推理时间计算

class InferenceTimeCompute:
    """
    推理时间计算的简化实现
    """
    def __init__(self, model, max_thinking_tokens=100000):
        self.model = model
        self.max_thinking_tokens = max_thinking_tokens
        
    def adaptive_thinking(self, problem, difficulty_score):
        """
        根据问题难度动态调整思考时间
        """
        # 简单问题：少量思考
        if difficulty_score < 0.3:
            thinking_budget = 1000
            num_samples = 1
        # 中等问题：适度思考
        elif difficulty_score < 0.7:
            thinking_budget = 10000
            num_samples = 3
        # 困难问题：深度思考
        else:
            thinking_budget = self.max_thinking_tokens
            num_samples = 6  # o3高效模式
        
        return self.think_and_solve(
            problem, 
            thinking_budget, 
            num_samples
        )
    
    def think_and_solve(self, problem, budget, samples):
        """
        执行思考过程
        """
        solutions = []
        
        for _ in range(samples):
            thinking_chain = []
            tokens_used = 0
            
            while tokens_used < budget:
                # 生成思考步骤
                thought = self.generate_thought(problem, thinking_chain)
                thinking_chain.append(thought)
                tokens_used += len(thought)
                
                # 检查是否找到解决方案
                if self.is_solution_found(thought):
                    break
                    
                # 检查是否需要回溯
                if self.should_backtrack(thought):
                    thinking_chain = self.backtrack(thinking_chain)
            
            solutions.append(self.extract_solution(thinking_chain))
        
        # 选择最佳解决方案
        return self.select_best_solution(solutions)

# 数学竞赛问题（AIME 2024）
problem = """
找出最小的正整数n，使得n! + (n+1)! + (n+2)! 
是一个完全平方数。
"""

# 传统模型：直接尝试
traditional_model_response = "让我计算...n = 3"  # 错误

# o1模型：中等思考
o1_response = """
[内部思考10k代币]
让我系统地分析...
首先，我注意到 n! + (n+1)! + (n+2)! = n!(1 + (n+1) + (n+1)(n+2))
= n!(1 + n + 1 + n² + 3n + 2) = n!(n² + 4n + 4) = n!(n+2)²
因此需要n!是完全平方数...
[继续推理]
答案：n = 3
"""

# o3模型：深度思考
o3_response = """
[内部思考100k代币，多路径探索]
[尝试路径1...发现问题...回溯]
[尝试路径2...验证...成功]
经过详细分析和验证，答案是n = 3
置信度：99.9%
"""

def assess_problem_difficulty(problem):
    """
    评估问题难度以决定推理时间
    """
    indicators = {
        'multi_step': check_multi_step(problem),
        'abstract': check_abstraction_level(problem),
        'constraint_complexity': analyze_constraints(problem),
        'domain_expertise': check_domain_requirements(problem)
    }
    
    # 根据指标计算难度分数
    difficulty = sum(indicators.values()) / len(indicators)
    
    # 决定推理策略
    if difficulty > 0.8:
        return {
            'mode': 'deep_thinking',
            'thinking_tokens': 100000,
            'samples': 1024,  # o3低效模式
            'estimated_cost': '$50-100'
        }
    elif difficulty > 0.5:
        return {
            'mode': 'moderate_thinking',
            'thinking_tokens': 10000,
            'samples': 6,  # o3高效模式
            'estimated_cost': '$5-10'
        }
    else:
        return {
            'mode': 'quick_thinking',
            'thinking_tokens': 1000,
            'samples': 1,
            'estimated_cost': '<$1'
        }

def cost_benefit_analysis(problem_value, solution_accuracy):
    """
    推理时间计算的成本效益分析
    """
    strategies = {
        'standard': {
            'cost': 0.1,
            'accuracy': 0.4,
            'time': 1
        },
        'o1_moderate': {
            'cost': 5,
            'accuracy': 0.7,
            'time': 10
        },
        'o3_efficient': {
            'cost': 20,
            'accuracy': 0.85,
            'time': 30
        },
        'o3_intensive': {
            'cost': 100,
            'accuracy': 0.95,
            'time': 300
        }
    }
    
    # 计算每个策略的ROI
    best_strategy = None
    best_roi = -float('inf')
    
    for name, metrics in strategies.items():
        expected_value = problem_value * metrics['accuracy']
        roi = (expected_value - metrics['cost']) / metrics['cost']
        
        if roi > best_roi:
            best_roi = roi
            best_strategy = name
    
    return best_strategy, best_roi