Answer:
Explanation:
We start from the bottom-most and rightmost internal node of min Heap and then heapify all internal modes in the bottom-up way to build the Max heap.
To build a heap, the following algorithm is implemented for any input array.
BUILD-HEAP(A)
heapsize := size(A)
for i := floor(heapsize/2) downto 1
do HEAPIFY(A, i)
end for
Convert the given array of elements into an almost complete binary tree.
Ensure that the tree is a max heap.
Check that every non-leaf node contains a greater or equal value element than its child nodes.
If there exists any node that does not satisfy the ordering property of max heap, swap the elements.
Start checking from a non-leaf node with the highest index (bottom to top and right to left).
Yes , it’s true. In a known-plaintext attack (kpa), the cryptanalyst can only view a small portion of encrypted data, and he or she has no control over what that data might be.
The attacker also has access to one or more pairs of plaintext/ciphertext in a Known Plaintext Attack (KPA). Specifically, consider the scenario where key and plaintext were used to derive the ciphertext (either of which the attacker is trying to find). The attacker is also aware of what are the locations of the output from key encrypting. That is, the assailant is aware of a pair. They might be familiar with further pairings (obtained with the same key).
A straightforward illustration would be if the unencrypted messages had a set expiration date after which they would become publicly available. such as the location of a planned public event. The coordinates are encrypted and kept secret prior to the event. But when the incident occurs, the attacker has discovered the value of the coordinates /plaintext while the coordinates were decrypted (without knowing the key).
In general, a cipher is easier to break the more plaintext/ciphertext pairs that are known.
To learn more about Plaintext Attack click here:
brainly.com/question/28445346
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Answer:
I'll happy to help
Explanation:
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