Brain Teaser No 3

quote:

There isn't a squadron of UH-64 Apache attack helicopters lurking perchance?

No helicopters, no bandits, just nice friendly nomads with their goats and camels and Mercedes Benz cars (but you can't use their fuel!!)

Cheers

Don

Don,
My answer to cross the 800 mile desert is 10 loads. Is this correct?
Steve

quote:

Don,
My answer to cross the 800 mile desert is 10 loads. Is this correct?
Steve

We only need to cross the desert one way ie we don't need to come back.

The number of loads can include fractions/decimals eg 7.642 loads (this isn't the right answer) but if somebody said that this would effectively mean 8 FULL loads, I wouldn't quibble (8 full loads isn't the answer either, BTW)

We are also looking for the MINIMUM number of loads required..........

Hope this all helps.

Cheers

Don

Is it 5 loads? (4.8 loads rounded up...)

quote:

Originally posted by munch:
ONE . Regards English. Born in the same Hospital as Mick from the Stones.

And I went to the same school as Brian Jones.

Big World.

English?

Cobb(l)ers!

quote:

Is it 5 loads? (4.8 loads rounded up...)

Not according to my calculations. Sorry.

Cheers

Don

4 loads

Trip 1
100 miles dump 30 litres and return home

Trip 2
100 miles refuel to full
then to 300 miles to dump 10 litres
return to 100 miles and use 10 litres from dump to return home

Trip 3
200 miles, dump 10 litres and return home

Trip 4
100 miles refuel to full tank;
200 miles refuel to full tank;
300 miles refuel to full tank and complete journey

Peter,

That is a dam good effort, very nicely explained, and logistically neat and tidy. Proper Logistics Corps............but

You can do it with less........

Cheers

Don

3 loads.

Load 1 - Travel to 100 miles cache 30 gals and return

Load 2 - Travel to 100 miles cache 30 gals and return.

Total at 100 miles = 60 gals.

Load 3 - Travel to 100 miles, collect 10 gals to fill tank.

Total at 100 miles now 50 gals.

Travel to 200 miles and cache 30 gals.

Return to 100 miles and collect 50 gals. Travel to 200 miles and collect 10 gals. Cache 30 gals at 300 miles

Total now at 300 miles = 30 gals
Total now at 200 miles = 20 gals

Return to 200 miles and collect 20 gals. Stop at 300 miles and collect 30 gals and.......and......run out at 700 miles. Bugger!

I'm off to bed!

Mark

quote:

Stop at 300 miles and collect 30 gals and.......and......run out at 700 miles. Bugger!

but a nice start, sort of getting the old brain into gear........

Cheers

Don

Hint

Ask yourself, how far the truck would be able to travel across the desert if it only had access to two loads of fuel.

Cheers

Don

I would love to see the answer that is less than four.

As long as the two borders are parallel and you can't use the edge of the desert to refuel

I have racked what brains remain in my head and also find 4 loads the solution. I would also love to see the solution of less than 4.
Definitely a good one, it had me scratching notes throughout the day.
As to 2 loads of fuel I can only cover 600 miles...
This will haunt me until I find it.

Thanks Don!

rOb

quote:

As to 2 loads of fuel I can only cover 600 miles...

o'dear.....is that all?????

I agree its a good one.....therefore even more satisfying when you figure it out, which someone will....

Cheers

Don

quote:

As long as the two borders are parallel and you can't use the edge of the desert to refuel

No tricks....the desert is infinely wide, and you need to go straight across for 800 miles....

....and yes, you can definitely do it with less than 4 loads.

Cheers

Don

quote:

Ask yourself, how far the truck would be able to travel across the desert if it only had access to two loads of fuel.

666 miles?

Paul

Don,
Is it simply 1.6 loads? The total number of gallons required to travel the distance, not place fuel caches?

rOb

If we focus on minimizing fuel usage - not minimizing the number of trips, I think of the following approaches.


First an approach which spreads a gallon/10miles cache all over the distance. That means that on the next run you can cover that stretch-with-caches without using up the fuel in the tank(you can choose to put the caches at the end of each stretch but this visualizes nicely).

Once the first 300 miles are filled with constant cache density you can cross the last 500 miles on your own.

trip 1: fill the first stretch(stretch1) 500/3 miles =~166 with caches, and return.

trip 2: using up the caches on stretch1, fill up stretch2 of (500-stretch1)/3 is ~111 miles with cache and return.

trip 3: fill up stretch1 again

trip 4: use up stretch1 and stretch2 to fill up stretch3, witch length (500-stretch1-stretch2)/3 ~ 74

If stretch1,2,3 have cachecs that is enough to bridge the first 300miles.

trip5: fill up stretch1 again

trip6: fill up stretch2 again - which uses up stretch 1.

trip7: fill up stretch1 again

trip8: go for it.

total fuel usage: between 7*50 and 8* 50 gallons = (nearly) 400 gallons. That's a lot, and a lot of it is spent on stretch1 which is crossed 15 times. Worst solution yet.


Learning from this I try to take all the fuel with me. Imagine you only advance a mile at the time, going to and fro to carry all the fuel.

The last 500 miles can be done on one fuel tank.
In order to advance with between one and two fuel tanks you use 3 gallons per 10 miles.(go-return-go)
In order to advance with three loads you use 5 gallons per 10 miles.
In order to advance four loads you use 7 gallons per 10 miles.

We can do the last 500 miles with 50 gallons.
We can bridge the extra 500/3 miles before that if we have two loads at the start of that distance.
we can do an extra 500/5 miles before that if we have three loads to start at the start of this distance.
And for the very first 33.3miles we have to shuttle 4 loads at 7 gallons per 10 miles, or half a tank. The first stretch is only covered 7 times, and it is much shorter as well.

So we can do it with about 174 gallons or 3 and a half loads. It looks like the absolute minimum to me.

There appears to be no limit to the distance that can be covered this way(cost no object) because the sum 1+ 1/3 + 1/5 +1/7 +... does not top out.

Regarding the second part of the puzzle, I calculate that there is a maximum distance that can be covered and it is 1250 miles.

Steve D

quote:

Sorry it can sill be done with one,

reminds me of that old joke:-

Biology teacher: "there are 67 ways in which a man and woman can make love...."

Boy's voice from back of class: "68!"

Biology teacher (trying to ignore the voice): "there are 67 ways in which a man and woman can make love...."

Boy's voice from back of class: "68!"

Biology teacher (again trying to ignore the voice): "there are 67 ways in which a man and woman can make love, first, the 'missionary position', in which the man lies on top of the woman......."

Boy's voice from back of class: "69!"

Cheers

Don

quote:

I calculate that there is a maximum distance that can be covered and it is 1250 miles.

Whilst that might be close to the "practical" limit (depends how you define "practical") it isn't the mathematical limit.......sorry

Cheers

Don

Alexander has effectively cracked it. Well done.

You just need to tidy up a few loose figures (decimal rounding-off sort of thing) and you can show that 3 and 7/15 loads will get you across.

I havem't figured out how you "proove" this is the "minimum", but appart from Munch, I haven't ever found any body who could confidently claim a lower figure.

Cheers

Don

There and back

So how many loads to get there AND back, across an 800-mile wide desert?

Cheers

Don

The initial premise is ambiguous ; those that have said '4' loads have based this on the statement in the premise that we were dealing with 'loads' as a minimum and no part 'loads'.

Our calculations do though show there is fuel not used, left in the desert.

However allowing part loads and finding the minimum is also the same method as finding the maximum possible distance and requires advance maths (integration and differentiation)