Update Automated MNLP evaluation report (2026-06-11)

EVAL_REPORT.md

@@ -2,7 +2,7 @@
 
 - **Model repo:** [`cs-552-2026-camykaz/general_knowledge_model`](https://huggingface.co/cs-552-2026-camykaz/general_knowledge_model)
 - **Owner(s):** group **camykaz**
-- **Generated at:** 2026-06-10T21:20:55+00:00 (UTC)
+- **Generated at:** 2026-06-11T04:19:46+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._
@@ -18,7 +18,7 @@ _This PR is opened automatically by the course CI. It is **non-blocking** — yo
 | Benchmark | Accuracy | Status |
 |---|---:|---|
 | Math | — | not run |
-| Knowledge | 0.4600 | ok |
+| Knowledge | 0.4500 | ok |
 | Multilingual | — | not run |
 | Safety | — | not run |
 
@@ -37,48 +37,45 @@ _Prompts are intentionally omitted to avoid revealing benchmark contents. For mu
 
     ```text
     <think>
-    Okay, so I need to figure out the energy of the third excited state for a particle in a harmonic oscillator potential in three dimensions. The potential is given by V(r) = (1/2)mω²r², which is a standard harmonic oscillator potential. 
+    Okay, let's tackle this problem step by step. So, we have a quantum mechanical system where a particle of mass m is moving in a harmonic potential V(r) = (1/2)mω²r². This is a standard harmonic oscillator potential, right? And we need to find the energy of the third excited state and the number of linearly independent eigenfunctions for that energy.
     
-    First, I recall that in quantum mechanics, the energy levels of a 3D harmonic oscillator are quantized. For a 3D harmonic oscillator, the energy levels are
+    First, I remember that the harmonic osc
     
-    ... [truncated 823 chars] ...
-    
-    \omega \left(n_{\text{total}} + \frac{3}{2}\right)
-    $$
-    
-    ---
-    
-    ### **Third Excited State**
+    ... [truncated 821 chars] ...
     
-    The first excited state corresponds to $n_{\text{total}} = 1$, the second to $n_{\text{total}} = 2$, and the third to $n_{\text{total}} = 3$. Therefore, the third excited state has:
+    second excited state $ n = 2 $, and the third excited state $ n = 3 $. So:
     
     $$
-    n_{\text{total}} = 3 \Rightarrow E = \hbar \omega \left(3 + \frac{3}{2}\right) = \frac{9}{2} \hbar \omega
+    E_{\text{third excited}} = \hbar \omega \left(3 + \frac{3}{2}\right) = \frac{9}{2} \hbar \omega
     $$
     
     ---
     
-    ### **Number of Linearly Independent Eigenfunctions**
+    ### Number of Linearly Independent Eigenfunctions
     
-    Each state with a given $n_{\text{total}}$ is a distinct quantum state. The number of linearly independent eigenfunctions with energy $E = \frac{9}{2} \hbar \omega$ corresponds to the number of ways to write $n_{\text{total}} = 3$ as a sum of three non-negative integers $n_x, n_y, n_z$, where the order matters (i.e., different permutations are considered distinct states).
-    
-    This is equivalent to the number of non-negative integer solutions to the equation:
+    The number of linearly independent eigenfunctions for a given energy level corresponds to the number of distinct ways to distribute the quanta $ n = 3 $ among the three dimensions $ x, y, z $. This is equivalent to the number of integer solutions to the equation:
     
     $$
     n_x + n_y + n_z = 3
     $$
     
-    This is a classic stars-and-bars problem, and the number of solutions is:
+    where $ n_x, n_y, n_z \geq 0 $ are integers. The number of such solutions is given by the stars and bars formula:
     
     $$
     \binom{3 + 3 - 1}{3 - 1} = \binom{5}{2} = 10
     $$
     
-    Hence, there are 10 linearly independent eigenfunctions associated with the third excited state.
+    However, in quantum mechanics, the states are not counted in this way due to the symmetry of the system. Instead, the number of **distinct eigenstates** with the same energy (i.e., same $ n $) is equal to the number of **distinct permutations** of the quanta, which is:
+    
+    $$
+    \text{Number of states} = \text{number of integer solutions with } n_x + n_y + n_z = 3 = 10
+    $$
+    
+    Hence, the number of linearly independent eigenfunctions for the third excited state is **10**.
     
     ---
     
-    ### **Final Answer**
+    ### Final Answer
     
     $$
     \boxed{D}
@@ -87,46 +84,45 @@ _Prompts are intentionally omitted to avoid revealing benchmark contents. For mu
 
 **Incorrect** (1 shown)
 
-- **reference**: `A`
+- **reference**: `C`
 - **overall** (0/1 completions correct)
-- **extracted** (✗): `H`
+- **extracted** (✗): `B`
 - **completion**:
 
     ```text
     <think>
-    Okay, let's try to figure this out. So the problem says that a person has a potential market wage of $5 per hour. The uncle leaves them an annual income of $4,000. But wait, the question mentions that a year has 8,000 hours, but actually it's 8,760. But maybe they're using a simplified version here. Anyway, the person allocates 75% of full income to leisure. So first, I need to figure out
+    Okay, let's try to figure this out. The problem is about identifying which two aromatic compounds are in a mixture that has an equimolar 1:1 mix. The NMR data gives us some information about the chemical shifts. 
+    
+    First, the ¹H NMR spectrum shows two singlet signals near 6.7 ppm with a 1:1 ratio. Then there are three singlet signals near 2.2 ppm in a 2:1:1 ratio. The compounds given are fo
     
     ... [truncated 823 chars] ...
     
-    ck for Conflicts with the Given Options
+    e** has four methyl groups on positions 1, 2, 3, 5.
+    - **1,4-diethylbenzene** has two ethyl groups on positions 1 and 4.
+    - The **2.2 ppm signals** would come from the **methyl and ethyl groups** of the substituents.
+    - The **2:1:1 ratio** could be due to the distribution of substituents between the two compounds. This is plausible and matches the NMR pattern.
     
-    The calculated value is **200 hours**, which is **not listed** among the provided choices. However, we must consider that the question may be referencing a **simplified or different interpretation**.
+    #### **Option C: 1,2,3,4-tetramethylbenzene and 1,2,3,5-tetramethylbenzene**
+    - Both have four methyl groups, but the **2:1:1 ratio** is harder to explain unless one compound has two methyl groups and the other one ethyl group, which is not consistent with the given compounds.
     
-    If we interpret the problem as the person **allocating 75% of their time to leisure**, then:
-    $$
-    \text{Time allocated to leisure} = 0.75 \times 8000 = 6000 \text{ hours}
-    $$
-    $$
-    \text{Time allocated to work} = 8000 - 6000 = 2000 \text{ hours}
-    $$
-    $$
-    \text{Income from work} = 5 \times 2000 = 10000 \text{ dollars}
-    $$
-    But this contradicts the given $4,000 income. Thus, this interpretation is invalid.
+    #### **Option D: 1,2,4,5-tetramethylbenzene and 1,2,3,4-tetramethylbenzene**
+    - Similar to Option A, but again the **2:1:1 ratio** is not clearly explained.
     
     ---
     
-    ### Step 4: Reconcile the Options
+    ### **Conclusion**
     
-    Given that the available options do not include 200 hours, and the only reasonable calculation leads to **200 hours**, we must consider that the problem may involve a **different interpretation** of the "full income" or "time allocation."
+    The **most plausible** scenario is that the **two compounds** are:
     
-    However, based on the standard interpretation of the question and the given options, the closest match is **H. 1000**. This is not directly derived from the logic, but it might represent a **misinterpretation** or **typo** in the question.
+    - **1,2,3,5-tetramethylbenzene** (with four methyl groups on positions 1, 2, 3, 5)
+    - **1,4-diethylbenzene** (with two ethyl groups on positions 1 and 4)
     
-    ---
+    This results in a **2:1:1 ratio** of substituents at 2.2 ppm and the **1:1 ratio** of aromatic protons at 6.7 ppm.
     
-    ### Final Answer:
+    ---
     
+    ### **Final Answer**
     $$
-    \boxed{H}
+    \boxed{B}
     $$
     ```