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

EVAL_REPORT.md

@@ -2,16 +2,22 @@
 
 - **Model repo:** [`cs-552-2026-vibe-trainers/math_model`](https://huggingface.co/cs-552-2026-vibe-trainers/math_model)
 - **Owner(s):** group **vibe-trainers**
-- **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:** [`37abb85`](https://huggingface.co/cs-552-2026-vibe-trainers/math_model/commit/37abb85a14cf832bc2b6ecb7ff5edd20213c56b3)
+- **Message:** Update chat_template.jinja
+- **Committed:** 2026-06-04T20:08:52+00:00
+
 ## Summary
 
 | Benchmark | Accuracy | Status |
 |---|---:|---|
-| Math | 0.3900 | ok |
+| Math | 0.4100 | ok |
 | Knowledge | — | not run |
 | Multilingual | — | not run |
 | Safety | — | not run |
@@ -24,54 +30,142 @@ _Prompts are intentionally omitted to avoid revealing benchmark contents. For mu
 
 **Correct** (1 shown)
 
-- **reference**: `194`
-- **overall** (2/8 completions correct)
-- **extracted** (✓): `194`
+- **reference**: `(3,2,5)`
+- **overall** (3/8 completions correct)
+- **extracted** (✓): `(3, 2, 5)`
 - **completion**:
 
     ```text
-    We are given a $194 \times 194$ board, and we are to find the **least positive integer $k$** such that we can choose $k$ cells to mark, and there exists **exactly one tiling** of the board using $1 \times 2$ and $2 \times 1$ dominoes, where **every domino contains at most one marked cell**.
+    We are given the problem:
+    
+    > Determine all ordered triples $(p, m, n)$ where $p$ is a prime greater than 2 and $m, n$ are positive integers with $m \le n$, such that $m^{p-1} + n$ and $n^{p-1} + m$ are both powers of $p$.
     
     ---
     
-    ### Step 1: Understand the problem
+    ### Step 1: Understand the conditions
     
-    - The board is $194 \times 194$, so it has $194^2 = 37636$ cells.
+    We are told:
     
-    ... [truncated 822 chars] ...
+    - $p$ is a **prime greater than 2**, so $p \ge 3$.
+    - $m, n$ are **positive integers** with $m \le n$.
+    - $m^{p-1} + n$ a
     
-    ose we mark $k$ cells. We want **exactly one perfect matching** (tiling) of the board such that **each domino contains at most one marked cell**.
+    ... [truncated 823 chars] ...
     
-    This is equivalent to: **Find the smallest $k$ such that there is exactly one perfect matching in the graph where each edge is allowed to contain at most one marked cell**.
+    $m = 3, n = 4$:
     
-    ---
+    $$
+    3^6 + 4 = 733, \quad 4^6 + 3 = 4099
+    $$
+    
+    Not powers of 7.
+    
+    Try $m = 2, n = 6$:
+    
+    $$
+    2^6 + 6 = 70, \quad 6^6 + 2 = 46658
+    $$
+    
+    Not powers of 7.
     
-    ### Step 4: Minimum number of marked cells
+    Try $m = 1, n = 6$:
     
-    We want **only one perfect matching** in the graph, and **each domino (edge in the matching) contains at most one marked cell**.
+    $$
+    1^6 + 6 = 7 = 7^1, \quad 6^6 + 1 = 46657
+    $$
+    
+    Only one is a power of 7.
+    
+    Try $m = 2, n = 7$:
+    
+    $$
+    2^6 + 7 = 65, \quad 7^6 + 2 = 117649 + 2 = 117651
+    $$
+    
+    Not powers of 7.
+    
+    Try $m = 3, n = 5$:
+    
+    $$
+    3^6 + 5 = 733, \quad 5^6 + 3 = 15628
+    $$
     
-    This is equivalent to: **Find the smallest $k$ such that the graph has only one perfect matching, and the matching is such that each edge contains at most one marked cell**.
+    Not powers of 7.
     
-    This is a known result in combinatorics.
+    So for $p = 7$, no valid triples.
     
     ---
     
-    ### Step 5: Use of known results
+    ### Step 3: Try $p = 3$ again
+    
+    We already found:
+    
+    $$
+    (p, m, n) = (3, 2, 5)
+    $$
+    
+    Check if there are more:
     
-    It is a known result in combinatorics that the **minimum number of marked cells** needed to ensure **exactly one tiling** of the board with **each domino containing at most one marked cell** is:
+    Try $m = 2, n = 5$:
     
     $$
-    \boxed{194}
+    2^2 + 5 = 9 = 3^2, \quad 5^2 + 2 = 27 = 3^3
     $$
     
-    This is because the board is $194 \times 194$, and to ensure that **only one tiling** is possible, we need to **mark at least one cell per row and column** to prevent multiple tilings.
+    ✅ Both are powers of 3.
+    
+    Try $m = 1, n = 2$:
+    
+    $$
+    1^2 + 2 = 3 = 3^1, \quad 2^2 + 1 = 5
+    $$
+    
+    Only one is a power of 3.
+    
+    Try $m = 1, n = 3$:
+    
+    $$
+    1^2 + 3 = 4, \quad 3^2 + 1 = 10
+    $$
+    
+    Not powers of 3.
+    
+    Try $m = 2, n = 3$:
+    
+    $$
+    2^2 + 3 = 7, \quad 3^2 + 2 = 11
+    $$
+    
+    Not powers of 3.
+    
+    Try $m = 2, n = 4$:
+    
+    $$
+    2^2 + 4 = 8, \quad 4^2 + 2 = 18
+    $$
+    
+    Not powers of 3.
+    
+    Try $m = 2, n = 6$:
+    
+    $$
+    2^2 + 6 = 10, \quad 6^2 + 2 = 38
+    $$
+    
+    Not powers of 3.
+    
+    So the only valid triple for $p = 3$ is:
+    
+    $$
+    (p, m, n) = (3, 2, 5)
+    $$
     
     ---
     
-    ### ✅ Final Answer:
+    ### Final Answer:
     
     $$
-    \boxed{194}
+    \boxed{(3, 2, 5)}
     $$
     ```
 
@@ -79,11 +173,11 @@ _Prompts are intentionally omitted to avoid revealing benchmark contents. For mu
 
 - **reference**: `$g(x)=c, g(x)=\lceil x \rceil, g(x)=\lfloor x \rfloor$`
 - **overall** (0/8 completions correct)
-- **extracted** (✗): `h(x) = x`
+- **extracted** (✗): `h(x) = 0`
 - **completion**:
 
     ```text
-    We are given a functional equation involving a mapping $ h : \mathbb{Q} \to \mathbb{Z} $, and the condition:
+    We are given a functional equation:
     
     $$
     h(x) = h\left( \frac{h(bx - a) + a}{b} \right)
@@ -91,99 +185,94 @@ _Prompts are intentionally omitted to avoid revealing benchmark contents. For mu
     
     for all $ x \in \mathbb{Q} $, $ a \in \mathbb{Z} $, and $ b \in \mathbb{N} $.
     
+    We are to find all functions $ h : \mathbb{Q} \to \mathbb{Z} $ satisfying this equation.
+    
     ---
     
-    ### Step 1: Understand the structure of the equation
+    ### Step 1: Try simple candidate functions
     
-    Let’s rewrite the equation for clarity:
+    Let’s try a simple form of $ h(x) $, such as:
     
     $$
-    h(x) = h\left( \frac{h(bx - a) + a}{b} \right)
+    h(x) = kx + c
     $$
     
-    Thi
-    
-    ... [truncated 821 chars] ...
+    for some constants $ k,
     
-    ### Step 5: Try $ n = 0 $
+    ... [truncated 823 chars] ...
     
-    If $ n = 0 $, then $ h(x) = c $, a constant function.
+    , since $ b \in \mathbb{N} $, this implies $ c = 0 $ (since $ k = 0 $ or $ k = 1 $, and $ b > 0 $).
     
-    Plug into the original equation:
+    So the only possible candidate is:
     
     $$
-    h(x) = h\left( \frac{h(bx - a) + a}{b} \right)
+    h(x) = 0
     $$
     
-    If $ h(x) = c $, then:
-    
-    $$
-    c = h\left( \frac{c + a}{b} \right) = c
-    $$
-    
-    So this is valid for any $ c \in \mathbb{Z} $.
-    
     ---
     
-    ### Step 6: Try $ n = 1 $
+    ### Step 2: Verify that $ h(x) = 0 $ satisfies the equation
     
-    If $ n = 1 $, then $ h(x) = x + c $.
-    
-    Let’s check if this works.
-    
-    Then:
+    Let’s check:
     
     $$
-    h(bx - a) = bx - a + c
+    h(x) = 0
     $$
-    
-    Then:
-    
     $$
-    \frac{h(bx - a) + a}{b} = \frac{bx - a + c + a}{b} = \frac{bx + c}{b} = x + \frac{c}{b}
+    h(bx - a) = 0
+    $$
+    $$
+    \frac{h(bx - a) + a}{b} = \frac{0 + a}{b} = \frac{a}{b}
     $$
-    
-    Then:
-    
     $$
-    h\left( \frac{h(bx - a) + a}{b} \right) = h\left( x + \frac{c}{b} \right) = x + \frac{c}{b} + c
+    h\left( \frac{a}{b} \right) = 0
     $$
     
-    Now compare with $ h(x) = x + c $:
+    So the equation becomes:
     
     $$
-    h(x) = x + c \quad \text{and} \quad h\left( \frac{h(bx - a) + a}{b} \right) = x + \frac{c}{b} + c
+    h(x) = 0 = 0
     $$
     
-    So unless $ \frac{c}{b} = 0 $, these are not equal.
-    
-    Thus, this only works if $ c = 0 $, and then $ h(x) = x $.
+    which is true.
     
     ---
     
-    ### Step 7: Final check
+    ### Step 3: Are there other possible functions?
+    
+    Suppose $ h(x) $ is not identically zero. Then there exists some $ x \in \mathbb{Q} $ such that $ h(x) \ne 0 $. Let’s suppose $ h(x) = c \ne 0 $ for some $ x \in \mathbb{Q} $.
     
-    Let’s test $ h(x) = x $ in the original equation:
+    Then:
     
     $$
-    h(x) = h\left( \frac{h(bx - a) + a}{b} \right)
+    h(bx - a) = c
+    $$
+    $$
+    \frac{h(bx - a) + a}{b} = \frac{c + a}{b}
+    $$
+    $$
+    h\left( \frac{c + a}{b} \right) = c
     $$
     
-    If $ h(x) = x $, then:
+    This implies:
     
     $$
-    x = \frac{h(bx - a) + a}{b} = \frac{bx - a + a}{b} = x
+    h\left( \frac{c + a}{b} \right) = c
     $$
     
-    So this works.
+    But this must hold for all $ a \in \mathbb{Z} $, and for all $ b \in \mathbb{N} $. So $ h $ must be constant on all rational numbers of the form $ \frac{c + a}{b} $, which is a dense set in $ \mathbb{Q} $, unless $ h $ is constant on the entire domain.
+    
+    Thus, the only possible solution is:
+    
+    $$
+    \boxed{h(x) = 0}
+    $$
     
     ---
     
-    ### ✅ Final Answer:
+    ### Final Answer:
     
     $$
-    \boxed{h(x) = x}
+    \boxed{h(x) = 0}
     $$
-    
-    is the only solution.
     ```