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Checklist VsOSh AI 2026 Final Stage, Tour 1 (theory) · D task

Laser

Russian title: Лазер

Find the smallest size of a non-zero convolution kernel that annihilates every possible "heat map" produced by laser beams reflecting in a mirrored rectangular chamber.

  • Combinatorics / linear algebra (convolution kernels)
  • Russian original · English translation

The task

The floor of a mirrored chamber is a 1000 × 3000 grid with temperature sensors at the 1001 · 3001 grid nodes. 300 devices sit at cell centres; each emits one beam along a diagonal of its cell and absorbs beams arriving from the opposite node. Beams reflect from the walls until absorbed; every sensor a beam passes raises its reading by 1 °C (by 2 °C where the beam reflects). The final readings form a 1001 × 3001 heat map A.

An s × s kernel K is called determining if the convolution B = A ⋆ K is identically zero for every possible heat map A while K itself is not zero. Find the smallest s for which a determining kernel exists.

Abridged and translated by SOTA from the official Russian materials. The official statement has the exact rules, and it wins wherever this summary differs.

In English

This task was published in Russian. SOTA translated its 4 files into English on 17 September 2026.

Read the task statement in English 699 words

Laser

English translation by SOTA – AI Community of the Russian original. Organisers who would like this translation removed can email [email protected].

All-Russian School Olympiad in Informatics 2025–2026, Final Stage
“Artificial Intelligence” profile, Tour 1, Moscow, 23 March 2026
Task D

The floor of a rectangular chamber with mirrored walls has the shape of a 1000×30001000 \times 3000 grid rectangle. Temperature sensors are placed at the grid nodes (100130011001 \cdot 3001 sensors in total). Initially, all sensors show a temperature of 0C0^\circ\mathrm{C}.

At the centres of some cells there is one device each that emits laser beams. Each device emits a laser beam along a diagonal of the cell at whose centre it is located (that is, in one of the four possible directions, towards one of the nodes of that cell). The same device absorbs beams arriving from the opposite node. The device does not affect beams travelling in the other three directions.

In total, 300 devices are installed, each emitting one beam. Each beam is reflected specularly from the walls (on hitting a corner, the beam is reflected back in the opposite direction) until it is absorbed by one of the devices.

A laser beam, on "visiting" a temperature sensor, increases its reading by 1C1^\circ\mathrm{C}. If the beam is reflected at the point where a sensor is located, it increases the reading of that sensor by 2C2^\circ\mathrm{C} at once. We call the heat map the matrix AA of size 1001×30011001 \times 3001 whose elements are equal to the final readings of the corresponding temperature sensors.

To check that the sensor readings are correct, the values obtained are analysed as follows. A square matrix KK of size s×ss \times s, called a kernel of size ss, is chosen, and then the convolution of the matrix AA with the kernel KK is computed, that is, the matrix B=AKB = A \star K of size (1002s)×(3002s)(1002 - s) \times (3002 - s), where

B[i][j]=u=0s1v=0s1A[i+u][j+v]K[u][v],0i1001s,0j3001s.B[i][j] = \sum_{u=0}^{s-1} \sum_{v=0}^{s-1} A[i+u][j+v]\,K[u][v], \qquad 0 \leq i \leq 1001 - s, \quad 0 \leq j \leq 3001 - s.

We call a kernel KK determining if, for any heat map AA, all elements of the matrix BB are equal to zero, while not all elements of the kernel KK are equal to 0. For what smallest value of ss does a determining kernel of size ss exist?

The figure below shows how a launched beam changes the temperature at the sensors until it is absorbed.

[Figure: see the original statement (page 2 of the PDF). Two panels, each titled "Chamber of 4×12 cells, T=24", with the legend "trajectory", "reflections", "device"; the axes are x (0 to 12) and y (0 to 4), and each node on the beam's path is labelled with its final reading. Left panel: the device at the centre of the cell (11.5, 0.5) emits its beam towards the node (11, 0). The beam travels (11, 0) → (7, 4) → (3, 0) → (0, 3) → (1, 4) → (5, 0) → (9, 4) → (12, 1) and then returns to the device, which absorbs it. The reflection nodes (11, 0), (7, 4), (3, 0), (0, 3), (1, 4), (5, 0), (9, 4), (12, 1) read 2; the nodes (4, 1) and (8, 3), where the path crosses itself, read 2; every other node on the path reads 1. Right panel: the device at the centre of the cell (11.5, 0.5) emits its beam towards the node (11, 1). The beam travels (11, 1) → (8, 4) → (4, 0) → (0, 4); the corner (0, 4) sends it back along the same path to (11, 1); it passes through the device's cell to the corner (12, 0), is reflected back and is absorbed by the device. The corners (0, 4) and (12, 0) read 2, the reflection nodes (8, 4) and (4, 0) read 4, and every other node on the path reads 2.]

Translated by SOTA. The Russian original is the official version and wins wherever the two differ. The two example figures are described in words in the statement; see page 2 of the original PDF for the images. If you organise this olympiad and would like the translation removed, email [email protected] and we will take it down.

At a glance

You get
Statement with a figure showing one beam's effect on the sensors.
You submit
Written solution (proof or answer with justification), graded by the criteria in the official solutions file.
Scoring
Points awarded by the published criteria (partial credit for stated intermediate results; deductions for gaps).
Rules
  • Written tour; no computer-based submission.
Format
Final stage, Tour 1 (theoretical), Moscow, 23 March 2026; grades 9–11; individual written tour.

Details

Year
2026, Moscow, Russia
Round
Final Stage, Tour 1 (theory) · D task
Language
Russian; English translation by SOTA
License
Not stated by the source