Answer:
When the text contain the ASCII code (Which contains the character) is the correct answer of the above question.
Explanation:
ASCII is used for the Unicode for the computer system. The full form of this is the American standard code of information interchange. It is a code that is used by the computer system for their personal use. It is because no one can hack the information from the system. If anyone can do so then they can get that text which has no mean.
The above question asked about the condition in which Unicode output is the same from plain text. So the answer is ASCII code because the ASCII code and the Unicode are the same. The only difference between them is the representation of bits.
In regards to the rules of compaction, the IPv6 address 2001:0db8:0000:0000:0000:ff00:0012:3456 could also be written as 2001:db8::ff00:12:3456.
<h3>What is IPv6?</h3>
The IPv6 address is known to be a form of Internet Layer protocol made for packet-switched internetworking and it helps to give an end-to-end datagram movement in course of multiple IP networks.
Note that, In regards to the rules of compaction, the IPv6 address 2001:0db8:0000:0000:0000:ff00:0012:3456 could also be written as 2001:db8::ff00:12:3456.
Learn more aboutIPv6 address from
brainly.com/question/5296366
#SPJ1
The computer declined the process for the password
Procedure SumEvenOdd(stdin: array [0..100] of integer)
var
i, sum_even, sum_odd: integer;
begin
for i := 0 to 100 do begin
if stdin[i] < 0 then
break;
if stdin[i] mod 2 = 0 then //even number
sum_even := sum_even + stdin[i]
else
sum_odd := sum_odd + stdin[i];
ShowMessage('sum of even is ' + IntToStr(sum_even) + ' ' + 'sum of odd is' + IntToStr(sum_odd) ) ;
end;
end
Answer:
Encryption method security:-
- Encryption method representing each alphabetic character as an integer between 0 and 25.
- Now, apply RSA algorithm for each integer is not an efficient secure encryption method.
Explanation:
- Let consider alphabetic characters set {A, B... Z} .
- Each alphabetic character forms a set of message block values M={0,1,2...,25).
- From encryption method the corresponding cipher text block value C = {0e mod n,1e mod n, 2e mod n,..., 25e mod n}.
- Then, compute the cipher text with knowledge of Bob's public key;
- If the public key is generated, then possibility to decrypt the cipher text to plain text.
- The generated public key is easily calculated by everyone and security is lost.
- Therefore, the encryption method is not secure.
Encryption method is not secure, Most efficient attack against this method:-
- Now, compute the encryption message with the function Me mod n for all the possible values of M.
- This is the most efficient attack against the scheme in encryption method.
- Then, create a look-up table with cipher text as an index to the corresponding plain text as a value for appropriate location in the table.