# Prime numbers Computationally-intensive numerical problems are not usually undertaken with q due to the looping nature of the solutions. However, there are neat, fast solutions using q, and they give us scope to practice writing efficient q functions – avoiding traditional loops – but also using iterators where loops are unavoidable.

A prime is a positive integer with exactly two factors: itself and one

## Test for prime

A simple and inefficient method to find if a number $$x$$ is prime would be to divide it by all positive integers less than $$x$$ and then see if any of the results are integers.

Although far from optimal this can be done very simply in q with the mod and til keywords.

Note that we drop 0 and 1 from the list til x as there is no point in checking x mod these. If x divides by a number y where y<x, then x mod y=0, hence min of this list of binaries will be 0b. If there is no 0b in the list we can certify this as a prime number.

q)isprimeSLOW:{min x mod 2_til x}
q)isprimeSLOW 13
1
q)isprimeSLOW 14
0
q)isprimeSLOW 1010101
0
q)l where isprimeSLOW each l:100+til 10
101 103 107 109


To speed this up exponentially we can check values only up to the square root of x. This makes mathematical sense as if a number x has a factor greater than sqrt x it must also have a factor less than sqrt x. For example, 27 divides by 3 and 7, if we are checking 3 there is no need to check 7.

q)isprime:{min x mod 2_til 1+floor sqrt x}

q)/ We can quickly check much larger numbers:
q)isprime 10101010101
0

q)/ Find which are prime from a list of numbers:
q)l where isprime each l:20202020202+til 100
20202020203 20202020231 20202020233 20202020273 20202020299


Feel free to try timing the previous vs the following function to see the speed-up when it comes to large numbers.

q)\ts isprimeSLOW 10101010101
2982031 549755814080
q)\ts isprime 10101010101
25 4194496


We can also add a conditional wrapper around our isprime function to give 1b if 2 or 3, and 0b if x<2.

q)/ isprime does not work for 0 1 2 3
q)isPrime:{$[x in 2 3;1;x<2;0;isprime x]} q)where isPrime each til 10  ## Sum primes Q1.1 Sum all the primes below two million https://projecteuler.net/problem=10 We can solve this quite simply by running isPrime on each of the list til 2000000. As we know there is no point testing even numbers other than 2 we can take these out. q)\ts 0N!sum a where isPrime each a:2,1+2*til 1000000 142913828922 5860 32778544  ## Nth prime Q1.2 Find the 10,001st prime number https://projecteuler.net/problem=7 We could run the isPrime function on a long list of numbers to generate a long list of primes, and then take the 10,001st entry of the prime list. Let’s try to do better. Break this down into the following iterator-based steps. 1. Use the isPrime function to find the next prime given a number 2. Run that function 10,000 times on initial argument of 2 With the While iterator (/) we can extend isPrime to keep adding 2 to an odd number, until isPrime is true. q)nextprime:{(not isPrime@)(2+)/x}(2+)@ q)nextprime 5 / adds two; immediately finds a prime 7 q)nextprime 7 / add two until finds a prime 11 q)nextprime 119 / 121 123 125: none is prime 127  Applying nextprime to an even number will cause it to run forever. Again, a conditional wrapper prevents this. q)nextPrime:{nextprime x-1 0 x mod 2} q)nextPrime each 2 3 4 3 5 5  We can test this with the Scan iterator (\) to generate lists of primes. ## Next prime Find the next 10 primes after 2 q)10 nextPrime\2 2 3 5 7 11 13 17 19 23 29 31  Find all primes between 1,000,000 and 1,000,100 Function will overshoot by one, and include the initial value, so we drop these. q)-1_1_{x<1000100}nextPrime\1000000 1000003 1000033 1000037 1000039 1000081 1000099 q)l where isPrime each l:1000000+til 100 / equivalent using just isPrime 1000003 1000033 1000037 1000039 1000081 1000099  Now let’s use nextPrime to find the 10,001th prime with an Over (/). q)\ts show 10000 nextPrime/2 104743 185 17280  Reducing a given number to a list of its prime factors is a natural next step, and we can use the previous section’s functions to to build an efficient prime factorization function. Let’s first sketch an outline of one algorithm to go about this: 1. Start with L:() empty list; append primes to this 2. Given an input x, check if x is prime; if it is, exit appending x to L 3. Generate a list l, up to floor sqrt x, omitting 0 and 1, of divisors of x, as we do in isPrime 4. Redefine l as just the primes in l 5. Redefine x as x % prd l 6. Append l to L and repeat steps 2 onwards on the new x q)/Example q)L:() q)x:100 /not prime so step 2 does nothing q)l:l where 0=x mod l:2_til 1+floor sqrt x q)l:l where isPrime each l q)L,:l;L 2 5 q)x:x%prd l q)/Repeat q)/x is now 10 so step 2 still does nothing q)l:l where 0=x mod l:2_til 1+floor sqrt x q)l:l where isPrime each l q)L,:l;L 2 5 2 q)x:x%prd l q)/x is now 5 so we add it to L and we are finished. Cast to match L's type q)L,"j"$x
2 5 2 5


As this is a repeating algorithm we should already be thinking how we can pull this together using an iterator. The following steps allow us to convert the above code into a prime factorization function:

1. Consider the input x as a list (enlist x)
2. Each loop produces the list of found prime factors with the remaining compound number (x%prd l) last in the list
3. Apply the looping function on just the last term of the list
4. Use the Converge iterator (/) so function runs until the result stops changing
5. Lastly, add a wrapper with the casting and remove 1 (1 appears sometimes otherwise, e.g. 30)

This is a good test of writing concise q. It is possible write the whole function as a single expression.

primeFactors:{"j"\$except[;1] {(-1_x),l,last[x]%prd l@:where isPrime each l@:where 0=last[x] mod l:2_til 1+floor sqrt last x}/[enlist x]}

q)primeFactors 121
11 11
q)primeFactors 100000000000
2 5 2 5 2 5 2 5 2 5 2 5 2 5 2 5 2 5 2 5 2 5


## Largest prime factor

Find the largest prime factor of the number 600851475143

https://projecteuler.net/problem=3

q)\ts show max primeFactors 600851475143
6857
18 33554976


Let’s try a harder question.

The first two consecutive numbers to have two distinct prime factors are:

14 = 2 × 7
15 = 3 × 5


The first three consecutive numbers to have three distinct prime factors are:

644 = 2² × 7 × 23
645 = 3 × 5 × 43
646 = 2 × 17 × 19


## Distinct prime factors

Find the first four consecutive integers to have four distinct prime factors each.
What is the first of these numbers?

https://projecteuler.net/problem=47

q)next4PFs:{(not 4=count distinct primeFactors@)(1+)/x}(1+)@
q)\ts show first {not all first[x]=x-til 4}{1_x,next4PFs last x}/4#1
134043
2889 4195744


## Author Noah Attrup is a senior kdb+ developer in New York City.