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Just because you love it when I post mathematical puzzles, and also because I got two books packed full of them for Xmas
Here is the problem:
You have 12 coins, each one indistinguishable from the other in terms of size and appearance, and all but one of them weighing the same. That is to say, 11 of them have the same weight, whilst one of them is either lighter or heavier than the rest.
You also have a pair of scales that measures accurately. Your quest is to find which coin is the oddly weighted one, and to determine whether it is heavier or lighter than the rest of the coins.
Here is the catch (and the puzzle):
You are only allowed to use the scales, and are only allowed 3 measurements in order to solve the puzzle.
By three measurements, I mean putting X number of coins on one side of the scales, and Y number on the other, and seeing whether the scales tilt to one side, or remain balanced. What groupings you put the coins into to measure them is totally up to you (so you could measure 11 coins against 1...but I wouldn't!), but after you have recorded three measurements, you have to reveal which coin has a different weight and whether that coin is heavier or lighter than the others.
a) If za = zb, discard za and zc.
b) If za ≠ zb, discard zb and zd
We now have 2 groups with 1 coin each. za and zb, with weighted groups for each scenario
outcome A = zc is heavier and zd is lighter.
outcome B = za is heavier and zb is lighter.
(3rd weigh)
A) Weigh zb and zd.
If zb = zd. z1 is heavier and the anomaly.
Else = zb is the anomaly
B) Weigh za and zc
If za = zc. Zb is the anomaly
Else = za is the anomaly.
Scanario 2
(remember we have 3 groups of 4, X, Y and Z.)
(first weigh)
If X = Y, Z = anomaly group (shown above).
If X = Heavier (assuming for this example, just switch heavy with light here on in for the rest of the solution if otherwise), break X into 1,2,3,4 and Y into 5,6,7,8.
Separate into groups of 3, remember what group they were in but mix them up.
Group A = 156
Group B = 278
Group C = 34
(second weigh)
If A = B, Group C is anomaly group (solve this now by weighing one remaining coin against one known average coin, remember 3 and 4 were in the heavy group in this scenario. If scale is even the other coin is the anomaly, if 7 is light or heavy again it is the anomaly).
If A(156) = heavier and B = 278 is lighter, and since we already know that group A, which 1 is in, was the heavy side last time, then either i)1 is the Heavy coin or ii)7 or 8 are the light coin.
(Third weigh)
i) Weigh 7 against 8.
If 7 = 8, 1 is the anomaly
if 7 = heavier, 8 is the anomaly
If 8 = heavier then 7 is the anomaly
Is this correct Adrian?
First (failed) attempt:
*EDIT* The following did not work. One out of all of the possible outcomes was impossible to solve.
Mate, this is harder to type out than it is to solve!
]Split the 12 coins into 3 groups of 4: X, Y, Z
(1) Put X and Y on scale
a) If Weight = level discard x and y
b) If x = heavier replace y with z
c) if y = heavier replace x with z
(2)
(a) Split Z into groups of 1: Z1 and Z2 , place z1 and z2 on scale
(b) You have group x, y and z each with 4 coins. You know that X is heavier than Y: Weigh X and z
(ba) If Scales = level Y is anomaly group
(bb) if x = heavier then Y=Z and Z is the anomaly group.
etc
c) Procedure B, just replace X with Y...
(3)
(a) Replace the lightest scale's coin with any previous coins
(aa) If scales = level, the coin you removed from the lighter scale is the anomaly
(ab) If the other side stays heavier, you know the coin on the heavy side of the scale is the anomaly
(b and c)
By now we also know if we are looking for a heavy coin or a light coin
We have 4 coins left.
Split the group into 3 groups of 3, recycling coins where needed.
Group 1 has 2 Original 1 new
Group 2 has 1 original 1 new
Group 3 has 1 original 1 new
This is where it fails, if group 1 is heavier then you must weight again to differentiate between the two. I'll work
In step 2a) please explain how you can split a group of 4 (Z) into 3 groups and make an accurate measurement. At some point surely you'll be weighing a group of 1 against a group of 2, and that won't tell you anything as the group of 2 will most likely always weigh more.
For instance, say you have the Z group: A, B, C, D.
You split it into Z1, Z2, and Z3 as follows:
Z1 = A
Z2 = B
Z3 = C, D
Let's suppose A is actually the odd coin, and it is heavier than the rest.
Weighing A against B (the second weighing) means that you don't get any information, since either A or B could be the odd coin (since the odd coin could also be lighter!). Weighing A against both C and D is a risk since it might not tell you anything. For example, assume the coin regular weight is 1, and A is a heavier coin at 1.5.
On the scale, the side with A on it would still go up when the other side has C and D (since the total weight is 2). Thus, you cannot say anything about the weight of coin A, since if A is of normal weight, the same tilt occurs.
Then...well, you don't have anymore weighings left.
Let A,B,C,D,E,F,G,H,I,J,K,L be the set of 12 coins. Let L have a heavier weight.
X = A,B,C,D
Y = E,F,G,H
Z = I,J,K,L
(1) X vs Y => Balance. (1st weighing)
==> a) Discard X and Y.
(2)
a) Split Z into groups of 2:
Z1 = I,J
Z2 = K,L
Place them on the scale (2nd weighing).
Z1 is the lighter side.
(3)
a) Replace the lightest scale's coin (there are two of them aren't there???) with any previous coins. I'll assume you mean replace one of the two coins.
Z1 = A,J (replacing I with A).
Z1 vs Z2 => Z2 is heavier. (3rd weighing).
Now you are stuck with Z2, and no way of knowing which is the heavier coin.
Also, what if the coin was a lighter one, and in step 3a you replace a normal coin with a normal coin? This means you have no idea whether the odd coin is on the heavy side, or the other coin on the lighter side.
January 20, 2010 at 10:06 pm (This post was last modified: January 20, 2010 at 10:10 pm by theVOID.)
I was in the process of typing it all out when you all responded, i got it done on paper in doodles in a minuet or so, took a long time to make sense of my abstract imagery.
Second attempts:
Split into 3 groups of 4, (X, Y, Z)
Scanario 1
(1st weigh)
If X = Y, Z = Anomaly
Split z in 4. za, zb, zc, zd.
(2nd weigh)
Weigh Za and Zb
a) If za = zb, discard za and zc.
b) If za ≠ zb, discard zb and zd
We now have 2 groups with 1 coin each. za and zb, with weighted groups for each scenario
outcome A = zc is heavier and zd is lighter.
outcome B = za is heavier and zb is lighter.
(3rd weigh)
A) Weigh zb and zd.
If zb = zd. za is heavier and the anomaly.
Else = zc is the anomaly
B) Weigh za and zc
If za = zc. Zb is the anomaly
Else = za is the anomaly.
Scanario 2
(remember we have 3 groups of 4, X, Y and Z.)
(first weigh)
If X = Y, Z = anomaly group (shown above).
If X = Heavier (assuming for this example, just switch heavy with light here on in for the rest of the solution if otherwise), break X into 1,2,3,4 and Y into 5,6,7,8.
Separate into groups of 3, remember what group they were in but mix them up.
Group A = 156
Group B = 278
Group C = 34
(second weigh)
If A = B, Group C is anomaly group (solve this now by weighing one remaining coin against one known average coin, remember 3 and 4 were in the heavy group in this scenario. If scale is even the other coin is the anomaly, if 7 is light or heavy again it is the anomaly).
If A(156) = heavier and B = 278 is lighter, and since we already know that group A, which 1 is in, was the heavy side last time, then either i)1 is the Heavy coin or ii)7 or 8 are the light coin.
(Third weigh)
i) Weigh 7 against 8.
If 7 = 8, 1 is the anomaly
if 7 = heavier, 8 is the anomaly
If 8 = heavier then 7 is the anomaly
Is this correct Adrian?
I hope it is... I've been racking my brain over this one, haven't done any real algebra in quite some time.
(January 20, 2010 at 8:34 pm)Eilonnwy Wrote: Um, I think you're measuring them wayyy more than 3 times.
You can only weigh the coins 3 times, switching coins or groups counts as a weigh in.
You can remove all the coins from the scales, change the stacks and then weigh them again.
The other reason why it seemed like so many is because i have to account for all potential scenarios as there is an element of chance regarding where the anomaly is amongst the coins you weigh at any given moment.
Your scenario 2 works fine, but scenario 1 is flawed.
Suppose that Z = 1,2,3,4 and you've shown that X = Y in the 1st weighing.
Z1 = 1
Z2 = 2
Z3 = 3
Z4 = 4
Suppose that you weigh Z1 against Z2 (2nd weighing) and they are equal.
So you know that the odd coin is either in Z3 or Z4. You then weight Z3 against a known "good" coin and it is equal. All this leaves you with is that Z4 is the anomaly, but you don't know whether it is heavier or lighter.
I might have misunderstood your method though, since you seem to have made mistakes in labelling (for instance, if za = zb I don't understand why you'd discard za and zc...surely discarding za and zb would be better?)
If possible, could you write it in a form of "If X=Y GOTO (4)" or something like that. A step by step algorithm that points you in the right direction. Yours is all over the place at the moment :S
Woah, i was just waiting for you to shut me down on scenario 2! I'm surprised 1 was wrong, but i'm stoned so i sorta fleshed it out as i thought about it, but i just had a bong, feeling like more math, strangely
12 Coins
1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 , 12
*weigh 1234 and 5678*
Part A
If '1,2,3,4' = '5,6,7,8'
then '9,10,11,12' = Anomaly
*weigh ''7,8,9,10' and '12,11,1,2'*
if '7,8,9,10' > '12,11,1,2'
then '9,10' = Anomaly
*weigh 9,8*
if 9 = 8
then 8= anomaly
if 9 > 8
then 9 = anomaly
Part B
*weigh 1234 and 5678*
If '1,2,3,4' = '5,6,7,8'
then '9,10,11,12' = Anomaly
odcast:: 
