In mathematics, the sieve of Pritchard is an algorithm for finding all prime numbers up to a specified bound. Like the ancient sieve of Eratosthenes, it has a simple conceptual basis in number theory. It is especially suited to quick hand computation for small bounds. Whereas the sieve of Eratosthenes marks off each non-prime for each of its prime factors, the sieve of Pritchard avoids considering almost all non-prime numbers by building progressively larger wheels, which represent the pattern of numbers not divisible by any of the primes processed thus far. It thereby achieves a better asymptotic complexity, and was the first sieve with a running time sublinear in the specified bound. Its asymptotic running-time has not been improved on, and it deletes fewer composites than any other known sieve. It was created in 1979 by Paul Pritchard. Since Pritchard has created a number of other sieve algorithms for finding prime numbers, the sieve of Pritchard is sometimes singled out by being called the wheel sieve (by Pritchard himself) or the dynamic wheel sieve. == Overview == A prime number is a natural number that has no natural number divisors other than the number 1 and itself. To find all the prime numbers less than or equal to a given integer N, a sieve algorithm examines a set of candidates in the range 2, 3, …, N, and eliminates those that are not prime, leaving the primes at the end. The sieve of Eratosthenes examines all of the range, first removing all multiples of the first prime 2, then of the next prime 3, and so on. The sieve of Pritchard instead examines a subset of the range consisting of numbers that occur on successive wheels, which represent the pattern of numbers left after each successive prime is processed by the sieve of Eratosthenes. For i > 0, the ith wheel Wi represents this pattern. It is the set of numbers between 1 and the product Pi = p1 · p2 ⋯ pi of the first i prime numbers that are not divisible by any of these prime numbers (and is said to have an associated length Pi). This is because adding Pi to a number does not change whether it is divisible by one of the first i prime numbers, since the remainder on division by any one of these primes is unchanged. So W1 = {1} with length P1 = 2 represents the pattern of odd numbers; W2 = {1,5} with length P2 = 6 represents the pattern of numbers not divisible by 2 or 3; etc. Wheels are so-called because Wi can be usefully visualized as a circle of circumference Pi with its members marked at their corresponding distances from an origin. Then rolling the wheel along the number line marks points corresponding to successive numbers not divisible by one of the first i prime numbers. The animation shows W2 being rolled up to 30. It is useful to define Wi → n for n > 0 to be the result of rolling Wi up to n. Then the animation generates W2 → 30 = {1,5,7,11,13,17,19,23,25,29}. Note that up to 52 − 1 = 24, this consists only of 1 and the primes between 5 and 25. The sieve of Pritchard is derived from the observation that this holds generally: for all i > 0, the values in Wi → (p2i+1 − 1) are 1 and the primes between pi+1 and p2i+1. It even holds for i = 0, where the wheel has length 1 and contains just 1 (representing all the natural numbers). So the sieve of Pritchard starts with the trivial wheel W0 and builds successive wheels until the square of the wheel's first member after 1 is at least N. Wheels grow very quickly, but only their values up to N are needed and generated. It remains to find a method for generating the next wheel. Note in the animation that W3 = {1,5,7,11,13,17,19,23,25,29} − {5 · 1 , 5 · 5} can be obtained by rolling W2 up to 30 and then removing 5 times each member of W2.This also holds generally: for all i ≥ 0, Wi+1 = (Wi → Pi+1) − {pi+1 · w | w ∈ Wi}. Rolling Wi past Pi just adds values to Wi, so the current wheel is first extended by getting each successive member starting with w = 1, adding Pi to it, and inserting the result in the set. Then the multiples of pi+1 are deleted. Care must be taken to avoid a number being deleted that itself needs to be multiplied by pi+1. The sieve of Pritchard as originally presented does so by first skipping past successive members until finding the maximum one needed, and then doing the deletions in reverse order by working back through the set. This is the method used in the first animation above. A simpler approach is just to gather the multiples of pi+1 in a list, and then delete them. Another approach is given by Gries and Misra. If the main loop terminates with a wheel whose length is less than N, it is extended up to N to generate the remaining primes. The algorithm, for finding all primes up to N, is therefore as follows: Start with a set W = {1} and length = 1 representing wheel 0, and prime p = 2. As long as p2 ≤ N, do the following: if length < N, then extend W by repeatedly getting successive members w of W starting with 1 and inserting length + w into W as long as it does not exceed p · length or N; increase length to the minimum of p · length and N. repeatedly delete p times each member of W by first finding the largest ≤ length and then working backwards. note the prime p, then set p to the next member of W after 1 (or 3 if p was 2). if length < N, then extend W to N by repeatedly getting successive members w of W starting with 1 and inserting length + w into W as long as it does not exceed N; On termination, the rest of the primes up to N are the members of W after 1. === Example === To find all the prime numbers less than or equal to 150, proceed as follows. Start with wheel 0 with length 1, representing all natural numbers 1, 2, 3...: 1 The first number after 1 for wheel 0 (when rolled) is 2; note it as a prime. Now form wheel 1 with length 2 × 1 = 2 by first extending wheel 0 up to 2 and then deleting 2 times each number in wheel 0, to get: 1 2 The first number after 1 for wheel 1 (when rolled) is 3; note it as a prime. Now form wheel 2 with length 3 × 2 = 6 by first extending wheel 1 up to 6 and then deleting 3 times each number in wheel 1, to get 1 2 3 5 The first number after 1 for wheel 2 is 5; note it as a prime. Now form wheel 3 with length 5 × 6 = 30 by first extending wheel 2 up to 30 and then deleting 5 times each number in wheel 2 (in reverse order), to get 1 2 3 5 7 11 13 17 19 23 25 29 The first number after 1 for wheel 3 is 7; note it as a prime. Now wheel 4 has length 7 × 30 = 210, so we only extend wheel 3 up to our limit 150. (No further extending will be done now that the limit has been reached.) We then delete 7 times each number in wheel 3 until we exceed our limit 150, to get the elements in wheel 4 up to 150: 1 2 3 5 7 11 13 17 19 23 25 29 31 37 41 43 47 49 53 59 61 67 71 73 77 79 83 89 91 97 101 103 107 109 113 119 121 127 131 133 137 139 143 149 The first number after 1 for this partial wheel 4 is 11; note it as a prime. Since we have finished with rolling, we delete 11 times each number in the partial wheel 4 until we exceed our limit 150, to get the elements in wheel 5 up to 150: 1 2 3 5 7 11 13 17 19 23 25 29 31 37 41 43 47 49 53 59 61 67 71 73 77 79 83 89 91 97 101 103 107 109 113 119 121 127 131 133 137 139 143 149 The first number after 1 for this partial wheel 5 is 13. Since 13 squared is at least our limit 150, we stop. The remaining numbers (other than 1) are the rest of the primes up to our limit 150. Just 8 composite numbers are removed, once each. The rest of the numbers considered (other than 1) are prime. In comparison, the natural version of Eratosthenes sieve (stopping at the same point) removes composite numbers 184 times. == Pseudocode == The sieve of Pritchard can be expressed in pseudocode, as follows: algorithm Sieve of Pritchard is input: an integer N >= 2. output: the set of prime numbers in {1,2,...,N}. let W and Pr be sets of integer values, and all other variables integer values. k, W, length, p, Pr := 1, {1}, 2, 3, {2}; {invariant: p = pk+1 and W = Wk ∩ {\displaystyle \cap } {1,2,...,N} and length = minimum of Pk,N and Pr = the primes up to pk} while p2 <= N do if (length < N) then Extend W,length to minimum of plength,N; Delete multiples of p from W; Insert p into Pr; k, p := k+1, next(W, 1) if (length < N) then Extend W,length to N; return Pr ∪ {\displaystyle \cup } W - {1}; where next(W, w) is the next value in the ordered set W after w. procedure Extend W,length to n is {in: W = Wk and length = Pk and n > length} {out: W = Wk → {\displaystyle \rightarrow } n and length = n} integer w, x; w, x := 1, length+1; while x <= n do Insert x into W; w := next(W,w); x := length + w; length := n; procedure Delete multiples of p from W,length is integer w; w := p; while pw <= length do w := next(W,w); while w > 1 do w := prev(W,w); Remove pw from W; where prev(W, w) is the previous value in the ordered set W before w. The algorithm can be initialized with W0 instead of W1 at the minor complication of making next(W, 1) a special case when k = 0. This a
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