Professor Smith has been studying the knights, knaves, and liars in their villages, and is currently living among them. You and your guide (who is a knight) approach a fork in the road and see five people standing in a line facing you. Your guide tells you there is one person he knows to be a knight, one person he knows to be a liar, one person he knows to be a knave, one he doesn't know at all, and Professor Smith. They said:

A: I am a knight.
B: I am a knight.
C: I am a knave.
D: I am a knave.
E: I am a knight.

A: E is a knave.
B: A is a knave.
C: D is a liar.
D: C is a knave.
E: B is a knight.

A: D's first statement is a lie.
B: C's first statement is a lie.
C: A's second statement is a lie.
D: B's third statement is true.
E: C's second statement is true.

A: D is Professor Smith.
B: C is not Professor Smith.
C: I am Professor Smith.
D: A is Professor Smith.
E: I am not Professor Smith.

Which one is Professor Smith? Remember: Knights always tell the truth. Liars always lie. Knaves alternate between truths and lies. Professor Smith is one of these three types, but you don't know which.

Again, we start off with assigning fake ids.

Assume A is a knight. That make B a liar(1st, 2nd statements false) and E a liar. B claims C's first statement is false, E claims C's 2nd is true. They are both lying,and that makes C a knave yet C double-lied for statement 2 and 3 if so. Hence A cannot be a knight.

Assume A a knave, then of cos A's 1st statment is a lie. That makes E a knave (A's 2nd  statement), D a knave (A's 3rd statement), B a knight(E's 2nd statement) and C a liar (E and B's 3rd statement). By luck, this case seem to have no contradiction and D's professor smith, who's a knave.

Anyway, we assume the final case that A's a liar. then B's a liar(same case in assuming A knight) and E a liar. C's a knave. D is then not a liar since he point out C's a knave. Being  a knight or , his first statement cannot be a lie, yet it is! Being a knave, his first statement cannot be true, yet it is ! D has no id then, contradiction. Hence this case is also proved invalid.

Ans: D(which i once thought E)'s professor Smith.

(i misthought that the last statement is the 5th and thought E's the professor...proves me a careless jerk)

 

 

Edited on October 12, 2005, 10:58 am

Posted by Terence on 2005-10-12 10:53:18