20131005, 12:50  #1 
Apr 2012
2·47 Posts 
New flame war idea ??
Hello
Some time ago I was trying some new ideas to quickly factorize numbers. Such as rewriting known factorization relations to work directly with the binary coefficients of a number instead of the whole number, and applying statistical heuristics. I was trying to implement a functional method many years ago but it got too complicated for me and finally I found some other authors trying some of those ideas and the results were not so good. Today I want to post something different I found, maybe not useful but at least nice graphics. I haven't developed it further. If you plot Mod[N, k] for all k you get a nice graph like shown at the first picture. (the remainder of the division). In this example N=272123, but you get similar graphs for other numbers. It seems that there is a pattern. 
20131005, 12:51  #2 
Apr 2012
94_{10} Posts 
I you plot the difference of two consecutive remainders you get something much clearer:
I don't know why I can't load several pictures at once. This is Mathematica notation, and graphics. Last fiddled with by skan on 20131005 at 12:58 
20131005, 12:53  #3 
Apr 2012
2×47 Posts 
And if you represent ListPlot[Differences[Mod[N, k], 2] is much clearer, there is a simple pattern.
You need to zoom in to notice that the graph is right, the value alternates between the curves. We could plot an approximate curve from a few random divisors and then save time by only trying numbers near this curves? or watching the areas where the results diverges... Is this already implicit in any other method? Is this just stupid? I have more strange graphs The two vertical lines represent the values of the known divisors of 272123. You can notice a gap there, this gap could also be useful. Regards Last fiddled with by skan on 20131005 at 12:56 
20131005, 13:52  #4  
Nov 2003
2^{2}×5×373 Posts 
Quote:
"binary coefficients of a number": Numbers do not have coefficients. "statistical heuristics": meaningless gibberish. If you want to discuss mathematics you have to learn to use standard terminology and standard definitions. Quote:
What is a "functional method"? Quote:
The first thing you need to do is learn some math. You don't know any. Take a course in number theory. Take one in abstract algebra. Learn the difference (topologically speaking) between R and Q. You do not seem to understand what a function is. You do not seem to understand what a curve is. Quote:
topology. Tell us why you think this graph is relevant to finding k such that N mod k = 0. Do you imagine that there is some relationship between N mod k and N mod (k+1)? Tell us what you think it is. What "patterns" do you think are there? Tell us. Describe them algebraically. Stop babbling. 

20131005, 15:40  #5  
Apr 2012
2·47 Posts 
Quote:
First of all English is not my native language, it's difficult to use the proper words. When I say "binary coefficients of a number" I mean the coefficients of the number when you express it in base 2. I know they aren't "curves" in the strict sense of the word, that's why I said that you could "approximate" or "fit" a curve to that points. Anyway, I was sharing my ideas, maybe they are not good but that doesn't imply you are a rude with me. I don't what relations would arise from here that's why I share the graphics, maybe somebody else find them interesting. Last fiddled with by skan on 20131005 at 15:41 

20131005, 17:48  #6 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
3^{2}×7×149 Posts 
Let's move this discussion to Misc.Math, and by definition, it will be immunized* from further "real math" criticism.
By skan's own admission, this is not factoring, just some "graphics which maybe somebody else find interesting". 
20131006, 04:59  #7  
"Richard B. Woods"
Aug 2002
Wisconsin USA
2^{2}×3×641 Posts 
Quote:
This isn't your classroom. Stop treating this forum as though it were. Your response was inappropriate, rude and uncouth. You have an enormous "blind spot" about other people and about proper verbal behavior. Quote:
Quote:
Quote:
But you are blind to that which is obvious to others. Last fiddled with by cheesehead on 20131006 at 05:05 

20131006, 05:26  #8  
"Richard B. Woods"
Aug 2002
Wisconsin USA
2^{2}×3×641 Posts 
Quote:
Since 272123 = 503 * 541, Mod [272123,k] should equal k for k in the range [0,502]. The graph should be a straight line with a slope of 1 for k between 0 and 502. My conclusion is that your first graph is not simply Mod[N, k], but something else. I think I could guess what it actually is, given enough time, but not yet. What is the ycoordinate for k=100, 200, 300, 400, 500, 600, 700, 800, 900 and 1000? Last fiddled with by cheesehead on 20131006 at 05:30 

20131006, 05:48  #9  
"Richard B. Woods"
Aug 2002
Wisconsin USA
2^{2}·3·641 Posts 
Quote:
I think all other visual patterns can be traced back to knowing the factors of N ... but the Mod[] calculations to produce the graph already have revealed those. In other words: in order to produce these graphs, one has to do enough computation to already have revealed the factors of N before the graph is completed. The patterns shown in the graphs reveal nothing beyond what will already have been known by doing those computations. Quote:
Quote:
I recall looking, when I was young, at some graphs similar to yours and not yet knowing why the patterns were predictable. (In other words: been there, done that. And I actually do have a mathematics tshirt, but it's not about graphs.) Asking about new patterns you notice is just part of learning. But once in a while, after you've learned a lot, you may notice a pattern no one else has actually ever noticed or explained, and it will be your own new discovery when that happens. Quote:
Last fiddled with by cheesehead on 20131006 at 06:20 

20131006, 06:27  #10 
May 2013
East. Always East.
6BF_{16} Posts 
+1 Cheesehead. I won't be saying very much on the matter as I recently withdrew from a different fight with someone who similarly rhymes with glasshole and have no interest in starting another one.
To the OP, in particular, don't let this one bother you too much. People here are nice. Mister (Doctor, yes?) Silverman is the exception, not the rule. 
20131006, 11:31  #11 
Romulan Interpreter
Jun 2011
Thailand
10010010010100_{2} Posts 
We are now in the misc math subforum. I really appreciate post #8. However I believe that both #7 and #10 posts were TOTALLY uncalled for. Please stop! (it will end with bans, and I don't want to lose a chess player from my team, again! )
On the other hand, I know that Mr. Silverman was ill, I am happy that he is well again, and biting... Last fiddled with by LaurV on 20131006 at 11:34 
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