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[Bogieman]

Hi all. I know I'm a bit late to the thread, but I only just got the card and worked quite hard on it. I thought I'd share my results.

We know the answer is as stated above. But why? Ollie says his solution relies upon the cars waiting in the desert for their companions and donating fuel as the companions make their return journey. We now know, however, that this constitutes a "cache of fuel" (or so say the folks at MC).

Therefore, for anyone who's interested (and because I was quite bored tonight), I've posted a PDF of my own workings. The "spare fuel" shown relates to the amount of fuel in the tank plus spares kept on board each car.

For each step of the journey, I've subtracted 2 fuel points, assuming that a given car spends one getting there and reserves one to get back again. Finally, the arrow underneath each pit-stop represents the redistribution of fuel from the left-most car in the table to all other cars. The left-most car then returns immediately, hence not having to wait around in the desert as a "fuel cache." The final car makes it as far as the final stop before it has no fuel to carry on (but having reserved enough to get back again, of course).

I know this will be old news to most of you, but I'm hoping that putting my own explanation into this format will help someone out there understand the problem better.

EDIT: Had to remove the real answer from the above text... The PDF is a spoiler, obviously, so don't look if you don't want to know.

[Hituro]

No thank you Bogieman, I appreciate your solution. I too couldn't see why the answer Ollie used was any more correct than the other rejected one, so I appreciate a way at arriving at the same answer that makes more sense, thanks!

[GasparLewis]

Well, since I can't read Bogieman's on the account of my Acrobat being bugged out (though it's very likely the same thing), here is mine:

[Daffydil]

OK, maybe I'm just a bit tired but I can't think of what your "cars as caches" solution is BoardBloke

Any chance of an explanation?

[Ariadni]

must admit I think the card could have been worded better... or am I just being pedantic?

for instance, if the drivers cant leave their cars, how do they refuel? they must have extremely long arms!

isn't a one-gallon tank a stupidly small size?

[groovygirl20]

[Mustavagander]

I dont get this I tried working it from the other end and my distance is rather farther. Im open to objection though

[Phant]

I too did it "the Neill way", so as some people seem interested in this I shall do my best to explain.

First of all, Mustavagander's method is correct only if opened cans can be exchanged between cars (and if "car caches" are allowed). However, the wave 0.5 card states only unopened tins may be transfered so the actual answer would be less than this. (In ollie's transcription his second "extra gallon" should read "unopened")

So assuming cars can be used as caches...

[Scott]

I just got this card. Rather late, me. I have a bit of a complaint. I assumed the wording "..fForty miles to the gallon, which is also the capacity of .. their tanks" meant they have 40 gallon tanks, AND 40 miles to the gallon. Which, I can tell you, allows the cars to go much much fFarther. Like 1640 miles.

3 solves used up. *grumble* honestly, who ever heard of a car with a 1 gallon tank?[/spoiler]

[almagest]

No waiting must include no waiting at the start, otherwise it is easy to get over 360 (by having some cars start later to refuel the leader on his way back).

But I still do not see how one proves that 360 is maximal.

Any ideas?

[almagest]

A heuristic argument is that the key to success is to get the total burn rate for fuel down as fast as possible. Initially it is 10 gals/hr with 90 gals available, so all fuel will be burnt in 9 hrs giving a max range of 180, unless the rate is reduced.

The only way to get the rate down is to transfer fuel asap so that the car transferring can retire. The solution gives one car stopping (back at base) every 2 hours and it looks impossible to get the rate down faster than that.

But I still cannot see a proof that 360 is optimal ...

[almagest]

Proving that there must be an optimal solution in which each car makes just one turn and transfers fuel only at the turn, would be a big step forward.

[doctorclark]

I used a little bit different of strategy than the posted solves:

The card states that "the cars can't wait around in the middle of the desert," which would suggest that they can wait around at the start, before the desert begins. Using this logic I came up with the following (sorry, almagest, I couldn't do better than others even using this reasoning):

[almagest]

Doctorclark. It is not hard to go more than 360 if you allow waiting at the start.
For example, A, B, C, D set off together. After 80 miles, D transfers 2 gals to each of A, B, C (leaving him with 2 gals) and returns. After another 80 miles, C transfers 2 gals to each of A, B (leaving him with 4 gals) and returns. After another 80 miles B transfers 2 gals to A (leaving B with 6 gals) and returns. A is now 240 miles out with a full tank. He drives another 160 miles (leaving him with 6 gals) and then turns around (having reached the 400 mile point) and drives 240 miles back, so that he runs out 160 miles from home. So we need to use the other cars to get him 4 gals at that moment. That is easy. They can set off together to reach the 160 mile point at the same time as A. They each have 6 gals left and need 4 gals to get back, so have 2 gals to spare. So we have overkill.
Since we know the answer is 360 we conclude that waiting around at the start is not allowed!

[doctorclark]