20030908, 02:15  #1 
Aug 2003
Snicker, AL
3BF_{16} Posts 
Finding the square root of a large mersenne number
I am trying to figure out how long it would take to find the square root of say a 10 million digit number. I am currently testing exponent 33,349,067. What would be the fastest code method of determining the square root? I think I can show this in math pseudo better than I can explain it.
(a 10 million digit number) q = (2^333490671) (the square root of q) srq = sqrt (q) (the remainder after subtracting the square from the number) rmd = q  (srq^2) My interest is based on the current method of trial factoring. If I have the math down correctly, the number of potential factors of a Mersenne number is severely limited. Would a squares based algorithm potentially use fewer steps to find factors than the current method shown on http://www.mersenne.org/math.htm ? Thanks, Fusion 
20030908, 02:56  #2 
Aug 2002
3·5^{2}·7 Posts 
A good first approximation to sqrt(2^333490671) would be 1.414*2^166724533

20030908, 02:59  #3 
Aug 2002
525_{10} Posts 
A good first approximation to sqrt(2^333490671) would be 1.414*2^16674533
Why can't we edit posts in this forum? 
20030908, 04:49  #4 
Aug 2003
Upstate NY, USA
2×163 Posts 
they've disabled editting for regular users to try to deal with speed issues from what i understand

20030908, 06:25  #5 
"Mike"
Aug 2002
2×4,019 Posts 
Yes, please check the message preview before you commit! :)

20030908, 07:15  #6  
"Richard B. Woods"
Aug 2002
Wisconsin USA
2^{2}×3×641 Posts 
Re: Finding the square root of a large mersenne number
Quote:


20030908, 12:53  #7 
Aug 2003
Snicker, AL
1677_{8} Posts 
Cheesehead,
Yes, but no. Fermat understood something important about finding the factorization of a number. What he did was organize a number into a matrix then analyze the matrix for its prime constituents. In its current implementation it would be useless on mersenne numbers. I think a mersenne specific optimization could be coded. Thats the reason I am trying to figure out the time required to find the square root of a very large number. What would qualify as a mersenne specific optimization? To start with it would only permit factors of the form 2kp+1. Fusion 
20030908, 16:18  #8 
Aug 2003
Upstate NY, USA
146_{16} Posts 
factors can only be congruent to 1 or 7 mod 8

20030908, 16:23  #9  
Bamboozled!
"πΊππ·π·π"
May 2003
Down not across
24624_{8} Posts 
Quote:
Try implementing it, by all means, but test it on small Mp first. You will find it useless for p as small as 500, where we can already factor Mersenne numbers by other algorithms. The run time get worse, much worse, as p increases. Paul 

20030909, 00:14  #10 
Feb 2003
2·3·29 Posts 
Fusion_power 
Here's a fairly straightforward square root algorithm I wrote for the java.math.BigInteger class as it doesn't have one already... I'm sure there are faster algorithms, but this one is quadratic, which isn't too bad. [code:1] public BigInteger sqrt(BigInteger n) { int currBit = n.bitLength() / 2; BigInteger remainder = n; BigInteger currSquared = BigInteger.ZERO.setBit(2*currBit); BigInteger temp = BigInteger.ZERO; BigInteger toReturn = BigInteger.ZERO; BigInteger potentialIncrement; do { temp = toReturn.setBit(currBit); potentialIncrement = currSquared.add(toReturn.shiftLeft(currBit+1)); int cmp = potentialIncrement.compareTo(remainder); if (cmp < 0) { toReturn = temp; remainder = remainder.subtract(potentialIncrement); } else if (cmp == 0) { return temp; } currBit; currSquared = currSquared.shiftRight(2); } while (currBit >= 0); return toReturn; } [/code:1] Some timings: [code:1] 2^33349  1 sqrt takes 1109 ms 2^333490  1 sqrt takes 161891 ms [/code:1] on my Athlon XP, which extrapolates to 67 weeks for 2^33349067  1, so I would recommend finding a better sqrt algorithm. 
20030909, 04:28  #11 
6809 > 6502
"""""""""""""""""""
Aug 2003
101Γ103 Posts
22414_{8} Posts 
Could I suggest the old fashioned method?
Get that program that will expand 2^333490671 to its decimal equivalent. Then write (I think that I could do it, but don't quote me) and run a simple program that takes the square root the by hand method. The only critical part should be knowing if it has an even or odd number of digits. Si o No 
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