Reasoning ...
OK. Each point in the straight line journey must be traversed an odd number of times. Thus if you could carry enough gas to get across, you would traverse each point only once.
Logically then, to minimize the total distance driven you want to maximize those points traversed only once, since this is the most efficient. Once this is done, you want to maximize those points that are traversed 3 times; then maximize those points traversed 5 times, etc.
Now, if you can get to the middle and have a full tank, you can get across, and you've maximized the distance traversed only once.
So how do you get to the middle and have a full tank? Well, if you can reach a point A, which is 1/6 (= 1/2 * 1/3) of the way before the middle (1/2 - 1/6 = 1/3 of the way across) and if you can have 2 loads of gas there, then you can get to the middle with 1 tank. Like this:
Travel from point A to middle, use 1/3 tank of gas
Deposit 1/3 yank in the middle
Use remaining 1/3 tank to get back to Point A.
Travel from point A to middle, use 1/3 tank of gas
You have 2/3 tank, added to the 1/3 you left beofre is 2 tanks.
OK so far?
SO how do you get to point A with 2 tanks? Well if you can reach point B which is 1/10 (= 1/2 * 1/5) of the way before point A (1/3 - 1/10 = 7/30 of the way across) with 3 tanks, then you can get to point A with 2 tanks. Like this.
Travel from point B to point A, use 1/5 tank;
Deposit 3/5 tank at point A;
Return to point B using the remaining 1/5 tank;
Travel from point B to point A a second time, use 1/5 tank;
Deposit 3/5 tank at point A;
Return to point B using the remaining 1/5 tank;
Travel from point B to point A a third time, use 1/5 tank;
You arrive at point A with 4/5 tank, in addition to the 6/5 tank (2 * 3/5) that you've previously deposited, making 2 full tanks at point A.
So how do you get to point B with 3 tanks? Well, if you can reach point C which is 1/14 (= 1/2 * 1/7) of the way before point B (7/30 - 1/14 = 17/105 of the way across) with 4 tanks, then you can get to point B with 3 tanks. Like this:
Travel from C to B, use 1/7 tank;
Deposit 5/7 tank at B;
Return to C using the remaining 1/7 tank.
Travel from C to B a second time, use 1/7 tank;
Deposit 5/7 tank at B; (total left = 2 * 5/7 = 10/7);
Return to C using the remaining 1/7 tank.
Travel from C to B a third time, use 1/7 tank;
Deposit 5/7 tank at B; (total left = 3 * 5/7 = 15/7)
Return to C using the remaining 1/7 tank.
Travel from C to B a fourth time, use 1/7 tank;
You have 6/7 tank, added to the 15/7 tank is 21/7 or 3 full tanks.
So how do you get to point C with 4 tanks? Well, if you can reach point D which is 1/18 (= 1/2 * 1/9) of the way before point C (17/105 - 1/18 = 67/642 of the way across) with 5 tanks, then you can get to point C with 4 tanks. Like this:
Etc etc etc ....
Thid looks like an infinite series, but you can cut it after a couple of more iterations. The answer is about 7 + tankfulls.
Did that help? (I hope I didn't make any arithmetic errors in this .. basically the method is correct).
Works for me. You can actually go 1.01+ times the length of the desert with 8 loads.
Pjnbarb, I remember seeing an answer similar to yours somewhere. I would say you've got the optimal solution. Well done.
Try, I'm in training as we speak old mate. This year I'm goin' for gold.
If I'm successful I promise to retire and take up a career in pure math riddles, OK?
"This year I'm goin' for gold."
Adrian, fair dinkum mate we should all have goals.
"If I'm successful I promise to retireĀ
"
I am betting on you, but personnel reckon you retired long ago. :wink:
