Question

我想知道是否可以使用python中的列表来解决Josepheus问题。

简单来说，约瑟夫斯问题就是找到一个圆形排列的位置，如果使用事先已知的跳过参数来处理执行，这将是安全的。

例如：给定循环安排，例如[1,2,3,4,5,6,7]和跳过参数3，人员将按照3,6,2,7,5,1的顺序执行，而职位4将是安全的。

我一直试图使用列表解决这个问题一段时间了，但索引位置对我来说变得棘手。

 a=[x for x in range(1,11)]
 skip=2
 step=2
 while (len(a)!=1):
   value=a[step-1]
   a.remove(value)
   n=len(a)
   step=step+skip
   large=max(a)
   if step>=n:        
      diff=abs(large-value)
      step=diff%skip
   print a

使用代码段更新了问题，但我不认为我的逻辑是正确的。

Answer 1

很简单，你可以使用list.pop(i)删除每个受害者（并获取他的ID）。然后，我们只需要担心包装索引，你只需将跳过的索引mod作为剩余囚犯的数量即可。

那么，问题解决方案变为

def josephus(ls, skip):
    skip -= 1 # pop automatically skips the dead guy
    idx = skip
    while len(ls) > 1:
        print ls.pop(idx) # kill prisoner at idx
        idx = (idx + skip) % len(ls)
    print 'survivor: ', ls[0]

测试输出：

>>> josephus([1,2,3,4,5,6,7], 3)
3
6
2
7
5
1
survivor:  4

Answer 2

In [96]: def josephus(ls, skip):
    ...:     from collections import deque
    ...:     d = deque(ls)
    ...:     while len(d)>1:
    ...:         d.rotate(-skip)
    ...:         print(d.pop())
    ...:     print('survivor:' , d.pop())
    ...:     

In [97]: josephus([1,2,3,4,5,6,7], 3)
3
6
2
7
5
1
survivor: 4

如果您不想计算索引，可以使用deque数据结构。

Answer 3

看起来更糟，但对于初学者来说更容易理解

np.pad(img, ((row,row), (col,col)), mode)

Answer 4

我的解决方案使用了我在此处网上找到的数学技巧：https://www.youtube.com/watch?v=uCsD3ZGzMgE 它使用二进制方式写圈中的人数和幸存者的位置。结果相同，代码更短。

代码是这样的：

numar_persoane = int(input("How many people are in the circle?\n")) #here we manually insert the number of people in the circle

x='{0:08b}'.format(int(numar_persoane)) #here we convert to binary

m=list(x) #here we transform it into a list

for i in range(0,len(m)): #here we remove the first '1' and append to the same list

    m.remove('1')

    m.append('1')

    break

w=''.join(m) #here we make it a string again

print("The survivor sits in position",int(w, 2)) #int(w, 2) makes our string a decimal number

Answer 5

这是我对你问题的解决方案：

# simple queue implementation<ADT>
class Queue:
    def __init__(self):
        self.q = []
    def enqueue(self,data):
        self.q.insert(0,data)
    def dequeue(self):
        self.q.pop()
    def sizeQ(self):
        return len(self.q)
    def printQ(self):
        return self.q


lists = ["Josephus","Mark","Gladiator","Coward"]
to_die = 3
Q = Queue()
# inserting element into Q
for i in lists:
    Q.enqueue(i)
# for size > 1 
while Q.sizeP() > 1:
    for j in range(1,3): 
# every third element to be eliminated
         Q.enqueue(Q.dequeue())
    Q.dequeue()
print(Q.printQ())

Answer 6

def Last_Person(n):
    person = [x for x in range(1,n+1)]
    x = 0
    c = 1
    while len(person) > 1:
        if x == len(person) - 1:
            print("Round ", c, "- Here's who is left: ", person, "Person ", person[x], "killed person", person[0])
            person.pop(0)
            x = 0
            c = c+1
        elif x == len(person) - 2:
            print("Round ", c, "- Here's who is left: ", person, "Person ", person[x], "killed person", person[x + 1])
            person.pop(x+1)
            x = 0
            c = c + 1
        else:
            print("Round ", c, "- Here's who is left: ", person, "Person ", person[x], "killed person", person[x + 1])
            person.pop(x + 1)
            x = x + 1
            c = c + 1
    print("Person", person[x], "is the winner")

Last_Person(50)

Answer 7

人数n和人数k的总数，表示k-1个人被跳过，圈子中第k个人被杀。

def josephus(n, k): 
              if (n == 1): 
                  return 1
              else: 
                  return (josephus(n - 1, k) + k-1) % n + 1
        n = 14
        k = 2
        print("The chosen place is ", josephus(n, k))

Answer 8

如果您只是在寻找最终结果，这是一个简单的解决方案。

def JosephusProblem(people):
  binary = bin(people)  # Converting to binary
  winner = binary[3:]+binary[2]  # as the output looks loke '0b101001'. removing 0b and adding the 1 to the end
  print('The winner is',int(winner,2))  #converting the binary  back to decimal

如果您要查找此代码背后的数学运算，请查看以下视频： Josephus Problem(youTube)

使用python中的列表的“约瑟夫问题”

8 个答案: