Hash Table and IPv6

Question

Consider an initially empty hash table of length 10. Following set of keys are inserted using open addressing with hash function h(k) = kmod 10 and linear probing.

The number of different insertion sequence of the key values using the given hash function and linear probing will result in the hash table shown in above __________.

I am getting 168.

---

Which of the following is true?

A) In IPv6 broadcasting is used to send packet to every host inside a network.

B) In IPv6 for Path MTU discovery ICMP protocol is used.

C) Flooding always guaranteed to send the packet from source to destination with minimum number of hops.

D) both b and c.

I think it is B).

Comments

1st one will be 36×8=288.

---

For the second question I think option D.

Flooding guarantees minimum hops.

Reference:

https://gateoverflow.in/188103/flooding

---

For 1st one:-

Case 1:-  When 77 come with $91, 2,13,24$ i.e. before 12, 62. So, no of ways to arrange this is 120.

Case 2:- When 77 come between 12 and 62 then 91, 2, 13, 24 can be arranged in 24 ways, So, for this case 24 ways.

Case 3:- When 77 come after 12 and 62 then 91, 2, 13, 24 can be arranged in 24 ways, So, for this case 24 ways.

In total $120+ 24+ 24 = 168$ ways.

---

1)i am getting 288

2) both B) and C) are correct, flooding always guarantee shortest path.

---

91 is independent in each case , i think that is wrong in your case.

---

I agree with $288$ and both $b)$ and $c)$
Sending packet to all links will guarantee that it would have also sent to one path which will be reported by shortest path algorithm therefore it will be ensured by flooding as well.

---

Duplicates of 1^st question:

https://gateoverflow.in/188101/hashing-test-series

https://gateoverflow.in/243402/hashing

https://gateoverflow.in/293487/me-test

---

2^nd question:

https://gateoverflow.in/201061/me-test-series

https://gateoverflow.in/188103/flooding

Answer 1

for the first one graph will be 

2,13,24,12,62,82

permute the (2,13,24) => 6 ways

i.e

2,13,24,12,62,82

2,24,13,12,62,82

13,2,24,12,62,82

13,24,2,12,62,82

24,13,2,12,62,82

24,2,13,12,62,82

Remaining are 77 and 91

for 77

__2__13__24__12__62__82 => 6 ways

for 91 

__77__2__13__24__12__62__82__ => 8 ways

Total possibilities=6*6*8=288