It is a supply problem more than a road problem. Each day won seems small, but it forces you to pay in advance for increasingly expensive logistics.
The messenger and the provisions
Riddle statement
Between a messenger's house and his destination there are seven days of travel. At the end of each day there is a house where he can sleep and leave provisions.
The messenger can only carry food for four days.
He must reach the destination and return home without ever running out of food.
What is the minimum number of days of march he needs in total?
Show solution
Solution
Answer: you need 128 days of walking.
Let's call $T(n)$ the minimum number of days of walking necessary to go to a destination located $n$ days away and return to the starting point.
If the destination is only one day away, the answer is clear:
One day to go and another to return.
Now let's think about a destination located $n$ days away. The first house is one day away from departure. If the messenger manages to convert that first house into a new, well-supplied base, then from there he has exactly the same problem, but with a distance of $n-1$ days.
The question is how much it costs to convert that first house into a useful base. So that the remaining trip can be made from there and return, the necessary provisions for the entire $n-1$ day plan must be prepared there. And, furthermore, the messenger must have been able to go back and forth between the initial house and that first house while transporting them.
As he can carry food for four days, each round trip between two consecutive houses consumes two days of food and still allows net provisions to be transferred to the next point. In this way, preparing the next base costs as much as the trip that will later be made from it.
Therefore, by adding one more day of distance, the total cost doubles:
Starting from $T(1)=2$, gets:
Like this that the minimum is 128 days of walking.
The essential idea is that each new house is not just a passing point: it must be prepared as if it were the new starting point. This preparation forces us to repeat, backwards, all the work necessary for the remaining section. That is why growth is not linear, but rather by doubling.