python经典趣味24点游戏程序设计

  • Z时代
  • 2024-01-10
  • 分类：综合

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])

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