[Guest]

You’re a counter-spy at an embassy party. You know that 1 of the 51 guests is an enemy spy. You know two things about the spy:

1) The spy knows everyone in the room.

2) Nobody knows the spy.

You are an experienced spy, so it only takes you 1 minute to approach a guest and ascertain if they know some other guest. A bomb will explode if you don’t find the spy in 60 minutes; your boss tells you it is possible to do it in exactly 50min. How?

Keep in mind the 2 things you know about the spy.

[Guest]

You ask random people if they know everyone else in the party. The only one who does is the spy.

[Guest]

You ask random people if they know everyone else in the party. The only one who does is the spy.

To ask 1 person if they know everyone else at the party, it would take 50 minutes (1 minute for each person). If you were to repeat that for every guest it would take 50*50=2500 minutes in the worst case - by which time I'm sure everyone would have gone home!

It can be done in exactly 50 minutes.

Edited July 22, 2008 by bezza

[Guest]

two questions.

1. What nationality is the enemy spy

2. What nationality is the counter-spy????

[Guest]

That's not important. Assume nothing else about the spy. He can be found in 50 minutes given the facts that:

1) The spy knows everyone in the room.

2) Nobody knows the spy.

Being a very good spy, you will always find whether person A knows person B in 1 minute.

The best way is to use the facts to eliminate potential spies.

Edited July 22, 2008 by bezza

[Guest]

pair everyone up into 25 pairs (one left over).

ask each person in each pair if they know the other person.

if one person knows the other person but the other doesn't know him/her, then that person is the spy (if none of the pairs are of this case then the leftover person who isn't paired is probably the spy).

[Guest]

You're thinking along the right lines, rjsghk107. But 1 guest may know another guest, without being known back (it's not a transitive relation). I.e. someone senior like the ambassador will probably be known by everyone at the party, but will not know everyone back.

[Guest]

y cant the spy lie?

[Guest]

you pair everyone up (excluding the leftover) and ask only one person (person A) in each pair if they know the other person (person B).

-if person A knows person B then you eliminate person B from your suspect list.

-if person A doesn't know person B then you eliminate person A from your suspect list.

after you do this you will have eliminated the number of suspects by half plus the one leftover. 26 suspects.

You repeat this process again and again until you only have one possible spy.

this takes 50 minutes if you calculate it

Edited July 22, 2008 by rjsghk107

[Guest]

Good going rjsghk107!

[Guest]

You have to ask each person to introduce you to the one that you haven't been introduced with, after asking them whether they know everyone at the party. Who ever tells that they know everyone is a spy.

[Guest]

At the party, announce a contest to get everyone to write a list of known names in the room (of course with their name as identification) within 10 minutes. For the next 10 mins go through the list. The winner is the spy.

[unreality]

can't the spy lie, like pieman said, and pretend not to know people? lol ;D

[Guest]

You’re a counter-spy at an embassy party. You know that 1 of the 51 guests is an enemy spy. You know two things about the spy:

1) The spy knows everyone in the room.

2) Nobody knows the spy.

You are an experienced spy, so it only takes you 1 minute to approach a guest and ascertain if they know some other guest. A bomb will explode if you don’t find the spy in 60 minutes; your boss tells you it is possible to do it in exactly 50min. How?

Keep in mind the 2 things you know about the spy.

you ask each of the fifty people (in a separate room for a max. time of 1 minute.) "how many people do you know in this room"? If they say "49", they are the spy. This process (I bet) would even take less than fifty minutes.