经典趣味24点游戏程序设计(python)

一、游戏玩法介绍：

    24点游戏是儿时玩的主要益智类游戏之一，玩法为：从一副扑克中抽取4张牌，对4张牌使用加减乘除中的任何方法，使计算结果为24。例如，2,3,4,6，通过( ( ( 4 + 6 ) - 2 ) * 3 )  = 24，最快算出24者剩。

二、设计思路：

    由于设计到了表达式，很自然的想到了是否可以使用表达式树来设计程序。本程序的确使用了表达式树，也是程序最关键的环节。简要概括为：先列出所有表达式的可能性，然后运用表达式树计算表达式的值。程序中大量的运用了递归，各个递归式不是很复杂，大家耐心看看，应该是能看懂的

    

    表达式树：

        表达式树的所有叶子节点均为操作数(operand)，其他节点为运算符(operator)。由于本例中都是二元运算，所以表达式树是二叉树。下图就是一个表达式树

　　

    具体步骤：

    1、遍历所有表达式的可能情况

    遍历分为两部分，一部分遍历出操作数的所有可能，然后是运算符的所有可能。全排列的计算采用了递归的思想

    

#返回一个列表的全排列的列表集合
def list_result(l):
    if len(l) == 1:
        return [l]
    all_result = []
    for index,item in enumerate(l):
        r = list_result(l[0:index] + l[index+1:])
        map(lambda x : x.append(item),r)
        all_result.extend(r)
    return all_result

　2、根据传入的表达式的值，构造表达式树

　　由于表达式树的特点，所有操作数均为叶子节点，操作符为非叶子节点，而一个表达式(例如( ( ( 6 + 4 ) - 2 ) * 3 )  = 24) 只有3个运算符，即一颗表达式树只有3个非叶子节点。所以树的形状只有两种可能，就直接写死了

　　

#树节点
class Node:

    def __init__(self, val):
        self.val = val
        self.left = None
        self.right = None

def one_expression_tree(operators, operands):
    root_node = Node(operators[0])
    operator1 = Node(operators[1])
    operator2 = Node(operators[2])
    operand0 = Node(operands[0])
    operand1 = Node(operands[1])
    operand2 = Node(operands[2])
    operand3 = Node(operands[3])
    root_node.left = operator1
    root_node.right =operand0
    operator1.left = operator2
    operator1.right = operand1
    operator2.left = operand2
    operator2.right = operand3
    return root_node

def two_expression_tree(operators, operands):
    root_node = Node(operators[0])
    operator1 = Node(operators[1])
    operator2 = Node(operators[2])
    operand0 = Node(operands[0])
    operand1 = Node(operands[1])
    operand2 = Node(operands[2])
    operand3 = Node(operands[3])
    root_node.left = operator1
    root_node.right =operator2
    operator1.left = operand0
    operator1.right = operand1
    operator2.left = operand2
    operator2.right = operand3
    return root_node

3、计算表达式树的值

　　也运用了递归

　　

#根据两个数和一个符号，计算值
def cal(a, b, operator):
    return operator == '+' and float(a) + float(b) or operator == '-' and float(a) - float(b) or operator == '*' and  float(a) * float(b) or operator == '÷' and float(a)/float(b)

def cal_tree(node):
    if node.left is None:
        return node.val
    return cal(cal_tree(node.left), cal_tree(node.right), node.val)

4、输出所有可能的表达式

　　还是运用了递归

def print_expression_tree(root):
    print_node(root)
    print ' = 24'

def print_node(node):
    if node is None :
        return
    if node.left is None and node.right is None:
        print node.val,
    else:
        print '(',
        print_node(node.left)
        print node.val,
        print_node(node.right)
        print ')',
        #print ' ( %s %s %s ) ' % (print_node(node.left), node.val, print_node(node.right)),

5、输出结果

三、所有源码

　　

#coding:utf-8
from __future__ import division

from Node import Node


def calculate(nums):
    nums_possible = list_result(nums)
    operators_possible = list_result(['+','-','*','÷'])
    goods_noods = []
    for nums in nums_possible:
        for op in operators_possible:
            node = one_expression_tree(op, nums)
            if cal_tree(node) == 24:
                goods_noods.append(node)
            node = two_expression_tree(op, nums)
            if cal_tree(node) == 24:
                goods_noods.append(node)
    map(lambda node: print_expression_tree(node), goods_noods)




def cal_tree(node):
    if node.left is None:
        return node.val
    return cal(cal_tree(node.left), cal_tree(node.right), node.val)


#根据两个数和一个符号，计算值
def cal(a, b, operator):
    return operator == '+' and float(a) + float(b) or operator == '-' and float(a) - float(b) or operator == '*' and  float(a) * float(b) or operator == '÷' and float(a)/float(b)

def one_expression_tree(operators, operands):
    root_node = Node(operators[0])
    operator1 = Node(operators[1])
    operator2 = Node(operators[2])
    operand0 = Node(operands[0])
    operand1 = Node(operands[1])
    operand2 = Node(operands[2])
    operand3 = Node(operands[3])
    root_node.left = operator1
    root_node.right =operand0
    operator1.left = operator2
    operator1.right = operand1
    operator2.left = operand2
    operator2.right = operand3
    return root_node

def two_expression_tree(operators, operands):
    root_node = Node(operators[0])
    operator1 = Node(operators[1])
    operator2 = Node(operators[2])
    operand0 = Node(operands[0])
    operand1 = Node(operands[1])
    operand2 = Node(operands[2])
    operand3 = Node(operands[3])
    root_node.left = operator1
    root_node.right =operator2
    operator1.left = operand0
    operator1.right = operand1
    operator2.left = operand2
    operator2.right = operand3
    return root_node

#返回一个列表的全排列的列表集合
def list_result(l):
    if len(l) == 1:
        return [l]
    all_result = []
    for index,item in enumerate(l):
        r = list_result(l[0:index] + l[index+1:])
        map(lambda x : x.append(item),r)
        all_result.extend(r)
    return all_result

def print_expression_tree(root):
    print_node(root)
    print ' = 24'

def print_node(node):
    if node is None :
        return
    if node.left is None and node.right is None:
        print node.val,
    else:
        print '(',
        print_node(node.left)
        print node.val,
        print_node(node.right)
        print ')',

if __name__ == '__main__':
    calculate([2,3,4,6])