| The Trip around the Moon |
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| Method 1: So I started by looking at dropping a can of fuel at location 1 (L1), where L0 is the home base. It takes two units of fuel (one in the tank and one to drop). You go two spaces, drop it, turn around, and go back to the base. The way I wrote this on my paper was
P1 = D1, 0 = 2 which means �to place a can of fuel at L1� (P1), "you deposit the can at L1" (D1), "and return to the base" (0). The initial fill up is assumed, so that whole process takes two cans of fuel. Next, I tried it for L2. It took me a minute to realize that I can leave more than one can of fuel at any given location. Once I saw that important fact, the rest came pretty simply. P2 = P1, P1, [R1, D2, R1, 0] = 6 The brackets are shown to simplify the calculations. Each P1 takes 2 units, and then bracketed actions, setting one at L2 (and remembering to include the initial fueling), take an additional two units of fuel. Notice that both units that were left at location one were used up in the process (the refueling at L1, denoted as R1, occurs twice), so the only fuel existing outside of home base when the car returns is sitting at L2. In the same manner, P3 = P2, P2, [R2, D3, R2, 0] = 14 P4 = P3, P3, P1, P1, [R1, R3, D4, R3, R1, 0] = 34 By symmetry, P1 = P9, P2 = P8, P3 = P7, and P4 = P6. Now for the actual solution � To make it all the way around, cans of fuel must be left at locations 2, 4, 6, and 8. This takes 6 + 34 + 34 + 6 = 80 units to do, and then counting the unit of fuel in the gas tank to initiate the trip, you end up with 81 units of fuel required to perform the round trip. |