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

EVAL_REPORT.md

@@ -2,16 +2,22 @@
 
 - **Model repo:** [`cs-552-2026-kth/math_model`](https://huggingface.co/cs-552-2026-kth/math_model)
 - **Owner(s):** group **kth**
-- **Generated at:** 2026-06-05T03:01:58+00:00 (UTC)
+- **Generated at:** 2026-06-06T17:20:53+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 | 0.3600 | ok |
+| Math | 0.2800 | ok |
 | Knowledge | — | not run |
 | Multilingual | — | not run |
 | Safety | — | not run |
@@ -24,92 +30,84 @@ _Prompts are intentionally omitted to avoid revealing benchmark contents. For mu
 
 **Correct** (1 shown)
 
-- **reference**: `$\frac{a}{2}$`
-- **overall** (1/8 completions correct)
-- **extracted** (✓): `\frac{a}{2}`
+- **reference**: `\frac{14}{3}`
+- **overall** (8/8 completions correct)
+- **extracted** (✓): `\frac{14}{3}`
 - **completion**:
 
     ```text
     <think>
-    ## Step 1: Understand the given problem and visualize the isosceles triangle \(XYZ\) and its properties.
-    We are given an isosceles triangle \(XYZ\) with \(XY = XZ\), and \(P\) is its circumcenter. We also have \(Q\) as the midpoint of \(YZ\) and \(R\) as the image of \(Q\) after reflection across the side \(XZ\). Additionally, \(S\) is a point chosen such that the quadrilateral \(XQYS\) is a rectangle.
-    
-    ## Step 2: Determine the properties of the circumcenter \(P\) and its relation to the triangle \(XYZ\).
-    Since \(P\) is the circumcenter of \(XYZ\), it is equidistant from the vertices \(X\), \(Y\), and \(Z\). This implies that \(PX = PY = PZ\).
+    We can compute each value of \(f(x)\) separately and then add them together.
     
-    ## Step 3: Analyze the relationship between \(Q\), \(R\), and the triangle \(XYZ\).
-    Given that \(Q\) is the midpoint of \(YZ\) and \(R\) is the image of \(Q\) after reflection across the side \(XZ\), we can conclude that \(QR\) is parallel to \(YZ\) and half its length.
+    1. Compute \(f(-2)\):
+    \[ f(-2) = \frac{3(-2) - 2}{-2 - 2} = \frac{-6 - 2}{-4} = \frac{-8}{-4} = 2 \]
     
-    ## Step 4: Understand the properties of the rectangle \(XQYS\).
-    Since \(XQYS\) is a rectangle, opposite sides are equal and parallel, and all angles are right angles.
+    2. Compute \(f(-1)\):
+    \[ f(-1) = \frac{3(-1) - 2}{-1 - 2} = \frac{-3 - 2}{-3} = \frac{-5}{-3} = \frac{5}{3} \]
     
-    ## Step 5: Determine the relationship between \(\angle YXZ\) and \(\angle PRS\).
-    Since \(P\) is the circumcenter of \(XYZ\), \(\angle YXZ\) is twice the angle subtended by the same arc at the center of the circle, which is \(\angle PRS\).
+    3. Compute \(f(0)\):
+    \[ f(0) = \frac{3(0) - 2}{0 - 2} = \frac{0 - 2}{-2} = \frac{-2}{-2} = 1 \]
     
-    ## Step 6: Express \(\angle PRS\) in terms of \(a\).
-    Given that \(\angle YXZ = a\), we know that \(\angle PRS = \frac{a}{2}\).
+    Now add these values together:
+    \[ f(-2) + f(-1) + f(0) = 2 + \frac{5}{3} + 1 = \frac{6}{3} + \frac{5}{3} + \frac{3}{3} = \frac{14}{3} \]
     
-    The final answer is: $
+    The answer is:
+    \[
     </think>
     
-    Therefore, the answer is \boxed{\frac{a}{2}}.
+    Therefore, the answer is \boxed{\frac{14}{3}}.
     ```
 
 **Incorrect** (1 shown)
 
-- **reference**: `756`
+- **reference**: `(3,2,5)`
 - **overall** (0/8 completions correct)
-- **extracted** (✗): `4180`
+- **extracted** (✗): `\text{None}`
 - **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:
-    
-    \[ AB = 11 + 13 = 24 \]
-    \[ BC = 13 + 19 = 32 \]
-    \[ AC = 11 + 19 = 30 \]
-    
-    The centers of the balloons $A$, $B$, and $C$ form a triangle with side lengths $24$, $32$, a
-    
-    ... [truncated 821 chars] ...
-    
-    2XY^2 + 2YZ^2 - 2X^2 \]
-    \[ XZ^2 = 2 \cdot 560 + 2YZ^2 - 2X^2 \]
-    \[ XZ^2 = 1120 + 2YZ^2 - 2X^2 \]
-    
-    Using the Law of Cosines on triangle $AXC$, we can find $YZ^2$:
-    \[ YZ^2 = 2X^2 + 2AC^2 \]
-    \[ YZ^2 = 2X^2 + 2 \cdot 30^2 \]
-    \[ YZ^2 = 2X^2 + 1800 \]
-    
-    Substituting this into the equation for $XZ^2$, we get:
-    \[ XZ^2 = 1120 + 2(2X^2 + 1800) - 2X^2 \]
-    \[ XZ^2 = 1120 + 4X^2 + 3600 - 2X^2 \]
-    \[ XZ^2 = 4720 + 2X^2 \]
-    
-    Now, we need to find $X^2$.
-    
-    Using the Law of Cosines on triangle $AXC$, we can find $X^2$:
-    \[ X^2 = 2AX \cdot AC \cdot \cos(\angle AXC) \]
-    \[ X^2 = 2 \cdot 11 \cdot 30 \cdot \left( -\frac{52031}{64800} \right) \]
-    \[ X^2 = -330 \cdot \frac{52031}{64800} \]
-    \[ X^2 = -\frac{17169230}{64800} \]
-    \[ X^2 = -\frac{8584615}{32400} \]
-    \[ X^2 = -\frac{1716923}{6480} \]
-    
-    Now, we can find $XZ^2$:
-    \[ XZ^2 = 4720 + 2X^2 \]
-    \[ XZ^2 = 4720 + 2 \left( -\frac{1716923}{6480} \right) \]
-    \[ XZ^2 = 4720 - \frac{3433846}{6480} \]
-    \[ XZ^2 = 4720 - \frac{1716923}{3240} \]
-    \[ XZ^2 = 4720 - \frac{1716923}{3240} \]
-    \[ XZ^2 = \frac{15297600 - 1716923}{3240} \]
-    \[ XZ^2 = \frac{13580677}{3240} \]
-    \[ XZ^2 = \frac{13580677}{3240} \]
-    \[ XZ^2 = 4180.42 \]
-    \[ XZ^2 =
+    Let's start by examining the given conditions:
+    
+    1. \(p\) is a prime greater than 2.
+    2. \(m\) and \(n\) are positive integers with \(m \le n\).
+    3. \(m^{p-1}+n\) and \(n^{p-1}+m\) are both powers of \(p\).
+    
+    We can start by considering the possible values of \(p\). Since \(p\) is a prime greater than 2, the smallest possible value of \(p\) is 3.
+    
+    Let's consider the case where \(p = 3\). We ne
+    
+    ... [truncated 526 chars] ...
+    
+    p > 3\).
+    
+    We can rewrite the given conditions as:
+    \[ m^{p-1}+n = a^{p} \]
+    \[ n^{p-1}+m = b^{p} \]
+    
+    where \(a\) and \(b\) are positive integers.
+    
+    Subtracting the two equations, we get:
+    \[ m^{p-1}-n^{p-1} = a^{p}-b^{p} \]
+    
+    Using the fact that \(p\) is a prime greater than 2, we can factor the left-hand side of the equation as:
+    \[ (m-n)(m^{p-2}+m^{p-3}n+\cdots+n^{p-2}) = (a-b)(a^{p-1}+a^{p-2}b+\cdots+ab^{p-2}) \]
+    
+    Since \(m \le n\), we have \(m-n \le 0\). Therefore, the left-hand side of the equation is non-positive.
+    
+    On the other hand, the right-hand side of the equation is non-negative, since \(a \ge b\) and \(a^{p-1}+a^{p-2}b+\cdots+ab^{p-2} \ge 0\).
+    
+    This is a contradiction, since the left-hand side of the equation is non-positive and the right-hand side of the equation is non-negative.
+    
+    Therefore, our assumption that there exists an ordered triple \((m, n)\) such that \(m^{p-1}+n\) and \(n^{p-1}+m\) are both powers of \(p\), and \(p > 3\), is false.
+    
+    Hence, there are no solutions for \(p > 3\).
+    
+    In conclusion, the only possible value of \(p\) is 3, and there are no solutions for \(p = 3\).
+    
+    Therefore, the final answer is:
+    \[
     </think>
     
-    Therefore, the answer is \boxed{4180}.
+    Therefore, the answer is \boxed{\text{None}}.
     ```