Youve had the unfortunate fate to be captured by an insane e

You’ve had the unfortunate fate to be captured by an insane eugenicist who is on a quest to eliminate people who cannot solve his logic puzzle. You are in a room with finitely many other people, and let’s say the total number of people is some number n.

The eugenicist tells you that in an hour he will put you into one long line, and then have his minions place red or blue cones on everyone’s head. In this way, you will be able to tell the color of the cone of everyone in front of you, but you will not be able to see the color of your own cone or the color of the cone of anyone behind you.

He now says the each person will be allowed to say either RED or BLUE (with no in inflection, mind you he will detect this and immediately decide you are unfit to live). If you say the color of your own cone, you’re allowed to live. Otherwise... not so lucky.

Naturally, during this hour you have the opportunity to talk to one another and strategize, but do note he can hear everything you’re saying and will do everything in his power to make you lose. Also, we unfortunately can’t use time to our advantage: we are prompted to say RED or BLUE at exactly ten seconds following the previous person in line.

What us the minimum number of people the team can save? Prove that you can’t do better.

Solution

the below answer is applicable if started from the 50th person.if started from the 1st person the it wolud be difficult for the persons to decide the color because they couldnot see the color of cones behind them

let suppose the number of persons be 50

At-most 49 persons can be saved and the 50th person has 50-50 chances of being unlucky
The idea is that every person counts number of red cones in front of him.

50th person says red if the number of red cones is even. He may or may not be saved, but he coneys enough information to save 49th person.

The 49’th person decides his answer on the basis of answer of 50’th person’s answer. There are following possibilities and 49’th person can figure out color of his cone in every case.

If 50’th person said ‘Red’ (There must have been even number of red cones in front of him)
a) If 49’th person sees even number of red cones in front of him, then his color is blue.
b) If 49’th person sees odd number of red cones in front of him, then his color is red.

If 50’th person said ‘Blue’ (There must have been odd number of redcones in front of him)
a) If 49’th person sees even number of red cones in front of him, then his color is Red.
b) If 49’th person sees odd number of red cones in front of him, then his color is Blue.

The 48’th person decides his answer on the basis of answer of 49’th person’s answer and uses same logic.

Similarly other persons from 47 to 1 are saved