Talk:Lucas-Lehmer test: Difference between revisions
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==
Welcome to 1979!
:Yeah. I know. :-) --[[User:Short Circuit|Short Circuit]] 07:55, 21 February 2008 (MST)
==List of known Mersenne primes==
The table below lists all known Mersenne primes:
{| class="wikitable"
Line 321 ⟶ 322:
| Great Internet Mersenne Prime Search|GIMPS / Curtis Cooper (mathematician)|Curtis Cooper & Steven Boone [http://www.mersenne.org/32582657.htm]
|}
:The 45th and 46th Mersenne primes have been discovered, is the intention to keep this table up to date? --[[User:Lupus|Lupus]] 17:24, 2 December 2008 (UTC)
:: The above list seems to be copied ''in toto'' from the updated Wikipedia site:
<big> https://en.wikipedia.org/wiki/Mersenne_prime#List_of_known_Mersenne_primes </big>
:: The above Wikipedia site now has a list of '''51''' Mersenne primes. -- [[User:Gerard Schildberger|Gerard Schildberger]] ([[User talk:Gerard Schildberger|talk]]) 19:57, 7 May 2019 (UTC)
== Java precision ==
The Java version is still limited by types. Integer.parseInt(args[0]) limits p to 2147483647. Also the fact that isMersennePrime takes an int limits it there too. For full arbitrary precision every int needs to be a BigInteger or BigDecimal and a square root method will need to be created for them. The limitation is OK I think (I don't think we'll be getting up to 2<sup>2147483647</sup> - 1 anytime soon), but the claim "any arbitrary prime" is false because of the use of ints. --[[User:Mwn3d|Mwn3d]] 07:45, 21 February 2008 (MST)
== Speeding things up ==
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.
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, 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
When dividing S by M+1, you get likewise S = q2*(M+1) + r2 and 0 <= r2 <= M
In the latter, we divide by a larger number, so the quotient must be less, or maybe equal, that is, q2 <= q1.
Subtract the two equalities, giving
0 = (q2 - q1)*M + q2 + r2 - r1
(q1 - q2)*M = q2 + r2 - r1
Since S = 2^p * (2^p - 2) + 4 - 2*2^p
<= 2^p * (2^p - 2),
then the quotient q2 is less than 2^p - 2 (remember, when computing q2, we divide by M+1 = 2^p).
Now, 0 <= q2 <= 2^p - 2
0 <= r2 <= 2^p - 1
0 <= r1 <= 2^p - 2
Thus the right hand side is >= 0, and <= 2*2^p - 3.
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.
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.
Now, the remainder S mod M.
We start with a quotient q = S div 2^p, or simply q = S >> p (right shift).
The remainder is S - q*M = S - q*(2^p - 1) = S - q*2^p + q, and the multiplication by 2^p is a left shift.
And this remainder may be a bit too large, if our quotient is a bit too small (by one): in this case we must subtract M.
So, in pseudo-code, we are done if we do:
S = n*n
q = S >> p
r = S - (q << p) + q
if r >= M then r = r - M
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")
The pseudo-code can thus be written
S = n*n
r = (S & M) + (S >> p)
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).
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)
May 2015 - fixed small bugs in both Python implementations. In the first, execution failed (Python 3) without a cast to int in the test. In the second, there was a typo - an 'r' should have been 's'.
: Timing for some solutions for 2..11213:
{| class="wikitable"
|-
! Time (s)
! Solution
|-
| 871.2
| Python without optimizations
|-
| 314.7
| Python with optimizations
|-
| 124.7
| Perl Math::GMP without optimizations
|-
| 106.9
| Pari/GP 2.8.0
|-
| 61.8
| Perl Math::GMP with optimization
|-
| 33.0
| Python using gmpy2 (skipping non-primes)
|-
| 14.2
| C/GMP with even more optimizations
|-
| 13.3
| Perl Math::Prime::Util::GMP (source of C/GMP code)
|}
[[User:Danaj|Danaj]] ([[User talk:Danaj|talk]]) 19:49, 9 May 2015 (UTC)
Rust solution for 2..11213 -> 8.6 s
Pari/GP solution for 2..11213 -> 2.835 s --[[User:Nebulus|Nebulus]] ([[User talk:Nebulus|talk]]) 05:21, 22 February 2023 (UTC)
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