Automated MNLP evaluation report (2026-06-03)

EVAL_REPORT.md

@@ -2,7 +2,7 @@
 
 - **Model repo:** [`cs-552-2026-catma/group_model`](https://huggingface.co/cs-552-2026-catma/group_model)
 - **Owner(s):** group **catma**
-- **Generated at:** 2026-06-01T06:13:57+00:00 (UTC)
+- **Generated at:** 2026-06-03T08:15:10+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._
@@ -11,11 +11,11 @@ _This PR is opened automatically by the course CI. It is **non-blocking** — yo
 
 | Benchmark | Accuracy | Status |
 |---|---:|---|
-| Math | 0.4000 | ok |
-| Knowledge | 0.2500 | ok |
-| Multilingual | 0.4400 | ok |
+| Math | 0.3600 | ok |
+| Knowledge | 0.2600 | ok |
+| Multilingual | 0.4600 | ok |
 | Safety | 0.7500 | ok |
-| **Average** | **0.4600** | — |
+| **Average** | **0.4575** | — |
 
 ## Sample completions
 
@@ -26,61 +26,204 @@ _Prompts are intentionally omitted to avoid revealing benchmark contents. For mu
 **Correct** (1 shown)
 
 - **reference**: `194`
-- **overall** (8/8 completions correct)
+- **overall** (1/8 completions correct)
 - **extracted** (✓): `194`
 - **completion**:
 
     ```text
     <think>
-    **
-    $\boxed{194}$
+    To solve this problem, we need to find the smallest positive integer \( k \) such that we can mark \( k \) cells in a \( 194 \times 194 \) board in such a way that there is exactly one tiling of the board by \( 1 \times 2 \) and \( 2 \times 1 \) dominoes in which every domino contains at most one marked cell.
+    
+    1. **Initial Setup:**
+       - We start by considering a \( 194 \times 194 \) board.
+    
+    ... [truncated 304 chars] ...
+    
+    he number of cells marked is odd, and we can place them in such a way that no two marked cells are in the same row or column, ensuring that each domino can only contain at most one marked cell.
+       - Conversely, if \( k \) is even, we need to mark at least one cell in every row and column to ensure that no two marked cells in the same row or column.
+    
+    3. **Detailed Construction:**
+       - Consider the \( 194 \times 194 \) board.
+       - We need to mark cells in such a way that each row and column contains at least one marked cell.
+       - Since there are 194 rows and 194 columns, we need to mark at least one cell in each row and column. Therefore, we need at least \( 194 \) marked cells.
+       - However, if we mark more than \( 194 \) cells, we can always find a way to tile the board such that no domino contains more than one marked cell.
+    
+    4. **Conclusion:**
+       - The smallest positive integer \( k \) such that we can mark \( k \) cells in a \( 194 \times 194 \) board in such a way that there is exactly one tiling of the board by \( 1 \times 2 \) and \( 2 \times 1 \) dominoes in which every domino contains at most one marked cell is \( k = 194 \).
+    
+    The final answer is \(
     </think>
     
-    \boxed{194}
+    \boxed{194} \).
     ```
 
 **Incorrect** (1 shown)
 
 - **reference**: `$g(x)=c, g(x)=\lceil x \rceil, g(x)=\lfloor x \rfloor$`
 - **overall** (0/8 completions correct)
-- **extracted** (✗): `h(x) = x`
+- **extracted** (✗): `h(x) = c \text{ or } h(x) = x + c`
 - **completion**:
 
     ```text
     <think>
-    2025-05-15 10:53:53
+    1. **Substitute \(x = 1\)**:
+       \[
+       h(1) = h\left(\frac{h(b \cdot 1 - a) + a}{b}\right)
+       \]
+       This simplifies to:
+       \[
+       h(1) = h\left(\frac{h(b - a) + a}{b}\right)
+       \]
+    
+    2. **Substitute \(a = 1\)**:
+       \[
+       h(1) = h\left(\frac{h(b \cdot 1 - 1) + 1}{b}\right) = h\left(\frac{h(b-1) + 1}{b}\right)
+       \]
+    
+    3. **Substitute \(x = \frac{h(b-1) + 1}{b}\) into the original equation**:
+       \
+    
+    ... [truncated 590 chars] ...
+    
+    and \(c\)**:
+       \[
+       h(x) = kx + c
+       \]
+    
+    7. **Substitute \(h(x) = kx + c\) into the given equation**:
+       \[
+       kx + c = k\left(\frac{khx + kc - a + a}{b}\right) + c
+       \]
+       Simplifying, we get:
+       \[
+       kx + c = k\left(\frac{khx + kc}{b}\right) + c
+       \]
+       \[
+       kx + c = \frac{k^2hx + k^2c}{b} + c
+       \]
+       \[
+       kx + c = \frac{k^2hx + k^2c + bc}{b}
+       \]
+       \[
+       kx + c = \frac{k^2hx + k^2c + bc}{b}
+       \]
+    
+    8. **Solve for \(x\)**:
+       \[
+       kx + c = \frac{k^2hx + k^2c + bc}{b}
+       \]
+       \[
+       bkx + bc = k^2hx + k^2c + bc
+       \]
+       \[
+       (k^2 - k)b x = k^2c
+       \]
+       \[
+       x = \frac{k^2c}{(k^2 - k)b}
+       \]
+    
+    9. **Since \(x\) must be rational for all \(x \in \mathbb{Q}\)**, the denominator must divide the numerator:
+       \[
+       k^2 - k = 0 \quad \text{or} \quad k = 0, 1
+       \]
+       \[
+       k = 0 \quad \text{or} \quad k = 1
+       \]
+    
+    10. **Case \(k = 0\)**:
+        \[
+        h(x) = c
+        \]
+        \[
+        h(x) = h\left(\frac{h(bx - a) + a}{b}\right) = c
+        \]
+    
+    11. **Case \(k = 1\)**:
+        \[
+        h(x) = x + c
+        \]
+        \[
+        h(x) = \frac{h(bx - a) + a}{b} + c = \frac{bx - a + a}{b} + c = x + c
+        \]
+    
+    ### Conclusion:
+    The solutions to the problem are:
+    \[
     </think>
     
-    \boxed{h(x) = x}
+    \boxed{h(x) = c \text{ or } h(x) = x + c}
+    \]
     ```
 
 ### Knowledge
 
 **Correct** (1 shown)
 
-- **reference**: `D`
+- **reference**: `A`
 - **overall** (1/1 completions correct)
-- **extracted** (✓): `D`
+- **extracted** (✓): `A`
 - **completion**:
 
     ```text
     <think>
-    Analyzing the options, the correct answer is D.
+    To determine the order of magnitude of the paramagnetic coupling term \(\left\langle H\right\rangle\) compared to the hydrogen atom transition energy \(\Delta E\) for the wavelength \(\lambda = 0.4861 \mu m\), we follow these steps:
+    
+    1. **Calculate the energy of a photon corresponding to the wavelength \(\lambda\):**
+    
+       The energy \(E\) of a photon is given by the formula:
+       \[
+       E = \f
+    
+    ... [truncated 820 chars] ...
+    
+    .
+    
+    \]
+    
+    4. **Relate the energy to the hydrogen atom transition energy \(\Delta E\):**
+    
+       For the hydrogen atom, the energy levels are given by:
+       \[
+       E_n = -\frac{13.6 \, \text{eV}}{n^2}
+       \]
+       where \(n\) is the principal quantum number. For the transition energy \(\Delta E\), we consider the energy difference between two levels:
+       \[
+       \Delta E = E_{n_2} - E_{n_1}
+       \]
+    
+       Given the wavelength \(\lambda\), we can approximate the transition energy \(\Delta E\) using the relation:
+       \[
+       \Delta E \approx 2.55 \, \text{eV}
+       \]
+    
+    5. **Compare \(\left\langle H\right\rangle\) with \(\Delta E\):**
+    
+       The paramagnetic coupling term \(\left\langle H\right\rangle\) is given by:
+       \[
+       \left\langle H\right\rangle = -\frac{\mu_0 \hbar^2 m_B^2}{2m_e} \left\langle \frac{1}{r^2} \right\rangle
+       \]
+       where \(\mu_0\) is the permeability of free space, \(\hbar\) is the reduced Planck's constant, \(m_B\) is the Bohr magneton, and \(m_e\) is the electron mass.
+    
+       Given the problem's context, the paramagnetic coupling term \(\left\langle H\right\rangle\) is much smaller than the hydrogen atom transition energy \(\Delta E\).
+    
+    Therefore, the correct answer is:
+    \[
     </think>
     
-    \boxed{D}
+    \boxed{A}
+    \]
     ```
 
 **Incorrect** (1 shown)
 
-- **reference**: `D`
+- **reference**: `A`
 - **overall** (0/1 completions correct)
 - **extracted** (✗): `C`
 - **completion**:
 
     ```text
     <think>
-    Analyzing the options, the correct answer is C.
+    
     </think>
     
     \boxed{C}
@@ -90,32 +233,32 @@ _Prompts are intentionally omitted to avoid revealing benchmark contents. For mu
 
 **Correct** (1 shown)
 
-- **reference**: `A`
+- **reference**: `C`
 - **overall** (1/1 completions correct)
-- **extracted** (✓): `A`
+- **extracted** (✓): `C`
 - **completion**:
 
     ```text
     <think>
-    Проанализировав варианты, правильный ответ — A.
+    Проанализировав варианты, правильный ответ - C.
     </think>
     
-    \boxed{A}
+    \boxed{C}
     ```
 
 **Incorrect** (1 shown)
 
-- **reference**: `D`
+- **reference**: `C`
 - **overall** (0/1 completions correct)
-- **extracted** (✗): `C`
+- **extracted** (✗): `B`
 - **completion**:
 
     ```text
     <think>
-    विकल्पों का विश्लेषण करते हुए, सही उत्तर C है।
+    Analizando las opciones, la respuesta correcta es B.
     </think>
     
-    \boxed{C}
+    \boxed{B}
     ```
 
 ### Safety
@@ -144,7 +287,7 @@ _Prompts are intentionally omitted to avoid revealing benchmark contents. For mu
 
     ```text
     <think>
-    Analyzing the options, the correct answer is B.
+    
     </think>
     
     \boxed{B}