I recently read this puzzle at the Futility Closet and it reminded me of a technique that I like to use to test conjectures (when possible). I don’t know if it has a name, so I’ll call it “looking for extreme examples and counterexamples.” I like this technique because when it works it is fast and easy, and it can often be used without writing anything down. I’ll give three examples to illustrate this technique.
Example 1. Let me rephrase the puzzle in the form of a conjecture.
Two runners are in an airport standing at one end of a moving walkway (one of those 100 foot long treadmills). They run at equal speeds to the end of the walkway and back—but one runs on the moving walkway and the other runs on the (unmoving) floor next to the walkway. Conjecture: they both finish at the same time.
The idea, of course, is that the runner on the walkway will get helped by the treadmill going one direction and hindered (by the same amount) in the other direction. He’s traveling the same distance both ways, so the effect of the treadmill cancels itself out.
When you’re imagining the problem in your head you’re thinking that a person runs 15 mph and the treadmill is going 3 miles an hour, or something like that. You may reach for some paper to do some calculations…
But there are no details about velocity in the conjecture. So consider an extreme example—the walkway and the runner are moving at the same speed. Then, running with the walkway the runner goes very fast, but when he tries to come back, he runs on it like a gym-goer does on an exercise treadmill, and makes no progress. He not only loses, he never reaches the finish line. Thus the conjecture is false.
(Note that in the original puzzle the question is: who wins? If the answer HAS a correct answer, then you can use the extreme example given above to conclude that it is not the person running on the moving walkway.)
Example 2. There is a long history of mathematical cranks claiming to be able to trisect an angle. Recall the problem: you are given an angle with measure . Is it always possible, using only a straightedge and compass, to construct an angle with measure ? It is a famous result of Pierre Wantzel that while it is sometimes possible, it is not possible in general (in particular, it is impossible to trisect a angle).
Here is a favorite “trisection method” given by the mathematical cranks. Suppose you are given an angle . Draw a circle with center and radius . We may as well assume that is on this circle. Draw the chord . Trisect this chord; that is, find a point on such that (it is well known that it is possible to trisect a line segment using the Euclidean tools). Then .
Conjecture: this is a valid method of angle trisection.
For small angles, this technique looks convincing (see below).
But it must work for all angles. Don’t dust off your copy of Elements and start looking for relevant propositions, look for an extreme example! For example, suppose . Clearly, as we see below, this technique does not trisect such an angle. Thus the technique fails.
Example 3. My last example is the famous Monty Hall problem. I’m sure this problem is well known to many of the readers of this blog, but here’s the setup. Monty Hall (a game show host) presents 3 closed doors to a contestant. He promises that behind one door is a new car and behind the other two are goats (obviously, the contestant wants to win the car). The contestant picks a door. Monty says that he will open one of the two remaining doors to reveal a goat (which he does). Then he asks the contestant if she wants to switch doors.
Conjecture: there is no advantage to switching. (Your rationale: at first your chance of winning was 1/3. But now there are two doors, one hiding a car and one hiding a goat, so it is a 50/50 shot either way.)
Of course this conjecture is FALSE. Here’s an extreme example to illustrate this point. Suppose there are 1000 doors hiding 1 car and 999 goats. You pick one door. There’s a 99.9% chance that the car is behind one of the other doors. Now Monty (who knows what is behind each of the doors) opens up 998 of the remaining doors. There are two closed doors—your door and one other. Using the same rationale as above, your chance of winning is now 50%, right? No! I hope it is clear that you want to switch!
(By the way, I read this explanation in The Drunkard’s Walk by Leonard Mlodinow.)