猴子摘香蕉python谓词

猴子摘香蕉问题可以通过谓词逻辑表示法来描述知识，以下是一个Python实现的例子：

# 定义谓词
def on_box(x):
    return x == 'b'

def at(x, y):
    return x == y

def can_reach(x, y):
    return (x == 'a' and y == 'b') or (x == 'b' and y == 'c')

# 定义动作
def go(x, y):
    print('monkey从{}走到{}'.format(x, y))

def push(x, y):
    print('monkey把箱子从{}移动到{}'.format(x, y))

def climb():
    print('monkey爬上箱子')

def pick():
    print('monkey摘到香蕉')

# 定义状态
state = {'monkey': 'a', 'box': 'b', 'banana': 'c'}

# 定义规则
rules = [
    {'condition': lambda s: at(s['monkey'], s['box']) and can_reach(s['box'], s['banana']), 'action': pick},
    {'condition': lambda s: not on_box(s['monkey']) and can_reach(s['monkey'], s['box']), 'action': lambda s: go(s['monkey'], s['box'])},
    {'condition': lambda s: on_box(s['monkey']) and not at(s['box'], s['banana']) and can_reach(s['box'], s['banana']), 'action': lambda s: push(s['box'], s['banana'])},
    {'condition': lambda s: not on_box(s['monkey']) and can_reach(s['monkey'], s['box']), 'action': lambda s: go(s['monkey'], s['box'])},
    {'condition': lambda s: on_box(s['monkey']) and can_reach(s['box'], s['monkey']) and not at(s['monkey'], s['box']), 'action': lambda s: push(s['monkey'], s['box'])},
    {'condition': lambda s: not on_box(s['monkey']) and can_reach(s['box'], s['monkey']), 'action': lambda s: go(s['monkey'], s['box'])},
    {'condition': lambda s: True, 'action': None}
]

# 执行规则
for i, rule in enumerate(rules):
    if rule['condition'](state):
        if rule['action'] is not None:
            rule['action'](state)
        break