Mathematica

Concepts of Mathematics programmed in Ruby.

Table Of Contents:

• Factorial
  
• Factors
  
• Fibonacci
  
• Kaprekar
  
• Prime & Composite
  
• Ratio
  

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Factorial

❗ Problem Details

What is factorial?

Why 0! = 1

Should return the factorial for any integer input which is greater than 0.

⌨️ Example input:

19

💻 Expected output:

121645100408832000

✔️ Solution Details

Logic

1. Check user input, n, exit program for negative input.
2. Initiate a variable, total, which will hold the factorial of n with value 1 as 0! is equal to 1.
3. Initiate a loop which will initiate from integer 1 to n where for each iteration the varaible total will be updated with the product of the number its iterating at, i, and total, where total holds the value of previous iteration.

Ruby provides a funtion Math.gamma which returns the approximation factorial of n-1 for any n greater than 0. I have also used this function in the program to compare the results with the logic code I created. The answer of this method changes slightly compared to my logic as we give high input numbers as gamma is an approximation.

The solution of this problem can be found in factorial/program.rb file.

📋 Local Execution

Clone this repo, navigate to the factorial/ directory and run the following commands in Terminal:

ruby program.rb

This will run the given code file. Enter appropraite input to get desired output.

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Factors

❗ Problem Details

What are factors?

Should return list of factors in ascending order for any integer input which is greater than equal to 1.

⌨️ Example input:

100

💻 Expected output:

[1, 2, 4, 5, 10, 20, 25, 50, 100]

✔️ Solution Details

Logic

1. Check user input, n, exit program for negative input.
2. If user input is 1 then end the program with the output 1.
3. If user input is greater than 1 then initiate an array factors which will hold all the numbers that are factors of n. We know for any number n, 1 and n itself are common factors, hence we initiate array with these 2 factors.
4. Since we know that prime numbers have no factors other than 1 and n itself, we check for it with the help of prime library and end the program.
5. Now, for composite numbers we initiate a loop which will initiate from integer 2 (since we know 1 is a factor for any n where n > 0) to n-1 (with same logic) where for each iteration it will check the following:
   1. Stop the loop if the loop variable factor is already in our factors array. This means its calculation is already been done and also for the numbers till n.
      
   2. If number isn't in factors then we check if n is fully divisible by factor. If it is fully divisible it will return remainder 0. And then we append the factor in factors. After that we would also like to append another factor second_factor to factors list for which factor was fully divisible. To find second_factor, we do n / factor. Before adding second_factor to list we check if its equal to factor, because for many integers n such as 36 where its square root is 6, it has both factor and second_factor as equal to 6. If both factors are equal we dont add second_factor to factors to avoid duplicate factor and this also means that no new factors of n will be found after, so we stop the loop. If both factors aren't equal then we append to factors.
      
   3. If factor isn't fully divisible then we move on to next iteratiom.
      
6. Finally, we have our factors array of n. We show the list to the user by arranging numbers in ascending order with .sort method.

The solution of this problem can be found in factors/program.rb file.

📋 Local Execution

Clone this repo, navigate to the factors/ directory and run the following commands in Terminal:

ruby program.rb

This will run the given code file. Enter appropraite input to get desired output.

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Fibonacci

❗ Problem Details

What is fibonacci?

Should return whether given positive integer by user, n, is fibonacci number or not. If its a Fibonacci number then show its index number in the fiobancci series. And if its not a fibonacci number then show the nearest fibonacci numbers both before and after the number n.

The program will calculate fibonacci numbers for positive integers only and not negative.

⌨️ Example input:

23

3

💻 Expected output:

Output for 23

Number entered is not a fibonacci number.
Its previous nearest fibonacci number is 21.
And next nearest fibonacci number is 34.

Output for 3

Success! Number you entered is a fibonacci number.
Its index in fibonacci sequence is 4.

✔️ Solution Details

Logic

1. Perform calculation only for integer input where n > 0.
2. Initialize an array, fibonacci_seq, which will hold the fibonacci numbers and initiate it with values of 0 & 1 so we can calculate next fibonacci numbers.
3. Initiate a loop where in each iteration it performs addition of last 2 fibonacci numbers in fibonacci_seq array to find next fiobanacci number. Break the loop if the fibonacci number calculated in this iteration is greater than the input number n. However if its less than n then append the fibonacci number to fibonacci_seq. Break the loop if the fibonacci number is equal to n.
4. Will continue iteration (Step 3) until output of each iteration is equal to or greater than n.
5. If user entered a fibonacci number then show output of its index in the fibonacci_seq. Otherwise show the nearest fibonacci numbers both before and after n.

The solution of this problem can be found in fibonacci/program.rb file.

📋 Local Execution

Clone this repo, navigate to the fibonacci/ directory and run the following commands in Terminal:

ruby program.rb

This will run the given code file. Enter appropraite input to get desired output.

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Kaprekar

❗ Problem Details

What are Kaprekar numbers?

List of Kaprekar Numbers

Should return whether given positive integer by user, n, is kaprekar number or not.

⌨️ Example input:

9

113

💻 Expected output:

Output for 9

9 is a Kaprekar number.

Output for 113

113 is not a Kaprekar number.

✔️ Solution Details

Logic

1. Perform calculation only for integer input where n > 0.
2. Take Square of n and save in a variable n_sqr.
3. Convert n_sqr to an array, arr.
4. We want our length of elements in arr to be an even number so it can be easily grouped into 2 where each group will be of equal lengths. So for odd length we input 0 before the number so it doesn't change its value and we have even length. So for n_sqr whose value is 121 whose length is 3 and after adding 0 we have 0121 which is equal to 121.
5. Count the number of elements in arr or in other words count the number of digits of n_sqr and save in arr_count.
6. Initialize a variable, total which will calculate the sum of groups with 0.
7. Now we calculate the number of elements of arr we need to group for addition. We know that we want 2 groups and since arr_count is even it will easily divide into 2 and give us the number of members we will have for each group. Store this number in members var.
8. Now we slice the list where each group has members number of elements of arr with the help of .each_slice method. Each iteration of this loop will take a group (iteration var) of these members, join the elements and add them to out total var.
9. If total == n then its kaprekar else it isn't.

The solution of this problem can be found in kaprekar/program.rb file.

📋 Local Execution

Clone this repo, navigate to the kaprekar/ directory and run the following commands in Terminal:

ruby program.rb

This will run the given code file. Enter appropraite input to get desired output.

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Prime and Composite

❗ Problem Details

What are Prime numbers?

What are Composite Numbers?

Should return whether given positive integer by user, n, is prime or composite number.

⌨️ Example input:

12

7

💻 Expected output:

Output for 12

12 is a composite number.

Output for 7

7 is a prime number.

✔️ Solution Details

Logic

1. Perform calculation only for integer input where n >= 0.
2. End program for negative integer input.
3. 0 & 1 are 2 numbers are neither prime nor composite so for input n with value of 0 or 1, end program.
4. We also know that any number which is fully divisble by 2 or in other words if its even, then it wont be a prime number but composite. But 2 is the only even number that is prime so for n == 2 its prime number.
5. Now to check if a odd number is prime or not we need to run loop from 3 to n-1 because we know n is a factor of itself and we dont start from 2 because no odd number is fully divisible by 2.
6. We can loop through each number from 3 to n but what if n was a very large odd number? Looping through each number and checking if its fully divisible would be take time and put load on memory. To avoid this we only loop from 3 to n * 0.5 because we know that factors can be found if there are halfway through only, so looping through all the way to n isn't practical.
7. For any n greater than 5, loop from 3 to n * 0.5. And for any n less than 5 loop from 3 to n - 1.
8. Now in each loop we see if n is divisible by factor (loop var). If it is then its a composite number and we exit the program.
9. It will continue to loop until it finds a factor that fully divides n.
10. If it can't find factors then n is a Prime number.

The solution of this problem can be found in prime_composite/program.rb file.

📋 Local Execution

Clone this repo, navigate to the prime_composite/ directory and run the following commands in Terminal:

ruby program.rb

This will run the given code file. Enter appropraite input to get desired output.

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Ratio

❗ Problem Details

Should return simplest form of rational number for any integers input by user, in the form of n1/n2 where n1 can be any integer whereas n2 can also be any integer except 0.

⌨️ Example input:

-91/-73

-21/98

💻 Expected output:

Output for -91/-73

The greatest common divisor is 1 of -91/-73.
And it is already in its simplest form.

Output for -21/98

The greatest common divisor is 7 of -21/98.
And its simplest form is -3/14.

✔️ Solution Details

Logic

1. Take input from user in a rational form.
2. Check if n2 is not 0. If it is then exit the program since dividing any integer by 0 will lead to infinity.
3. To convert the fraction input to simplest form we need to first find n1 & n2 Greatest Common Divisor (GCD). Ruby provides .gcd method to find out.
4. GCD calculation will raise error for unexpected string input. We will wrap begin-rescue to GCD calculation and gently infrom user and exit the program.
5. After finding GCD of n1/n2, we will check if gcd == 1. If it is, it means that input fraction is already in its simplest form. We give output to the user and stop the program.
6. For gcd other than 1, divide both n1 & n2 by gcd to find other factor that which multiplied by gcd returns n1 or n2. Store the value in new variable called simple_n1 & simple_n2 respectively.
7. Convert simple_n1 & simple_n2 variables to rational numbers and store them in new var simplest_form.
8. Show results to user.

The solution of this problem can be found in ratio/program.rb file.

📋 Local Execution

Clone this repo, navigate to the ratio/ directory and run the following commands in Terminal:

ruby program.rb

This will run the given code file. Enter appropraite input to get desired output.

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