The puzzle that is being expressed is a puzzle about
crossing a long distance, given certain "fuel" resources and a limited carrying capacity. Here on Puzzling, we have it as A camel transporting bananas, but there are also other statements of the problem.
Specifically, one possible problem statement is:
You want to travel between two cities which are five miles apart, but you can only carry three cans of fuel at a time. Conveniently, each can of fuel allows you to travel exactly one mile. Both cities have as much fuel as you want, but there's none in between them. You also have friends at the destination city who are willing to help you, but they have the same carrying capacity. How can you get to the other city, spending as little fuel total as possible?
The diagram shows
units of fuel (represented by black dots), and the farthest distance they could currently travel if they used all of those units of fuel to travel right.
In the first version, time is on the horizontal axis, and the two cities are the bottom and top of the image; in the second, the distance is left-to-right, and time flows downwards. I will use the second for my explanation.
At the start,
the left group sets out with three cans of fuel, which I've colored red, blue, and green. With these, their "reach" - the furthest they could potentially go using the fuel they currently have - is the 3-mile mark.
They travel one mile, using up their red can. Now they could theoretically travel two more miles...
...but instead they drop their blue can of fuel. This decreases their 'reach' to the 2-mile marker instead of the 3-mile marker.
And now they use the green unit of fuel to go back. Once they're halfway back, you can see their "reach" cross the blue line - they're at a distance of 0.5 miles, so they could just barely get back to their fuel can. But instead, they continue leftwards, making it back to the original city just as they run out of fuel.
On the other side, a friend does the same thing - starting at the 5-mile mark, they could theoretically reach the 2-mile mark. But instead, they go one unit forward; they drop a can of fuel, redirecting one "potential mile" into physical form; and then they just barely make it back to the city on the right.
Finally, it's time to make the crossing. The first group sets out with three cans of fuel once again, and so their reach is to the three-mile mark.
After using their orange can to travel, they pick up the blue can they left earlier - this increases their reach by 1.
Undeterred, they continue on - first, spending their green and cyan cans, as originally planned. Then, they use the blue can of fuel to get to the 4th mile mark... and just barely reach the black can that the other side left for them, which they can now spend to complete the journey!
For the bonus question, the problem has been modified slightly in that:
the capacity is 4 units instead of 3, the distance is 6 miles instead of 5, and instead of trying simply to get one group across, the two groups must swap sides.
In this version, the two groups drop two cans a mile away from their starting position, instead of one can. The white circles look the same, but you can see that each one has two corresponding "pickup points" later on.
You might think 5⅓ is correct because
that would be the maximum distance for a single group going it alone, with only one "preparation trip": the strategy would be to go forward 1⅓ miles, drop 1⅓ cans, then go back 1⅓ miles. Then you'd fill up again, travel 1⅓ miles forward, refill at the drop point, and continue for 4 more miles.
But, as the diagram shows, there's a better strategy if two groups are trying to do this at the same time - if each of them use one can placed by the other, they can both make a 6-mile journey instead of a 5⅓-mile one.
