小学奥数只能理解，不能证明
需要使用图论中的“广度优先搜索”（BFS）或者“状态空间分析”的方法。虽然这些方法超出了小学奥数的范围，但它们提供了一个严谨的数学证明。



广度优先搜索是图论中的一种遍历算法，用于找到从起点到终点的最短路径。应用到这里，就是在状态图中找到从
(8,0,0) 到
(4,0,4) 的最短路径。

这个提可以当成信奥题目了
一下是chatgpt给的，厚颜无耻直接用了别人函数
# -*- coding: utf-8 -*-
"""
Spyder Editor

This is a temporary script file.
"""
# We will implement a BFS algorithm to search through all possible states
# and find the shortest path that divides the 8 liters of water into two
# containers with exactly 4 liters each.

from collections import deque

def bfs_water_jug():
    # Initial state: (8 liters in main jug, 0 liters in 3L jug, 0 liters in 5L jug)
    initial_state = (8, 0, 0)
    # Goal states: (4, 0, 4) or (4, 4, 0)
    goal_states = {(4, 0, 4), (4, 4, 0)}
    
    # Define the queue for BFS
    queue = deque([(initial_state, 0)])  # (state, steps)
    
    # Visited states to avoid cycles
    visited = set()
    visited.add(initial_state)
    
    # All possible capacities
    max_main = 8
    max_3l = 3
    max_5l = 5
    
    while queue:
        (main, jug3, jug5), steps = queue.popleft()
        
        # Check if we've reached the goal
        if (main, jug3, jug5) in goal_states:
            return steps
        
        # Generate all possible next states by pouring water
        next_states = []
        
        # Pour water from main jug to 3L jug
        if main > 0 and jug3 < max_3l:
            transfer = min(main, max_3l - jug3)
            next_states.append((main - transfer, jug3 + transfer, jug5))
        
        # Pour water from main jug to 5L jug
        if main > 0 and jug5 < max_5l:
            transfer = min(main, max_5l - jug5)
            next_states.append((main - transfer, jug3, jug5 + transfer))
        
        # Pour water from 3L jug to main jug
        if jug3 > 0 and main < max_main:
            transfer = min(jug3, max_main - main)
            next_states.append((main + transfer, jug3 - transfer, jug5))
        
        # Pour water from 5L jug to main jug
        if jug5 > 0 and main < max_main:
            transfer = min(jug5, max_main - main)
            next_states.append((main + transfer, jug3, jug5 - transfer))
        
        # Pour water from 3L jug to 5L jug
        if jug3 > 0 and jug5 < max_5l:
            transfer = min(jug3, max_5l - jug5)
            next_states.append((main, jug3 - transfer, jug5 + transfer))
        
        # Pour water from 5L jug to 3L jug
        if jug5 > 0 and jug3 < max_3l:
            transfer = min(jug5, max_3l - jug3)
            next_states.append((main, jug3 + transfer, jug5 - transfer))
        
        # Empty the 3L jug
        if jug3 > 0:
            next_states.append((main, 0, jug5))
        
        # Empty the 5L jug
        if jug5 > 0:
            next_states.append((main, jug3, 0))
        
        # Empty the main jug
        if main > 0:
            next_states.append((0, jug3, jug5))
        
        # Fill the 3L jug
        if jug3 < max_3l:
            next_states.append((main, max_3l, jug5))
        
        # Fill the 5L jug
        if jug5 < max_5l:
            next_states.append((main, jug3, max_5l))
        
        # Add valid next states to the queue
        for state in next_states:
            if state not in visited:
                visited.add(state)
                queue.append((state, steps + 1))
    
    return -1  # In case no solution is found (which shouldn't happen)

# Running the BFS algorithm to find the minimum steps required.
min_steps = bfs_water_jug()
min_steps




【 在 sym 的大作中提到: 】
: 【 以下文字转载自 XiTiYanJiu 讨论区 】
: 发信人: sym (海洋宇宙), 信区: XiTiYanJiu
: 标  题: 水桶分水问题
: ...................
--
修改:lurh FROM 222.70.57.*
FROM 222.70.57.*