**Reference:-** https://freshlybuilt.com/100-doors-problem-using-python/

### Problem Statement

This is a general form of 100 doors problem.Here you have n doors.All doors are initially closed.m persons walk through all the doors and toggle them(open if close and vice-versa).

Person 1 walks and toggle(open) all doors.

Person 2 walks and toggle each door that is a multiple of 2,i.e.,2nd,4th,6th,8th,…

Person 3 walks and toggle each door that is a multiple of 3,i.e.,3rd,6th,9th,12th,…

………………………..

…………………………

Person m walks and toggle each door that is multiple of m.

**Which doors are open in the end?**

### Solution:-

The solution is somewhat similar to the 100 doors problem.Each door that has odd no. of walks will be opened and rests will be closed.Now if there are m walks then all door no. which are less than m and a perfect square will be opened.But for other doors ,i.e., where door no. is greater than n the scenario will change and some more doors will be added to the list.

**For Example:-**

Let there are 5 doors and 3 persons.

Then Person 1 will open each door.

Person 2 will close doors that are even,i.e. door 2 and door 4.

Person 3 will close door 3.

**In this case,the answer is 1 and 5.**

### Implementation in Python:-

#Here 0 represents that door is close and 1 represents that door is open. def toggle(n): #closes door if they are open and opens if they are closed. if doors[n]==1: doors[n]=0 else: doors[n]=1 n=int(input('Enter the no. of doors available')) m=int(input('Enter the no. of persons')) doors=[0]*n#Initializing n doors with 0.(They are closed initially) for i in range(1,m+1):#Iterating through m times j=i while j<=n: toggle(j-1) j=j+i for x in range(n): if doors[x]==1: print(x+1,end=" ")

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