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How would I do this in 1 loop since I need to loop again and close every second door? Do they want me to loop through the program 100 times? Should I be using pointers ?
Yes, you should loop through the program 100 times if you want to simulate this behavior.
But if you want to know the final condition(Open/Close) then you can have better algorithm:
As every perfect square number only have odd number of factor, if number is perfect square then final condition of door is open otherwise door is close.
If you are interested see perfect square number and Why perfect squares only have odd numbers of factor .
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So I'm trying to use fork and pipe in order to find the prime numbers from 1-35. I'm having trouble understanding how to store my prime numbers without overwriting them in future steps. For example, say I have the number 2, and I decide it is prime, how do I store that value as prime and store values like 3, 5, 7 as the same variable? I think I'm mostly having trouble understanding how to use pipe and fork recursively. Thanks.
I tried creating a parent process that writes in the numbers 1-35 and a child process that takes said number and compares it to the numbers that were previously decided as prime, and then printing the result. This did not work.
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I am writing a Monte Carlo simulation in which I take "measurements" of the magnetization every 1 correlation time. I want to have 3 for loops. The first is to change the temperature and inside it (with constant temperature) 2 other for loops in which I run e.g 10 correlation times, so 10 measurements of M. My question is, how can I switch the txt file in which I write (with fprintf) M, for different temperatures? Shape of the 3 loops I want
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Time Complexity of Juggling algorithm for array rotation(Suppose 'd' times) is computed as O(n), where n is the size of the array. But for any number of rotation(i.e. for any value of 'd'), the algorithm runs exactly for n times. So, shouldn't the time complexity of the algorithm be "Theta(n)" ? It always loops for n times in any case.If not, can anyone provide a test case where it doesn't run for n times?
It is unclear what you ask, but if we look at https://www.geeksforgeeks.org/array-rotation/ we see that it is described as O(n) time but if we want to rotate zero steps it could be done in O(1) time, so it doesn't always take n times - i.e. Theta(n) would be wrong; but O(n) is correct.
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how to find all prime numbers between 1 and 10^9 , i know we can use Sieve_of_Eratosthenes for smaller range, but what when range is too large equivalent to 10^6 ?
Up to 10^9 is not really a big deal. First, only look at odd numbers (because there is only one even prime). Second, use a bit array, so you only need 500 million bits or about 62 Megabyte. Even straightforward code should do that in a few seconds at most.
If you go further, you'd do a sieve for numbers from 1 to 10^9, then from 10^9 + 1 to 2 * 10^9 and so on. Above 10^13 it gets interesting and you need to put a bit more effort into it.
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How would I write a function that contains a nested for-loop but is only order(n)? Im not sure if i need to use recursion or not.
If the inner for loop is a constant number of loops rather than a variable number of loops, while the outer loop is a variable number of loops (or vice versa) the time complexity is O(n*C) where C is a constant, which just means O(n) (since big O notation is only concerned with growing factors).