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This one is particularly for solarflare, but open to everyone else (he asked for a new one, sorry ;)).
You have 9 coins with 2 false ones. The false ones are the same make, meaning that they have the same weight and are either both lighter or both heavier than a standard coin. Find out in four weighings which two coins are false and whether they are lighter or heavier than standard coins.
It's a hard one, I have to admit ;)
Wow, we are really beginning to become corrupt here, aren't we...Quote:
Originally posted by Yves M
You have 9 coins with 2 false ones.
I know you're expecting a brilliant answer immediately from me :rolleyes:, but I'm actually busy right now, so I'll come look at it later today, maybe tomorrow... ;)
No we just really like jokers :pQuote:
Originally posted by solarflare
Wow, we are really beginning to become corrupt here, aren't we...
Exactly, you have all the pressure on you ;)Quote:
I know you're expecting a brilliant answer immediately from me
Call me "treasurer" (that is to say, I solved the coins problem)
First:
Yves and others, I thank You for your reliance. ;)
Second: Please, Yves, don't post answer for task about 9 coins this week. Now I haven't enough time, but I hope to find it, and solve this task by this week:rolleyes: :)
Good luck, dimm_coder. solarflare has already nearly solved it correctly. The idea is there but there are a few small mistakes.
Argggg ! It's actually not solvable in four weighings in the general case :( Sorry, so the riddle with 2 false out of 9 is not solvable. You can do it in five weighings though.
Arg ;)
Really? I thought for sure that... oh well. Can you prove that it can't be done in 4? 'cause I thought I had it. Sometime this weekend I'll check it out.Quote:
Originally posted by Yves M
Argggg ! It's actually not solvable in four weighings in the general case :( Sorry
Highly probable, it could be just a typo, or maybe I messed up one of the weighings... a program like that is gonna be annoying to debug, its one huge series of "if" statements!Quote:
Originally posted by Yves M
solarflare has ... a few small mistakes.
By the way, here is the solution to the problem with 12 coins, one of which is false and you don't know whether it's lighter or heavier.
It's an HTML file and I hope it's clear enough :)
Of course there are many different ways of solving it, this is only one.
Well, it's not easy to prove. But in case pile A is lighter (or heavier) in the first weighing, you are left with 24 states. BUT these states can't be partioned into 8, 8, 8 or 9, 9, 6 states by any weighing. So you have to make do with 10, 7, 7 and since 10 > 3^2 you require one weighing more.Quote:
Originally posted by solarflare
Really? I thought for sure that... oh well. Can you prove that it can't be done in 4? 'cause I thought I had it. Sometime this weekend I'll check it out.
Incorrect, I believe. They can be divided into 8,8,8. More info when I pm u in a sec. (so dimm_coder doesnt see)Quote:
Originally posted by Yves M
in case pile A is lighter (or heavier) in the first weighing, you are left with 24 states. BUT these states can't be partioned into 8, 8, 8 or 9, 9, 6 states by any weighing. So you have to make do with 10, 7, 7 and since 10 > 3^2 you require one weighing more.
The 2 of 9 coins puzzle is certainly doable in 4 weighings, what I'm now wondering is whether or not it can be done in 4 predefined weighings ;)
Of course you are right ;)
Ok, so it is solvable, sorry for the Args ;)