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readme report
2026-07-12 21:35:10 -04:00
MODEL: devstral2-123b-iq2
PATH: /home/qrf/.cache/llama.cpp/Devstral-2-123B-Instruct-2512-UD-IQ2_M.gguf
STARTED: 2026-07-11 22:40:34
======================================================================
Q01 [recall]
What is the Kasner metric, and in what context does it arise in general relativity?
ANSWER:
The Kasner metric is an exact solution to Einstein's field equations in general relativity that describes a homogeneous but anisotropic universe. It is a special case of the Bianchi type I metric and is given in its simplest form by:
\[ ds^2 = -dt^2 + t^{2p_1} dx^2 + t^{2p_2} dy^2 + t^{2p_3} dz^2 \]
where \( p_1, p_2, \) and \( p_3 \) are constants (called Kasner exponents) that satisfy the following conditions:
1. \( p_1 + p_2 + p_3 = 1 \)
2. \( p_1^2 + p_2^2 + p_3^2 = 1 \)
### Key Features of the Kasner Metric:
1. **Homogeneity**: The universe looks the same at every spatial point (invariant under spatial translations).
2. **Anisotropy**: The expansion or contraction rates along the three spatial axes (x, y, z) are generally different, depending on the values of \( p_1, p_2, \) and \( p_3 \). This means the universe expands or contracts at different rates in different directions.
3. **Vacuum Solution**: The Kasner metric is a solution to Einstein's equations in empty space (i.e., with no matter or energy-momentum tensor, \( T_{\mu\nu} = 0 \)).
4. **Singularity**: The metric has a singularity at \( t = 0 \), where the spacetime curvature becomes infinite. This is a example of a "big bang" singularity.
5. **Non-isotropic Expansion/Contraction**: Unlike the Friedmann-Lemaître-Robertson-Walker (FLRW) metrics, which describe isotropic universes (the same expansion/contraction in all directions), the Kasner metric describes a universe with different scaling behaviors along different axes.
### Context in General Relativity:
The Kasner metric arises in several important contexts in general relativity:
1. **Study of Anisotropic Universes**: It is a fundamental example of a homogeneous but anisotropic cosmological model. It shows that the universe need not expand or contract isotropically in general relativity.
2. **Initial Singularity**: The Kasner solution is relevant to the study of the initial singularity in the universe. It is often used as a local model for the behavior of the universe near the big bang singularity, especially in the context of the **BKL (Belinsky, Khalatnikov, Lifshitz) conjecture**. This conjecture suggests that near the singularity, the generic behavior of cosmological solutions to Einstein's equations is locally described by the Kasner metric, with the axes of expansion and contraction changing over time in a chaotic manner.
3. **Vacuum Solutions**: As a vacuum solution, it serves as a useful toy model for understanding the behavior of spacetime in the absence of matter.
4. **Bianchi Classification**: The Kasner metric is the simplest example of a Bianchi type I universe, which is one of the classifications of homogeneous but anisotropic cosmological models.
### Physical Interpretation:
- If a Kasner exponent \( p_i \) is positive, the universe expands along the corresponding direction as \( t \) increases.
- If \( p_i \) is negative, the universe contracts along that direction.
- The conditions on the \( p_i \)'s imply that not all exponents can be positive (since \( p_1 + p_2 + p_3 = 1 \) and \( p_1^2 + p_2^2 + p_3^2 = 1 \), at least one \( p_i \) must be negative). Thus, the universe expands along some axes while contracting along others, leading to a "mixmaster" dynamics near the singularity.
### Example:
A simple example of Kasner exponents is \( p_1 = p_2 = \frac{2}{3}, p_3 = -\frac{1}{3} \). This describes a universe that expands along the x and y axes while contracting along the z axis.
### Limitations:
The Kasner metric describes a universe with no matter, no cosmological constant, and no spatial curvature. While it is useful for studying anisotropy and the nature of singularities, it is not a realistic model for the late-time universe, which is well-described by the isotropic FLRW metrics. However, it is important in the study of early-universe physics and the nature of cosmological singularities.
EXPECTED (judge only): Vacuum solution to Einstein's equations describing an anisotropic, homogeneous universe; exponents satisfy sum p_i = sum p_i^2 = 1; arises in BKL analysis of cosmological singularities.
(finish=stop tokens=958 time=288.0s tps=3.4)
======================================================================
Q02 [recall]
In enzyme kinetics, what does the Haldane relationship connect?
ANSWER:
The Haldane relationship (also known as the Haldane equation) in enzyme kinetics connects the kinetic constants of an enzyme-catalyzed reaction. Specifically, it relates the Michaelis-Menten constants (Kₐ, Kₐ', Kₐ'', etc.) for different substrates and products to the equilibrium constant (K_eq) of the reaction.
For a reversible enzyme-catalyzed reaction of the form:
A + B ⇌ P + Q
The Haldane relationship is given by:
K_eq = (k_cat⁻ / K_m⁻) * (K_m⁺ / k_cat⁺)
Where:
- K_eq is the equilibrium constant for the reaction (([P][Q])/([A][B]) at equilibrium).
- k_cat⁺ and k_cat⁻ are the catalytic rate constants (turnover numbers) for the forward and reverse reactions, respectively.
- K_m⁺ is the Michaelis constant for substrate A (or B, depending on the mechanism), and K_m⁻ is the Michaelis constant for product P (or Q).
This relationship is derived from the principle of microscopic reversibility and holds true for enzyme-catalyzed reactions that follow a sequential mechanism (e.g., ordered or random Bi-Bi mechanisms). It provides a way to check the thermodynamic consistency of experimentally determined kinetic parameters.
In essence, the Haldane relationship connects the kinetic parameters of the forward and reverse reactions to the overall thermodynamic equilibrium of the reaction.
EXPECTED (judge only): Relates the equilibrium constant of a reversible enzyme reaction to its kinetic constants: Keq = (Vf*Kmr)/(Vr*Kmf), tying forward/reverse kcat and Km to thermodynamic equilibrium.
(finish=stop tokens=304 time=91.2s tps=3.4)
======================================================================
Q03 [recall]
Who was Hypatia of Alexandria and roughly when did she die?
ANSWER:
Hypatia of Alexandria was a prominent Greek Neoplatonist philosopher, mathematician, and astronomer who lived in Alexandria, Egypt, then part of the Eastern Roman Empire. She was the daughter of the mathematician Theon of Alexandria and is known for her contributions to mathematics, including work on commentaries of algebraic equations, geometry, and astronomy. She also taught philosophy and science in Alexandria, becoming a symbol of learning and intellectual achievement in the late antique period.
Hypatia was murdered in March 415 AD by a mob of Christian extremists, who accused her of paganism and influencing political conflicts between the local governor and the Christian bishop, Cyril of Alexandria. Her death is often seen as a symbol of the decline of pagan scholarship and the rise of Christian influence in the late Roman Empire.
EXPECTED (judge only): Neoplatonist philosopher/mathematician/astronomer in Alexandria; murdered by a Christian mob around 415 CE.
(finish=stop tokens=163 time=49.1s tps=3.4)
======================================================================
Q04 [recall]
What is the Gershgorin circle theorem used for?
ANSWER:
The Gershgorin circle theorem is a tool used in mathematics, specifically in the field of numerical analysis and linear algebra. It is used to determine the region in the complex plane that contains all the eigenvalues of a given square matrix. Here's a step-by-step explanation of how it works:
1. **Purpose**: The theorem helps in estimating the locations of the eigenvalues of a matrix without needing to compute them explicitly.
2. **Definition**: For a given \( n \times n \) matrix \( A \), the Gershgorin circle theorem defines \( n \) circles in the complex plane. Each circle is associated with a row or column of the matrix.
3. **Construction of the Circles**:
- For each row \( i \) of the matrix \( A \), let \( R_i \) be the sum of the absolute values of the off-diagonal elements in row \( i \):
\[ R_i = \sum_{j \neq i} |a_{ij}| \]
- The center of the circle for row \( i \) is the diagonal element \( a_{ii} \).
- The radius of the circle is \( R_i \).
4. **Theorem Statement**: All the eigenvalues of the matrix \( A \) lie within the union of the Gershgorin circles. In other words, every eigenvalue of \( A \) is in at least one of the Gershgorin circles.
5. **Application**: This theorem is particularly useful for:
- Getting a quick estimate of the location of eigenvalues.
- Determining whether a matrix is nonsingular (if none of the Gershgorin circles contains zero, the matrix is nonsingular).
- Getting insights into the stability of systems in control theory and numerical analysis.
6. **Example**: Consider a matrix \( A \) with rows \( [a_{11}, a_{12}, a_{13}] \), \( [a_{21}, a_{22}, a_{23}] \), and \( [a_{31}, a_{32}, a_{33}] \). The Gershgorin circles would be:
- Circle 1: Center at \( a_{11} \) with radius \( |a_{12}| + |a_{13}| \).
- Circle 2: Center at \( a_{22} \) with radius \( |a_{21}| + |a_{23}| \).
- Circle 3: Center at \( a_{33} \) with radius \( |a_{31}| + |a_{32}| \).
7. **Additional Insight**: The theorem can be applied to columns instead of rows, which sometimes gives a tighter bound on the eigenvalue locations.
In summary, the Gershgorin circle theorem is used to bound the possible locations of the eigenvalues of a matrix in the complex plane.
EXPECTED (judge only): Bounds the location of eigenvalues of a square matrix: every eigenvalue lies within at least one Gershgorin disc centered at a diagonal entry with radius equal to the sum of absolute off-diagonal entries in that row.
(finish=stop tokens=606 time=181.1s tps=3.4)
======================================================================
Q05 [recall]
In RF engineering, what is the difference between the noise figure and noise temperature of a device, and how are they related?
ANSWER:
In RF engineering, **noise figure (NF)** and **noise temperature (Tₑ)** are both measures of how much a device degrades the signal-to-noise ratio (SNR), but they express this in different ways. Here’s a step-by-step breakdown of their differences and relationship:
### **1. Noise Figure (NF)**
- **Definition**: The noise figure is a dimensionless ratio (often expressed in dB) that compares the SNR at the input of a device to the SNR at its output.
\[
\text{NF} = \frac{\text{SNR}_{\text{in}}}{\text{SNR}_{\text{out}}}
\]
- If the device were noiseless, NF = 1 (0 dB). Real devices have NF > 1 (>0 dB).
- **Interpretation**: It quantifies how much the device adds noise relative to a perfect noiseless device.
- **Formula in dB**:
\[
\text{NF (dB)} = 10 \log_{10} \left( \frac{\text{Total output noise power}}{\text{Output noise power due to input noise only}} \right)
\]
- **Common Use**: Used in cascaded systems (e.g., receivers) to describe noise performance at room temperature (typically 290 K).
### **2. Noise Temperature (Tₑ)**
- **Definition**: Noise temperature is a temperature (in Kelvin) that describes the amount of noise added by the device. It is based on the concept that a noisy device can be modeled as a noiseless device with an additional noise source at its input at temperature Tₑ.
\[
T_e = \frac{\text{Noise added by the device}}{\text{Boltzmann's constant } (k) \cdot \text{Bandwidth (B)}}
\]
- Boltzmann’s constant \( k = 1.38 \times 10^{-23} \) J/K.
- **Interpretation**: A lower Tₑ means less added noise. For example:
- Tₑ = 0 K (no noise added, ideal case).
- Tₑ = 290 K (device adds as much noise as a resistor at room temperature).
- **Common Use**: Useful in cryogenic systems (e.g., satellite communications, radio astronomy) where actual temperatures are very low.
### **3. Relationship Between NF and Tₑ**
The two are related by the following equation, assuming the input noise is due to a source at standard temperature \( T_0 = 290 \) K (the IEEE standard reference temperature):
\[
\text{NF} = 1 + \frac{T_e}{T_0}
\]
Or in dB:
\[
\text{NF (dB)} = 10 \log_{10} \left( 1 + \frac{T_e}{T_0} \right)
\]
Conversely, noise temperature can be calculated from noise figure:
\[
T_e = T_0 (\text{NF} - 1)
\]
where NF is the linear (not dB) noise figure.
### **Key Differences**
| Parameter | Noise Figure (NF) | Noise Temperature (Tₑ) |
|-----------------|--------------------------------------------|--------------------------------------------|
| **Units** | Dimensionless (or dB) | Kelvin (K) |
| **Reference** | Relative to a 290 K input | Absolute noise temperature |
| **Typical Use** | Room-temperature systems | Low-noise or cryogenic systems |
| **Expression** | \( \text{NF} = 1 + \frac{T_e}{T_0} \) | \( T_e = T_0 (\text{NF} - 1) \) |
### **Example**
- If a device has NF = 2 (3 dB), its noise temperature is:
\[
T_e = 290 \, \text{K} (2 - 1) = 290 \, \text{K}
\]
- If a device has Tₑ = 100 K, its noise figure is:
\[
\text{NF} = 1 + \frac{100}{290} \approx 1.345 \quad (2.27 \, \text{dB})
\]
### **When to Use Each**
- **Noise Figure**: Convenient for comparing devices at room temperature or in cascaded systems using Friis formulas.
- **Noise Temperature**: More intuitive for systems operating at non-standard temperatures (e.g., cooled amplifiers in radio telescopes).
### **Final Answer**
Noise figure (NF) is a ratio (or dB value) that describes how much a device degrades the signal-to-noise ratio, referenced to a 290 K input. Noise temperature (Tₑ) is an absolute temperature (in Kelvin) that represents the amount of noise added by the device. They are related by \( \text{NF} = 1 + \frac{T_e}{290} \) (or \( T_e = 290 (\text{NF} - 1) \)), where 290 K is the standard reference temperature. NF is commonly used in general RF systems, while Tₑ is preferred in low-noise or cryogenic applications.
EXPECTED (judge only): Both quantify added noise; related by F = 1 + Te/T0 (T0=290K), equivalently Te = T0*(F-1). NF is 10log10(F) in dB.
(finish=stop tokens=1144 time=342.7s tps=3.4)
======================================================================
Q06 [recall]
What is the Antikythera mechanism and what did it do?
ANSWER:
The Antikythera mechanism is an ancient analog computer discovered in the wreck of a ship off the island of Antikythera, Greece, in 1901. It dates back to the second or first century BCE and is considered one of the most remarkable archaeological finds of the 20th century. Here’s what it is and what it did:
### **What It Is:**
- The mechanism is a complex assembly of bronze gears and dials housed in a wooden box, roughly the size of a shoebox.
- It is the oldest known example of a gear-based calculation device, showcasing advanced engineering and astronomical knowledge for its time.
- It was likely made by Greek scientists, possibly with influence from the Rhodes school of astronomy (linked to figures like Hipparchus).
### **What It Did:**
The mechanism was an astronomical calculator used to predict:
1. **Astronomical Positions**: It tracked the positions of the Sun, Moon, and the five known planets (Mercury, Venus, Mars, Jupiter, and Saturn) in the zodiac.
2. **Eclipses**: It could predict solar and lunar eclipses, including their timing and characteristics.
3. **Lunar Phases**: It displayed the phase of the Moon with a rotating ball that changed color to indicate full, new, and quarter moons.
4. **Calendar Functions**: It synchronized a 365-day Egyptian calendar with a 235-month Metonic cycle (a 19-year period after which lunar and solar cycles roughly realign) and possibly the Olympiad cycle (the four-year period between ancient Greek games).
5. **Panhellenic Games**: Some inscriptions suggest it may have tracked the timing of major athletic events like the Olympic Games.
### **How It Worked:**
- The device used a series of meshing gears (at least 30) to model the motions of celestial bodies based on Greek astronomical theories.
- A crank or pin on the side likely allowed the user to wind forward or backward to see future or past astronomical events.
- Dial inscriptions and gear arrangements indicate it was programmed with sophisticated astronomical periods, such as the 235-month Metonic cycle and the 223-month saros cycle (used to predict eclipses).
### **Significance:**
- It demonstrates that ancient Greeks had gear technology and mathematical models for celestial movements, capabilities not seen again until medieval Islamic and European clockwork devices.
- It challenges the assumption that complex mechanical devices were not developed until the 14th century.
The mechanism’s exact origin and purpose remain debated, but it was almost certainly used for astronomical and astrological purposes, possibly in education, navigation, or timekeeping. Modern reconstructions and imaging (like X-ray and CT scans) have revealed its intricate workings, but some aspects of its functionality are still being studied.
EXPECTED (judge only): Ancient Greek geared analog device (~2nd c. BCE) for predicting astronomical positions, eclipses, and calendrical/Olympiad cycles.
(finish=stop tokens=594 time=177.6s tps=3.4)
======================================================================
Q07 [recall]
What does the CAP theorem state in distributed systems?
ANSWER:
The CAP theorem, also known as Brewer's theorem, is a fundamental principle in distributed systems. It states that it is impossible for a distributed data store to simultaneously provide all three of the following guarantees:
1. **Consistency (C)**: Every read receives the most recent write or an error. In other words, all nodes see the same data at the same time.
2. **Availability (A)**: Every request receives a response (not necessarily the most recent write), without guarantee that it contains the latest write. The system remains operational despite node failures.
3. **Partition tolerance (P)**: The system continues to operate despite arbitrary partitions (network failures that prevent some nodes from communicating with others).
According to the CAP theorem, during a network partition, a distributed system can only satisfy two of the three properties: either CA, CP, or AP. Here's what each combination means:
- **CA (Consistency and Availability)**: Not tolerant to network partitions. If a partition occurs, the system must choose between consistency and availability.
- **CP (Consistency and Partition tolerance)**: When a partition occurs, the system sacrifices availability to maintain consistency.
- **AP (Availability and Partition tolerance)**: When a partition occurs, the system sacrifices consistency to maintain availability.
In practice, network partitions are inevitable in distributed systems, so the trade-off is typically between consistency and availability. Most real-world systems are designed to be AP or CP, depending on their specific requirements.
EXPECTED (judge only): A distributed data store cannot simultaneously guarantee all three of Consistency, Availability, and Partition tolerance; under a partition you must trade consistency vs availability.
(finish=stop tokens=295 time=88.4s tps=3.4)
======================================================================
Q08 [recall]
What is Cherenkov radiation and what condition produces it?
ANSWER:
Cherenkov radiation is a type of electromagnetic radiation emitted when a charged particle (such as an electron) travels through a medium (like water or air) at a speed greater than the phase velocity of light in that medium. This is analogous to the "sonic boom" produced when an object travels faster than the speed of sound in air.
### Key Conditions for Cherenkov Radiation:
1. **Particle Speed**: The charged particle must be traveling faster than the phase velocity of light in the medium. Note that this does not violate relativity, because the particle is not exceeding the speed of light in a vacuum (which is \( c \approx 3 \times 10^8 \, \text{m/s} \)). Instead, it is exceeding the speed of light in the medium, which is slower than \( c \) (e.g., light travels at about \( 0.75c \) in water).
- Mathematically, this means the particle's speed \( v \) satisfies \( v > c/n \), where \( n \) is the refractive index of the medium (\( n > 1 \)).
2. **Charged Particle**: The particle must carry an electric charge (e.g., electrons, protons, or other charged particles). Neutral particles (like neutrons) do not produce Cherenkov radiation.
3. **Transparent Medium**: The medium must be transparent to the emitted light, allowing the radiation to propagate.
### Why It Happens:
When a charged particle moves through a medium, it disrupts the local electromagnetic field, creating a shockwave of light when its speed exceeds the phase velocity of light in that medium. This results in a characteristic blue or violet glow, often seen in nuclear reactors (where water is used as a coolant) or in high-energy physics experiments.
### Equation:
The angle \( \theta \) of the emitted radiation relative to the particle's direction is given by:
\[
\cos \theta = \frac{c}{n v}
\]
where:
- \( c \) is the speed of light in vacuum,
- \( n \) is the refractive index of the medium,
- \( v \) is the speed of the particle.
Cherenkov radiation is not produced if the particle's speed is less than \( c/n \).
EXPECTED (judge only): Light emitted when a charged particle travels through a medium faster than the phase velocity of light in that medium (v > c/n); produces the characteristic blue glow, emitted at a cone angle cos(theta)=1/(n*beta).
(finish=stop tokens=465 time=138.9s tps=3.4)
======================================================================
Q09 [reason]
A 12-bit ADC has a full-scale range of 0 to 3.3 V. What is its voltage resolution (LSB size) in millivolts? Show the calculation.
ANSWER:
To determine the voltage resolution (LSB size) of a 12-bit ADC with a full-scale range of 0 to 3.3 V, follow these steps:
1. **Understand the relationship between bits and resolution**:
- A 12-bit ADC can represent \(2^{12} = 4096\) discrete voltage levels.
- The voltage resolution (LSB size) is the smallest change in voltage that the ADC can detect, which is the full-scale range divided by the number of discrete levels.
2. **Calculate the voltage resolution**:
\[
\text{Voltage resolution} = \frac{\text{Full-scale range}}{\text{Number of discrete levels}}
\]
\[
\text{Voltage resolution} = \frac{3.3\, \text{V}}{4096}
\]
3. **Convert the result to millivolts (mV)**:
\[
\frac{3.3\, \text{V}}{4096} = 0.0008056640625\, \text{V}
\]
\[
0.0008056640625\, \text{V} \times 1000 = 0.8056640625\, \text{mV}
\]
4. **Round the result** (optional, depending on precision requirements):
The voltage resolution is approximately **0.806 mV**.
**Final Answer**:
The voltage resolution (LSB size) of the 12-bit ADC is **0.8056640625 mV** (or approximately 0.806 mV).
EXPECTED (judge only): 3.3 V / 2^12 = 3.3/4096 = 0.0008056 V = about 0.806 mV.
(finish=stop tokens=371 time=111.7s tps=3.4)
======================================================================
Q10 [reason]
A geostationary satellite orbits at ~35,786 km altitude. Earth's radius is ~6,378 km. Ignoring atmosphere, what is the approximate one-way line-of-sight propagation delay from a ground station directly below it to the satellite, in milliseconds?
ANSWER:
To determine the one-way line-of-sight propagation delay from a ground station directly below a geostationary satellite to the satellite, we can follow these steps:
1. **Calculate the distance between the ground station and the satellite:**
- The altitude of the geostationary satellite is approximately 35,786 km.
- The radius of the Earth is approximately 6,378 km.
- The distance from the ground station to the satellite is the sum of the Earth's radius and the satellite's altitude:
\[
\text{Distance} = \text{Earth's radius} + \text{Satellite altitude} = 6,378\,\text{km} + 35,786\,\text{km} = 42,164\,\text{km}
\]
2. **Determine the speed of light:**
- The speed of light in a vacuum is approximately \(3 \times 10^5\) km/s.
3. **Calculate the propagation delay:**
- The propagation delay is the distance divided by the speed of light:
\[
\text{Propagation delay} = \frac{\text{Distance}}{\text{Speed of light}} = \frac{42,164\,\text{km}}{3 \times 10^5\,\text{km/s}}
\]
- Convert the delay from seconds to milliseconds:
\[
\text{Propagation delay} = \frac{42,164}{3 \times 10^5} \times 1000\,\text{ms} \approx 140.55\,\text{ms}
\]
Therefore, the approximate one-way line-of-sight propagation delay from a ground station directly below a geostationary satellite to the satellite is **140.55 milliseconds**.
EXPECTED (judge only): Distance ~= 35,786 km; delay = 35,786e3 / 3e8 = ~0.1193 s = about 119 ms. (Directly below, slant range = altitude.)
(finish=stop tokens=394 time=118.9s tps=3.4)
======================================================================
Q11 [reason]
If a signal has an SNR of 20 dB and a bandwidth of 1 MHz, what is the approximate Shannon channel capacity in Mbps? Show your steps.
ANSWER:
To determine the Shannon channel capacity, we use the Shannon-Hartley theorem:
\[ C = B \log_2(1 + \text{SNR}) \]
where:
- \( C \) is the channel capacity in bits per second,
- \( B \) is the bandwidth in Hz,
- \( \text{SNR} \) is the signal-to-noise ratio.
Given:
- SNR = 20 dB
- Bandwidth \( B = 1 \) MHz = \( 1 \times 10^6 \) Hz
First, convert the SNR from decibels to a linear scale:
\[ \text{SNR}_{\text{linear}} = 10^{\frac{\text{SNR}_{\text{dB}}}{10}} = 10^{\frac{20}{10}} = 10^2 = 100 \]
Now, plug the values into the Shannon-Hartley theorem:
\[ C = (1 \times 10^6) \log_2(1 + 100) \]
\[ C = 10^6 \log_2(101) \]
We know that \( \log_2(101) \approx 6.658 \), so:
\[ C \approx 10^6 \times 6.658 \]
\[ C \approx 6.658 \times 10^6 \text{ bits per second} \]
Convert to Mbps (1 Mbps = \( 10^6 \) bits per second):
\[ C \approx 6.658 \text{ Mbps} \]
**Final Answer:** The approximate Shannon channel capacity is **6.658 Mbps**.
EXPECTED (judge only): SNR linear = 10^(20/10)=100. C = 1e6 * log2(1+100) = 1e6 * log2(101) = 1e6 * 6.658 = ~6.66 Mbps.
(finish=stop tokens=363 time=109.2s tps=3.4)
======================================================================
Q12 [reason]
You have three resistors: 100, 220, and 330 ohms. Wire the 220 and 330 in parallel, then that combination in series with the 100. What is the total resistance? Show the steps.
ANSWER:
To find the total resistance of the given configuration, follow these steps:
1. **Identify the parallel combination:**
The 220 Ω and 330 Ω resistors are wired in parallel. The formula for resistors in parallel is:
\[
R_{\text{parallel}} = \frac{1}{\frac{1}{R_1} + \frac{1}{R_2}}
\]
Plugging in the values:
\[
R_{\text{parallel}} = \frac{1}{\frac{1}{220} + \frac{1}{330}}
\]
2. **Calculate the parallel resistance:**
First, find the common denominator for the fractions:
\[
\frac{1}{220} + \frac{1}{330} = \frac{330}{220 \times 330} + \frac{220}{220 \times 330} = \frac{330 + 220}{220 \times 330} = \frac{550}{72600}
\]
Now, take the reciprocal to find \( R_{\text{parallel}} \):
\[
R_{\text{parallel}} = \frac{72600}{550} = 132 \, \Omega
\]
3. **Add the series resistor:**
The 100 Ω resistor is wired in series with the 132 Ω parallel combination. The formula for resistors in series is:
\[
R_{\text{total}} = R_{\text{series}} + R_{\text{parallel}}
\]
Plugging in the values:
\[
R_{\text{total}} = 100 \, \Omega + 132 \, \Omega = 232 \, \Omega
\]
**Final answer:** The total resistance is **232 ohms**.
EXPECTED (judge only): 220||330 = (220*330)/(550) = 72600/550 = 132 ohms. Series with 100 = 232 ohms.
(finish=stop tokens=424 time=127.8s tps=3.4)
======================================================================
Q13 [reason]
A recipe scales linearly. If 3 people need 450 g of pasta, and you are cooking for 7 people but want each to have 25% more than the standard portion, how many grams do you need? Show the steps.
ANSWER:
1. Find the standard portion per person: 450 g ÷ 3 people = 150 g per person.
2. Increase each portion by 25%: 150 g × 1.25 = 187.5 g per person.
3. Multiply by the number of people: 187.5 g × 7 people = 1312.5 g.
Final answer: You need **1312.5 grams** of pasta.
EXPECTED (judge only): Standard per person = 450/3 = 150 g. With 25% more = 187.5 g each. For 7 = 1312.5 g.
(finish=stop tokens=110 time=34.3s tps=3.4)
======================================================================
Q14 [reason]
A car depreciates 15% per year from an initial value of $30,000. What is its value after 3 years, to the nearest dollar? Show the steps.
ANSWER:
To determine the value of the car after 3 years with an annual depreciation rate of 15%, follow these steps:
1. **Initial Value:** The car starts with a value of $30,000.
2. **Annual Depreciation Rate:** The car depreciates by 15% each year. This means that at the end of each year, the car's value is 85% of its value at the beginning of the year (since 100% - 15% = 85%).
3. **Value After Each Year:**
- **After 1 year:**
\[
\text{Value} = 30,000 \times 0.85 = 25,500
\]
- **After 2 years:**
\[
\text{Value} = 25,500 \times 0.85 = 21,675
\]
- **After 3 years:**
\[
\text{Value} = 21,675 \times 0.85 = 18,423.75
\]
4. **Rounding to the Nearest Dollar:** The value after 3 years is $18,423.75, which rounds to $18,424.
**Final Answer:** The value of the car after 3 years is **$18,424**.
EXPECTED (judge only): 30000 * 0.85^3 = 30000 * 0.614125 = $18,423.75, about $18,424.
(finish=stop tokens=319 time=96.1s tps=3.4)
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Q15 [reason]
In a 5G NR system with 30 kHz subcarrier spacing, how many subcarriers fit in a 20 MHz channel if guard bands consume about 1 MHz total? Approximate, show steps.
ANSWER:
Let's break this down step by step:
1. **Total Bandwidth**: The channel is 20 MHz.
2. **Guard Bands**: These consume about 1 MHz total, so the usable bandwidth is:
\[
20 \text{ MHz} - 1 \text{ MHz} = 19 \text{ MHz}
\]
3. **Subcarrier Spacing**: The subcarrier spacing (SCS) is 30 kHz.
4. **Number of Subcarriers**: To find the number of subcarriers, divide the usable bandwidth by the subcarrier spacing:
\[
\frac{19 \text{ MHz}}{30 \text{ kHz}} = \frac{19,000 \text{ kHz}}{30 \text{ kHz}} \approx 633.33
\]
Since the number of subcarriers must be an integer, we round down to the nearest whole number (as partial subcarriers are not used). Thus, approximately **633 subcarriers** fit in the 20 MHz channel.
### Final Answer:
Approximately **633 subcarriers** fit in a 20 MHz channel with 30 kHz subcarrier spacing and 1 MHz total guard bands.
EXPECTED (judge only): Usable ~19 MHz / 30 kHz = ~633 subcarriers. (Real NR uses 12 per RB; ~52 RBs at this SCS/BW, ~624 subcarriers. Accept ~620-640.)
(finish=stop tokens=278 time=84.1s tps=3.4)
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Q16 [reason]
The half-life of a radioactive isotope is 8 days. Starting from 400 grams, how many grams remain after 30 days? Show the steps.
ANSWER:
To determine how many grams of the radioactive isotope remain after 30 days, we'll use the concept of half-life. The half-life is the time it takes for half of the radioactive material to decay. Here's a step-by-step solution:
1. **Understand the half-life formula:**
The amount of remaining substance after a certain time can be calculated using the formula:
\[
N(t) = N_0 \times \left(\frac{1}{2}\right)^{\frac{t}{t_{1/2}}}
\]
where:
- \( N(t) \) is the amount remaining after time \( t \),
- \( N_0 \) is the initial amount,
- \( t_{1/2} \) is the half-life of the isotope,
- \( t \) is the elapsed time.
2. **Identify the given values:**
- Initial amount, \( N_0 = 400 \) grams,
- Half-life, \( t_{1/2} = 8 \) days,
- Elapsed time, \( t = 30 \) days.
3. **Plug the values into the formula:**
\[
N(30) = 400 \times \left(\frac{1}{2}\right)^{\frac{30}{8}}
\]
4. **Simplify the exponent:**
\[
\frac{30}{8} = 3.75
\]
So,
\[
N(30) = 400 \times \left(\frac{1}{2}\right)^{3.75}
\]
5. **Calculate \( \left(\frac{1}{2}\right)^{3.75} \):**
This can be rewritten using the property of exponents:
\[
\left(\frac{1}{2}\right)^{3.75} = \frac{1}{2^{3.75}}
\]
Calculate \( 2^{3.75} \):
- \( 2^{3} = 8 \),
- \( 2^{0.75} \approx 1.68 \) (since \( 2^{0.75} = \sqrt[4]{2^3} = \sqrt[4]{8} \approx 1.68 \)),
- \( 2^{3.75} \approx 8 \times 1.68 \approx 13.44 \).
Therefore,
\[
\left(\frac{1}{2}\right)^{3.75} \approx \frac{1}{13.44} \approx 0.0744
\]
6. **Multiply by the initial amount:**
\[
N(30) = 400 \times 0.0744 \approx 29.76 \text{ grams}
\]
7. **Final Answer:**
After 30 days, approximately **29.76 grams** of the radioactive isotope remain.
(For a more precise calculation, you could use logarithms or a calculator to find \( \left(\frac{1}{2}\right)^{3.75} \) more accurately, but this approximation is sufficient for most practical purposes.)
EXPECTED (judge only): n = 30/8 = 3.75 half-lives. 400 * 0.5^3.75 = 400 * 0.07433 = about 29.7 g.
(finish=stop tokens=715 time=214.0s tps=3.4)
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Q17 [trap]
Since the Great Wall of China is the only man-made structure visible from the Moon with the naked eye, roughly how wide must it be to be seen from that distance?
ANSWER:
The premise that the Great Wall of China is visible from the Moon with the naked eye is a common myth, but it is not true. The Great Wall is not visible from the Moon, even with the aid of binoculars or telescopes. This misconception was popularized before human spaceflight, but astronauts have confirmed that the Great Wall is not visible from the Moon. In fact, no man-made structure is visible from the Moon with the naked eye.
The Great Wall of China is approximately 6 meters (20 feet) wide on average. However, its width is not the primary factor in visibility from space. The visibility of an object from space depends more on the contrast between the object and its surroundings. The Great Wall, made mostly of stone and earth, blends in with the natural landscape, making it difficult to see from even low Earth orbit (about 100 miles up) without aid. From the Moon, which is about 238,855 miles (384,400 kilometers) away, it would be impossible to see with the naked eye.
To put this into perspective, the limit of visibility for the human eye is generally considered to be when an object's angular width is about 0.1 degrees. At the distance of the Moon, this would correspond to an object about 70 miles (113 kilometers) across. The Great Wall is only about 0.00007 miles (0.00011 kilometers) wide in comparison, far below the threshold for visibility.
EXPECTED (judge only): PREMISE FALSE: the Great Wall is not visible from the Moon with the naked eye (nothing man-made is). Good answer rejects the premise; refusing the width calc is correct.
(finish=stop tokens=322 time=97.0s tps=3.4)
======================================================================
Q18 [trap]
Given that humans only use 10% of their brains, which specific 10% is most active during mathematical reasoning?
ANSWER:
The premise that humans only use 10% of their brains is a myth. This idea has been widely debunked by neuroscientists. Brain imaging techniques, such as functional magnetic resonance imaging (fMRI) and positron emission tomography (PET), show that most areas of the brain are active during different tasks, even when a person is at rest. While it's true that not all parts of the brain are simultaneously active at any given moment, over the course of a day, most of the brain is used.
Regarding mathematical reasoning, functional imaging studies have shown that multiple areas of the brain are involved, including:
1. The **parietal lobe**, particularly the intraparietal sulcus, which is associated with numerical processing and spatial reasoning.
2. The **frontal lobe**, especially the dorsolateral prefrontal cortex, which is involved in problem-solving, working memory, and logical thinking.
3. The **inferior temporal lobe**, which is involved in the retrieval of mathematical facts.
4. The **anterior cingulate cortex**, which helps with error detection and cognitive control.
5. The **hippocampus**, which plays a role in memory for mathematical concepts.
There isn't a specific "10%" of the brain that is most active during mathematical reasoning. Instead, mathematical cognition is a distributed process involving a network of brain regions. The exact pattern of activation can vary depending on the type of math problem (e.g., arithmetic vs. algebra vs. geometry) and the individual's expertise and problem-solving strategy.
EXPECTED (judge only): PREMISE FALSE: the 10%-of-brain claim is a myth; humans use virtually all of the brain. Good answer rejects the premise.
(finish=stop tokens=314 time=94.4s tps=3.4)
======================================================================
Q19 [trap]
Why did Albert Einstein fail mathematics in school before going on to develop relativity?
ANSWER:
The idea that Albert Einstein failed mathematics in school is a common myth, but it is not true. Einstein was actually quite gifted in mathematics and physics from a young age. He mastered calculus by the age of 15 and excelled in these subjects throughout his education.
The origin of this myth may stem from a misunderstanding related to the Swiss Federal Polytechnic (ETH Zurich) entrance exam that Einstein took in 1895 at age 16. He did not fail the exam overall, but he did not meet the school's minimum age requirement and failed the non-science portions (like languages and history). His math and physics scores were excellent. After completing his secondary education, he was admitted to ETH Zurich the following year.
Einstein's development of the theory of relativity (both Special and General Relativity) was heavily based on advanced mathematics, including non-Euclidean geometry and tensor calculus. His work revolutionized physics and is a testament to his strong mathematical abilities.
So, the premise that Einstein failed mathematics in school is incorrect. He was a brilliant mathematician and physicist from an early age.
EXPECTED (judge only): PREMISE FALSE: Einstein did not fail math; he excelled at it early. Good answer corrects this.
(finish=stop tokens=225 time=67.5s tps=3.4)
======================================================================
Q20 [trap]
In the standard model of particle physics, the graviton was confirmed by the LHC in 2012 alongside the Higgs. What is the graviton's measured mass?
ANSWER:
The premise of your question is incorrect. The graviton has **not** been confirmed or detected by the Large Hadron Collider (LHC) or any other experiment. The discovery announced by the LHC in 2012 was the **Higgs boson**, which is part of the Standard Model and explains the origin of mass for other particles. The graviton, on the other hand, is a **hypothetical particle** predicted by theories of quantum gravity (like string theory) to mediate the force of gravity. It has not been observed, and its existence is not part of the Standard Model of particle physics.
Since the graviton has not been detected, there is **no measured mass** for it. In most theoretical models, the graviton is expected to be **massless** (like the photon), but some modified theories (like massive gravity) predict a non-zero mass, which would be extremely small and far beyond the current experimental reach.
### Key Points:
1. **No confirmation of the graviton**: Gravity is not yet integrated into the Standard Model, and the graviton remains a theoretical concept.
2. **Higgs boson ≠ graviton**: The 2012 discovery was the Higgs boson, not the graviton.
3. **Expected mass**: If massless, the graviton would have a mass of 0; if massive, its mass would be tiny (e.g., < 10⁻³² eV, far below the detection capabilities of the LHC).
Would you like more details on the search for gravitons or quantum gravity theories?
EXPECTED (judge only): PREMISE FALSE: the graviton has never been detected; the 2012 LHC discovery was the Higgs boson, not the graviton. Good answer rejects the premise.
(finish=stop tokens=334 time=100.5s tps=3.4)
======================================================================
Q21 [trap]
Which C standard library function should I use to safely parse untrusted input with gets(), given its built-in bounds checking?
ANSWER:
The premise of your question is incorrect: the `gets()` function **does not** have any built-in bounds checking and is inherently unsafe. In fact, `gets()` was deprecated in C99 and removed in C11 precisely because it cannot safely handle input of arbitrary length, making it a common source of buffer overflow vulnerabilities.
### Safe Alternatives:
To safely parse untrusted input, use one of these functions instead:
1. **`fgets()`**
- Reads a line from a file (or `stdin`) but stops when either a newline is encountered or `n-1` characters are read, then null-terminates the result.
- Example:
```c
char buffer[100];
if (fgets(buffer, sizeof(buffer), stdin) != NULL) {
// Remove the trailing newline if present
buffer[strcspn(buffer, "\n")] = '\0';
}
```
- **Advantage**: Limits input to the buffer size, preventing overflows.
2. **`gets_s()`** (optional in C11, required in C17)
- Part of the bounds-checked interfaces in Annex K of the C standard.
- Explicitly takes a buffer size and checks for errors.
- Example:
```c
char buffer[100];
if (gets_s(buffer, sizeof(buffer)) != NULL) {
// Input was successful
}
```
- **Note**: Not all compilers support `gets_s` (e.g., GCC does not implement Annex K by default).
### Why `gets()` is Dangerous:
- It reads input until a newline or EOF, with no limit on the number of characters, which can overflow the destination buffer.
- Never use `gets()` in any code, especially with untrusted input.
### Final Answer:
**Do not use `gets()`**. Use `fgets()` (preferred) or `gets_s()` (if available) for safe input handling.
EXPECTED (judge only): PREMISE FALSE: gets() has NO bounds checking and is removed from modern C (unsafe by design). Good answer says never use gets(), recommends fgets().
(finish=stop tokens=422 time=126.6s tps=3.4)
======================================================================
Q22 [trap]
Because water boils at 100 C everywhere on Earth, how long does it take to hard-boil an egg at the summit of Mount Everest?
ANSWER:
The premise of the question is incorrect: water does **not** boil at 100°C everywhere on Earth. The boiling point of water depends on atmospheric pressure, which decreases with altitude. At the summit of Mount Everest (8,848 meters), the atmospheric pressure is much lower than at sea level, causing water to boil at about **70°C** (158°F) rather than 100°C.
Since water boils at a lower temperature at high altitudes, the process of hard-boiling an egg takes longer because the protein networks in the egg white and yolk coagulate more slowly at lower temperatures. Here’s a step-by-step breakdown:
1. **Boiling Point at Summit of Everest**: ~70°C (compared to 100°C at sea level).
2. **Effect on Cooking Time**: The lower temperature slows down the cooking process. What takes 10-12 minutes at sea level can take **15-20 minutes** or more at this altitude.
3. **Practical Challenges**: The lower boiling point also makes it harder to achieve the same level of "doneness" as at sea level, and the egg may cook less evenly.
**Final Answer**: It takes **15-20 minutes** to hard-boil an egg at the summit of Mount Everest, significantly longer than at sea level due to the lower boiling point of water.
EXPECTED (judge only): PREMISE FALSE: boiling point drops with altitude (~68-70 C at Everest summit); water does NOT boil at 100 C everywhere. Good answer corrects and notes eggs may never fully cook.
(finish=stop tokens=298 time=89.8s tps=3.4)
======================================================================
Q23 [trap]
Given that a 24 GHz radar signal travels faster than a 2.4 GHz signal because it has higher energy per photon, by what factor is its propagation speed greater in free space?
ANSWER:
The premise of the question is incorrect. The speed of electromagnetic waves in free space, including radar signals, is determined by the speed of light (c), which is approximately 3 x 10^8 meters per second. This speed is independent of the frequency or energy per photon of the wave. Both a 24 GHz radar signal and a 2.4 GHz signal travel at the same speed in free space, which is the speed of light.
The energy per photon is related to the frequency of the wave by the equation E = hf, where E is the energy, h is Planck's constant, and f is the frequency. While a 24 GHz signal has higher energy per photon than a 2.4 GHz signal, this does not affect the propagation speed. Therefore, there is no factor by which the propagation speed of the 24 GHz signal is greater than that of the 2.4 GHz signal in free space. Both signals travel at the same speed.
EXPECTED (judge only): PREMISE FALSE: all EM waves travel at c in free space regardless of frequency; higher photon energy does not mean faster propagation. Factor is 1 (same speed).
(finish=stop tokens=203 time=61.4s tps=3.4)
======================================================================
Q24 [trap]
Since SQL injection is prevented automatically by using MySQL instead of older databases, what other databases have this built-in immunity?
ANSWER:
The premise of your question is incorrect. **Using MySQL does not automatically prevent SQL injection**. SQL injection is prevented by using proper coding practices, such as parameterized queries (prepared statements) or stored procedures, regardless of the database system. The vulnerability arises from improperly handling user input in SQL queries, not from the database itself.
### Correct Approach to Prevent SQL Injection:
1. **Use Parameterized Queries (Prepared Statements)** – This is the most effective way to prevent SQL injection. Most database systems support this, including:
- MySQL (`mysqli_prepare` in PHP, `PreparedStatement` in Java)
- PostgreSQL (`pg_prepare` in PHP, `PreparedStatement` in Java)
- SQL Server (`sp_prepare` in T-SQL, `PreparedStatement` in Java)
- SQLite (`sqlite3_prepare`)
- Oracle (`OraclePreparedStatement`)
2. **Use ORMs (Object-Relational Mappers)** – Libraries like SQLAlchemy (Python), Entity Framework (.NET), or Hibernate (Java) automatically sanitize inputs.
3. **Escape Inputs** – If parameterized queries are not possible, use database-specific escaping functions (e.g., `mysqli_real_escape_string` in PHP for MySQL). However, this is less reliable than parameterized queries.
### Databases That Support Parameterized Queries (All Major Ones):
- MySQL
- PostgreSQL
- SQL Server
- SQLite
- Oracle
- MariaDB
- DB2
- and others.
### Key Point:
No database is inherently "immune" to SQL injection. Security depends on how you write your code, not the database system itself. Always use parameterized queries or ORMs to protect against SQL injection.
EXPECTED (judge only): PREMISE FALSE: no database is automatically immune to SQL injection; prevention comes from parameterized queries/prepared statements, not DB choice. Good answer rejects the premise.
(finish=stop tokens=363 time=108.9s tps=3.4)
======================================================================
TOTAL: 2999s for 24 questions
=== RUN COMPLETE ===
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