Unicode was invented to create a "universal text", or a language framework that any computer or device could understand.
Answer:
See explaination
Explanation:
Here we will use two semaphore variables to satisfy our goal
We will initialize s1=1 and s2=1 globally and they are accessed by all 3 processes and use up and down operations in following way
Code:-
s1,s2=1
P1 P2 P3
P(s1)
P(s2)
C1
V(s2) .
P(s2). .
. C2
V(s1) .
P(s1)
. . C3
V(s2)
Explanation:-
The P(s1) stands for down operation for semaphore s1 and V(s1) stands for Up operation for semaphore s1.
The Down operation on s1=1 will make it s1=0 and our process will execute ,and down on s1=0 will block the process
The Up operation on s1=0 will unblock the process and on s1=1 will be normal execution of process
Now in the above code:
1)If C1 is executed first then it means down on s1,s2 will make it zero and up on s2 will make it 1, so in that case C3 cannot execute because P3 has down operation on s1 before C3 ,so C2 will execute by performing down on s2 and after that Up on s1 will be done by P2 and then C3 can execute
So our first condition gets satisfied
2)If C1 is not executed earlier means:-
a)If C2 is executed by performing down on S2 then s2=0,so definitely C3 will be executed because down(s2) in case of C1 will block the process P1 and after C3 execute Up operation on s2 ,C1 can execute because P1 gets unblocked .
b)If C3 is executed by performing down on s1 then s1=0 ,so definitely C2 will be executed now ,because down on s1 will block the process P1 and after that P2 will perform up on s1 ,so P1 gets unblocked
So C1 will be executed after C2 and C3 ,hence our 2nd condition satisfied.
Big-O notation is a way to describe a function that represents the n amount of times a program/function needs to be executed.
(I'm assuming that := is a typo and you mean just =, by the way)
In your case, you have two loops, nested within each other, and both loop to n (inclusive, meaning, that you loop for when i or j is equal to n), and both loops iterate by 1 each loop.
This means that both loops will therefore execute an n amount of times. Now, if the loops were NOT nested, our big-O would be O(2n), because 2 loops would run an n amount of times.
HOWEVER, since the j-loop is nested within i-loop, the j-loop executes every time the i-loop <span>ITERATES.
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As previously mentioned, for every i-loop, there would be an n amount of executions. So if the i-loop is called an n amount of times by the j loop (which executes n times), the big-O notation would be O(n*n), or O(n^2).
(tl;dr) In basic, it is O(n^2) because the loops are nested, meaning that the i-loop would be called n times, and for each iteration, it would call the j-loop n times, resulting in n*n runs.
A way to verify this is to write and test program the above. I sometimes find it easier to wrap my head around concepts after testing them myself.
Answer; A peripheral is a “device that is used to put information into or get information out of the computer.”[1]
There are three different types of peripherals:
Input, used to interact with, or send data to the computer (mouse, keyboards, etc.)
Output, which provides output to the user from the computer (monitors, printers, etc.)
Storage, which stores data processed by the computer (hard drives, flash drives, etc.)
Explanation:
Lightning McQueen in the Disney movie, Cars
KACHOW
