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Andrei [34K]
4 years ago
13

The following relation schema can be used to register information on the repayments on micro loans.

Computers and Technology
1 answer:
Kruka [31]4 years ago
6 0

Answer:

Check the explanation

Explanation:

Answer a):

From the given information, we can derive the following set of functional dependencies;

Borrower_id → name [given: every borrower is identified with an unique id]

Borrower_id → address [given: each borrower has only one address]

Borrower_id, Requestdate → loanamount [given: more than one loan cannot be requested by a single borrower]  

Borrower_id, requestdate, repayment_date →repayment_amount [given: a borrower can make multiple repayments on a single day, but not on a single loan]

From the above set of FDs, it is evident that we can uniquely identify all the other attributes of Repayment table, if we know the values of (borrower_id, requestdate, repayment_date). That is,

Borrower_id, requestdate, repayment_date →name, address, loanamount, repayment_amount.

Hence, attributes (Borrower_id, requestdate, repayment_date) together forms a candidate key.

Answer b):

Is the given relation Repayment is in 1NF?  

Yes. It has a key. Hence, we can make unique identification of records.

Is the given relation is in 2NF?  

No. We have the following partial key dependencies.  

1. We can easily derive name and address of every borrower if we know the borrower_id from the FDs Borrower_d →name, and Borrower_id →address.

2. We can derive the loanamount if we know borrower_id, and requestdate from the FD Borrower_id, Requestdate → loanamount.

Hence, the relation Repayment is not in 2NF. To convert it into a 2NF relation, we can decompose Repayment into the following relations;

Borrower (Borrower_id, Name, Address)

Borrower_loan (Borrower_id, Requestdate, Loanamount)

Repayment (Borrower_id, Requestdate, Repayment_date, Repayment_amount)

From the derived FDs, we know that all these tables are in 2NF.

Are these tables in 3NF?  

Yes. There are no transitive dependencies present in the above tables’ set of functional dependencies. Hence, we would say that all these tables are in 3NF.

Are these tables in BCNF?  

Yes. There are no more than one candidate keys present in the above set of tables. Hence the following decomposed tables are in Boyce-Codd Normal Form.

Borrower (Borrower_id, Name, Address)

Borrower_loan (Borrower_id, Requestdate, Loanamount)

Repayment (Borrower_id, Requestdate, Repayment_date, Repayment_amount)

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Answer:

Explanation:

The python code for the question is attached in the image below.

The objective here is to write a code in python with a function called has_duplicates that takes a string parameter and returns True if the string has any repeated characters. Otherwise, it should return False.

SEE BELOW FOR THE CODE.

alphabet = "abcdefghijklmnopqrstuvwxyz"

test_dups = ["zzz","dog","bookkeeper","subdermatoglyphic","subdermatoglyphics"]

test_miss = ["zzz","subdermatoglyphic","the quick brown fox jumps over the lazy dog"]

def histogram(s):

   d = dict()

   for c in s:

       if c not in d:

           d[c]=1

       else:

           d[c] += 1

   return d

def has_duplicates():

   for string in test_dups:

       dictionary=histogram(string)

       duplicates=False

       for ch in dictionary.keys():

           if dictionary[ch] > 1:

               duplicates=True

               break

       if duplicates==True:

           print(string,"has duplicates")

       else:

           print(string,"has no duplicates")

def missing_letters(string):

   answer=""

   for ch in alphabet:

       if ch not in string:

           answer=answer+ch

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print("__________________________________________")

print(" Calling has_duplicates() function")

print("__________________________________________")

has_duplicates()

print("\n______________________________________________")

print("Calling missing_letters() function in for loop")

print("______________________________________________")

for i in test_miss:

   answer=missing_letters(i)

   if len(answer)>=1:

       print(i,"is missing letters",answer)

   else:

       print(i,"uses all the letters")

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Explanation:

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