Consider an ordinary binary heap data structure with n elements that supports the instructions Insert and Extract-Min in O(log n) worst- case time. Give a potential function Φ such that the amortized cost of Insert is O(log n) and the amortized cost of Extract-Min is O(1), and show that it works.

It has been proved from the explanation below that the amortized cost of Insert is O(log n) and the amortized cost of Extract-Min is O(1),

Step-by-step explanation:

First of all, let's look at what "Extra-min" does;

It does the following ;

- Extract the root, (O1) actual time

- Replaces it with the last element x in the heap and repair the heap down or up.

- It takes O (Logn) time which is the depth of the heap.

Now, to get O(1) amortized time for extra-min, the last element "x" must have enough potential for repairing the heap and this potential must be in order of its depth.

Since we have K elements and so the weight of the kth element is WK in the heap as it's depth. So we can say;

WK = log k - - - - - - - - - eq (1)

Thus, the potential function Φ of the heap Hn for the n elements can be defined as ΦHn = Σ(WK) for k = 1 up till n. And so ΦHn will surely be greater than 0.

Thus; ΦHn = n(Logn)

Now let's check for insert ;

- pay actual cost of insert (logn)

-raise potential of the new elements

Now, using extra-min; O(1) amortized. To repair the heap, we'll use the potential of log(n) for the last element x as mentioned earlier.

In summary, when we use insert, potential increases and this can be afforded by paying insert more while when extra-min, potential decreases by a right moment of the cost of repairing.

Thus, we clearly see that Insert is O(log n) amortized and Extract-Min is O(1) amortized.

Linear

Step-by-step explanation:

If you graph this, you start at 1500 on the y-axis. Every time the x-axis increases by 1, the y-axis increases by 200. This forms a straight line, making it exponential since it increases by a constat amount.