weigh 1,2,5 against 3,6,*
If 1,2,5<3,6,* then either 5 is light or 3 is heavy. Weigh either against "" to determine.
If 1,2,5>3,6, then either (1,2) is heavy or 6 is light. Weigh 1 against 2 to determine. (1=2 then 6 is light, otherwise heavier coin is counterfeit.

Follow the same procedure for case where 1,2,3,4<5,6,7,8

Thank you, thank you very much.

7

Dang, Enright, ya beat me to it.

8

CKDext,

A friend of mine and I were working this puzzle in a bar several years ago, and the waitress wondered why we were writing on so many napkins. We told her about the puzzle and in her effort to help, her first comment was “What denomination where the coins?”!!!

Those chicks from the Jersey shore… I tell ya…

9

A bit off topic…

When I took a “Philosophy of Logic” in college, our teacher gave this problem as extra credit. He said that no one ever gets figures it out, and he offered an “A” for the course to anyone who did.

I figured it out. He backed out of the offer.

Anyways, as an interesting side note to a side note, he said that even the smartest of the science types (especially computer scientists) are baffled by this problem. The explanation he gave was that when you do the weighing, you are getting up to three pieces of information about each coin (not lighter that others, not heavier than others, same as others.) Most people think of these problems in “binary” manner, a “yes or no” or “same or different” type of thing.

10

Nice job Enright! I tried my hand at this problem a few years ago and utterly failed to solve it. I did find a solution on the Internet that could find the minimum number of weighings to find the counterfeit in any given number of coins. One interesting feature was it didn’t rely on information from previous weighings to determine what to weigh in the following ones. I wish I could find that link again.

11

This exact puzzle is also mentioned (in two different forms) in the novel With a Tangled Skein, by Piers Anthony. At the outset, you can determine that it’s possible, because you get three trits (base-3 bits) of information, which is enough to distinguish between 3[sup]3[/sup] = 27 different cases, and you only need to distinguish between 24 (12 coins times two possibilities on the weight).

12

This puzzle is often posed on the usenet group rec.puzzles; here is the entry from the rec.puzzles FAQ, with a simpler-to-remember solution:
[/quote]
One solution is to label the coins with the letters from FAKE MIND CLOT and weigh the coins: MA DO – LIKE, ME TO – FIND, FAKE – COIN. Logic will now suffice to find the odd coin. For instance, if the results are left down, balance, and left down, then coin “A” is heavy.
[/quote]
A more detailed solution for an arbitrary number of coins is included in the rec.puzzles archive; perhaps that’s the site you recalled, Greg.

13

Here you will find a non-branching solution using trits.

14

Wow, I always just solved it the old fashioned way. My point of view was always that a balance scale in this case can give you three pieces of information at once. If you put ALL the coins on the scale at once, then you’re limiting yourself to two pieces of information, whereas dividing into three group allows you to make an inference about the items you haven’t weighed.

It took a while for me to understand what the above mentioned web site was saying to do, but once I got it, it made sense. So the formula for determining the maximum number of coins you can use given X weighings on a balance scale is (3^X-3)/2. Just think, if you’re given just 4 weighings on a balance scale you could solve this puzzle for 39 coins!

15

First put half (6) on one side and half (6) on the other… one side will weigh more. Next split the heavy side in half and weigh 3 one side and 3 on the other… again one side will weigh more. Next take the heavy side and weigh any two (2) of the remaining three (3). If one weighs more than the other, that is the heavy rock/coin. If they weigh the same then the one you did not weigh is the heavy rock/coin. That took three (3) weighings.

16

Jesus, Boby Drake,
I would appreciate if you would give us here at the Straight Dope a little more credit than that. Either that or you need to learn to read better. The OP clearly says that the coin is either heavier or lighter. As you clearly pointed out, there’s not much difficulty in finding the coin if you know it’s the heaviest one.

Now I’ve got one for you… There are three common words in the English language that end in “gry”. Hungry and Angry are two of them…

excuse me while I get out of here while it’s still safe!

17

But you don’t know if the rogue coin is heavier or lighter than the others, so that doesn’t work.

Interestingly, you don’t have to find out if the rogue coin is heavier or lighter, so you could start with 13 coins. Use Enright3’s solution for any 12 coins, but if all the weighings are equal, then the 13th coin is the one!

18

Confession time: I got this puzzle from a MUD (text-based multiplayer game). The punishment for selecting the wrong object is instant death, and the best algorithm I could come up with left me with a 50/50 shot at the end. You have saved me many deaths. Also, thanks to others for searching/posting links to solutions, even those who misread my OP :smiley: