Question

How many moves are required to solve the Towers of Hanoi problem when we have 3 disks?

Group of answer choices

5

6

8

4

7

After the following statements execute, what item is at the front of the queue?

QueueInterface<String> zooDelivery = new LinkedQueue<>();
zooDelivery.enqueue("lion");
zooDelivery.enqueue("tiger");
zooDelivery.enqueue("cheetah");
String next = zooDelivery.dequeue();
next = zooDelivery.dequeue();
zooDelivery.enqueue("jaguar");
zooDelivery.enqueue("cat");

Group of answer choices

"lion"

"tiger"

"cheetah"

"jaguar"

"cat"

When a recursive method does not include a base case, calling the method results in

Group of answer choices

A. iterative recursion

B. stack overflow

C. infinite recursion

D. A and B

E. B and C

In StackInterface, the method pop()

Group of answer choices

removes and returns the item that was pushed last on the stack

returns the item that was pushed last on the stack but does not remove it

removes and returns the item that was pushed first on the stack

returns the item that was pushed first on the stack but does not remove it

None of the above

Answer 1

Question 1.

If we have 3 disks then the number of moves required to solve the Tower of Hanoi is 7.

Lets assume we have 3 rods: A, B, C and 3 disks marked 1, 2, 3 respectively.

Then, below are the step by step manipulation to solve Tower of Hanoi

Move disk 1 from rod A to rod B  // step 1
Move disk 2 from rod A to rod C  // step 2
Move disk 1 from rod B to rod C  // step 3
Move disk 3 from rod A to rod B  // step 4
Move disk 1 from rod C to rod A  // step 5
Move disk 2 from rod C to rod B  // step 6
Move disk 1 from rod A to rod B  // step 7

Question 2.

Lion is the correct answer.

Now lets, line by line see what happened in the java code.

enqueue: adds an element to the end of linked list

dequeue: removes the most recently added element on the linked list

lion - > Null //zooDelivery.enqueue("lion");

lion -> tiger -> Null //zooDelivery.enqueue("tiger");

lion -> tiger -> cheetah - > Null //zooDelivery.enqueue("cheetah");

lion -> tiger -> Null //zooDelivery.dequeue();

lion - > Null //zooDelivery.dequeue();

lion -> jaguar -> Null //zooDelivery.enqueue("jaguar");

lion -> tiger -> cheetah - > Null //zooDelivery.enqueue("cat");

but we can see the front of the queue doesn't get affected from the above code and it still remains as lion.

Question 3.

Option E. B and C, is the correct answer

When a recursive call is made without a base case the code doesn't stop and infinite recursion happens until the heap memory is full and then stack overflow error occurs.

Question 4.

Option 1. removes and returns the item that was pushed last on the stack