Let us imagine, for simplicity, that each traveller can carry enough supplies to travel three units of distance. How far into the desert can we advance a pilgrim, leaving him with a whole 3-unit supply, and returning the entire support team to base? We can get him one unit into the desert with one helper, who can just get back to base before supplies run out. Suppose we can get him n units into the desert using k people. Here is a procedure for getting him n+1 units in. Launch a double expedition, of 2k people. This safely gets 2 people n units. Now they do a one-unit transfer, leaving the pilgrim topped up (as required), but the helper has only one unit left. The helper can clearly be "rescued" by a sufficiently elaborate cascade of rescue parties departing from the base at two-time-unit intervals. For some reason I am unable to work out the exact numbers needed. Just tired, I guess. On Tue, Jul 3, 2012 at 1:34 PM, Victor Miller <victorsmiller@gmail.com>wrote:
Bill, That's pretty good, in that your recursion is quite simple. I may be asking for the impossible but I thought it would be nice if there was some quick a a priori argument. Similarly, in the jeep problem, is there a convincing reason why the distance that the jeep can travel as a function of n (the number of trips) goes to infinity also goes to infinity without bringing in the whole argument using the harmonic series. This would be somewhat analogous to the difference between proving that there are infinitely many primes (say using Euclid's proof -- which has a minimum of computation) and proving the prime number theorem (though the difference in difficulty here is not even close to the difference in difficulty with the problem with primes).
Victor
On Tue, Jul 3, 2012 at 12:24 PM, Cordwell, William R <wrcordw@sandia.gov
wrote:
Let "excess" be the amount beyond the amount required to return to base.
If the group travels together for a distance of 1/3, then at that point half of the group can transfer half of their remaining supplies and completely replenish the packs of the other half, while still retaining enough to return to base. Starting at that point, the remainder of the group, each of which has excess of 2/3, can travel 1/3 of the excess distance = 2/9, at which point half of the remainder transfer enough to fill the packs of the remaining half (a quarter of the original) and return home.
So, we have that the remainder of the group (after consecutive halvings) can travel 1/3 + 2/9 + 4/27 + ... -> 1.
I'm not sure that that counts as non-computational, though.
-----Original Message----- From: math-fun-bounces@mailman.xmission.com [mailto: math-fun-bounces@mailman.xmission.com] On Behalf Of Victor Miller Sent: Tuesday, July 03, 2012 10:00 AM To: math-fun Subject: [EXTERNAL] [math-fun] Travelers through the dessert
After listening to Car Talk this weekend, I was inspired to create the following question:
We have an inexhaustible supply of *travelers* each of whom can carry up to 1 day's worth of rations (we measure the distance in days of travel time). One of them, designated the pilgrim, wants to set off across the dessert from an oasis, aided by a cohort of n other travelers -- the helpers. Each of his helpers can travel a certain distance, transfer some of his rations to the remaining travelers (pilgrim plus remaining helpers) and turn back to the oasis (and thus needs to keep enough rations to allow him to get there). After the last helper leaves, the pilgrim keeps going. When he runs out of rations a magical genie appears and creates a new oasis for him. Let L[n] denote the longest distance (in days) that the pilgrim can travel using n helpers. It's clear that L[n] <= 2 since the last helper to leave can't be more than 1 day's distance away from the oasis. It seems intuitively clear that as n ---> infinity that L[n] ----> 2. Indeed one can work out a recursion for L[n] and solve for it explicitly. However, is there a simple, non-computational proof of this fact which doesn't involve solving for L[n]? _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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