Problem Set 6(Non-Isomorphic Numbers)

Non-Isomorphic Numbers

Q.Given ‘n’ integers, write an algorithm and the subsequent Python code to print all numbers that are Non-isomorphic to the first number. Two numbers with non-distinct digits ( numbers in which digits are repeated)  are said to be non-isomorphic if they have the same number of digits in it and the sets of positions  having the same  digits are different. For example:

• 12321 is not non-isomorphic to 83538 .

Both the  numbers are of length five.  In 12321, positions 2 and 4 have the same digit 2, Positions 1 and 5 have same digit 1. Hence the set of positions having the same digit are {{1,5},{2,4}}. Similarly , for the number 83538, the set of positions having the same digit is {{1,5}, {2,4}}. Since the sets of positions having the same digit are same , the numbers are not non-isomorphic

• 1232 is non-isomorphic to 2342 because set of positions having same digits for 1232 is {{2, 4}} and for 2342 is {{1, 4}} . The sets of positions having same digits are different. Positions are numbered from left to right starting from 1.
• 12 is not non-isomorphic to 10.

Write a function to check whether two  numbers are non-isomorphic. If none of the numbers are non-isomorphic then print ‘No non-isomorphic’.

Input Format

First line contains the number of elements, n

Next ‘n’ line contains the numbers

Output Format

Print first number in the first line

Next few lines contain the numbers that are non-isomorphic to first number.

If the none of the numbers are not non-isomorphic, then Print “No non-isomorphic

Input:

the number n, n numbers

Processing:

def isomorphic(number):

    number_list = list(map(int,str(number)))

    length = len(number_list)

    common_pos = []

    temp_pos = []

    for i in range(len(list(set(number_list)))):

        if number_list.count(number_list[i]) >= 2:

            for j in range(len(number_list)):

                if number_list[i] == number_list[j]:

                    temp_pos.append(j)

            common_pos.append(temp_pos)

            temp_pos = []

    return length,common_pos

final_num_list = [number]

count = 0
final_list = []

first_number = isomorphic(number)

for i in range(n-1):

    temp = int(input())

    if first_number != isomorphic(temp) and first_number[0] == isomorphic(temp)[0] and isomorphic(temp)[1]:
        if final_list == []:
            final_list.append(number)
        final_list.append(temp)

        count += 1

Output:

Display the numbers that are isomorphic to first number if none are present then print No non-isomorphic

Program:

def isomorphic(number):

    number_list = list(map(int,str(number)))

    length = len(number_list)

    common_pos = []

    temp_pos = []

    for i in range(len(list(set(number_list)))):

        if number_list.count(number_list[i]) >= 2:

            for j in range(len(number_list)):

                if number_list[i] == number_list[j]:

                    temp_pos.append(j)

            common_pos.append(temp_pos)

            temp_pos = []

    return length,common_pos

n = int(input())

number = int(input())

final_num_list = [number]

count = 0

first_number = isomorphic(number)
final_list = []

for i in range(n-1):

    temp = int(input())

    if first_number != isomorphic(temp) and first_number[0] == isomorphic(temp)[0] and isomorphic(temp)[1] != []:

        if final_list == []:

            final_list.append(number)

        final_list.append(temp)

        count += 1

if count == 0:

    print('No non-isomorphic')
else:
    for i in final_list:
        print(i)

Algorithm:

Step1. Define the function isomorphic with one parameter number

Step1.1 convert the number to individual digits and assign it to number_list and assign length as the number of digits of the number

Step1.2 repeat till i is less than the number of unique digits

Step1.2.1 check whether the ith digit in the number_list is repeated if yes find all the positions and add these positions to temp_pos list and append temp_pos to common_pos

Step1.3 return the number of digits and common_pos list

Step2. Get n i.e. the number of numbers to be entered

Step3. Get the first number and call the isomorphic function with the first number and store the value that is returned in first_number variable

Step4. Repeat till i is less than n-1

Step4.1 get the number and check if the value returned when the number is called with isomorphic function with first_number is not same and also check if both the numbers have the same number of digits if both conditions are satisfied display the number.

Step4.2 increment count by 1 which was initially zero

Step5. If count is equal 0 then display no non-isomorphic

Step6.End