Update Automated MNLP evaluation report (2026-05-23)

EVAL_REPORT.md

@@ -2,7 +2,7 @@
 
 - **Model repo:** [`cs-552-2026-flab/math_model`](https://huggingface.co/cs-552-2026-flab/math_model)
 - **Owner(s):** group **flab**
-- **Generated at:** 2026-05-22T03:04:12+00:00 (UTC)
+- **Generated at:** 2026-05-23T14:37:35+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._
@@ -11,7 +11,7 @@ _This PR is opened automatically by the course CI. It is **non-blocking** — yo
 
 | Benchmark | Accuracy | Status |
 |---|---:|---|
-| Math | 0.3200 | ok |
+| Math | 0.3400 | ok |
 | Knowledge | — | not run |
 | Multilingual | — | not run |
 | Safety | — | not run |
@@ -24,62 +24,72 @@ _Prompts are intentionally omitted to avoid revealing benchmark contents. For mu
 
 **Correct** (1 shown)
 
-- **reference**: `194`
-- **overall** (3/8 completions correct)
-- **extracted** (✓): `194`
+- **reference**: `\frac{448}{15625}`
+- **overall** (4/8 completions correct)
+- **extracted** (✓): `\frac{448}{15625}`
 - **completion**:
 
     ```text
     </think>
     
-    The minimum number of marked cells \(k\) is \(\boxed{194}\).
+    The probability that exactly 4 out of the 7 islands contain treasure is given by the binomial distribution:
     
-    **Reasoning:**
-    
-    1. **Upper bound:**  
-       If \(k \le 193\), then the board can be tiled with \(97\) dominoes (each covering two cells). Since each domino can contain at most one marked cell, at most \(97\) of these dominoes can be chosen to contain a marked cell. The remaining \(194 - 97 = 97\) dominoes can be tiled without any
-    
-    ... [truncated 646 chars] ...
-    
-    inoes do not contain a marked cell.  
-    
-       - **Case \(n = 194\):**  
-         Split the board into \(97\) disjoint \(2 \times 2\) squares. In each square, choose one cell to mark and mark the other cell of the same domino. This uses \(97\) marked cells.  
-         Tile the board with dominoes:  
-         - In each \(2 \times 2\) square, tile it with two vertical dominoes (or two horizontal dominoes).  
-         - Combine the vertical dominoes from adjacent squares to form a larger horizontal domino, and combine the horizontal dominoes to form a larger vertical domino.  
-         The result is a single \(194 \times 194\) board tiled by \(97\) dominoes (each covering two cells). The chosen marked cell belongs to exactly one domino, and all other dominoes contain no marked cells.  
-         This tiling uses exactly one marked cell, so \(k = 1\). Hence \(k = 97\) is attainable.
+    \[
+    P(X = 4) = \binom{7}{4} \left(\frac{1}{5}\right)^4 \left(\frac{4}{5}\right)^3 = 35 \cdot \frac{1}{625} \cdot \frac{64}{125} = \frac{2240}{78125}.
+    \]
     
-       The argument generalises to any even \(n\): split the board into \(n/2\) disjoint \(2 \times 2\) squares, mark one cell in each square, and tile with dominoes as above. For odd \(n\), the same construction works by choosing one cell in each \(2 \times 3\) block.
+    Simplifying the fraction by dividing numerator and denominator by 5 gives \(\frac{448}{15625}\).
     
-    3. **Conclusion:**  
-       The minimum possible value of \(k\) is \(97\).
+    Thus, the required probability is \(\boxed{\frac{448}{15625}}\).
     ```
 
 **Incorrect** (1 shown)
 
-- **reference**: `104`
+- **reference**: `$g(x)=c, g(x)=\lceil x \rceil, g(x)=\lfloor x \rfloor$`
 - **overall** (0/8 completions correct)
-- **extracted** (✗): `34`
+- **extracted** (✗): `h(x)=c\ \text{for some integer }c`
 - **completion**:
 
     ```text
     </think>
     
-    The tetrahedron \(WXYZ\) is isosceles with two equal sides \(WX = YZ = \sqrt{41}\), \(WY = XZ = \sqrt{80}\), and \(WZ = XY = \sqrt{89}\). The common distance from the interior point \(P\) to the four faces is \(d\). By symmetry and the relation \(d = \frac{V}{S}\) (where \(V\) is the volume and \(S\) the total surface area), we compute \(V = \frac{22\sqrt{85}}{15}\) and \(S = \frac{44\sqrt{5}}{5}\). Thus
-    
+    We need to determine all functions \(h:\mathbb{Q}\to\mathbb{Z}\) satisfying  
     \[
-    d = \frac{V}{S} = \frac{22\sqrt{85}}{15} \div \frac{44\sqrt{5}}{5} = \frac{22\sqrt{85}}{15} \cdot \frac{5}{44\sqrt{5}} = \frac{11\sqrt{85}}{66\sqrt{5}} = \frac{\sqrt{17}}{6}.
+    h(x)=h\!\left(\frac{h(bx-a)+a}{b}\right)
+    \qquad\text{for all }x\in\mathbb{Q},\;a\in\mathbb{Z},\;b\in\mathbb{Z}^+\ .
     \]
     
-    Hence \(a = 11\), \(b = 17\), \(c = 6\). The condition \(a\) and \(c\) relatively prime is satisfied, and \(b\) is square‑free. Therefore
+    **Step 1.  Constant solutions.**  
+    If \(h(x)=c\) for all \(x\in\mathbb{Q}\) and any integer \(c\), then the equation holds because both sides are \(c\). Hence all constant functions are sol
+    
+    ... [truncated 110 chars] ...
     
+    \(h(x)\neq h(bx-a)+a\) (otherwise the constant solution would also work).  
+    
+    Take \(x=0\). For any \(a,b\in\mathbb{Z}^+\),  
     \[
-    a + b + c = 11 + 17 + 6 = 34.
-    \]
+    h(0)=h\!\left(\frac{h(b\cdot0-a)+a}{b}\right)=h\!\left(\frac{-h(a)+a}{b}\right).
+    \]  
+    If \(h(0)=c\), then \(\frac{-h(a)+a}{b}=c\) for all \(a,b\in\mathbb{Z}^+\), i.e., \(-h(a)+a = bc\).  
+    If \(b=1\) this gives \(a-h(a)=c\), so \(h(a)=a-c\); then \(h(0)=0-c=-c\).  
+    Thus \(h(a)=a-c\) for all \(a\in\mathbb{Z}\).  
+    
+    Now use \(h(a)=a-c\) in the original equation. Let \(b=1\). For any \(x\in\mathbb{Q}\),  
+    \[
+    h(x)=h\!\left(\frac{h(x-a)+a}{1}\right)=h(h(x-a)+a).
+    \]  
+    Since \(h(h(x-a)+a)=h(x-a)+a\), we obtain \(h(x)=h(x-a)+a\) for all \(x\in\mathbb{Q}\), \(a\in\mathbb{Z}\).  
+    
+    Take any \(a\in\mathbb{Z}\). For \(x=0\) this gives \(h(0)=h(-a)+a\).  
+    For \(x=a\) we have \(h(a)=h(a-a)+a=h(0)+a\), so \(h(a)=h(0)+a\).  
+    Thus \(h(0)=h(a)\) for all \(a\in\mathbb{Z}\). Hence \(h\) is constant.  
+    
+    Therefore the only non‑constant solutions are the constant functions.
+    
+    **Conclusion.**  
+    All functions \(h:\mathbb{Q}\to\mathbb{Z}\) satisfying the given equation are exactly the constant functions \(h(x)=c\) for some integer \(c\).
     
     \[
-    \boxed{34}
+    \boxed{h(x)=c\ \text{for some integer }c}
     \]
     ```