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Wrote a little additional code and apparently all the failures were strictly products that had 3, 5, 7, or 11 as their lowest factor
Extended it to the first 2000 primes (first 2 million product of two unique non-trivial factors).
As predicted, only products of 3, 5, 7, and 11 failed the test. All others past.
The implication is the factors that fail to expand into larger products with bigger factor trees, never extend beyond 3, 5, 7, and 11. And consequently all other products would adhere to this solution.
In the case of brute forcing prime factorization, this defacto cuts the search space in half, or at least your *outer* loop is never going to be greater than your lowest factor divided by 2.
Also I haven't really explored the implications of the larger factor tree when you happen to come across the correct i value. Nor if there are earlier points in the search could terminate before beginning factorization.
It's remains entirely impractical at the moment, but it was still a fun little curiosity.
It's too early for this horseshit lmao
Wouldn't it be interesting if you could find some underlying purpose for prediction based on this
Nihil7562880dI know people from the way-back think this is what you do with a computer,
But honestly mate, if it ain't got/serves an API and/or stores selfies - it's useless :P
@Frederick the size of the products tend to grow extremely fast too.
Like too big for modern techniques like ECM to factor quickly.
I'm hoping though that some overlapping properties between the original products and the derived products, might allow for some interesting approaches.
@Wisecrack settled issues out west
Got my knob polished by 5 different women about 2 years ago
Getting lots of exercise
Considering taking a trip somewhere sooner rather than later to accomplish another task
Overall sick of traveling pointessly and seeing the same content in person and online
Don't eat leafy greens lol
This also sounds like something I did heh
Are you self taught?
Like from the very beginning of professional level or from after school years?
I think I have literally witnessed you grow.
Asking for readings.
Bringing them up in conversation.
And bit by bit making more and ever more sense.
What are you studying?
What's your learning path?
I think I'm quite impressed, anyway.
@scor thanks. Rough childhood so I didnt really start learning till a few years ago. Wont waste your time with the whole life story.
Self taught. Just now, very slowly, picking up linear algebra. Would love to explore math and algorithms full time one day but I dont have the means or credentials for university.
Learning path is anyone's guess with the state of the world the way it is.
Gonna just keep grinding and rain- manning by night.
I want to eventually explore statistics. Did you know our ideas of mean, mode, and average have geometric equivalents? And I was thinking those measures of central tendency are all derived from *euclidean* geometry, but what about measures derived from non-standard geometry? Is there any applications there? Any utility or interesting connections to other types of math, or new interpretations of old problems?
In other news you need to have your beyond compare license returned to you or you'll crack the fucking pro edition !
@AvatarOfKaine they were paranoid iran would some how reverse engineer the chip the emotion engine ran on, and put it into nukes or some crazy-stupid shit like that.
Almost as absurd as making people take their shoes off at the airport because of one lunatic mad at the cost of those little liquor bottles on flights.
Well, yes, I've been taught and flabbergasted about such cross functionalities in maths.
Especially in higher education, despite staying in the financial mathematics sphere.
Gosh I wish I could have spent more time with all that.
Then, I lived in Germany that time, where education follows highly typified patterns. I guess one would need a well typified culture to be allowed a properly structured learning environment, if not self imposed.
"the financial mathematics sphere."
Very cool. Was it tedious or exciting?
Did you get to solve interesting problems?
"I guess one would need a well typified culture to be allowed a properly structured learning environment, if not self imposed."
Thats good you had that opportunity.
What I do is I just keep going until I bump up against the ceiling of
what I know. Then I ask "what do I need to know to do this next thing?"
And then I go and learn it. Or just enough to get things done.
What kind of systems have you worked on? (if thats not asking too much)
Made a few additional modifications to what I already posted today and the results have factors that are always either a-smooth or, if not, b-smooth.
The two solutions are :
In addition to being smooth, all the prime factors in the new product are trivial.
I have to write a factorization test to confirm this, but its what I'm seeing.
Reducing from 9**i to 4**i also helped to reduce the precision requirements.
I don't have access to the resources to confirm it holds for the products of more than the first two thousand primes though.
@atheist, @hitko, @Fast-Nop any input on relevance to things like TWIRL and TWINKLE, esp. b-smooth factors? You guys are far and away more familiar with math than I am.
@Fast-Nop lol, the smiley face gives it away.
"once burned, twice shy" I suppose.
But the results appear real for a change.
I was hoping for someone that had a better understanding of the general number field sieze in relation to smooth numbers and why exactly the latter is relevant to factorization.
@Wisecrack As for the latter, my guess is that if you want to factorise a large number N and can determine that there is no prime factor greater than B, then you have already excluded the search space between B and sqrt(N). That would help if the two prime factors are several orders of magnitude away from each other.
If you repeat that process and lower B until you fail, then you have a search range where a prime factor could probably be hiding.
@Fast-Nop "As for the latter, my guess is that if you want to factorise a large number N and can determine that there is no prime factor greater than B, then you have already excluded the search space between B and sqrt(N). "
So, if I'm understanding you correctly, I would be looking for a variation of the formula where [a] was the *smallest* factor in the new larger factor tree?
Keep in mind too, 9^i his fails for any products of 3, 5, 7, 11.
And 4^i fails for any products of 3 to 19.
Question: is what you're showing here, that the approach fails when theres a sufficient difference between b and a? Laptops currently updating so I cant check your results at the moment.
hitko258978d@Wisecrack Basically if B is smaller than about 3^A (see previous comment for the exact value), you get a variation of modular exponentiation for prime exponent A, which should hold as long as A is a prime number greater than 3.
If B is instead larger than about 3^A, the absolute value in your formula somehow messes the whole thing up and the equation no longer holds.
Wisecrack665278d@hitko I just figured out what you did (really only because you told me). Thank you hitko for excising the abs from the code.
Noticed the original exponent (i) has been removed.
What was your thinking on that? And in its original form, do you see any method of lowering the requirement to find a/2?
Say to a/4, or a/8 or some n other than 2?
Wisecrack665278d@hitko also I'm only passingly familiar with modular exponentiation (because, I'm embarrassed to admit, I just read the wikipedia article.)
If modular expotentiation is easy, but finding the modulus is hard, how does this relate to the original product and the product produced by this--how does it relate to these two products sharing (a) as a common factor?
hitko258977d@Wisecrack What I did only works because 9^n is the same as 3^(2n), therefore 9^n can be replaced by 3^((2n + 1) - 1), and we know that n=floor(a/2) is equivalent to a=2n+1 when (a) is a prime number. Nothing as simple can be done for floor(a/4) or floor(a/8), and nothing similar works if (a) can be an even number.
The thing with modulo is that (v^x)%c is the same as ((((v%c)*v)%c)*v)%c ... Now, in your case (v) equals 3, which is a prime number, and (c) is also a prime number. Let's assume c=7:
1%7 = 1
(1*3)%7 = 3
(3*3)%7 = 2
(2*3)%7 = 6
(6*3)%7 = 4
(4*3)%7 = 5
(5*3)%7 = 1
(1*3)%7 = 3
As you can see, the result is 1 exactly when x=k*(c-1), for any non-negative integer k. If I plug this into your original expression, I get 2*(1) - 2 = 0.
If either c or v aren't prime, you'll get to 1 in fever than c-1 steps, and the result of (v^(c-1))%c probably won't be 1.
Wisecrack665275d"This only works because a = 2n+1"
What if we used something other than a/2?
For example what if we defined our primes as say 3n+1?
What would be the implications of that? How would the math change?
Sorry, had not found the time to properly respond to your comment here.
Though you really shall have a look into this prime factor functions and mathematical distinguishing of number areals :
I personally assume it's more discreet in German. So here you go