python经典趣味24点游戏程序设计

Python  /  管理员 发布于 7年前   145

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_nodedef 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-8from __future__ import divisionfrom Node import Nodedef 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_nodedef 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_resultdef 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])

以上就是本文的全部内容，希望对大家的学习有所帮助，也希望大家多多支持。