Wheel factorization: Efficient?

[science_man_88]

you all like starting arguments that never end and Pi I asked for your suggestions 2 of your posts back have a list ready yet ?

[3.14159]

Originally Posted by axn

What I mean is, the expected number of candidates to search for a prime is dependent only on the average candidate size and the sieve depth. For a factor-rich base, you would naturally start with less candidates (before sieving) as compared to a factor-deficient base.

Less candidates? Why is it that factor-rich bases can barely eliminate 9 in 10 in the sieving process, while prime or factor-deficient bases can easily eliminate 14 in 15 or 19 in 20?

Originally Posted by axn

The numbers such as "1 in 7 candidates left" are irrelevant -- what is relevant is the aposteriori probability, after sieving, of a candidate yielding prime.

Irrelevant? They're irrelevant? Having fewer candidates = Better odds of finding a prime among the *remaining* candidates, as the obvious composites have all been eliminated.

[science_man_88]

Originally Posted by 3.14159

Less candidates? Why is it that factor-rich bases can barely eliminate 9 in 10 in the sieving process, while prime or factor-deficient bases can easily eliminate 14 in 15 or 19 in 20?



Irrelevant? They're irrelevant? Having fewer candidates = Better odds of finding a prime among the *remaining* candidates, as the obvious composites have all been eliminated.

yet you claim my idea is crap and it has some of that in it as well who falls now.

[CRGreathouse]

Originally Posted by 3.14159

Why is it that factor-rich bases can barely eliminate 9 in 10 in the sieving process, while prime or factor-deficient bases can easily eliminate 14 in 15 or 19 in 20?

You should be able to answer this. (Also, the difference between the two should be larger, unless you're thinking of a less-rich version of "factor-rich" than I'm thinking of.)

Originally Posted by 3.14159

Having fewer candidates = Better odds of finding a prime among the *remaining* candidates, as the obvious composites have all been eliminated.

False.

[3.14159]

Originally Posted by CRGreathouse

False.

If you were assuming I was talking about changing the amount of primes found: Strawman. Never said anything about the amount of primes to be found. Point invalidated.

Else, disregard the above.

[3.14159]

Originally Posted by CRGreathouse

I wasn't.

.. How would it be equally difficult to find a prime when there are less candidates to test? (Please, present us with your brilliant explanation.)

[kar_bon]

Originally Posted by 3.14159

.. How would it be equally difficult to find a prime when there are less candidates to test? (Please, present us with your brilliant explanation.)

Because there're less primes, too!

Read about Nash weight.

[3.14159]

Originally Posted by kar_bon

Because there're less primes, too!

In a fixed k-range, there are less primes after sieving? So, tell me, how does sieving accidentally kick out primes? X 2

[CRGreathouse]

Originally Posted by 3.14159

In a fixed k-range, there are less primes after sieving? So, tell me, how does sieving accidentally kick out primes?

For someone who throws out strawmen arguments so often, I'd think you would be more careful about saying that sort of thing.

[3.14159]

Originally Posted by CRGreathouse

For someone who throws out strawmen arguments so often, I'd think you would be more careful about saying that sort of thing.

Strawmen? What strawmen did I make? I distorted nothing! He stated that in plain text!

[kar_bon]

Originally Posted by 3.14159

In a fixed k-range, there are less primes after sieving? So, tell me, how does sieving accidentally kick out primes? X 2

Less primes to find among the remaining candidates!

Example: k*2^n-1

n-range: 600000-1000000
sieved to p=26*10^12
k=337: 2618 candidates left
k=315: 32276 candidates left

Primes in that range found:
k=337: none
k=315: 3

BTW: k=337 got no prime for 172000<n<2600000!