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Vlad [161]
2 years ago
6

What are the limits of hashing and verifying that files haven’t changed?

Computers and Technology
1 answer:
earnstyle [38]2 years ago
5 0

The hashing function can take any number of key-value pairs and there is no specific limit to it.

<h3>What is hashing?</h3>

Hashing is a file-based algorithm for producing a fixed-length bit string value. A file is essentially a collection of data blocks. The length of the data is reduced by hashing to a fixed number or key that represents the original string.

When hashing is employed, the hash function may plot all of the keys and values to what the real size of the table is, demonstrating that the hashing function can take any number of key-value pairs with no restriction.

However, if the passwords are hashed in encryption, recovering the passwords is extremely difficult.

Thus, the hashing function can take any number of key-value pairs and there is no specific limit to it.

Learn more about the hashing here:

brainly.com/question/13106914

#SPJ1

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Dahasolnce [82]

Answer:

cpu

Explanation:

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3 years ago
Imagine a typical website that works as a storefront for a business, allowing customers to browse goods online, place orders, re
Gwar [14]

Answer: See explanation

Explanation:

Following the information given in the question, the testing process for the website will include testing the links that are on the site.

Another that ng to test is to check if the menus and the buttons are working properly. Furthermore, the layout should be ensured that it's consistent as well as the ease with which the website can be used.

5 0
3 years ago
Where is the thesis statement usually located in research paper?
Ugo [173]
The thesis statement is typically located in the last sentence of the first paragraph of a research paper.
7 0
3 years ago
2.24 Write the relevant Python expression or statement, involving a list of numbers lst and using list operators and methods for
miskamm [114]

Answer:

lst = [11, 603, 99, 7, 582, 1];

print("The index of the middle element: " + str(int(len(lst)/2)))

print("The middle element: " + str(lst[int(len(lst)/2)]))

lst.sort()

sorted_lst = lst.reverse()

print("The list in descending order: " + str(lst))

first_number = lst.pop(0)

lst.append(first_number)

print("New lst: "+ str(lst))

Explanation:

- Initialize a list

- Find the middle index, int(len(lst)/2)

- Find the middle element, lst[int(len(lst)/2)]

- Sort the list in descending order (First sort the list in ascending order than reverse it)

- Get the first element using <em>pop</em> method, and append it to the end using <em>append</em> method

- - - - -

a) 0 == 1 == 2 --> Always evaluates false

b) 2 + (3 == 4) + 5 == 7 --> Evaluates true

c) 1 < -1 == 3 > 4 --> Always evaluates false

3 0
3 years ago
Given a collection of n nuts and a collection of n bolts, arranged in an increasing order of size, give an O(n) time algorithm t
kari74 [83]

Answer:

See explaination

Explanation:

Keep two iterators, i (for nuts array) and j (for bolts array).

while(i < n and j < n) {

if nuts[i] == bolts[j] {

We have a case where sizes match, output/return

}

else if nuts[i] < bolts[j] {

this means that size of nut is smaller than that of bolt and we should go to the next bigger nut, i.e., i+=1

}

else {

this means that size of bolt is smaller than that of nut and we should go to the next bigger bolt, i.e., j+=1

}

}

Since we go to each index in both the array only once, the algorithm take O(n) time.

8 0
3 years ago
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