Update Automated MNLP evaluation report (2026-06-11)

EVAL_REPORT.md

@@ -2,7 +2,7 @@
 
 - **Model repo:** [`cs-552-2026-mnlplus/math_model`](https://huggingface.co/cs-552-2026-mnlplus/math_model)
 - **Owner(s):** group **mnlplus**
-- **Generated at:** 2026-06-10T21:20:55+00:00 (UTC)
+- **Generated at:** 2026-06-11T04:19:46+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._
@@ -17,7 +17,7 @@ _This PR is opened automatically by the course CI. It is **non-blocking** — yo
 
 | Benchmark | Accuracy | Status |
 |---|---:|---|
-| Math | 0.4500 | ok |
+| Math | 0.4200 | ok |
 | Knowledge | — | not run |
 | Multilingual | — | not run |
 | Safety | — | not run |
@@ -30,68 +30,92 @@ _Prompts are intentionally omitted to avoid revealing benchmark contents. For mu
 
 **Correct** (1 shown)
 
-- **reference**: `1012`
-- **overall** (5/8 completions correct)
-- **extracted** (✓): `1012`
+- **reference**: `99774`
+- **overall** (1/8 completions correct)
+- **extracted** (✓): `99774`
 - **completion**:
 
     ```text
     <think>
-    Okay, so I need to solve this problem where there are 2024 positive real numbers a₁, a₂, ..., a₂₀₂₄. The conditions are that each a_{i+1} is at least twice the previous a_i, so for each i from 1 to 2023, we have a_{i+1} ≥ 2a_i. Also, a₀ is defined as 0. The goal is to find the maximum possible value of the sum from i=1 to 2023 of (a_i - a_{i-1}) divided by (a_{i+1} - a_i). 
+    Alright, let's try to solve this cryptarithm problem. The equation given is:
     
-    First, let me
+    HATER + HEART = EARTH
+    
+    We need to find the unique 5-digit integer EARTH where E, A, R, T, H are digits (with E and H being non-zero). The digits don't have to be distinct, but leading digits can't be zero. Let's break this down step by step.
+    
+    First, let's write out the addition vertically to visualize the columns
     
     ... [truncated 823 chars] ...
     
-    re, c_i = a_{i+1} - c_{i+1} = a_i. Therefore, c_{i+1} = a_{i+1} - c_{i+1} = a_i. Therefore, c_{i+1} = a_i, but a_i = c_i. Therefore, c_{i+1} = c_i. Therefore, all c_i are equal. But in our example, c₁=1, c₂=1, c₃=2, c₄=4, etc. This contradicts the conclusion that c_{i+1} = c_i. Therefore, there must be an error in this reasoning.
+    + c2. But c2=1, so 16 +1=17, which is R +10*c3=7 +10*1=17. Correct. Then, the thousands place is T + E + c3=7 +9 +1=17, which is T +10*c4=7 +10*1=17. Correct. The ten-thousands place is H + H +c4=4 +4 +1=9. Correct. The units and tens places are correctly handled. So the final result is 99774, which is EARTH=99774. Therefore, the solution is H=4, E=9, A=9, R=7, T=7.
+    
+    But let's check the original problem statement: "Determine the unique 5-digit integer EARTH for which the cryptarithm HATER + HEART = EARTH is valid."
+    
+    So EARTH=99774. But wait, in HATER, the digits are H=4, A=9, T=7, E=9, R=7. So HATER=4 9 7 9 7=49797.
+    
+    HEART=4 9 9 7 7=49977.
+    
+    Adding them:49797 +49977=99774. Yes, that's correct. So the unique solution is EARTH=99774.
     
-    Wait, let's take the example step by step. a₀=0, a₁=1. Then c₁=1. a₂=2. c₂=2 - 1=1. a₃=4. c₃=4 - 2=2. a₄=8. c₄=8 - 4=4. a₅=16. c₅=16 - 8=8. So in this case, c₁=1, c₂=1, c₃=2, c₄=4, c₅=8, etc. So c₁=c₂=1, c₃=2=2^1, c₄=4=2^2, c₅=8=2^3, etc. So after c₁=c₂, the differences double each time. Therefore, the general formula is c₁=c₂=1, c₃=2, c₄=4, ..., c_{k}=2^{k-3} for k ≥ 3. Therefore, in this case, the sum S = 0/1 + 1/1 + 1/2 + 2/4 + 4/8 + ... + 2^{2022}/2^{2023} = 0 + 1 + 0.5 + 0.5 + 0.5 + ... + 0.5. The number of terms after the first is 2023 - 1 = 2022 terms, each contributing 0.5. Therefore, the sum is 1 + 0.5*2022 = 1 + 1011 = 1012. 
+    But let's check if there are other possible solutions. Earlier, we had H=4, E=9, c3=1, c4=1, leading to A=9, R=7, T=7. Is there any other possibility?
     
-    So even though c_i doubles after the first two terms, the sum still results in 1012. Therefore, the maximum sum is indeed 1012. 
+    We considered E=9, since c3=1 requires E=9. What if c3=0? Earlier, we thought E=0, but E cannot be 0. So E must be 9. Therefore, the only solution is E=9, H=4, A=9, R=7, T=7.
     
-    Therefore, the answer is \boxed{1012}.
+    Hence, the unique 5-digit integer is 99774.
     
     **Final Answer**
-    \boxed{1012}
+    The unique 5-digit integer is \boxed{99774}.
     </think>
     
-    \boxed{1012}
+    \boxed{99774}
     ```
 
 **Incorrect** (1 shown)
 
 - **reference**: `60750, 854130, 854970`
 - **overall** (0/8 completions correct)
-- **extracted** (✗): `854130`
+- **extracted** (✗): `854970`
 - **completion**:
 
     ```text
     <think>
-    Okay, so I need to solve this problem where there's a function h: N → N\{1} such that for every natural number n, h(n) + h(n+1) = h(n+2)h(n+3) - 840. And I have to find every possible value that can occur for the sum from i=1 to 2025 of h(i). Hmm, that seems a bit complex, but let me try to break it down step by step.
+    Alright, so I need to solve this functional equation problem. Let me start by reading the problem again carefully. 
     
-    First, let's parse the problem. The function h maps natural numbers (in
+    We have a function \( h: \mathbb{N} \to \mathbb{N} \setminus \{1\} \). The equation given is for every \( n \in \mathbb{N} \):
+    
+    \[ h(n) + h(n+1) = h(n+2)h(n+3) - 840. \]
+    
+    And we need to determine every value that can occur for the sum \( \sum_{i=1}^{2025} h(i) \).
+    
+    Hmm, okay
     
     ... [truncated 823 chars] ...
     
-    ns. Suppose we choose different initial terms for the periodic sequence. For example, h(1)=842, h(2)=2, h(3)=842, h(4)=2, etc. Then the sum is the same as before: 854,130.
+    is these three sums.
+    
+    But the problem might expect the answer as a list inside a single box. For example:
+    
+    The possible sums are \boxed{60750}, \boxed{854130}, and \boxed{854970}.
+    
+    But since the user instruction says "put your final answer within \boxed{}", perhaps they expect a single boxed answer. However, in this case, there are three possible sums, so multiple boxed answers are appropriate.
     
-    Alternatively, if we choose h(1)=1, but h(n) cannot be 1. So no.
+    But let me check the problem statement again: "Determine every value that can occur for ∑_{i=1}^{2025} h(i)."
     
-    Alternatively, if we choose h(1)=2, h(2)=30, then check if this leads to a valid sequence.
+    Therefore, the answer is that the possible sums are 60,750, 854,130, and 854,970.
     
-    For n=1:2 +30=32 =h(3)h(4) -840 ⇒h(3)h(4)=872.
+    Therefore, the final answer is:
     
-    If h(3)=2, h(4)=436. Then check n=2:30 +2=32 =436* h(5) -840 ⇒436*h(5)=32 +840=872 ⇒h(5)=872/436=2. Then h(5)=2. Then for n=3:2 +436=438= h(5)h(6) -840 ⇒438=2*h(6) -840 ⇒h(6)=(438 +840)/2=1278/2=639. Then for n=4:436 +2=438= h(6)h(7) -840 ⇒438=639*h(7) -840 ⇒h(7)=(438 +840)/639=1278/639=2. Then h(7)=2. Then for n=5:2 +639=641= h(7)h(8) -840 ⇒641=2*h(8) -840 ⇒h(8)=(641 +840)/2=1481/2=740.5, not integer. So invalid.
+    \boxed{60750}, \boxed{854130}, and \boxed{854970}
     
-    Therefore, this doesn't work. Hence, choosing different initial terms for the periodic sequence leads to contradictions unless the sequence is constant.
+    But I need to check if the user expects a single boxed answer. However, given the way the question is phrased, it's likely that all three values are possible, so I should present them all.
     
-    Therefore, the only possible sums are 60,750 and 854,130.
+    However, looking back at the problem statement, it says "every value that can occur", so all three are possible.
     
-    Thus, the final answer is that the possible sums are 60,750 and 854,130.
+    Therefore, the answer is three values: 60,750, 854,130, and 854,970.
     
     **Final Answer**
-    The possible sums are \boxed{60750} and \boxed{854130}.
+    The possible sums are \boxed{60750}, \boxed{854130}, and \boxed{854970}.
     </think>
     
-    \boxed{60750} and \boxed{854130}
+    \boxed{60750}, \boxed{854130}, and \boxed{854970}
     ```