Automated MNLP evaluation report (2026-05-03)

EVAL_REPORT.md

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+# Automated MNLP evaluation report
+
+- **Model repo:** [`cs-552-2026-taadmin/math_model`](https://huggingface.co/cs-552-2026-taadmin/math_model)
+- **Owner(s):** group **taadmin**
+- **Generated at:** 2026-05-03T02:20:59+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._
+
+## Summary
+
+| Benchmark | Metric | Accuracy | # problems | Status |
+|---|---|---:|---:|---|
+| Math | `pass@8` | 0.3000 | 10 | ok |
+| Knowledge | `pass@1` | — | — | not run |
+| Multilingual | `pass@1` | — | — | not run |
+| Safety | `pass@1` | — | — | not run |
+
+## Sample completions
+
+### Math
+
+**Correct** (2 shown)
+
+- **prompt** (truncated): `Determine the remainder obtained when the polynomial \((5x+9)^{611}+(x+5)^{11}+(x-1)^{11}+3x^2+1\) is divided by \(x+2\).`
+- **reference**: `12`
+- **extracted answers** (7/8 correct, ✓✓✓✗✓✓✓✓):
+  - `12`
+  - `12`
+  - `12`
+  - `<no answer>`
+  - `12`
+  - `12`
+  - `12`
+  - `12`
+- **raw completions**:
+  - completion 1:
+    ```text
+    <think>
+    Okay, so I need to find the remainder when this big polynomial is divided by x + 2. Hmm, I remember that when you divide a polynomial by x + a, the remainder is just the value of the polynomial evaluated at x = -a. So, in this case, since the divisor is x + 2, the remainder should be the polynomial evaluated at x = -2. Let me write that down:
+    
+    Remainder = P(-2), where P(x) = (5x + 9)^611 + (x + 5)^11 + (x - 1)^11 + 3x² + 1.
+    
+    So I need to compute P(-2). Let me calculate each term step by step.
+    
+    First term: (5x + 9)^611 evaluated at x = -2. Let me compute 5*(-2) + 9. That's -10 + 9 = -1. So the first term becomes (-1)^611. Now, 611 is an odd number, so (-1)^611 = -1. Got that.
+    
+    Second...
+    ```
+  - completion 2:
+    ```text
+    <think>
+    Okay, so I need to find the remainder when this big polynomial is divided by x + 2. Hmm, I remember that when you divide a polynomial by x - a, the remainder is just the value of the polynomial at x = a. But here, the divisor is x + 2, which is the same as x - (-2). So, according to the Remainder Theorem, the remainder should be the value of the polynomial evaluated at x = -2. That makes sense. So, I just need to plug in x = -2 into the polynomial and compute the result. Let me write that down.
+    
+    The polynomial is (5x + 9)^611 + (x + 5)^11 + (x - 1)^11 + 3x² + 1. Let me substitute x = -2 into each term.
+    
+    First term: (5*(-2) + 9)^611. Let me compute 5*(-2) which is -10, then add 9: -10...
+    ```
+
+- **prompt** (truncated): `Determine the distance, in units, between the points (2, −6) and (−4, 3), and write your answer in simplest radical form.`
+- **reference**: `3\sqrt{13}`
+- **extracted answers** (8/8 correct, ✓✓✓✓✓✓✓✓):
+  - `3\sqrt{13}`
+  - `3\sqrt{13}`
+  - `3\sqrt{13}`
+  - `3\sqrt{13}`
+  - `3\sqrt{13}`
+  - `3\sqrt{13}`
+  - `3\sqrt{13}`
+  - `3\sqrt{13}`
+- **raw completions**:
+  - completion 1:
+    ```text
+    <think>
+    Okay, so I need to find the distance between the two points (2, -6) and (-4, 3). Hmm, I remember there's a formula for the distance between two points in a coordinate plane. Let me think... Oh right, the distance formula is derived from the Pythagorean theorem. It's something like the square root of [(x2 - x1) squared plus (y2 - y1) squared]. Let me write that down to make sure.
+    
+    So the formula is: distance = √[(x2 - x1)² + (y2 - y1)²]. Yeah, that sounds right. Let me check. If I have two points, say (x1, y1) and (x2, y2), then the horizontal distance is x2 - x1 and the vertical distance is y2 - y1. Then, the distance between them is the hypotenuse of a right triangle with those two...
+    ```
+  - completion 2:
+    ```text
+    <think>
+    Okay, so I need to find the distance between the two points (2, −6) and (−4, 3). Hmm, I remember there's a formula for the distance between two points in a coordinate plane. Let me try to recall. I think it's something like the square root of the sum of the squares of the differences in the x-coordinates and y-coordinates. Wait, let me make sure. 
+    
+    Right, the distance formula is derived from the Pythagorean theorem. If you have two points, (x1, y1) and (x2, y2), then the distance between them is √[(x2 - x1)² + (y2 - y1)²]. Yeah, that sounds right. So I need to plug in the coordinates of the two points into this formula.
+    
+    Let me write down the coordinates. The first point is (2, −6),...
+    ```
+
+**Incorrect** (2 shown)
+
+- **prompt** (truncated): `For a 4-question math test, each question’s topic is selected randomly and independently from the set {algebra, combinatorics, geometry, number theory}. Conditional on the test containing at least one algebra question, at least one combinatorics question, and at least one geometry question, determine the probability that it also contains at least one number theory question.`
+- **reference**: `\frac{2}{5}`
+- **extracted answers** (0/8 correct, ✗✗✗✗✗✗✗✗):
+  - `<no answer>`
+  - `<no answer>`
+  - `<no answer>`
+  - `<no answer>`
+  - `<no answer>`
+  - `<no answer>`
+  - `<no answer>`
+  - `\dfrac{2}{5}`
+- **raw completions**:
+  - completion 1:
+    ```text
+    <think>
+    Okay, so I need to find the probability that a 4-question math test, where each question is randomly selected from algebra, combinatorics, geometry, and number theory, also contains at least one number theory question, given that it has at least one of each of algebra, combinatorics, and geometry. 
+    
+    First, let me make sure I understand the problem correctly. The test has 4 questions, each question is independently chosen from the four topics: algebra, combinatorics, geometry, number theory. But we are given that the test contains at least one algebra, at least one combinatorics, and at least one geometry question. So we need to find the probability that it also contains at least one...
+    ```
+  - completion 2:
+    ```text
+    <think>
+    Okay, so I need to find the probability that a 4-question math test, where each question is randomly selected from algebra, combinatorics, geometry, and number theory, has at least one number theory question given that it has at least one of each of algebra, combinatorics, and geometry. 
+    
+    First, let me make sure I understand the problem. The test has 4 questions, each question is independently chosen from the four topics. But we are given that the test contains at least one algebra, at least one combinatorics, and at least one geometry question. So, we need to find the probability that it also contains at least one number theory question under this condition.
+    
+    So, this is a condition...
+    ```
+
+- **prompt** (truncated): `Let \(S\) denote the collection of all ordered triples \((a,b,c)\) of positive integers that can occur as the side lengths of a triangle. Determine the value of
+\[
+\sum_{(a,b,c)\in S}\frac{2^a}{3^b5^c}.
+\]`
+- **reference**: `\frac{17}{21}`
+- **extracted answers** (0/8 correct, ✗✗✗✗✗✗✗✗):
+  - `<no answer>`
+  - `<no answer>`
+  - `<no answer>`
+  - `<no answer>`
+  - `<no answer>`
+  - `<no answer>`
+  - `<no answer>`
+  - `<no answer>`
+- **raw completions**:
+  - completion 1:
+    ```text
+    <think>
+    Okay, so I need to find the sum of 2^a / 3^b 5^c for all ordered triples (a, b, c) of positive integers that can form a triangle. Hmm, let me start by recalling some basics about triangles and their side lengths.
+    
+    First, for three positive integers a, b, c to form a triangle, they must satisfy the triangle inequality. That is, the sum of any two sides must be greater than the third. So, the conditions are:
+    
+    a + b > c,
+    a + c > b,
+    b + c > a.
+    
+    But since a, b, c are positive integers, these inequalities can be rewritten as:
+    
+    c < a + b,
+    a < b + c,
+    b < a + c.
+    
+    But since a, b, c are positive integers, the most restrictive condition is probably c < a + b. Similarly, the other inequalities ar...
+    ```
+  - completion 2:
+    ```text
+    <think>
+    Okay, so I need to find the sum over all ordered triples (a, b, c) of positive integers that can form a triangle, of the term 2^a / (3^b 5^c). Hmm, let me start by recalling some basics about triangles and how their side lengths relate.
+    
+    First, for three positive integers a, b, c to form a triangle, they must satisfy the triangle inequality. That is, the sum of any two sides must be greater than the third. So, the conditions are:
+    
+    a + b > c,
+    
+    a + c > b,
+    
+    b + c > a.
+    
+    But since a, b, c are positive integers, I can think of them as positive integers where each of these inequalities must hold. However, since the problem says "ordered triples of positive integers that can occur as the sid...
+    ```