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A natural number is said to be prime if it is only divisible by itself and 1. In short, a prime number has only two factors which are 1 and the number itself. Numbers which are not prime are called composite numbers
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A prime number can only be written as the product of two numbers For example, consider 3.
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The following diagram shows the prime numbers between 1 and 100 Check out this table to find some interesting facts about prime numbers, which will be discussed in the next section
In the above program, the “for” loop is iterating 2 to n/2 times where “n” is the input number We check for each number up to n/2 if it divides the number If any multiple of n is found, update the value of temp and return “Not Prime” else “Prime”.
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Since the loop is iterating 2 to n/2, the time complexity for the worst case will be O(n), where n is the input element.
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The program is not using any extra auxiliary space, only fixed space is being used Hence, the space complexity is O(1).
You may ask why we are repeating the loop up to n/2 instead of n Why leave the other half? Let’s understand it with the help of example Consider the factors of the integer 12 The factors are 1, 2, 3, 4, 6, 12 case). This is true for all integers
Now consider a prime integer that can be found in the first half as one factor, as you can notice that the second half has only one factor and that is the number itself.
In the following program, we have implemented “while” loop instead of “for” loop The logic is similar to the previous program
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The loop in the program runs from 2 to n/2, so, the time complexity will be O(n) for the worst case. Here, n is the input element
In this program, only constant space is used for some variables Hence, the space complexity is O(1).
The function has only one “for” loop that runs from 2 to n/2 So, in the worst case, the time complexity will be O(n).
No extra space is used here Functions only use constant space to store variables Hence, the space complexity will be O(1).
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This is an iterative method for finding prime numbers Here, a recursive function find_Prime is created that simply checks if num is divisible by a number. If so, it returns false otherwise the recursive call is made This process is repeated until any multiple numbers are found
The function is calling itself repeatedly n times until a factor is found or a condition is met. Hence, the time complexity is O(n).
The n calls made by the recursive function will be stored on the stack which will consume space in memory. Hence, the space complexity of the iterative method will be O(n).
In the above program, we have used an optimized solution to check whether a number is prime or not. Here, instead of looping over the number itself, we are looping it up to its square root (√n). This is because the smallest factor of a number (greater than 1) cannot be greater than the square root of the number For example, the smallest factor of 64 is 2 which is not greater than √64
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This is because the “for” loop is iterating from 2 to the square root of (√n), where n is the input element.
No extra space is used in the program Only a constant auxiliary space is used to store variables and iterate over space
In the above program, we print the sum of prime numbers in the given range The start and end points are read by the user We have a nested iteration for loop The outer loop runs from left to right and the inner loop runs up to n/2 times for each iteration of the outer loop.
(Here, “num” is the input number, “left” and “right” are the start and end points of the given range). We used a nested “for” loop where the outer loop is iterating from “left” to “right”, the inner loop is iterating from 2 to n/2. So, the outer loop runs for (right-left) time, and the inner loop repeats “num” times,
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Step 2: Create a boolean array isPrime and initialize all its elements to 1 (assuming all elements are prime at the beginning).
Step 3: If an element k is equal to 1 (or true), then mark all its properties from k2 to greater than 0.
STEP 4: Repeat STEP 2 for all identified (1 or equal to true) elements until the square root of the number is reached.
The above method is based on the theory of Eratosthenes We have defined an array of type boolean whose elements are initially true True means we assume that initially, all elements are prime If a number is updated to false, it will not be a prime number We start with 2 and iterate through a loop that will mark all properties of 2 that are greater than or equal to its square until false. We will repeat this process up to √num After that, all elements that are unsigned (or true) will be prime numbers
The Sum Of Digits Of All Numbers From 1 To 300 Is Equal To
The time complexity of identifying all non-prime numbers is assumed to be constant To find the time complexity for running the for now loop, the equation becomes,
By solving this equation using the formula for harmonic progression and Taylor series expansion, following Euler’s formula and simplifying, the final result can be determined which is equal to n*log(log(n)).
The space consumed by the boolean array isPrime is declared to be the same size as num Hence, the time complexity is O(n), where n is the input element
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To conclude, in this article, you have learned about multiple ways to write a C program for prime numbers You started with a quick introduction to prime numbers and some interesting facts about prime numbers
Moving on, in C programs for prime numbers article you have seen various techniques for checking a prime number using for loops, while loops, functions, recursion, optimize method etc. The most popular method is that of Sives of Erastosthenes You’ll see the algorithm, pseudocode, C program, time, and space complexity for each method we’ve discussed.
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Ravikiran AS works as a Research Analyst He is always a keen geologist to learn latest technologies He is proficient with Java programming language, BIG
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