Update Automated MNLP evaluation report (2026-06-08)

EVAL_REPORT.md

@@ -2,22 +2,22 @@
 
 - **Model repo:** [`cs-552-2026-flab/math_model`](https://huggingface.co/cs-552-2026-flab/math_model)
 - **Owner(s):** group **flab**
-- **Generated at:** 2026-06-06T17:20:53+00:00 (UTC)
+- **Generated at:** 2026-06-08T04:40:54+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:** [`15ef83b`](https://huggingface.co/cs-552-2026-flab/math_model/commit/15ef83b40416375e2db11e5e1171011ce45adcf1)
-- **Message:** push_best_a025_step250
-- **Committed:** 2026-06-05T01:17:14+00:00
+- **Commit:** [`e38dc63`](https://huggingface.co/cs-552-2026-flab/math_model/commit/e38dc632b04087730c509fa3a27efe6e93f3daa5)
+- **Message:** grpo-dsr-sft-8g-ckpt50-budgetRL
+- **Committed:** 2026-06-07T18:47:01+00:00
 
 ## Summary
 
 | Benchmark | Accuracy | Status |
 |---|---:|---|
-| Math | 0.2400 | ok |
+| Math | 0.3800 | ok |
 | Knowledge | — | not run |
 | Multilingual | — | not run |
 | Safety | — | not run |
@@ -30,71 +30,68 @@ _Prompts are intentionally omitted to avoid revealing benchmark contents. For mu
 
 **Correct** (1 shown)
 
-- **reference**: `1600`
-- **overall** (8/8 completions correct)
-- **extracted** (✓): `1600`
+- **reference**: `279`
+- **overall** (2/8 completions correct)
+- **extracted** (✓): `279`
 - **completion**:
 
     ```text
-    The given expression is \(441 + 2 \times 21 \times 19 + 361\).  
-    Compute each term:
-    - \(441 = 21^2\),
-    - \(2 \times 21 \times 19 = 798\),
-    - \(361 = 19^2\).
+    <think>
+    Okay, so I need to figure out how many positive integers n less than 1000 satisfy f(n) = n. The function f(n) is defined by taking the usual base-10 digits of n, finding the smallest integer b that's strictly larger than every digit in n, then interpreting the same digits as a base-b number. For example, if n is 72, the digits are 7 and 2, so the smallest b larger than both is 8. Then 72 i
     
-    Adding them:  
-    \(21^2 + 798 + 19^2 = 441 + 798 + 361 = 1600\).
+    ... [truncated 822 chars] ...
     
-    Thus, \(x = 1600\).
+    single-digit numbers satisfy \( f(n) = n \).
+       - There are 9 such numbers.
     
-    \boxed{1600}
+    2. **Two-digit numbers (10-99):**
+       - For a two-digit number \( n = 10a + b \) where \( a \) and \( b \) are digits, the maximum digit is \( m \). Then \( b = m + 1 \).
+       - For \( f(n) = n \), we need \( 10a + b = a(m + 1) + b \). This simplifies to \( 10a = a(m + 1) \) which implies \( m = 9 \).
+       - Two-digit numbers with maximum digit 9 are \( 19, 29, \ldots, 99 \), totaling 18 numbers.
+    
+    3. **Three-digit numbers (100-999):**
+       - For a three-digit number \( n = 100a + 10b + c \) where \( a, b, c \) are digits, the maximum digit is \( m \). Then \( b = m + 1 \).
+       - For \( f(n) = n \), we need \( 100a + 10b + c = a(m + 1)^2 + b(m + 1) + c \). This simplifies to \( 100a + 10b = a(m + 1)^2 + b(m + 1) \).
+       - For \( m = 9 \), \( b = 10 \), and the equation holds for all three-digit numbers with at least one digit 9. The number of such three-digit numbers is 252.
+    
+    ### Total Count:
+    - Single-digit numbers: 9
+    - Two-digit numbers: 18
+    - Three-digit numbers: 252
+    
+    Summing these, we get \( 9 + 18 + 252 = 279 \).
+    
+    Thus, the number of positive integers \( n < 1000 \) that satisfy \( f(n) = n \) is \(\boxed{279}\).
     ```
 
 **Incorrect** (1 shown)
 
-- **reference**: `4`
+- **reference**: `75`
 - **overall** (0/8 completions correct)
-- **extracted** (✗): `6^{1/3}(6^{1/3} - 1)`
+- **extracted** (✗): `<no answer>`
 - **completion**:
 
     ```text
-    The given constraint can be rewritten as:
-    $$
-    \sum_{\text{cyc}} (x-1)^3 = 4,
-    $$
-    where the sum is cyclic over $x, y, z$. Let $a = x-1$, $b = y-1$, $c = z-1$; then $a, b, c \ge -1$ (since $x, y, z \ge 0$) and we have
-    $$
-    a^3 + b^3 + c^3 = 4.
-    $$
-    
-    Define $f(a,b,c) = a^2 + b^2 + c^2 - (a+b+c)$. We want to maximize $f$ under the constraint $a^3 + b^3 + c^3 = 4$ with $a, b, c \ge -1$.
+    <think>
+    Okay, so I need to solve this geometry problem about a parallelogram. Let me start by understanding the given information and what's being asked.
     
-    **Step 1: Symmetry a
+    First, the problem states that in parallelogram PQRS, angle S is 60 degrees. PS is 2 units long, and PQ is sqrt(3) + 1 units. N is the midpoint of PS, so that means PN = NS = 1. Then, there's a point E on segment NQ such that RE bisects angle R.
     
     ... [truncated 823 chars] ...
     
-    /3}$, $c = 0$, $g(-2^{1/3}) = 2(4) + 0 - 2(-2^{1/3}) = 8 + 2^{4/3} \approx 10.464$.  
-    - For $a = -1.5$, $c \approx 2.0$, $g(-1.5) = 2(2.25) + 4 - 2(-1.5) - 2 = 4.5 + 4 + 3 - 2 = 9$.  
-    - The derivative $g'(a)$ is positive for $a < -1$, so $g$ increases from $a = -\infty$ to $a = -1$, reaching a maximum at $a = -1$.
+    t the problem involves sqrt(3) and 60 degrees, the answer might be 30 degrees, 45 degrees, or 60 degrees. However, our calculation shows approximately 51 degrees, which is close to 45 degrees but not exact.
+    
+    Alternatively, maybe there's a geometric insight I'm missing. Let me think again.
     
-    Comparing all these values, the largest possible $g$ is $10.464$ when $a = -2^{1/3}$, $b = -2^{1/3}$, $c = 0$.
+    Given that angle R is 120 degrees, and RE bisects it into two 60-degree angles. If we can show that triangle REQ is a 30-60-90 triangle, then angle REQ would be 30 degrees. Alternatively, if triangle EQN is equilateral, but that seems unlikely.
     
-    **Step 6: Back to $x, y, z$.**  
-    Recall $a = x-1$, $b = y-1$, $c = z-1$.  
-    - $a = -2^{1/3}$ implies $x = 1 - 2^{1/3} \approx -0.26$ (but $x$ must be nonnegative, so this is invalid).  
-    - $a = -1$ gives $x = 0$, $y = 0$, $c = \sqrt[3]{6} \approx 1.817$.  
-    - $c = 0$ gives $z = 1$, and then $x = y = -1$ which are not allowed.
+    Alternatively, since N is the midpoint of PS, and PS is 2, so PN = NS =1. Maybe there's a symmetry or using properties of midpoints and angle bisectors.
     
-    Thus the only valid configuration that achieves the maximum is $x = y = 0$, $z = \sqrt[3]{6}$, because any other attempt either yields a nonnegative $x$ or violates the nonnegativity of all variables.
+    Alternatively, consider triangle PNS. Since N is the midpoint of PS, and PN =1, PS=2. If we can relate this to other parts of the parallelogram.
     
-    **Step 7: Compute the maximum value.**  
-    $$
-    x^2 + y^2 + z^2 - x - y - z = 0^2 + 0^2 + (\sqrt[3]{6})^2 - 0 - 0 - \sqrt[3]{6} = 6^{2/3} - 6^{1/3} = 6^{1/3}(6^{1/3} - 1).
-    $$
+    Alternatively, since PQ = sqrt(3)+1, and PS=2, maybe using the Law of Cosines in triangle PQS or something.
     
-    This is the largest possible value.
+    Alternatively, since the problem involves angle bisectors and midpoints, maybe using coordinate geometry to find the required angle.
     
-    $$
-    \boxed{6^{1/3}(6^{1/3} - 1)}
-    $$
+    But given the time I've spent and the calculations leading to approximately 51 degrees, which is close to 45 degrees but
     ```