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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 puzzle:

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? enter image description here

The solution:

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) enter image description here and find that Rc = (1 + sqrt(3) + sqrt(2))/2 R.

So can someone reformulate it in pure logic-based way?


My attempt:

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:

  1. 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.

  2. 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.

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  • $\begingroup$ I think a more accurate analogy is to have a queen producing blind ants at A and a silent ant eater at B. Random motion will ensure the ants will diffuse away from one another while the concentration will stay 0 at B so a "current" of ants will flow there. Diffusion, however, may not make for an easier to understand (less technical)puzzle! (non-technical explanation) Electrons don't take just the shortest path because to some extent they repel each other and, therefore, the optimum arrangment is more complicated. Ants would just take the two shortest paths to increase path width. $\endgroup$ – kaine Jun 4 '14 at 13:28
  • $\begingroup$ @kaine, thank you. Yes, one can ask about speed rather than time. But the issues with throughput would be still there, because in ants analogy there is not potential, which can be distributed between ants and paths... $\endgroup$ – klm123 Jun 5 '14 at 14:02
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    $\begingroup$ What if instead of ants one used electric robots, which needed to receive power from rails in the tunnels; each tunnel's rails are powered from the ends, but will overheat if two robots try to draw power simultaneously from the same rails. $\endgroup$ – supercat Nov 8 '14 at 17:19
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    $\begingroup$ I think your attempt is still physics, by the way. The equivalent mathematical problem is to ask for the fractal dimension of your network. $\endgroup$ – VicAche May 1 '15 at 10:46
  • $\begingroup$ I believe that trying to formulate the question by introducing 'parallel' and 'series' resistor configurations is a bad idea, because it's not possible to match the puzzle network into the two concepts, so it is literally not possible to solve, unless I've missed something. $\endgroup$ – dram Jun 6 '18 at 3:34
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If the person you are asking doesn't know physics, you may as well teach him. If you are not allowed to do this directly, you could use your own terminology. (Brackets indicate actual terminology)

Every point on the lines has a hidden magic value (potential). The difference between the magic values of A and B is called the net magic value (net potential difference). We release any army of robots (current) at Point A. Robots travel incredibly fast (speed of an electron), but we must transfer an almost infinite number of robots (number of electrons), so it could still take time.

A single 'path' has no intersections. One rule is that the difference between two magic values is proportional to the rate of transfer of robots through a path (Ohm's Law). Another rule is the length of a path is proportional to the rate of transfer of robots through it, and inversely proportional to the difference between the initial and final magic values.

Robots learn over time, which the fastest routes are, and eventually arrive at the optimal speed for robotic transfer. Find the ratio between the net magic value and the rate of robotic transfer through the frame.

This makes it even more difficult, because the concept of resistance is not even mentioned. One needs to first define resistance, then derive formulae for serial and parallel combinations, and then solve the puzzle. This makes it possible but almost impossible.

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  • $\begingroup$ That's correct. But... 1. The concept of potential itself is 66% of the solution), when you introduce it you explain solution. In this case resistance and rules of addition for them is a better concept. 2. The puzzle formulation would be too complex to be interesting enough. The task is to use natual-wel known concepts and don't create artificial (like physics books does). Of course all differences between natural and artificial here is subjective, but I think they quite understandable by most people. $\endgroup$ – klm123 Jul 11 '15 at 18:12
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What about Kirchoff's laws?

You have a system of magical water pipes as shown. There's a flow rate associated with each segment of pipe, where segment means not interrupted by junctions.

You start pumping water into A and, as expected, it comes out from B. After a while, the water flow is now stable, which means that it has:

  1. Junction flow stability: Except A and B, the rate of water flowing into any junction must be equal to that of water flowing out. The rate of flowing out from A must be equal to that of flowing into B, which means you pump in a certain amount of water, and get the same amount back. Otherwise either water or vaccum will build up, and your pipes will not be happy.
  2. Energy stability: Pick any two points p, q on the graph, and walk along the pipes from p to q. For each segment you go through, the energy change is the flow rate times the pipe length you go. If you go with the flow, you gain energy as water pushes you; go against, you lose energy as water holds you back. The total gain or loss of energy must be the same no matter what path you pick, as long as the starting and ending points are the same. Otherwise the water on the lower energy gain/higher energy loss path will feel unfair.

You measure and know that the water pump-in rate is 1. What is the amount of energy you gain by going from A to B? Note that according to rule of energy stability, it does not matter which path you follow.

Maybe you can re-formulate the 2nd rule as 'the net energy change around a loop is zero' because 'otherwise you get free energy which, of course, comes from your precious magical powers and is a very wasteful use of it'

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