Follow-up: The Stork and The Frog
Amusingly, this problem has exactly the same solution as the proof that there are as many rational numbers as there are counting numbers. And the proof generalizes: one stork can catch three frogs, or ten or fifty.
Here are some bonus problems:
- The stork can catch the frog even if it can start at any rational number and hop any fixed rational distance each step.
- However, if the frog can start at any real number or hop any real distance, the stork has no strategy that guarantees a catch. This is, in effect, the same as proving that the real numbers are not countable.
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rmjarvis said,
May 6, 2007 at 12:25 pm
This problem is basically the same as the proof that the rational numbers are countable. In the same kind of way, we can prove that the set of all possible frog hops are countable.
Define the frogs motion as x(n) = a + b*n, where a is the starting position, b is the hop size and n is the number of hops taken.
So any given frog motion is characterized by the pair (a,b). Furthermore, we can sort these pairs by the following criteria:
First by the sum (|a| + |b|).
Then, within the pairs with the same (|a|+|b|), by a.
Finally, within those with the same (|a|+|b|) and the same a, by b.
So the first several frog motions are:
(0,0)
(-1,0) (0,-1) (0,1) (1,0)
(-2,0) (-1,-1) (-1,1) (0,-2) (0,2) (1,-1) (1,1) (2,0)
(-3,0) (-2,-1) (-2,1) (-1,-2) (-1,2) (0,-3) (0,3) (1,-2) (1,2) (2,-1) (2,1) (3,0)
etc.
The strategy is then simple. In the first second, when n=0, the stork tries out the first frog motion (0,0). At n=0, this motion is at x=0, so the stork tries that spot. If it doesn’t get the frog, then in the second second it tries the next pair (-1,0), which at n=1 is at x=-1. Then (0,-1),n=2 is x=-2. Then (0,1),n=3 is x=3. Then (1,0),n=4 is x=1. Then (-2,0),n=4 is x=-2. And so on.
Since all possible frog motions are enumerated in this list, eventually the stork will try the one that actually corresponds to the frogs motion.
Note that the proof that rational numbers are enumerable follows the same strategy where a,b are the numerator and denominator instead. And in that case, you can skip the b=0 possibility. I wasn’t sure whether the statement of the problem implied that b=0 was allowed or not for this problem, but if not then you just remove those pairs from the above list and the proof still holds.
Shawn said,
November 19, 2011 at 9:11 pm
What about the possibilities where b = ±1/2, ±sqrt2/2, ±sqrt3/2, ±sqrt5/2, ±sqrt6/2… ±sqrt(n)/2, ±cube root(2)/2, ±cube root(3)/2 … ±nth root(n)/2, ±nth root(n)/3, ±nth root(n)/4… ±nth root(n)/n… n^[n/n] … n^[-n/n] … pi, 2pi, 3pi, 4pi … npi, pi^2, pi^3, pi^4, pi^5 … pi^n. (I’m guessing that we’re going to assume that the frog hops completely horizontally, so we can ignore complex numbers of the form a + bi where b ? 0) For the stork to guarantee that he can catch the frog, we would have to assume that he can make an infinite number of attempts in his lifetime.