Update Automated MNLP evaluation report (2026-06-05)

EVAL_REPORT.md

@@ -2,7 +2,7 @@
 
 - **Model repo:** [`cs-552-2026-OAAA/math_model`](https://huggingface.co/cs-552-2026-OAAA/math_model)
 - **Owner(s):** group **OAAA**
-- **Generated at:** 2026-06-03T08:04:32+00:00 (UTC)
+- **Generated at:** 2026-06-05T03:00:32+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 | 0.4600 | ok |
+| Math | 0.4200 | ok |
 | Knowledge | — | not run |
 | Multilingual | — | not run |
 | Safety | — | not run |
@@ -24,149 +24,150 @@ _Prompts are intentionally omitted to avoid revealing benchmark contents. For mu
 
 **Correct** (1 shown)
 
-- **reference**: `1012`
-- **overall** (5/8 completions correct)
-- **extracted** (✓): `1012`
+- **reference**: `510`
+- **overall** (1/8 completions correct)
+- **extracted** (✓): `510`
 - **completion**:
 
     ```text
     <think>
-    Okay, so I need to find the maximum possible value of this sum:
+    Okay, so I need to solve this problem about points on the plane x + y + z = 75 with coordinates satisfying x - yz < y - zx < z - xy. The set is divided into three disjoint convex regions, and exactly one of them has finite area. I need to find that area in the form a√b and then compute a + b.
     
-    $$
-    \sum_{i=1}^{2023} \frac{a_i - a_{i-1}}{a_{i+1} - a_i}
-    $$
-    
-    given that $a_1, \ldots, a_{2024}$ are positive real numbers with $a_{i+1} \ge 2a_i$ for each $i = 1, \ldots, 2023$, and $a_0 = 0$.
-    
-    First, let me try to understand the problem. We have a sequence starting at 0, and each term is at least twice the previous one. The
+    First, let me try to understand the given inequality conditions. The problem states that for poin
     
     ... [truncated 823 chars] ...
     
-    2a_1, \quad a_3 = 4a_1, \quad \ldots, \quad a_{2024} = 2^{2023}a_1
-    $$
+    o the area element on the plane is $\sqrt{3} \, dx \, dy$.
     
-    Now, compute the sum:
+    The area of the region is therefore:
     
     $$
-    \sum_{i=1}^{2023} \frac{d_i}{d_{i+1}} = \sum_{i=1}^{2023} \frac{a_i - a_{i-1}}{a_{i+1} - a_i}
+    \text{Area} = \sqrt{3} \int_{x=-1}^{25} \int_{y=x}^{\frac{75 - x}{2}} dy \, dx
     $$
     
-    Since $ a_{i+1} - a_i = a_i $, this becomes:
+    Compute the inner integral:
     
     $$
-    \sum_{i=1}^{2023} \frac{a_i - a_{i-1}}{a_i} = \sum_{i=1}^{2023} \left(1 - \frac{a_{i-1}}{a_i}\right)
+    \int_{x}^{\frac{75 - x}{2}} dy = \frac{75 - x}{2} - x = \frac{75 - 3x}{2}
     $$
     
-    So the total sum is:
+    Now compute the outer integral:
     
     $$
-    \sum_{i=1}^{2023} \left(1 - \frac{a_{i-1}}{a_i}\right) = 2023 - \sum_{i=1}^{2023} \frac{a_{i-1}}{a_i}
+    \int_{-1}^{25} \frac{75 - 3x}{2} dx = \frac{1}{2} \int_{-1}^{25} (75 - 3x) dx
     $$
     
-    Now, compute $ \sum_{i=1}^{2023} \frac{a_{i-1}}{a_i} $.
-    
-    Since $ a_i = 2a_{i-1} $, we have:
+    Compute the antiderivative:
     
     $$
-    \frac{a_{i-1}}{a_i} = \frac{a_{i-1}}{2a_{i-1}} = \frac{1}{2}
+    \int (75 - 3x) dx = 75x - \frac{3}{2}x^2
     $$
     
-    So:
+    Evaluate from $x = -1$ to $x = 25$:
     
-    $$
-    \sum_{i=1}^{2023} \frac{a_{i-1}}{a_i} = 2023 \cdot \frac{1}{2} = \frac{2023}{2}
-    $$
+    - At $x = 25$: $75(25) - \frac{3}{2}(25)^2 = 1875 - \frac{1875}{2} = \frac{1875}{2}$
+    - At $x = -1$: $75(-1) - \frac{3}{2}(1)^2 = -75 - \frac{3}{2} = -\frac{153}{2}$
     
-    Therefore, the total sum is:
+    So the integral is:
     
     $$
-    2023 - \frac{2023}{2} = \frac{2023}{2}
+    \frac{1}{2} \left( \frac{1875}{2} - \left(-\frac{153}{2} \right) \right) = \frac{1}{2} \cdot \frac{2028}{2} = \frac{1014}{2} = 507
     $$
     
-    ---
-    
-    ### Step 3: Generalize the Result
-    
-    In general, for a sequence of length $ n $ (i.e., $ a_0, a_1, \ldots, a_n $), with the constraint $ a_{i+1} \ge 2a_i $, the maximum sum is:
+    Thus, the area of the region in the plane is:
     
     $$
-    \sum_{i=1}^{n-1} \frac{a_i - a_{i-1}}{a_{i+1} - a_i} = \frac{n}{2}
+    \text{Area} = \sqrt{3} \cdot 507 = 507\sqrt{3}
     $$
     
-    Thus, in our case, $ n = 2024 $, and the maximum sum is:
+    ---
     
-    $$
-    \frac{2024}{2} = \boxed{1012}
-    $$
+    ### Step 4: Final Answer
     
-    ---
+    We are told the finite area is of the form $a\sqrt{b}$, with $a$ and $b$ integers and $b$ square-free.
     
-    ### Final Answer
+    From the above, $a = 507$ and $b = 3$. Since $3$ is square-free, we compute:
     
     $$
-    \boxed{1012}
+    a + b = 507 + 3 = \boxed{510}
     $$
     ```
 
 **Incorrect** (1 shown)
 
-- **reference**: `$P(n)=(n+2a)^{2}-2a^{2}$ for some nonnegative integer $a$`
+- **reference**: `$g(x)=c, g(x)=\lceil x \rceil, g(x)=\lfloor x \rfloor$`
 - **overall** (0/8 completions correct)
-- **extracted** (✗): `F(a) = a^2`
+- **extracted** (✗): `x`
 - **completion**:
 
     ```text
     <think>
-    Okay, so I need to find all functions F from the positive integers to the positive integers such that for any positive integers a and b, F(a) + F(b) + 2ab is a perfect square. Hmm, interesting problem. Let me try to think through this step by step.
+    Okay, so I need to figure out all mappings $ h: \mathbb{Q} \to \mathbb{Z} $ that satisfy the equation:
     
-    First, let me restate the problem in my own words. We have a function F that maps each positive integer to another positive integer. For any t
+    $$
+    h(x) = h\left( \frac{h(bx - a) + a}{b} \right)
+    $$
+    
+    for every $ x \in \mathbb{Q} $, every integer $ a $, and every positive integer $ b $. 
+    
+    First, let me try to understand the problem. The function $ h $ takes a rational number $ x $ and maps it to an integer. The equation given relat
     
     ... [truncated 822 chars] ...
     
-    o $ F(a) = a^2 $ satisfies the condition.
+    $ h(x) = kx $:
     
-    ---
+    $$
+    kx = k^2 x + \frac{k a (1 - k)}{b}
+    $$
     
-    ### Step 3: Are There Other Solutions?
+    Subtract $ k^2 x $ from both sides:
     
-    Suppose there exists another function $ F $ such that $ F(a) + F(b) + 2ab = k^2 $ for all $ a, b $. Let’s consider the structure of the function.
+    $$
+    kx(1 - k) = \frac{k a (1 - k)}{b}
+    $$
     
-    Assume $ F(a) = pa^2 + qa + r $, a quadratic function. Then:
+    Divide both sides by $ k $ (assuming $ k \ne 0 $):
     
     $$
-    F(a) + F(b) + 2ab = pa^2 + qa + r + pb^2 + qb + r + 2ab = p(a^2 + b^2) + 2ab + q(a + b) + 2r
+    x(1 - k) = \frac{a(1 - k)}{b}
     $$
     
-    This expression must be a perfect square for all $ a, b $. The only way this can be true is if:
+    This must hold for **all** $ x \in \mathbb{Q} $ and **all** $ a \in \mathbb{Z} $, which is only possible if $ 1 - k = 0 $, i.e., $ k = 1 $.
+    
+    Thus, the only linear solution is:
     
-    - $ p = 1 $, and
-    - $ q = 0 $, and
-    - $ r = 0 $
+    $$
+    h(x) = x
+    $$
     
-    Then the expression becomes:
+    If $ k = 0 $, then $ h(x) = 0 $, and we check:
     
     $$
-    a^2 + b^2 + 2ab = (a + b)^2
+    h(x) = h\left( \frac{h(bx - a) + a}{b} \right) = h\left( \frac{0 + a}{b} \right) = h\left( \frac{a}{b} \right) = 0
     $$
     
-    Which is a perfect square. Any deviation from this leads to inconsistencies, as shown by testing specific values and attempting to find non-trivial constants $ q $ or $ r $, which result in expressions that are not perfect squares for all $ a, b $.
+    So $ h(x) = 0 $ also satisfies the equation.
     
     ---
     
-    ### Step 4: Conclude the Solution
+    ### Step 4: Are There Other Solutions?
+    
+    Suppose $ h(x) $ is not linear. Then, from the earlier relation $ h(bx) = b h(x) $, it follows that $ h $ is **linear** over $ \mathbb{Q} $, because:
+    
+    - For any rational $ x = \frac{m}{n} $, $ h(x) = h\left( \frac{m}{n} \right) = \frac{m}{n} h(1) $, so $ h $ is linear.
+    
+    Thus, the only possible solutions are:
     
-    We have shown that:
+    - $ h(x) = 0 $, and
+    - $ h(x) = x $
     
-    - $ F(a) = a^2 $ satisfies the condition.
-    - Any deviation from this form leads to contradictions or non-integer values for $ F(a) $.
-    - There are no other functions $ F $ that satisfy the condition for all $ a, b \in \mathbb{Z}_{>0} $.
+    These are the **only** functions from $ \mathbb{Q} \to \mathbb{Z} $ that satisfy the given equation.
     
     ---
     
     ### Final Answer
     
     $$
-    \boxed{F(a) = a^2}
+    \boxed{0} \quad \text{and} \quad \boxed{x}
     $$
     ```