ab77ef No.7 [View All]
General RSA thread. The next thread will have a properly formatted OP.
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c64999 No.808
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408cbf No.809
>>807
no idea. I was looking for relationships between the records, and that number came out.
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1d0be2 No.810
>>796
No worries, we have a great baker! Teach, just keep whiteboarding everything. Work, brother the next bake is in good hands.
>>782
>>793
PMA rocking it.
>>791
Baker, nice pics!
>>797
(0,1)(1,1)(2,1)(3,1) etc.
(0,2)
(0,3)
(0,4)
etc.
Patterns.
KEK RULES!
E
K
R
U
L
E
S
!
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1d0be2 No.811
>>801
I'm honored, thanks! Truth and Justice/Equity.
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1d0be2 No.812
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408cbf No.813
>>802
>>793
any chance you can plot one of these to explore your theory on x?
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c64999 No.814
>>811
…And if you see Topolanon This one is for him! Glad you guys like them. When I see a good one I will make it up and post it., Keep up the good work. I have been on the edge of my seat for a while now. If you need any back-end support let me know. I will do what I can.
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1d0be2 No.815
>>814
Sorry new ID, is this Mr. E?
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8e12ae No.816
>>813
X is what I've been plotting by, instead of T. It shows you can move on a 3D linear path through cells.
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c64999 No.817
>>815
Nope NOT Mr E. Just an anonymous Hobo. IDK how to trip/namefag properly? Its limited? Tried here and post didn't go through.
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0aea66 No.818
>>797
The multiple of 12 observation seems to be a robust conclusion for a of 5 or above. Between the 1*cc and aa*bb cases, the difference in x works out algebraically to a^2-1, or (a+1)(a-1).
Since a is odd, because it's prime, both a+1 and a-1 are even, for two factors of 2. And one of a-1, a, or a+1 has to be a multiple of 3, and it can't be a because again, it's prime.
So a^2-1 has to be a multiple of 2*2*3 = 12. Neat.
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8e12ae No.819
>>548
>>554
What happens when you substitute X for T.
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c64999 No.820
Good luck tonight ! KEK be with you…
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408cbf No.821
>>818
Wow! Great explanation!
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1d0be2 No.822
>>817
Well shit, thanks for the memes.
In return, here you go.
When go got to post, go to the name field.
Enter yourname##password. Simple as that.
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1d0be2 No.824
>>821
>>818
Wow, Autist Alert!! Great Explanation. WhoTF is this? Good job, anon.
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0aea66 No.825
>>818
Actually the same logic can show that one of a+1 or a-1 has to be a multiple of 4, too. So the differences in x ought to all be divisible by 24.
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8e12ae No.826
Render of E0 with X for T.
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8e12ae No.827
While the diagonal lines are what pops out, the pattern stems from the E0,N1 column. X increases by 2 each T, and extends down N. It would probably fill in each N with a deep render.
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408cbf No.828
>>825
I believe they were. My original tests above include an xdiffby24 variable for that exact reason.
12 seemed more reasonable as an assumption.
Now, however, can you help find a formula for the actual diff?
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1d0be2 No.829
>>826
MA, thanks. These visuals are excellent.
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8e12ae No.830
>>828
From what I can tell, for any (E,N) X has an initial value, and a fixed growth value. X will start at the initial value, and += the growth value for every T.
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0aea66 No.831
>>828
a^2-1 is the actual difference in x values. Or did you mean something else?
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8e12ae No.832
Nope, I lied. Whatever X starts as is how it increments in the cell. This is true for the entire grid (please confirm) and why I initially began using X as T.
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1d0be2 No.833
Hehe. Let's solve for n. :)
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8e12ae No.834
Here's a picture for Topolanon. Tree of life!
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1d0be2 No.835
Baker! I know you got us. New bread soon?
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1d0be2 No.836
>>834
Topol is the shit. I love that faggot. Nice pic, MA!
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1d0be2 No.837
>>834
Wow, MA! Tree of life indeed. Crazy beautiful. Thx!!!
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1d0be2 No.838
>>829
>>832
Yup. T is derived from x, at least for row 1. To other row n's we have no clue yet.
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ab77ef No.840
(You) me what I should add to the batter.
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ab77ef No.841
>>840
Also send me some fancy math pics. Have been too busy to make more of the ones like in the OP, but I have a lot of them planned.
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8e12ae No.842
>>838
>>838
What I'm suggesting is that T=T*X, for all cells. There is something special about these lines.
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4f6d1b No.844
>>834
>>836
I'll… do a board thing? I guess? However that works. I just got of a … hoooours long discord chat.
Thank you for the support ^_^ I freakin' love y'all.
So, having achieved Super Rainbow Dash status… it's…
ADVENTURE TIME!!!
https://youtu.be/_50_-8r2XUA
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8e12ae No.845
>>842
Horizontal lines are actually fixed N,D,X, and A, and are the incrementing of B. The 3D lines the incrementing of A.
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04a37c No.850
>>777
chk'd!i! ..and inching closer!
>>800
lulz, multipersona anons.
>>803
Hilarious TG!
>>806
Agreed, was going to do a search earlier on topic of "algorithm for squaring massive numbers", but was in the general.
>>797
>>807
gut tells me 12 is a key to this. Why I posted that 12X spiral (w/ primes only on 4 rays, so eliminates 3/4 of options) and then the Squares with 3 entrances on each side (temple layouts). But that's just me letting intuition flow (which i trust)
>>818
gotta read that again, interesting.
>>834
Nice!!
>>815
wasn't me.
>>842
That's awesome MA!
>>459
^^ This was me. Probably doxxed myself using word 'folks' anyway.
Was trying not to overdo the name and tripfagging unless had something mathy to offer. Have had some other immediate priorities so haven't been able to focus and digest everything to level needed to actually be contributing.
Finally got resume updated and should be writing more cover letters tonight instead of hanging with you fags!
VQC, please take note, that this collective effort is a real testament to individuals showing faith: strong or unshakeable belief in something, esp without proof or evidence.
Ok, went ahead and came up with a query for a quick search, here's the second result in the list, has several algorithms including pseudo code, pdf here:
https:/ /www.hindawi.com/journals/jam/2014/107109/
Efficient Big Integer Multiplication and Squaring Algorithms for Cryptographic Applications
2014, Shahram Jahani, Azman Samsudin, and Kumbakonam Govindarajan Subramanian
School of Computer Sciences, Universiti Sains
Abstract
Public-key cryptosystems are broadly employed to provide security for digital information. Improving the efficiency of public-key cryptosystem through speeding up calculation and using fewer resources are among the main goals of cryptography research. In this paper, we introduce new symbols extracted from binary representation of integers called Big-ones. We present a modified version of the classical multiplication and squaring algorithms based on the Big-ones to improve the efficiency of big integer multiplication and squaring in number theory based cryptosystems. Compared to the adopted classical and Karatsuba multiplication algorithms for squaring, the proposed squaring algorithm is 2 to 3.7 and 7.9 to 2.5 times faster for squaring 32-bit and 8-Kbit numbers, respectively. The proposed multiplication algorithm is also 2.3 to 3.9 and 7 to 2.4 times faster for multiplying 32-bit and 8-Kbit numbers, respectively. The number theory based cryptosystems, which are operating in the range of 1-Kbit to 4-Kbit integers, are directly benefited from the proposed method since multiplication and squaring are the main operations in most of these systems.
Did some searches on Fermat primes and other stuff early on as well. You all were starting to generate some patterns earlier to what I found, would have to look back.
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ab77ef No.851
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04a37c No.853
>>851
Thank you Baker, tasty tasty and what a Be-You-tee-Full day!
Louis *BAKER*
https:/ /youtu.be/_AymCWgXQaA
You said I love you like a rainbow
I know what that means
You know I love you just the same
I could never let go of you again
You said I love you like the rain
And the sound when it hits the roof
I don't know what I would do
If I couldn't be close to you again
My skin feels like it's thickening
This cold-hearted world is sinking in
We've seen enough to know how deep we're in
But with you I feel
But with you I feel free
We've come through the darkest shades of blue
To a new and brighter point of view
Who knows where this rainbow will take us to
As long as we're together
You said you love me like no other
I know what that means
You know I love you just the same
I could never let go of you again
You said you love me like the colour
In the earth and in the sky
When we spoke of what we'd been through
Tears fell softly on you and I
My skin feels like it's thickening
This cold-hearted world is sinking in
We've seen enough to know how deep we're in
But with you I feel
But with you I feel free
We've come through the darkest shades of blue
To a new and brighter point of view
Who knows where this rainbow will take us to
As long as we're together
But with you I feel free
We've come through the darkest shades of blue
To a new and brighter point of view
Who knows where this rainbow will take us to
As long as we're together
Because with you I feel free
Because with you I feel free
Because with you I feel free
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cb5395 No.854
Problem is to solve for n where a and b are prime. in record (0,2,c^2,x,a^2,b^2) you can solve for n in terms of a and b.
x=a^2(b^2-1)
2an=x^2
n=a^2(b^2-1)^2
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cb5395 No.855
S>>854
sorry
n=a^2(b^2-1)/2
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cb5395 No.856
>>854
Sorry again
record is (0,n(c^2),c^2,x,a^2,b^2) so you can solve n for squared record in terms of square of original primes.
n(c^2)= a^2(b^2-1)/2
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cb5395 No.857
To be clearer
For record (0,n(c^2),c,x(c^2),a^2,b^2)
for primes a,b
x(c^2)^2=2a^2n(c^2)
x(c^2)=a^2(b^2-1)
n(c^2) =a^2(b^2-1)/2
using x(c^2) and n(c^2) to distinguish from x and n for record (e,n,c,x,a,b)
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cb5395 No.858
Fucked up!
n(c^2)=a^2(b^2-1)^2/2
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ab77ef No.859
>>858
When does this equation work?
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cb5395 No.860
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64cf3f No.862
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64cf3f No.863
>>850
Here is one for you Mr E Melange! The Memes must flow
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a96637 No.864
>>863
Thank you new Hobo friend! LOVE them, and perfect for our mission actually!
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0d970b No.877
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9ccc3e No.880
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