Function to find the next prime number...

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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'.

采纳的回答

Basil C.
Basil C. 2018-2-19
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
  4 个评论
Walter Roberson
Walter Roberson 2019-12-21
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
Hicham Satti 2020-8-31
%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

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更多回答(12 个)

Arafat Roney
Arafat Roney 2020-5-11
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
Walter Roberson 2018-2-19
nextprime() was added to the Symbolic Toolbox in R2016b.
In releases before that,
feval(symengine, 'nextprime', sym(100))

Siddharth Joshi
Siddharth Joshi 2020-4-25
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
  4 个评论
Walter Roberson
Walter Roberson 2021-7-28
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
Stephen23 2021-7-28
"The code would have been better as"
... and ultimately simplifies down to this.

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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.
  2 个评论
Walter Roberson
Walter Roberson 2020-6-21
you test if k is prime but you increment n
SAKSHI CHANDRA
SAKSHI CHANDRA 2020-7-22
your control statement defines k but there is nothing related in the block statements which would check
~isprime(k)

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MD SADIQUE IQBAL
MD SADIQUE IQBAL 2020-7-17
unction n = next_prime(n)
k = n+1;
while ~isprime(k)
n = n+1;
end
end
  2 个评论
Stephen23
Stephen23 2020-7-17
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
SAKSHI CHANDRA 2020-7-22
your control statement defines k but there is nothing related in the block statements which would check
~isprime(k)

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SAKSHI CHANDRA
SAKSHI CHANDRA 2020-7-22
function k = nxt_prime(n)
k=n+1;
while ~isprime(k)
k=k+1;
end
end

Ravindra Pawar
Ravindra Pawar 2020-8-13
编辑:Ravindra Pawar 2020-8-13
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

shweta s
shweta s 2020-8-14
%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

Hicham Satti
Hicham Satti 2020-8-31
%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
  3 个评论
Hicham Satti
Hicham Satti 2020-9-8
why the other codes answers are not deleted ??
Rik
Rik 2020-9-8
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).

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Pragyan Dash
Pragyan Dash 2020-9-19
function k = next_prime(n)
while (~isprime(n + 1))
n = n + 1;
end
k = n + 1;
end

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
  1 个评论
Walter Roberson
Walter Roberson 2021-6-26
Is this efficient? If you are currently at 15, is there a point in testing 16?

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Dikshita Madkatte
Dikshita Madkatte 2021-7-14
编辑:Dikshita Madkatte 2021-7-14
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
  1 个评论
Rik
Rik 2021-7-14
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?

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