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The characters of the string K R P C S N Y T J M are inserted into a hash table of size 10 using hash function
h(x) = ((ord(x) - ord(A) + 1)) mod 10
If linear probing is used to resolve collisions, then the following insertion causes collision
  • a)
    Y
  • b)
    C
  • c)
    M
  • d)
    P
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
The characters of the string K R P C S N Y T J M are inserted into a h...
The hash table with size 10 will have index from 0 to 9. hash function = h(x) = ((ord(x) - ord(A) + 1)) mod 10 So for string K R P C S N Y T J M: K will be inserted at index : (11-1+1) mod 10 = 1 R at index: (18-1+1) mod 10 = 8 P at index: (16-1+1) mod 10 = 6 C at index: (3-1+1) mod 10 = 3 S at index: (19-1+1) mod 10 = 9 N at index: (14-1+1) mod 10 = 4 Y at index (25-1+1) mod 10 = 5 T at index (20-1+1) mod 10 = 0 J at index (10-1+1) mod 10 = 0 // first collision occurs. M at index (13-1+1) mod 10 = 3 //second collision occurs. Only J and M are causing the collision. Final Hash table will be:
0    T
1    K
2    J
3    C
4    N
5    Y
6    P
7    M
8    R
9    S
So, option (C) is correct.
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Most Upvoted Answer
The characters of the string K R P C S N Y T J M are inserted into a h...
Collision occurs when two or more elements are hashed to the same index in the hash table. In this case, the hash function h(x) = ((ord(x) - ord(A) + 1)) mod 10 is used to determine the index for each character. Let's analyze the given string and the hash function to understand why option 'C' causes a collision.

Given string: K R P C S N Y T J M

Hash function: h(x) = ((ord(x) - ord(A) + 1)) mod 10

Step 1: Calculate the index for each character using the hash function.

Character 'K':
ord('K') = 75
h('K') = ((75 - 65 + 1) mod 10) = 1

Character 'R':
ord('R') = 82
h('R') = ((82 - 65 + 1) mod 10) = 8

Character 'P':
ord('P') = 80
h('P') = ((80 - 65 + 1) mod 10) = 6

Character 'C':
ord('C') = 67
h('C') = ((67 - 65 + 1) mod 10) = 3

Character 'S':
ord('S') = 83
h('S') = ((83 - 65 + 1) mod 10) = 9

Character 'N':
ord('N') = 78
h('N') = ((78 - 65 + 1) mod 10) = 4

Character 'Y':
ord('Y') = 89
h('Y') = ((89 - 65 + 1) mod 10) = 5

Character 'T':
ord('T') = 84
h('T') = ((84 - 65 + 1) mod 10) = 4

Character 'J':
ord('J') = 74
h('J') = ((74 - 65 + 1) mod 10) = 0

Character 'M':
ord('M') = 77
h('M') = ((77 - 65 + 1) mod 10) = 3

Step 2: Check for collision.

Looking at the calculated indices, we can see that characters 'N' and 'T' both hash to index 4. This means that there is a collision between these two characters.

Therefore, the correct answer is option 'C' because inserting the character 'M' causes a collision with the character 'N' at index 4.
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Community Answer
The characters of the string K R P C S N Y T J M are inserted into a h...
The hash table with size 10 will have index from 0 to 9. hash function = h(x) = ((ord(x) - ord(A) + 1)) mod 10 So for string K R P C S N Y T J M: K will be inserted at index : (11-1+1) mod 10 = 1 R at index: (18-1+1) mod 10 = 8 P at index: (16-1+1) mod 10 = 6 C at index: (3-1+1) mod 10 = 3 S at index: (19-1+1) mod 10 = 9 N at index: (14-1+1) mod 10 = 4 Y at index (25-1+1) mod 10 = 5 T at index (20-1+1) mod 10 = 0 J at index (10-1+1) mod 10 = 0 // first collision occurs. M at index (13-1+1) mod 10 = 3 //second collision occurs. Only J and M are causing the collision. Final Hash table will be:
0    T
1    K
2    J
3    C
4    N
5    Y
6    P
7    M
8    R
9    S
So, option (C) is correct.
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Consider a simple communication system where multiple nodes are connected by a shared broadcast medium (like Ethernet or wireless). The nodes in the system use the following carrier-sense the medium access protocol. A node that receives a packet to transmit will carrier-sense the medium for 5 units of time. If the node does not detect any other transmission in this duration, it starts transmitting its packet in the next time unit. If the node detects another transmission, it waits until this other transmission finishes, and then begins to carrier-sense for 5 time units again. Once they start to transmit, nodes do not perform any collision detection and continue transmission even if a collision occurs. All transmissions last for 20 units of time. Assume that the transmission signal travels at the speed of 10 meters per unit time in the medium. Assume that the system has two nodes P and Q, located at a distance d meters from each other. P starts transmitting a packet at time t = 0 after successfully completing its carrier-sense phase. Node Q has a packet to transmit at time t = 0 and begins to carrier-sense the medium. The maximum distance d (in meters, rounded to the closest integer) that allows Q to successfully avoid a collision between its proposed transmission and P’s on going transmission is _______ .Note -This was Numerical Type question.

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The characters of the string K R P C S N Y T J M are inserted into a hash table of size 10 using hash functionh(x) = ((ord(x) - ord(A) + 1)) mod 10If linear probing is used to resolve collisions, then the following insertion causes collisiona)Yb)Cc)Md)PCorrect answer is option 'C'. Can you explain this answer?
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