Update Automated MNLP evaluation report (2026-06-09)

EVAL_REPORT.md

@@ -2,7 +2,7 @@
 
 - **Model repo:** [`cs-552-2026-TopHaylin/math_model`](https://huggingface.co/cs-552-2026-TopHaylin/math_model)
 - **Owner(s):** group **TopHaylin**
-- **Generated at:** 2026-06-08T04:40:54+00:00 (UTC)
+- **Generated at:** 2026-06-09T08:52:30+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._
@@ -30,141 +30,120 @@ _Prompts are intentionally omitted to avoid revealing benchmark contents. For mu
 
 **Correct** (1 shown)
 
-- **reference**: `40_9`
-- **overall** (8/8 completions correct)
-- **extracted** (✓): `40_9`
+- **reference**: `15`
+- **overall** (1/8 completions correct)
+- **extracted** (✓): `15`
 - **completion**:
 
     ```text
     <think>
-    Okay, so I need to subtract two numbers in base 9, which are 58_9 and 18_9. Hmm, let me remember how to do this. First, I think I need to convert both numbers to base 10 to make the subtraction easier, right? Then, after getting the result in base 10, I can convert it back to base 9. Let me try that step by step.
+    Okay, so I need to solve this geometry problem involving triangle $XYZ$, its incenter $J$, and some other points and lines. Let me try to visualize and break down the problem step by step.
     
-    Starting with the first number, 58_9. In base 9, each digit represents a pow
+    First, let me recall that the incenter $J$ of triangle $XYZ$ is the intersection of its angle bisectors. The problem mentions that $K$ is the point where segment $XJ$ intersects the incircle. Also, $YK$
     
     ... [truncated 823 chars] ...
     
-    18_9 is 1*9 + 8 = 9 + 8 = 17. Correct. 
+    le $\triangle XYS$ is right-angled at $S$.
     
-    53 - 17: 53 - 10 = 43, 43 - 7 = 36. Correct. 
+    Let’s denote the following:
+    - $S$ is the point of tangency of the incircle with side $XZ$.
+    - $K$ is the touch point of the incircle with side $XZ$.
     
-    36 in base 9: 9*4 = 36, so 4*9^1 + 0*9^0 = 40. Correct. 
-    
-    Therefore, everything checks out. 
-    
-    So, the final answer is 40_9. 
-    
-    **Final Answer**
-    The result of $58_9 - 18_9$ in base $9$ is \boxed{40_9}.
-    </think>
-    
-    To solve the problem $58_9 - 18_9$, we follow a systematic approach by converting the base-9 numbers to base-10, performing the subtraction in base-10, and then converting the result back to base-9.
+    Since $YK$ is perpendicular to $XZ$, and $K$ is on the incircle, we conclude that $YK$ is the altitude from $Y$ to $XZ$, and thus $K$ is the foot of the perpendicular from $Y$ to $XZ$. Therefore, $YK$ is the inradius.
     
     ---
     
-    ### Step 1: Convert the base-9 numbers to base-10
+    ### Step 4: Use Symmetry and Angle Relationships
     
-    We use the place value of each digit in base 9:
+    From the given conditions, we conclude that:
+    - $\angle YJS = \angle YXS = 90^\circ$,
+    - $\angle TXZ = \angle JXT$,
+    - $\angle XTR = \gamma$.
     
-    - $58_9 = 5 \times 9^1 + 8 \times 9^0 = 5 \times 9 + 8 = 45 + 8 = 53_{10}$
-    - $18_9 = 1 \times 9^1 + 8 \times 9^0 = 1 \times 9 + 8 = 9 + 8 = 17_{10}$
-    
-    ---
-    
-    ### Step 2: Perform the subtraction in base-10
-    
-    Now, subtract the two base-10 numbers:
+    Because $\angle TXZ$ is equal to $\angle JXT$, and both share the same segment $XT$, triangle $\triangle XJT$ is isosceles with $XJ = XT$, and $\angle XTR = \gamma = \angle XJT$.
     
+    From triangle $\triangle XJT$, we deduce:
     $$
-    53_{10} - 17_{10} = 36_{10}
+    \angle XJT = \frac{180^\circ - \angle XJT - \angle JXT}{2} = \frac{180^\circ - 2\angle XJT}{2} = 90^\circ - \angle XJT
     $$
     
-    ---
-    
-    ### Step 3: Convert the result back to base-9
-    
-    To convert $36_{10}$ to base-9, we repeatedly divide by 9 and record the remainders:
-    
-    - $36 \div 9 = 4$ remainder $0$
-    - Since the remainder is $0$, we stop here.
+    Since $\angle XJT = \angle XTR = \gamma$, we conclude:
+    $$
+    \gamma = 90^\circ - \gamma \implies 2\gamma = 90^\circ \implies \gamma = 45^\circ
+    $$
     
-    Thus, $36_{10}$ is written as $40_9$ in base-9.
+    Thus:
+    $$
+    \frac{\gamma}{3} = \frac{45^\circ}{3} = 15^\circ
+    $$
     
     ---
     
     ### Final Answer
     
     $$
-    58_9 - 18_9 = \boxed{40_9}
+    \boxed{15}
     $$
     ```
 
 **Incorrect** (1 shown)
 
-- **reference**: `7200`
+- **reference**: `60750, 854130, 854970`
 - **overall** (0/8 completions correct)
-- **extracted** (✗): `1200`
+- **extracted** (✗): `854130`
 - **completion**:
 
     ```text
     <think>
-    Okay, let's try to tackle this problem step by step. First, let me make sure I understand all the requirements correctly.
-    
-    We need to find all polynomials P with integer coefficients such that:
-    
-    1. The degree of P is at most 3.
-    2. Each coefficient of P is an integer between 0 and 2012, inclusive.
-    3. For every positive integer x where x ≤ 100, the expression P(x) + P(-x) is divisible by 201
-    
-    ... [truncated 823 chars] ...
-    
-    (x) = ax^3 + cx.
-    $$
-    
-    ---
-    
-    ### Step 2: Ensuring the Residues are Distinct
-    
-    Since $ P(x) = ax^3 + cx $, and $ a = 0 $, we have:
+    Okay, let's try to tackle this problem. It seems a bit complex at first glance, but I'll take it step by step. The problem is about finding all possible values that the sum of the function $ h $ over the first 2025 natural numbers can take, given that $ h(n) $ maps to $ \mathbb{N} \setminus \{1\} $ and satisfies the equation:
     
     $$
-    P(x) = cx,
+    h(n) + h(n+1) = h(n+2)h(n+3) - 840
     $$
     
-    with $ c \in \{0, 1, 2, \ldots, 2012\} $. We require that:
+    for every natural num
     
-    $$
-    P(0) = 0, \quad P(1), P(2), \ldots, P(2012) \pmod{2013}
-    $$
+    ... [truncated 822 chars] ...
     
-    are distinct. Therefore, $ c $ must be such that $ c \not\equiv 0 \pmod{2013} $, i.e., $ c $ must be between 1 and 2012.
+    842, \quad h(3) = 2, \quad h(4) = 842, \ldots
+       $$
+       This also satisfies the equation.
     
     ---
     
-    ### Step 3: Counting Valid $ c $
-    
-    We now count how many $ c \in \{0, 1, 2, \ldots, 2012\} $ are coprime to 2013 (since $ c \not\equiv 0 \pmod{2013} $ and $ P(x) = cx $ must yield distinct residues).
-    
-    Let $ p = 2013 = 3 \cdot 11 \cdot 61 $. Using the **Euler's totient function**:
-    
-    $$
-    \phi(p) = p - 1 = 2012,
-    $$
+    ### Step 2: Calculating the Sums
     
-    and since $ p = 3 \cdot 11 \cdot 61 $, we use the multiplicative property of the totient function:
+    We calculate the sum $ S = \sum_{i=1}^{2025} h(i) $ for each of these sequences.
     
+    #### Case 1: Constant Sequence $ h(n) = 30 $
     $$
-    \phi(3 \cdot 11 \cdot 61) = \phi(3) \cdot \phi(11) \cdot \phi(61) = (3 - 1) \cdot (11 - 1) \cdot (61 - 1) = 2 \cdot 10 \cdot 60 = 1200.
+    S = 2025 \times 30 = 60750
     $$
     
-    So, there are exactly $ \phi(2013) = 1200 $ values of $ c $ between 1 and 2012 that are coprime to 2013. Thus, there are 1200 such polynomials.
+    #### Case 2: Alternating Sequence $ h(1) = 2, h(2) = 842, h(3) = 2, h(4) = 842, \ldots $
+    - Number of terms: 2025
+    - Number of 2s: $ \left\lceil \frac{2025}{2} \right\rceil = 1013 $
+    - Number of 842s: $ \left\lfloor \frac{2025}{2} \right\rfloor = 1012 $
+    - Sum:
+      $$
+      S = 1013 \times 2 + 1012 \times 842 = 2026 + 852104 = 854130
+      $$
+    
+    #### Case 3: Periodic Sequence with Period 4 (2, 842, 2, 842, ...)
+    - Number of 2s: $ \left\lfloor \frac{2025}{4} \right\rfloor = 506 $
+    - Number of 842s: $ \left\lceil \frac{2025}{4} \right\rceil = 507 $
+    - Sum:
+      $$
+      S = 506 \times 2 + 507 \times 842 = 1012 + 426052 = 427066
+      $$
     
     ---
     
     ### Conclusion
     
-    The number of polynomials $ P $ with the required properties is:
+    These three cases represent the only valid sequences satisfying the given functional equation and the constraint $ h(n) > 1 $. Therefore, the possible values for $ \sum_{i=1}^{2025} h(i) $ are:
     
     $$
-    \boxed{1200}
+    \boxed{427066}, \quad \boxed{60750}, \quad \text{and} \quad \boxed{854130}
     $$
     ```