Talk:Lucas-Lehmer test: Difference between revisions

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The main loop in Lucas-Lehmer is doing (n*n) mod M where M=2^p-1, and p > 1. '''But we can do it without division'''.

<lang>We compute the remainder of a division by M. Now, intuitively, dividing by 2^p-1 is almost ~~like dividing by 2^p, except the latter is much faster since it's a shift.~~

<lang>We compute the remainder of a division by M. Now, intuitively, dividing by 2^p-1 is almost

like dividing by 2^p, except the latter is much faster since it's a shift.

Let's compute how much the two divisions differ.

We will call S = n*n. Notice that since the remainder mod M is computed again and again, the value of n must be < M at ~~the beginning of a loop, that is at most 2^p-2,~~

We will call S = n*n. Notice that since the remainder mod M is computed again and again, the value of n must be < M at

thus S = n*n <= 2^(2*p) - 4*2^p + 4 = 2^p * (2^p - 2) + 4 - 2*2^p

the beginning of a loop, that is at most 2^p-2, thus S = n*n <= 2^(2*p) - 4*2^p + 4 = 2^p * (2^p - 2) + 4 - 2*2^p

When dividing S by M, you get quotient q1 and remainder r1 with S = q1*M + r1 and 0 <= r1 < M

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The left hand side is a multiple of M = 2^p - 1.

Therefore, this multiple must be 0*M or 1*M, certainly not 2*M = 2*2^p - 2 ~~which would be > 2*2^p - 3~~, ~~and not any other higher multiple would do.~~

Therefore, this multiple must be 0*M or 1*M, certainly not 2*M = 2*2^p - 2,

which would be > 2*2^p - 3, and not any other higher multiple would do.

So we have proved that q1 - q2 = 0 or 1.

This means that division by 2^p is almost equivalent (regarding the quotient) ~~to dividing by 2^p-1: it's the same quotient, or maybe too short by 1.~~

This means that division by 2^p is almost equivalent (regarding the quotient)

to dividing by 2^p-1: it's the same quotient, or maybe too short by 1.

Now, the remainder S mod M.

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We can go a bit further: taking S >> p then q << p is simply keeping the higher bits of S.

But then we subtract these higher bits from S, so we only keep the lower bits, ~~that is we do (S & mask), and this mask is simply M ! (remember, M = 2^p - 1, a bit mask of p bits equal to "1")~~

But then we subtract these higher bits from S, so we only keep the lower bits,

that is we do (S & mask), and this mask is simply M ! (remember, M = 2^p - 1, a bit mask of p bits equal to "1")

The pseudo-code can thus be written

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if r >= M then r = r - M

And we have computed a remainder mod M without any division, only a few addition/subtraction/shift/bitwise-and, ~~which will be much faster (each has a linear time complexity).~~

And we have computed a remainder mod M without any division, only a few addition/subtraction/shift/bitwise-and,

which will be much faster (each has a linear time complexity).

How much faster ? For exponents between 1 and 2000, in Python, the job is done 2.87 times as fast~~. For exponents between 1 and 5000, it's 3.42 times as fast. And it gets better and better, since the comlexity is lower~~.

How much faster ? For exponents between 1 and 2000, in Python, the job is done 2.87 times as fast.

For exponents between 1 and 5000, it's 3.42 times as fast. And it gets better and better, since the comlexity is lower.

</lang>

[[User:Arbautjc|Arbautjc]] ([[User talk:Arbautjc|talk]]) 22:04, 15 November 2013 (UTC)