This is kinda driving me crazy…
Does someone have a clue.
there are total 12 eggs,
only 1 have different weight (don’t know more or less on weight)using a balance and measure 3 times only,
how do I found out the egg which is different in weight??
First weight in: 6 on one side of the balance, 6 in the other: side with the most weight gets chosen for the next round
Second weight in: 3 on each side: side with the most weight will be used in the next round
Third weight in:
Logic takes over, put only one egg in each side of the balance, put the last egg next to the balance.
If both eggs weight the same, the egg next to the balance is the one with the most weight, if not, the balance will tell you.
- welcome to the boards
- It “helps” to put a more descriptive title on your posts
- The answer: I thought I had the answer right off the bat until I reread the problem and it said you don’t know if the egg is lighter or heavier. That one threw me for a loop. Took a few minutes after that. It’s a fairly complicated one. OK, so don’t look down unless you’re certain you want to know
Divide the eggs up into four groups of three. Balance groups 1 and 2 together.
If they’re not equal, the egg is in one of those six. Take two eggs from group 1 and two from group 2. If those balance, the bad egg is one of the two remaining. Take one egg that you know is good and balance it against one of the two questionable eggs. If it balances, the other one is bad. If it doesn’t balance, that one is bad.
If the 2&2 groups do not balance, take one of the groups and balance those two eggs against each other. If they don’t balance, compare one of them against another egg, as per the instructions in the preceding paragraph.
If groups 1 and 2 do balance, the abnormal egg must be in the other six. Divide groups 3&4 into two eggs apiece. Repeat the steps given in the paragraph above regarding 2 eggs.
Using these steps, you should be able to find the abnormal egg no matter where it’s located in three balances or less.
GIGO I don’t think that’s gonna do it. I think you’ve assumed that the defective egg is heavier than the rest and that is not given by the puzzle definition.
First let me introduce a lemma for simplicity: Single Weighing Lemma: You can determine with a single weighing which of three eggs is defective if you know the nture of the defect (heavier or lighter). Place two eggs on the scale. If they balance you know the other egg is the defective one. If they do not, you know which is defective since you know the nature of the defect.
OK.
Take four eggs and balance against another four eggs. They either balance, A or do not, B.
A: You now have eight eggs of known regular weight. Mark them as such, say R. Take three of the unknown eggs and weight them against three of the R eggs. If they balance you know the irregular egg is the one remaining, unweighed egg and the single weighing left will suffice to determine its nature, i.e. heavy or light. If they do not balance you have three eggs of known defect and by the SWL can determine the solution.
B: You have four candidates for heavier, mark them H, and four candidates for lighter, mark them L. Place two H’s and one L on one side of the scale and the remaining two H’s and an R on the other side.
1. If they balance you know the defective egg is among the three uninvolved L’s. Use the SWL.
2. If the HHL side goes down the defective egg is one of those two H’s and the remaining weighing will suffice to sniff it out.
3. If the HHR side goes down you know it is either one of those two H’s or the L on the other side. Place the two suspect H’s on the scale. If they balance it’s the L from the other side. If they don’t balance, well I think you know what that means.
QED
Enderw24 the problem is that your solution doesn’t determine the nature of the defect, only which egg is defective.
You’re right, of course, but the OP didn’t ask for the nature of the defect.
Yep, I need to read the OP.s carefully next time.
When you said,
you don’t continue on to say what to do if they do balance. Then you have two eggs of unknown quality with no weighings remaining. At least, I think so…
I went back and read the OP, and you’re right about that point you just posted, though. 
My rep really does exceed me. Disregard that last post…
MonkeyMensch,
one of the problems in your explanation is that you go through this entire process and then refer back to the SWL in the final step. But the thing about the SWL is that it assumes you know whether the egg is heavier or lighter, which you don’t know.
Further, if you did know it, you wouldn’t have to trouble yourself with anything complicated at all. Divide the eggs up into groups of 4. Balance them against each other. If they match, divide the last group of 4 into groups of 2 and balance, narrow down to 1 and balance. Voila. Your answer.
But you can’t go under the assumption that you don’t know the nature of the bad egg and then explain your answer as if you did know. That doesn’t make sense.
So now I’m completely confused. Do you agree with my answer or not?
Boy if you think that I confuse you, you ought to see hwat I do to myself…
I think I was tackling the problem of determining the egg and the nature of its defect all along. As you pointed out that was not inluded in the OP’s problem.
In the final step, B, if I read you right, I use the SWL on three eggs that were determined in the first weighing to be light if they are defective at all. The second weighing, if balanced, determines the defective egg to be among them. That condition staisfies the SWL.
And I don’t agree with your answer, even as pertains the OP. Your first weighing determines which group of six contains the defective egg. Your second weighing determines if it is in a group of two, or a group of four. Say that the two-vs.-two weighing doesn’t balance and the bad egg is therefore in that group of four. What is the last weighing operation to determine which egg is irregular?