Posting this again since you forgot again
A picture will probably look better than unformatted text, but until then at least this means you don't have to wait for 8 751-post threads to load.
Repost
>"The solution is a decision tree. The first decision is whether remainder is zero. The second decision is whether the remainder is odd or even. What's the third decision?"
When did he say that? Link?
>What's the third decision?
At the top of the decision tree, if e = 0, it's a square, so it's not an RSA number (right?). If e != 0, we need to find n or x, which apparently relies on whether it's odd or even. In the second decision, whether it's even or odd determines what t is, but then to calculate t you need to know x, and if you knew x you'd know a. So do we know any other calculations that require knowing if e is even or odd? Otherwise, should we be using this to find relationships?
>you increase your D by 4
Are you sure that 4 is constant or do you think maybe the reason it's not working yet for big numbers is that the offset (maybe this is what Chris was talking about) changes depending on the cell? How did you get 4? I noticed last thread when I was analyzing c^2 in the 0 column (nobody else seemed to care about what I was doing so I don't know if anyone will know what I'm talking about) that when c is the square of a semiprime (so the square of what we want to factor), the infinite set starting from (a, b) = (1, c^2) has a beginning d value and it increases at a constantly increasing rate throughout the infinite set (i.e. for (0, 200), d = 21, 44, 69, 96, 125 - this is +23, +25, +27, +29), but the start values and increase values change depending on the cell. The value of x in these sets increases just with the one increase value for each set (same example, x = 20, 40, 60, 80 etc) and the start value and increase changes from cell to cell too. I don't know how the beginning value or the increase values are determined, but if we're using a decision tree instead of iterating through an infinite set's values, perhaps the next decision in the decision tree relates to finding the beginning value and increase values or either d or x. Of course I say this without knowing how you found those original values. If I'm right about this, I have a feeling it has something to do with the relationship between (a, b) = (1, c) and (a, b) = (1, c^2) and something to do with the relationship between t and x. Maybe it would be worth graphing a few. That's a little bit of an educated shot in the dark though.
>(1)
Looks like you have some secret admirers, baker. I'm actually personally surprised at how many of us are still here. I thought more of us would have given up. As unrelatedly fitting as Q's message a couple hours ago about nobody being asleep tonight is considering how much progress we seem to suddenly be making, I'm off to bed.
Where did you notice it, so I can look into it?
Whoever that was did it when we were blindly looking for patterns by changing the colours and criteria for bitmap generation. It isn't necessarily going to help exactly the way it looks, but obviously Chris did reply to it saying that it was "where the fun really begins". Isn't there some formula for calculating the golden ratio exactly instead of just rounding to 1.618?
>>3434
Oh, you were the one who was emailing me in December. I wasn't sure which one of you it was.
I made the pdf but I'm not the one making the image version.
As someone who has always thought Flat Earth was bullshit but could see there was some weird shit going on (like planes not being allowed to fly over the poles, the heavy protection around the poles and Joe Biden etc going there after the US election) I think Agartha makes perfect sense. Since something like this would be provable with video etc when the time comes, and Q said the 40,000ft view would be too much for 99% of people who would just outright deny it if it was told to them, imagine how crazy the 40,000ft view really is and where this would even fit into it.
I'm the one who posted >>3413 and I think it would pay to mention that I looked at the places CA was talking about in (e, 1) and they seem to always follow the pattern he mentioned (for whatever unknown reason). That said, I actually don't completely understand what CA is doing, so what I said is possibly still important, just in a different context than calculating an infinite set of ds and xs given (e, 1). We just don't know what that context is yet, and I don't really have the time to do anything other than lurk for the rest of the day. If we're using a decision tree rather than iteration then it potentially becomes important when we link (e, 1) to (e, n). Currently, from what I understand (and I might be wrong), CA's method iterates through (e, 1) to find the correct x for (e, n).
Some of Chris' clues directly relate to (e, 1), and since we're working on that at the moment, I'll point a few out:
>At the correct element in the grid at (e, 1) where the value of a at that element equals the na we want, if you subtract 2d+1 from na then you are in the negative half of the grid in terms of e and the value at the same element in the first row will be (n-1)a
So does that mean either the next or one of the next parts of the decision tree relates to na?
>You're going to use column e=1 to show whether the conjecture about Fermat primes is true during this
Has anyone looked for the Fermat numbers in (e, 1) yet?
>A prime appears once in a column. This is SIMPLE to calculate. This calculation is important. I call n for a prime number big_n or N
>The product of two primes appears twice. Once the same as the prime, which is easy to calculate and the second time, this is the n that we are after.
We already know most of this with (a, b) = (1, prime) etc, but this sounds like it's part of the decision tree based on how sequentially Chris describes it.
>How many times does the product of three distinct primes appear?
I looked up the phrase "three distinct primes" and I found this: https://en.wikipedia.org/wiki/Sphenic_number
He wouldn't have mentioned it if it wasn't important, so does anyone know anything about Spehnic numbers? I haven't seen anyone discussing them. He said this right after the above two lines about the product of two primes, so maybe if we can find one prime and we can find the product of three primes we can use those methods to find the product of two primes.
>The relationship between cells 2n to the left or right in (e, 1)
Has this come up in relation to the decision tree yet?
>Extend the cells at (e, 1) into negative x
Has anyone done that yet? He's said a bunch of times that listing everything we know (every rule, every positive and negative value of every integer, etc) would make the answer obvious.
>REMEMBER, take d from all values of d[t] at (e,1) and there is a known patter of (n-1) as factor in these values of d[t]-d that is different (increasingly) from the pattern of factors of n in a[t]. It is THIS that gives the offset that is used to solve the problem and thus get the cell at (e,1) to do all the work for you.
This seems like the most important of all things to focus on based on where we are right now.
Also this might be a stupid thing to consider but has anyone tried contacting Motti Milgrom and asking if he remembers emailing with a guy called Chris Curtis 10 years ago?
It would surprise me if he was using his real name to give people on 8ch the knowledge necessary to crack RSA. I tried looking in various places on the internet for someone called Chris Curtis from Auckland, New Zealand out of curiosity at one point and while I didn't put much effort in, I didn't find anything.
CA found a way to calculate any d and x in (e, 1) cells >>3393 although it doesn't work as it's meant to because to find the correct x value at the moment we're iterating.
There's a newer version of the map >>3453
Chris hasn't been here and said anything in a while. Someone messaged him on Twitter, though, and he pointed out some things about the decision tree. The first question in the decision tree is whether e = 0. The second is whether e is odd or even (which relates to CA's thing). We're looking for the third decision at the moment. There are a lot of things you'll see in the map that seem like they relate to decisions on a decision tree, possibly some of the ones I pointed out >>3466 here, but we don't know what the third decision is yet. Other than that, there have been lots of discoveries that didn't go anywhere. I might have missed something but that's what happened last thread and this thread so far off the top of my head.
Sounds like another decision for the decision tree. It also cuts the time in half if we were to iterate. Nice work. If n is even and d is even, that means (a+b) % 4 == 0, right? Because it means (a+b)/2 is an even number too. That means if n is even and d is odd that (a+b) % 4 != 0. Vice versa for odd n values. Chris said something about mod 4, right? I don't remember what it was so I don't remember if it applies to this situation.
If we know the parity of d and n then we know the parity of (a+b) too, as explained >>3512 here. a = d - x. If d and x are both odd or both even, a is even. If d is either odd or even and x is the opposite, a is odd. If a is even and (a+b) is odd, b is odd. If a is even and (a+b) is even, b is even. If a is odd and (a+b) is odd, b is even. If a is odd and (a+b) is even, b is odd. If (a+b) is odd then c is even (oddeven or evenodd). If (a+b) is even then c could be odd or even (oddodd=odd or eveneven=even). So if we know the parity of x, we know the parity of every variable. I'm too tired to remember if we know how to find the parity of x yet.
Okay I don't know if we actually do have the parity of d and n yet just rereading my post from earlier. As I said I'm a bit sleep-deprived. But if we do, all the logic in this post still checks out.
I might put a program together that takes in a public key and outputs the equivalent private key. I'll have a gap where the decision tree code will eventually go. That way, once we're there, we don't have to figure out how keys work in code at the last minute (and so far it seems quite complicated). Nobody has any objections to Java still, yes?
>(am I missing anyone)
hi
Oh I already know that stuff. I was the one who wrote those guides for the lurkers who didn't know anything several threads ago and then stopped because I didn't understand the grid well enough at the time myself. I'm just talking about how you're meant to import keys into Java and get the modulus and exponent out of them. If you've already got that working I won't need to, but I think I can figure it out if you haven't.
I can't wait for the crazy shit Chris said we would do when we're ready to turn this into a company or whatever he called it, but, I've got to be honest, causing mass world hysteria sounds quite entertaining too, even if we're also affected ourselves. I bet CNN etc will try to act like they have any idea what they're talking about and act like they're super smart while saying absolutely nothing of substance (like the "hacker known as 4chan" thing), and all those YouTubers will try to make sense of it while the ones who use their accounts to teach people math will fumble around with it for a while.
Actually, now that I think about it, people might actually kill themselves over this when crypto goes under. A few did when it tanked recently. Fuck. I'm definitely still on board because this is inevitable and if anyone should do it it should be people like us who have good intentions, but I point this out because it should hopefully discourage anyone from taking credit for it publicly.
By the way, >>3577
Man, you keep getting trips all the time. Were you working on a Java version or did you want someone else to?
Yeah, I don't know if you've thought about it much, but consider >>3645 this. Once it's all over the media and everything people might be able to get away with it. I've been trying to think of ways to use the grid to create music, too.
>Wouldn't call myself a musician, I just mess around with synths tbh
There are people who consider themselves musicians who don't even know how to mess with synths, so I hope you don't say that out of pessimism. I'm a little more invested in it than that. I have a thing recorded and mixed professionally, and one of the song names is a subtle reference to Q. It's not out yet, but hopefully it's subtle enough that all 5 people who will ever hear it don't think I'm an evil internet hacker nazi who masturbates to anime.
The grid is meant to have many uses, so it might be used to connect the logic of factoring numbers with elliptic curve cryptography. Also, one of the early crumbs said something like "you factor a number with two input variables d and e". The whole time this was meant to revolve around d and e.
So I guess that means there's an easier way than the way we have right now? Maybe it reduces the number of leaves to involve n or x in the math somehow.
It's not the end of the world if we don't figure this out right away. Chris did say he would pop in before the end of the week, remember.
So far it is O(log n) in the worst case but that's based on the number itself and not the bits in c. Chris said we had 95% of the information we need, didn't he? If he did then I really have no idea how we get to log of the number bits in c if he has somehow told us that. Hopefully we're up the point at which he can tell us that and we'll understand (it seems like it).
There's no other context that suggests that they're talking about our Chris.
Oh okay, I learned a new meaning of the word identity then.
I'm not aware of anyone having thought about this yet but in this image a point is plotted whenever c is a square. So the line in the middle is (0, n) and every other point has a relationship in terms of e and n with (0, n).
Another interesting thing: this is the same idea (looking for relationships between the e=0 column and other cells) but with x as the x axis instead of e and n still the y axis. Is this maybe the start of an elliptic curve before it diverges? The second image is to make the shape more obvious.
I don't see you giving any useful insights, (2). Starting arguments is only going to cause the conversation to go off-track even more. Not all of us have any useful ideas right this second. A couple slightly irrelevant posts isn't stopping anyone from working and collaborating.
>DOW
Did you see it was at 666 points the other day?
The gradient of each of those lines increases by 1 each time (rise/run = 2/2, 4/2, 6/2, 8/2, etc). What's an X chain pattern?
>d of f solves for x in row 1
Here's the bitmap with e and f either side of the zero line on the x axis and d on the y axis.
I hope this is actually useful and not just pretty pattern shit again, but I did e&f by x both for square cs and not based on your post and found something interesting. In the non-square one, you can kind of see a diagonal grid in the f space within the grid we're already looking at.
It's a little irritating that we can't just edit our posts on this website. I can but that's only because I control the BO account. I get what you mean about the x chain thing. Do you think there's any way to mathematically predict these points? Just looking at this image, there are obvious patterns in the valid/invalid n cells (the first one being two valid, two invalid over and over, the second being one invalid, two invalid etc) but they seem to get a lot more complicated the further down you go. If we could find a way to predict the pattern of a particular line, I don't know that it would help with the tree but we'd be able to find something.
>How do we use it? Any ideas?
I'm not sure that the thing I posted was entirely useful just since all of the variables used in the image come from c (if c's a square, as well as d, e and f). It shows a linear relationship between d and f, but that doesn't give us n or x.
>AA
Yeah, I'm still here every day even if I don't always have ideas. This is an important board so someone has to keep it from being overrun by shills.
Have we considered using f in the tree somehow? That might be a stupid idea but I'm just thinking out loud. Obviously it revolves around input variables d and e, like Chris said, but if the tree either terminates with x or x+n, maybe the thing that determines whether it's one or the other is another variable that comes from input variables d and e (i.e. f).
If when this is implemented we just plug numbers in and get numbers out, how are we going to know whether the number it spits out is x or x+n?
I've been meaning to keep going with that thread explaining the occult but nobody seemed all that interested and my life has been quite hectic. Plus, a lot of you keep using Christian themes in the things you say, and the Bible says occultism is inherently wrong (you're all entitled to your opinions but as someone who used to be Christian I think that's complete horseshit), so I don't want to step on anyone's toes for no good reason. If you read what I wrote about astral projection in that other thread and you are interested I guess I could keep going.
Noticing the part of this image at the top where it seems like there's a solid pattern - do we have a maximum number at which the tree doesn't work anymore? Maybe that section of the image shows where it works and lower is where it doesn't work. This is a blind guess, though, since I don't know what value of d it is when that pattern stops in this image.
That was the original point of the board: to more eloquently organize everything, have a list of thread archives, etc. Then /cbts/ turned to shit. We've just sort of been winging the structure of threads and everything so far. Of course you're more than welcome to put something like that together, and I could sticky it if everyone would like. I don't really have the time to go through everything myself at the moment. My life is quite busy and stressful lately (not that anyone needs my life story but one of the things is that I'm moving out of my parents' house quite soon) so while I can easily lurk and maybe make some bitmap images occasionally, someone else is going to have to get all of that information together. I did put some pdfs together early on for the people who knew nothing about programming or how the grid worked. There's a thread somewhere. Maybe that information would come in handy too.
Do keep in mind, the reason we left /cbts/ is because a human started doing dumb human stuff that interfered with the function and culture of the board, and I'm also a human. Maybe I don't think I'm going to start doing anything shitty, but I can't imagine anyone who controls this kind of thing ever does, and sometimes they do. You all have the responsibility of keeping me in check.
I didn't think to put the Twitter DMs that some of us have posted screenshots of into that pdf, but the images in the archived webpages don't work. It would be a good idea to get those messages together (maybe with their context in time too if possible), if everyone who has been messaging Chris on Twitter is reading this and could post them again.
That makes me think; how are we going to explain this to the world when it's done? Should we try to make some kind of long and concise text-based explanation, or should we turn it into a video with graphics and someone brave's voice, or what? I kind of just went with the text-based thing when I was trying to help the lurkers who can't code etc however long ago that was but it would be better to discuss and organize it rather than just having someone like me decide that they should do it all.
It's going to be really funny when mainstream news anchors try to explain how it happened. Every so often when a big tech thing happens I look for the reaction from TV news places just to watch them bumble around not knowing what the fuck they're talking about, e.g. when they talk about darknet drug markets, or the hacker known as 4chan.
>people like me who have no earthly idea what you're talking about
This has been my thought since the start. The whole point of this disclosure is to make people aware that we have a fundamental misunderstanding of how mathematics works and to hopefully change the world for the better. When it is ready, it'll most likely seem like something only scholarly PhDs can grasp to everyone who hasn't been following along like us, at least at a surface glance. If we could explain everything from the ground up as simply as possible, it would greatly help in making this common knowledge instead of just being something that gets studied at high levels in universities behind literal and figurative closed doors. It's just a question of whether it would be enough to type it all out or if we'd need to put videos together (and if we put videos together instead of typing it all that implies that there wouldn't be a lot of reading involved, which implies that one of us has to record their voice, potentially de-anonymizing themselves), or if we could have a big infographic (it would have to be huge and there would likely be a fuck ton of text anyway). Any ideas?
Don't mind me, just posting 6 more times so the thread's full and I can sticky the next one.
Another post
Different words so I don't trigger flood detection
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