Yes, but you know that each bag has different color of sockets… sorry if I was not clear.
You don’t see them but feel with hands that there is two different bags, so two different colors , but you don’t know what color.
No problem for me to say that Edoink is right, because I was not enough clear at the beginning in the real process.
That part was clear. No problem.
But if you can actually see the bags, or know that the one on the left (say) is always the one on the left, then selecting a socket to match a previous colour is a certainty. If you can’t see the bags, or they get moved at random such that you never know which bag is which, then there is absolutely no certainty whatsoever as to the colour you are about to select.
Nobody was critisising you. merely seeking clarity.
Yes, I think I , from memory, have not explained well the problem. Very sorry.
Now I remember it well and you will be both disappointed. But I honestly say that I made a mistake and remember the right problem now: You are blind, but there is only one bag, not two. 50 red and 50 black sockets, all mixed of.
All same but only one bag.
I should have not join this thread…
Yep, in that case the answer is 3 for both questions.
Do stay on the thread, it’s fun.
Thanks, and sorry also.
sisisisisisigarsisansisisisansigar
What does it mean correctly written in French ? I wrote in phonetic, without spaces.
Do you want the response ?
I’m still looking at it occasionally, trying to work out where the sis fit together.
Ok, no problem.
One a bit difficult. I couldn’t found . However I am not a scientist.
I will give you the exact problem this time.
Diamants
Vous devez retrouver le faux dans un lot de 9 diamants.
Vous savez qu’il est moins lourd que les autres.
Vous disposez d’une balance à balancier, et vous avez le droit à seulement deux pesées.
Comment retrouvez vous le faux ?
You have to find the false Diamond among 9 diamonds. You know it’s lighter vs the others.
You use a pendulum balance and you can do only 2 weightings.
How will you find the false?
Put three diamonds in each side of the scale. If one side is lighter, then it has the false diamond. If not, then the false diamonds is in the other three which weren’t weighed.
Take the set of three diamonds we know has the lighter one, and balance one in each side of the scale. If they are the same, the one not on the scale is the only one that can be lighter. If one is lighter than the other, it’s the one.
Nice teaser FR and nice solution Eoink.
Try 12 diamonds. (And a pendulum balance)
One is either lighter or heavier than the other 11.
What is the minimum number of weighings to find the odd diamond and determine light or heavy.
Response :
On choisi au hasard 3 diamants que l’ont met de chaque côtés de la balance.
- Si les 6 sont vrais, la balance s’aligne, il nous reste donc trois diamants.
- Si la balance s’incline d’un côté ,ils nous reste toujours les trois du côté le plus léger.
Avec les trois restant il suffit d’en mettre 1 de chaque coté: - Si la balance s’aligne le restant est le faux.
- Sinon le plus léger sera le faux.
Is it that?
Oui.
3 I think Don. I can see 2 methods, there may be more.
A)
Balance 4 coins on each side. If one set is lighter, then the the lighter diamond is there, if not as before it’s in the other set.
From the set of 4 with the lighter diamond, balance 2 against 2.
From the lighter set of 2, balance 1 against 1.
B)
Balance 3 against 3. If one set of 3 is lighter, it has the lighter diamond.
If 1st weighing was equal, balance the remaining coins 3 against 3.
From the lighter set of 3, as before balance 1 against 1.
Hi Eoink,
3 is the correct answer, and putting 4 diamonds on each side as per your method A, is a good start, but …
… you don’t yet know whether the odd diamond is lighter or heavier than the other 11.
Oops, failed at the first hurdle, read the bleeding question properly. I’ll have a think.