Function to find the next prime number...

I have a vector of size 1*152.Now i want to find the next prime number of every number present in the vector..

Ex: My vector is a=[2 4 7 8] i want the output as [2 5 7 11]..i.e., if the number is a prime then that number will be the output i.e., like 2 and 7 in the given example...

I tried using nextprime like below it gives the following error:

case 1:

>>nextprime(sym(100))
Undefined function 'nextprime' for input arguments of type 'sym'.

case 2:

>> nextprime(3)
Undefined function 'nextprime' for input arguments of type 'double'.

Answer 1

[Basil C.]

Method 1 This functionality does not run in MATLAB and can be used only via MuPAD Notebook Interface.

• To create an MuPAD interface use the following code

 mupad
nb = allMuPADNotebooks

Then a interface screen shall pop up where you can proceed by using the nextprime(num) function.

Method 2

• You could also create a user defined function to compute the next prime number. This function takes only a non-negative integers as an argument

 function p = nextprime(n)
if (isprime(n))
    p=n;
else
    while(~isprime(n))
        n=n+1;
    end
    p=n;
end
end

[Stephen23]

The if is not required:

function n = nextprime(n)
n=n+1;
while ~isprime(n)
  n=n+1;
end
end

>> n = 1;
>> n = nextprime(n)
n =  2
>> n = nextprime(n)
n =  3
>> n = nextprime(n)
n =  5
>> n = nextprime(n)
n =  7
>> n = nextprime(n)
n =  11
>> n = nextprime(n)
n =  13
>> n = nextprime(n)
n =  17

[emmanuel adewumi]

Just a little improvement to the functions written above worked.

 function Q = nextprime(n)
if (isprime(n))
    Q=n+1;
else
    while(~isprime(n))
        n=n+1;
    end
    Q=n;
end
 end

[Walter Roberson]

No. Consider nextprime(3) . isprime(3) is true, so your code would return the non-prime 4.

function Q = nextprime(n)
    if (isprime(n))
        n=n+1;
    end
    while(~isprime(n))
        n=n+1;
    end
    Q=n;
end

However, this can be simplified down to Stephen's code of always adding 1 to n first

[Hicham Satti]

%That will run very well
function k=next_prime(n)
if ~n>0 || n~=fix(n) || ~isscalar(n)
    error ('Tap a integer positif scalar argument');
else
    if isprime(n+1)
        k=n+1;
    else
        j=n+1;
        while ~isprime(j)
            k=j+1;
            j=j+1;
        end
    end
end

Answer 2

[Arafat Roney]

function k=next_prime(n)
i=n+1;
if(isprime(i))
    k=i;
else
    while(~isprime(i))
        i=i+1;
    end
    k=i;
end
end

[Walter Roberson]

Your if is not needed, you can go directly to the while,

[THIERNO AMADOU MOUCTAR BALDE]

thanks

[Kartik rao]

thanks

Answer 3

[Walter Roberson]

nextprime() was added to the Symbolic Toolbox in R2016b.

In releases before that,

feval(symengine, 'nextprime', sym(100))

Answer 4

[Siddharth Joshi]

function k = next_prime(n)
if (~isscalar(n) || n<1 || n ~= fix(n))
    error('n should be positive scalar ineger!!!')
else 
    p=-1;
    while p<=0
        n=n+1;
        p=isprime(n)
    end
    k=n
end 
end

k = next_prime(79)

k =

83

[Chaitanya Milampure]

might be a stupid doubt, but why does p<0 and p<=0 make a big difference?

[Stephen23]

"might be a stupid doubt, but why does p<0 and p<=0 make a big difference?"

Because the author of this code did not understand how to handle logical data in a simpler way. Instead they obfuscated the simple logical condition behind some numeric comparisons: look at what p value is being used for, and what values it can have.

Once the various bugs and "features" are ironed out, all of these answers boil down to the same concept (which is easy to implement in just a few lines, even if not the most efficient approach to finding prime numbers).

[Walter Roberson]

isprime() returns 0 (false) or 1 (true). Comparing that as < 0 is going to be false except the first time due to the initialization of p=-1 .

The code would have been better as

function k = next_prime(n)
if (~isscalar(n) || n<1 || n ~= fix(n))
    error('n should be positive scalar ineger!!!')
else 
    p=false;
    while ~p
        n=n+1;
        p=isprime(n);
    end
    k=n;
end 
end

[Stephen23]

"The code would have been better as"

... and ultimately simplifies down to this.

Answer 5

[Buwaneka Dissanayake]

% what's wrong with this? it take too long to run & fail
function n = next_prime(n)
    k = n+1;
    while ~isprime(k)
        n = n+1;
    end
end

what's wrong with this? it take too long to run & fail.

[Walter Roberson]

you test if k is prime but you increment n

[SAKSHI CHANDRA]

your control statement defines k but there is nothing related in the block statements which would check

~isprime(k)

Answer 6

[MD SADIQUE IQBAL]

unction n = next_prime(n)

k = n+1;

while ~isprime(k)

n = n+1;

end

end

[Stephen23]

Fails every basic test:

>> next_prime(1)
ans =  1
>> next_prime(2)
ans =  2
>> next_prime(3) % infinite loop, stop using ctrl+c
>> next_prime(4)
ans =  4
>> next_prime(5) % infinite loop, stop using ctrl+c
>> next_prime(6)
ans =  6
>> next_prime(7) % infinite loop, stop using ctrl+c

I can see the pattern... it is a very big hint as to what the bug is. As is reading this thread.

[SAKSHI CHANDRA]

your control statement defines k but there is nothing related in the block statements which would check

~isprime(k)

Answer 7

[SAKSHI CHANDRA]

function k = nxt_prime(n)
k=n+1;
while  ~isprime(k)
  k=k+1;
end
end

Answer 8

[Ravindra Pawar]

function k = next_prime(n) %function definition

while ~isprime(n+1) %if n+1 is prime we are out of for loop else loop restarts

n = n+1;

end

k = n+1;

end

Answer 9

[shweta s]

%to find the next prime no.
function p = next_prime(n)
if (isprime(n))
    p=n+1;
else
    while(~isprime(n))
    n=n+1;
    end
    p=n;
end
end

[Sai Krishna Praneeth Duggirala]

fails when n=3

[Rik]

@Sai Krishna Praneeth Duggirala And that is why you should be smart if you want to cheat from this page.

[Walter Roberson]

Fails for any prime except 2.

Answer 10

[Hicham Satti]

%That will run very well
function k=next_prime(n)
if ~n>0 || n~=fix(n) || ~isscalar(n)
    error ('Tap a integer positif scalar argument');
else
    if isprime(n+1)
        k=n+1;
    else
        j=n+1;
        while ~isprime(j)
            k=j+1;
            j=j+1;
        end
    end
end

[Rik]

Same question here as with your other answers: why are you posting solutions to homework questions? What does it teach?

Since this exact solution has been posted before in this exact thread I will delete this answer if you don't respond to that question.

[Hicham Satti]

why the other codes answers are not deleted ??

[Rik]

Because I'm just one person trying to clean up thread like this. And you didn't answer my question (neither here, nor on the other next_prime thread).

Answer 11

[Pragyan Dash]

function k = next_prime(n)
    while (~isprime(n + 1))
        n = n + 1;
    end
    k = n + 1;
end

Answer 12

[Malgorzata Frydrych]

function k= next_prime(n)
if ~isscalar(n) || n<=0 || mod(n,1)~=0;
    error('number should be a positive integer scalar')
end
k=0;
while ~isprime(k)
    n=n+1;
    k=n;
end
end

[Walter Roberson]

Is this efficient? If you are currently at 15, is there a point in testing 16?

Answer 13

[Dikshita Madkatte]

function k=next_prime(n)
if  ~n>0 || n~=fix(n) || ~isscalar(n);
    fprintf('n should be positive interger')
end
i=n+1;
if (isprime(i))
    k=i;
else 
    while(~isprime(i))
        i=i+1;
    end
    k=i;
end
end

[Rik]

A few remarks:

fprintf is not an error. Your code will still run after it fails the check.

You can increment i by two, since 2 is the only even prime, and the while loop will not be reached if n is 1.

You forgot to write documentation for your function. What is this going to teach? Why should it not be deleted?