At a competition, a group of 16 members is divided into 4 teams of 4 members each. The procedure of the competition is as follows:
Each team will be seated in a different room and after which every one will be given a cap either red or blue, i.e., total of 16 caps out of which 8 are red and 8 are blue and one from each room has to shout one of these words: Yes/No/OK and only once. They have to shout loud enough so that all other teams of their group can hear.
Then, one from each group(all 4 together) has to go to another room where they should tell the number of red and blue caps each of the other teams wore.
Before the competition started, all 16 of them are asked to devise a strategy together for every team to accomplish the task successfully.
They are also told that they should not take the caps off their heads and should not talk to members of other teams until the competition is over.
Can you find out the strategy they used? It is known that all the 3 words have been shouted at least once and one of the guys said,"In this case, anyone one of four of us could have told you the distribution of hats even if we came here individually"?
Assume all their voices are indistinguishable.
Submitted by Praneeth
There are 2 strategies which will work to accomplish the task (without the last condition).
Its not said that they don't know the colour of their own cap. So, assume every person knows his cap's colour.
Let R: No. of red caps in a team
B: No. of blue caps in a team
R > B: Yes Possibilities: (3,1) (4,0)
R = B: OK Possibilities: (2,2)
R < B: No Possibilities: (1,3) (0,4)
If R > B and 4 Reds: The person with red cap goes to other room.
If R > B and 3 Reds and 1 Blue: The person with blue cap goes to other room.
Same is the case with R < B.
Consider the case of shouts:
(a)Yes, No, No, OK
(b)No, Yes, Yes, OK (These can be in any order)
(a) No => B ≥ 3, OK => B=2
2 No's + 1 OK => Total No. of Blue caps ≥ 8
So, in this case we can say that in both the "No" cases
number of blue caps is 3 from which every one can say the colours of other teams' caps. Same with the case (b).
Yes: (R,B) = (4,0) or (0,4)
OK: (R,B) = (2,2)
No: (R,B) = (3,1) or (1,3)
Any person with the cap with max. count in that room will go to other room.
But this has no unique combination.
This won't satisfy the last given condition that any one of four when gone to other room individually can tell the other teams' caps' distribution.
So, it's the strategy(1) they devised.
Note: These Yes, No, OK words can be assigned in any order you wish.