OPhO 2026, Единствен кръг — Задача 11. Diffractive Cryptography
2026 Online Physics Olympiad · 21 август 2026 г.
Last updated
Last updated: 22nd August 03:35 UTC
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Corrections (all times are in UTC)
- Q8: Cleared up wording (21/08 04:40 and 04:52)
- Q3: improved wording (21/08 08:49)
- Q19: Improved wording (21/08 09:42)
- Q16: Attempts reset (21/08 22:00), Error bound increased (22/8 02:30)
- Q1: Improved wording (22/08 03:35)
Указания
Format. The OPhO is a team-based competition (3 members at most per team). Unlike previous years, the 2026 OPhO consists of a single round held from August 21, 12 am (UTC), to August 23, 11:59 pm (UTC). The contest is made up of three parts, all running concurrently over the full contest window:
- Short Answer — problems requiring numerical answers, submitted on the OPhO portal;
- Long Answer — problems requiring full written solutions, submitted on the OPhO portal;
- Simulation — computational problems requiring teams to run a provided executable locally, with solutions submitted on the OPhO portal.
All teams have access to all three parts for the entire duration of the contest. Submissions are made on the OPhO portal https://opho.physoly.org/ using the team login, and may be made at any time within the contest window.
Short Answer. Teams submit numerical answers on the portal, with up to three attempts per problem. Every answer submission must be accompanied by a file upload, on the portal, of the working used to obtain that answer. This working will not be graded, but submissions without it will not be accepted. The base score for each problem depends on the number of teams that successfully answer it, each incorrect attempt multiplies the base score by a factor of 0.8, and there is no time factor. The score for an individual problem follows:
where represents whether a team got it on the first, second, or third try, represents the number of teams that answer the problem correctly, and is the sum of across all teams that have solved the problem. The base score is independent of problem number, but expect questions to be ordered roughly by difficulty. A team’s raw Short Answer total is the sum of these values, and all raw totals are then scaled linearly such that the top team receives 55 points.
Long Answer. Teams submit full solutions on the portal. Rubrics are predetermined, and solutions will be reviewed to determine the proper number of points to give. The points breakdown is given per problem, for a total of 35 points. It is recommended that participants write their solutions in LaTeX. Handwritten solutions (or a combination of both) are accepted too; if participants have more than one photo of a handwritten solution (jpg, png, etc.), it is required that they be organized in the correct order in a pdf before submitting.
Simulation. Teams will need to be able to run a provided executable locally. Solutions are submitted on the portal and will be graded against predetermined rubrics. The points breakdown is given per problem, for a total of 10 points.
Scoring. A team’s cumulative score is the sum of its scaled Short Answer score (out of 55 points), Long Answer score (out of 35 points), and Simulation score (out of 10 points), for a maximum score of 100 points.
Permitted Resources. Use of the internet is allowed, including textbooks, published papers, calculators, and computer algebra systems. However, the use of AI tools of any kind is strictly prohibited. This includes, but is not limited to, large language models (e.g. ChatGPT, Claude, Gemini), AI coding assistants, AI-generated search summaries, and any other AI-powered assistance.
Dishonest Behavior. Please refrain from cheating in any form. This includes, but is not limited to, creating alternate accounts to gain additional attempts, sharing answers with other teams, posting problems online, utilizing AI tools as described above, or employing any other method to gain an unfair advantage.
We have previously caught numerous teams engaging in such practices, and pursuing dishonest tactics provides no genuine sense of security. Any teams found to be acting maliciously will be notified via email, along with the opportunity to appeal. Failing to provide a valid appeal will result in a ban from participating in any future Physoly competitions. Our utmost priority is to ensure an enjoyable contest experience for all participants, and we remain committed to preventing any disruption to the fun for others.
Awards and Results. The top competitors will be recognized on the website. No cash prizes will be awarded this year.
List of Constants
- Proton mass,
- Neutron mass,
- Electron mass,
- Avogadro’s constant,
- Universal gas constant,
- Boltzmann’s constant,
- Electron charge magnitude,
- 1 electron volt,
- Speed of light,
- Universal Gravitational constant,
- Solar Mass
- Acceleration due to gravity,
- 1 unified atomic mass unit,
- Planck’s constant,
- Permittivity of free space,
- Coulomb’s law constant,
- Permeability of free space,
- Magnetic constant,
- 1 atmospheric pressure,
- Wien’s displacement constant,
- Stefan-Boltzmann constant,
Условие
The five images below shows five diffraction patterns. The central image is a diffraction pattern from an aperture which resembles one mathematical binary operation , while the rest represents the English alphabet. The possible alphabetical apertures are the following letters, with the corresponding assigned digits: ,, ..., . The font of the aperture is sans serif, that is, written such that the edges of the letter is a simple straight edge. To encrypt the message, the corresponding digits of each apertures are concatenated, then operated by the binary operation. Determine the result of the encrypted message up to three significant figures. (e.g. if the apertures are AC×KM, it represents ).
Do not solve this by using external tools to apply the Fourier Transform.


Отговори
0.855
KQ/MG = 1117/1307
Решение
For the first and fourth diffraction pattern, we eliminate any curvy characters, leaving A,H,K,M, and X. Since there are four straight fringes, we must have four edges/slits in the direction perpendicular to the fringes, this leaves only K and M left, since H contains no diagonals, and both A and X doesn’t contain a vertical line. Finally, we notice the lack of a double slit secondary pattern within in the first image , in contrast to the fourth image which shows a clear single+double slit pattern. Hence the first and the fourth aperture is K and M, respectively.
As for the second and the fourth, we observe concentric fringes, therefore we only consider curvy alphabets, being C,G,O, and Q. Since both pattern exhibits an additional perpendicular linear fringes, we have that there must be either a rectangle, or two perpendicular edges in the pattern, eliminating O and C. Finally, the orientation of the second pattern is rotated, which must correspond to Q, while the fifth corresponds to G.
For the central operation, it must be the division symbol, as it is the only character with circular curvature. Therefore, the final result is
Оригинал в Архива: problems.pdf · официални решения: solutions.pdf