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Topic: A Variation of the Camel Problem (Read 554 times) |
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raprap
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A Variation of the Camel Problem
« on: Jul 14th, 2005, 8:08pm » |
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A variation on the camel problem
Yeah Yeah Yeah you know the one about the beat up old camel that could only carry 1000 figs, but ate a fig a furlong, and Hajiib had 5000 figs to deliver for sale across the 1000 furlong desert and wanted to maximize his profits..
But when you consider the problem  you’ve only solved one aspect of the problem. That is to maximize the figs delivered to the far side.
Say what if Hajiib is trying to get to the princes Ezeramelda’s hot to trot boudoir and he wants to get across using just enough figs. What is that minimum number?
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 Yeah! Well it’s 1000. Yeah! Well that IS kinda obvious, but hey gimme a break. Hajiib was thinking with his little head.
 But say if Hajiib wanted to get back across the 1000 furlong expanse.  Yeah that’s the ticket.
That’s a variation on the camel problem. And Hajiib wants to minimize his fig fuel.
Also, for fellow porcine puzzlers Hajiib’s camel problem is open to other variations. Say one that might consider the consequence of an all fig diet.
Rap
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raprap
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Re: A Variation of the Camel Problem
« Reply #1 on: Jul 17th, 2005, 8:08pm » |
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The solution to the traditional camel problem.
Hajib has 5000 figs on the near edge of the 1000 furlong desert to deliver as many as possible to the far side. So the first step is to get as many figs into the desert as far as possible.
The first step takes a 1000 figs into the desert and back, stashing a portion of the load. The portion is 1000-2x where x is the number of furlongs into the desert the first step goes. Since the camel capacity is 1000 figs, delivering the initial 5000 figs takes four round trips and one one-way trip, and the stash is 4000 figs. The distance is 4(2x)+x=1000 furlongs and the fig stash is 1000/9 furlongs into the desert. The second step is the same, but now it takes 3 round trips and a one way trip to advance the fig stash, which is being consumed by the camel. This step also reduces the load by 1000 figs.
The advance? 3*2x+x=1000 or x=1000/7 furlongs and now Hajiib has a stash of 3000 figs 1000(1/9+1/7) furlongs into the desert.
Repeat with 3000 figs. 2 round trips and 1 one-way eating 1000 figs. Now Hajiib has 2000 figs 1000(1/9+1/7+1/5) furlongs into the desert
Once more 2000 figs. 1 round trip and 1 one way eating 1000 figs. Finally Hajiib has 1000 figs 1000(1/9+1/7+1/5+1/3) furlongs into the desert.
This is 248/315 of the total way across, leaving 67/315 to cover in a single one-way trip. 67/315 of 1000 furlongs is 213 furlongs leaving Hajiib 787 figs to vend on the far side.
The other problem is to get Hajiib back home with a minimum number of figs.
Rap
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