Car

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Jayraj
Posts: 15
Joined: Sun Oct 21, 2018 11:06 am UTC

Car

Postby Jayraj » Sat Nov 24, 2018 3:35 am UTC

We are in the NEW and technology has reached its peak.

ALSET, a renowned car company, has decided to advertise for its new Hover Electric Car prototype named ALSET HE-1.

The car can travel both ways, front and back, at the same speed and can reverse and continue at same speed within no time.

Fuel transfer can be done spontaneously once 2 cars are side by side through a lightning bolt.

On a huge perfectly round track, a fully charged car can go half way.

The goal of this exhibition is to showcase the car's capabilities, thus having a car complete the entire track without stopping.

No cars are allowed to stop on the track except for the start/finish point.

You have a number of cars at your disposal and all cars have to refuel at the starting point.

Can you determine the minimum number of cars needed to achieve this and how?

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ThirdParty
Posts: 337
Joined: Wed Sep 19, 2012 3:53 pm UTC
Location: USA

Re: Car

Postby ThirdParty » Mon Nov 26, 2018 3:42 am UTC

Here's one way to do it with 3 cars. Not maximally efficient or anything (probably this could be done without ever using any of the three cars' full tank capacity), but since I'm confident that it can't be done with 2 cars, this solution is good enough.
Spoiler:
t0: A is at 0° with 100% fuel, B is at 0° with 100% fuel, C is at 0° with 100% fuel
t1: A is at 45° with 75% fuel, B is at 45° with 75% fuel, C is at 45° with 75% fuel. C gives 25% fuel to each of A and B.
t2: A is at 90° with 75% fuel, B is at 90° with 75% fuel, C is at 0° with 0% fuel. B gives 25% fuel to A; C refuels.
t3: A is at 135° with 75% fuel, B is at 45° with 25% fuel, C is at 0° with 100% fuel.
t4: A is at 180° with 50% fuel, B is at 0° with 0% fuel, C is at 0° with 100% fuel. B refuels.
t5: A is at 225° with 25% fuel, B is at -45° with 75% fuel, C is at -45° with 75% fuel. C gives 25% fuel to B.
t6: A is at 270° with 0% fuel, B is at -90° with 75% fuel, C is at 0° with 25% fuel. B gives 50% fuel to A; C refuels.
t7: A is at 315° with 25% fuel, B is at -45° with 0% fuel, C is at -45° with 75% fuel. C gives 25% fuel to B.
t8: A is at 360° with 0% fuel, B is at 0° with 0% fuel, C is at 0° with 25% fuel.

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jaap
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Re: Car

Postby jaap » Mon Nov 26, 2018 10:06 am UTC

This puzzle has been discussed here before too, as part of the Crossing the desert thread.

Tirear
Posts: 31
Joined: Fri Feb 05, 2016 5:42 pm UTC

Re: Car

Postby Tirear » Mon Nov 26, 2018 4:47 pm UTC

ThirdParty wrote: 3 cars. Not maximally efficient

Optimization:
Spoiler:
Let's look at the middle trip of Car C. It travels some distance x, transfers enough fuel to Car B to travel x distance, and travels x distance back to refuel. Car B splits the fuel with Car A, letting them each travel an extra x/2 distance. This shortens the third trip of Car C by x/2, and another x/2 on the way home. In total car C travels an extra 2x distance now to avoid travelling x distance later. The most efficient solution is therefore to minimize x. As it turns out, x=0 can work, sending Car B by itself to meet with Car A. Car B reaches -90° at 50% fuel, and splits it with Car A. This gets them to -45°/315°. Car C spends 25% fuel to get there, and they each spend 25% to get to the finish line with no fuel remaining. We started with 3 tanks and refueled from empty twice, so total fuel consumption is 500%.

If early investment is bad, could skimping on the first trip of Car C help? Car C travels x distance, gives each other car enough fuel for x distance each, and returns x distance to refuel. As a result, Car B travels some extra y distance, transfers an extra y distance worth of fuel to Car A, and returns an extra y distance. Those obviously add up to the x distance worth of fuel it got from Car C. In total, Cars B and C travel 8y distance to give Car A enough fuel to travel an extra 4y distance. This means Car B needs to travel 4y less distance to meet up with Car A, and 4y less distance to return. That already recovers the initial investment, and Car C also has savings to consider. So the best solution is to maximize y, which your initial solution already did.


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