- 11 Feb '07 02:29You are in a car that will use 10L of fuel to travel round a racetrack. Prove or disprove that there is a starting point that will enable you to travel the entire track for every possible distribution of fuel around the track. Only the exact amount of fuel required will be present and the car starts with no fuel in the tank (so you will have 0 fuel at the end).
- 11 Feb '07 16:34

Do you mean*Originally posted by XanthosNZ***You are in a car that will use 10L of fuel to travel round a racetrack. Prove or disprove that there is a starting point that will enable you to travel the entire track for every possible distribution of fuel around the track. Only the exact amount of fuel required will be present and the car starts with no fuel in the tank (so you will have 0 fuel at the end).**

Prove or disprove that for every possible distribution of fuel around the track there is a starting point that will enable you to travel the entire track. - 11 Feb '07 19:28

So all you have to do is have the fuel in a small ditch dug around the track and it's all the same level so there will be an even level all around the track. With a scoop sweeping up just enough fuel to start and keep the vehicle running it wouldn't matter where on the track the vehicle starts and where on the track the fuel is stored since its equal all the way round.*Originally posted by XanthosNZ***You are in a car that will use 10L of fuel to travel round a racetrack. Prove or disprove that there is a starting point that will enable you to travel the entire track for every possible distribution of fuel around the track. Only the exact amount of fuel required will be present and the car starts with no fuel in the tank (so you will have 0 fuel at the end).** - 11 Feb '07 20:49

why would i want to do that? that is not fun like most puzzles, it is work*Originally posted by XanthosNZ* - 12 Feb '07 01:03

Firstly, this is a thought puzzle, there is no need to imagine a system for picking up the fuel. Secondly, I'm asking for a proof/disproof that there is always a starting point that will allow a full rotation of the track no matter what distribution the fuel is in.*Originally posted by sonhouse***So all you have to do is have the fuel in a small ditch dug around the track and it's all the same level so there will be an even level all around the track. With a scoop sweeping up just enough fuel to start and keep the vehicle running it wouldn't matter where on the track the vehicle starts and where on the track the fuel is stored since its equal all the way round.**

For example, imagine 8L is at point A and 2L is at point B (0.5 rotations away). Obviously the car must start at either A or B (anywhere else has no fuel) and obviously it must be A as from B he would be unable to reach A.

So, either (a) Construct a distribution of fuel such that no full circuit is possible OR (b) Define a method by which for every distribution of fuel you can find a working starting point. - 12 Feb '07 01:22 / 1 edit

I think it's safe to assume there is no internal fuel storage on the car otherwise it could just pick up every fuel cannister and add it to it's internal store. So with no on-board storage 1L=1/10 rotation. If we assume a circular track, 1L=36 degrees. If there is no onboard storage and the cannisters were all 1L spaced 3 degrees apart and you had to discard the container even though it had fuel and go to the next, you would only go 66 degrees and run out of fuel. If you had on-board storage, you lose no fuel and it would not matter where the fuel was placed. So which is it for this problem?*Originally posted by XanthosNZ***Firstly, this is a thought puzzle, there is no need to imagine a system for picking up the fuel. Secondly, I'm asking for a proof/disproof that there is always a starting point that will allow a full rotation of the track no matter what distribution the fuel is in.**

For example, imagine 8L is at point A and 2L is at point B (0.5 rotations away). Obviousl ...[text shortened]... Define a method by which for every distribution of fuel you can find a working starting point. - 12 Feb '07 01:23

The car can hold as much fuel as it needs to.*Originally posted by sonhouse***I think it's safe to assume there is no internal fuel storage on the car otherwise it could just pick up every fuel cannister and add it to it's internal store. So with no on-board storage 1L=1/10 rotation. If we assume a circular track, thats 1L=36 degrees. If there is no onboard storage then if the cannisters were all 1L and spaced 3 degrees or so apart t ...[text shortened]... d storage, it wouldn't matter since no fuel would be wasted. So which is it for this problem?** - 12 Feb '07 01:39 / 1 edit

Under those conditions, an onboard fuel tank which just gets portions from a cannister the driver picks up all the way, it seems to me you can find a starting point for which you cannot make it all the way around and a starting point for which you can make it all the way round, it all depends on the distribution. For instance, if the fuel was bunched up together 3 degrees apart and you start at the last fuel cannister you would never make it all the way round. If you started from the second fuel can on to the end you would not make it around.*Originally posted by XanthosNZ***The car can hold as much fuel as it needs to.**

So you have to start with the first container in line. If you had 1L cans every 36 degrees it would not matter at which can you started and it would be like my example of the ditch around the track so you would get around the track from every starting point, starting points presumable being where each can is located. I think because of those two examples it is always possible to find at least one location where you could make it around the track with any distribution of fuel.

A little harder to picture is uneven distributions with a system like 1,2,3 amd 4 L containers spaced unevenly, that might set up a condition where you cannot get all the way round. Like 1L at zero degrees and 2 L at 72 degrees, 3L at 144 degrees and 4 L at 324 degrees there may not be any point at which you can get all the way around. Those last numbers were just examples, I did not work it out. - 12 Feb '07 02:16

Stop rambling. Either solve the problem or get off the pot.*Originally posted by sonhouse***Under those conditions, an onboard fuel tank which just gets portions from a cannister the driver picks up all the way, it seems to me you can find a starting point for which you cannot make it all the way around and a starting point for which you can make it all the way round, it all depends on the distribution. For instance, if the fuel was bunched up tog ...[text shortened]... ch you can get all the way around. Those last numbers were just examples, I did not work it out.** - 12 Feb '07 04:09 / 5 edits
*Originally posted by XanthosNZ* - 12 Feb '07 04:10 / 4 editsHere's my stab:

Let the fuel distribution around the track be g(t), and the rate of fuel consumption be R. The mileage the car gets out of the amount of fuel picked up between any two points (n-1) and (n) is:

int(g(t)/R)dt from (t(n-1)..t(n))

And the distance actually traveled is:

delta L = t(n)-t(n-1)

Since the car must get more mileage out of its fuel than the distance it needs to travel (otherwise it runs out of gas), we have:

int(g(t)/R)dt from (t(n-1)..t(n)) >= t(n)-t(n-1)

Starting from t0, and breaking the trip up into "n" trips, we have

int(g(t)/R)dt from (t(0)..t(1)) + int(g(t)/R)dt from (t(1)..t(2)) + .. + int(g(t)/R)dt from (t(n-1)..t(n)) >= t(1)-t(0) + t(2)-(t1) + .. t(n)-t(n-1)

This simplifies to:

int(g(t)/R)dt from (t(0)..t(n)) >= t(n)-t(0)

int(g(t))dt from (t(0)..t(n)) >= R*(t(n)-t(0))

The left side of this equation is the total amount of fuel picked up, which is 10L, and the right side is the amount of fuel required, which is also equal to 10L.

Now, the assumption that there is no starting point "t(A)" such that int(g(t)/R)dt from (t(A)..t(A+1)) >= t(A+1)-t(A) cannot be true because we would have:

int(g(t)/R)dt from (t(A)..t(A+1)) < = t(A+1)-t(A)

int(g(t)/R)dt from (t(A+1)..t(A+2)) < = t(A+2)-t(A+1)

etc..

int(g(t)/R)dt from (t(n-1)..t(n)) < = t(n)-t(n-1)

Summing these sequences and simplifying we get:

int(g(t))dt from (t(A)..t(n)) < = R*(t(n)-t(A))

Which means that there is not enough fuel to make the trip, contrary to the problem statement. This contradiction applies to each set of 2, 3, ..n consecutive micro-trips as follows (using sets of 2 as an example):

int(g(t)/R)dt from (t(A)..t(A+2)) < = t(A+2)-t(A)

int(g(t)/R)dt from (t(A+1)..t(A+3)) < = t(A+3)-t(A+1)

int(g(t)/R)dt from (t(A+2)..t(A+4)) < = t(A+4)-t(A+2)

etc..

int(g(t)/R)dt from (t(n-2)..t(n)) < = t(n)-t(n-2)

Summing these sequences and simplifying, we get:

int(g(t))dt from (t(A)..t(n)) < = t(n)-t(A)

which we have already established as being contrary to the problem statement. Therefore, the assumption that there is no set of consecutive micro-trips which can be traveled without running out of fuel must be false, and therefore car must always be able to make the trip given the judicious selection of a starting location.