Program to find prime numbers in MATLAB - matlab

Question

I wrote some code to display the prime numbers between 2 and a user-chosen number, following the pseudocode on wikipedia. I am not sure why this does not work, as I have the increments correct following Erastothenes' Sieve. Please help me.
I have tried changing the bounds but this did not work.
There are no errors, but it returns the wrong output. If I enter 10, it returns 2, 3, 4, 5, 6, 7, 8, 9, 10.
n=input("Enter an upper limit: ");
nums= 2:n;
p=2;
for i = p:sqrt(n)
for j = (i^2):i:sqrt(n)
nums(j) = 0;
end
end
for k = 1:n-1
if nums(k) ~= 0
disp(nums(k))
end
end

Answer 1

You can use the primes function in MATLAB for this
N = 10; % upper limit
p = primes(N); % List of all primes up to (and including) N
With one step less automation, you could use another in-built isprime
p = 1:N; % List of all numbers up to N
p( ~isprime( p ) ) = []; % Remove non-primes
Finally without using built-ins, we can address your code!
I assume you're referring to this pseudocode for the Sieve of Eratosthenes on Wikipedia.
Input: an integer n > 1.
Let A be an array of Boolean values, indexed by integers 2 to n,
initially all set to true.
for i = 2, 3, 4, ..., not exceeding √n:
if A[i] is true:
for j = i2, i2+i, i2+2i, i2+3i, ..., not exceeding n:
A[j] := false.
Output: all i such that A[i] is true.
I'll follow it step by step, pointing out differences to your code:
n = 10;
A = [false; true(n-1,1)]; % array of true Booleans, first element (1) is not prime
% But I've included a first element to make indexing easier.
% In your code, you were using index 'i' which was incorrect, as your array started at 2.
% Two options: (1) take my approach and pad the array
% (2) take your approach and using indices i-1 and j-1
for ii = 2:sqrt(n)
if A(ii) == true % YOU WERE MISSING THIS STEP!
for jj = ii^2:ii:n % YOU ONLY LOOPED UNTIL SQRT(n)!
A(jj) = false;
end
end
end
p = find(A);
disp(p)
This outputs the expected values.
Note that, at the end of the manual looping method, A is equivalent to isprime(1:n), mirroring my earlier suggestions.

Answer 2

There is two mistakes in your code:
The multiple should be check until n and not sqrt(n)
Since your nums vector start with 2 and not 1, if you want to
access the right value you need to use nums(j-1) = 0
So:
n=100
nums= 2:n;
p=2;
for i = p:sqrt(n)
for j = (i^2):i:n
nums(j-1) = 0;
end
end
for k = 1:n-1
if nums(k) ~= 0
disp(nums(k))
end
end
Noticed that you can skip one for loop using a modulo, it's probably not faster than the previous solution since this code create a logical index that include each prime that already been found.
n = 100
nums= 2:n;
for i = 2:sqrt(n)
nums(mod(nums,i)==0 & nums != i) = [];
end
nums.'
I simply delete the value in nums that can be divided by x but not x.