Comments on: Follow-up: The Stork and The Frog http://mathfactor.uark.edu/2007/05/follow-up-the-stork-and-the-frog/ The Math Factor Podcast Site Fri, 08 Aug 2014 12:52:06 +0000 hourly 1 https://wordpress.org/?v=4.9.25 By: Shawn http://mathfactor.uark.edu/2007/05/follow-up-the-stork-and-the-frog/comment-page-1/#comment-895 Sun, 20 Nov 2011 02:11:38 +0000 http://mathfactor.uark.edu/2007/05/06/follow-up-the-stork-and-the-frog/#comment-895 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.

]]> By: rmjarvis http://mathfactor.uark.edu/2007/05/follow-up-the-stork-and-the-frog/comment-page-1/#comment-70 Sun, 06 May 2007 18:25:38 +0000 http://mathfactor.uark.edu/2007/05/06/follow-up-the-stork-and-the-frog/#comment-70 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.]]> 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.

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