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vlabodo [156]
3 years ago
15

Write a python program to read four numbers (representing the four octets of an IP) and print the next five IP addresses. Be sur

e to verify that the second, third and fourth numbers are between 0 and 255. The first one should be between 1 and 255.
Computers and Technology
1 answer:
sdas [7]3 years ago
8 0

Answer:

first_octet = int(input("Enter the first octet: "))

second_octet = int(input("Enter the second octet: "))

third_octet = int(input("Enter the third octet: "))

forth_octet = int(input("Enter the forth octet: "))

octet_start = forth_octet + 1

octet_end = forth_octet + 6

if (1 <= first_octet <= 255) and (0 <= second_octet <= 255) and (0 <= third_octet <= 255) and (0 <= forth_octet <= 255):

   for ip in range(octet_start, octet_end):

       forth_octet = forth_octet + 1

       if forth_octet > 255:

           forth_octet = (forth_octet % 255) - 1

           third_octet = third_octet + 1

           if third_octet > 255:

               third_octet = (third_octet % 255) - 1

               second_octet = second_octet + 1

               if second_octet > 255:

                   second_octet = (second_octet % 255) - 1

                   first_octet = first_octet + 1

                   if first_octet > 255:

                       print("No more available IP!")

                       break

       print(str(first_octet) + "." + str(second_octet) + "." + str(third_octet) + "." + str(forth_octet))

else:

   print("Invalid input!")

Explanation:

- Ask the user for the octets

- Initialize the start and end points of the loop, since we will be printing next 5 IP range is set accordingly

- Check if the octets meet the restrictions

- Inside the loop, increase the forth octet by 1 on each iteration

- Check if octets reach the limit - 255, if they are greater than 255, calculate the mod and subtract 1. Then increase the previous octet by 1.

For example, if the input is: 1. 1. 20. 255, next ones will be:

1. 1. 21. 0

1. 1. 21. 1

1. 1. 21. 2

1. 1. 21. 3

1. 1. 21. 4

There is an exception for the first octet, if it reaches 255 and others also reach 255, this means there are no IP available.

- Print the result

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wlad13 [49]

Answer:

  1. def check_subset(l1, l2):
  2.    status = False
  3.    count = 0
  4.    if(len(l1) > len(l2)):
  5.        for x in l2:
  6.            for y in l1:
  7.                if x == y:
  8.                    count += 1
  9.        if(count == len(l2)):
  10.            return True  
  11.        else:
  12.            return False
  13.    else:
  14.        for x in l1:
  15.            for y in l2:
  16.                if x==y:
  17.                    count += 1
  18.        if(count == len(l1)):
  19.            return True  
  20.        else:
  21.            return False
  22. print(check_subset([1,4,6], [1,2,3,4,5,6]))
  23. print(check_subset([2,5,7,9,8], [7,8]))
  24. print(check_subset([1, 5, 7], [1,4,6,78,12]))

Explanation:

The key idea of this solution is to create a count variable to track the number of the elements in a shorter list whose value can be found in another longer list.

Firstly, we need to check which list is shorter (Line 4). If the list 2 is shorter, we need to traverse through the list 2 in an outer loop (Line 5) and then create another inner loop to traverse through the longer list 1 (Line 6).  If the current x value from list 2 is matched any value in list 1, increment the count variable by 1. After finishing the outer loop and inner loop, we shall be able to get the total count of elements in list 2 which can also be found in list 1. If the count is equal to the length of list 2, it means all elements in the list 2 are found in the list 1 and therefore it is a subset of list 1 and return true (Line 10-11) otherwise return false.

The similar process is applied to the situation where the list 1 is shorter than list 2 (Line 15-24)

If we test our function using three pairs of input lists (Line 26-28), we shall get the output as follows:

True

True

False

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