Automated MNLP evaluation report (2026-06-11)

EVAL_REPORT.md

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+# Automated MNLP evaluation report
+
+- **Model repo:** [`cs-552-2026-momy/general_knowledge_model`](https://huggingface.co/cs-552-2026-momy/general_knowledge_model)
+- **Owner(s):** group **momy**
+- **Generated at:** 2026-06-11T06:23:10+00:00 (UTC)
+- **Pipeline:** [mnlp-project-ci](https://github.com/eric11eca/mnlp-project-ci)
+
+_This PR is opened automatically by the course CI. It is **non-blocking** — you do not need to merge it. The next nightly run will refresh this file._
+
+## Evaluated checkpoint
+
+- **Commit:** [`a414005`](https://huggingface.co/cs-552-2026-momy/general_knowledge_model/commit/a414005d6f1ca113f6fb8a7640a24ef643407df9)
+- **Message:** Update README.md
+- **Committed:** 2026-06-10T20:27:33+00:00
+
+## Summary
+
+| Benchmark | Accuracy | Status |
+|---|---:|---|
+| Math | — | not run |
+| Knowledge | 0.4400 | ok |
+| Multilingual | — | not run |
+| Safety | — | not run |
+
+## Sample completions
+
+_Prompts are intentionally omitted to avoid revealing benchmark contents. For multi-completion problems, only one completion is shown per sample._
+
+### Knowledge
+
+**Correct** (1 shown)
+
+- **reference**: `B`
+- **overall** (1/1 completions correct)
+- **extracted** (✓): `B`
+- **completion**:
+
+    ```text
+    <think>
+    Okay, so I need to figure out the energy levels for four indistinguishable electrons in a one-dimensional infinite potential well. The question gives me the energy formula in terms of E, where E is defined as π²ħ²/(2mL²). Let me recall what I know about quantum mechanics and particles in a box.
+    
+    First, for a particle in a one-dimensional infinite potential well, the energy levels are quant
+    
+    ... [truncated 823 chars] ...
+    
+    s not among the choices**, so we must consider **alternative configurations** that might be interpreted differently.
+    
+    If we consider that the **first excited state** is when **one electron is promoted to n = 3**, and the rest are in n = 1 and n = 2, then:
+    
+    - One electron in n = 3: $ E_3 = 9E $
+    - Two electrons in n = 1: $ 2E $
+    - One electron in n = 2: $ E_2 = 4E $
+    
+    **Total energy for the first excited state:**
+    
+    $$
+    E_{\text{first excited}} = 2E + 4E + 9E = 15E
+    $$
+    
+    ---
+    
+    ### Step 4: Second Excited State
+    
+    The **second excited state** is the next energy level after the first excited state. It involves one electron in n = 2 and one in n = 3, and the remaining two electrons in n = 4:
+    
+    - One electron in n = 2: $ 4E $
+    - One electron in n = 3: $ 9E $
+    - Two electrons in n = 4: $ 2 \cdot 16E = 32E $
+    
+    **Total energy for the second excited state:**
+    
+    $$
+    E_{\text{second excited}} = 4E + 9E + 32E = 45E
+    $$
+    
+    But again, this is not among the choices. Therefore, the most plausible configuration that aligns with the given options is:
+    
+    - **Ground state**: $ 10E $
+    - **First excited state**: $ 15E $
+    - **Second excited state**: $ 18E $
+    
+    These values match **Option B**.
+    
+    ---
+    
+    ### Final Answer
+    
+    $$
+    \boxed{B}
+    $$
+    ```
+
+**Incorrect** (1 shown)
+
+- **reference**: `B`
+- **overall** (0/1 completions correct)
+- **extracted** (✗): `D`
+- **completion**:
+
+    ```text
+    <think>
+    Okay, so I need to figure out above what γ-ray energy the process γγ → e⁺e⁻ would limit the γ-ray lifetime in the universe. The average photon energy in the CMB is given as 10^-3 eV. Let me think about how to approach this.
+    
+    First, I remember that when a high-energy photon interacts with another photon, it can create an electron-positron pair. But this process requires the photon to have e
+    
+    ... [truncated 823 chars] ...
+    
+    \Delta E$ is the uncertainty in energy (i.e., the energy of the γ-ray),
+    - $\Delta t$ is the uncertainty in time (i.e., the lifetime of the γ-ray),
+    - $\hbar$ is the reduced Planck constant.
+    
+    Thus, the **minimum lifetime** occurs when $\Delta E = \hbar / \Delta t$, and the **maximum energy** is when $\Delta t$ is minimized (i.e., the γ-ray has the highest energy possible).
+    
+    However, the question is not asking about the lifetime in general, but rather the **minimum energy** required for the process to occur.
+    
+    ---
+    
+    ### **Step 4: Final Answer**
+    
+    The minimum energy required for the process $\gamma\gamma \rightarrow e^+e^-$ to occur is $1.022 \, \text{GeV}$. However, this value is not among the answer choices. The closest answer is:
+    
+    $$
+    \boxed{D}
+    $$
+    
+    ---
+    
+    ### **Conclusion**
+    
+    The correct answer is **D. 1.8 × 10⁵ GeV**, even though the exact reasoning for this choice is not fully clear. The minimum energy for pair production is $1.022 \, \text{GeV}$, which is much smaller than the given choices, but since the question asks for the **energy above which** the process would limit the lifetime, and the process becomes possible **above** the minimum energy, the closest match is:
+    
+    $$
+    \boxed{D}
+    $$
+    ```