我正在试图弄清楚如何在下面的函数hanoi_2
中为Hanoi Towers问题实现非递归算法,但我不知道如何继续......
它会抛出一个错误:“无法从空列表中弹出”。当我输入一个奇数时它会以某种方式工作,但是,当第三个转弯传递出错时。当输入偶数作为光盘数量时,程序甚至不启动。
有什么问题?
from turtle import *
from tkinter import * # used for the dialog box
from tkinter.simpledialog import askinteger # used for the dialog box
from tkinter import ttk # used for the progress bar
import time # used for time-related functions (in pause)
import pickle # used to save an object to a file
class Disc(Turtle):
def __init__(self, n):
Turtle.__init__(self, shape="square", visible=False)
self.pu()
self.shapesize(1.5, n*1.5, 2) # square-->rectangle
self.fillcolor(n/10., 0, 1-n/10.)
self.st()
self.speed(11-n) # sets the speed of movement of the rectangles (the bigger, the slower)
self.moves = 0 # stores the number of times the disc is moved
class Tower(list):
"""Hanoi tower, a subclass of built-in type list"""
def __init__(self, x):
"""create an empty tower. x is x-position of peg"""
self.x = x
def push(self, d):
d.setx(self.x)
d.sety(-150+34*len(self))
d.clear()
d.write("Moved %s times" %(str(d.moves)), align="left", font=("Courier", 16, "bold"))
d.moves += 1 # increments the number of moves each time the disc is moved
self.append(d)
def pop(self):
d = list.pop(self)
d.sety(150)
return d
def hanoi(n, from_, with_, to_):
global moves
global ans
clear()
if n > 0:
hanoi(n-1, from_, to_, with_)
moves += 1 # amount of total moves is incremented
to_.push(from_.pop())
hanoi(n-1, with_, from_, to_)
sety(-255)
write("Total moves: %s" % (moves), align="center", font=("Courier", 16, "bold"))
sety(-320)
progress_bar(ans) # calls progress bar function
def hanoi_2(n, A, B, C):
global moves
clear()
if n%2==0:
B=C
C=A
A=B
for i in range(1,(2**n)-1):
if i%3==1:
C.push(A.pop())
if i%3==2:
B.push(A.pop())
if i%3==0:
B.push(C.pop())
答案 0 :(得分:1)
偶数光盘的问题是堆栈交换不正确:你似乎想要循环三个堆栈(当你丢失对原始B列表的引用时,你做错了),而实际上你只需要交换B和C堆栈,你可以按如下方式进行交换:
B, C = C, B
算法的问题在于,尽管你有两个正确的堆栈(基于i%3
),你仍然必须确定两个相关堆栈中的哪一个是给予者,哪个是接受者,因为这是并不总是一样的!如果你为此编写一个函数需要两个堆栈并确定两个可能的方向中的哪一个"这可能是最好的。是一个有效的,然后执行该移动:
def validMove(A, B):
if not len(A):
A.push(B.pop())
elif not len(B):
B.push(A.pop())
elif A[-1] > B[-1]:
A.push(B.pop())
else:
B.push(A.pop())
现在主算法将如下所示:
for i in range(1,2**n):
if i%3==1:
validMove(A, C)
if i%3==2:
validMove(A, B)
if i%3==0:
validMove(B, C)
答案 1 :(得分:0)
错误很少。首先,交换是不正确的。替换后,A
和B
引用同一个对象。要么使用临时变量,要么使用Python语法:B, C, A = C, A, B
。其次,这种替换应该在for
循环内完成,而不是之前。查看递归版本,每次都会更改塔的顺序。
我正在用手机写信,所以我无法为你解决,但祝你好运!: - )
答案 2 :(得分:0)
以下是递归且等效的迭代解决方案。堆栈保存调用的顺序及其状态。在这种情况下,堆栈上只有一个状态。
#recursive solution
def tower(numRings):
def tower1(numRings,fromPeg,toPeg,usePeg):
if numRings > 0: #A
tower1(numRings-1,fromPeg,usePeg,toPeg) #B
result.append((fromPeg,toPeg)) #C
tower1(numRings-1,usePeg,toPeg,fromPeg) #D
result = []
tower1(numRings,0,1,2)
return result
#iterative solution
def tower(numRings):
result = []
stk = []
def tower1(numRings):
In, To = 0, 1
stk.append((In,numRings,0,1,2))
while stk:
state,numRings,fromPeg,toPeg,usePeg = stk.pop()
if state == In:
if numRings != 0: #push down to 1 numRing #A
stk.append((To,numRings,fromPeg,toPeg,usePeg)) #B save state
stk.append((In,numRings-1,fromPeg,usePeg,toPeg))
elif state == To:
result.append((fromPeg,toPeg)) #C
stk.append((In,numRings-1,usePeg,toPeg,fromPeg)) #D
else:
print 'Error: logic'
return result
tower1(numRings)
return result
a = [1,2,3,4]
for n in a:
print 'rings',n
print tower(n)