I am thinking how to reformulate a physical puzzle in a way that it becomes purely logic based (doesn't uses physical lows and can be solved by a person, who knows nothing about physics).
The electric circuit in the image is made from the infinite number of wire-frame-squares. The next square is exactly sqrt(2) smaller than the previous. The resistance of used wire is R per one side of the largest square. What is electrical resistance between A and B?
Solution explores symmetry with respect to the diagonal AB line. Let's imagine one runs current from A to B. Potential of A is 1V, potential of B is 0V. 1. All symmetric points will have the same potential, therefore they can be connected together. 2. If we consider the second diagonal (let's call it CD, so square would be ACBD) we can note that all points on CD line have same way to point A and to point B. That means they all will have 0.5V potential and can be connected of disconnected freely. 3. Now we can make and equivalent circuit (see the picture) and find that Rc = (1 + sqrt(3) + sqrt(2))/2 R.
So can someone reformulate it in pure logic-based way?
Megamind created a micro robotic ants. Once upon a time he asked the ants to move a ton of sand through his anthill (from the entrance A to exit B) as quickly as possible. Anthill is a complex system of passes (see the figure, the side of the big square - 1 m). There are a lot of ants, they are very smart and always able to find the best approach to the problem. However, the throughput of passes is very limited - one ant can't overtake the other one. How long will it take for ants to complete the task? It is known that for simple straight passage from A to B the ants could complete the task in sqrt(2) days?
Here I am trying to simulate series and parallel resistors connections. If two passes with lengths L combined in series configuration it will be a path 2L, which will take twice more time to pass the sand through. If two passes with lengths L combined in parallel configuration it will be a 2 times wider path, which will take twice less time to pass the sand through.
But there are critical issues with this formulation:
I don't know how to express "limited throughput" strictly and naturally. If ants are points (case A) they can be as close as needed to each other and basically became one point and use only one path. If ants have some length (case B), then it will take time to fill all paths and free them.
If we consider parallel connection following a series connection of 3 resistors: 1 || 1 + 1 the resistance would be 0.5+1=1.5. Mean while in case (B) ants will wait at the correction, so "resistance" will be 1 + 1 = 2.