Physics Cup – 2018, Problem 1

Problem 1

• Problem 1 with the first hint and the results after one week
• Problem 1 with two hints and the results after two weeks
• Problem 1 with three hints and the results after three weeks
• Problem 1 with four hints and the results after four weeks

To begin with, there were few students who managed to find the exact value for the resistance, and the best solution bonus is distributed between these submissions.
If you are interested how one can find an exact answer, you can continue reading this paragraph, otherwise skip it – you won’t be needing anything mentioned later in this paragraph neither at IPhO nor at EuPhO. Exact solutions are based on a mathematical method which is way beyond IPhO Syllabus: discrete two-dimensional Fourier transform. The basic idea is to consider currents I_nm through resistors (n and m counting rows and columns) or voltages V_nm at nodes as amplitudes of a Fourier series. In that case, the Fourier series yields periodic functions I(x,y) or V(x,y). Kirhoff’s laws would yield in terms of I_nm or V_nm an infinite set of equations, but in terms of I(x,y) or V(x,y) this infinite set of equations is neatly simplified and becomes solvable; what remains to do is to find the Fourier amplitudes based on the known functions of x and y. This works in the same way as any linear integro-differential equation including a convolution can be solved via taking Fourier transform, because Fourier transform of a convolution of two functions is a product of the corresponding Fourier transforms. In the case of a regular square lattice, the mathematical calculations are somewhat simpler than in the case of this essentially honeycomb lattice. What makes a square lattice also easier to analyse is that one can find papers dealing with such lattice, although most of these deal with the case of equal resistances (so that one would still need to carry this procedure over to the case of non-equal resistances).

The best solution awards [bonus factor exp(1/3)] goes to:
Navneel Singhal who found the resistance exactly, all by hand, down to finding all the needed integrals analytically.
Prathyush Poduval who also fond the exact resistance, but used numerical integration at the last step
Siddharth Tiwary who took another approach to finding the exact resistance – by numerical simulations on square lattice; the result is slightly less accurate as the value converges to the exact one fairly slowly with increasing lattice size. Meanwhile, he found (and explained nicely how) good upper and lower bounds using the dissipation minimum theorem.

Apart from these, here are few more solutions which are of interest (and awarded with a bonus factor of 1.1).
Satoshi Yoshida calculated lower bound by converting the honeycomb lattice into a square lattice
Konstantine Gagnidze solution what I had in mind when I made the problem
Dylan Toh proving a relationship between horizontal and vertical nearest-neighbour resistances and calculating lower bond for the vertical nearest-neighbour vertices as the upper bound for horizontal nearest-neighbours
Tóbiás Marozsák short-circuiting more than necessary, but using nicely Delta-connections.

A short summary of the numerical values and typical mistakes. If we cut everything except for three rows of resistors, we obtain 44.911… Ω; almost everyone who submitted a solution has got this part correctly done. If we short-circuit whole horizontal lines, the first above A, and the first below B, we obtain 36.67…Ω. so, the ratio of these two values is 1.22, which is a surprisingly small value! If we short-circuit more than needed, for instance short-circuit also the line passing through B, we obtain 22.459…Ω, which very narrowly satisfies the requirement. The main mistake was short-circuiting some parts, and then cutting (or forgetting about) other parts of the circuit; in that case, the net effect of the procedure cannot be known – was it increasing or reducing the resistance.

And here are the results. Number of fully correct solutions: 21. Names in italic correspond to unofficial participants (they get their deserved speed bonus, but do not advance the count for the next speed bonuses