Dhaka Power Distribution Company Limited

Post: Technical (Computer technology)
Subject code: 155; Mark: 80
1.(a) Find the output of this program.
#include
int main() {
int n = 4;
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= n - i; j++) {
printf(" ");
}
for (int j = 1; j <= i; j++) {
printf("*");
}
printf("\n");
}
return 0;
}

The hash function is given as h(k)=k mod 13.
The table size is 13, meaning we have indices from 0 to 12.

Step-by-Step Insertion Process
Key: 10
Hash value: h(10) = 10 mod 13 = 10. Insert 10 at index 10.
Key: 3
Hash value: h(3) = 3 mod 13 = 3. Insert 3 at index 3.
Key: 6
Hash value: h(6) = 6 mod 13 = 6. Insert 6 at index 6.
Key: 16
Hash value: h(16) = 16 mod 13 = 3. Index 3 is occupied. Use linear probing and insert 16 at index 4.
Key: 17
Hash value: h(17) = 17 mod 13 = 4. Index 4 is occupied. Use linear probing and insert 17 at index 5.
Key: 19
Hash value: h(19) = 19 mod 13 = 6. Index 6 is occupied. Use linear probing and insert 19 at index 7.
Final Hash Table

IndexValue
0
1
2
33
416
517
66
719
8
9
1010
11
12

Summary
The final hash table after inserting all keys using linear probing is as follows:

[ – , – , – , 3 , 16 , 17 , 6 , 19 , – , – , 10 , – , – ]
1.(b) Class C has some features. Class D uses all features of Class C and has some extra features of its own. Identify the concept used here in Object-Oriented Programming (OOP) and explain with an example.

The hash function is given as h(k)=k mod 13.
The table size is 13, meaning we have indices from 0 to 12.

Step-by-Step Insertion Process
Key: 10
Hash value: h(10) = 10 mod 13 = 10. Insert 10 at index 10.
Key: 3
Hash value: h(3) = 3 mod 13 = 3. Insert 3 at index 3.
Key: 6
Hash value: h(6) = 6 mod 13 = 6. Insert 6 at index 6.
Key: 16
Hash value: h(16) = 16 mod 13 = 3. Index 3 is occupied. Use linear probing and insert 16 at index 4.
Key: 17
Hash value: h(17) = 17 mod 13 = 4. Index 4 is occupied. Use linear probing and insert 17 at index 5.
Key: 19
Hash value: h(19) = 19 mod 13 = 6. Index 6 is occupied. Use linear probing and insert 19 at index 7.
Final Hash Table

IndexValue
0
1
2
33
416
517
66
719
8
9
1010
11
12

Summary
The final hash table after inserting all keys using linear probing is as follows:

[ – , – , – , 3 , 16 , 17 , 6 , 19 , – , – , 10 , – , – ]
2.(a) Let A = {1,2,3,4,5,6,7,8}. If five integers are selected from A, prove that at least one pair of selected integers has a sum of 9. Use the Pigeonhole Principle.

The hash function is given as h(k)=k mod 13.
The table size is 13, meaning we have indices from 0 to 12.

Step-by-Step Insertion Process
Key: 10
Hash value: h(10) = 10 mod 13 = 10. Insert 10 at index 10.
Key: 3
Hash value: h(3) = 3 mod 13 = 3. Insert 3 at index 3.
Key: 6
Hash value: h(6) = 6 mod 13 = 6. Insert 6 at index 6.
Key: 16
Hash value: h(16) = 16 mod 13 = 3. Index 3 is occupied. Use linear probing and insert 16 at index 4.
Key: 17
Hash value: h(17) = 17 mod 13 = 4. Index 4 is occupied. Use linear probing and insert 17 at index 5.
Key: 19
Hash value: h(19) = 19 mod 13 = 6. Index 6 is occupied. Use linear probing and insert 19 at index 7.
Final Hash Table

IndexValue
0
1
2
33
416
517
66
719
8
9
1010
11
12

Summary
The final hash table after inserting all keys using linear probing is as follows:

[ – , – , – , 3 , 16 , 17 , 6 , 19 , – , – , 10 , – , – ]
2.(b) Simplify using K-map: F(A,B,C) = A'BC' + A'BC + AB'C' + AB'C + ABC' + ABC

The hash function is given as h(k)=k mod 13.
The table size is 13, meaning we have indices from 0 to 12.

Step-by-Step Insertion Process
Key: 10
Hash value: h(10) = 10 mod 13 = 10. Insert 10 at index 10.
Key: 3
Hash value: h(3) = 3 mod 13 = 3. Insert 3 at index 3.
Key: 6
Hash value: h(6) = 6 mod 13 = 6. Insert 6 at index 6.
Key: 16
Hash value: h(16) = 16 mod 13 = 3. Index 3 is occupied. Use linear probing and insert 16 at index 4.
Key: 17
Hash value: h(17) = 17 mod 13 = 4. Index 4 is occupied. Use linear probing and insert 17 at index 5.
Key: 19
Hash value: h(19) = 19 mod 13 = 6. Index 6 is occupied. Use linear probing and insert 19 at index 7.
Final Hash Table

IndexValue
0
1
2
33
416
517
66
719
8
9
1010
11
12

Summary
The final hash table after inserting all keys using linear probing is as follows:

[ – , – , – , 3 , 16 , 17 , 6 , 19 , – , – , 10 , – , – ]
3.(a) Create an example of a trigger.

The hash function is given as h(k)=k mod 13.
The table size is 13, meaning we have indices from 0 to 12.

Step-by-Step Insertion Process
Key: 10
Hash value: h(10) = 10 mod 13 = 10. Insert 10 at index 10.
Key: 3
Hash value: h(3) = 3 mod 13 = 3. Insert 3 at index 3.
Key: 6
Hash value: h(6) = 6 mod 13 = 6. Insert 6 at index 6.
Key: 16
Hash value: h(16) = 16 mod 13 = 3. Index 3 is occupied. Use linear probing and insert 16 at index 4.
Key: 17
Hash value: h(17) = 17 mod 13 = 4. Index 4 is occupied. Use linear probing and insert 17 at index 5.
Key: 19
Hash value: h(19) = 19 mod 13 = 6. Index 6 is occupied. Use linear probing and insert 19 at index 7.
Final Hash Table

IndexValue
0
1
2
33
416
517
66
719
8
9
1010
11
12

Summary
The final hash table after inserting all keys using linear probing is as follows:

[ – , – , – , 3 , 16 , 17 , 6 , 19 , – , – , 10 , – , – ]
3.(b) Consider the following tables: Customer(customerID, name), Accounts(accountID, customerID), Orders(orderID, accountID, orderAmount). Write an SQL query to display customerID, name, and total order amount of all customers whose total order amount is greater than 5000.

The hash function is given as h(k)=k mod 13.
The table size is 13, meaning we have indices from 0 to 12.

Step-by-Step Insertion Process
Key: 10
Hash value: h(10) = 10 mod 13 = 10. Insert 10 at index 10.
Key: 3
Hash value: h(3) = 3 mod 13 = 3. Insert 3 at index 3.
Key: 6
Hash value: h(6) = 6 mod 13 = 6. Insert 6 at index 6.
Key: 16
Hash value: h(16) = 16 mod 13 = 3. Index 3 is occupied. Use linear probing and insert 16 at index 4.
Key: 17
Hash value: h(17) = 17 mod 13 = 4. Index 4 is occupied. Use linear probing and insert 17 at index 5.
Key: 19
Hash value: h(19) = 19 mod 13 = 6. Index 6 is occupied. Use linear probing and insert 19 at index 7.
Final Hash Table

IndexValue
0
1
2
33
416
517
66
719
8
9
1010
11
12

Summary
The final hash table after inserting all keys using linear probing is as follows:

[ – , – , – , 3 , 16 , 17 , 6 , 19 , – , – , 10 , – , – ]

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