MH
There are only two hard things in computer science: cache invalidation, naming things, and offbyone errors
Saurabh Pillai wrote:(Did I take correct route?)
if you weigh 3 against three, and one side is lighter, you only need two weighings.
Both of the fake coins are in the lighter side.
kri shan wrote:If one fake coin is one side and another fake coin is another side during my first weighing
like (2 original coins + 1 fake coin one side and 2 original coins + 1 fake coin another side).
Then can we start from begining ?
MH
J2EE Architect/Developer
Henrique Ordine wrote:That's a lot more than 3 weighings.
J2EE Architect/Developer
Henrique Ordine wrote:Nice try.
I stopped counting after that.
Henrique Ordine wrote:Nice try.
1:A
2:B
3:C
4:D
I stopped counting after that.
J2EE Architect/Developer
J2EE Architect/Developer
Henrique Ordine wrote:I still have doubts about the last solution to the previous problem.
Are they assuming that if A<B then A is one of the lighter coins and B is not? If so, they said that the lighter coins don't necessarily need to have the same weight, so how do you know that they're not both the lighter coins. <br /> <br /> Also, are they assuming that if then they must be both original coins? If so, how do they know that they're not both lighter coins that coincidentally weigh the same weight?>
1: AB
2: CD
If we have identified two coins as lighter, we are done (e.g. A < B and C < D).
If we have identified one coin as lighter
3: EF to find the 'other" lighter coin.
Otherwise (all scales balanced to date):
3: AC
If balanced, then the two 'lighter' coins are E,F
If A<C, then the coins are A,B
otherwise the coins are C,D
It sure was nice of your sister to lend us her car. Let's show our appreciation by sharing this tiny ad:
the value of filler advertising in 2020
https://coderanch.com/t/730886/filleradvertising
