Update Automated MNLP evaluation report (2026-05-20)

EVAL_REPORT.md

@@ -2,7 +2,7 @@
 
 - **Model repo:** [`cs-552-2026-kth/math_model`](https://huggingface.co/cs-552-2026-kth/math_model)
 - **Owner(s):** group **kth**
-- **Generated at:** 2026-05-19T03:22:52+00:00 (UTC)
+- **Generated at:** 2026-05-20T05:49:46+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 | 0.3300 | ok |
+| Math | 0.2900 | ok |
 | Knowledge | — | not run |
 | Multilingual | — | not run |
 | Safety | — | not run |
@@ -24,69 +24,127 @@ _Prompts are intentionally omitted to avoid revealing benchmark contents. For mu
 
 **Correct** (1 shown)
 
-- **reference**: `\frac{7}{4}`
-- **overall** (2/8 completions correct)
-- **extracted** (✓): `\frac{7}{4}`
+- **reference**: `4`
+- **overall** (1/8 completions correct)
+- **extracted** (✓): `4`
 - **completion**:
 
     ```text
     <think>
-    The red line has a direction vector of $\begin{pmatrix} -7 \\ b \end{pmatrix}$.
+    Let's rewrite the given equation as:
+    \[ (x^3 - 3x^2 + 3x) + (y^3 - 3y^2 + 3y) + (z^3 - 3z^2 + 3z) = 4 \]
     
-    The slope of the red line is $\frac{b}{-7}$.
+    Factor the terms:
+    \[ (x - 1)^3 + (y - 1)^3 + (z - 1)^3 = 4 \]
     
-    The slope of the red line is also equal to $\frac{4-3}{-5-(-1)} = \frac{1}{-4} = -\frac{1}{4}$.
+    Now, use the fact that for any real numbers $a$, $b$, and $c$:
+    \[ a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca) \]
     
-    Therefore, we can set up the equation:
-    \[ \frac{b}{-7} = -\frac{1}{4} \Rightarrow b = \frac{-7}{-4} \Rightarrow b =
+    Apply this to our equation:
+    \[ (x - 1)^3 + (y - 1)^3 + (z - 1)^3 - 3(x - 1)(y - 1)(z
+    
+    ... [truncated 64 chars] ...
+    
+    (x - 1)^2 + (y - 1)^2 + (z - 1)^2 - (x - 1)(y - 1) - (y - 1)(z - 1) - (z - 1)(x - 1)) = 4 \]
+    
+    \[ (x + y + z - 3)((x - 1)^2 + (y - 1)^2 + (z - 1)^2 - (xy - x - y + 1) - (yz - y - z + 1) - (zx - z - x + 1)) = 4 \]
+    
+    \[ (x + y + z - 3)((x - 1)^2 + (y - 1)^2 + (z - 1)^2 - xy - yz - zx + x + y + z - 3) = 4 \]
+    
+    \[ (x + y + z - 3)(x^2 + y^2 + z^2 - 2x - 2y - 2z + 3 - xy - yz - zx + x + y + z - 3) = 4 \]
+    
+    \[ (x + y + z - 3)(x^2 + y^2 + z^2 - xy - yz - zx - x - y - z) = 4 \]
+    
+    Now, we want to find the largest possible value of:
+    \[ x^2 + y^2 + z^2 - x - y - z \]
+    
+    Let $a = x + y + z - 3$ and $b = x^2 + y^2 + z^2 - xy - yz - zx$. The equation becomes:
+    \[ ab = 4 \]
+    
+    We want to maximize $b$, which is equivalent to minimizing $a$.
+    
+    Since $a$ and $b$ are nonnegative, the minimum value of $a$ is 0, which occurs when $b$ is at its maximum value.
+    
+    When $a = 0$, we have:
+    \[ b = 4/a = 4/0 \]
+    
+    However, this is not possible since $b$ cannot be infinity.
+    
+    The next possible value of $a$ is 1, which occurs when $b$ is at its maximum value.
+    
+    When $a = 1$, we have:
+    \[ b = 4/a = 4/1 = 4 \]
+    
+    Therefore, the largest possible value of $x^2 + y^2 + z^2 - x - y - z$ is:
+    \[
     </think>
     
-    Therefore, the answer is \boxed{\frac{7}{4}}.
+    Therefore, the answer is \boxed{4}.
     ```
 
 **Incorrect** (1 shown)
 
-- **reference**: `$g(x)=c, g(x)=\lceil x \rceil, g(x)=\lfloor x \rfloor$`
+- **reference**: `(3,2,5)`
 - **overall** (0/8 completions correct)
-- **extracted** (✗): `0`
+- **extracted** (✗): `\text{None}`
 - **completion**:
 
     ```text
     <think>
-    Let's use the given functional equation to derive conditions for $h$.
+    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 \(p\).
+    
+    Since \(m^{p-1}+n\) and \(n^{p-1}+m\) are both powers of \(p\), we can write:
+    \[ m^{p-1}+n = p^a \]
+    \[ n^{p-1}+m = p^b \]
+    
+    where \(a\) and \(b\) are positive integers.
+    
+    Subtracting the tw
+    
+    ... [truncated 347 chars] ...
+    
+    f \(a = b\), and the left-hand side is equal to 0.
+    
+    So, we have:
+    \[ m^{p-1} - n^{p-1} = 0 \]
     
-    Given:
-    \[ h(x) = h\left(\frac{h(bx - a) + a}{b}\right) \]
+    This implies that \(m^{p-1} = n^{p-1}\), which means that \(m = n\).
     
-    For any $a \in \mathbb{Z}$ and any positive integer $b$, we want to find $h$ that satisfies this equation for all $x \in \mathbb{Q}$.
+    Substituting \(m = n\) into the original equations, we get:
+    \[ m^{p-1}+m = p^a \]
+    \[ m^{p-1}+m = p^b \]
     
-    Substitute $x = 0$ into the given equation:
-    \[ h(0) = h\left(\frac{h(-a) + a}{b}\right) \]
+    Since \(m^{p-1}+m\) is a power of \(p\), we can write:
+    \[ m^{p-1}+m = p^c \]
     
-    Since $h$ maps to $\mathbb{Z}$, $h(-a)$ must also be an integer.
+    where \(c\) is a positive integer.
     
-    Now, let's find a general form for $h(x)$.
+    Now, we can factor out \(m\) from the left-hand side of the equation:
+    \[ m(m^{p-2}+1) = p^c \]
     
-    Assume $h(x) = nx + c$ for some integers $n$ and $c$.
+    Since \(p\) is a prime greater than 2, we know that \(p\) is odd.
     
-    Substitute this form into the given equation:
-    \[ nx + c = n\left(\frac{n(bx - a) + a + c}{b}\right) + c \]
+    Therefore, \(m\) must be a power of \(p\), say \(m = p^d\), where \(d\) is a positive integer.
     
-    Simplify and equate coefficients:
-    \[ nx + c = \frac{n^2b^2x - nab^2 + na + nc}{b} + c \]
+    Substituting \(m = p^d\) into the equation, we get:
+    \[ p^d(p^{p-2}+1) = p^c \]
     
-    Comparing coefficients of $x$ and constant terms:
-    \[ n = \frac{n^2b^2}{b} \Rightarrow n = nb^2 \Rightarrow n = b^2n \]
+    Since \(p^d\) is a power of \(p\), we can cancel out \(p^d\) from both sides of the equation:
+    \[ p^{p-2}+1 = p^{c-d} \]
     
-    \[ c = \frac{na - nab^2 + na + nc}{b} + c \Rightarrow bc = 2na - nab^2 + na + nc \]
+    Now, we can see that \(p^{p-2}+1\) is not a power of \(p\), unless \(p-2 = 0\), which means \(p = 2\).
     
-    Since $n$ and $c$ are integers, the only way this equation can hold for all $a$ and $b$ is if $n = 0$ and $c = 0$.
+    But we are given that \(p\) is a prime greater than 2, so this case is not possible.
     
-    Thus, the only possible form for $h(x)$ is $h(x) = 0$ for all $x \in \mathbb{Q}$.
+    Therefore, there are no ordered triples \((p, m, n)\) that satisfy the given conditions.
     
-    Therefore, the solution is:
-    \[ h(x) =
+    So, the answer is:
+    \[
     </think>
     
-    Therefore, the answer is \boxed{0}.
+    Therefore, the answer is \boxed{\text{None}}.
     ```