Logic Masters Deutschland e.V.

Puzzle link: Play on SudokuPad.

Rules: Normal Sudoku rules apply.

Each circle is an “emitter” from which a stream of electrons is emitted with “energy” E, where E is the clue in the top left corner. Electrons propagate on a straight path in arrow direction until they reach the first cage, which is a “detector”.

Digits on a path of electrons between emitter and detector represent “potentials” V_n, where n = 1,...,k numbers the cells with 1 being the first and k the last index number. The “output” of cell n, O_n, is calculated from the “input” into cell n, I_n, as follows. I₁ is the digit in the emitter. Then, for each n, if V_n ≤ E, we set O_n = I_n, meaning that all electrons are transmitted through that cell because their energy is above the potential barrier; if V_n > E, we set

O_n = I_n-(V_n-E+2),

meaning that only a fraction of electrons can “tunnel” through that cell. For the next cell's input, I_n+1 = O_n. The output of the last cell, O_k, must occur in the detector.

For example, assume a stream of electrons from an emitter containing digit 6 and an energy of 4 passing through a two-cell path with digits 1 and 5. Thus, we have I₁ = 6 and I₂ = O₁ = 6 because V₁ = 1 ≤ E = 4. Next, we have O₂ = O₁-(5-4+2) = 3 because V₂ = 5 > E = 4, and 3 is the digit in the detector.

Example: The following image provides a fully solved example on a 4x4 grid. You may solve the example for yourself here on SudokuPad.

Your feedback, ratings and comments are highly appreciated. Have fun!

Background: In classical physics, if we throw a ball at a solid wall, it cannot spontaneously occur on the other side (unless the wall is destroyed). In quantum mechanics, however, there is a small probability that this happens and that the particle “tunnels” through to the other side. The wider the barrier (i.e., the “wall”), the lower the probability for tunneling. Also, the higher the barrier (in energy), the lower the transmission.

Quantum tunneling is relevant in many phenomena and applications. The Nobel Prize in Physics 2025 was awarded jointly to John Clarke, Michel H. Devoret and John M. Martinis “for the discovery of macroscopic quantum mechanical tunnelling and energy quantisation in an electric circuit”.

There is a physical reason for the “peculiar” arithmetic rules in the puzzle, including the “+2” term. In the given form, they describe the tunneling probability correctly, even quantitatively, with the following interpretation. Assume that the energies E and V_j are given in eV (electronvolt), the width of one cell is 7.75 nm, and the numbers I and O are decadic-logarithmic electron fluxes. Then, for 10^I incident electrons, between 10^O-1 and 10^O electrons tunnel through a rectangular barrier. In the example from the puzzle rules, out of 10⁶ = 1 million (I = 6) incident electrons, between 100 and 1000 of them (O = 3 for the upper bound) tunnel to the output. One can calculate an exact number, but since Sudoku operates only with integer values, the result is correct to an order of magnitude, i.e., it lies within the given bounds.

These results can be derived by starting with the transmission coefficient T for the quantum tunneling probability in the Wentzel–Kramers–Brillouin (WKB) approximation, which is given by

T = exp{-2d sqrt[2m(V-E)/ℏ²]},

where d is the width of the barrier, m the electron mass, V the height of the potential barrier, E the electron energy, and ℏ the reduced Planck constant. Although the logarithm of T depends on the square root (sqrt) of V-E, it can be shown that an upper bound is given by the linear relation of the puzzle rules using the parameters and scaling factors quoted above and for V-E taking integer values from 1 to 7 eV. In reality, the transmission is not quite 1 even for V_n ≤ E, but this is ignored in the puzzle.