fd3a9d No.10193 [View All]
The virtual quantum computer (VQC) is a grid made of constructably infinite elements that follow a known pattern. It is indexed using e, n, and t, where e is the column, n is the row, and t is the specific element in the cell.
The grid in its entirety serves as the superposition. The required input parameters to collapse the superposition are d and e, which are trivial to calculate for all c that is the difference of two squares. When the integers that are the difference of two squares are arranged into the grid and their corresponding properties are shown, a pattern emerges that shows a path to calculate the factors of c instead of searching for them. This pattern can be understood using only the basic operations of arithmetic and sqrt.
Glossary
Look-up
A pattern used to calculate the factors of c, like a value look-up table.
Column
All cells for a given e
Row
All cells for a given n
Cell
All entries for a given e,n (not to be confused with an entry itself.)
Entry; record; element
A set of variables corresponding to a factorization for a given c. The legend to read entries is {e:n:d:x:a:b} (e, n, t) = c
Example: {1:5:12:7:5:29} (1, 5, 4) = 145
ab record; nontrivial factorization
The element that contains the factorization of c that is not 1*c, hence, nontrivial.
1c record; trivial factorization
The element generated from setting a=1 and b=c
Mirror element
The element in -f corresponding to an element in e, in the context of a given c.
Sieve
A sieve is an algorithm for factoring integers arising out of number theory in the 1900s, most notably from Carl Pomerance.
Smoothness
A number is described as smooth if it is composed of small prime factors, opposed to large ones.
Sqrt Tree
The sqrt tree is a structure created by recursively taking d and e starting with c, creating a tree with several to many branches. (Changed from remainder to sqrt because there is more than one tree).
Triangle Tree
The triangle tree is a similar structure but uses the T-1 function instead of sqrt.
Functions
na transform
A movement from a record in (e, n) into (e,1) where n becomes 1 and a becomes a times the n of the (e,n) record. It has also been used to refer to moving n*a records down in a cell.
Pell(n)
The nth Pell number function. Can be calculated recursively or formulaically using the square root of 2.
ST(n)
The nth square triangular number. A square triangular number is a number that is both a square and a triangle number. The Pell function is used to calculate square triangular numbers.
T(n)
The triangle number function.
Example: T(7) = the 7th triangle number
T-1(n), inverse T
The inverse triangle number function.
Example: T(7th triangle number) = 7
Variables
a and b are, to reiterate, the factors of c. a is the smaller factor of c, and b is the larger one.
d is the integer square root of c.
e is the remainder of taking the integer square root of c. Unless c is a perfect square, a remainder will be left over.
i is the root of the large square. It is equal to (d+n).
j is the root of the small square. it is equal to (x+n).
n is what you add to d to be exactly halfway between a and b, and it is the root of the large square, so it takes you from d to the large square.
x is what you add to a to make d. When added to n it makes the root of the small square.
f is what you add to c to make a square. (e is what you subtract from c to make the square below it, f adds to make the square above c.)
t is the third coordinate in the VQC, it is a function of x.
q is a product created by multiplying successive primes until the product is above d.
u is the triangle base of (x+n)^2. 8 times the triangle number of u plus one is (x+n)^2 for c with odd x+n.
h was a variable used to quantify families of numbers. The way to calculate it is currently unknown.
When capitalized versions of the variables are used in comparison to lowercase versions, the capitalized versions refer to the variable's value for the trivial record, and the lowercase variables refer to the values for the nontrivial record. Sometimes these trivial uppercase variables are referred to with "Big" preceding the letter.
{e:N:d:X:A:B} (e, N, T) is the trivial element.
{e:n:d:x:a:b} (e, n, t) in this context is the nontrivial element, the prime factorization of c.
Since under this rule D and E would refer to the same variable as d and e and thus be redundant, D and E have been defined as references to elements.
D and D2 refer to the two elements whose d[t] values d from c is between in row one.
Since N and n are references to specific values, R is used to refer to an element pair whose a[t] values are the estimates of n provided by D and D2.
E refers to any element whose values are a starting estimate of the factorization of c (ie i0, j0, n0, x0, etc).
____________________________
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2695a8 No.10909
Furthermore, if log_10 instead of log_e is used to calculate the amount of operations on c, the (singular) possible bit length of a for RSA100 is 165. If log_2 is used, the possible bit values are [288, 247, 206, 165, 124, 85, 42]. Thus more calculations are needed to establish a bit-accurate estimation.
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5be6c4 No.10910
>>10907
<Are you going to post another algorithm that fits these criteria then?
-pic related-
ffs.
>>10908
MOND you say…
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000000 No.10911
> Jan is a faggot who LARPed
> Jan leaves the board like the bitch he is (After being constantly called out)
> Suddenly, out of the blue M appears (After 2.5+ years)
> M is like Jan, except M is not a faggot
> M continues off where Jan left
Are you guys shitting me? Jan was and is a shill. Embrace, Extend and Extinguish. Embrace the grid (The grid is real, but VQC is a faggot << JAN/M ON DISCORD), Extend the grid (Let's throw more shitty math at it << THIS IS WHERE JAN / M IS), Extinguish (If you don't understand XYZ then you might as well leave << Where he will be in a few weeks).
How is Jan extinguishing the board? Simple, he is throwing oooooold math at the problem. Shit you guys can't wrap your head around because you didn't spend 10 years memorizing idiotic symbols and force yourself to remember the 5 axioms of Euclid.
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5be6c4 No.10912
>>10911
That's nice.
Unless you can give some explicit proof that "M" is Jan, then plz no run off the newcomers.
Don't be niggers.
No glowing.
Make Terry Proud Again!
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1fb3e1 No.10913
>>10911
I know who wrote this lol.
>>10912
Follows a pattern. Although M is definitely a better avatar or real person. Who knows.
>>10876
M - The prime sin of Jan was this: insulting other people's honest ideas, work, and exploration and making anons feel like shit for trying to help. He was belittling people's work and claiming to have the solution. It was super fucked up. I avoided the board for months.
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cef271 No.10914
>>10911
If you have any issues understanding summations or their rules (it is always a very fun story recalling the idea of Gauss deriving the first rule of summations in front of his schoolteacher), I'm more than willing to help or explain the process. I can explain them using C (I'm old school). There is no such dichotomy of productive or unproductive mathematics to use for investigating numbers. Mathematics, like programming is a tool and its symbols represent processes; it is the programmer or mathematician's job to write insightful theorems, or, should he fail (I hope nobody fails here), it is not the fault of the tool.
>>10912
I appreciate your sympathy.
>>10913
It's not a good way to spend one's time or mood trying to decipher who is who on a platform like this. I would just judge each post by the maths therein, the logic involved and whether it is sound, and whether that poster is polite. Mathematics doesn't lie; ideas can be dissected decisively. As for a character, it's better not to. That is why I am not concerning myself with such things.
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1fb3e1 No.10915
>>10914
>It's not a good way to spend one's time or mood trying to decipher who is who on a platform like this
Actually, it's vitally important, M. That's why we have names and trust here.
Personally, I love your elegant approach.
Keep working!
Math(s) don't lie.
Find the solution with your equations, and you will be honored.
We're simply hardened by shills and many dead ends.
Your formulas don't impress us, straight up.
Neither does the scholarly attitude.
It comes off as talking down to us lads who have been here since the beginning of CBTS.
That being said, you have clearly studied the threads more closely than any other newcomer since the inception of this board. So, you're either legit, or the French incarnation of Jan lol.
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5be6c4 No.10916
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5be6c4 No.10917
>>10915
Also, it could just be Chris fucking around.
As usual. ;)
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51896f No.10918
>>10915
I am not here to impress nor condescend. If something gives you that impression, it is not me. Why study the threads and become involved? I may have misjudged the age of those here, but it really seemed like a place that attracts the youth and the youthful to learn mathematics with no prerequisites. "Breaking RSA is a consequence, not an aim." If VQC became a place where more people joined in the future to study maths, it would be a wonderful thing. There are a lot of people in university who do mathematics coursework, but there is a dwindling amount of people who love maths. I have a been a part of many maths forums online, and there is something special here that isn't found there (plus all of the forums that I was fond of have shut down). So if VQC becomes a place where people love maths and more people come to learn to love maths, I want to be part of it.
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5be6c4 No.10919
>>10918
I'm chill with you bein' here.
Might wanna thicken the ol' skinz if you plan on sticking around though.
The others can get all sorts up in arms at the drop of a hat.
But they mean well.
WELCOME ABOARD!
Whether you're Jan/Jean, another one of Chris's alter personas, or actually someone new…
I'm just here for the maths. And by that I mean I'm the muse of the group, so let's get back to the maths and away from the draaaaaama drama drama.
And theeeeeeeeeen?
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ccfa01 No.10920
I'm going to analyze a hint I was thinking about. I can't recall which post number it was at, but it was something along the lines of "[using the grid solution] a number can be identified as a square faster than O(log n)." Over the course of VQC, I have seen two unique square root algorithms (if these don't link properly I will repost):
>>4306
>>8679
The first method (the original method posted in /cbts/) starts with an initial "high" guess of d = c / 2 and an initial "low" of 0. At the start of the loop, d becomes c / 4, and then is squared. If (c / 4)^2 is less than c, then it becomes the "low," otherwise it becomes the "high." In the subsequent iterations, a midpoint (low + high) / 2 is calculated and then squared, with the low and high being based off of which is too high and too low. This repeats until low and high are within 1 of each other, which means the value of the square root has been found. Finally the high value is squared and then the right value is returned based on checking both. This method is known as the bisection or binary search method. It is slower than the new method as it only gains one bit of accuracy on each iteration (corresponding to one binary digit each time).
The second method converges to d faster and corresponds to Newton's method. Although derived in calculus, Newton's method works well over the integers, it just terminates before becoming a floating point number and the remainder is calculated. Under Newton's method, the problem of square roots is written as:
f(x) = x^2 - c where c is c from the grid, and x is an independent variable. Without going into derivatives, it will suffice that we use two simple rules to explain Newton's method:
The derivative of x^2 = 2x; the derivative of a constant is zero. Thus the derivative of f (f') =
f'(x) = 2x
Newton's method is:
x1 = x0 - ( f(x) / f'(x) )
Evaluating this with the formulas,
x1 = x0 - ( (x0^2 - c) / (2x0) )
Which when multiplying both sides by 2x0, distributing the negative sign and then dividing by 2x0 again becomes the simpler form:
x1 = (x0 + (c / x0)) / 2
If we set the initial x0 to the bitlength of c multiplied by 2, then we arrive at the same code presented in the second method.
How can this be applied to factoring, if at all? As it turns out, factorization can be expressed using the same type of function but with two input variables:
f(x, y) = (x + y)(x - y) - c = 0
However a multivariate root finding function is complex (literally), so I won't go into that unless it could be efficient.
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5db6fa No.10921
>>10911
Funny how every time someone suggests M might be Jan he avoids giving a yes or a no (also funny how Jan always did that with everything else)
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6ae8c9 No.10922
>>10921
That which is asserted without proof can be can also be dismissed without evidence.
I think if we all jump up and accuse every single person who comes to help us out as being Jan we're going to find ourselves all alone pretty soon. I'm usually just observing but this kind of behavior is pretty common with a group of people on the decline.
Just a warning from an oldfag.
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5be6c4 No.10924
YouTube embed. Click thumbnail to play.
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3a1d72 No.10925
>>10922
Completely agree.
>>10911
You're fucking retarded please shut the fuck up about Jan for god sakes holy shit I'm just trying to follow along. Post math instead of drama or kill yourself and shut the fuck up
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5be6c4 No.10927
>>10907
Totally not related to anything, but did you ever figure out if Syntax Highlighting could be toggled?
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491c2c No.10928
>>10922
Hello Oldfag.
>That which is asserted without proof can be can also be dismissed without evidence.
>Just a warning from an oldfag.
Very True. Thanks Oldfag.
>>10925
Also very true. We're just tired of being fucked with.
>>10920
According to VQC, we're looking for a new way to view number families. Old ways can hint towards that, but we need new ideas. Newton won't help us on this mission.
If you can't tell yet anons, I like to stir the pot when I'm bored. Please don't take it personally. It's like DJT tweeting insults, it makes the day more interesting. And makes me laugh when people get all pissed off on Discord. At least we're not bored, right?
So who has ideas? Shall we work?
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752b86 No.10929
>>10927
That's a question for Ron. I don't see any new settings.
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5be6c4 No.10930
>>10929
kek, that's why I was asking.
I'd put in the request before 8kun was operational and was told he'd look into it and that it maybe might habben. Ah well.
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752b86 No.10931
I'm about to make thread #18. Should I redo the text? I feel like it could be stripped down and made a bit more coherent. For example the relevance of Pell numbers and triangle numbers and so on is never mentioned in the block of text, and even among those functions there isn't any mention of the functions themselves, just that they exist. Most of the code probably doesn't get used anymore, and only a handful of patterns are mentioned too. Also for example with this line:
>D and D2 refer to the two elements whose d[t] values d from c is between in row one.
That was one of Jan's ideas that was never brought up anywhere else but in his workings (I hadn't even noticed it there before). Even if I was to edit the OP text it would all still exist in these previous threads. What do people think about me redoing the OP text?
>>10930
I remember you mentioning it. There are probably only one or two other boards on this website that even use code anyway (I've only really been on this board and occasionally /qr/ to find Chris for the last few years so I don't know anymore), so I doubt it's a priority.
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5be6c4 No.10932
>>10931
Make it the bestest bread text ever!
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1aab70 No.10933
>>10922
Not the words I would use, but this is precisely why I didn't spend any more time acknowledging it. Claiming I am here to stop you all from progressing by giving you more mathematical tools is logically vacuous.
>>10921
I'm not Jan, are we finished?
>>10931
I don't think it should be removed just because he made it. I haven't analyzed the hint myself but I know other people have used the elements d is between in n=1.
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1aab70 No.10934
>>10920 (cont'd)
We have two equations which describe the unknowns under two problems, roots of square and roots of two numbers multiplied together (factorization):
f(x) = x^2 - c
f(x, y) = (x+y)(x-y) - c
This problem has been represented as finding roots from two independent variables before, in Get_Remainder_2dnm1(base, d, n , f) >>4594. Since d and f are constant here, it is equivalent to the two unknowns of f(x, y) form.
These representations may seem simple, but they allow you to learn something about the problem that is not as obvious from viewing it as many solution variables. Firstly (I have not seen this described here yet), it is that the problem of finding the base of a square is 2-dimensional, but the problem of finding the bases of a rectangle is 3-dimensional. In the first equation, f(x) is y, but in the second equation, f(x) is z with x and y inputted. While not a complete proof, it is evident that if you try to make factorization a univariate f(x), you will get an unknown root instead of 0, as in
f(x) = (x+n)^2 = (d+n)^2 - c
hence the problem is shown to still be of two independent variables.
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67b224 No.10935
Pseudo-code. Demonstration part 1. Even d, odd e. Branching.
1. Convert to binary as a string
2. Trim Right zeros (divide by 2 until odd)
3. Find square root d and remainder e
4. GCD(Larger(d,e),Smaller(d,e))
5. Branch on e_bin[e_bin.len], case 1 (where e is odd) and d_bin[d_bin.len], case 2
6. Case last digit of c 9, then dpn must 0 (9+1), 4(9+5), not six as no square ends in seven, meaning xpn ends in either 1 or 5
7. Create matrix of dp1,nm1,fm1
8. Factorise dp1 and fm1
Why is it obvious that for RSA_100 that d+1, n-1, f-1, x and (x+n)-1 are divisible by 5?
Why is it obvious that for RSA_100 that (x+n)(x+n) - 1 is divisible by 40 (8T + 1)?
Why is it obvious that for RAS_100 that (d+n) mod 100 = 0?
Why is this so hard? Is it due to the requirement to work with numbers at a certain scale to see patterns emerge?
Are you working in a domain that will not show it?
Should you be working with RSA numbers that have been determined?
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67b224 No.10936
Timeline alignment.
Work with RSA100.
The problem has always been masked at low integers.
Otherwise it would have been solved.
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67b224 No.10937
You are closing in on the solution to the formerly intractable multi variate problem.
Numbers are not a line.
They are families.
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67b224 No.10938
Good Bless President Donald Trump and the United States of America.
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8686cf No.10939
Device switch.
When the time comes, take the victory.
You have EARNED it.
Deus Ex Machina.
The ghost(God) in the machine helped, but the work, the hard work was yours.
The ghost in the machine is not me, I just pass it on.
This is only the Beginning.
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8686cf No.10940
Q Timestamps have authentication.
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8587c8 No.10941
>>10935
Hello! Looks interesting, I'm writing a C implementation of this now. Here's my CMakeList.txt for including the C arbitrary precision libraries (if my previous code doesn't compile remove the & in front of a_prod_t, I was accidentally calling it from C++).
cmake_minimum_required(VERSION 3.16)
project(Mathematics C)
set(CMAKE_C_STANDARD 99)
add_executable(Mathematics main.c)
include_directories(
"/usr/lib/x86_64-linux-gnu/"
"/usr/include/")
target_link_libraries(
Mathematics
"/usr/lib/x86_64-linux-gnu/libgmp.so"
"/usr/lib/x86_64-linux-gnu/libgmpxx.so"
"/usr/lib/x86_64-linux-gnu/libmpfr.so")
The folders are the default install locations for GMP and MPFR, but may vary. The header files need to be in the include directories and the shared libraries in the linker.
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5be6c4 No.10942
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dd5c5b No.10943
>>10935
Why no trip?
>>10933
>I don't think it should be removed just because he made it. I haven't analyzed the hint myself but I know other people have used the elements d is between in n=1.
Having just that one and two or three others seems a bit weird when there are a ton of them, doesn't it? I was going to get rid of all of them and just point to the grid patterns thread (that pattern is in there so it's not like I'm just doing this because of Jan).
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5be6c4 No.10944
>>10943
So that no one accuses him of larping as Jan? XD
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dd5c5b No.10945
>>10944
You would think if he wanted to prove the validity of his trip after burning all the others that he would keep using it (or at least the name VQC).
>>10935
Thinking out loud.
>Even d, odd e
That means x is odd, f is even, and c/a/b are odd. n's parity could be either or but could be determined with BigN.
>GCD(Larger(d,e),Smaller(d,e))
That gcd is either going to be one of the factors or 1, so that would terminate the loop.
>Branch on e_bin[e_bin.len], case 1 (where e is odd) and d_bin[d_bin.len], case 2
I don't even know if these two concepts are linked, so this is just an idea, but if we're using branches based on d and e, assuming the size of these numbers has something to do with the search space, this would get rid of at least half like he said >>10407 here. I wouldn't have a clue how these branches link together with (e,1) and (f,1) though. t values maybe?
>Factorise dp1 and fm1
If we're factoring decreasingly-sized branch numbers then we still need a way to link the factors of the branch to the factors of its parent. That would explain all the times you've said that this is recursive though. I think this might be related to the triangle method since that's based on nn+2d(n-1)+f-1, you're saying we're meant to factor f-1, and then later in this post you mention RSA_100's 8T+1 being divisible by 40 (which you mentioned when you introduced the triangle method). That's the only thing I can think of that would make any of this relevant, so I could be wrong.
>Case last digit of c 9, then dpn must 0 (9+1), 4(9+5), not six as no square ends in seven, meaning xpn ends in either 1 or 5
>dpn must 0
Equal zero? End in zero? This poster might not have a trip but Chris always seemed to accidentally not include an important word in his explanations here or there so this probably is him, lol. Just using a smaller number to try to understand this wording, c3139 (43*73) has an even d and an odd e, and its d+n is 58, so I really don't know what this is supposed to mean.
>not six as no square ends in seven
So this has to do with a square minus one maybe?
>meaning xpn ends in either 1 or 5
c3139's x+n is 15, so that's a start.
At first read this doesn't really elucidate anything for me personally. I know the stuff about specifically using RSA_100 will require a bit more effort than this post of course, but there are a lot of different concepts to consider links between regardless.
Chris, if you come back any time soon, would you mind telling us whether the pattern in these pictures is relevant to anything?
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5be6c4 No.10946
>>10945
Yeah, but based on your (and others') past reactions, he probably has a better chance of getting his point across if he doesn't do either.
If Chris needs to/feels like it, he can always namefag.
Or stand out as only he can.
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5b013e No.10947
>>10945
Supposing there is a common factor or group of factors (multiplied together as one) between n-1, f-1, and d+1, does that tell you a way to find the index of n-1 any faster than searching? As an example, if they all had 3, could you go 3 elements up in t at once instead of one up at once?
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dd5c5b No.10948
>>10947
That could certainly be worth looking into. I haven't done anything with the RSA 100 stuff yet but it would be odd if whatever pattern he's referring to in that number that causes all of those numbers to have a common factor greater than one only worked in this one case, not only because that's a lot of numbers to share a common factor but also because then why would he bring it up?
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7967c0 No.10949
Here is the output of my implementation in C:
RSA100 = 1522605027922533360535618378132637429718068114961380688657908494580122963258952897654000350692006139
RSA100 is 330 bits long (1111 ns)
RSA100 = 39020571855401265512289573339484371018905006900194^2 + 61218444075812733697456051513875809617598014768503 (27 ms)
[d+1, n-1, f-1] = [39020571855401265512289573339484371018905006900195, 14387588531011964456730684619177102985211280935, 16822699634989797327123095165092932420211999031883]
The attached image is a matrix of the prime factors of [d + 1, n - 1 and f - 1] respectively (the final factor of f-1 is 146775299345680308269945210247021217361). If desired I can post my C code or images with this matrix for other known RSA numbers.
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f3753f No.10950
>>10949
Looks like you factored f-3.
f-1 should be 16822699634989797327123095165092932420211999031885
5 x 19 x 4488130363987 x 11079829100092583 x 3561013012572873623
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45aeae No.10951
>>10950
>>7388
Thanks for the correction; looks like you can move up or down by in/decrements of the common factor (5 in this case).
>>10723 was also insightful, I'm wondering if this can be rewritten to separate all an and bn's in (e, 1) and (-f, 1), which would give you a way to see which rows are defined without branching to a factoring algorithm.
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5be6c4 No.10953
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effedd No.10954
>>10935
>6. Case last digit of c 9, then dpn must 0 (9+1), 4(9+5), not six as no square ends in seven, meaning xpn ends in either 1 or 5
I looked into this since it's stylistically hard to decipher. n's parity determines what numbers (d+n)(d+n) and (x+n)(x+n) are going to end with was what he should have said. The parity of e and d don't have an effect (not pictured is odd d even e but that was the same as even d odd e).
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effedd No.10955
>>10954
I fucked up the picture, that should say odd e even d.
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dd5c5b No.10956
>>10935
I made a full tree with the branching he mentioned here and all possible [d+1,n-1,|f|+1] matrices and their factorizations. No idea what any of this is meant to accomplish.
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dd5c5b No.10957
>>10956
Here are the binary a[t]s for the same branching, just as an idea.
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effedd No.10958
I've been thinking about the "lookup x", if anyone remembers that. At the time nobody seemed to have any idea what it meant to have an x value in (e,1) or (f,1) that would lead to the solution that wouldn't just be the x we were already looking for (thereby rendering it meaningless to apply it to (e,1) and (f,1) since we'd already have an unknown that would give us the solution). It's a lookup x that's meant to either give us na or n-1. I'm thinking maybe this refers to the binary chunk thing. It's the only thing that's come up since he introduced the concept of a lookup x that seems to give it any meaning (a lookup x that gives us the element where na or n-1 is a chunk at the end of a[t] in binary). I've run some tests looking for na or n-1 in binary at the end of a[t] in elements other than the six we were using before and I haven't run into any that it doesn't work for yet, but the ones it works for where it isn't one of the six known elements, I haven't seen anything yet that would give a path to this x value. Here are two examples (I had to use numbers in the thousands or these tests would have taken hours (RSA_100 would have taken at most ten octovigintillion years with this algorithm, so maybe not that one) - also there are other examples that found n-1 but these both happen to be na):
(123,222,521) = {123:222:3482:1041:2441:4967}, f=-6842, c=12124447, u=631, i=3704, j=1263
The x values for the six known elements are (e,1) 12120965, 3481, 12127929 and (f,1) 12107038, 3480, 12114003
Found na at (f,1) x=11659973 === (146580344905683,1,5823023) = {146580344905683:1:141105366169899:11646045:141105354523854:141105377815946}, f=-135630387434116, c=19910724361941423765278575884, u=5823022, i=141105366169900, j=11646046
12107038 - 11659973 = 447065, so the solution element is 223533 elements below the (f,1) a[t]=c*(BigN-1) element
(3012,276,472) = {3012:276:2555:942:1613:4049}, f=-2099, c=6531037, u=608, i=2831, j=1218
The x values for the six known elements are (e,1) 6528482, 2554, 6533592 and (f,1) 6528481, 2555, 6533592
Found na at (e,1) x=6292398 === (3012,1,3146200) = {3012:1:19797142589106:6292398:19797136296708:19797148881506}, f=-39594285175201, c=391926854693394617149882248, u=3146199, i=19797142589107, j=6292399
6528482 - 6292398 = 236084, so the solution element is 118042 elements below the (e,1) a[t]=c*BigN element
So this is just an idea but if it's possible to find the difference between these t values and one of the known t values, maybe through this branching method or something else, that'd probably solve it. But I haven't gone through a ton of examples, so I don't know if it works every time. It makes a lot of sense assuming this binary thing is what he was talking about in February and March though.
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c1582f No.10959
>>10958
AA. Great post.
> It's a lookup x that's meant to either give us na or n-1.
>I've run some tests looking for na or n-1 in binary at the end of a[t] in elements other than the six we were using before and I haven't run into any that it doesn't work for yet
Was it the binary patterns that revealed the (n-1) and (na) values? Your post is vague about how you arrived at your results.
>It makes a lot of sense assuming this binary thing is what he was talking about in February and March though.
I have questions about how the sorting occurs for a binary search. Please post all your results for each test, and I'll take the time to analyze them in binary to look for matching chunks. M's breakdown needs to be analyzed in binary too.
>>10949
This was also a great post, working on the clues given. Thanks M. Let's analyze in binary to see if anything pops up.
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effedd No.10960
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effedd No.10961
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