Automated MNLP evaluation report (2026-06-11)

EVAL_REPORT.md

@@ -0,0 +1,131 @@
+# Automated MNLP evaluation report
+
+- **Model repo:** [`cs-552-2026-kth/math_model`](https://huggingface.co/cs-552-2026-kth/math_model)
+- **Owner(s):** group **kth**
+- **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:** [`f3af117`](https://huggingface.co/cs-552-2026-kth/math_model/commit/f3af117f54f6320d85a2714fdd5565d50205e7ad)
+- **Message:** Upload tokenizer
+- **Committed:** 2026-05-28T17:08:49+00:00
+
+## Summary
+
+| Benchmark | Accuracy | Status |
+|---|---:|---|
+| Math | 0.3300 | ok |
+| Knowledge | — | not run |
+| 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._
+
+### Math
+
+**Correct** (1 shown)
+
+- **reference**: `-\frac{1}{21}`
+- **overall** (4/8 completions correct)
+- **extracted** (✓): `-\frac{1}{21}`
+- **completion**:
+
+    ```text
+    <think>
+    The equation of the curve is \(y = x^2 + \frac{2}{x}\). The straight line intersects this curve at three distinct points.
+    
+    Let the equation of the line be \(y = mx + b\), where \(m\) is the slope and \(b\) is the y-intercept.
+    
+    Substitute \(y = mx + b\) into the equation of the curve:
+    \[ mx + b = x^2 + \frac{2}{x} \Rightarrow x^2 - mx - \frac{2}{x} - b = 0 \]
+    
+    Since two of the intersection
+    
+    ... [truncated 823 chars] ...
+    
+    }\) into (1) to find \(b\):
+    \[ -6\left(\frac{286}{21}\right) - b = -\frac{107}{3} \Rightarrow -\frac{1716}{21} - b = -\frac{107}{3} \Rightarrow b = -\frac{107}{3} + \frac{1716}{21} \Rightarrow b = \frac{-107 \cdot 7 + 1716}{21} \Rightarrow b = \frac{-749 + 1716}{21} \Rightarrow b = \frac{967}{21} \]
+    
+    Now that we have \(m = \frac{286}{21}\) and \(b = \frac{967}{21}\), we can substitute these values into the equation of the line:
+    \[ y = \frac{286}{21}x + \frac{967}{21} \]
+    
+    To find the \(x\)-coordinate of the remaining point, substitute \(y = mx + b\) into the equation of the curve:
+    \[ \frac{286}{21}x + \frac{967}{21} = x^2 + \frac{2}{x} \Rightarrow x^2 - \frac{286}{21}x - \frac{2}{x} - \frac{967}{21} = 0 \]
+    
+    Multiply through by \(21x\) to clear the fractions:
+    \[ 21x^3 - 286x^2 - 42 - 967x = 0 \Rightarrow 21x^3 - 286x^2 - 967x - 42 = 0 \]
+    
+    Since \(x = 6\) and \(x = 7\) are solutions, we can factor out \((x - 6)\) and \((x - 7)\) from the equation:
+    \[ (x - 6)(x - 7)(21x + 1) = 0 \]
+    
+    This gives us the possible values of \(x\): \(x = 6\), \(x = 7\), and \(x = -\frac{1}{21}\).
+    
+    So, the \(x\)-coordinate of the remaining point is:
+    \[
+    </think>
+    
+    Therefore, the answer is \boxed{-\frac{1}{21}}.
+    ```
+
+**Incorrect** (1 shown)
+
+- **reference**: `756`
+- **overall** (0/8 completions correct)
+- **extracted** (✗): `100`
+- **completion**:
+
+    ```text
+    <think>
+    We know that the distance between the centers of the two balloons is equal to the sum of their radii. Let's denote the centers of the balloons with radii $11$, $13$, and $19$ as $A$, $B$, and $C$ respectively. Then, we have:
+    
+    \[ AB = 11 + 13 = 24 \]
+    \[ BC = 13 + 19 = 32 \]
+    \[ AC = 11 + 19 = 30 \]
+    
+    Since the centers of the balloons lie on one side of the plane, the triangle $ABC$ is a trian
+    
+    ... [truncated 34 chars] ...
+    
+    $. We can use Heron's formula to find the area of the triangle $ABC$:
+    
+    \[ s = \frac{24 + 32 + 30}{2} = 43 \]
+    \[ \text{Area} = \sqrt{s(s - 24)(s - 32)(s - 30)} = \sqrt{43 \cdot 19 \cdot 11 \cdot 13} = 43 \cdot 11 \cdot \sqrt{13} \]
+    
+    Now, we need to find the area of the triangle $XYZ$. Since the plane cuts the balloons in three congruent circular cross-sections, the triangle $XYZ$ is similar to the triangle $ABC$. Let's denote the distance between the centers $X$ and $Z$ as $d$. Then, we have:
+    
+    \[ XY = \frac{24}{3} = 8 \]
+    \[ YZ = \frac{32}{3} \]
+    \[ XZ = d \]
+    
+    Using the similarity of the triangles, we can write:
+    
+    \[ \frac{XY}{AB} = \frac{YZ}{BC} = \frac{XZ}{AC} \]
+    
+    Substituting the values, we get:
+    
+    \[ \frac{8}{24} = \frac{\frac{32}{3}}{32} = \frac{d}{30} \]
+    
+    Simplifying, we get:
+    
+    \[ \frac{1}{3} = \frac{1}{3} = \frac{d}{30} \]
+    
+    Solving for $d$, we get:
+    
+    \[ d = 30 \cdot \frac{1}{3} = 10 \]
+    
+    Now, we can find the area of the triangle $XYZ$:
+    
+    \[ \text{Area}_{XYZ} = \frac{1}{3} \cdot \text{Area}_{ABC} = \frac{1}{3} \cdot 43 \cdot 11 \cdot \sqrt{13} = \frac{43 \cdot 11}{3} \cdot \sqrt{13} \]
+    
+    Finally, we can find $XZ^2$:
+    
+    \[ XZ^2 = d^2 = 10^2 =
+    </think>
+    
+    Therefore, the answer is \boxed{100}.
+    ```