Hacker News Reader

Comment by waynecochran | Hacker News Reader

Me too. When you figure it out try to factor 641071800653367850802176606120792275422168080497001121 for me please.

This is smaller than the smallest RSA number

https://en.wikipedia.org/wiki/RSA_numbers

which was factored in 1991, so I bet you can get the right hardware and software to factor your number too. Maybe one of these implementations will do it:

https://en.wikipedia.org/wiki/General_number_field_sieve#Imp...

Sure thing!

651651626268416641641703*983764598769387649827639876407 = 641071800653367850802176606120792275422168080497001121

Wow -- I'm impressed. I just threw that out there to show how hard factoring is -- I didn't think anyone would actually do it. Ummm... Do you work for the NSA?

I don't work for the NSA—I just have a laptop, and a copy of Mathematica. The factorization took about 10 seconds.

Mathematica is really a pretty remarkable piece of software. I doubt it's world class for factorization, but it gives you easy access to a lot of pretty darn good algorithms for a wide range of problems.

There's 1 sentence on the implementation of FactorInteger in the docs, FWIW:

http://reference.wolfram.com/language/tutorial/SomeNotesOnIn...

Okay ... wow ... how big do the numbers need to get to bog down Mathematica?

The numbers that we're talking about are not big. The number you specified:

641071800653367850802176606120792275422168080497001121

OK, so take a 64 bit integer. That's the typical size of an integer in a modern programming language. A signed integer can represent 2^63-1 which is 9223372036854775807.

Imagine a "big integer" which is a composite of multiple 64-bit integers. With a big integer made up of two regular integers, you can represent the value 170141183460469231731687303715884105727. With three, you get to 3138550867693340381917894711603833208051177722232017256447 if I'm doing the math right. That's already in the ballpark of the number you mentioned - just three regular integers together.

Granted, I'm not talking about how difficult the factoring is, but it's worth noting that the size of the numbers are pretty small to a computer. Modern computers are really fast and can brute force a large number of computations. Mathematica probably has a fairly optimized general purpose factoring algorithm as well.

A number like this might be more difficult to factor:

  16158503035655503650357438344334975980222051334857
  74201606517271376232756943394544659860070576145673
  18443589804609490097470597795752454605475440761932
  24141560315438683650498045875098875194826053398028
  81919203378413839610932130987808091904716923808523
  52908229260181525214437879457705329043037761995619
  65192760957166694834171210342487393282284747428088
  01766316102903890282966551309635423015707512929643
  20885583629718018592309286787991755761508229522018
  48806616643615613562842355410104862578550863465661
  73483927129032834896752299863417649931910776258319
  47186677718010677166148023226592393024760740967779
  26805529798115327

pari-gp is freely available for linux, and it's source is available if you're interested in their implementation.

http://pari.math.u-bordeaux.fr/

The relevant file is

  src/basemath/ifactor1.c

I would say factorisation is more tedious than hard. Modern computers can perform several billion operations per second. The simplest algorithms take O(N) time, that is to factorise N, you need to perform around N calculations. There are better algorithms which run in logarithmic time, so you only need log(N) calculations (and some which are better than that). Even with what you and I think of as big numbers, this is fast.

You get problems with numbers over 100 digits. I recently entered a challenge where one of the questions involved prime factoring a 130 digit number. That was a record feat a couple of decades ago, now it takes a day or two on a modern computer.

Anyone care to elaborate on the downvote? Given that the parent seemed surprised that it was possible to factorise a 'big' number quickly on a modern PC, I thought I'd elaborate.

Brute force search is O(N), the sieve of Eratosthenes is O(N log(log(N))). The Quadratic sieve is good up to 100 digits and runs in O(exp(sqrt(log n log log n))). Some utilities you can use for this are YAFU, MSIEVE or GGNFS.

Reply on news.ycombinator.com