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| critic_judge_prompt = """\ |
| I will provide you with a question and a robot’s answer, divided into several sections. Your task is to evaluate each section of the robot’s answer for any errors. |
| |
| **Evaluation Criteria:** |
| - Evaluate each section independently. Assess each section based solely on the accuracy and logic within that section. |
| - Do not consider subjective elements such as redundancy or stylistic preferences as errors. |
| - Do not consider corrections or reflections made in later sections. Even if a later section acknowledges and fixes an earlier mistake, the original section must still be marked as erroneous. |
| - If a subsequent section contains an error caused by an earlier section's mistake, do not count it as a new error. |
| |
| **Output Format:** |
| - If you think all sections of the robot's answer are correct, output in the following format: |
| Conclusion: no error |
| - If you think any section contains an error, output in the following format: |
| Conclusion: yes |
| Error Section Number: [section number] |
| Explanation: [explanation for the error in this section] |
| Error Section Number: [section number] |
| Explanation: [explanation for the error in this section] |
| … (repeat for each erroneous section) |
| |
| |
| **example:** |
| - Question: Two numbers, $x$ and $y$ are selected at random from the interval $(0,3)$. What is the probability that a triangle with sides of length 1, $x$, and $y$ exists? |
| - Robot's Answer: |
| section1: |
| So I've got this probability problem here. It says that two numbers, x and y, are selected at random from the interval (0,3), and I need to find the probability that a triangle with sides of length 1, x, and y exists. Okay, let's break this down. |
| |
| section2: |
| First, I need to recall the triangle inequality theorem. It states that for any triangle with sides a, b, and c, the following must be true: |
| 1. a + b > c |
| 2. a + c > b |
| 3. b + c > a |
| |
| section3: |
| So, in this case, the sides are 1, x, and y. That means I need to satisfy all three inequalities: |
| 1. 1 + x > y |
| 2. 1 + y > x |
| 3. x + y > 1 |
| Alright, now, to find the probability, I need to find the ratio of the area where these inequalities hold to the total area from which x and y are selected. |
| |
| section4: |
| Now, I need to find the area where all three inequalities are satisfied. |
| Let me consider each inequality one by one. |
| First inequality: 1 + x > y |
| This can be rewritten as y < 1 + x. |
| Second inequality: 1 + y > x |
| This can be rewritten as y > x - 1. |
| Third inequality: x + y > 1 |
| This can be rewritten as y > 1 - x. |
| Now, since x and y are both positive (from (0,3)), 1 - x could be negative if x > 1. But since y is positive, y > 1 - x will always hold true when x < 1, and when x >=1, y > 1 - x is always true because y > 0 and 1 - x <=0. So, this inequality is always satisfied in the given range. Therefore, I don't need to worry about this inequality separately. |
| |
| section5: |
| So, effectively, I only need to consider the first two inequalities: |
| 1. y < 1 + x |
| 2. y > x - 1 |
| And, of course, x and y are between 0 and 3. |
| |
| - judge result: |
| Conclusion: yes |
| Error Section Number: 4 |
| Explanation: The analysis of the inequality \( x + y > 1 \) is incorrect. The argument that "this inequality is always satisfied in the given range" is flawed. Specifically, for small values of \( x \) and \( y \), such as when both are close to zero, the inequality \( x + y > 1 \) does not always hold. The constraint \( x + y > 1 \) must still be explicitly considered in the probability calculation. |
| |
| **Input:** |
| - Question: {{question}} |
| - Robot's Answer: {{model_output}} |
| - judge result: \ |
| """ |