Automated MNLP evaluation report (2026-06-11)

EVAL_REPORT.md

@@ -2,17 +2,23 @@
 
 - **Model repo:** [`cs-552-2026-thinking-tokens/general_knowledge_model`](https://huggingface.co/cs-552-2026-thinking-tokens/general_knowledge_model)
 - **Owner(s):** group **thinking-tokens**
-- **Generated at:** 2026-06-04T22:19:52+00:00 (UTC)
+- **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:** [`e1898f3`](https://huggingface.co/cs-552-2026-thinking-tokens/general_knowledge_model/commit/e1898f3acea9101ef00b63b9fb7392e9e67af3b1)
+- **Message:** upload prompt model
+- **Committed:** 2026-06-08T18:27:23+00:00
+
 ## Summary
 
 | Benchmark | Accuracy | Status |
 |---|---:|---|
 | Math | — | not run |
-| Knowledge | 0.1500 | ok |
+| Knowledge | 0.4700 | ok |
 | Multilingual | — | not run |
 | Safety | — | not run |
 
@@ -24,53 +30,114 @@ _Prompts are intentionally omitted to avoid revealing benchmark contents. For mu
 
 **Correct** (1 shown)
 
-- **reference**: `C`
+- **reference**: `D`
 - **overall** (1/1 completions correct)
-- **extracted** (✓): `C`
+- **extracted** (✓): `D`
 - **completion**:
 
     ```text
     <think>
-    Okay, let's try to figure out this problem. So the question is about a computer program where integers are represented using 3 bits. When they add 5 and 3, the result is 0. The options are about possible reasons for this outcome. Let's break it down step by step.
+    Okay, let's try to figure out this quantum mechanics problem. So, the question is about a particle in a three-dimensional isotropic harmonic oscillator potential. The potential is given by V(r) = (1/2)mω²r², which makes sense because that's the form for a harmonic oscillator potential, right? The force is Hooke's law, so that's consistent with the potential being quadratic in r.
     
-    First, I remember that in computer systems, integers are usually stored using binary, and when you add numbers, there's the pos
+    The quest
     
-    ... [truncated 822 chars] ...
+    ... [truncated 823 chars] ...
     
-    that the sum is larger than 3 bits, which leads to overflow. So E might not be the answer.
+    ic harmonic oscillator. The energy levels of such a system are quantized and given by the formula:
     
-    F. Calculated in binary, not decimal – but the program's output is 0. If the program uses binary addition, then yes, but how does that lead to 0? Maybe if there's an overflow, but the problem states the result is 0. So F's option is a possibility, but does it explain why 5 + 3 = 0?
+    $$
+    E = \left(n + \frac{3}{2}\right) \hbar \omega
+    $$
     
-    G. Underflow – no, underflow would be a negative number. So not relevant here.
+    where $ n $ is the total number of quanta (i.e., the sum of the quantum numbers in each direction: $ n_x + n_y + n_z $).
     
-    H. 2's complement and negative number – but 8 in 3 bits is 000, which is positive. So 2's complement would represent negative numbers, but 8 is positive. So H is incorrect.
+    The **third excited state** corresponds to $ n = 3 $, since the ground state is $ n = 0 $, and the first excited state is $ n = 1 $, the second excited state is $ n = 2 $, and the third excited state is $ n = 3 $.
     
-    I. Lossy compression – that's not related to binary addition.
+    Thus, the energy of the third excited state is:
     
-    J. Syntax error – the question says the program outputs 0. If there's a syntax error, maybe the program tried to add numbers in a different way, but the problem states the program's output is 0, so maybe that's not the issue. But syntax error would cause a runtime error, but the output is 0. Maybe not J.
+    $$
+    E = \left(3 + \frac{3}{2}\right) \hbar \omega = \frac{9}{2} \hbar \omega
+    $$
     
-    So the main issue is that the sum of 5 and 3 in 3-bit integers is 8, which is outside the range. So overflow occurs, and the result wraps to 0. Therefore, the correct answer is C.
-    </think>
+    For the number of linearly independent eigenfunctions associated with this energy level, we consider the number of ways to distribute $ n = 3 $ quanta among the three directions $ x, y, z $. This is equivalent to finding the number of non-negative integer solutions to the equation:
     
-    \boxed{C}
+    $$
+    n_x + n_y + n_z = 3
+    $$
+    
+    The number of such solutions is given by the combinatorial formula:
+    
+    $$
+    \text{Number of solutions} = \binom{n + 2}{2} = \binom{3 + 2}{2} = \binom{5}{2} = 10
+    $$
+    
+    Therefore, the number of linearly independent eigenfunctions is 10.
+    
+    ---
+    
+    **Final Answer:**  
+    $$
+    \boxed{D}
+    $$
     ```
 
 **Incorrect** (1 shown)
 
 - **reference**: `B`
 - **overall** (0/1 completions correct)
-- **extracted** (✗): `<no answer>`
+- **extracted** (✗): `C`
 - **completion**:
 
     ```text
     <think>
-    Okay, let's tackle this problem. So, we have a bosonic field X coupling to fermions through a term involving the fermion fields ψ_f and the boson X. The Lagrangian is given as L = -λ f f̄ (ῡ + X) ψ_f, and the mass of X is 6 GeV. We need to find the kinematically allowed decay modes of X at leading order.
+    Okay, let's try to figure out this problem. So, the question is about what product is formed when 3,4-dimethylhexanedial is treated with a series of reagents. The choices are A, B, C, D. Let me break it down step by step.
+    
+    First, I need to recall what each of these reagents does. Let's go through each step one by one.
     
-    First, I remember that in particle physics, when a heavy particle like X decays, it c
+    The first reagent is KOH, H2O, THF, Heat. I remember that KOH is a stro
     
     ... [truncated 823 chars] ...
     
-    e the Lagrangian again.
+    arboxylic acid.
+    
+    ---
+    
+    ### **Step 3: Reaction with PCC, CH₂Cl₂ (Oxidation)**
+    
+    - **Reagent:** PCC is a **reagent for oxidizing alcohols to ketones**.
+    - **Effect:** The ester is now a **secondary alcohol**, and PCC oxidizes it to a **ketone**.
     
-    The Lagrangian is -λ f f̄ (ῡ + X) ψ_f. So maybe f is a fermion field, f̄ is its antiquark, (ῡ + X) is the sum of the up quark field and the boson X, multiplied by ψ_f. Wait, maybe the coupling is between X and the fermion fields. So for each fermion field ψ_f, the term is -λ f f̄ (ῡ + X) ψ_f. So the coupling is between X and the product of two fermion fields. So when X decays, the decay products would have to be pairs of fermions, each of which is part of the product f f̄. But since X has a mass of 6 GeV, and the fermions here are part of the product f f̄, which would require that each fermion has mass at least the mass of X. But wait, the mass of X is 6 GeV, but the quark masses are much higher. Wait, maybe the coupling here is actually between X and the quark fields, and the problem is mistyped? Because if the coupling is between X and quarks, then X would decay into pairs of quarks, each with mass ~ GeV. But the quark masses are typically around 5 GeV (like up, down, charm, strange, top). So 5 GeV quark pairs would have a total mass of 10 GeV. But the boson X has mass 6 GeV, so the invariant mass of the decay products would be at least 6 GeV. But a quark
+    **Product after Step 3:**  
+    - The ester is now a **ketone**, with the oxygen atoms from the ester group forming a double bond with the carbonyl group.
+    
+    ---
+    
+    ### **Step 4: Reaction with O₃, H₂O₂ (Ozonolysis)**
+    
+    - **Reagent:** Ozonolysis cleaves the double bonds in the molecule.
+    - **Effect:** The ketone is cleaved into **carboxylic acid groups**.
+    
+    ---
+    
+    ### **Final Product Analysis**
+    
+    Now, we consider the **structure** of the final product:
+    
+    - The original compound is **3,4-dimethylhexanedial**, which has:
+      - Two methyl groups on carbons 3 and 4.
+      - An aldehyde group on carbon 6.
+    - After hydrolysis, the aldehyde becomes a carboxylic acid.
+    - After Grignard addition and oxidation, the molecule becomes a **ketone**.
+    - Ozonolysis cleaves the ketone into **carboxylic acid groups**.
+    
+    The final product is a **carboxylic acid** with the structure:  
+    **3,4-dimethyl-5,6-dioxooctanal**, which matches **Option C**.
+    
+    ---
+    
+    ### ✅ Final Answer:
+    
+    $$
+    \boxed{C}
+    $$
     ```