You have 8 coins. 1 coin weighs less than the others. You can

>>8836823
Wouldn't you need three weighings?

>>8836823
4 coins in each
Get the lighter pile
Put 2 in each
Get the lighter pile
Put 1 in each
The lighter pIle is the coin

Each weighing can give three results (left heavier, right heavier, equal weight) so you have to put a third of the coins on each side.

>>8836823
With my hands.

That's just a statement anon

Weigh 3 coins on each side, if they balance,t he two missing coins contain the lighter coin.
If not, the side with less has it, which means you can weigh one on each side. The one if even, the one taken from the pile is the lighter. If uneven, the lighter coin is the lighter coin.

>>8836843
it was stated that you can only do 3 weighings

File: thingken.png (53KB, 256x256px) Image search: [Google]

53KB, 256x256px

>>8836823
Here's an interesting one. You have 8 coins, and you know that one of the coins has a defect that makes it either lighter or heavier, but you do not know which. Can you come up with a scheme that lets you identify the defect coin, as well as if it is lighter or heavier with 3 weighings?

set two aside and weigh 3v3, if it's balanced then you know the lighter coin is in your removed pile and if it's not balanced then you can do the 1v1 with the 3 coins on the scale to reveal the lighter coin

>>8836857
You can even do this with 12 coins.

>>8836823
Weigh ABC vs DEF. case 1: ABC is lighter than DEF.

In this case we can simply weigh A and B, if they are balanced C is lighter.

If ABC DEF is balanced simply weigh G and H

>>8836867
Fuck me, posted the wrong one, you do indeed have 12 coins. I felt smart when I figured it out, it's pretty nifty

>>8836857
>>8836867
>>8836895
it's the same problem with different ratios

4v4 -> 2v2 -> 1v1

>>8836853
Can you count?

>>8836901
The solution is a little more tricky than just 4 -> 2 -> 1

>>8836853
Meant to be
>>8836856

>>8836924
Point out the flaw in his reasoning

>>8836924
Yes.
First weighing
Outcomes:
-Even => one of the two non weighted are the coin, so those would be weighted second and the outcome would determine the coin
-uneven => second weighing reveals the coin because if it's even then the coin not weighted is the coin, if it's uneven it's apparent.

>>8836925
>>8836928

>>8836933
Those are three weighings.

>>8836939
are you retarded?

3v3 -> 1v1

>>8836934
inb4 complaints that I didn't use [math]/LaTeX[/math]
I'm too lazy
>>8836939
It's two weighings. I'm fairly poor at explaining though.
3v3 outcomes = even = one of the two not on the scale is the coin in which case you use the scale to weigh the two coins not weighted. If the scale is uneven, you would weigh 1 v 1 from the lighter pile. Which would mean that if the scale is even, the remaining coin is the lighter coin, if it is uneven, the lighter side of the scale contains the lighter coin.

>>8836939
The way he described it I understand it as being the same scheme as >>8836869 aka 2 weighings

>>8836949
You are right.

>>8836823
Weigh 3 coins on each side.
If the weight is equal, weigh the other 2.
If it's not, weigh any two coins from the least massive 3 coins. If they are equal again you know the one you didn't pick is the right coins.

>>8836853
Seems right.

>>8836823
This question is asked as a part of an "Introduction to algorithms" class (or something like that) in high school and the next task is about finding a general way of finding the lighter coin in a set of n coins in O(log3(n)), amirite?

>>8836945
Lol 4v4 > 2v2 > 1c1 Are you retarded?

>>8837419
>brainlet reading comprehension: the post