I have a brute-force algorithm that takes an input n, and displays the first n prime numbers. The code works, but I also have to answer this question:
Give the number of test items that would be required to assure correctness.
Is there a way to calculate this? Would I technically need to test every possible int value (all 2 billion possibilities)?
If you have a number n, and need to check if it is a prime, there are more efficient ways than brute force.
One way would be to use the Miller-Rabin Primality Test. The Miller-Rabin test either finds proof that a number is a composite, or it does not such find proof (it does not directly prove that a number is a prime). So the scheme would be:
run Miller-Rabin at most k times (or until it found that the number is a composite)
if Miller-Rabin claims it is a possible prime, perform a brute force check
Miller-Rabin runs as follows. Obviously, you need test only for odd n, and so n - 1 is even, and you can write n - 1 = 2sd for some positive s and positive odd d. Randomly choose an a from the range (0, n - 1). If ad ≠ 1 | n and a2rd ≠ -1 | n, then n is a composite.
If n is a composite, the probability that k iterations of Miller-Rabin will not prove it so, is less than 4-k. Note that by the Prime Number theorem, primes are scarce.
The computational-complexity of k applications of Miller-Rabin, even with a naive implementation, is, is O(k log3(n)).
Related
I have searched for ways to determinate if a number is prime or not, but most ways are either probabilistic (Miller Rabin) or for numbers smaller than 64 bits.
The other solution would be to use the brute force method with a few improvements or the sieves, but neither of those is very efficient when the numbers go above the 64 bit threshold.
What you are looking for does not exist. There is no simple deterministic primality test that works always for all ranges of integers.
You already know about the Miller-Rabin test. It can be made deterministic on particular ranges; see here or here for details. If you assume the Riemann Hypothesis, then n is prime if n is an a-SPRP (a Miller strong pseudoprime) for all integers a with 1 < a < 2(log n)². A similar and somewhat better test is the Baillie-Wagstaff test; it is not deterministic, but no failures are known.
For numbers n up to 2128, it's not too hard to factor n − 1 and use a Pocklington test to prove primality. You can use trial division, or Pollard rho, or ECM to perform the factorization. There are also tests (BLS75) that can prove primality based on a partial factorization. Larger n can also be proved prime using a Pocklington test, though sometimes the factorization becomes difficult.
For n up to about 101000, a fast ECPP prime test is not unreasonable, though for the larger numbers in that range it might take a while. Beyond that, unless your number has some special form, you're pretty much out of luck.
I will assume that what you want is a provably correct answer, rather than avoiding randomness altogether.
Run a few rounds of the Miller-Rabin primality test. If this fails, you know the number is composite, and you're done.
Factorize n-1. For this, simplest is the Pollard's rho algorithm. If that's not fast enough, use the Quadratic Sieve.
Check whether the factors are prime, using the same approach recursively. If they are composite, continue factorizing them.
Use the Lucas Primality Test: try to find a generator of the multiplicative group modulo n of order n-1. Pick a random number a, check that a^(n-1) = 1 (mod n), and that a^((n-1)/p) ≠ 1 (mod n) for all prime factors p of n-1. If this is true, a is a generator, and n is provably a prime number, so you are done.
If n is prime, the probability of success in finding a generator is (1-1/p1)(1-1/p2)... where p1, p2, ... are the distinct prime factors of n-1. This is at least 1 / O(log log n). So after O(log log n) attempts you should succeed in proving that n is prime.
If you keep failing in proving n is prime, go back to step 1. Maybe it's composite after all.
Bear in mind I'm almost a complete noob at complexity theory.
I was reading about how AKS Primality shows that numbers of size n can be shown to be prime or composite in polynomial time. Given that, does that imply finding all prime numbers less than a number n is also doable in polynomial time and thus the algorithm runs in FP. Additionally, does this imply that counting all primes less than n is not in #P?
well. What aks says it that primality testing for a number n is O(b^k) where b = log2(n) and k is some integer.
So, if your questions is is listing all the primes from 1 to n also O(b^k). Then the answer is trivially no because the number of primes less than n is O(n/logn). Therefore you would need O(n/log(n) ) just to list them
If your question is does there exist a k such that the complexity of listing all the primes less than n is in O(n^k).
Then the answer is trivially yes because the most trivial form of sieve is O(n^(1.5)).
If anything is unclear please let me know
Section 1.2.6 of SICP describes an algorithm for Fermat prime testing as follows (my own words):
To test whether n is prime:
Choose a random integer a between 1 and n-1 inclusively.
If a^n %n = a, then n is probably prime.
The part I'm getting stuck on is the fact that we allow a = 1 because in this case, regardless of our choice of n (prime or not), the test will always pass.
You're right; there's no reason to choose a = 1. That being said, the statistical distance between the uniform distribution on [1, n-1] and the uniform distribution on [2, n-1] is O(1/n), so when n is very large (large enough that you don't just want to do trial division), the practical impact is very small (remember that this is already a probabilistic test, so a good number of other choices of a won't work either).
The text you link to actually says (emphasis mine):
Fermat's Little Theorem: If n is a prime number and a is any positive integer less than n, then a raised to the nth power is congruent to a modulo n.
(Two numbers are said to be congruent modulo n if they both have the same remainder when divided by n. The remainder of a number a when divided by n is also referred to as the remainder of a modulo n, or simply as a modulo n.)
If n is not prime, then, in general, most of the numbers a < n will not satisfy the above relation. This leads to the following algorithm for testing primality: Given a number n, pick a random number a < n and compute the remainder of a^n modulo n. If the result is not equal to a, then n is certainly not prime. If it is a, then chances are good that n is prime. Now pick another random number a and test it with the same method. If it also satisfies the equation, then we can be even more confident that n is prime. By trying more and more values of a, we can increase our confidence in the result. This algorithm is known as the Fermat test.
Up until now it never says to actually pick 1. It does later on though. I think that's a mistake, although not a big one. Even if it's true for a given value, you should test multiple values to be sure.
The pseudocode on Wikipedia uses [2, n - 1] as the range for example. You should probably use this range in practice (although the Fermat test isn't really used in practice, since Miller-Rabin is better).
This is the pseudo code for calculating integer factorisation took from CLRS. But what is the point in calculating GCD involved in Line 8 and the need for doubling k when i == k in Line 13.? Help please.
That pseudocode is not Pollard-rho factorization despite the label. It is one trial of the related Brent's factorization method. In Pollard-rho factorization, in the ith step you compute x_i and x_(2i), and check the GCD of x_(2i)-x_i with n. In Brent's factorization method, you compute GCD(x_(2^a)-x_(2^a+b),n) for b=1,2, ..., 2^a. (I used the indices starting with 1 to agree with the pseudocode, but elsewhere the sequence is initialized with x_0.) In the code, k=2^a and i=2^a+b. When you detect that i has reached the next power of 2, you increase k to 2^(a+1).
GCDs can be computed very rapidly by Euclid's algorithm without knowing the factorizations of the numbers. Any time you find a nontrivial GCD with n, this helps you to factor n. In both Pollard-rho factorization and Brent's algorithm, one idea is that if you iterate a polynomial such as x^2-c, the differences between the values of the iterates mod n tend to be good candidates for numbers that share nontrivial factors with n. This is because (by the Chinese Remainder Theorem) iterating the polynomial mod n is the same as simultaneously iterating the polynomial mod each prime power in the prime factorization of n. If x_i=x_j mod p1^e1 but not mod p2^e2, then GCD(xi-xj,n) will have p1^e1 as a factor but not p2^e2, so it will be a nontrivial factor.
This is one trial because x_1 is initialized once. If you get unlucky, the value you choose for x_1 starts a preperiodic sequence that repeats at the same time mod each prime power in the prime factorization of n, even though n is not prime. For example, suppose n=1711=29*59, and x_1 = 4, x_2=15, x_3=224, x_4=556, x_5=1155, x_6=1155, ... This sequence does not help you to find a nontrivial factorization, since all of the GCDs of differences between distinct elements and 1711 are 1. If you start with x_1=5, then x_2=24, x_3=575, x_4=401, x_5=1677, x_6=1155, x_7=1155, ... In either factorization method, you would find that GCD(x_4-x_2,1711)=GCD(377,1711)=29, a nontrivial factor of 1711. Not only are some sequences not helpful, others might work, but it might be faster to give up and start with another initial value. So, normally you don't keep increasing i forever, normally there is a termination threshold where you might try a different initial value.
If you need to generate primes from 1 to N, the "dumb" way to do it would be to iterate through all the numbers from 2 to N and check if the numbers are divisable by any prime number found so far which is less than the square root of the number in question.
As I see it, sieve of Eratosthenes does the same, except other way round - when it finds a prime N, it marks off all the numbers that are multiples of N.
But whether you mark off X when you find N, or you check if X is divisable by N, the fundamental complexity, the big-O stays the same. You still do one constant-time operation per a number-prime pair. In fact, the dumb algorithm breaks off as soon as it finds a prime, but sieve of Eratosthenes marks each number several times - once for every prime it is divisable by. That's a minimum of twice as many operations for every number except primes.
Am I misunderstanding something here?
In the trial division algorithm, the most work that may be needed to determine whether a number n is prime, is testing divisibility by the primes up to about sqrt(n).
That worst case is met when n is a prime or the product of two primes of nearly the same size (including squares of primes). If n has more than two prime factors, or two prime factors of very different size, at least one of them is much smaller than sqrt(n), so even the accumulated work needed for all these numbers (which form the vast majority of all numbers up to N, for sufficiently large N) is relatively insignificant, I shall ignore that and work with the fiction that composite numbers are determined without doing any work (the products of two approximately equal primes are few in number, so although individually they cost as much as a prime of similar size, altogether that's a negligible amount of work).
So, how much work does the testing of the primes up to N take?
By the prime number theorem, the number of primes <= n is (for n sufficiently large), about n/log n (it's n/log n + lower order terms). Conversely, that means the k-th prime is (for k not too small) about k*log k (+ lower order terms).
Hence, testing the k-th prime requires trial division by pi(sqrt(p_k)), approximately 2*sqrt(k/log k), primes. Summing that for k <= pi(N) ~ N/log N yields roughly 4/3*N^(3/2)/(log N)^2 divisions in total. So by ignoring the composites, we found that finding the primes up to N by trial division (using only primes), is Omega(N^1.5 / (log N)^2). Closer analysis of the composites reveals that it's Theta(N^1.5 / (log N)^2). Using a wheel reduces the constant factors, but doesn't change the complexity.
In the sieve, on the other hand, each composite is crossed off as a multiple of at least one prime. Depending on whether you start crossing off at 2*p or at p*p, a composite is crossed off as many times as it has distinct prime factors or distinct prime factors <= sqrt(n). Since any number has at most one prime factor exceeding sqrt(n), the difference isn't so large, it has no influence on complexity, but there are a lot of numbers with only two prime factors (or three with one larger than sqrt(n)), thus it makes a noticeable difference in running time. Anyhow, a number n > 0 has only few distinct prime factors, a trivial estimate shows that the number of distinct prime factors is bounded by lg n (base-2 logarithm), so an upper bound for the number of crossings-off the sieve does is N*lg N.
By counting not how often each composite gets crossed off, but how many multiples of each prime are crossed off, as IVlad already did, one easily finds that the number of crossings-off is in fact Theta(N*log log N). Again, using a wheel doesn't change the complexity but reduces the constant factors. However, here it has a larger influence than for the trial division, so at least skipping the evens should be done (apart from reducing the work, it also reduces storage size, so improves cache locality).
So, even disregarding that division is more expensive than addition and multiplication, we see that the number of operations the sieve requires is much smaller than the number of operations required by trial division (if the limit is not too small).
Summarising:
Trial division does futile work by dividing primes, the sieve does futile work by repeatedly crossing off composites. There are relatively few primes, but many composites, so one might be tempted to think trial division wastes less work.
But: Composites have only few distinct prime factors, while there are many primes below sqrt(p).
In the naive method, you do O(sqrt(num)) operations for each number num you check for primality. Ths is O(n*sqrt(n)) total.
In the sieve method, for each unmarked number from 1 to n you do n / 2 operations when marking multiples of 2, n / 3 when marking those of 3, n / 5 when marking those of 5 etc. This is n*(1/2 + 1/3 + 1/5 + 1/7 + ...), which is O(n log log n). See here for that result.
So the asymptotic complexity is not the same, like you said. Even a naive sieve will beat the naive prime-generation method pretty fast. Optimized versions of the sieve can get much faster, but the big-oh remains unchanged.
The two are not equivalent like you say. For each number, you will check divisibility by the same primes 2, 3, 5, 7, ... in the naive prime-generation algorithm. As you progress, you check divisibility by the same series of numbers (and you keep checking against more and more as you approach your n). For the sieve, you keep checking less and less as you approach n. First you check in increments of 2, then of 3, then 5 and so on. This will hit n and stop much faster.
Because with the sieve method, you stop marking mutiples of the running primes when the running prime reaches the square root of N.
Say, you want to find all primes less than a million.
First you set an array
for i = 2 to 1000000
primetest[i] = true
Then you iterate
for j=2 to 1000 <--- 1000 is the square root of 10000000
if primetest[j] <--- if j is prime
---mark all multiples of j (except j itself) as "not a prime"
for k = j^2 to 1000000 step j
primetest[k] = false
You don't have to check j after 1000, because j*j will be more than a million.
And you start from j*j (you don't have to mark multiples of j less than j^2 because they are already marked as multiples of previously found, smaller primes)
So, in the end you have done the loop 1000 times and the if part only for those j's that are primes.
Second reason is that with the sieve, you only do multiplication, not division. If you do it cleverly, you only do addition, not even multiplication.
And division has larger complexity than addition. The usual way to do division has O(n^2) complexity, while addition has O(n).
Explained in this paper: http://www.cs.hmc.edu/~oneill/papers/Sieve-JFP.pdf
I think it's quite readable even without Haskell knowledge.
the first difference is that division is much more expensive than addition. Even if each number is 'marked' several times, it's trivial when compared with the huge number of divisions needed for the 'dumb' algorithm.
A "naive" Sieve of Eratosthenes will mark non-prime numbers multiple times.
But, if you have your numbers on a linked list and remove numbers taht are multiples (you will still need to walk the remainder of the list), the work left to do after finding a prime is always smaller than it was before finding the prime.
http://en.wikipedia.org/wiki/Prime_number#Number_of_prime_numbers_below_a_given_number
the "dumb" algorithm does i/log(i) ~= N/log(N) work for each prime number
the real algorithm does N/i ~= 1 work for each prime number
Multiply by roughly N/log(N) prime numbers.
It was a time when I was trying to find an efficient way of finding sum of primes less than x:
There I decided to use N by N square table and started checking if numbers with a unit digits in [1,3,7,9]
But Eratosthenes Method of prime made it a little easier: How
Let you want to know if N is prime or Not
You started finding factorization. So you will realize when N is factorized
when you divide N with the highest factor quotient will be less.
So, You take a number: int(sqrt(N)) = K(say) divides N you get somewhat same and close number to K
Now let's say you divide N with u<K, but if "U" is not prime and one of the prime factors of U is V(prime) then will obviously be less than U (V<U) and V will also divide N
then
why not divide and check if N is prime or not by DIVIDING 'N' WITH ONLY PRIMES LESS THAN K=int(sqrt(N))
Number of times for which loop Keeps executing = π(√n)
This is how the brilliant idea of Eratosthenes starts taking pictures and will start giving you intuition behind this all.
Btw using the Sieve of Eratosthenes one can find sum of primes less than a multiple of 10.
because for a given column you just check need to check their unit digits[1,3,7,9] and for how many times a particular unit digit is repeating.
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