This is boring, I know, but I need a little help understanding an implementation of the Sieve of Eratosthenes. It's the solution to this Programming Praxis problem.
(define (primes n)
(let* ((max-index (quotient (- n 3) 2))
(v (make-vector (+ 1 max-index) #t)))
(let loop ((i 0) (ps '(2)))
(let ((p (+ i i 3)) (startj (+ (* 2 i i) (* 6 i) 3)))
(cond ((>= (* p p) n)
(let loop ((j i) (ps ps))
(cond ((> j max-index) (reverse ps))
((vector-ref v j)
(loop (+ j 1) (cons (+ j j 3) ps)))
(else (loop (+ j 1) ps)))))
((vector-ref v i)
(let loop ((j startj))
(if (<= j max-index)
(begin (vector-set! v j #f)
(loop (+ j p)))))
(loop (+ 1 i) (cons p ps)))
(else (loop (+ 1 i) ps)))))))
The part I'm having trouble with is startj. Now, I can see that p is going to be cycling through odd numbers starting at 3, defined as (+ i i 3). But I don't understand the relationship between p and startj, which is (+ (* 2 i i) (* 6 i) 3).
Edit: I understand that the idea is to skip previously sifted numbers. The puzzle definition states that when sifting a number x, sifting should start at the square of x. So, when sifting 3, start by eliminating 9, etc.
However, what I don't understand is how the author came up with that expression for startj (algebraically speaking).
From the puzzle comments:
In general, when sifting by n, sifting starts at n-squared because all the previous multiples of n have already been sieved.
The rest of the expression has to do with the cross-reference between numbers and sieve indexes. There’s a 2 in the expression because we eliminated all the even numbers before we ever started. There’s a 3 in the expression because Scheme vectors are zero-based, and the numbers 0, 1 and 2 aren’t part of the sieve. I think the 6 is actually a combination of the 2 and the 3, but it’s been a while since I looked at the code, so I’ll leave it to you to figure out.
If anyone could help me with this, that'd be great. Thanks!
