Answer to Question #278255 in Python for Chiarra

Question

The Josephus' problem is notoriously known. For those who are not familiar with the original problem: from among n people, numbered 1, 2, ..., n, standing in circle every m^th is going to be executed and only the life of the last remaining person will be saved. Joseph was smart enough to choose the position of the last remaining person, thus saving his life to give us the message about the incident. For example when n = 6 and m = 5 then the people will be executed in the order 5, 4, 6, 2, 3 and 1 will be saved.INPUTEach line contains of 3 integers n, m and p.  n is the number of people to be executed.  m is the interval on how the people are to be executed.  p is the starting position of the execution.OUTPUTPrint the number of the person who will be saved from the execution

Answer 1

# Python code for Josephus Problem

def Josh(person, k, index):


   

  # Base case , when only one person is left


  if len(person) == 1:


    print(person[0])


    return


   

  # find the index of first person which will die


  index = ((index+k)%len(person))


   

   # remove the first person which is going to be killed


  person.pop(index)


   

  # recursive call for n-1 persons


  Josh(person,k,index)

 
# Driver Program to test above function

n = 14 # specific n and k  values for original josephus problem


k = 2


k-=1   # (k-1)th person will be killed

 

index = 0

 
# fill the person vector

person=[]


for i in range(1,n+1):


  person.append(i)

 
Josh(person,k,index)