## Philosophy Puzzles

The Island of Knights and Knaves, First Problem: C says that B lies, so they cannot both be telling the truth; in fact one must be telling the truth (a knight) and the other lying (a knave). Suppose B is the knight; then what he says must be true. So if B is a knight, then A said that there is exactly one knight among the three. Is A a knight? No, that's not possible, because A's statement would have to be true then, but A and B would both be knights, so it would in fact be false. Could A's statement be false? In that case, A and C would be knaves and B a knight, so there would be exactly one knight among the three, and A's statement would be true, which is impossible, since A is supposed to be a knave. Thus on the supposition that B is a knight, either supposition about A's statement is impossible. The only way of avoiding contradiction is to conclude that B's statement is false. Hence B is a knave and C is a knight. Nothing can be determined about A.

*The Island of Knights and Knaves, Second problem:* The only thing you can conclude is that the author of this problem is no knight. It is impossible for an inhabitant of the island of knights and knaves to say, "Either I am a knave or two plus two equals five." If a knight were to say it, it would be false, and a knight does not says anything false. If a knave said it, it would be true, and a knave says nothing true. So the only thing to conclude is that a person lies if he or she tells you that he encountered such a thing on an island he or she describes as I have described the island of knights and knaves.

*The Surprise Test:* (Solution of W. V. Quine) The students are ready to conclude at the end of the argument that there will be no test. If they are willing to consider that a possibility at the end of the argument, they should consider it one at the beginning. When they conclude that if the test has not occurred by the end of class on Thursday it must occur on Friday, they are overlooking the possibility that there will be no test at all. All they can conclude before Friday's class is that either the test will occur on Friday or there will be no test. (Things might be different if the puzzle were set on the Island of Knights and Knaves and the students knew the Professor was a knight.)

*The Problem of the Light Switch:* (Solution of Paul Benacerraf) Before noon the light switch will be utterly destroyed by the forces needed to flip it on and off at exponentially increasing speeds, so there is no point in asking whether it will be on or off. Of course, you may protest that you were not thinking of an actual physical light switch. You thought this was a logical problem about a light switch not subject to those laws of physics that make the switching described impossible. Very well, in that case the answer is that nothing in the problem tells us what state the switch will be in at noon. The problem specifies the state of the switch at every instant before noon, but not the state at noon. In the actual world, we expect that if we know the state of an object at every moment before noon, we shall be able to determine its state at noon. But that's because actual switches obey the laws of physics, and we have thrown them away for our imaginary example.

*The problem of the Super Bullet:* The bullet will miss the armor plate. The supposition that it will hit leads to a contradiction: since it's a Super Bullet, it will penetrate the plate, but since it's a Super Strong Armor Plate, it won't. Hence the bullet will not hit the plate.

Questions for further thought: Why can't the army hit the plate with the bullet? Don't they have good marksmen? Knowing how much is at stake, don't you think they'd put their best shots on the task? When the first shot misses, wouldn't they try again from very close range? Two feet, say, or even two inches? How could they miss from two inches away?

*The Monkey:* Let x = the age of the monkey when its mother was three times as old as it was.

Then 3x = the mother's age then.

2x = the difference in their ages.

9x = the age of the monkey when it will be three times as old as etc.

4½ x = the mother's age when she was half as old as the monkey will be etc.

2½ x = the monkey's age then.

So 5x = the mother's age now,

And 3x = the monkey's age now.

Hence, 5x + 3x = 8 years, and

x = 1 year.

So the mother is 5 years old, and hence the monkey (and the weight) weigh 5 lb.

Now let r = the weight of the rope.

Then

r + 5 lb. = 3/2 * (5 + 5 - 5)

r = 2½ lb.

Hence, finally,

Length of the rope = 2½ lb. / 4 ounces/foot = 10 feet.

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