Planes and Island...

Wdido · April 28th, 2005, 9:43 pm

Facts:You are in an island with an unlimited number of planes and an unlimited amount of fuelEach plane can travel half way around the world on a full tank of fuelPlanes can only land/take off from the islandYou are required to make a plane go around the world using the least number of planesAssumptions:You cannot crash a plane - all have to return safely to the islandEach plane can refuel another mid-air instanteneouslyFuel weight has no impact on speedQuestion : Minimum planes for one of them to travel the world back to the island...I found 6! Can anyone do better?

Wdido · April 28th, 2005, 10:43 pm

I just got to 5...This is a UBS riddle by the way, if that might increase the number of posts..And I think it is optimal (as we cant do less)

jimmy · April 29th, 2005, 11:17 am

got the job done with 4 planes

jimmy · April 29th, 2005, 11:39 am

actually 3 if you can re-use (and re-fuel) the same plane...

exotiq · April 29th, 2005, 4:18 pm

QuoteOriginally posted by: jimmyactually 3 if you can re-use (and re-fuel) the same plane...Please tell us your solution/technique.I also get 3 if the planes do not have to go at the same speed, but I assumed that they did.

Wdido · April 29th, 2005, 5:00 pm

Here you are.I dont see how u can do it with less planes.The fuel is only on the island and can be carried by the planes.No speed assumptions can help us, we just want to get one plane travel around the world.Can you post your solutions as well?

Wdido · April 29th, 2005, 5:07 pm

Unless u say that planes C and D anre the same that already participated in the first part,that case it is 3, I agree.This is what u mean??I didnt even see that coming, ur completely right!Thanks.

exotiq · April 29th, 2005, 8:24 pm

Here's the method I came up with for 3 planes: assume at full speed it takes a plane 24 hours to go around the world, the island is on the equator and the prime meridian and planes only fly along the equator:1. Three planes leave the island at 0:00, going west, on a full tank of fuel.2. At 45 degrees west, 3:00, plane 3 gives planes 1 and 2 each 1/4 tank of fuel (so now they are full), returns to the island empty at 6:00, refuels and heads out west again at 8:00.3. At 120 degrees west, 8:00, plane 2 gives plane 1 exactly 1/3 tank of fuel (so 1 is now full), and then plane 1 heads back east.4. At 60 degrees west, 12:00, plane 3 gives plane 2 exactly 1/3 tank of fuel (just in time), so both 2 and 3 make it back to the island empty at 16:00. Plane 2 refuels and immediately heads east.5. Plane 1 makes it to 300 degrees west (60 degrees east) at 20:00, just in time for plane 2 to give it 1/3 of fuel for both of them to make it back by midnight.

Wdido · April 29th, 2005, 11:03 pm

I trust you for the calculations My solution is also possible, I ve also had other possible solutions,But we never thought of using the same planes to go east, hence my error...Thanks a lot.

exotiq · April 30th, 2005, 1:44 am

I'd like to prove that 3 is the minimum number but don't know if the reason is as simple as this: If there are only two planes, then no matter what the second plane does, it will not be able to get the first plane farther than 90 degrees west having a full tank of gas at that point, because any point beyond that the second plane would not be able to make a round trip by itself, let alone give any fuel. If the first plane has a full tank at 90 degrees west, that will carry it to 90 degrees east, but the second plane would not be able to meet it there with any fuel it could give the first plane...

Aaron · April 30th, 2005, 3:55 pm

Brilliant solution, and I agree that two is impossible. Even if the second plane were infinitely fast (or if the first plane could hover or float indefinitely without using fuel) all it could do is get the first plane infitestimally close to the 90-degree point with a full tank of fuel; and it can only meet it infitestimally beyond the 270 degree point. If the planes could fly 180.0001 degrees on a tank of fuel, you could consider a solution like this. Even then there would be the question of whether it's fair to assume different speeds for the planes.

exotiq · April 30th, 2005, 4:26 pm

Also, and I am sure this is absurd for to what the brainteaser's author intended, all of these solutions assume that a constant amount of fuel is burned per unit of distance travelled. For example, even though fuel weight may have no impact on speed, more fuel could be consumed when there is more fuel in the tank than when it is near empty, or similarly, the plane could use more fuel near the island, where it is accelerating and slowing down, and less on the other side, when it could become like a glider and coast on relatively little fuel.

hanzotutu · May 6th, 2005, 6:32 am

QuoteOriginally posted by: exotiqHere's the method I came up with for 3 planes: assume at full speed it takes a plane 24 hours to go around the world, the island is on the equator and the prime meridian and planes only fly along the equator:1. Three planes leave the island at 0:00, going west, on a full tank of fuel.2. At 45 degrees west, 3:00, plane 3 gives planes 1 and 2 each 1/4 tank of fuel (so now they are full), returns to the island empty at 6:00, refuels and heads out west again at 8:00.3. At 120 degrees west, 8:00, plane 2 gives plane 1 exactly 1/3 tank of fuel (so 1 is now full), and then plane 1 heads back east.4. At 60 degrees west, 12:00, plane 3 gives plane 2 exactly 1/3 tank of fuel (just in time), so both 2 and 3 make it back to the island empty at 16:00. Plane 2 refuels and immediately heads east.5. Plane 1 makes it to 300 degrees west (60 degrees east) at 20:00, just in time for plane 2 to give it 1/3 of fuel for both of them to make it back by midnight.from step2 to step3, plane1 and 2 travel 75 degrees (or 5 hrs) west. How could 1/3 tank make plane1 full?

user12 · May 9th, 2005, 12:21 pm

I think we have lost plane 2 around the 75 degree west anyway. If it should fly from 45 to 120 and back to 60 on a 2/3 tank.

adgyboy · May 11th, 2005, 10:41 am

3 planes is possible but your solution wasn't optimalIf three leave together, at one fourth of a pi-rad, one of them gives them each 1/4 of a tank full, leaving the two others full and keeping just enough to go back.Then at 1/2, one of the two gives 1/4 to the other leaving him full, and goes back with his half tankNow the plan can go all the way to 3/2, and we have enough time to get there.The first plane back leaves with a full tank at time 1, the round the planet plane and it meet at 3/2, and they split the fuel with 1/4 each. At that time, the other plane leaves with a full tank. They all meet at 7/4 and share 1/4 of a tank each, and they make it back!!more generally, if n leave at the same time, it is optimal for one to turn back every 1/(n+1).