MATLAB: Can someone help with the code? (about factores primes)

codeMATLABprojecteuler

Hi, i'm beginner in this language

It seems be simple. What is the largest prime factor of the number 600851475143 ?

i try that, what is wrong?

n=600861475143
prime=2
while n~=1
      prime=prime+1
      if mod(n,prime)==0      
            n=n/prime
end
      end

Best Answer

Trivial.

factor(600861475143)
ans =
   3   3   3   3   60943  121721
max(factor(600861475143))
ans =
   121721

Ok, I suppose you are not allowed to use factor. But, perhaps you need to look at the actual factors of that number, and think about why your code failed.

What happens when there are repeated factors? You will find the first factor of 3 there, but then you increment prime. So you never found those second, third, or fourth factors of 3. At the very least, once you find a prime factor, you need to see if it is a multiple factor of your number.

As bad, what would have happened with your code had your teacher given you this number instead: 600861475147

This last number is prime. How many iterations of your loop would be needed to identify that it is indeed prime? 600 BILLION iterations of that loop (essentially 600 billion trial divides) would be needed before it would give up. (A modulus computation is essentially equivalent to a division.)

The point is, do you need to test for a factor that is larger than sqrt(n)? Think about it. For any number n, how many prime factors can the number n have that are larger than sqrt(n)?

Related Solutions

MATLAB: Next prime number using While loops

There have now been multiple answers to this homework, all of which are fairly equivalent. (The response by Shubham Sangle has a lot of good points.) Let me offer a few ideas.

A good code to solve for the next prime would try to avoid calling isprime as much as possible, because calls to isprime are expensive. Of course, this depends on how large a number you are looking at

isprime(10000000000037)
ans =
  logical
   1
   
timeit(@() isprime(10000000000037))
ans =
            0.009439529724

10000000000037 is close to as large a number you can use isprime on, working in double precision. If a single test for primality costs roughly 0.01 seconds (at least on my computer, your mileage may vary) then how long will it take to find the next prime after some number?

As it turns out, 10000000000037 is the next prime that exceeds 1e13. After that, we see primes at 10000000000051, 10000000000099, 10000000000129, 10000000000183, 10000000000259...

So out as far as 1e13, the spaces between primes seem to be running around 30 to 70, sometimes more, sometims less. That means, if you wanted to find the next prime after one of those primes, you might need to test perhaps as many as 70 numbers, or more. (However, you might get unlucky, if you find yourself at the beginning of a rather large hole between primes.)

If a single test by isprime takes 0.01 seconds out there, then your code for nextprime might take as much as one second of CPU time. That is not too bad, as long as one second is acceptable to you. Even so, there are some tricks you might employ.

The idea I will suggest here relies on the concept of k-rough numbers. Thus a rough number is one that has no small prime divisors. A k-rough number is one that has no prime divisors less than k. Some definitions will allow k as the smallest divisor, whereas I won't lose any sleep over the distinction.

https://en.wikipedia.org/wiki/Rough_number

A test in advance for small divisors is not a bad thing. For example, you can rule out all even numbers in your search. And you can rule out all multiples of 3, 5, 7, etc. That is, suppose I picked an arbitary large integer. What is the probability that it happens to be divisible by some small primes? (I call this pre-test a partial sieve.) Thus, 50% of the time, some large number will be divisible by 2. But 2/3 of the time (67%), it will be divisible by EITHER 2, OR 3. That means we could exclude 67% of the numbers, merely by knowing in advance if they were divisuble by EITHER 2, or by 3.

100*(1 - prod(1 - 1./[2 3]))
ans =
          66.6666666666667
100*(1 - prod(1 - 1./[2 3 5 7 11 13 17 19]))
ans =
          82.8975977582789
100*(1 - prod(1 - 1./primes(100)))
ans =
          87.9682709525065
100*(1 - prod(1 - 1./primes(1000)))
ans =
          91.9034736493157

So, if we excluded all numbers that are divisible by any small prime below 1000, then we could reduce the number of calls to isprime by almost 92%. We can carry this pretty far of course. So, if we pretest for divisibility by all primes below 1e8, the exclusion rate goes down to only roughly 97%.

100*(1 - prod(1 - 1./primes(1e8)))
ans =
          96.9520278389414

That is important when a single call to isprime costs 30 minutes of CPU time. (This only happens if you are looking at numbers with many thousands of decimal digits though, using perhaps the sym version of isprime.)

As such, a decent version of nextprime might look like this:

function nextp = nextprime(N)
  % solve for the next prime that comes immediately after N
  
  % insure that N is itself an integer.
  if N == ceil(N)
    % then N is an integer already. We need to start looking at the NEXT
    % prime number though, so we need to start our search at N+1.
    N = N + 1;
  else
    % N was not an integer at all. So ceil will suffice to increase N
    N = ceil(N);
  end
  
  % is N even? If so, then since we need ony test odd numbers, since
  % even numers are not prime, then we can increment it to an odd number.
  % however, just in case N was so small that the next prime would have
  % been 2, special case that.
  if N < 2
    nextp = 2;
    return
  elseif rem(N,2) == 0
    % we only need to look at odd numbers after this.
    N = N + 1;
  end
  
  % Set up a partial sieve. Here, I'll set the sieve to check for divisibility
  % by primes that do not exceed 1000. We don't need to check for divisibility
  % by 2 though, since we will just step around all even numbers anyway, and
  % we have already excluded the case where N was less than 2. So we never
  % need to worry about the only even prime at this point.
  partialsieve = primes(1000);
  partialsieve(1) = [];
  
  % finally, this is just a while loop. The test in the while becomes a no-op,
  % because we break out when a prime gets hit. Use a while instead of a for
  % because we have no idea how long the search will go. It does take a very long
  % time before the intervening distance between two primes gets very large, so
  % the while loop will terminate before long.
  while true
    if ~all(rem(N,partialsieve))
      % no need to test this candidate
      N = N + 2;
    else
      if isprime(N)
        nextp = N;
        % just break out of the loop. No need to even reset flag.
        break
      else
        N = N + 2;
      end
    end
  end
end

This code is actually pretty efficient, allowing us to exclude tests with isprime for 92% of the numbers we might otherwise try. In fact, it looks like it had to make only one call to isprime in this example run:

timeit(@() nextprime(1e13))
ans =
            0.009934246724

I can make that claim, since that is roughly the time that MATLAB took for ONE call to isprime out there. And in this next test, there should have been only 5 calls to isprime.

timeit(@() nextprime(10000000000183))
ans =
            0.047855878724

The nextprime code above is efficient, since it is MUCH faster to test for divisibility by even a rather lengthy list of integers using the rem test, then it it to make one call to isprime out there.

How long will the search for the next prime require? Luckily, the primes are scattered randomly, but it takes a seriously long time before the inter-prime stride becomes terribly large. At the same time, it can be shown this stride can be arbitrarily large if you go out far enough.

As you can see below, it generally will not take that many tests before the next prime is found, but it will help if you can avoid some of those tests with careful code.

>> max(diff(primes(1e6)))
ans =
   114
>> max(diff(primes(1e7)))
ans =
   154
>> max(diff(primes(1e8)))
ans =
   220
>> max(diff(primes(1e9)))
ans =
   282

Will this all be significant for large numbers? Yes, very much so. If you plot the maximum interprime stride, as a function of the size of the numbers we are looking at, you will find that on a plot with logged x axis, the curve should be roughly linear.

semilogx(10.^[6 7 8 9],[114 154 220 282],'-o')
grid on

I'll just fit a linear polynomial to it, as:

interprimepoly = polyfit([6 7 8 9],[114 154 220 282],1)
interprimepoly =
                        57                      -235

This will be adequate for an extrapolation, since I don't care if it is only wildly approximate.

polyval(interprimepoly,[13, 100000])
ans =
                       506                   5699765

As such, out in the vicinity of 1e13, we may possibly be pushed to test several hundred numbers to see if they are prime. If we can intelligently exclude perhaps 90% of those tests, then we can cut the time required by up to 90%.

Further out, looking at numbers with tens or hundreds of thousands of decimal digits, the importance of avoiding millions of calls to isprime will be quite significant.

MATLAB: How can i add a prime counting function to this

n = zeros(1,N);
n(p) = 1;
c = cumsum(n);

Best Answer

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