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Old Wednesday, March 17, 2010
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Default Logic Puzzle

You meet a group of 6 persons, U,V,W,X and Z...Who speak to you as follows:

U says: None of is a knight.
V says: At least three of us are knights.
W says: At most three of us are knights.
X says: Exactly five of us are knights.
Y says: Exactly two of us are knights.
Z says: Exactly one of us is a knight.

you have to sort out that which are knights(true) and which are knaves (liar)?
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Old Wednesday, March 17, 2010
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If U is a knight then his statement is false, so U is a knave because Its not the case there are 0 knights.

If Z is knight, then W statement must be false. If W's statement is true, W is a knight too, along with Z, so Z's statement cannot possibly be true. So, Z is a knave.

If X is a knight, then there are 5 knights. So that means there is only one knave. Since we have that U and Z as a knave, we now have a total of 2 knaves. So X is a knave too.

If V is a knight then there are at least knights. Since U, X and Z are knaves that means that the remaining are knights. But then Y would be lying . So Y is a knave. This is a contradiction. So V is a knave

If W is a knave then there are at least 4 knights. But we have that U, V,X and Z are knaves. W is a knight.

If Y is a knave, then there is only one knight. But then Z would be telling the truth and hence would be a knight. We already know that Z is a knave so this is a contradiction.

W and Y and knights.
U, V, X, and Z are knaves
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