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Insert the characters of the string K R P C S N Y T J M into a hash table of size 10. Use the hash function
h(x) = ( ord(x) – ord("a") + 1 ) mod10
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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Insert the characters of the string K R P C S N Y T J M into a hash ta...
(a) 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. (b) 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
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Most Upvoted Answer
Insert the characters of the string K R P C S N Y T J M into a hash ta...
- 65 ) % 10.

First, we need to calculate the hash value for each character using the given hash function:

- h(K) = (ord(K) - 65) % 10 = (75 - 65) % 10 = 1
- h(R) = (ord(R) - 65) % 10 = (82 - 65) % 10 = 7
- h(P) = (ord(P) - 65) % 10 = (80 - 65) % 10 = 5
- h(C) = (ord(C) - 65) % 10 = (67 - 65) % 10 = 2
- h(S) = (ord(S) - 65) % 10 = (83 - 65) % 10 = 8
- h(N) = (ord(N) - 65) % 10 = (78 - 65) % 10 = 3
- h(Y) = (ord(Y) - 65) % 10 = (89 - 65) % 10 = 4
- h(T) = (ord(T) - 65) % 10 = (84 - 65) % 10 = 9
- h(J) = (ord(J) - 65) % 10 = (74 - 65) % 10 = 0
- h(M) = (ord(M) - 65) % 10 = (77 - 65) % 10 = 2

Note that we subtracted 65 from the ASCII code of each character to get a value between 0 and 25, which is then modulo 10 to get a value between 0 and 9.

Next, we can insert each character into the hash table based on its hash value:

Index | Hash value | Character
------|-----------|----------
0 | 0 | J
1 | 1 | K
2 | 2 | C
3 | 3 | N
4 | 4 | Y
5 | 5 | P
6 | - |
7 | 7 | R
8 | 8 | S
9 | 9 | T

We can see that the characters with the same hash value (C and M) are stored in the same slot of the hash table using separate chaining, which means that each slot contains a linked list of elements that have the same hash value. In this case, the slot at index 2 contains both C and M.
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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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Insert the characters of the string K R P C S N Y T J M into a hash table of size 10. Use the hash functionh(x) = ( ord(x) – ord("a") + 1 ) mod10If 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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