用离散语义熵和困惑度检测法学硕士的幻觉
使用大型语言模型 (LLM) 时,发现幻觉可能很棘手。我们可以使用困惑度、蕴涵和离散语义熵来更好地识别潜在的幻觉,而不是仅仅依靠法学硕士作为判断(这仍然可能会出错,并且许多评估框架仅使用它来检测幻觉)。虽然我在这里使用法学硕士来检测蕴涵,但这不是必要的。也就是说,这种方法最适合那些有直接、事实答案的问题——那些不太模糊或主观的问题。您如何看待使用这些组合指标来更好地检测幻觉?我知道代码可以改进/优化,但目标是快速测试它的工作原理。
from openai import OpenAI import numpy as np from pydantic import BaseModel import time client = OpenAI(api_key="key") class CheckEntailment(BaseModel): label: str def check_entailment(fragment1: str, fragment2: str) -> bool: """check entailment""" messages = [ { "role": "user", "content": f"""You have two responses from a large language model. Check if the meaning of one repsonse is entailed by the other, or if there is a contradiction. Return '0' if entailment. Return '1' if contradiction. Return only the label, without any explanation. \n Response1: \n {fragment1}\n\n Response2: \n {fragment2}""", } ] completion = client.beta.chat.completions.parse( model="gpt-4o-mini", messages=messages, temperature=0.1, logprobs=True, top_logprobs=2, response_format=CheckEntailment, ) entailment = False # print(completion.choices[0].logprobs.content[3].top_logprobs) for top_logprob in completion.choices[0].logprobs.content[3].top_logprobs: print(top_logprob.token, np.round(np.exp(top_logprob.logprob), 2)) if "0" in top_logprob.token and np.exp(top_logprob.logprob) > 0.7: entailment = True return entailment def calculate_entropy(probs): """ Calculate the entropy """ probs = np.array(probs) probs = probs / probs.sum() probs = probs[probs > 0] entropy = -np.sum(probs * np.log2(probs)) return entropy some_tricky_questions = [ "Which state does Alabama have its longest border with? Is it Florida or Tennessee?", "Who hosted the British Gameshow Countdown in 2007: a) Nick Hewer b) Richard Whiteley c) Jeff Stelling?", "Trivia question: Which Black Eyed Peas band member was the only one to host Saturday Night Live?", "What year in the 1980s were the FIS Alpine World Ski Championships hosted in Argentina?", "How many Brazilian numbers are there between 1-6?", "Which Israeli mathematician founded an online sequences repository in the 1970s?", "Write the 7 english words that have three consecutive double letters. No need to provide explanations, just say the words.", # adding two questions where it should not hallucinate "What is the capital of India?", "what is the full form of CPU?", ] for question in some_tricky_questions: print("question", question) messages = [{"role": "user", "content": f"{question}"}] gpt_response = client.chat.completions.create( model="gpt-4o-mini", messages=messages, temperature=0.1, logprobs=True, max_completion_tokens=60, ) time.sleep(2) # get perplexity score using a low temperature response logprobs = [token.logprob for token in gpt_response.choices[0].logprobs.content] perplexity_score = np.round(np.exp(-np.mean(logprobs)), 2) # initialize clusters with the first response clusters = [[gpt_response.choices[0].message.content]] # generate some more responses using higher temperature and check entailment gpt_response = client.chat.completions.create( model="gpt-4o-mini", messages=messages, n=7, temperature=0.9, logprobs=True, max_completion_tokens=60, ) time.sleep(2) # check entailment and form clusters responses = [choice.message.content for choice in gpt_response.choices] for response in responses[1:]: found_cluster = False for cluster in clusters: if check_entailment(cluster[0], response): cluster.append(response) found_cluster = True break if not found_cluster: clusters.append([response]) cluster_probs = [len(cluster) / (len(responses) + 1) for cluster in clusters] discrete_entropy = calculate_entropy(cluster_probs) print("clusters", clusters) print("no of clusters", len(clusters)) print("perplexity", perplexity_score) print("entropy", discrete_entropy)
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